Galois Theory, Second Edition

Galois

Monographs

Theory

S

David A. Cox

r

Second Edition

GALOIS THEORY

PURE AND APPLIED MATHEMATICS

A Wiley Series of Texts, Monographs, and Tracts

Founded by RICHARD COURANTEditors Emeriti: MYRON B. ALLEN III, DAVID A. COX, PETER HILTON,HARRY HOCHSTADT, PETER LAX, JOHN TOLAND

A complete list of the titles in this series appears at the end of this volume.

GALOIS THEORYSecond Edition

David A. CoxDepartment of MathematicsAmherst CollegeAmherst, MA

WILEY & SONS, INC., PUBLICATION

Copyright © 2012 by John Wiley & Sons, Inc. All rights reserved.

Published by John Wiley & Sons, Inc., Hoboken, New Jersey.Published simultaneously in Canada.

No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form orby any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except aspermitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the priorwritten permission of the Publisher, or authorization through payment of the appropriate per-copy fee tothe Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax(978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission shouldbe addressed to the Permissions Department, John Wiley & Sons, Inc., Ill River Street, Hoboken, NJ07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission.

Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts inpreparing this book, they make no representation or warranties with respect to the accuracy orcompleteness of the contents of this book and specifically disclaim any implied warranties ofmerchantability or fitness for a particular purpose. No warranty may be created or extended by salesrepresentatives or written sales materials. The advice and strategies contained herein may not besuitable for your situation. You should consult with a professional where appropriate. Neither thepublisher nor author shall be liable for any loss of profit or any other commercial damages, includingbut not limited to special, incidental, consequential, or other damages.

For general information on our other products and services please contact our Customer CareDepartment within the United States at (800) 762-2974, outside the United States at (317) 572-3993 orfax (317) 572-4002.

Wiley also publishes its books in a variety of electronic formats. Some Content that appears in print,however, may not be available in electronic formats. For more information about Wiley products, visitour web site at www.wiley.com.

Library of Congress Cataloging-in-Publication Data:

Cox, David A.Galois theory / David A. Cox. 2nd ed.

p. cm.Includes bibliographical references and index.ISBN 978-I-i 18-07205-9 (cloth)

1. Galois theory. I. Title.QA2i4.C69 20125l2'.32—dc23 2011039044

Printed in the United States of America.

10 9 8 7 6 5 4 3 2 1

To my family,

even the cats

CONTENTS

Preface to the First Edition xvii

Preface to the Second Edition Xxi

Notation xxiii

1 Basic Notation xxiii

2 Chapter-by-Chapter Notation xxv

PART I POLYNOMIALS

1 Cubic Equations 3

1.1 Cardan's Formulas 4

Historical Notes 8

1.2 Permutations of the Roots 10

A Permutations 10

B The Discriminant 11

C Symmetric Polynomials 13

VII

Viii CONTENTS

Mathematical Notes 14

Historical Notes 14

1.3 Cubic Equations over the Real Numbers 15

A The Number of Real Roots 15

B Trigonometric Solution of the Cubic 18

Historical Notes 19

References 23

2 Symmetric Polynomials 25

2.1 Polynomials of Several Variables 25

A The Polynomial Ring in n Variables 25

B The Elementary Symmetric Polynomials 27


2.2 Symmetric Polynomials 30

A The Fundamental Theorem 30

B The Roots of a Polynomial 35

C Uniqueness 36


Historical Notes 38

2.3 Computing with Symmetric Polynomials (Optional) 42

A Using Mat hematica 42

B Using Maple 44

2.4 The Discriminant 46


Historical Notes 50

References 53

3 Roots of Polynomials 55

3.1 The Existence of Roots 55


Historical Notes 61

3.2 The Fundamental Theorem of Algebra 62


Historical Notes 67

References 70

CONTENTS ix

PART II FIELDS

4 Extension Fields 73

4.1 Elements of Extension Fields 73

A Minimal Polynomials 74

B Adjoining Elements 75


Historical Notes 79

4.2 Irreducible Polynomials 81

A Using Maple and Mat hematica 81

B Algorithms for Factoring 83

C The Schönemann—Eisenstein Criterion 84

D Prime Radicals 85

Historical Notes 87

4.3 The Degree of an Extension 89

A Finite Extensions 89

B The Tower Theorem 91


Historical Notes 93

4.4 Algebraic Extensions 95


References 98

5 Normal and Separable Extensions 101

5.1 Splitting Fields 101

A Definition and Examples 101

B Uniqueness 103

5.2 Normal Extensions 107

Historical Notes 108

5.3 Separable Extensions 109

A Fields of Characteristic 0 112

B Fields of Characteristic p 113

C Computations 114


5.4 Theorem of the Primitive Element 119



References 123

X CONTENTS

6 The Galois Group 125

6.1 Definition of the Galois Group 125


6.2 Galois Groups of Splitting Fields 130

6.3 Permutations of the Roots 132



6.4 Examples of Galois Groups 136

A ThepthRootsof2 136

B The Universal Extension 138

C A Polynomial of Degree 5 139



6.5 Abelian Equations (Optional) 143


References 146

7 The Galois Correspondence 147

7.1 Galois Extensions 147

A Splitting Fields of Separable Polynomials 147

B Finite Separable Extensions 150

C Galois Closures 151


7.2 Normal Subgroups and Normal Extensions 154

A Conjugate Fields 154

B Normal Subgroups 155



7.3 The Fundamental Theorem of Galois Theory 161

7.4 First Applications 167

A The Discriminant 167

B The Universal Extension 169

C The Inverse Galois Problem 170


7.5 Automorphisms and Geometry (Optional) 173

A Groups of Automorphisms 173

B Function Fields in One Variable 175

C Linear Fractional Transformations 178

CONTENTS Xi

D Stereographic Projection 180


References 188

PART III APPLICATIONS

8 Solvability by Radicals 191

8.1 Solvable Groups 191


8.2 Radical and Solvable Extensions 196

A Definitions and Examples 196

B Compositums and Galois Closures 198

C Properties of Radical and Solvable Extensions 198


8.3 Solvable Extensions and Solvable Groups 201

A Roots of Unity and Lagrange Resolvents 201

B Galois's Theorem 204

C Cardan's Formulas 207


8.4 Simple Groups 210



8.5 Solving Polynomials by Radicals 215

A Roots and Radicals 215

B The Universal Polynomial 217

C Abelian Equations 217

D The Fundamental Theorem of Algebra Revisited 218


8.6 The Casus Irreducbilis (Optional) 220

A Real Radicals 220

B Irreducible Polynomials with Real Radical Roots 222

C The Failure of Solvability in Characteristic p 224


References 227

9 Cyclotomic Extensions 229

9.1 Cyclotomic Polynomials 229

A Some Number Theory 230

B Definition of Cyclotomic Polynomials 231

CONTENTS

C The Galois Group of a Cyclotomic Extension 233


9.2 Gauss and Roots of Unity (Optional) 238

A The Galois Correspondence 238

B Periods 239

C Explicit Calculations 242

D Solvability by Radicals 246



References 254

10 Geometric Constructions 255

10.1 Constructible Numbers 255



10.2 Regular Polygons and Roots of Unity 270


10.3 Origami (Optional) 274

A Origami Constructions 274

B Origami Numbers 276

C Marked Rulers and Intersections of Conics 279



References 288

11 Finite Fields 291

11.1 The Structure of Finite Fields 291

A Existence and Uniqueness 291

B Galois Groups 294



11.2 Irreducible Polynomials over Finite Fields (Optional) 301

A Irreducible Polynomials of Fixed Degree 301

B Cyclotomic Polynomials Modulo p 304

C Berlekamp's Algorithm 305


References 310

CONTENTS XIII

PART IV FURTHER TOPICS

12 Lagrange, Galois, and Kronecker 315

12.1 Lagrange 315

A Resolvent Polynomials 317

B Similar Functions 320

C The Quartic 323

D Higher Degrees 326

E Lagrange Resolvents 328


12.2 Galois 334

A Beyond Lagrange 335

B Galois Resolvents 335

C Galois's Group 337

D Natural and Accessory Irrationalities 339

E Galois's Strategy 341


12.3 Kronecker 347

A Algebraic Quantities 347

B Module Systems 349

C Splitting Fields 350


References 356

13 Computing Galois Groups 357

13.1 Quartic Polynomials 357



13.2 Quintic Polynomials 368

A Transitive Subgroups of S5 368

B Galois Groups of Quintics 371

C Examples 376

D Solvable Quintics 377



13.3 Resolvents 386

A Jordan's Strategy 386

B Relative Resolvents 389

XiV CONTENTS

C Quartics in All Characteristics 390

D Factoring Resolvents 393


13.4 Other Methods 400

A Kronecker's Analysis 400

B Dedekind's Theorem 404


References 410

14 Solvable Permutation Groups 413

14.1 Polynomials of Prime Degree 413



14.2 Impnmitive Polynomials of Prime-Squared Degree 419

A Primitive and Imprimitive Groups 419

B Wreath Products 421

C The Solvable Case 424



14.3 Primitive Permutation Groups 429

A Doubly Transitive Permutation Groups 429

B Affine Linear and Semilinear Groups 430

C Minimal Normal Subgroups 431

D The Solvable Case 433



14.4 Primitive Polynomials of Prime-Squared Degree 444

A The First Two Subgroups 444

B The Third Subgroup 446

C The Solvable Case 450



References 462

15 The Lemniscate 463

15.1 Division Points and Arc Length 464

A Division Points of the Lemniscate 464

B Arc Length of the Lemniscate 466

CONTENTS XV



15.2 The Lemniscatic Function 470

A A Periodic Function 471

B Addition Laws 473

C Multiplication by Integers 476


15.3 The Complex Lemniscatic Function 482

A A Doubly Periodic Function 482

B Zeros and Poles 484



15.4 Complex Multiplication 489

A The Gaussian Integers 490

B Multiplication by Gaussian Integers 491

C Multiplication by Gaussian Primes 497



15.5 Abel's Theorem 504

A The Lemniscatic Galois Group 504

B Straightedge-and-Compass Constructions 506



References 513

A Abstract Algebra 515

A.l Basic Algebra 515

A Groups 515

B Rings 519

C Fields 520

D Polynomials 522

A.2 Complex Numbers 524

A Addition, Multiplication, and Division 524

B Roots of Complex Numbers 525

A.3 Polynomials with Rational Coefficients 528

A.4 Group Actions 530

A.5 More Algebra 532

A The Sylow Theorems 532

CONTENTS

B The Chinese Remainder Theorem 533

C The Multiplicative Group of a Field 533

D Unique Factorization Domains 534

B Hints to Selected Exercises 537

C Student Projects 551

References 555

A Books and Monographs on Galois Theory 555

B Books on Abstract Algebra 556

C Collected Works 556

Index 557

PREFACE TO THE FIRST EDITION

Galois theory is a wonderful part of mathematics. Its historical roots date back to thesolution of cubic and quartic equations in the sixteenth century. But besides helpingus understand the roots of polynomials, Galois theory also gave birth to many of thecentral concepts of modem algebra, including groups and fields. In addition, there isthe human drama of Evariste Galois, whose death at age 20 left us with the brilliantbut not fully developed ideas that eventually led to Galois theory.

Besides being great history, Galois theory is also great mathematics. This is dueprimarily to two factors: first, its surprising link between group theory and the rootsof polynomials, and second, the elegance of its presentation. Galois theory is oftendescribed as one of the most beautiful parts of mathematics.

This book was written in an attempt to do justice to both the history and the powerof Galois theory. My goal is for students to appreciate the elegance of the theory andsimultaneously have a strong sense of where it came from.

The book is intended for undergraduates, so that many graduate-level topics arenot covered. On the other hand, the book does discuss a broad range of topics,including symmetric polynomials, angle trisections via origami, Galois's criterionfor an irreducible polynomial of prime degree to be solvable by radicals, and Abel'stheorem about ruler-and-compass constructions on the lemniscate.

A. Structure of the Text. The text is divided into chapters and sections. We usethe following numbering conventions:

xvii

XViii PREFACE TO THE FIRST EDITION

• Theorems, lemmas, definitions, examples, etc., are numbered according to chapterand section. For example, the third section of Chapter 7 is called Section 7.3. Thissection begins with Theorem 7.3.1, Corollary 7.3.2, and Example 7.3.3.

• In contrast, equations are numbered according to the chapter. For example, (4.1)means the first numbered equation of Chapter 4.

Sections are sometimes divided infonnally into subsections labeled A, B, C, etc. Inaddition, many sections contain endnotes of two types:

• Mathematical Notes develop the ideas introduced in the section. Each idea isannounced with a small black square.

• Historical Notes explain some of the history behind the concepts introduced in thesection.

The symbol u denotes the end of a proof or the absence of a proof, and denotesthe end of an example.

References in the text use one of two formats:

• References to the bibliography at the end of the book are given by the author'slast name, as in [Abel]. When there is more than one item by a given author, weadd numbers, as in [Jordani] and [Jordan2].

• Some more specialized references are listed at the end of the chapter in whichthe reference occurs. These references are listed numerically, so that if you arereading Chapter 10, then [1] means the first reference at the end of that chapter.

The text has numerous exercises, many more than can be assigned during anactual course. Some of the exercises can be used as exam questions. Hints toselected exercises can be found in Appendix B.

The algebra needed for the book is covered in Appendix A. Students should readSections A. 1 and A.2 before starting Chapter 1.

B. The Four Parts. The book is organized into four parts. Part I (Chapters 1 to 3)focuses on polynomials. Here, we study cubic polynomials, symmetric polynomialsand prove the Fundamental Theorem of Algebra. In Part II (Chapters 4 to 7), the focusshifts to fields, where we develop their basic properties and prove the FundamentalTheorem of Galois Theory. Part III is concerned with the following applications ofGalois theory:

• Chapter 8 discusses solvability by radicals.

• Chapter 9 treats cyclotomic equations.

• Chapter 10 explores geometric constructions.

• Chapter 11 studies finite fields.

Finally, Part IV covers the following further topics:

• Chapter 12 discusses the work of Lagrange, Galois, and Kronecker.

• Chapter 13 explains how to compute Galois groups.

• Chapter 14 treats solvability by radicals for polynomials of prime power degree.

• Chapter 15 proves Abel's theorem on the lemniscate.

PREFACE TO THE FIRST EDITION XiX

C. Notes to the Instructor. Many books on Galois theory have been stronglyinfluenced by Artin's thin but elegant presentation [Artin]. This book is different. Inparticular:

• Symmetric polynomials and the Theorem of the Primitive Element are used toprove some of the main results of Galois theory.

• The historical context of Galois theory is discussed in detail.

These choices reflect my personal preferences and my conviction that students needto know what an idea really means and where it came from before they can fullyappreciate its elegance. The result is a book which is definitely not thin, though Ihope that the elegance comes through.

The core of the book consists of Parts I and II (Chapters 1 to 7). It should bepossible to cover this material in about 9 weeks, assuming three lectures per week.In the remainder of the course, the instructor can pick and choose sections from PartsIII and IV. These chapters can also be used for reading courses, student projects, orindependent study.

Here are some other comments for the instructor:

• Sections labeled "Optional" can be skipped without loss of continuity. I sometimesassign the optional section on Abelian equations (Section 6.5) as part of a take-home exam.

• Students typically will have seen most but not all of the algebra in Appendix A.My suggestion is to survey the class about what parts of Appendix A are new tothem. These topics can then be covered when needed in the text.

• For the most part, the Mathematical Notes and Historical Notes are not used inthe subsequent text, though I find that they stimulate some interesting classroomdiscussions. The exception is Chapter 12, which draws on the Historical Notes ofearlier chapters.

D. Acknowledgments. The manuscript of this book was completed during aMellon 8 sabbatical funded by the Mellon Foundation and Amherst College. I amvery grateful for their support. I also want to express my indebtedness to the authorsof the many fine presentations of Galois theory listed at the end of the book.

I am especially grateful to Joseph Fineman, Walt Parry, Abe Shenitzer, and JerryShurman for their careful reading of the manuscript. I would also like to thankKamran Divaani-Aazar, Harold Edwards, Alexander Hulpke, Teresa Krick, BarryMazur, John McKay, Norton Starr, and Siman Wong for their help.

The students who took courses at Amherst College based on preliminary versionsof the manuscript contributed many useful comments and suggestions. I thank themall and dedicate this book to students (of all ages) who undertake the study of thiswonderful subject.

DAVID A. Cox

May 2004, Amherst, Massachusetts

PREFACE TO THE SECOND EDITION

For the second edition, the following changes have been made:

• Numerous typographical errors were corrected.

• Some exercises were dropped and others were added, a net gain of six.

• Section 13.3 contains a new subsection on the Galois group of irreducible separablequartics in all characteristics, based on ideas of Keith Conrad.

• The discussion of Maple in Section 2.3 was updated.

• Sixteen new references were added.

• The notation section was expanded to include all notation used in the text.

• Appendix C on student projects was added at the end of the book.

I would like thank Keith Conrad for permission to include his treatment of quarticsin all characteristics in Section 13.3. Thanks also go to Alexander Hulpke for his helpin updating the references to Chapter 14, and to Takeshi Kajiwara and Akira lino forthe improved proof of Lemma 14.4.5 and for the many typos they found in preparingthe Japanese translation of the first edition. I also appreciate the suggestions madeby the reviewers of the proposal for the second edition.

I am extremely grateful to the many readers who sent me comments and typosthey found in the first edition. There are too many to name here, but be assured thatyou have my thanks. Any errors in the second edition are my responsibility.

xxi

Xxii PREFACE TO THE SECOND EDITION

Here is a chart that shows the relation between the 15 chapters and the 4 parts ofthe book:

Part IV

Information about the book, including typo lists, can be found at

http : //www. cs. amherst edu/-dac/galois . html

As always, comments and corrections are welcome.

December 2011, Amherst, Massachusetts

DAVID A. Cox

Part

Part II

Part

NOTATION

1 BASIC NOTATION

Standard Rings and Fields. We use the following standard notation:

7L ring of integers,

Q field of rational numbers,

R field of real numbers,

C field of complex numbers.

Sets. We use the usual notation for union U and intersection fl, and we define

A\B= {xEASI = the number of elements in a finite set S.

We write A c B to indicate that A is a subset of B. (Some texts write A ç B for anarbitrary subset and reserve A c B for the case when A is strictly smaller than B. Wedo not follow this practice.) Thus A B if and only if A C B and B C A. Finally,given sets A and B, their Cartesian product is

A xB= {(a,b) a eA,b EB}.

XXIII

XXiV NOTATION

Functions. A function f : A —÷ B is sometimes denoted x '—* f(x). Also, a one-to-one onto map (a one-to-one correspondence) is often written

IfS is any set, then the identity map

15:S—*S

is defined by s '-+s for s e S. Also, given f : A —* B, we have:

:A0—*B restrictionofftoAocA,

f(Ao) = {f(a) a Ao} image ofA0 CA under f,

f1(Bo) = {a EA If(a) E Bo} inverse image of B0 C Bunderf.

The Integers. For integers a,b,n E Z with n > 0, we define:

alb b is an integer multiple of a,

a b b is not an integer multiple of a,

nla—b,gcd(a, b) greatest common divisor of a, b,

lcm(a, b) least common multiple of a, b,(Z/nZ)*I Euler

The Complex Numbers. Properties of C arereviewed in Section A.2. Also:

Re(z), Im(z) real and imaginary parts of z E C,

Izi complex conjugate and absolute value of z C,

= cosO + isinO Euler's formula,

z = IzIe'° polar representation of z C,

= primitive nth root of unity

S' = {e'°I

0 R} unit circle inC R2.

Groups. Basic properties of groups are reviewed in Section A. 1. Also:

o(g) order of an element g G,

(5) subgroup generated by S C G,

gH, Hg left and right cosets of subgroup H C G,

G/H quotient of group G by normal subgroup H,

5,, symmetric group on n letters,

A,, alternating group on n letters,

D2,, dihedral group of order 2n,

sgn(a) sign of a E Sn,

kernel and image of group homomorphism

CHAPTER-BY-CHAPTER NOTATION XXV

Rings. Basic properties of rings are reviewed in Section A. I. Also:

Im(p) kernel and image of ring homomorphism

(ri,... , r,,) ideal generated by r1, . . . , r1, ER,R/I quotient of ring R by ideal I,R* group of units of a ring R.

2 CHAPTER-BY-CHAPTER NOTATION

Here we list the notation introduced in each chapter of the text, followed by the pagenumber where the notation is defined. Many of these items appear in the index, whichlists other important pages where the notation is used.

Chapter 1 Notation.

w = primitive cube root of unity 6

D q2 + 4p3/27 quantity appearing in Cardan's formula 11

= —27D discriminantofy3+py+q 12

Chapter 2 Notation.

F[xi,... polynomial ring in variables Xi,. . . over F 26

deg(f) total degree of f E F[xi,... ,x,,] \ {O} 26

F(xi, . . . field of rational functions in Xi,... ,x,, over F 26

cr,, elementary symmetric polynomials 27

symmetric polynomial built from - 33

+• - universal polynomial of degree n 37

universal discnminant and its square root 46

discriminant off E F [x] 47

Chapter 3 Notation.

F c L L is an extension field ofF 58

Chapter 4 Notation.

nth cyclotomic polynomial 75

EL 75

subfield generated by F and . . EL 76

= Z/pZ finite field with p elements, p prime 84

[L: Fl degree of field extension F C L 89

Q field of algebraic numbers 96

XXVI NOTATION

Chapter 5 Notation.

Res(f,g,x) resultant of f,g E F[xI 115

Chapter 6 Notation.

Gal(L/F) Galois group of extension F CL 125

one-dimensional affine linear group 137

universalextension 138

Chapter 7 Notation.

LH fixed field of H C Gal(L/F) 147

aK conjuate field of F C K C L for o E Gal(L/F) 154

NG(H) normalizer of subgroup H C G 159

GL(2, F) general linear group of F2 178

PGL(2, F) projective linear group of F2 179

F,C FU{oo},CU{oo} 180

S2 unit sphere in JR3 180

Rot(S2) rotation group of S2 181

Chapter 8 Notation.

K1K2cL compositumofKi,K2CL 198

+ a(/3) + + Lagrange resolvent 203

Chapter 9 Notation.

= primitive nth root of unity 229

(x) nth cyclotomic polynomial 229

ef = p—i factorizationofp— l,p prime 238

Hf C (Z/pZ)* unique subgroup of order f 238

Lf C fixed field corresponding to Hf 238

(f, f-period, primitive element of Lf 240

Chapter 10 Notation.

St' field of constructible numbers 257

field of Pythagorean numbers 265

Fm = + 1 Fermat number 270

C field of origami numbers 277

CHAPTER-BY-CHAPTER NOTATION XXVIi


= finite field with q pfl elements, p prime 293

Frobenius automorphism of Fq 294

GL(n, F) general linear group of F" 296

Nm number of monic degree m irreducible f E IF1, [x] 301

Möbius function 302


0(x) resolvent polynomial of EL = F(xi, 317

H(p) C Sn isotropy group of EL = F(xi,... 318

x1 + C'x2 + C2'x3 + Lagrange resolvent 328

s(y) Galois resolvent 335

V = + primitive element used by Galois 336

algebraic quantities used by Kronecker 348


Of(y) Ferrari resolvent of quartic f 358

OJ-(y) sextic resolvent of quintic f 373

ef(Y) general resolvent off 387

D1 (y) quadratic resolvent off, replaces discriminant 390

D(f), D'(f) roots of D1(y) 390

SL(n, F) special linear group of F" 396

PGL(n, F) projective linear group of F" 396

PSL(n, F) projective special linear group of F" 396

(y) Kronecker resolvent 401


A B wreath product of groups A and B 421

AGL(n, JF'q) n-dimensional affine linear group 430

A['L(n, IFq) n-dimensional affine semilinear group 431

S(T) symmetry group of set T 433

M1, M2, M3 subgroups of from Section 14.4 444,445,450


one-half of arc length of lemniscate 466

r = p(s) Abel's lemniscatic function, s E R 467

n-division polynomials, n > 0 in Z 476

XXViii NOTATION

Z[iI

482

487

490

492

complex lemniscatic function, z e C

Weierstrass p-functionring of Gaussian integers/3-division polynomials, /3 E 7L[i]

PART I

POLYNOMIALS

The first three chapters focus on polynomials and their roots.We begin in Chapter 1 with cubic polynomials. The goal is to derive Cardan 's

formulas and to see how the permutations of the roots influence things.Then, in Chapter 2, we learn how to express the coefficients of a polynomial as

certain symmetric polynomials in the roots. This leads to questions about describingall symmetric polynomials. We also discuss the discriminant.

Finally, in Chapter 3, we show that all polynomials have roots in a possibly largerfield. We also prove the Fundamental Theorem of Algebra, which asserts that theroots of a polynomial with complex coefficients are complex numbers.

CHAPTER 1

CUBIC EQUATIONS

The quadratic formula states that the solutions of a quadratic equation

ax2+bx+c=O,

are given by

—b+'./b2—4ac(1.1)

2a

In this chapter we will consider a cubic equation

ax3+bx2+cx+d=O,

and we will show that the solutions of this equation are given by a similar thoughsomewhat more complicated formula. Finding the formula will not be difficult, butunderstanding where it comes from and what it means will lead to some interestingquestions.

Galois Theory, Second Edition. By David A. Cox 3Copyright © 2012 John Wiley & Sons, Inc.

4 CUBIC EQUATIONS

1.1 CARDAN'S FORMULAS

Given a cubic equation ax3 + bx2 + cx + d = 0 with a 0, we first divide by a torewrite the equation as

x3+bx2+cx+d=0, b,c,dEC,

where b/a, c/a, and d/a have been replaced with b, c, and d, respectively. Observethat x3 + bx2 + cx + d is a monic polynomial and that reducing to the monic case hasno effect on the roots.

The next step is to remove the coefficient of x2 by the substitution

bx=y—

The binomial theorem implies that

2 2 b b22 2b b2x =y =y

x3

so that

0 =x3 +bx2+cx+d2b2 b3 22b b2 b

=(y —by

If we collect terms, then we can write the resulting equation in y as

y3+py+q=0,

where

b2p = - — + C,

(1.2)2b3 bc

q = - + d.

You will verify the details of this calculation in Exercise 1.We call a cubic of the form y3 + py + q = 0 a reduced cubic. If we can find

the roots Yi ,Y2,Y3 of the reduced cubic, then we get the roots of the original cubicx3 + bx2 + cx + d 0 by adding —b/3 to each y,.

To solve y3 + py + q 0, we use the substitution

(1.3) y=z__?_.3z

CARDAN'S FORMULAS 5

This change of variable has a dramatic effect on the equation. Using the binomialtheorem again, we obtain

2 3 2 3

3 3 2P (P\ i'P\ p py =z Z

Combining this with (1.3) gives

y3 + py + q = 0 is equivalent to the equation

(1.4)

This equation is the cubic resolvent of the reduced cubic y3 + py + q = 0.

At first glance, (1.4) might not seem useful, since we have replaced a cubicequation with one of degree 6. However, upon closer inspection, we see that thecubic resolvent can be written as

=0.

By the quadratic formula (1.1), we obtain

z3

so that

(1.5)

Substituting this into (1.3) gives a root of the reduced cubic y3 + py + q, and thenx = y — b/3 is a root of the cubic x3 + bx2 + cx + d.

However, before we can claim to have solved the cubic, there are several questionsthat need to be answered:

• By setting y3 + py + q = 0, we essentially assumed that a solution exists. Whatjustifies this assumption?

• A cubic equation has three roots, yet the cubic resolvent has degree 6. Why?

• The substitution (1.3) assumes that z 0. What happens when z = 0?

• y3 +py+q has coefficients in C, since b,c,d E C. Thus (1.5) involves squareroots and cube roots of complex numbers. How are these described?

The first bullet will be answered in Chapter 3 when we discuss the existence of roots.The second bullet will be considered in Section 1.2, though the ultimate answer will

6 CUBIC EQUATIONS

involve Galois theory. For the rest of this section, we will concentrate on the last twobullets. Our strategy will be to study the formula (1.5) in more detail.

First assume that p 0 in the reduced cubic y3 + py + q. By Section A.2, everynonzero complex number has n distinct nth roots when n e Z is positive. In (1.5),the ± in the formula indicates that a nonzero complex number has two square roots.Similarly, the cube root symbol denotes any of the three cube roots of the complexnumber under the radical. To understand these cube roots, we use the cube roots ofunity 1, from Section A.2. We will write as w. Recall that

and that given one cube root of a nonzero complex number, we get the other two cuberoots by multiplying by w and w2.

We can now make sense of (1.5). Let

I 4p3q2 +

denote a fixed square root of q2 + 4p3/27 e C. With this choice of square root, let

Zi

denote a fixed cube root of (— q + + 4p3/27). Then we get the other two cuberoots by multiplying by w and w2. Note also that p 0 implies that Zi 0 and thatZi is a root of the cubic resolvent (1.4). It follows easily that if we set

Z2 =3zi

then

(1.6) YIZI+Z2Z13Z1

is a root of the reduced cubic y3 + py + q.To understand Z2, observe that

33_3( P\3_ P3Z1Z2 — Zj

—

An easy calculation shows that

1 / I 4p3 \ 1 / I 4p3'\ 1 / I 4p3 \ p3

Since Zi 0, these formulas imply that

3 1/ I 4p3

CARDAN'S FORMULAS 7

Hence Z2 = —p/3zi is a cube root of (—q — + so that

(1.7) zi = and Z2 =

are cube roots with the property that their product is —p/3.From (1.6), we see that Yi = Zi +Z2 is a root ofy3 +py+q when zi and Z2 are

the above cube roots. To get the other roots, note that (1.6) gives a root of thecubic whenever the cube roots are chosen so that their product is —p/3 (be sure youunderstand this). For example, if we use the cube root WZI, then

W2Z2 Z1Z2 =

shows that Y2 = WZI + w2z2 is also a root. Similarly, using the cube root W2ZI showsthat = w2z1 + WZ2 is a third root of the reduced cubic.

By (1.7), it follows that the three roots of y3 + py + q = 0 are given by

Y2=

+

provided the cube roots in (1.7) are chosen so that their product is —p/3. These areCardan 'S formulas for the roots of the reduced cubic y3 + py + q.

Example 1.1.1 For the reduced cubic y3 + 3y + 1, consider the real cube roots

and

Their product is —1 = —p/3, so by Cardan's formulas, the roots of y3 + 3y + 1 are

Yi=

Note that is real. In Exercise 2 you will show that y2 and y3 are complex conjugatesof each other.

Although Cardan's formulas only apply to a reduced cubic, we get formulas forthe roots of an arbitrary monic cubic polynomial x3 + bx2 + cx + d E C [xl as follows.

8 CUBIC EQUATIONS

The substitution x y b/3 gives the reduced cubic y3 + py + q =0, where p and qare as in (1.2). If zi and Z2 are the cube roots in Cardan's formulas fory3 + py + q 0,then the roots of x3 + bx2 + cx + d = 0 are given by

bXi = +Zi +Z2,

b 2Z2,

b 2 ZI+WZ2,

where Zi and Z2 from (1.7) satisfy Z1Z2 = —p/3. Our derivation assumed p 0, butthese formulas give the correct roots even when p = 0 (see Exercise 3).

We will eventually see that Cardan's formulas make perfect sense from the pointof view of Galois theory. For example, the quantity under the square root in (1.5) is

,132 P

q

Up to a constant factor, this is the discriminant of the polynomial y3 + py + q. Wewill give a careful definition of discriminant in Section 1.2, and Section 1.3 will showthat the discriminant gives useful information about the roots of a real cubic.

Here is an example of a puzzle that arises when using Cardan's formula.

Example 1.1.2 The cubic equation y3 = 0 has roots y = 0, all of whichare real. When we apply Cardan's formulas, we begin with

To pick a specific value for Zi, notice that (—i)3 = i, so that we can take Zi = —i.

Thus Z2 = —p/3zi = i, since p = —3. Then Cardan's formulas give the roots

Yi = —i+i=0,

Y2 = W(i) +w2(i) =

y3 = w2(—i) +w(i) =

(You will verify the last two formulas in Exercise 4.)

The surprise is that Cardan's formulas express the real roots of y3 — 3y in terms ofcomplex numbers. In Section 1.3, we will prove that for any cubic with distinct realroots, Cardan's formulas always involve complex numbers.

Historical Notes

The quadratic formula is very old, dating back to the Babylonians, circa 1700 B.C.Cubic equations were first studied systematically by Islamic mathematicians such as

CARDANS FORMULAS 9

Omar Khayyam, and by the Middle Ages cubic equations had become a popular topic.For example, when Leonardo of Pisa (also known as Fibonacci) was introduced toEmperor Frederick II in 1225, Fibonacci was asked to solve two problems, the secondof which was the cubic equation

x3+2x2+ lOx= 20.

Fibonacci's solution was

22 7 42 33 4 40x =1++ + + + +

In decimal notation, this gives x = 1.368808107853..., which is correct to 10 decimalplaces. Not bad for 787 years ago!

Challenges and contests involving cubic equations were not uncommon duringthe Middle Ages, and one such contest played a crucial role in the development ofCardan's formula. Early in the sixteenth century, Scipio del Ferro found a solutionfor cubics of the form x3 + bx = c, where b and c are positive. His student Rondoknew this solution, and in 1535, Florido challenged Niccolö Fontana (also known asTartaglia) to a contest involving 30 cubic equations. Working feverishly in preparationfor the contest, Tartaglia worked out the solution of this and other cases, and wenton to defeat Florido. In 1539, Tartaglia told his solution to Girolamo Cardan (orCardano), who published it in 1545 in his book Ars Magna (see [2]).

Rather than present one solution to the cubic, as we have done here, Cardan'streatment in Ars Magna requires 13 cases. For example, Chapter XIV considersx3 +64 = I and Chapter XV does x3 + 6x2 = 40. The reason is that Cardanprefers positive coefficients. However, he makes systematic use of the substitutionx = y — b/3 to get rid of the coefficient of x2, and Cardan was also aware that complexnumbers can arise in solutions of quadratic equations.

Numerous other people worked to simplify and understand Cardan's solution. In1550, Rafael Bombelli considered more carefully the role of complex solutions (seeSection 1.3), and in two papers published posthumously in 1615, François Viète (orVieta, in Latin) introduced the substitution (1.3) used in our derivation of Cardan'sformulas and gave the trigonometric solution to be discussed in Section 1.3.

In addition to the cubic, Ars Magna also contained a solution for the quarticequation due to Lodovico (or Luigi) Ferrari, a student of Cardan's. We will discussthe solution of the quartic in Chapter 12.

Exercises for Section 1.1

Exercise 1. Complete the demonstration (begun in the text) that the substitution x = y — b/3transforms x3 + bx2 + cx+ d into y3 + py + q, where p and q are given by (1.2).

Exercise 2. In Example 1.1.1, show that Y2 and are complex conjugates of each other.

Exercise 3. Show that Cardan's formulas give the roots of y3 + py + q when p = 0.

Exercise 4. Verify the formulas for y2 and y3 in Example 1.1.2.

10 CUBIC EQUATIONS

Exercise 5. The substitution x = y — b/3 can be adapted to other equations as follows.(a) Show that x = y — b/2 gets rid of the coefficient of x in the quadratic equation x2 + bx+ c =

0. Then use this to derive the quadratic formula.(b) For the quartic equation x4 + bx3 + cx2 + dx + e = 0, what substitution should you use to

get rid of the coefficient of x3?(c) Explain how part (b) generalizes to a monic equation of degree n.

Exercise 6. Consider the equation x3 + x —2 = 0. Note that x = 1 is a root.(a) Use Cardan's formulas (carefully) to derive the surprising formula

3 2/7 3 2/5= V

1 + = + and use this to explain the result of part (a).

Exercise 7. Cardan's formulas, as stated in the text, express the roots as sums of two cuberoots. Each cube root has three values, so there are nine different possible values for the sumof the cube roots. Show that these nine values are the roots of the equations y3 + py + q = 0,

y3+wpy+q"O, and y3+w2py+q= 0, where as usualw =

Exercise 8. Use Cardan's formulas to solve y3 + 3wy + 1 = 0.

1.2 PERMUTATIONS OF THE ROOTS

In Section 1.1 we learned that the roots of x3 + bx2 + cx + d = 0 are given by

b= +Zi +Z2,

b 2(1.8) Z2,

b 2X3 = +W Zi +WZ2,

where Zi and Z2 are the cube roots (1.7) chosen so that ZIZ2 = —p/3. We also knowthat Zi is a root of the cubic resolvent

(1.9)

and in Exercise 1 you will show that Z2 is also a root of (1.9). The goal of this sectionis to understand more clearly the relation between xi, x2 , x3 and Zi , Z2. We will learnthat permutations, the discriminant, and symmetric polynomials play an importantrole in these formulas.

A. Permutations. We begin by observing that we can use (1.8) to express Zi , Z2 interms of x1 ,x2,x3. We do this by multiplying the second equation by w2 and the thirdby w. When we add the three resulting equations, we obtain

x1 +w2x2+wx3 = +(1+w+w2)Z2.

PERMUTATIONS OF THE ROOTS 11

However,w is arootofx3 —1 = (x— l)(x2+x+1), whichimplies 1+w+w2 =0.Thus the above equation simplifies to

Xi +w2x2 + wx3 = 3zi,

so that

Zi = +W2x2+wx3).

Similarly, multiplying the second equation of (1.8) by w and the third by w2 leads tothe formula

2

This shows that the roots Zi and Z2 of the cubic resolvent can be expressed in termsof the roots of the original cubic. However, zi and Z2 are only two of the six roots of(1.9). What about the other four? In Exercise I you will show that the roots of thecubic resolvent (1.9) are

2 2Zi, Z2, WZ2, W Z1, W Z2,

and that these roots are given in terms of x1 ,x2,x3 by

zi = +w2x2+wx3),

Z2 = +W2X3 +WX2),

(1 10)WZi =

= +wx2),

w2z2 = x1 +wx3).

These expressions for the roots of the resolvent all look similar. What lies behindthis similarity is the following crucial fact: The six roots of the cubic resolvent areobtained from Zi by permuting x1 ,x2,x3. Hence the symmetric group S3 now entersthe picture.

From an intuitive point of view, this is reasonable, since labeling the roots Xi ,x2,x3simply lists them in one particular order. If we list the roots in a different order, thenwe should still get a root of the resolvent. This also explains why the cubic resolventhas degree 6, since 1S31 = 6.

B. The Discriminant. We can also use (1.10) to get a better understanding of thesquare root that appears in Cardan's formulas. If we set

(1.11)

then we can write Zi and Z2 as

zi =(1.12)

Z2 =

12 CUBIC EQUATIONS

We claim that D can be expressed in terms of the roots xi , , x3. To see why, notethat the above formulas imply that

However, (A. 15) gives the factorization

(1.13) = (zi —z2)(z1 —wz2)(z1 —W2Z2).

Using (1.10), we obtain

Zi —Z2 = +w2x2+wx3)— +wx2+w2x3)

=

=

where the last line uses — w = Similarly, one can show that

Z1 WZ2 = X3),(1.14)

V32 —iwi

Zi —W Z2 = —X2

(see Exercise 2). Combining these formulas with 4 — = and (1.13) easilyimplies that

(1.15) X2)(X1 —x3)(x2—x3).

If we square this formula for and combine it with (1.11), we obtain

(1.16) = —x2)2(xi —x3)2(X2—X3)2.

It is customary to define the discriminant of x3 + bx2 + cx + d to be

= (xl X2)(XI X3)2(X2 —X3)2.

Thus is the product of the squares of the differences of the roots. In this notationwe can write (1.16) as

431

(1.17)

Then (1.12) becomes

(1.18) zi and

Substituting this into (1.8), we get a version of Cardan's formulas which uses thesquare root of the discriminant.


The discriminant is also important in the quadratic case. By the quadratic formula,the roots of x2 + bx + c are

-b-VsX1

2and X2=

2

where = — 4c is the discriminant. This makes it easy to see that

and

Thus the discriminant is the square of the difference of the roots. In Chapter 2 wewill study the discriminant of a polynomial of degree n.

C. Symmetric Polynomials. We begin with two interesting properties of

A I= X2) —X3) —X3)

First suppose that we permute x1 ,x2 , X3 in this formula. The observation is that nomatter how we do this, we will still have the product of the squares of the differencesof the roots. This shows that is unchanged by permutations of the roots. In thelanguage of Chapter 2 we say that is symmetric in the roots , , X3.

Second, we can also express in terms of the coefficients of x3 + bx2 + cx + d.By (1.17), we know that = —4p3 — 27q2. However, we also have

b2p = — — + C,

(1.19)2b3 bc

q = - + d

by Exercise 1 of Section 1.1. If we substitute these into (1.17), then a straightforwardcalculation shows that

(1.20) = c2 + l8bcd—4c3 —4b3d—27d2

(see Exercise 3). When b = 0, it follows that x3 + cx + d has discriminant

= —4c3 — 27d2.

This will be useful in Section 1.3.The above formula expresses the discriminant in terms of the coefficients of the

original equation, just as the discriminant of x2 + bx + c = 0 is = b2 — 4c. TheFundamental Theorem of Symmetric Polynomials, to be proved in Chapter 2, willimply that any symmetric polynomial in x1 , x2 , x3 can be expressed in terms of thecoefficients b,c,d. In order to see why b,c,d are so important, note that if x1,x2,x3are the roots of x3 + bx2 + cx + d, then

x3 +bx2 +cx+d = (x—xi)(x—x2)(x—x3).

14 CUBIC EQUATIONS

Multiplying out the right-hand side and comparing coefficients leads to the followingformulas forb,c,d:

b = —(Xi +X2+X3),

(1.21) c=xIx2+xlx3+x2x3,d = —x1x2x3.

These formulas show that the coefficients of a cubic can be expressed as symmetricfunctions of its roots. The polynomials b, c, d are (up to sign) the elementarysymmetric polynomials of Xi ,x2,x3. These polynomials (and their generalization toan arbitrary number of variables) will play a crucial role in Chapter 2.

Mathematical Notes

One aspect of the text needs further discussion.

Algebra versus Abstract Algebra. High school algebra is very different from acourse on groups, rings, and fields, yet both are called "algebra." The evolution ofalgebra can be seen in the difference between Section 1.1, where we used high schoolalgebra, and this section, where questions about the underlying structure (why doesthe cubic resolvent have degree 6?) led us to realize the importance of permutations.Many concepts in abstract algebra came from high school algebra in this way.

Historical Notes

In 1770 and 1771, Lagrange's magnificent treatise Reflexions sur la resolutionalgebrique des equations appeared in the Nouvelles Mémoires de l'Academie royaledes Sciences et Belles-Lettres de Berlin. This long paper covers pages 205—421 inVolume 3 of Lagrange's collected works [Lagrange]. It is a leisurely account of theknown methods for solving equations of degree 3 and 4, together with an analysis ofthese methods from the point of view of permutations. Lagrange wanted to determinewhether these methods could be adapted to equations of degree � 5.

One of Lagrange's powerful ideas is that one should study the roots of a polynomialwithout regard to their possible numerical value. When dealing with functions of theroots, such as

= +W2X2+WX3)

from (1.10), Lagrange says that he is concerned "only with the form" of such expres-sions and not "with their numerical quantity" [Lagrange, Vol. 3, p. 385]. In modernterms, Lagrange is saying that we should regard the roots as variables. We will learnmore about this idea when we discuss the universal polynomial in Chapter 2.

We will see in Chapter 12 that many basic ideas from group theory and Galoistheory are implicit in Lagrange's work. However, Lagrange's approach fails whenthe roots take on specific numerical values. This is part of why Galois's work is soimportant: he was able to treat the case when the roots were arbitrary. The ideas of

CUBIC EQUATIONS OVER THE REAL NUMBERS 15

Galois, of course, are the foundation of what we now call Galois theory. This will bethe main topic of Chapters 4—7.


Exercise 1. Let z , Z2 be the roots of (1.9) chosen at the beginning of the section.(a) Show that Zi, Z2, WZi, WZ2, w2zi, W2Z2 are the six roots of the cubic resolvent.(b) Prove (1.10).

Exercise 2. Prove (1.14) and (1.15).

Exercise 3. Prove (1.20).

Exercise 4. We say that a cubic x3 + bx2 + cx + d has a multiple root if it can be written as(x — TI )2 (x r2). Prove that x3 + bx2 + cx + d has a multiple root if and only if its discriminantis zero.

Exercise 5. Since = —x2)2(XL —X3)2(x2 —x3)2, we can define the square root of tobe = (XI —X2)(XI —X3)(X2 —x3). Prove that an even pennutation of the roots takesto while an odd permutation takes to —v's. In Section 2.4 we will see that thisgeneralizes nicely to the case of degree n.

1.3 CUBIC EQUATIONS OVER THE REAL NUMBERS

The final topic of this chapter concerns cubic equations with coefficients in thefield R of real numbers. As in Section 1.1, we can reduce to equations of the formy3 + py + q =0, where p, q E R. Then Cardan's formulas show that the roots ,Y2,Y3lie in the field C of complex numbers. We will show that the sign of the discriminantof y3 + py + q = 0 tells us how many of the roots are real. We will also give anunexpected application of trigonometry when the roots are all real.

A. The Number of Real Roots. The discriniinant of y3 +py + q is

= (yi y3)2(y2 V3)2.

As we noted in the discussion following (1.20), can be expressed as

(1.22) = —4p3 — 27q2.

You will give a different proof of this in Exercise 1.For the rest of the section we will assume that the cubic y3 + py + q has distinct

roots Yi Y2 ,y3. It follows that the discriminant is a nonzero real number. We next

show that the sign of gives interesting information about the roots.

Theorem 1.3.1 Suppose that the polynomial y3 + py + q E R[y] has distinct rootsand discriminant 0. Then:(a) > 0 and only if the roots of y3 + py + q = 0 are all real.(b) < 0 if and only (fy3 +py+ q 0 has only one real root and the other two

roots are complex conjugates of each other

16 CUBIC EQUATIONS

Proof: First recall from Section A.2 that complex conjugation z satisfiesIt follows that ify1 is arootofy3+py+q= 0, then

so that yj is also a root. This proves the standard fact that the roots of a polynomialwith real coefficients either are real (if = or come in complex conjugate pairs

Ifyl,y2,y3 are all real and distinct, then = (yi —y2)2(yl —y3)2(y2 —y3)2 showsthat > 0. If the roots are not all real, then the above discussion shows that wemust have one real root, say and a complex conjugate pair, say Y2 and Write

Then =u—iv and

= ((yt —u) —iv)2((yi —u)+iv)2(2iv)2

= —4v2((yi _u)2+v2)2.

It follows that <0 when there is only one real root. This completes the proof. •In Exercises 2—5, we will sketch a different proof of Theorem 1.3.1 which uses

curve graphing techniques from calculus.We next apply the theory developed so far to Cardan's formulas

Y1 Zi+ Z2,

Y2 WZ1+W2Z2,

y3W2ZI+ WZ2,

where the cube roots

(1.23)= (— q + + and Z2 =

(— q— +

are chosen so that Z1Z2 = —p13.First, suppose that <0. Then Theorem 1.3.1 implies that y3 + py + q = 0 has

precisely one real root. Furthermore, by (1.22), we have

= —4p3 —27q2 <0.

Hence the square root + 4p3/27 is real, which means that we can take zi to bethe unique real cube root. Then Z1Z2 = —p/3 implies that Z2 is also the real cube root.It follows that

expresses the real root of y3 + py + q = 0 in terms of real radicals. Furthermore, inthe above formulas for y2 and y3, we see that y3 = since the cube roots are real and


= Thus we have a complete understanding of how Cardan's formulas workwhen the discriminant is negative.

However, the case when > 0 is very different. Here, y3 + py + q = 0 has threereal roots by Theorem 1.3.1. Since

= —4p3 — 27q2 > 0,

one value of the square root + 4p3/27 is

/+

==1

Using this and (1.23), we can write Zi and Z2 as the cube roots

3/1 ( 3/1 /and

This shows that Zi and Z2 are both nonreal complex numbers when > 0. You willprove in Exercise 6 that

(1.24)

Combining (1.24) with Cardan's formulas, we see that when > 0, the roots ofy3+py+qcanbe written

Yi= Zi+

Y2= WZi+W2ZI,

y3=W2Z1+ WZ1.

The root yi is real, since it is expressed as the sum of a complex number and itsconjugate. Furthermore, using w2 = one easily sees that

= w2zi and w2Zi =

so that y2 and y3 are also real, since they too are the sum of a complex number andits conjugate.

Notice that, unlike the case when <0, we no longer have a canonical choice ofZi —it is just one cube root of the complex number (— q + i Furthermore,we get Yi , Y2 , y3 by taking the three cube roots of this number and adding each to itsconjugate. This explains how Cardan's formulas work when > 0.

The puzzle, of course, is that we are using complex numbers to express the realroots of a real polynomial. Historically, this is referred to as the casus irreducibilis.We will have more to say about this below.

Example 1.3.2 In 1550, Rafael Bombelli applied Cardan's formulas to the cubic— l5y—4 = 0. This polynomial has discriminant _27(_4)2 =

18 CUBIC EQUATIONS

13068 > 0, so that all three roots are real. Bombelli noted that one root is y = 4 andused Cardan's formulas to show that

4 = lii

for appropriate choices of cube roots. To understand this formula, Bombelli notedthat (2+i)3 = 2+ lii and = 2— lii. Hence the cube roots in the aboveformula are 2+ i and 2 i, and their sum is clearly 4.

In Exercise 7 below, you will find the other two roots of the equation and explainhow Cardan's formulas give these two roots.

From the point of view of Cardan's formulas, complex numbers are unavoidablewhen > 0. But is it possible that there are other ways of expressing the roots whichonly involve real radicals? In Chapter 8 we will prove that when an irreducible cubichas real roots, the answer to this question is no—using Galois theory, we will seethat complex numbers are in fact unavoidable when trying to express the roots of anirreducible cubic with positive discriminant in terms of radicals.

B. Trigonometric Solution of the Cubic. Although complex numbers areunavoidable when applying Cardan's formulas to a cubic with positive discriminant,there is a purely "real" solution provided we use trigonometric functions rather thanradicals. This is the trigonometric solution of the cubic, due to Viète.

Our starting point is the trigonometric identity

cos(38) = 4cos38 — 3cos8,

which you will prove in Exercise 8. If we write this as 4cos38 —3 cosO — cos(30) = 0,

then ti = cos Oisa root of the cubic equation t3 — 3t — cos(30) 0. However, replacing8 with 8 + gives the same cubic polynomial, since cos(3(O + = cos(30). Itfollows that t2 = cos(O + is another root of 4t3 — 3t — cos(38) 0, and similarly,

= cos(8 + is also a root.In Exercise 9 you will show that the discriminant of 4t3 — 3t — cos(30) is given

by This is zero if and only if sin(30) = 0, which in turn is equivalent tocos(30) = ± 1. Thus cos(30) ± 1 implies that 4t3 — 3t — cos(38) has roots

(1.25) ti =cosO,

Hence 4t3 3t — cos(30) = 0 is a cubic equation with known roots. Viète's insightwas that by a simple change of variable, we can use this to solve any cubic equationwith positive discriminant. Here is his result.

Theorem 13.3 Let y3 + py + q = 0 be a cubic equation with real coefficients andpositive discriminant. Then p <0, and the roots of the equation are


where 0 is the real number defined by

1

_____

O=3COS

Proof: You will prove this in Exercise 10. •

In Exercise 11 you will explore how this relates to Cardan's formulas.

Historical Notes

When Cardan wrote Ars Magna in 1545, he and his contemporaries wanted to findreal roots of cubic equations. In fact, they worked almost exclusively with positiveroots, although they were aware of the existence of negative roots, which Cardancalled "false" or "fictitious." However, Cardan does use complex numbers in ChapterXXXVII when he considers the problem of dividing 10 into two parts so that theirproduct is 40. In modern notation this gives the equations x+y = 10 and xy = 40.Eliminating y, we get the quadratic equation

x2— lOx+40=0

with roots 5 ± After deriving this solution, Cardan says "Putting aside themental tortures involved, multiply 5 + by 5— making 25 — (—15)...Hence this product is 40." Cardan's conclusion is that "This truly is sophisticated"[2, pp. 219—220].

Cardan was also aware of Theorem 1.3.1, though he stated it in very differentterms. As an example of a cubic with three real roots, he considers x3 +9 — 1 2x, for

which he gives the "true" (i.e., positive) solutions 3 and — 1 and the "false"

(i.e., negative) solution

However, Cardan never applies his formulas to cubics like x3 +9 = 1 2x. Heonly considers cases where there is one real root, which can be expressed in termsof real radicals. Yet Cardan must have known that complex numbers appear in theradicals when the discriminant is positive. This is the casus irreducibilis ("irreduciblecase") mentioned above. According to [1], Tartaglia was also aware of the casusirreducibilis, and in fact delayed publication of his results because he was so troubledby it. This is part of the reason why Cardan's work appeared first.

One of the first people to comment directly on the casus irreducibilis was RafaelBombelli. In his book L'algebra, written around 1550 but not published until 1572,he treats this case in detail, including the formula

(1.26)

from Example 1.3.2. There we saw how Bombelli explained this formula by showingthat 2+lli= (2+i)3, so that (1.26)reduces to 4 = (2+i)+(2—i). Bombelli waspleased with this calculation and commented that

At first, the thing [equation (1.26)1 seemed to me to be based more on sophismthan on truth, but I searched until I found a proof.

20 CUBIC EQUATIONS

In working out this solution, Bombelli was the first to give systematic rules foradding and multiplying complex numbers. Exercise 12 will discuss another exampleof complex cube roots taken from Bombelli's work.

The moral is that cubic equations forced mathematicians to confront complexnumbers. For quadratic equations, one could pretend that complex solutions don'texist. But for a cubic with real roots, we've seen that Cardan's formula must involvecomplex numbers. So it is impossible to ignore complex numbers in this case. See thebooks [1] and [3J for more background and discussion on the discovery of complexnumbers.

We should also say a few words about Viète's trigonometric solution of the cubic.Once we realize that cos(30) = 4cos3O — 3 cosO gives a cubic equation with cos 6as a root, proving Theorem 1.3.3 is not that difficult. Viète was well aware ofsuch identities. For example, in 1593, Adrianus Romanus (also called Adriaen vanRoomen) posed the problem of finding a root of the equation

A = x45—45x43+945x4' — 12300x39+ 111 150x37—740259x35

+3764565x33— 14945040x31+46955700x29— 1 17679100x27

(1.27) +236030652x25—37865800x23+483841800x2' —488494125x19

+384942237x'7—232676280x'5+ 105306075x'3—345l207x'1

+781 1375x9— 1 138500x7+95634x5—3795x3+45x,

where

(1.28)

Viète solved this equation by noting that 2sin(45a) can be expressed as a polynomialof degree 45 in 2 sin whose coefficients match the right-hand side of (1.27). Itfollows that if A = thenx = is a root.

Viète also realized that (1.28) can be written

A = 2sin(7r/15) = 2sin(45 .ir/675),

which easily implies that one root of (1.27) is x = 2sin(7r/675). Using the trick of(1.25), we get the 44 additional solutions

j=1,...,44.

Viète listed only 23 roots, since he (like Cardan) wanted positive solutions. Never-theless, Viète's insight is impressive, and his solution of (1.27) makes it clear how hewas able to find the trigonometric solution of the cubic.



Exercise!. Letf(y) =y3+py+q= (y—yl)(y—y2)(y—y3), and set

= (yl —y2)2(yI •—y3) Y3).

The goal of this exercise is to give a different proof of (1.22).(a) Use the product rule to show that f' (yi) = (yi — y2) (yl — y3), where f' denotes the

derivative off. Also derive similar formulas forf' (y2) and f'(b) Conclude that = — f'(y' ) f' f' Be sure to explain where the minus sign comes

from.(c) The quadratic f' (y) = 3y2 + p factors as f' (y) = — a) (y — /3), where a = and

/3 = (when p >0, we let = Prove that = —27f(a)f(fl).(d) Use f(y)=y3+py+qand a= to show that

f(a) = (v

Similarly, show that =(e) By combining parts (c) and (d), conclude that = —4p3 — 27q2.

Exercise 2. Letf(y) = y3 + py + q. The purpose of Exercises 2—5 is to prove Theorem 1.3.1geometrically using curve graphing techniques. The proof breaks up into three cases cone-sponding to p > 0, p = 0, and p < 0. This exercise will consider the case p > 0.(a) Explain why <0.(b) Analyze the sign of f'(y), and show that f(y) is always increasing.(c) Explain why f(y) has only one real root.

Exercise 3. Next, consider the case p = 0.

(a) Explain why <0.(b) Explain why f(y) has only one real root.

Exercise 4. Finally, consider the case p < 0. In this case, f' (y) = 3y2 + p has roots a =and /3 = — which are real and distinct.

(a) Show that the graph of f(y) has a local minimum at a and a local maximum at /3. Thusf(a) is a local minimum value and f(fi) is a local maximum value. Also show thatf(a) <f(j3).

(b) Explain why f(y) has three real roots if f(a) and f(/3) have opposite signs and has onereal root if they have the same sign. Illustrate your answer with a drawing of the threecases that can occur.

(c) Conclude that f(y) has three real roots if and only if f(a) <0.(d) Finally, use part (c) of Exercise 1 to show that the roots are all real if and only if > 0.

Exercise 5. Explain how Theorem 1.3.1 follows from Exercises 2, 3, and 4. Notice thatthe quantity which appeared earlier in part (c) of Exercise 1, arises naturally inExercise 4.


Exercise 7. Example 1.3.2 expressed the root y = 4 of y3 — 15y —4 in terms of Cardan'sformulas. Find the other two roots, and explain how Cardan's formulas give these roots.

Exercise 8. Derive the trigonometric identity cos(39) = 4cos3O —3 cos 0 using cos(x + y) =cosxcosy—sinxsinyandcos2O+sin20 = 1.

22 CUBIC EQUATIONS

Exercise 9. When divided by 4, 4t3 3t — cos(39) gives t3 — — which is monic.Show that the discriminant of this polynomial is

Exercise 10. The goal of this exercise is to prove Theorem 1.3.3. Lety3 + py + q = 0 be acubic equation with positive discriminant. Consider the substitution y = Ar, which transformsthe given equation into A3t3 + Apt + q = 0.

(a) Show that Exercises 2 and 3 imply that p < 0.(b) The equation A3t3 + Apr + q = 0 can be written as

=0.

Show that this coincides with 4r3 — 3t — cos(30) = 0 if and only if

and

Note that is real and nonzero by part (a).(c) Use = —(4p3 +27q2) >0 to prove that

1<

(d) Explain how part (c) implies that the second equation of part (b) can be solved for 9. Alsoshow that > 0 implies that cos(30) :1:1.

(e) By (1.25), r1 = cosO, t2 = cos(8+ and r3 = cos(O+ are the three roots ofA3t3 + Apt + q = 0. Then show that the theorem follows by transforming this back to

Ar via part (b).

Exercise 11. Consider the equation 4r3 — 3t — cos(38) = 0, where cos(30) ±1. In (1.25),we expressed the roots in terms of trigonometric functions. In this exercise, you will studywhat happens when we use Cardan's formulas.(a) Show that Cardan's formulas give the root

= + isin(39) + .Vcos(39) isin(38).

(b) Explain why = (cos8 + isin9) is a value of + isin(38), and use this toshow that t1 is just cos 8.

(c) Similarly, show that Cardan's formulas also give the roots t2 and t3 as predicted by (1.25).

Exercise 12. Example 1.3.2 discusses Bombelli's discovery that + 1 ii = 2 + i. But not all

cube roots can be expressed so simply. This exercise will show that + is not of theform EZ.(a) Suppose that 4+ = (a+b',./iii)3 for some a,b E 7L. Show that this implies that

4=a3 —33ab2 and 1 =3a2b— 11b3.(b) Show that the equations of part (a) imply that b = ± 1 and a14. Conclude that the equation

4 + = (a + bvTii)3 has no solutions with a, b E Z.(c) Find a cubic polynomial_of the form x3 + px + q with p, q E Z which has the number

'V4—

In contrast to 'Y2 + 11 i = 2 + i, Bombelli was not certain that + was a complex

number. He calls + "another sort of cubic radical." Bombelli never deals with this

REFERENCES 23

radical by itself, but rather considers the sum + flu + v'iii, which is a root of thecubic equation found in part (c).

Exercise 13. Suppose that a quartic polynomial f = x4 + bx3 + cx2 + dx+ e in IR{x] has distinctroots xl ,x2 , , X4 E C. The discriminant off is defined by the equation

= (xl —X2)(X1 —X3)(X1 —X4)(X2 —X3)(X2 —X4)(X3 —X4).

The theory developed in Chapter 2 will imply that E 11k, and 0, since the x are distinct.Adapt the proof of Theorem 1.3.1 to show that

<0 x4+bx3+cx2+dx+e=Ohasexactlytworealroots.

Exercise 14. In Section 1.1, we discussed the equation x3 + 2x2 + lOx = 20 considered byFibonacci.(a) Show that this equation has precisely one real root. This is the root Fibonacci approxi-

mated so well.(b) Use Cardan's formulas and a calculator to work out numerically the three roots of this

polynomial.

Exercise 15. Use a calculator and Theorem 1.3.3 to compute the roots of the cubic equationy3 — 7y + 3 = 0 to eight decimal places of accuracy.

REFERENCES

1. I. 0. Bashmakova and G. S. Smirnova, The Beginnings and Evolution of Algebra, Englishtranslation by A. Shenitzer, MAA, Washington, DC, 1999.

2. G. Cardan, Ars Magna, Johann Petrieus, Nürnberg, 1545. English translation The GreatArt by T. R. Witmer, MIT Press, Cambridge, MA, 1968.

3. B. Mazur, Imagining Numbers (particularly the square root of minusfifteen), Farrar StrausGiroux, New York, 2003.

CHAPTER 2

SYMMETRIC POLYNOMIALS

The goal of this chapter is to provide some tools needed for our study of Galois theory.The basic result is that any polynomial unchanged under all possible permutationsof the variables can be expressed in terms of certain special polynomials called theelementary symmetric polynomials. After proving this, we will show how to computewith symmetric polynomials and discuss the discriminant mentioned in Chapter 1.

2.1 POLYNOMIALS OF SEVERAL VARIABLES

Galois theory often deals with polynomials of more than one variable, especiallywhen studying the roots of a polynomial. This section will introduce polynomials ofseveral variables and the elementary symmetric polynomials.

A. The Polynomial Ring in n Variables. Let x1, . . . be distinct formalsymbols called variables. A polynomial in x1,... with coefficients in a field F isa finite sum of terms, which are expressions of the form

cinF,


26 SYMMETRIC POLYNOMIALS

We call the product . . . a monomial, so that a term is an element of F timesa monomial. A term is nonzero if the constant is nonzero. The total degree of anonzero term . . . x' is the sum of its exponents a1 + +

We define F[xi, . .. , to be the set of all polynomials in Xi, . . . , with coefficientsin F. It is easy to see that F[xi, . . . is a ring under addition and multiplicationof polynomials. The total degree of a nonzero f e F[xi, . . . denoted deg(f), isthe maximum of the total degrees of the nonzero terms of f. Since F is an integraldomain, one can prove without difficulty that if f, g E F [Xi, . . . , are nonzero, then

(2.1) deg(fg) =deg(f)+deg(g).

It follows that F[xi,.. . is an integral domain. Note that deg(O) is not defined.Since F[xi,... is an integral domain, we can define its field of fractions

This is the field of rationalfunctions in n variables. Note that:

• Square brackets, as in F[xi,. . . refer to polynomials.• Parentheses, as in F (x1,... refer to quotients of polynomials.

A nonconstant polynomial in F[xi,. . . is irreducible over F if it is not aproduct of polynomials of strictly smaller total degree. We can factor polynomials inF[xi, . . . into irreducibles as follows.

Theorem 2.1.1 Let f F[xi,... be nonconstant. Then there are irreduciblepolynomials gi, . . . ,g,- E F[xi, . . . such that

Furthermore, if there is a second factorization off into irreducibles

f=hi...h5,

then r = s and the h 's can be permuted so that each h is a constant multiple of g,.

Pmof: See Corollary A.5.7 of Appendix A. U

In Section A.5, we define the general notion of a unique factorization domain, or

UFD. In this terminology, Theorem 2.1.1 states that F [x1,... , is a UFD.A useful property of F[xi,... is that evaluation is a ring homomorphism.

Suppose that we have a field F, a ring R containing F, and elements co,.. . , E R.

Then the evaluation map

is defined by

(2.2)

POLYNOMIALS OF SEVERAL VARIABLES 27

We have the following important result.

Theorem 2.1.2 Given a field F, a ring R containing F, and . . . , E R, theevaluation map (2.2) is a ring homomorphism F[xi,... —* R.

Proof: The proof is a tedious verification that

(fg)(ai,...

where f + g and fg are the sum and product of polynomials f and g.

Once we fix the field F, the variables Xi,... play two roles. At the beginningof the section, they were formal symbols used in the definition of polynomial. Buteach variable x, also has the ability to "take any value." In other words, XI,...can take arbitrary values in any ring R containing F. Be sure you understand howTheorem 2.1.2 makes this precise.

B. The Elementary Symmetric Polynomials. How do the roots of a monicpolynomial in x relate to its coefficients? To answer this question, we begin withcubic and quartic polynomials. Suppose that f = x3 + + a2x + a3 E F[x] hasroots EF. Then

f=If we multiply this out and compare coefficients, then the coefficients can be expressedin terms of the roots as

a1 = —(cr1

(2.3)

a3 =

(See also (1.21) in Section 1.2.) For n 4, a similar computation shows that iff =x4+aix3+a2x2+a3x+a4 E F[x] has roots

a1 = —(cvi

a2 =c + cx1c3 + + + a2a4 +

a3 =

a4 =

Up to sign, a! uses the sum of the roots, a2 takes the roots two a time, and a3 takesthem three at a time. We generalize this pattern as follows.


Definition 2.1.3 Let x1,... be variables over afield F. Then

02= XiXJ,

=

are the elementary symmetric polynomials. Thus . . . E F[xi,...

We will sometimes write a,- = . . . ,x,,). The following identity is one of thekey properties of the elementary symmetric polynomials.

Proposition 2.1.4 Let Xi,... be variables over a field F. Then, given anothervariable x, we have

(2.4) (X_xi)(X_xn)=f_aiXn_i+.+(_1)rxr+...+(_1)nan.

Proof: The proof follows by multiplying out the left-hand side of (2.4) and thencomputing the coefficient of each power of x. For example, the constant term is ob-viously the product of constant terms, namely (—xi)... = Similarly,the coefficient of x"' is easily seen to be —x1 — = —at.

For readers interested in the details of how this works in general, observe that wemultiply out (x — x1)... (x — as follows:

• For each of the n factors x — x,, choose either x or —x1.

• Take the product of these n choices.

• Sum these products over all possible ways of making the n choices.

It follows that the terms involving in (x — xi)... (xx exactly n — r times in the first bullet. This means choosing —x1

for the ii st, i2nd, ..., jrth factors and choosing x for the remaining n — r factors. Asdescribed in the second bullet, the product of these choices is

. . .=

. .

When we sum over all possible ways of making the n choices (as described in thethird bullet), it follows that the coefficient of in the left-hand side of (2.4) is

I<ii<.<i,<n

This completes the proof of the proposition. .

POLYNOMIALS OF SEVERAL VARIABLES 29

Proposition 2.1.4 has the following useful application. Suppose that a monicpolynomial! = + + +

a larger field L. This means that

=

However, since evaluation is a ring homomorphism (Theorem 2.1.2), we can evaluatethe identity (2.4) = = to obtain

+ . . .

These two formulas give the following corollary of Proposition 2.1.4.

Corollary 2.1.5 Let! = + a monic poly-nomial of degree n > 0 with coefficients in a field F. 1ff has roots ai,... , alarger field L, then the coefficients off are expressed in tenns of its roots as

ar = ,a,1)

forr=1,...,n. .Here is what happens when n = 3.

Example 2.1.6 If x3 + a1x2 + a2x + a3 has roots , cr2, a3, then Corollary 2.1.5implies that

a! = —o1(ai,a2,a3) = —(ai +a2+a3),a2=cT2(c!I,a2,a3) =alc!2+aIa3+a2a3,

a3 = —a3(al,a2,a3) = —a1a2a3,

in agreement with (2.3).

Mathematical Notes

There are two topics for us to discuss.

• Ideals in a Polynomial Ring. The text makes it seem that F[xi,... behaveslike the one-variable case studied in Section A. 1. However, once we start talkingabout ideals, some significant differences emerge. For example, Theorem A.1.17implies that F[x] is a PID. But as soon as the number of variables is two or more, notall ideals are principal. Exercise 1 will give a simple example.

In fact, F[xi,... has a rich supply of ideals when n > 2. These are relatedto solutions of simultaneous sets of polynomial equations, which is the subject ofalgebraic geometry. See [2] for an introduction to this area of mathematics.


• Coefficients as Polynomials. There are other ways to think about polynomials inseveral variables. For example, we can regardf E F[xi, . . . as a polynomial inwith coefficients in F{xi, . .. i.e.,

This is expressed more formally as F [XI,... = F [x1,... ii For instance,(2.4) takes place in F[xi, . . . See Exercise 2 for more examples.


Exercise 1. Show that (x,y) = {xg+yh g,h E F[x,y]} C F[x,y] is not a principal ideal inF [x,y}.

Exercise 2. Express each of the following polynomials as a polynomial in y with coefficientsthat are polynomials in the remaining variables.(a) x2y+3y2—xy2+3x+xy2+7x3y3.(b) (y—(xI+x2))(y—(xl+x3))(y—(x2+x3)).

Exercise 3. Given positive integers n and r with 1 � r � n, let be the number of ways ofchoosing r elements from a set with n elements. Recall that =(a) Show that the polynomial is a sum of terms.

(b) Show that a,(—a,...,—cs) =(c) Let f = (x+ cs)'1. Use part (b) and Corollary 2.1.5 to prove that

(x+a)'1=

where (g) = 1. This shows that the binomial theorem follows from Corollary 2.1.5.

2.2 SYMMETRIC POLYNOMIALS

We will consider polynomials in n variables Xi,... ,x,, over a field F.

Definition 2.2.1 A polynomial f E F[xi,.. . ,x,,] is symmetric

. .

for all permutations a in the symmetric gmup Sn.

A. The Fundamental Theorem. In Section 2.1, we defined the elementarysymmetric polynomials ai, . . . , To prove that these are symmetric in the abovesense, consider the identity

SYMMETRIC POLYNOMIALS 31

from Proposition 2.1.4. The product on the left-hand side is symmetric becausepermuting the x1 simply permutes the factors. Comparing this with the right-handside, it follows that a1,.. . , are symmetric.

Since cr1,... , are symmetric, any polynomial in Cr1, . . . , is also symmetric.The remarkable fact is that all symmetric polynomials arise in this way. Moreprecisely, we have the following Fundamental Theorem of Symmetric Polynomials.

Theorem 2.2.2 Any symmetric polynomial in F[xi,... can be written as a poly-nomial in at,... with coefficients in F.

Proof: We will follow (with a few changes) the argument given by Gauss in 1816in his second proof of the Fundamental Theorem of Algebra. The proof will involvean inductive process which requires that we order monomials in x1,...We will use graded lexicographic order, which is defined by

+ <b1 +

(2.5)and a1 <b1,

and a1 = b1 and a2 <b2,

or

We also define 4"• > .x' to mean <xv.To compare one monomial with another, one first computes the total degree of each

monomial, and when these are equal, one checks the two monomials one exponentat a time, starting with x1, to find the first which differs. For example,

(smaller total degree),

> (same total degree, equal x1 exponent,

greater X2 exponent).

An important property of graded lexicographic order is that there are at mostfinitely many monomials such that

(2.6) for fixed ai,...

This follows because (2.6) and (2.5) imply that a1 + + � b1 +... + (be sureyou understand this). Since N = a1 + is fixed and b � 0 for all i, we get theinequality

for all i. Hence there are only N + 1 possibilities for each b, which easily impliesthat (2.6) can hold for at most finitely many

We can apply graded lexicographic order to a nonzero polynomial as follows. Wesaw in Section 2.1 that such a polynomial is a sum of nonzero terms, each of which is a


nonzero element ofF times a monomial. Then the leading term is the greatest of thesemonomials—relative to (2.5)—times its coefficient. Thus any nonzero polynomialhas a leading term.

For example, the leading term of o2 = X1X2 +X1X3 + +X2X3x1x2. In other words, xIx2 > x1x3 when i < j and (i,j) (1,2). This follows bychecking the exponent of x1 (if i> 1) or X2 (if i = 1 and j > 2) in You willgeneralize this in Exercise 1 by showing that x1x2 is the leading term of the rthelementary symmetric polynomial

= . .xir.l<ll<<lr<fl

We are now ready to prove the theorem. Let f e F[xi, . . . be symmetric andnonzero with leading term

(2.7) cxr.

We claim that

(2.8) a1 � � >

To show this, suppose that <aa+i for some 1 i n — 1. The symmetry offimplies that interchanging x1 and gives the same polynomial. Since (2.7) is aterm off, it follows that

(2.9)

is also a term of f. To compare this with (2.7), note that both monomials have thesame total degree and the same exponents of x1, . . . ,x1_1. However, x, has exponent

in (2.9) and exponent a, in (2.7). Then a,+ > a implies that (2.9) is a term offgreater than (2.7) according to the order relation (2.5). Yet (2.7) is the leading termoff. This contradiction proves (2.8).

Now consider

(2.10) g =

a polynomial by (2.8). In Exercise 2, you will prove that the leading term ofa product is the product of the leading terms. Since the leading term of 0r is x1 Xr,

it follows that the leading term of g is

... (Xi .

. . .

(2.11) = —a,,+a,a,

.

This shows that f and cg have the same leading term. Hence fi = f — cg has astrictly smaller leading term according to the ordering defined in (2.5). Note that fiis symmetric, since f and g are.


Now repeat this process, starting with fi instead of f. Since fi is symmetric, ithas a leading term with coefficient c1 and exponents b1 � � As above, thiswill give an expression gi in the elementary symmetric polynomials such that fi andcigi have the same leading term. It follows that

f2 =fi —clgi =f—cg—cigi

has a strictly smaller leading term. Continuing in this way, we get polynomials

f, f1=f—cg, f2=f—cg—c1gi, f3=f—cg—clg1—c2g2,...,

where at each stage the leading term gets strictly smaller according to the orderdefined in (2.5). This process will terminate if we find some m with fm = 0, for thezero polynomial has no leading term. If, on the other hand, we never had fm = 0,then the above would give an infinite sequence of nonzero polynomials with strictlydecreasing leading terms. But we showed above that there are only finitely manymonomials strictly smaller than the leading term off. Hence the above process mustterminate.

However, once we have fm = 0 for some m, we obtain

f = cg + ... + cm_lgm_1

since fm = f — cg — cigl — — cm_lgm_I. Each gj is a product of the cr, to variouspowers, which proves that f is a polynomial in the elementary symmetric polynomi-als. This completes the proof. •

In Theorem 2.2.7 below, we will prove that the expression off as a polynomial inan is unique.

The proof of Theorem 2.2.2 can be turned into an algorithm for writing a givensymmetric polynomial in terms of the a1. For this purpose, we will use the notation

to denote the sum of all distinct monomials obtained from x' by permutingXi,... Here are some simple examples.

Example 2.2.3 One easily sees that

and

=

Also, has 12 terms instead of 24. This is because = for n = 4.Switching the last two variables gives the same monomial, yet x?x2 uses only thedistinct monomials we get by permuting the variables. It>


If x" is a term of a symmetric polynomial f E F[xi,... then

(2 12)= . . + a sum of terms involving monomials

different from those in

Do you see how we used this fact in the proof of Theorem 2.2.2?Here is an example of how to write a symmetric polynomial in terms of the a,.

Example 2.2.4 The polynomial in x1 ,x2,x3,x4 given by

f =

has 24 terms and is symmetric. In the first chapter of his 1782 book Meditationes[6], Edward Waring shows how to express f in terms of , a2, a3, a4.

His method is similar to what we did in the proof of Theorem 2.2.2. In this case, weproceed as follows (you will supply the details in Exercise 3):

Step 1. The leading term off is = so that (2.10) becomes

3—2 2—1 1—0 0a1 a2 a3 a4 = a1a2a3.

Furthermore, one can use a computer to show that

(2.13) aIa2a3 = + + +

Hence

fI = f— = —3>4x?x2x3x4 — —

Step 2. The leading term off' is —3x?x2x3x4, which gives

(2.14) = =

Thus

f2 = f — aIa2a3 + = —

Step 3. For f2, we have as leading term. Since

(2.15) =

we obtain

= f— ala2a3 + =

Step 4. The leading term of is and from

(2.16) a2a4 = >J4X1X2X3X4,

we see that

= f— ala2a3 + + — 4a2a4 = 0.


Conclusion. Since = 0, the process terminates and we obtain the formula

2 2f = a1 a2a3 — 3U1 — 3a3 + 4a204

expressing f in terms of the elementary symmetric polynomials.

In the exercises you will apply these methods to a variety of problems dealing withsymmetric polynomials. For readers interested in doing more substantial problems,Section 2.3 will explain how to compute with symmetric polynomials using Mapleand Mat hematica.

B. The Roots of a Polynomial. In Galois theory, symmetric polynomials areoften evaluated at the roots a1, . . . , a polynomial f e F [xl. The following resultwill be crucial.

Corollary 2.2.5 Let f F [x] be a monic polynomial of degree n > 0 with rootsin a larger field L. Then, given any symmetric polynomial p(xI,. . .

with coefficients in F, we have

Proof: The evaluation map F[xi, . . . —* L defined by p '—* p(ai,... ,a,1) is a ringhomomorphism by Theorem 2.1.2.

Since p is symmetric in x1, . . . Theorem 2.2.2 implies that p is a polynomialin the a,- with coefficients in F. Hence, when we evaluate at . . . , we see thatp(ai,. .. , a polynomial in the ar(ai,... , with coefficients in F.

Corollary 2.1.5 tells us that a,-(ai, . . . is, up to sign, a coefficient of f. Sincef E F[x] by hypothesis, we conclude that ar(ai,...,an) E F. The corollary nowfollows immediately from the previous paragraph. .

Here is an example of how Corollary 2.2.5 works.

Example 2.2.6 Suppose that f = x3 + +x+ 7 E Q[x] has roots , a2, a3 E C.Let g be the monic polynomial whose roots are a1 + a2, a1 + a3, and a2 + a3. Weclaim that g has coefficients in Q. To prove this, note that g can be written

g(x)= (x—(al+a2))(x—(aI+a3))(x—(a2+a3))

(2.17)X — (2a1 +2a2+2a3)x2

+ 3a1a2 + 3a,a3 + 3a2a3)x

—(ai +a2)(ai +a3)(a2 + a3).

The coefficients of (2.17) are symmetric polynomials evaluated at aI,a2,a3. Sincethe a1 are the roots of a polynomial with coefficients in Q, Corollary 2.2.5 impliesthat the coefficients of g are in Q. Hence g E Q[x].

We can also determine g explicitly. In general, if f = x3 + bx2 + cx + d has rootsaI,a2,a3, and g is the polynomial with roots +a2, ai +a3, and a2+a3, asdefined in (2.17), then the techniques of this section imply that

g(x) =x3+2bx2+(b2+c)x+bc—d


(see Exercise 4 for the details). For f = x3 + 2x2 + x + 7, it follows that

g(x) l)x+2. 1 —7=x3+4x2+5x—5

is the polynomial whose roots are the sums of distinct pairs of roots off.

C. Uniqueness. Every symmetric polynomial in x1, . . . can be written in termsof a!, . . . , by Theorem 2.2.2. We now prove that this expression is unique.

Theorem 2.2.7 A given symmetric polynomial can be expressed as a polynomial inthe elementary symmetric polynomials in only one way.

Pmof: We will use the polynomial ring F[ui,. . . , un], where u1,... , are newvariables. By Theorem 2.1.2, the map sending u, to E F [x1,. . . , defines a ringhomomorphism

In other words, if h = h(ui, . . . , is a polynomial in u1,. . . , with coefficients inF,thenip(h)

The image of is the set of all polynomials in the 01 with coefficients in F. Wedenote this image by

F[ai,... is a subring of F[xi, . . . In this notation, we can writeas a map

(2.18)

This map is onto by the definition of F . . , and uniqueness will be proved byshowing that is one-to-one. Be sure you understand this.

To prove that is one-to-one, it suffices to show that its kernel is {O}. Thuswe must show that if h is a nonzero polynomial in the u,, then h(ai,... gives anonzero polynomial in the x,. We will sketch the main idea of the argument and leavethe details for Exercise 5.

Let be a nonzero term of h. Applying gives and theargument of (2.11) shows that the leading term of this polynomial is

cx1 x2 x,,.

Since h is the sum of its terms, is the sum of the corresponding polynomialsc at', each of which has a leading term as displayed above. The crucial fact isthat the map

is one-to-one, so that the leading terms can't all cancel. Hence ço(h) can't be the zeropolynomial, and uniqueness follows. See Exercise 5 for the details. •


The proof of Theorem 2.2.7 constructs a ring isomorphism

(2.19)

where F[ui,... , is the polynomial ring in variables ui,... , u, '—+ Hencewe can regard at,.. , a,, as independent variables. This leads to the followinginteresting application.

Using the above variables UI,... , we call

(2.20)

the Universal polynomial of degree n (the reason for the signs will soon becomeclear). This name is justified because if f = + + + + E

is any monic polynomial of degree n, then the evaluation map sending to (— 1)'a1takes f to f. Thus the universal polynomial of degree n can be mapped to any monicpolynomial of degree n with coefficients in F.

We can construct the roots of f as follows. Under the isomorphism (2.19), thepolynomial (2.20) maps to — + . . . + (—1 But F . . , a,,] lies in thelarger ring F [XI,... , and in this ring, (2.4) gives the factorization

x" + = (x—xi).. .(x—xn)

Inotherwords,f —aix"t hasrootsx1,...,x,,.Because of this, we identify (2.20) with its image under (2.19) and call

(2.21)

the universal polynomial of degree n. Then f is not only universal in the above sensebut also has known roots, namely Xi,... ,X,,.

As mentioned in the Historical Notes to Section 1.2, Lagrange studied the roots of apolynomial without regard to their numerical value. For a monic polynomial of degreen > 0, this means considering its roots as variables x1,... ,x,,. The above discussionshows that in modem terms, Lagrange was studying the universal polynomial f.

Mathematical Notes

Let us discuss further two ideas that appeared in this section.

Algebraic Independence. The uniqueness proved in Theorem 2.2.7 implies inparticular that the map (2.19) is one-to-one. Hence there are no nontrivial polynomialrelations among the (since any such relation would give a nonzero element in thekemel). When this happens, we say that . . . , a,, are algebraically independent.Not all collections of polynomials in F [xi,... , are algebraically independent. SeeExercise 6 for an example.

• Symmetric Rational Functions. The polynomial ring F[xi,... sits insideF(xi,. . . ,x,,), the field of rational functions in Xi, . . . ,x,, with coefficients in F. In


this situation, one can ask which elements of F(xi,. .. are symmetric, i.e., areunchanged under all permutations of the variables. An example is

1 1 1—+—+...+—.X1 X2

Using a common denominator, one can express this as

+ X1X3 + + X1 =x1...xn 0-n.

More generally, one can show that any symmetric rational function in x1, . . .

is a rational function in the elementary symmetric polynomials. In other words,all symmetric elements of F(xi, . . . lie in the subfield F(o-i,... , of rationalfunctions in cri,... , This will be proved in Exercises 7 and 8.

In Chapters 6, 7, and 8, we will study

F(cri,...,an) C F(xi,...

from the point of view of Galois theory. We will see that the Galois group of thisfield extension is the symmetric group This in turn will enable us to determinewhen one can solve polynomials of degree n by radicals.

Historical Notes

Symmetric polynomials have been around for a long time. In 1629 Albert Girardpublished Invention nouvelle en l'algebre, which contains a clear description of theelementary symmetric polynomials. Girard also considers the power sums

Sr = +

In the notation used above, note that 5r = xc. Girard gives formulas for s1 , S2 ,53,54

in terms of the (see Exercise 17).In 1665—1666 Isaac Newton worked out many examples of symmetric polynomi-

als, expressing them in terms of the His 1707 book Arithmetica universalis showshow power sums relate to elementary symmetric polynomials. For r = 1, the relationis trivial, namely Si ai, and for r> 1, we have the Newton identities, which statethat

(222)SraiSr_I_a25r_2+..+(_1)ntrar

ifr>n.

Proofs of these identities can be found in [2, Ch. 7, §1], [3, pp. 62—63, 72—73], and[7,pp. 114—115].

As already noted, Waring's Meditationes from 1782 contains an im-plicit version of the algorithm used in the proof of Theorem 2.2.2, though in exampleshe often used clever shortcuts. The Fundamental Theorem of Symmetric Polynomi-als was widely known and used in the eighteenth century, though the first complete


proof is due to Gauss. He was also the first to raise the issue of uniqueness, and hisproof is the one we used.

One difference between Gauss's proof of Theorem 2.2.2 and ours is that he orderedhis polynomials differently. In [Gauss, Vol. III, p. 36], he says:

Dein e duobus terminis

et

priori ordinem altiorem tribuemus quam posteriori, si fit

vel a> a', vel a = a', /3 > a = a', /3 = /3',y > y', vel etc.

i.e. si e differentiis a — a', /3 — /3', -y — -y' etc. prima, quae non evanescit, positivaevadit.

Even though this is in Latin, the meaning is quite clear once one realizes that "vel...vel ... vel" means "either... or... or." This is now called lexicographic order. InExercise 9 you will use this order to prove Theorem 2.2.2.

Although our interest in symmetric polynomials is due to their importance inGalois theory, these polynomials also arise naturally in invariant theory, algebraiccombinatorics, and representation theory. A basic reference for symmetric polyno-mials is [4]. See also [Tignol, Chs. 4, 8] for the history of symmetric polynomials.


Exercise 1. Show that the leading term of ar iS XIX2 Xr.

Exercise 2. This exercise will study the order relation defined in (2.5). Given an exponentvector a = (al,. . . where each a1 � 0 is an integer, let xa denote the monomial

If a and /3 are exponent vectors, note that = Also, the leading term of a nonzeropolynomial f E Fki , . . . will be denoted LT(f).(a) Suppose that > and let be any monomial. Prove that >(b) andf(c) Let f,g E F[xi,. . . be nonzero. Prove that LT(fg) = LT(f)LT(g).

Exercise 3. Prove (2.1 3)—(2. 16). For (2.13), a computer will be helpful; the others can beproved by hand using the identity

Exercise 4. Let f = x3 + bx2 + cx + d E F[x] have roots al , a2, a3 in a field L containing F,and let g be the polynomial defined in (2.17). Show carefully that

g(x) =x3+2bx2+(b2+c)x+bc—d.

Exercise 5. This exercise will complete the proof of Theorem 2.2.7. Let h E F{ui, . . . , be

a nonzero polynomial. The goal is to prove that h(ai, . . . , is not the zero polynomial inX1,.. . ,Xn.


(a) If c is a term of h, then use Exercise 2 to show that the leading term of c .

(b) Show that (b1, . . . (1,1 + +bn,b2 + . . . . is one-to-one.(c) To see why h(o-i,...

, a,,) is nonzero, consider the term of h(ui, . . . , u,,) for which theleading term of c - - . is maximal. Prove that this leading term is in fact the leadingterm of h(ai,. . . , a,,), and explain how this proves what we want.

Exercise 6. Here is an example of polynomials which are not algebraically independent.Consider E F[xi ,x2J, and let 4 : F[ui , u2, u3] —* F[xi ,x21 be defined by

q5(u3)=4.

Show that is not one-to-one by finding a nonzero polynomial h E F [u i , U2, 143] such that= 0. (Using the notion of transcendence degree, one can show that any collection of

three or more elements in F[xi ,x2] is algebraically dependent. See, for example, [Jacobson,Vol. II, Sec. 8.12].)

Exercise 7. Given a polynomial f E F [XI, . . . , xn] and a permutation a E let a f denote thepolynomial obtained from f by permuting the variables according to a. Show that flaES a-fand a-f are symmetric polynomials.

Exercise 8. In this exercise, you will prove that if E F(xi,. . ,x,,) is symmetric, thenis a rational function in aI,. . . , an with coefficients in F. To begin the proof, we know that

= A/B, where A and B are in F[xi, . . . ,x,,]. Note that A and B need not be symmetric; onlytheir quotient = A/B is. Let

C= a-B,ES,,\ {e}

where we are using the notation of Exercise 7.(a) Use Exercise 7 to show that BC is a symmetric polynomial.(b) Then use the symmetry of = A/B to show that AC is a symmetric polynomial.(c) Use = (AC)/(BC) and Theorem 2.2.2 to conclude that p is a rational function in the

elementary symmetric polynomials with coefficients in F.

Exercise 9. In the Historical Notes, we gave Gauss's definition of lexicographic order.(a) Give a definition (in English) of lexicographic order.(b) In the proof of Theorem 2.2.2, we showed that graded lexicographic order has the property

that there are only finitely many monomials less than a given monomial. In contrast, thisproperty fails for lexicographic order. Give an explicit example to illustrate this.

(c) In spite of part (b), lexicographic order does have an interesting finiteness property.Namely, prove that there is no infinite sequence of polynomials fi that havestrictly decreasing leading terms according to lexicographic order.

(d) Explain how part (c) allows one to prove Theorem 2.2.2 using lexicographic order.Besides graded lexicographic order and lexicographic order, there are many other ways toorder monomials. See [2, Ch. 2, §2].

Exercise 10. Apply the proof of Theorem 2.2.2 to express in terms of al , a2, a3.

Exercise 11. Let the roots of y3 + 2y2 3y + 5 be a, 'y E C. Find polynomials with integercoefficients that have the following roots:(a) a/3, cry, and(b) a+l,8+l,and'y+l.(c) a2,/32,and-y2.


Exercise 12. Consider the symmetric polynomial f =(a) Prove that f has n! terms when a! ,.. . , are distinct.(b) (More challenging) Suppose that the exponents ai,. . . , break up into r disjoint groups

so that exponents within the same group are equal, but exponents from different groups areunequal. Let L denote the number of elements in the ith group, so that Li + + =Prove that the number of terms in f is

n!

For example, f = has Li = £2 = 2 and £3 = 1. It follows that f has5!/(2!2!l!) = 30 terms.

Exercises 13—16 will discuss some classic tricks for dealing with symmetric polynomials. Apolynomial g E F[xi,.. . is homogeneous of total degree d if every nonzero term of g hastotal degree d.

Exercise 13. Let g , E F [Xi,. . , x,,J be homogeneous of total degrees d1 , d2.

(a) Show that gig2 is homogeneous of total degree d1 + d2.(b) When is gi + g2 homogeneous?

Exercise 14. We define the weight of . . . to be + 2a2 + 3a3 + . . + na,.(a) Prove that of," is homogeneous and that its weight is the same as its total degree

when considered as a polynomial in xi,.. ,

(b) Letf E F [xi,. . . , be symmetric and homogeneous of total degree d. Show that f is alinear combination of products of weight d.

Exercise 15. Given a polynomial f e F[xi,. . . let deg1(f) be the maximal exponent of xwhich appears in f. Thus f = 4x2 +xi4 has deg1(f) = 3 and deg2(f) = 4.

(a) 1ff is symmetric, explain why the deg1 (f) are the same for i = 1,. . . , n.

(b) ...r') fori= l,...,n.Exercise 16. This exercise is based on [7, pp. 110—112] and will express the discriminant

= (xi —x2)2(xi —x3)2(x2 —x3)2 in terms of the elementary symmetric functions withoutusing a computer. We will use the terminology of Exercises 14 and 15. Note that is

homogeneous of total degree 6 and = 4 for i = 1,2,3.(a) Find all products of weight 6 and 4.

(b) Explain how part (a) implies that there are constants Li,. . . , Ls such that

= +L2aia2a3

(c) We will compute the by using the universal property of the elementary symmetricpolynomials. For example, to determine use the cube roots of unity 1, w, w2 to showthat x3 — 1 has discriminant —27. By applying the ring homomorphism defined by xi i—* 1,x2 w, X3 w2 to part (b), conclude that = —27.

(d) Show that x3 — x has roots 0, ± 1 and discnminant 4. By adapting the argument of part (c),conclude that £4 = —4.

(e) Similarly, use x3 2x2 + x to show that £5 = 1.(1) Next, note that x3 — 2x2 — x + 2 has roots ± 1,2, and use this (together with the known

values of , £4, £5) to conclude that £2 4L3 = 34.

(g) Finally, use x3 3x2 + 3x — Ito show £2 + 3L3 = 6. Using part (f), this implies £2 = 18,

£3 = —4 and gives the usual formula forOther examples illustrating this method can be found in [1, pp. 442—444].


Exercises 17—20 will study power sums Sr = 4+ ... + 4 and the Newton identities (2.22)discussed in the Historical Notes.

Exercise 17. Use the Newton identities (2.22) to express the power sums s2, S3, S4 in terms ofthe elementary symmetric polynomials 0!, 02, CT3, CT4.

Exercise 18. Suppose that complex numbers

3,

=5= 12.

E Zforalln >4. Alsocomputea4+/34+'y4.

Exercise 19. Suppose that F is a field of characteristic 0.(a) Use the Newton identities (2.22) and Theorem 2.2.2 to prove that every symmetric

polynomial in F[xi, . . . can be expressed as a polynomial in SI,...(b) Show how toexpress EF[xi,x2,x3,x4] asapolynomial in 51,52,53,54.

Exercise 20. Let F2 be the field with two elements. Show that in F2 [x1,. . . , it is impossibleto express 02 as a polynomial in SI,... when n 2.

2.3 COMPUTING WITH SYMMETRIC POLYNOMIALS (OPTIONAL)

The method described in Section 2.2 for expressing a given symmetric polynomialin terms of , ci,, is useful for simple problems, but can be cumbersome in morecomplicated situations. Fortunately, computer algebra programs such as Maple orMathematica make it relatively easy to represent symmetric functions in terms ofthe elementary symmetric polynomials. We will discuss briefly how these powerfulprograms can be used to manipulate symmetric polynomials. (One can use Mapleand Mathematica in other parts of Galois theory as well—see [Swallow].)

Although few readers will have access to both Maple and Mat hematica , we suggestreading both discussions in order to better appreciate the underlying ideas.

A. Using Mathematica. We begin by using Mathematica to write the discriminant

= (Xi —X2)2(Xi —x3)2(x2 —X3)2

from Section 1.2 in terms of the elementary symmetric polynomials. We can thinkof this in terms of the system of equations

= (Xi X2)2(XI —x3)2(x2 X3)2,

(2.23)X1 +X2+X3,

XIX2 +XjX3 +X2X3,

cT3 X1X2X3.

The idea is to eliminate x1 ,X2,X3 from these equations. This will give the desiredexpression for in terms of a1 , U3.

COMPUTING WITH SYMMETRIC POLYNOMIALS (OPTIONAL) 43

We tell Mathematica to do this elimination using the command

Eliminate [ {Delta== el ==

xl+x2+x3, e2==xl x2+xl x3+x2x3, e3==xl x2x3}, {xl,x2,x3}]

The output is

—e12e22 + 4e13e3 — 18e1 e2e3== — Delta — 4e23— 27e32,

which tells us that

= — + 4a?73 +

This agrees with (1.20) from Section 1.2 after the substitution b = C = a2, andd = —U3.

Using the Eliminate command is straightforward, though having to enter theelementary symmetric polynomials by hand can be time-consuming, especially whenthe number of variables is large. This can be avoided by using the Mathematicapackage SyminetricPolynomials that comes with the program. This package isloaded by

<<Algebra'SyininetricPolynomials'

Then the above computation can be done using the command

SyminetricReduction[{xl,x2,x3},{el,e2,e3}]

The output is the two-element list

{e12e22—4e23—4e13e3-f- 18e1e2e3—27e32,0}

where the second element, 0, tells us that is in fact symmetric, and the first elementis the polynomial expressing in terms of the To go directly to the first elementof this list, one could give the command

SymmetricReduction[{xl,x2,x3},{el,e2,e3}]

since in general Mathematica uses to extract the ith element of a list.Here is an example which illustrates one of the interesting things which can be

done with symmetric polynomials.

Example 2.3.1 Let a3 E C be the roots of y3 + 2y2 — 3y + 5. Our goal is touse Mathematica to find the cubic polynomial whose roots are ,

Let Yi ,Y2,Y3 be variables and define the polynomial f in Mathematica to be

f =


Note that the evaluation F—* takes f to the polynomial we want. If we multiplyout f, we get a polynomial whose coefficients are symmetric in Yi , Y2 y3. We expressthese in terms of the elementary symmetric polynomials using the Mathematicacommand

Do[Print[SymmetricReduction[Coefficient[f,y, 1],{yl,y2,y3},{el,e2,e3}]

This instructs Mathematica to print out the coefficients off expressed in terms of the

elementary symmetric polynomials, here denoted e 1, e2, e3. The output is

constant term —e33,

(2.24) coefficient of y : e23 —3 el e2 e3 + 3

coefficientofy2: —e13+3e1 e2—3e3.

The evaluation '—+ sends el '-+ —2, e2 i—* —3, e3 '—+ —5. Using (2.24), we seethat y3 + 41y2 + l38y + 125 is the polynomial with roots (4.

The formulas (2.24) imply that for any cubic polynomial, we can find a cubicpolynomial whose roots are the cubes of the given one. This is part of the universalaspect of the elementary symmetric polynomials.

B. Using Maple. As above, our first Maple computation will be to express the

discriminant = (Xi — x2)2 (x1 — x3)2(x2 x3)2 in terms of , Cr3. We will againuse the equations (2.23) to eliminate x1 ,x2,x3, which will give the desired expression

To do this in Maple, we proceed as follows. The last three lines of (2.23) give thepolynomials

(2.25) e1 X1 X2 —X3, e2 X1X2 X1X3 X2X3, e3 —X1X2X3

in C {x1 , , , , e2, e3]. These generate an ideal in this ring, and we eliminatex1 ,X2,X3 from by replacing all instances of xi +X2 +x3, XIX2 +X1X3 +X2X3, x1x2x3with e1, e2, e3, respectively. This operation can be thought of as the normal form of

with respect to the ideal generated by (2.25).The first step is to load the Maple package Groebner, which contains the com-

mands we need. This is done by

with(Groebner);

We next tell Maple to order the monomials in C [Xi , X2 , x3, ei , e2, e3] via

I := lexdeg([xl,x2,x3],[el,e2,e3]);

The monomial order T is specially designed for elimination and is more efficientthan the graded lexicographic order used in the proof of Theorem 2.2.2. We need to

COMPUTING WITH SYMMETRIC POLYNOMIALS (OPTIONAL) 45

specify a monomial order because the precise definition of "normal form" dependson how the monomials are ordered.

Once we have the monomial order T, we compute an intermediate object called aGröbner basis using the command

GB:= Basis([el-xl—x2—x3,e2—xl*x2—xl*x3—x2*x3,

e3—xl*x2*x31,T):

Note that Maple uses * for multiplication. We also used : to suppress output, since wedon't need to see the Gröbner basis. Roughly speaking, the Gröbner basis consists ofpolynomials that generate the same ideal as (2.25) and are optimized for the monomialorder T.

The final step is to compute the normal form. This is precisely what the Maplecommand NormalForm does:

* * GB, T);

This gives the output

—4 * e13 * e3 + 18e1 * e2 * e3 27* e32+ e12 * e22 —4 * e23

which agrees with our earlier computation.The Mathematjca command Eliminate described earlier uses a Gröbner basis

computation similar to what we did here. Gröbner basis methods can be applied to avariety of elimination problems. The full details can be found in [2].

Notice that once the monomial order T is defined and the Gröbner basis GBis computed, NormalForm can be used repeatedly to write a series of symmetricpolynomials in terms of the elementary symmetric polynomials.

Example 2.3.2 In Example 2.3.1, we used Mathematica to find a polynomial whoseroots were the cubes of the roots of y3 +2y2 — 3y+S. Let us redo thisexample using Maple. We first enter the polynomial

f := (y * (y * (y —

Then the Maple command

for i from 0 to 2 do print(NormalForm(coeff(f,y,i),GB,T)) od;

prints out the coefficients of f expressed in terms of the elementary symmetricpolynomials. As in (2.24), the result is

constant term :

coefficient ofy : —3 el e2 e3 + 3 e32 + e23,

coefficient ofy2: —3 e3 + 3 el e2 — e13,

and the evaluation el F-+ —2, e2 F—+ —3, e3 '—* —5 shows that the roots of y3 +41y2 +

Similar examples are given in the exercises.



Exercise!. Examples 2.3.1 and 2.3.2 showed that the roots of y3 +41y2 + l38y+ 125 are thecubes of the roots of y3 + 2y2 — 3y + 5. Verify this numerically.

Exercise 2. Use the method of Example 2.3.1 or 2.3.2 to find the cubic polynomial whoseroots are the fourth powers of the roots of the polynomial y3 + 2y2 3y + 5.

Exercise 3. Express in terms of the elementary symmetric polynomials. This examplewas first done by Newton around 1665.

Exercise 4. Given a cubic x3 + bx2 + cx + d, what condition must b, c, d satisfy in order thatone root be the average of the other two?

Exercise 5. Given a quartic x4 + bx3 + cx2 + dx + e, what condition must b, c, d, e satisfy inorder that one root be the negative of another?

Exercise 6. Find the quartic polynomial whose roots are obtained by adding 1 to each of theroots ofx4+3x2+4x-F7.

2.4 THE DISCRIMINANT

Given n > 2 variables Xi,... over a field F, the discriminant is

fi1) factors in this product. Furthermore, since — x3)2 =

(x, — (x1 — x1), we can rewrite the above formula as

[JI

This shows that if we permute the variables, then we still have the product of thedifferences of all distinct pairs of variables. Thus is symmetric in Xi,... ,

Theorem 2.2.2 implies that can be written as a polynomial in the elementarysymmetric polynomials a1,... In other words,

When n 3, the formulas of Section 1.2 (or the methods of Section 2.3) imply that

(2.26) — + — + 1 a2a3.

For general n an explicit formula for in terms of , will be given in theMathematical Notes.

The definition of shows that it has a square root in F[xi, . . . We define

[JI<i<j<n

THE DISCRIMINANT 47

We next describe how transforms under permutations.

Proposition 2.4.1 If a E then

a defined in (A.3), and is the polynomial obtainedfrom by permuting the variables Xi, . . . ,Xn according to a.

Proof: In 1841, Jacobi studied how transforms under a transposition (if). Hisargument, adapted to our notation, goes as follows. We can assume i <f. Thenobserve that there is E {+1, —1} such that

(2.27) [J (X1xm).

1<m

This follows because the factors appearing in the right-hand side are, up to sign, thefactors of For example, when k i, j, then

IXj—Xk, i<k,Xj—Xk=c

1_(xk_xj), k <1.

Combining all of these signs gives E = +1 in (2.27). Since the transposition (if)takes (x1 —xk)(xJ —Xk) to —xk) and doesn't affectx1 Xm for l,m i,j,we see that (2.27) implies that (if) . =

Now let a and write a as a product of transpositions, say a = -re. Thenthat

(2.28)

Since sgn(a) = (_l)t (be sure you understand why), the proposition follows. •

We next define the discriminant of a monic polynomial

f + + + + an E F[xI

of degree n � 2. As in Section 2.2, the universal polynomial

+ + . . . +

maps to f via Since is a symmetric polynomial, we can write

(2.29)

Then we define the discriminant off, denoted to be

(2.30) = ,(_1)la,,... E F.

Thus the evaluation a1 i—+ takes! to f and A to A(f).


We also define = 1 when f has degree 1. This will be useful later.

Example 2.4.2 Consider f x3 + bx2 + cx + d. We saw in (2.26) that

Since the evaluation is given by cr1 —b, ci2 c, and a3 '—* —d, we obtain

=

= —4c3 27(—d)2 + (—b)2c2 —4(—b)3(—d) + 18(—b)c(—d)

= —4c3 —27d2+b2c2 —4b3d+ l8bcd.

This agrees with the formula (1.20) found in Section 1.2.

In the case when we know the roots of a polynomial, we get the following formulafor its discriminant.

Proposition 2.43 Suppose that a monic polynomial f E F [xJ of degree n � 2 hasroots . . . , in afield L containing F. Then

[J (cx_cx)2

1�i<j—<n

Pmof: In F[xi,. . we know that = —x3)2. Now consider theevaluation map that takes x, to cx,. Since a ring homomorphism, thistakes to

fJ1<i<j<n

By Corollary 2.1.5, this equals

=

by the definition of

Mathematical Notes

There are several ideas in this section in need of further discussion.

• The Action of the Symmetric Group. This chapter used the action of on thepolynomial ring F{xi,... ,x,,]. Foro E andf E F[xi,... a.f is the polynomialobtained by permuting the variables according to This operation has the followingproperties:

a• (f + g) = a f + a g,(2.31) a.(fg) = (a.f)(a.g),

r.(af)=(ra).f

THE DISCRIMINANT 49

for E Sn and f,g E F[xi, . . . We have used these properties implicitlythroughout the chapter, and we will give a formal proof of (2.31) in Chapter 6. Thefirst two imply that f i—* a f is a ring homomorphism from F[xi, . . . to itself,and the last implies that (a, f) a f is a group action, as defined in Section A.4.

The Alternating Group. Let F be a field of characteristic different from 2.Proposition 2.4.1 implies that a. = sgn(a) Since —1 1 in F,

sgn(a)=1

Thus the alternating group is the subgroup of permutations that fixThis leads to the question of which other polynomials or rational functions are

fixed by The answer is as follows.

Theorem 2.4.4 Let F be afield of characteristic 2. 1ff E F(xi,... ,Xn) is invariantthen there are A,B E F(ai,... such that

f=A+Bf&

Furthermore, f E F [Xi,... , Xn] implies that A, B E F [a!,.. . , •

We will prove this theorem in Chapter 7. We will also explain how it relates to theGalois correspondence between subgroups of and subfields of F (Xi,... , whichcontainF(ai,...

• The Existence of Roots. Our discussion of discriminants raises an interestingquestion about roots. Given a monic polynomial f, our definition of involvesonly the coefficients of f. However, if we know the roots of f, then we get the simplerformula given by Proposition 2.4.3. This brings up the fundamental question: Doesevery polynomial in F [x] have roots in a possibly larger field? We will answer thisquestion in Chapter 3.

• Discriminant Formulas. The polynomial expressing in terms of a1,.. . , getsmore and more complicated as n increases. But if we use determinants, then we getseveral compact ways to represent both and We begin with Vandermonde '5formula for See [7, p. 56] for a proof.

Proposition 2.4.5

...x:—2

: :

Xi X2 Xpj

1 1 ... 1

The determinant on the right is called a Vandermonde determinant. •


In Exercise 1, YOU will use Proposition 2.4.5 to derive the following formula forintermsofthepowersumssr

S2n_2 S2n_3

S2n_3 S2n_4(2.32)

:

Sn_2 S0

In Section 5.3 we will discuss the relation between discriminants and resultants.This leads to a formula that uses the (2n — 1) x (2n — 1) matrix M defined by

1 n

—aI . —(n—l)aI fl

1 (n—2)a2 —(fl—1)aI

(n—2)a2 fl

—(n—l)ai

(fl—2)o2

.

(_

(—

1 columns n columns

where the empty spaces are filled in with zeros. In (5.13) we will see that

=

This gives an explicit representation of in terms of as,. . . ,

See [5] for further comments on computing discriminants.

Historical Notes

The discriminant can be represented as a polynomial in x1,... ,x,, and also asa polynomial in the elementary symmetric polynomials co,... The second ofthese came first, since the discriminant for n = 3 is implicit in Cardan's formulasfrom Chapter 1. By the 1 770s, Lagrange and Vandermonde knew the properties of

and for small n. For example, when n = 4, Lagrange explicitly stated that atransposition changes the sign of

The general form of the discriminant was defined independently by Cauchy in1815 and Gauss in 1816. Cauchy did this in his pioneering studies of the symmetricgroupS,,. For him, a polynomial was symmetric if it was unchanged by transpositions,so that the next class of functions to study were those which changed sign under a

THE DISCRIMINANT 51

transposition. In essence, Cauchy proved that if F has characteristic different from2 and f F[xi,.. . satisfies r . f —f for all transpositions r, then f =

B E F [al, . . . , In Exercise 2 you will show that this follows fromTheorem 2.4.4.

Cauchy also considered determinants, drawing on earlier work of Vandermondeand Laplace. He proved Proposition 2.4.5, though he mistakenly attributed it toVandermonde. In 1841 Jacobi gave the argument used to prove Proposition 2.4.1.

Gauss studied the discriminant in his second proof of the Fundamental Theoremof Algebra. His discussion of is surprisingly modern. Like us, he initially defines

as a polynomial in F[xi,... and then shows that it lies in F[ai,... Usingthe isomorphism

from (2.19) in Section 2.2, Gauss defines E F[ui,... to be thepolynomial such that = in F[xi, . . . Finally, given a monicpolynomial f E F[xI, Gauss defines just as we did in (2.30). His notation andterminology are different, but his treatment is virtually identical to ours.


Exercise 1. Let M be the n x n matrix appearing on the right-hand side of the Vandermondefonnula given in Proposition 2.4.5. Prove that (2.32) follows from the fact that M and itstranspose both have determinant

Exercise 2. Let F have characteristic 2, and let f E F{xi,. . . ,x,] satisfy y . f = —f for alltranspositions r E S,,. Prove that f = B E F[ai,. . . ,

Exercise 3. Let f = x2 + bx + c E F [x]. Use the definition of discriminant given in the text toshow that = b2 — 4c.

Exercise 4. Let f F[x] be monic, and assume that f = (x — a!) . . . (x — a,,) in some field Lcontaining F. Prove that 0 if and only if ai,. . . , a,, are distinct. This shows that f hasdistinct roots if and only if its discriminant is nonvanishing.

Exercise 5. Show that F [xi, . . . , x,,] is symmetric if and only if F is a field of character-istic 2.

Exercise 6. Exercise 5 showed how things can differ in a field of characteristic 2. Anotherexample comes from the quadratic formula, which doesn't apply over such fields because ofthe 2 in the denominator. This exercise will describe how to solve quadratic equations over afield F of characteristic 2.(a) Given b E F, we will assume there is a larger field F C L such that b = /32 for some E L.

Show that /3 is unique and that /3 is the unique root of x2 + b. Because of this, we denoteby

(b) Now suppose that f = x2 + ax + b is a quadratic polynomial in F [x] with a 0. Supposealso that f is irreducible over F, so that it has no roots in F. We will see in Chapter 3that f has a root a in a field L containing F. Prove that a cannot be written in the forma = u + where u, v, wE F.

(c) Part (b) shows that solving a quadratic equation with nonzero x-coefficient requires morethan square roots. We do this as follows. If b E F, let R(b) denote a root of x2 + x + b


(possibly lying in some larger field). We call R(b) and R(b) + 1 the 2-roots of b. Provethat the roots of x2 +x+ b are R(b) and R(b) + 1, and explain why adding ito the second2-root gives the first. Note that in characteristic 2, square roots behave as follows: Ifone square root is then we multiply by —ito get the other square root Incharacteristics 2, 2-roots work the same way, provided that we replace "multiply by —1"with "add 1."

(d) Show that the roots of f = x2 + ax-f- b, a 0, are aR(b/a2) and a(R(b/a2) + 1).It follows that when F has characteristic 2, then the roots of x2 + ax + b E F{x] are

a=0,

a field of characteristic 2.

Exercise 7. Explain how the third property of (2.31) was used (implicitly) in (2.28) in theproof of Proposition 2.4.1.

Exercise 8. As explained on page 37, we can regard F[cii,... ,cr,,] as a polynomial ring in thevariables , ci,,. In this exercise, you will prove that although factors in F(xi,. . ,x,,], itis irreducible in F[o-i,. . . , ci,,] when F has characteristic different from 2. To begin the proof,assume that = AB, where A, B E F[ai, . . , are nonconstant.(a) Using the definition of and unique factorization in F [Xi, . . . , x,,], show that A is divisible

(b) Given 1 <i<j<nandl <1<m<n,showthatthereisapermutationcrES,,suchthatci(i) = 1 and ci(j) = m.

(c) Use parts (a) and (b) to show that A is divisible by Xj — Xm for all 1 1 <m n.

(d) Conclude that A is a multiple of and that the same is true for B.

(e) Show that part (d) implies that A and B are constant multiples of and explain whythis contradictsA,B E F[o-i,...,o-,,j.

(1) Finally, suppose that F has characteristic 2. Prove that is not irreducible.

Exercise 9. For n = 4, the variables xi , , X3 , X4 have discriminant

= (xl X2)(Xi X3)(Xi —X4)2(X2 X3)(X2 —X4) (X3 X4)

Letyi X1X2 +X3X4, Y2 = x1x3 +X2x4, y3 = x1x4 +x2x3, and consider

0(y) = (y—yl)(y—y2)(y—y3).

This is a cubic polynomial in y. As in the text, the discriminant of 0 will be denoted

Show that = When we discuss Lagrange's work in Chapter 12, we will see that 0 isthe Ferrari resolvent, which plays an important role in the solution of the quartic equation.

Exercise 10. Let C, D E F [cii, . . . , be nonzero and relatively prime. This exercise willshow that C and D remain relatively prime when regarded as elements of F[xi,. . .

(a) Show that Dm are relatively prime in F [at, . . . , a,,] for any positive integer m.

(b) Suppose that p E F[xi, ,x,,] is a nonconstant polynomial dividing C and D. Prove thatcip divides C and D for alla E S,,.

(c) As in Exercise 7 of Section 2.2, let P = ci p. Show that P divides Cn! and Dn!,

and then use part (a) and Exercise 7 of Section 2.2 to obtain a contradiction.

REFERENCES 53

Exercise 11. Exercise 8 of Section 2.2 showed that if E F(xi,. . . is symmetric, thenE F(ai,. . . , o,,). In this exercise, you will refine this result as follows. Suppose thatE F(xi, . . . is symmetric, and write = A/B, where A,B e F[xi,. . . are relatively

prime. The claim is that A, B are themselves symmetric and hence lie in F[oi,. . . We canassume that A and B are nonzero.(a) Use the previous exercise and Exercise 8 of Section 2.2 to show that p = C/D where

(b) Show that AD = BC and then use unique factorization in F[xt, . . . , to show that A andB are constant multiples of C and D, respectively.

(c) Conclude thatA,B E F[cri,...,o-,,] as claimed.

REFERENCES

1. G. Chrystal, Algebra: An Elementary Text-Book, Fifth Edition, A. & C. Black, Lon-don, 1904. Seventh Edition by Chelsea, New York, 1964, reprinted by AMS Chelsea,Providence, RI, 1999.

2. D. Cox, J. Little, and D. O'Shea, Ideals, Varieties andAlgorithms, Third Edition, Springer,New York, Berlin, Heidelberg, 2007.

3. D. E. Dobbs and R. Hanks, A Modern Course on the Theory of Equations, PolygonalPublishing House, Passaic, NJ, 1980.

4. I. G. MacDonald, Symmetric Functions and Hall Polynomials, Oxford U. P., Oxford,1979.

5. J. MCKay, On computing discriminants, Amer. Math. Monthly 94 (1987), 523—527.

6. E. Waring, Meditationes algebraicir, English translation by D. Weeks, AMS, Providence,RI, 1991.

7. L. Weisner, Introduction to the Theory of Equations, Macmillan, New York, 1938.

CHAPTER 3

ROOTS OF POLYNOMIALS

This chapter will study the roots of a polynomial in one variable. We will firstshow that every nonconstant polynomial with coefficients in a field has roots in somepossibly larger field. Then, in the special case of a polynomial with coefficients inthe field C of complex numbers, we will show that the roots also lie in C.

3.1 THE EXISTENCE OF ROOTS

In this section, we will show that given a field F and a nonconstant polynomialf E F [xJ, there is a field L containing F which also contains all roots of f. We willmotivate our construction by considering the complex numbers C.

So far we have assumed the existence of the real and complex numbers. But ifwe're given just the real numbers R, how do we get C? There are several ways ofdoing this. For example, in 1835, Hamilton defined

C = {(a,b) a,b E

where addition and multiplication are given by

(a,b)+(c,d)= (a+b,c+d) and (a,b).(c,d) = (ac—bd,ad+bc).


56 ROOTS OF POLYNOMIALS

It is straightforward (though somewhat tedious) to verify that these operations makethe above set into a field with (1,0) as the multiplicative identity. Furthermore, theformula for multiplication implies that

(0, 1).(0, 1) = (—1,0) = —(1,0).

If we let 1 denote (1,0) and i denote (0, 1), then this equation becomes i2 = —1, andwe also have

(a,b)=a(l,0)+b(0,l)=a• 1In this way, we recover the usual description of C as the set of numbers of the forma + hi, where a,b IL

A very different definition of C was given by Cauchy in 1847. He worked in thepolynomial ring R[xI and defined

(3.1) [mod.x2+1]

to mean that x(x) e REx] have the same remainder on division by x2 + 1. Then,to simplify these congruences, he introduced the symbol i as follows:

the symbolic letter i, when substituted for the letter x in a polynomial f(x),indicates the value obtained, not by the polynomial f(x), but rather by theremainder of the algebraic division of f(x) by x2 + 1, when one attributes to xthe particular value i.

(See [Cauchy, p. 317].) This allowed Cauchy to replace (3.1) with the equivalentstatement

= x(i).

To illustrate how this works, Cauchy considered the polynomial

f(x) = (a+bx)(c+dx) =ac+bdx2+(ad+bc)x.

The remainder of f(x) on division by x2 + 1 is easily seen to be ac — bd + (ad + bc)x(be sure you see why), so that by the above quotation, f(i) is defined to be thesymbolic expression ac bd + (ad + bc)i. The same process, when applied to a + bxand c + dx, yields a + hi and c + di, respectively. From this Cauchy concluded that

(3.2) (a +bi)(c+di) = ac — bd+ (ad+bc)i.

Thus we have a symbolic construction of the complex numbers using remainders ofpolynomials in R[x] on division by x2 + 1.

From a modern point of view, we can explain Cauchy's construction as follows.In the notation of Section A. 1, x2 + 1 generates the ideal

in the ring R[x]. This gives the quotient ring

R[x]/(x2+1) = {g+(x2+l) IgER[x]},

THE EXISTENCE OF ROOTS 57

where we are using the coset notation of Section A. 1.Now, following Cauchy, we take E REx] and divide it by x2 + 1 to obtain

l)+a+bx

forauniqueq e R[x] anda,b e JR. Since cosets g+(x2 + l),h+(x2 + 1) in the ringJR[x]/(x2+ 1) are equal if and only ifg—h E (x2+ 1), we see that

1) =a+bx+(x2+ 1).

It follows that (3.1) is true if and only if and x give the same coset in R[x] / (x2 + 1).Since the remainder of (a + bx) (c + dx) on division by x2 + 1 is ac — bd + (ad + bc)x,we have

(a+bx+(x2+ 1))(c+dx+(x2+ 1))— (a+bx)(c+dx)+(x2+ 1)ac—bd+(ad+bc)x+(x2 + 1).

Hence Cauchy's construction of C is equivalent to the quotient ring R[x] / (x2 + 1).But we can do even better, for we can also interpret Cauchy's symbolic letter i as

the coset x + (x2 + 1). More precisely, if we identify 1 with 1 + (x2 + 1) and i withx+(x2+l),then

a+bx+(x2+ 1) =a 1and the symbolic multiplication (3.2) becomes the above multiplication of cosets.

However, interpreting Cauchy's construction as / (x2 + 1) gives only a ringstructure. In order for this quotient ring to be a field, we need (x2 + 1) to be a maximalideal. The following proposition will be useful.

Proposition 3.1.1 1fF is afield and f E F{x] is nonconstant, then the following areequivalent:(a) The polynomial f is irreducible over F.(b) The ideal (f) = {fg I g F[x] } is a maximal ideal.(c) The quotient ring F [x] / (f) is afield.

Pmof: The equivalence (b) (c) is Theorem A. 1.12 from Section A. 1. It remainsto prove (a) (b).

Suppose f is irreducible and I is an ideal of F[xI such that (f) C I C F[x]. ByTheoremA.1.l7fromSectionA.l,l= (g) forsomeg E F[x]. Thenfe (f) (g)implies that f = gh for some h E F[xl. Since f is irreducible, g or h must be constant.We will leave it as Exercise 1 for the reader to show that g constant implies I = F [xJ

and h constant implies I = (f). It follows that (f) is maximal.Conversely, suppose that (f) is maximal and let f = gh where g, h E F [x]. This

gives (f) c (g) c F[x]. Since is maximal, (g) must equal either (f) or F[x].In Exercise 1 you will show that the former implies that h is constant and the latterimplies that g is constant. Thus f is irreducible.

Since x2 + 1 is irreducible over R, Proposition 3.1.1 implies that R{x]/(x2 + 1) isa field. This completes our second construction of C.


One interesting feature of the above two constructions of C just given is thatneither contains R. This might seem contradictory, but consider the following:

• In Hamilton's construction, a complex number is an ordered pair (a, b) of realnumbers. In order for this definition of C to contain R, we must identify thenumber a E with the ordered pair (a, 0) E C.

• In the modem interpretation of Cauchy's construction, a complex number is acoset g + (x2 + 1). In order for this to contain IR, we must identify the numbera R with the coset a+ (x2 + 1) e C.

Both constructions give one-to-one homomorphisms R —* C that become inclusionsafter we identify R with its image in C. This motivates the following definition.

Definition 3.1.2 Given a ring homomorphism offields 'p : F —* L, we say that L is afield extension of F via 'p. We will usually identify F with its image

'p(F) = {'p(a) a E F} C L

and write F C L

Recall from Section A. 1 that by definition, a ring homomorphism maps 1 to 1.Using this, in Exercise 2 you will show that a ring homomorphism of fields 'p : F —* Lis automatically one-to-one and induces an isomorphism : F 'p(F). Hence oncewe identify F with 'p(F) C L via'p, we may regard F as a subfield of L. For the twoconstructions of C given above, this gives R C C, as desired.

Armed with this notion of a field extension, we can prove that every irreduciblepolynomial has a root in an extension field.

Proposition 3.1.3 1ff e F[xI is irreducible, then there is an extension field F C L=0.

Proof: Let I = (f), so that L = F [x] /1 is a field by Proposition 3.1.1. Furthermore,a E F gives the constant polynomial a EF[x], which in turn gives the coset a + I E L.Thus we get a natural map 'p : F —+ L. In Exercise 3 you will check that 'p is a

ring homomorphism, so that using the convention of Definition 3.1.2, we get a fieldextension F C L.

It remains to show that there is E L such that = 0. This is surprisinglyeasy. Motivated by Cauchy's symbolic construction, we set = x + I. To provethat f = a0f + + where a, E F. Then, recalling ouridentification of a E F with the coset a + I L, we have

=

= 0+!,

where the third equality uses the definition of addition and multiplication of cosets,and the last uses f + I = 0 + I f —0 EL Since 0 + I is the additive identity of L,we have = 0, as claimed. •


Recall the elementary fact that a E L is a root of a polynomial f E L[x] if and onlyif x — a is a factor off in L[x] (this is Corollary A. 1.15). Thus, to say that a field Lcontains all roots of f means that f factors as

where a1, . . . , E L. When this happens, we say that f splits completely over L.

Theorem 3.1.4 Let f E F[x] be a polynomial of degree n > 0. Then there is anextension field F C L such that f splits completely over L.

Pmof: We will prove this using induction on n = deg(f). If n = 1, then f =aox+a1, where 0 and ao,al E F. Setting L = F and a! = —al/ao implies thatf = ao(x — ai) and proves the theorem in this case.

Now suppose that deg(f) = n> 1 and that the theorem is true for n 1. Since Fis a field, F [x] is a UFD by Theorem 2.1.1. In particular, f has an irreducible divisorfi. If we apply Proposition 3.1.3 to fi E F[x], then we get an extension field F CF1

and an element a1 e F1 such that fi(ai) 0 in F1.Since fi is a factor of f, we also have f(ai) 0 in F1. As noted above, this

implies that x a1 is a factor off in F1 [x}. In other words,

f=(x—ai)g

for some g e F1 [xJ of degree n — 1. Applying our inductive hypothesis to g, we get afield extension F1 C L and elements a2,... , E L such that

g=ao(x—a2)...(x—an).

The displayed formulas for f and g show that f splits completely over L. U

Mathematical Notes

This section includes several ideas which are worthy of comment.

• Identifications. In Definition 3.1.2, we wrote a field extension F —+ L as F C Lby identifying F with This might seem like cheating, but it happens all thetime in mathematics. For example, consider Z c Q. Since Q is the field of fractionsof the integral domain Z, an element a/b E Q is the equivalence class

(3.3) = {(c,d) I c,d E Z, d 0,ad = bc}.

(See Exercise 4 for the details.) In particular, according to (3.3), an integer n E Zdoesn't equal the fraction n/i E Q, since n is an integer and n/i is an infinite set ofordered pairs of integers. Rather, we have the ring homomorphism

which sends n to n/ 1, and we write Z C Q by identifying Z with This is similarto what we did in the discussion preceding Definition 3.1.2.


• Construction of Extension Fields. Beginning students in algebra often havedifficulty with cosets and quotient rings. The key insight is that in a quotient ringR/I, elements of the ideal 1 become zero. This is because r e I gives the coset r + I,which equals the zero coset 0+!, since r 0 e I. Applying this to the situation ofProposition 3.1.3, f E (f) means that f+ (f) is zero in F[x]/(f). But f+ (f) is fapplied to a = x+ (f), so that a is a root off in

When f is irreducible, Proposition 3.1.3 also showed that L = F[x]/(f) is afield. But in practice, if we are given a nonzero coset g + (f), how do we find itsmultiplicative inverse in L? In Exercise 5 you will show the following:

• f and g are relatively prime, so thatAf +Bg = 1 for someA,B E F[x].• The multiplicative inverse of g + (f) in L = F [x] / (f) is the coset B +

While it is important to be able to manipulate cosets at an abstract level, it is alsooften useful to represent them concretely. This means coming up with a method forpicking a unique element—a coset representative—from each coset. In the case ofL = F / (f), we will show in Chapter 4 that if f has degree n, then every coset inF [xI / (f) can be written uniquely in the form

co + c1x+ + +

where c0,... , E F. The rough idea is that given a coset g + (f), we replace gwith its remainder on division by f, which is a polynomial of degree at most n — 1.

Furthermore, setting a = x + (f) as in the proof of Proposition 3.1.3, we can rewritethe above expression as

(3.4)

When F = R and f = x2 + 1, this is what Cauchy did in his construction of C.The idea of representing cosets by remainders can be applied to other quotient

rings as well. For example, the theory of Gröbner bases enables one to representelements of the quotient ring

uniquely by remainders (see [3, Ch. 5, §21).

• Construction of Splitting Fields. The proof of Theorem 3.1.4 constructs a fieldover which f e F [x] splits completely by iterating the quotient ring construction ofProposition 3.1.3. Hence elements of this field are cosets of cosets of cosets, etc.This seems very abstract until one remembers that in modern algebra, we don't carewhat the objects are; it is their behavior that counts. Since the field has the desiredbehavior, we are content.

In Chapter 5, we will give a refined version of Theorem 3.1.4 where L is chosento be the smallest extension of F over which f e F [x] splits completely. We will callthis a splitting field. We will show that splitting fields are unique up to isomorphism.

• The Complex Numbers. This section began with two constructions of the complexnumbers C. The one of greater interest to us was Cauchy's, which eventually led to


Proposition 3.1.3. The other construction of C, due to Hamilton, used ordered pairsof real numbers. This suggests using triples of real numbers, and Hamilton triedhard to define addition and multiplication so that such triples would form a field. Hedidn't succeed, but on October 16, 1843 he realized that this idea would work withquadruples (a, b, c, d) provided multiplication wasn't required to be commutative. Ifthe standard basis of R4 is denoted 1, i, j, k, then

(a,b,c,d) = al +bi+cj+dk,

and Hamilton defined multiplication so that 1 is the multiplicative identity and

i2=j2=k2=—1, ij==—ji=k, jk=—kj=i, ki=—ik=j.

These are the famous quaternions. They form a division ring, which is a noncom-mutative ring where every nonzero element has a multiplicative inverse.

Historical Notes

In solving the cubic and quartic equations, Cardan and Ferrari implicitly assumedthe existence of roots, just as we did in Chapter 1. Girard, in the early seventeenthcentury, was one of the first to assert the existence of roots, real or imaginary, though"imaginary root" did not have a clear meaning in his work. As people became morecomfortable with complex numbers, the existence of roots evolved into the existenceof complex roots, which come in complex conjugate pairs when the coefficients arereal. Thus the eighteenth-century version of the Fundamental Theorem of Algebraasserts that every nonconstant polynomial in R[x] factors into linear and quadraticfactors with coefficients in R. In Section 3.2, we will prove the equivalent statementthat every nonconstant polynomial in C[x] splits completely over C.

The first attempt to prove the Fundamental Theorem of Algebra was due toD'Alembert in 1746, and at roughly the same time Euler discovered an algebraicproof (still somewhat incomplete), to be discussed in the next section. Like Cardan,Euler implicitly assumed that the roots exist. In 1799, Gauss noted that Euler's proofin effect made the assumption that

every equation can be satisfied by a real value of the unknown, or by an imaginaryvalue of the form a + or by a value that is not subsumed under any form.

(See [Gauss, Vol. III, p. 14].) Gauss criticized this assumption as follows:

How these magnitudes of which we can form no idea whatsoever—these shadowsof shadows—are to be added or multiplied cannot be understood with the kindof clarity required by mathematics.

The main result of this section, Theorem 3.1.4, answers Gauss's criticism verynicely. Given f E of positive degree, we can regard f as lying in C[x}. ApplyingTheorem 3.1.4, we get an extension C C L where f splits completely over L. Then,as Gauss observes in the first part of the quote, each root of f either lies in R, inC, or in L. However, the roots in L are no longer "shadows of shadows" but ratherelements of a field which can be manipulated by the usual operations of algebra, justas Euler assumed they could.


Gauss's 1815 proof of the Fundamental Theorem of Algebra uses symbolic meth-ods to avoid assuming the existence of roots, though his actual construction wasquite different from what we did in Theorem 3.1.4. We will say more about Gauss'sargument in the next section.

In his 1847 construction of the complex numbers, Cauchy defined congruencesmodulo an arbitrary polynomial f. He also introduced sums similar to (3.4). How-ever, Cauchy did not recognize the importance of f being irreducible, which byProposition 3.1.1 is necessary if we want the quotient ring to be a field.

The general case of this construction is due to Kronecker. He developed anelaborate theory of algebraic quantities in his 1881—1882 treatise Grundzüge einerarithmetischen Theorie der algebraischen Grössen [Kronecker, Vol. II, pp. 237—3 87]and applied these ideas to the existence of roots in his 1887 paper Em Fundamentalsatzder aligemeinen Arithmetik [Kronecker, Vol. III, pp. 209—240]. His version ofTheorem 3.1.4 uses the language of congruences (rather than cosets) to construct anextension F C L in which f E F [xJ splits completely. In Chapter 12, we will see howKronecker drew on ideas of Lagrange and Galois to create L using a single quotient,rather than the sequence of quotients used in the proof of Theorem 3.1.4.


Exercise 1. This exercise is concerned with the proof of Proposition 3.1.1. Suppose thatf, g, h e F [x] are polynomials such that f is nonzero and f = gh. Also let I = (g).(a) Prove that g constant if and only if 1= F [x].(b) Prove that h constant if and only if! = (f).

Exercise 2. Let F and L be fields, and let cp: F —* L be a ring homomorphism as defined inSection A.l. Prove that is one-to-one and that we get an isomorphism : F

Exercise 3. Let ! C F[x] be an ideal, and define p F —÷ F[x]/I by = a +!. Prove

carefully that is a ring homomorphism.

Exercise 4. In your abstract algebra text, review the definition of the field of fractions of anintegral domain and verify that (3.3) is the correct definition of a/b for a, b E 74 b 0.

Exercise 5. Let f E F [x] be irreducible, and let g + (f) be a nonzero coset in the quotient ringL = F[x]/(f).(a) Show that f and g are relatively prime and conclude that Af + Bg = 1, where A, B are

polynomials in F[x].(b) Show that B + (f) is the multiplicative inverse of g + (f) in L.

Exercise 6. Apply the method of Exercise 5 to find the multiplicative inverse of the cosetl+x+ (x2 +x+ 1) in the field Q[x]/(x2 1).

3.2 THE FUNDAMENTAL THEOREM OF ALGEBRA

The Fundamental Theorem of Algebra asserts that every nonconstant f E C[x] splitscompletely over C. In other words,

for some ao,cti,. . . E C with ao 0.

THE FUNDAMENTAL THEOREM OF ALGEBRA 63

The following proposition shows that there are several different ways of statingthe Fundamental Theorem.

Proposition 3.2.1 The following are equivalent:(a) Every nonconstant f E C [x] has at least one root in C.(b) Every nonconstant f E C [x] splits completely over C.(c) Every nonconstant f E has at least one root in C.

Proof: For (a) (b), we use induction on n = deg(f). For the base case n = 1,

writing f ax + b = a(x — (—b/a)) shows that f splits completely over C.Now suppose that n> 1 and that our assertion is true for n — 1. If f E C[x]

has degree n, then assumption (a) implies that = 0 for some E C. ByCorollary A.l.15, this implies that f is divisible by x — Thus

f=for some g E C[xJ of degree n 1. By our inductive assumption, g splits completelyover C, and then the above equation shows that the same is true for f.

The implication (b) (c) is clear since C C. To prove (c) (a), we must showLet

(3.5)

denote the polynomial obtained by taking the complex conjugates of the coefficientsof f. In Exercise 1 you will prove that if f,g E C[x], then

fg =

Now leth=ffEC[xJ. Then

=jf = fj = ff = h

implies that h E R[xI. By (c), we can find E C such that = 0. Then f(cs)f(cs)0, so that f(a) = 0 or = 0. In the former case, e C is a root off, and in thelatter, Exercise 1 will show that e C is a root off. This completes the proof of theproposition. •

We next study polynomials of odd degree with real coefficients.

Proposition 3.2.2 Every f E lPjx] of odd degree has at least one root in R.

Proof: We will use the Intermediate Value Theorem (IVT) from calculus. We knowthat f E R[x] is continuous. Thus, if we can find M > 0 such that

(3.6) f(—M) <0 <f(M),

then the IVT, applied to f on the interval [—M,M1, will guarantee that f(c) = 0 forsome c E (—M,M).


Given f e REx] of odd degree, we can assume that f is monic by multiplying f bya suitable nonzero constant. Then

wheren is odd and a1,... ER. If we set

then

I + a2l + . . . +

where the first inequality uses the triangle inequality, the second uses M> 1, and thethird uses M> Iai I + Ia2l + . . . + It follows that

since the expression in parentheses has absolute value by (3.7). We also have

<W'

by a similar argument. Then

f(—M) = —M" + + . . . <0,

since n is odd and the expression in parentheses has absolute value <W'.Thus M satisfies (3.6). As noted above, the proposition follows. U

Finally, we note the following simple consequence of the quadratic formula.

Lemma 3.2.3 Every quadratic polynomial in C [xI splits completely over C.

Proof: Given f = ax2 + bx + c E C[x] with a 0, the quadratic formula impliesthat the roots off are given by

—b±i/b2—4ac2a

By Section A.2, every complex number has a square root in C. Hence the aboveroots are complex numbers, which shows that f splits completely over C. •

We can now prove the Fundamental Theorem of Algebra.

Theorem 3.2.4 Every nonconstant f E C[x] splits completely over C.

Proof: Our proof will follow a strategy due to Euler, together with a clever ideafirst used by Laplace. By Proposition 3.2.1, it suffices to prove that

(3.8) Every f E R[x] of degree n > 0 has at least one root in C.


We can write n uniquely in the form

n=2mk, kodd,m>O.

Euler's strategy is to prove (3.8) by induction on m. By Proposition 3.2.2, a polyno-mial of odd degree in REx] has a root in JR C C. Hence (3.8) is true when m = 0.

Now suppose that m > 0 and that (3.8) is true for m — 1. Take f E R{x] of degree= 2tmk, k odd. We can regard f as a polynomial in C[x], so that Theorem 3.1.4

implies that there is an extension C C L such that f splits completely over L. We willdenote the roots of f by cvi,..., EL.

Laplace's clever idea is to consider the following auxiliary polynomial. Pick areal number A, and set

gx(x)= [fI

This has degree — 1) = the number of distinct pairs of variables.We first claim that has coefficients in JR. To prove this, consider

(3.9) = ff (x— + )tx,xj).1<i<j<n

The identityx— (x1+x1)+ Ax,x3 =x—

shows that GA is a product indexed by pairs of distinct variables. It follows easilythat GA is unaffected by permutations of the x. Then multiplying out GA gives

I)

GA(X)=

p(xi,...

where the polynomials pj(xi, . . . are symmetric in Xi,... ,x,, since GA is. Alsonote that pj(Xi, . . . E RExi, . . . since A E JR. Then Corollary 2.2.5 implies thatpj(cvi,.. , E JR since cv1,..., cv,, are the roots off E REx]. We conclude that

We next compute the exponent of 2 in the degree of gA. Using n = 2mk, the degreeof is given by

Since m > 0, we can write this as

(3.10) deg(gA) = 2m_ik(2mk_ 1).

Furthermore, k odd and m > 0 imply that k(2mk — 1) is odd. Thus, even thoughhas larger degree than f, the exponent of 2 has been reduced by one.


It follows that for any real number A, our inductive assumption (3.8) applies toThis means that has a root inC. By definition, the roots of are aj + —

Thus one of these lies in C. In other words, for each A E R, we can find a pair i, jwith 1 <i <j <n such that

E C.

Note that the pair i, j depends on A—if we switch to a different value of A, we mightget a different pair. But as we vary over the infinitely many possible values of A,there are only finitely many possibilities for the corresponding pair i, j. This impliesthat there must exist A in R that use the same pair i, j. Thus

(3.11)

a1 — Aa1a3) — (aj + a3 — = — A)a,a3 e C,

and since A are real, it follows that a,a3 E C. Then a, + a3 — Aa,a3 E C impliesthat a, + a3 E C. Thus the sum and product of a3 are complex numbers.

Now consider the quadratic polynomial

By what we just proved, it has coefficients in C, so that its roots also lie in C byLemma 3.2.3. But the roots are clearly a, and a3. This proves that a1, a3 E C. Hencef has a complex root, which completes the proof of the theorem.

Mathematical Notes

As usual, this section has some interesting ideas to discuss.

• Proofs of the Fundamental Theorem. There are many proofs of the FundamentalTheorem of Algebra. Students often see a proof in a course on complex analysis,but there are also some lovely proofs which use topology. The book [4] discussesa variety of proofs of the theorem, including a version of the proof given here. Seealso [6] for another proof and references to some of the many other proofs in theliterature.

The proof of the Fundamental Theorem of Algebra given above is one of the more"algebraic" proofs. However, a closer inspection shows that our proof has three mainingredients:

• Every polynomial of odd degree in R[xI has a root in R (Proposition 3.2.2).• Every complex number has a square root in C (this gives Lemma 3.2.3).• Every polynomial splits completely over some extension field.

Of these three, only the last is purely algebraic. The proof of Proposition 3.2.2 usesthe IVT, and as shown in Exercise 2, square roots of complex numbers reduce tosquare roots of positive real numbers, which exist by the IVT (if you're unfamiliar


with this argument, do Exercise 3). Since the IVT depends on the completeness ofIft, one could argue that the Fundamental Theorem of Algebra is really a theorem inanalysis. See [1] for a discussion of these issues.

Once we have proved the main theorems of Galois theory, we will give an elegantproof of the Fundamental Theorem due to Artin in Theorem 8.5.9.

Algebraically Closed Fields. The Fundamental Theorem of Algebra leads to thefollowing definition.

Definition 3.2.5 A field F is called algebraically closed if every nonconstant poly-nomial in F [x] splits completely over F.

Theorem 3.2.4 shows that C is algebraically closed. We will see that there areother algebraically closed fields. In general, one can prove that every field has analgebraically closed extension.

• Real Closed Fields. Another approach to the question of algebraic versus analyticis given by the theory of real closed fields. The basic idea is to make the aboveproof as algebraic as possible. We know from Exercise 2 that the existence of squareroots of complex numbers follows directly from the existence of real square roots ofpositive real numbers. Then one defines a real closed field to be a field F that has thefollowing properties:

• F has an order relation > compatible with addition and with multiplication bypositive elements (an element a E F is positive if it satisfies a > 0).

• Every positive element of F has a square root in F.• Every polynomial of odd degree in F has a root in F.

The field of real numbers is the prototypical example of a real closed field, but it isnot the only one.

Given a real closed field F, we can adjoin i to F using the methods ofSection 3.1 (e.g., Cauchy's method). This gives a field F(i), and one can easily adaptthe proof of Theorem 3.2.4 to show that F(i) is algebraically closed. Details can befound in Exercises 4 and 5 (see [Jacobson, Vol. I, Sec. 5.11 for a complete treatment).There is also the related idea of aformally realfield, due to Artin and Schreier. Thesefields have an interesting relation to Hilbert's Seventeenth Problem and are discussedin [Jacobson, Vol. II, Ch. 111.

Historical Notes

In 1749 Euler attempted to prove the Fundamental Theorem of Algebra for f E lft[x]using induction on the exponent of 2 in deg(f). To give the flavor of his proof,consider the case when deg(f) = The idea is to write f as a product

(3.12) f=gh

where g, h have degree Euler did this by finding the equations satisfied by thecoefficients of g, h and then showing that they have real solutions. It follows that


coefficients of g and h can be chosen to be real. Once this is done, our inductiveassumption implies that g and h have roots in C.

Euler's proof has some major gaps, and in 1772 Lagrange wrote Sur la formedes racines imaginaires des equations [Lagrange, pp. 479—516] to make Euler'sargument more rigorous. Lagrange's proof is almost complete—the difficulty comeswhen some of the polynomials in the proof have multiple roots, which might causecertain denominators to vanish. Lagrange was well aware of this problem and gavesome very interesting arguments to deal with multiple roots. Many authors (including[Tignol] and [2]) accept Lagrange's argument as complete, though I think that somesubtle gaps still remain. See [1] and [2] for more on the history of all this.

We next turn to Gauss's 1815 proof of the Fundamental Theorem of Algebra,which appears in [Gauss, Vol. III, pp. 3 1—56] (see [5, pp. 292—306] for an Englishtranslation). The overall strategy of Gauss's argument is similar to what we did inTheorem 3.2.4, with one major exception: he never uses the roots a, off. He beginsinstead with the universal situation and defines

[f1<—i<j�n

This is similar to the polynomial G), defined in (3.9), except that u is now a variable.Gauss observes that z is a polynomial in x and u whose coefficients are symmetricin the x1. Hence the coefficients are polynomials in the o,. He then replaces each a,with a new variable u,. This gives a new polynomial

Thus Gauss is using the isomorphism R[ui, . . . , R[ai,... , which we provedin Section 2.2 using arguments taken from this paper of Gauss. Then, given

f = x" + aix"' + + a,, E

he uses the substitution u1 '—÷ (—1 to send to a polynomial

(3.13) Z=Z(x,u) ER[x,u].

In this way, Gauss gets an analog of (x) without knowing the roots off.From here, Gauss's argument departs from what we did in Theorem 3.2.4. One

difference is that he considers only monic polynomials with nonvanishing discrim-inant (to be called separable in Section 5.3). Other aspects of Gauss's proof arediscussed in [1].

In his proof, Gauss uses the methods of Lagrange, which apply to the universalpolynomial f studied in Chapter 2. Although these methods are powerful, they canbe cumbersome to use in practice. What we really need are methods which applydirectly to any field. This leads to the language of field extensions, which is the maintopic of the next chapter.



Exercise 1. Forf E C[x}, define f as in (3.5).(a) Show carefully that fg = forf,g E C[x].(b) Let E C. Show that = 0 implies that = 0.

Exercise 2. In Section A.2, we use polar coordinates to construct square (and higher) roots ofcomplex numbers. In this exercise, you will give an elementary argument that every complexnumber has a square root. The only fact you will use (besides standard algebra) is that everypositive real number has a real square root.(a) First explain why every real number has a square root in C.(b) Nowfixa+biE Forx,yE JR. showthattheequation (x+iy)2 =a+biis

equivalent to the equationsx2—y2=a, 2xy=b.

(c) Show that the equations of part (b) are equivalent to

2 a±V'a2+b2 b

2'

Also show that x 0 and that a ± '/a2 + b2 is positive when we choose the + sign in theformula forx2.

(d) Conclude that a + bi has a square root in C.

Exercise 3. Use the IVT to prove that every positive real number a has a real square root.

Exercise 4. A field F is an ordered field if there is a subset P C F such that:(a) P is closed under addition and multiplication.(b) For any a E F, exactly one of the following is true: a E P. a = 0, or —a E P.

One then defines a > b to mean a — b E P (so that P becomes the set of positive elements).From this, one can prove all of the typical properties of>. Now let F be an ordered field.Prove that —1 is not a square in F.

Exercise 5. Let F be a real closed field. As in the text, this means that F is an ordered field(see Exercise 4) such that every positive element of F has a square root in F and every f E F[x]of odd degree has a root in F.(a) Use Exercise 4 to show that x2 + 1 is irreducible over F. Then define F(i) to be the field

F [x] / (x2 + 1). By the Cauchy construction described in Section 3.1, elements of F(i)can be written a + bi for a, b E F.

(b) Show that every quadratic polynomial in F(i) splits completely over F(i).(c) Prove that F(i) is algebraically closed.

Exercise 6. Here is yet another way to state the Fundamental Theorem of Algebra.(a) Suppose that f(a) = 0, where f E and a E C. Prove that = 0.

(b) Prove that the Fundamental Theorem of Algebra is equivalent to the assertion that everynonconstant polynomial in R[x] is a product of linear and quadratic factors with realcoefficients.

Exercise 7. Prove that a field F is algebraically closed if and only if every nonconstantpolynomial in F[x] has a root in F.


REFERENCES

1. I. G. Bashmakova and A. N. Rudakov (with the assistance of A. N. Parshin and E. I.Slavutin), Algebra and Algebraic Number Theory, in Mathematics of the 19th Century:Mathematical Logic, Algebra, Number Theory, Probability Theory, edited by A. N.Kolmogorov and A. P. Yushkevich, English translation by A. Shenitzer, Birkhäuser,Boston, Basel, Berlin, 1992. Second Revised Edition, 2001.

2. I. G. Bashmakova and G. S. Smirnova, The Beginnings and Evolution of Algebra, Englishtranslation by A. Shenitzer, MAA, Washington, DC, 1999.


4. B. Fine and G. Rosenberger, The Fundamental Theorem of Algebra, Springer, New York,Berlin, Heidelberg, 1997.

5. D. E. Smith, A Source Book in Mathematics, Volume One, Ginn, Boston, New York, 1925.Reprint by Dover, New York, 1959.

6. D. Velleman, Another proof of the Fundamental Theorem of Algebra, Math. Mag. 70(1997), 216—217.

PART II

FIELDS

In the next four chapters, we shift our attention from polynomials to fields.We begin by developing the basic language of field extensions in Chapter 4. One

of the key concepts is the degree of an extension. We also consider the special roleplayed by irreducible polynomials.

Chapter 5 continues our study of fields by considering splitting fields, which arefields obtained by adjoining the roots of a given polynomial. This leads naturallyto the notion of a normal extension. Finally, we introduce the idea of separability,which for a polynomial means not having multiple roots.

We introduce the Galois gmup in Chapter 6, and we explain how it relates topermutations of roots in the case of a splitting field. We also give some nontrivialexamples and, in an optional section, discuss Abel's notion of an Abelian equation.

Finally, Chapter 7 defines the key ideas of a Galois extension and the Galoiscorrespondence. After stating and proving the Fundamental Theorem of GaloisTheory, we give some simple applications.

CHAPTER 4

EXTENSION FIELDS

This chapter will develop the language of algebraic extensions, which is needed toprove the main theorems of Galois theory. Recall from Chapter 3 that an extensionof a field F consists of a field L and a ring homomorphism

F —* L.

As before, we will identify F with its image 'p(F) in L. In this way, we will write afield extension as F C L.

4.1 ELEMENTS OF EXTENSION FIELDS

Given a field extension F C L, elements of the larger field can relate to the smallerfield in two different ways.

Definition 4.1.1 Let L be an extension of F, and let a E L. Then a is algebraicover F if there is a nonconstant polynomial f E F[x] such that f(a) = 0. If a is notalgebraic over F, then a is transcendental over F.


74 EXTENSION FIELDS

For example, E R is algebraic over Q, since is a root of x2 —2 e Q[x], and= E C is algebraic over Q, since it is a root off — 1 e Q[x]. The numbers

7r and e are transcendental over Q, though this is not easy to prove.

Example 4.1.2 To show that + is algebraic over Q, consider the polynomial

(x—

Multiplying this out gives x4 — lOx2 + 1. Thus + is the root of a nonconstantpolynomial in Q [x]. We will return to this example many times. 4>

In Section 4.4 we will generalize Example 4.1.2 by showing that if a, E L arealgebraic over F, then so are a + and Furthermore, in Exercise 1 you willshow that if a 0 is algebraic over F, then so is 1/a. This will imply that the set{a E L a is algebraic over F} is a subfield of L.

A. Minimal Polynomials. When a E L is algebraic over F, there may be manynonconstant polynomials in F [xl with a as a root. One of these polynomials isespecially nice.

Lemma 4.1.3 If a E L is algebraic over F, then there is a unique nonconstant monicpolynomial p E F[x] with the following two properties:(a) a is a root of p, i.e., p(a) = 0.

(b) 1ff E F[x] is any polynomial with a as a root, then f is a multiple of p.

Proof: Among all nonconstant polynomials in F [xJ with a as a root, there must beone of smallest degree. Pick one such polynomial and call it p. Multiplying by a

constant if necessary, we may assume that p is monic.This polynomial certainly satisfies (a). As for (b), suppose that f(a) = 0 for some

f E F [xl. The division algorithm from Section A. 1 gives us polynomials q, r E F [x]

such thatf=qp+r, r=O or deg(r)<deg(p).

Evaluating this equation at a gives

0 = f(a) = q(a)p(a) + r(a) = r(a),

where the last equality uses p(a) = 0. If r had strictly smaller degree than p. thiswould contradict the definition of p, and r = 0 follows. Thus p satisfies (b).

Finally, to prove uniqueness, suppose that another monic polynomial j3 satisfiedproperties (a) and (b). Then applying (b) for p to f = implies that p divides j3,

and reversing the roles of p and j3, we see that j3 divides p. Since these are monicpolynomials, it follows easily that j3 = p (see Exercise 2 for details).

It is customary to name the polynomial of Lemma 4.1.3 as follows.

Definition 4.1.4 Let a E L. If a is algebraic over F, then the polynomial p ofLemma 4.1.3 is called the minimal polynomial of a over F.

ELEMENTS OF EXTENSION FIELDS 75

Besides the characterization given in Lemma 4.1.3, there are other ways to thinkabout the minimal polynomial.

Proposition 4.1.5 Let E L be algebraic over F, and let p E F[xI be its minimalpolynomial. 1ff E F[x] is a nonconstant monic polynomial, then

f = p f is a polynomial of minimal degree satisfying

f is irreducible over F and f(a) = 0.

Proof: The first equivalence follows from the proof of Lemma 4.1.3. For the second,we prove that the minimal polynomial p is irreducible over F as follows. If p = gh,where g,h E F[xI have strictly smaller degree than p, then 0 = p(a) =would imply = 0 or = 0, which would contradict the first equivalence.

Conversely, suppose that f ff = ph with h E F[x]. Since f is irreducible and p is

nonconstant, h must be constant. Then f = p follows, since f and p are monic. .Here are some examples of minimal polynomials.

Example 4.1.6 The minimal polynomial of over Q is x2 —2. This follows fromthe irrationality of which implies that cannot be the root of a polynomial ofdegree 1 in Q [x].

Example 4.1.7 For + we showed in Example 4.1.2 that + is a rootof lOx2 + 1. But is this the minimal polynomial? By Proposition 4.1.5, this isequivalent to x4 lOx2 + 1 being irreducible over Q. For an explicit polynomial, theeasiest way to check for irreducibility is by computer. For example, the Mat hematicacommand

Factor [x"4- 10f2+1]

will produce the output x4 — lOx2 + 1, which means that the polynomial is irreducibleover Q. In Maple, the command would be

and again the output x4 lOx2 + 1 proves irreducibility over Q. Thus x4 — lOx2 + 1is the minimal polynomial of + In Section 4.2 we will say more about usingMathematica and Maple to check irreducibility.

Example 4.1.8 The minimal polynomial of over Q is called the nthcyclotomic polynomial and is denoted In Chapter 9 we will show thathas degree where is the Euler from number theory.

B. Adjoining Elements. We next show how to describe some interesting subringsand subfields of a given extension F C L. Given cd,... , e L, we define

76 EXTENSION FIELDS

Hence F . , consists of all polynomial expressions in L that can be formedusing , with coefficients in F. Then let

Thus , is the set of all rational expressions in the with coefficients inF. We can characterize , as follows.

Lemma 4.1.9 , is the smallest subfield of the field L containing F and

Proof: We leave it as Exercise 3 to show that F(ai,... , is a subfield of L. Thus,to prove the lemma, we must show that if K is a subfield of L containing F andcr1,. . . , then , c K. This is what "smallest" means in the statementof the lemma.

Suppose that K C L contains F and co,.. . , Since K is closed under mul-tiplication and addition, it follows that p(ai,... , a,) E K for any polynomial p EF[xi,. . . This shows that F[ai,... C K. Then F(cEi,... ,a,,) C K followsimmediately, since K is a field.

Since F(ai,... , cx,,) is a subfield of L containing F, we get extensions

F C F(ai,... ,a,,) C L.

We say that F(ai . . . ,a,,) is obtained from F by adjoining ,a,, E L. We canuse this to construct fields as follows.

Example 4.1.10 Consider the polynomial x4 —2 E Q [x}. Over the complex numbers,this factors as

x4 2

the x4 — 2 splits completely. We will see in Section 5.1that this is an example of a splitting field.

This field can be described more compactly as

(4.1)

To see why, let K andL = Q(i, Then K CL followsfrom Lemma 4.1.9, since L contains Q and For the opposite inclusion,note that

= E K.

Since K obviously contains Q and we have L C K, and (4.1) follows.


Lemma 4.1.9 also implies that we can adjoin elements to a field in stages. Moreprecisely, we have the following corollary.

Corollary4.1.11 1fF EL, then

F(ai, . . . = F(ai,... ,ar)(ar+i, . . . ,an)

foranyl

Proof: The field on the right is obtained by first adjoining ai, . . . to F to getthe field F(ai,...,ar) and then adjoining ar+I,... ,CE,1 to ,ar) to get thefield F(ai,... , , This field obviously contains F and the elements

. . . Then Lemma 4.1.9 implies that

F(ai,... C F(ai,... ,ar)(ar+i,...,an).

The opposite inclusion is similar and is left as Exercise 4. •

Here is a simple example of why this corollary is useful.

Example 4.1.12 Corollary 4.1.11 implies that

Q C c =

shows that we get Q adjoining toan extension this way will be very useful.

We next consider F(ai,. . . and F[ai,... when at,... are algebraicover F. We begin with the case of adjoining a single element.

Lemma 4.1.13 Assume that F C L is a field extension, and let a E L be algebraicover F with minimal polynomial p E F [xJ. Then there is a unique ring isomorphism

Fia]

F a the coset x + (p).

Proof: Consider the ring homomorphism : F[x] —+ L that sends h(x) E F[xI toh(a) E L. By definition, the image of is F{a]. As for the kernel, we claim that

= (p). To prove this, first note that g E F [x] implies that

ço(gp) = ço(g) cp(p) = g(a)p(a) = g(a)O =0.

This shows that (p) C For the other inclusion, suppose that f EThen f(a) = 0, which by part (b) of Lemma 4.1.3 implies that f is a multiple of p.Thus c (p), and = (p) follows.

Since we know the image and kernel of the Fundamental Theorem of RingHomomorphisms (Theorem A. 1.9) gives a ring isomorphism

F[x]/(p) F[a].

78 EXTENSION FIELDS

This isomorphism is the identity on F and maps the coset x + (p) to a. Its inverse isthe isomorphism described in the statement of the lemma.

Finally, uniqueness follows since a ring homomorphism defined on F[a] isuniquely determined by its values on F and a. .

This lemma shows how to represent F [al as a quotient ring. However, we alsoknow that the minimal polynomial p of a is irreducible (Proposition 4.1.5). As wesaw in Proposition 3.1.1, this implies that F[xI/(p) is a field. By Lemma 4.1.13, itfollows that F[a] is a field when a is algebraic over F. Hence we have proved partof the following proposition.

Proposition 4.1.14 Assume that F c L is afield extension, and let a E L. Then a isalgebraic over F if and only if F[a] = F(a).

Proof: When a is algebraic over F, the above paragraph shows that F[aI is a fieldcontaining F and a. Since F(a) is the smallest subfield of L containing F and a(Lemma 4.1.9), it follows that F(a) C F[a]. The opposite inclusion always holds,so that F(a) = F[a] when a is algebraic over F.

For the other implication, suppose that F[a] F(a). We may assume that a 0since 0 is obviously algebraic over F. Then 1/a E F(a) = F[a] implies that

1/a = a0 + a1a+ +amam

for some . . ,am E F. Thus

proving that a is algebraic over F. •

We next study what happens when we adjoin several algebraic elements to a field.

Proposition 4.1.15 Let F C L be afield extension, and let a1,... , E L be algebraicover F. Then

=F(ai,...,an).

Proof: By the argument used in the proof of Proposition 4.1.14, it suffices to provethat F[ai, . . . is a field. We will do this by induction on n. The case n = 1 iscovered by Proposition 4.1.14. Now suppose that n> 1 and that F[ai, . . . isa field. We know that f fas having coefficients in the larger field F [al, . . . , so that is algebraic overF[ai,. . . Then Proposition 4.1.14 implies that

is a field. We leave it as Exercise 5 to show that this equals F[ai,. . . .Here is an example of Proposition 4.1.15.


Example 4.1.16 Consider The above proposition shows that this equalsso that every element of is a polynomial in with ra-

2n 2n+1tional coefficients. Furthermore, since T' and = and similarlyfor powers of it follows easily that

(4.2) = {a +I

a,b,c,d E Q}.

In Section 4.3, we will show that the representation of elements of givenby (4.2) is unique.

Mathematical Notes

Let us discuss two of the ideas that have appeared in this section.

• The Structure of Fields. Consider a field of the form F(a,,... , In Propo-sition 4.1.15, we studied the case when the a, are all algebraic over F. The otherextreme is when the a, are not only transcendental over F but also algebraicallyindependent, which (as defined in Mathematical Notes to Section 2.2) means that the

satisfy no nontrivial polynomial relation with coefficients in F. In Exercise 6 youwill show that this implies that F(ai,... , is isomorphic to the field of rationalfunctions F(xi, . . . We call F(ai,... , a purely transcendental extension ofF in this case.

For the general case, a result of Steinitz says that a field L = F(ai,. . . , cia) canalways be written in the form

where m < n, , 13m are algebraically independent over F (so that F C K is purelytranscendental), and 'y1, . . . , are algebraic over K. A proof of this theorem can befound in [Jacobson, Vol. II, Sec. 8.121.

• Number Fields. A field of the form Q(ai,. .. , where a1, . .. ,

a numberfield. The fields appearing in Examples 4.1.10 and 4.1.16are number fields. These fields and their Galois theory occupy a central role inalgebraic number theory.

Historical Notes

Fields have been used implicitly ever since the discovery of addition, subtraction,multiplication, and division. Cardan's formulas, dating from the sixteenth century,use Q, R, and C. The field of rational functions in n variables arises naturally whenconsidering symmetric functions, and Lagrange used such fields (implicitly) in his1770 study of the roots of polynomials. Number fields also appeared around thistime. For example, Euler used the fields and to study problemsin number theory raised by Fermat.

The first reasonably general definition of F(ai, . . . , was given by Galois in1831, where he says the following:

80 EXTENSION FIELDS

One can agree to regard as rational all rational functions in a certain number ofdeterminate quantities, which are supposed to be known a priori. For example,one can choose a certain root of an integer, and regard as rational all rationalfunctions of the radical.When we agree in this way to regard certain quantities as known, we say that weADJOIN them to the equation that we are trying to solve.

(See [Galois, p.451.) This is why we say that is obtained from F byadjoining a i,... , On the other hand, Abel was the first person to understand thatF[a] = F(a) when a is algebraic over F (see [Abel, Vol. I, pp. 66—72]).

Abel, Galois, and their predecessors tended to work with explicitly constructedfields. The first truly "abstract" notion of field is due to Dedekind. In 1871, he gavethe following definition:

I call a system A of numbers a (not all zero) a field when the sum, difference,product and quotient of any two of numbers in A also belongs to A.

(See [3, p. 1071.) This is not completely general, for the "numbers" in this definitionare all complex. From our point of view, Dedekind is really defining a subfield of C.But his definition is modem in spirit, in that he allows any set (he says "system"because set theory was not fully established in 1871) that behaves nicely underaddition, subtraction, multiplication, and division. This is very different from hisgreat contemporary Kronecker, who took a more conservative view and only dealtwith fields that could be constructed explicitly in finitely many steps. It wasn't until1893 that Weber gave the first fully abstract definition of field. Weber's definition issimilar to the one in use today. A discussion of the evolution of the field concept canbe found in [7]. See also [8] for the evolution of the ring concept.


Exercise 1. Let a E L\{0} be algebraic over a subfield F. Prove that 1/a is also algebraicover F.

Exercise 2. Complete the proof of Lemma 4.1.3 by showing that if f and g are monicpolynomials in F [x] each of which divides the other, then f = g.

Exercise 3. Suppose that F C L is a field extension and that al,.. . , E L. Show thatF[ai,. . . , a subring of L and that F(csj, . . . ,a,) is a subfield of L.

Exercise 4. Complete the proof of Corollary 4.1.11 by showing that

F(ai,... ,ar)(ar+i,...,an) C

ExerciseS. Prove carefully that F[ai, . . . , ii [ar] = F[ai,. . . ,

Exercise 6. Suppose that F C L and that ai, . . . , E L are algebraically independent over F(as defined in the Mathematical Notes to Section 2.2). Prove that there is an isomorphism offields

F(xi,. . . is the field of rational functions in variables xi, . . . ,x,,.

Exercise 7. In the proof of Proposition 4.1.14, we used the quotient ring F [x] /(p) to showthat F[a] is a field when a is algebraic over F with minimal polynomial p E Fk]. Here, you

IRREDUCIBLE POLYNOMIALS 81

will prove that F[a] is a field without using quotient rings. Since we know that F[a] is a ring,it suffices to show that every nonzero element E has a multiplicative inverse inSo pick /3 0 in Then /3 = g

g and p are relatively prime in F{x].(b) By part (a) and the Euclidean algorithm, we have Ap+Bg = 1 for some A,B E F[x].

Prove that B(a) E F [cr] is the multiplicative inverse of g(a).Do you see how this exercise relates to Exercise 5 of Section 3.1?

Exercise 8. If a polynomial is irreducible over a field F, it may or may not remain irreducibleover a larger field. Here are examples of both types of behavior.(a) Prove that x2 —3 is irreducible over(b) In Example 4.1.7, we showed that x4 — lOx2 + 1 is irreducible over Q (it is the minimal

polynomial of = Show that x4 lOx2 + 1 is not irreducible over

4.2 IRREDUCIBLE POLYNOMIALS

Since minimal polynomials are irreducible, it should be clear that the notion ofirreducibility plays an important role in field theory. However, given an arbitrarypolynomial f E F [x], it may not be obvious that f is irreducible. How do we tell? Inthis section, we will discuss some ways of answering this question.

A. Using Maple and Mathematica. In the previous section we saw examplesof how Maple and Mathematica factor polynomials over Q into irreducibles. Theseprograms can also factor over number fields, which as in Section 4.1 are fields of theform with algebraic over Q.

We first describe how Maple factors polynomials over a number field. In Sec-tion 4.1 we used the factor command to show that x4 — lOx2 + 1 is the minimalpolynomial of + over Q. To study this polynomial over we use

sqrt(2));

which gives the result

1).

This implies in particular that x2 — — 1 and x2 + — 1 are irreducible overSimilarly, the command

[sqrt(2), sqrt(3)));

gives

(4.3)

This is the factorization of x4 — lOx2 + 1 overNot all number fields have such simple descriptions. For example, consider the

field + Since the minimal polynomial of + is x4 lOx2 + 1,

82 EXTENSION FIELDS

Maple would represent this algebraic number using the RootOf command. This isdone most conveniently via

alias(alpha =

which makes a root of x4 — lOx2 + 1. Then we can factor a polynomial poiy inQ[xj or using the command factor(poly,alpha). For example, if we let

f = — + 1, then we know that x — is a factor of f over Q(a). But what arethe other factors? Using the command

alpha);

we get the result

(4.4)

The surprise is that the polynomial factors completely. This has an interestingconsequence concerning the fields + and

In Maple, the factorization (4.4) takes place in Q [x] / (x4 I Ox2 + 1). To getsomething involving numbers, consider the map x + This induces anisomorphism

— 10x2 + 1)

and allows us to assume that = in (4.4).By comparing (4.3) and (4.4), we conclude that a3 — 1 Oa = ± — and then

an easy numerical calculation shows that a = +adding these two equations gives — 9a = and it follows that E Q(a) =

Then we also have = a — Then Lemma 4.1.9implies that

Since the opposite inclusion clearly holds (be sure you can explain why), we get

(4.5)

We can also do these computations in Mathematica. For example, factoringx4 lOx2 + 1 over is done by the command

-> {Sqrt[2]}]

and factoring over is done via

—> {Sqrt[2],Sqrt[3]}I

Finally, to work over the field generated by a root of an irreducible polynomial such

as x4 — lOx2 + 1, one sets

a = Root 1]


Then the command

—> {a}]

produces a result similar to (4.4), except that a is replaced with the ungainly expressionit represents. To get a nicer result, one should use the command

—> {a}] /. a —> b

which gives the result

(x—b)(x+b)(x— lOb+b3)(x+ lob—b3).

In general, Maple and Mathematica have roughly equivalent capabilities for comput-ing with algebraic numbers.

B. Algorithms for Factoring. The use of Maple and Mathematica to factorpolynomials over number fields implies the existence of an algorithm for doing so.To give the reader an idea of how factoring is done, we will describe an algorithmfor deciding whether f Z[x] is irreducible over Q. The key tool is Gauss's Lemma,which is Theorem A.3.2. We will use the following corollary of this result.

Corollary 4.2.1 1ff E 7L[x] has degree > 0 and is reducible (i.e., not irreducible)over Q, then f = gh where g, h E Z[x] have degrees strictly smaller than deg(f).

Proof: 1ff is reducible in Q[XJ, then f = gjh1, where ,h1 E Q[x] have degrees<deg(f). By Gauss's Lemma, there is Q such that g = and h = haveinteger coefficients. Then f = gh is the desired factorization. •

We now describe an algorithm to test the irreducibility of f Z[x]. Let n =deg(f) > 0. First note that if f(i) = 0 for some 0 i n — 1, then x — i is a factorof f and we can quit. Hence, when performing the algorithm, we may assume thatf(0),... , f(n — 1) are nonzero. Then create a set of polynomials as follows:

• FixanintegerO<d<n.• Fix divisors a0,... ,ad E Z of f(0),... ,f(d) E 7L.• Use the Lagrange interpolation formula from Exercise 1 to construct a polynomial

g E Q{x] of degree d such thatg(i) = a fori = 0,...,d.• Accept g if it has degree d and integer coefficients; reject it otherwise.

Doing this for all 0 <d < n and all divisors aoIf(0),. . . ,adlf(d) gives a set ofpolynomials g E Z[x].

Proposition 4.2.2 This set of polynomials g E Z[xJ is finite, and f is irreducible overQ and only it is not divisible by any of the polynomials in this set.

Proof: We are assuming that f(0),... , f(d) are nonzero, so that each f(i) has onlyfinitely many divisors. Hence there are only finitely many choices for ao,... ,ad E Z.Since g is uniquely determined by the it follows that there are only finitely manysuch g's.

84 EXTENSION FIELDS

To finish the proof, we will show that f is reducible if and only if it is divisible byone of these polynomials. One direction is obvious. For the other direction, supposethat f is reducible. By Corollary 4.2.1, f = gh, where g,h E Z[x] and g has degreed,O<d<n.

Then, for 0 < i <d, let a1 = g(i), and note that since f(i) = g(i)h(i).The Lagrange interpolation formula gives g E Q [xI of degree d with = a1 foro i d. Since g — has degree at most d and vanishes at the d + 1 numbers 0, . . . , d,

it must be the zero polynomial. Hence g = is on our list. U

Since there are known algorithms for factoring integers, there is an algorithm forcomputing the set of polynomials in g E Z[x] used in Proposition 4.2.2. Then dividingthese into our given polynomial f via the division algorithm gives an algorithm fordeciding whether f is irreducible over Q.

From a computational point of view, this algorithm is dreadful. The methods usedby Maple and Mathematica are much more efficient. The book [1] describes somegood algorithms for factoring polynomials over a number field.

C. The Schönemann—Eisenstein Criterion. While algorithms and computerscan be extremely helpful in computing examples of irreducible polynomials, thereare certain classes of polynomials that can be proved to be irreducible by traditionalmeans. Here, we will prove the Schönemann—Eisenstein irreducibility criterion.

Theorem 4.2.3 Let f = + + E Z[x] have degree n > 0. If there is a primep such that plan_i pIao, and p2{ao, then f is irreducible overQ.

Proof: By Corollary 4.2.1, if f is reducible over Q, then there are g,h e Z[x] ofdegree <n such that f = g h. Now consider the ring homomorphism Z[xI —+ IF,, [x]

defined by sending q = bmXm + bø 7L[x] to = [bm]Xm + [be] E lFp[xJ,where [b] E IF,, = Z/pZ is the congruence class modulo p of b E Z.

Then f = gh implies that = since plao. However, IF,, is afield, which means that unique factorization holds in IF,, [x]. Since p it followsthat = [a]xT and h = [b]xs, where [a] [b] = [ani and r + s = n.

If r = 0, then = [aJ and deg(g) > 0 would imply that the leading term of g is

divisible by p. Then f = gh would imply that the same is true for the leading term

off. Thus p implies that r> 0, and s > 0 follows similarly.But then = [a]xr for r> 0 implies that p divides the constant term of g, and the

same is true for the constant term of h, since s > 0. Since the constant term a0 of fis the product of the constant terms of g and h, it follows that p2 lao. This contradictsp2 and completes the proof. •

Here is a simple example to illustrate the Schönemann—Eisenstein criterion.

Example 4.2.4 Consider the polynomial

f=f+px+p, n�2, pprime.

The Schönemann—Eisenstein criterion for the prime p implies immediately that f is

irreducible over Q, no matter what n � 2 we choose.


The interesting feature of this example is that it cannot be done by Maple orMathematica. For a specific n and p. we could check irreducibility by computer(assuming n and p aren't too big), but standard computer algebra systems can't factorpolynomials with symbolic exponents. On the other hand, only very special polyno-mials satisfy the Schönemann—Eisenstein criterion. (If is a root of a polynomialsatisfying this criterion, then from the point of view of algebraic number theory, theextension Q C is totally ramified at p, which is a rather rare phenomenon.)

We can use the Schönemann—Eisenstein criterion to determine the minimal poly-nomial of the pth root of unity = where p is prime. Using

we see that is a root of = xPt + . . . + x + 1. This is called the pth cyclotomicpolynomial.

Proposition 4.2.5 = x + 1 is irreducible over Q when p is prime.

Proof: First observe that = — l)/(x — 1), so that

(x+ i)P (ç)xP_1 +...+ 1,

and then substituting this into the above formula for (x + 1) gives

(4.6) 1) + (p)

However, for 1 <r p — 1, the integer

(p'\ p! p(p—1)...(p—r+1)

— r!(p—r)! r!

is divisible by p, since p divides the numerator but not the denominator (rememberthat p is prime). Furthermore, note that p2 does not divide = p. Then (x + 1)is irreducible, since (4.6) satisfies the Schönemann—Eisenstein criterion.

From here it is easy to see that is irreducible, for a factorization (x) =g(x)h(x) in Q[xJ would imply 1) = g(x+ l)h(x+ 1). If g and h have degree<p l,then the same would be true for g(x + 1) and h(x+ 1), which would contradictthe irreducibility of 1). This completes the proof.

It follows that the minimal polynomial of over Q is +... + x + 1. InChapter 9 we will describe the minimal polynomial of for arbitrary n.

D. Prime Radicals. Given a prime p, our final task is to investigate when thepolynomial — a E F[x] is irreducible over F. Note that if a is a root of a, then

= a, so that the roots of — a are the pth roots of a. Here is our result.

86 EXTENSION FIELDS

Proposition 4.2.6 Let p be prime. Then f = — a E F [x] is irreducible over F ifand only if f has no roots in F.

Proof: One direction is obvious, for if f has a root E F, then x — E FIx] is

a factor of f by Corollary A. 1.15. Going the other way, we will assume that f isreducible and prove that f has a root in F.

We first study the roots off. By Theorem 3.1.4, there is a field F C L over whichf splits completely, say

(4.7)

f a root in F. Thus we may assume that ai 0. If we set

for 1 i p, then a implies that

= = 1.a

It follows that a, = where is a pth root of unity. Hence (4.7) can be written

(4.8)

Now suppose thatf = gh, where g,h E F[x] have degree r,s <p. We may assumethat g, h are monic by multiplying them by suitable constants if necessary. By f = gh

and unique factorization, g must be a product of r of the factors of (4.8). Afterrelabeling if necessary, we may assume that

Since the constant term of g lies in F, this implies that

EF, where(=(i.•.(r.

Note also that (P—iSince 0< r< p andp is prime, mr+np= 1 for some m,n E Z. Then

= çmarnr+nP = E F

since e F and a e F. It follows that (tma1 E F. Thus

((ma)P = ((P)maP =a

shows that (mai is a root of f = a lying in F. •

The pth roots of unity used in the above proof are more abstract than the roots ofunity constructed in Section A.2.


Here is an easy application of Proposition 4.2.6 that will be useful when we studythe casus irreducibilis in Chapter 8.

Example 4.2.7 Let F be a subfield of and p be an odd prime. Given a E F, wedefine to be the real pth root of a. Furthermore, since p is odd, is the onlyreal pth root of a (be sure you understand why). Then Proposition 4.2.6 implies that

a is irreducible over F if and only if F.

Historical Notes

The factorization algorithm for polynomials in Q[xI is due to Kronecker and waspart of his constructive approach to algebra. Precise references can be found in [4].

The Schönemann—Eisenstein criterion was published by Schönemann in 1846 andindependently by Eisenstein in 1850. Although it is often called the "Eisensteincriterion," Schönemann's name should be included, since he proved it first. See [41and [9, p. 254] for references to the original papers. See also [2].

The slick proof of the irreducibility of given in Proposition 4.2.5 is dueto Eisenstein. In Chapter 15 we will explore the fascinating mathematics that ledEisenstein to the irreducibility criterion.

Schönemann discovered the criterion in a very different context. He asked whethera polynomial that is reducible modulo p remains reducible modulo p2. His versionof the criterion states that polynomials of the form (x — a)" + pF(x), where a E Z,F [xj E Z[xJ, and p F (a), are always irreducible. You will prove this in Exercise 2,and in Exercise 3 you will use this to give another proof of Proposition 4.2.5.

The first proof of Proposition 4.2.5 is due to Gauss in 1799 as part of his studyof regular polygons in Disquisitiones [5]. We will say more about this in Chapters 9and 10. In 1818, Gauss gave an interesting application of Proposition 4.2.5. His sixthproof of quadratic reciprocity used congruences modulo + + x + 1, which inmodern terms means that he was working in the quotient ring

Earlier, Gauss had given a proof of quadratic reciprocity (his fourth, in 1811) using thepth root of unity + . . . +x+ 1 is irreducible, it is the minimalpolynomial of Combining this with Lemma 4.1.13 and Proposition 4.1.14 givesan isomorphism

= + . . .

It follows that Gauss's sixth proof of quadratic reciprocity is a version of the fourth,with complex numbers replaced by the above quotient ring.

Abel used many properties of radicals in his proof of the unsolvability of thegeneral quintic. In a manuscript written shortly before his death in 1829, Abelproved Proposition 4.2.6 in the special case when E F [Abel, Vol. II, p. 229],and the general case is due to Kronecker in 1879 [Kronecker, Vol. IV, pp. 75—76].An even more general version of Proposition 4.2.6 is the following 1901 theorem ofCapelli (see [Chebotarev, p. 294]).

88 EXTENSION FIELDS

Theorem 4.2.8 Let f = — a E F [xl. Then f is reducible over F and only if mhas a divisor d> 1 such that

I bd, bEF,ora= <

1_4c4, d=4,cEF.


Exercise 1. This exercise will study the Lagrange interpolation formula. Suppose that F is afield and that b0,. . . , bd,co,. . . ,ca E F, where b0,.. . , bd are distinct and d � 1. Then considerthe polynomial

g(x) = >cifJ E F[x].i=O

(a) Explain why deg(g) <d, and give an example for F = R and d = 2 where deg(g) <2.(b) Show that g(bt) = for i = 0,... ,d.(c) Let h be a polynomial in F[xl with deg(h) <d such that = for i = 0,... ,d. Prove

thath=g.

Exercise 2. This exercise deals with Schönemann's version of the irreducibility criterion.(a) Let f(x) = (x — a E Z and F(x) E Z[x] satisfy deg(F) � n and

Prove that f is irreducible over Q.(b) More generally, let g(x) e Z[x] be irreducible modulo p (i.e., reducing its coefficients

modulo p gives an irreducible polynomial in Then let f(x) = g(x)lz + pF(x),where F[x] E Z[x] and g(x) and F(x) are relatively prime modulo p. Also assume thatdeg(F) ndeg(g). Prove thatf is irreducible over Q.

Exercise 3. Use part (a) of Exercise 2 with a = ito give another proof of Proposition 4.2.5.

Exercise 4. For each of the following polynomials, use a computer to determine whether it isirreducible over the given field.(a) x4+x3+x2+x+2overQ.(b)

Exercise 5. Find the minimal polynomial of the 24th root of unity (24 as follows.(a) Factor x24 — I over Q.(b) Determine which of the factors is the minimal polynomial of (24•

Exercise 6. Let F be a finite field. Explain why there is an algorithm for deciding whetherf E F[x] is irreducible.

Exercise 7. For each of the following polynomials, determine, without using a computer,whether it is irreducible over the given field.(a) x3-l-x+loverF5.(b) x4+x+loverlF2.

Exercise 8. Let a E Z be a product of distinct prime numbers. Prove that — a is irreducibleover Q for any n 1. What does this imply about when n � 2?

Exercise 9. Let k be a field, and let F = k(t) be the field of rational functions in t withcoefficients ink. Then consider f = — t e F[x], where p is prime. By Proposition 4.2.6, fis irreducible provided we can show that f has no roots in F. Prove this.

THE DEGREE OF AN EXTENSION 89

4.3 THE DEGREE OF AN EXTENSION

When F is a subfield of a field L, there is one bit of structure that hasn't been usedyet. We know that L is an Abelian group under addition. Furthermore, since F C L,the ability to multiply elements of L implies that we can multiply elements ofF timeselements of L. This gives a scalar multiplication, and one can easily check that Lbecomes a vector space over F.

A. Finite Extensions. The above paragraph leads to the following definition.

Definition 4.3.1 Let F C L be afield extension.(a) L is afinite extension ofF if L is a finite-dimensional vector space over F.(b) The degree of L over F, denoted [L: F], is defined as follows:

[L F] —fdimFL, if L is a finite extension ofF,

oo, otherwise,

where dimFL is the dimension of L as a vector space over F.

Here is a simple example.

Example 4.3.2 For JR C C, the usual way of writing complex numbers as a + bishows that 1 and i form a basis of C as a vector space over JR. Thus [C : IR] = 2.

We can also characterize extensions of degree 1.

Lemma 4.3.3 An extension F C L has degree [L: F] = 1 and only F = L.

Proof: If [L : F] = 1, then any nonzero element of L, say 1 L, is a basis. ThusL = {a. 11 a E F} = F. The opposite implication is even easier and is omitted. •

In general, we compute the degree of an extension F C F(a) as follows.

Proposition 4.3.4 Suppose that F C L is an extension and a E L.(a) a is algebraic over F and only if [F(a) : F] <00.(b) Let a be algebraic over F. If n is the degree of the minimal polynomial of a over

F, then l,a,... form a basis of F(a) over F. Thus [F(a) F] = n.

Proof: First suppose that a is algebraic over F with minimal polynomial p, wheren = deg(p). We need to show that 1, a,. . . , form a basis of F(a) over F. SinceF(a) = F [a], every element of F(a) is of the form g(a) for some g e F[x]. Dividingg by p gives

g = qp+ao+alx+

where q E F[x] and a0,... F, and evaluating this at x a yields

(4.9)

90 EXTENSION FIELDS

since p(a) = 0. Thus 1, , I span over F. To show linear independence,suppose that

where E F. is E F[x]. Sincethe minimal polynomial p has degree n, this must be the zero polynomial. Hence

= 0 for all i, and linear independence is proved. Then : F] = n follows fromDefinition 4.3.1.

This proves part (b) of the proposition and also one implication of part (a). Itremains to consider the case when Fl < 00. If we let n = : F], then

is an n-dimensional vector space over F. This implies that any collection ofn + 1 elements of F(cx) is linearly dependent. In particular, 1, cr" arelinearly dependent over F. Hence there are as,.. . , e F, not all zero, such that

(4.10)

As in the previous paragraph, it follows that is a root of

(4.11)

which is nonzero, since the a's are not all zero. Hence is algebraic over F, and the

proposition is proved. •

This proposition implies that when the minimal polynomial of has degree n,

every E F(a) can be written uniquely in the form

In Exercise I you will use an argument similar to the proof just given to describeunique coset representatives for elements of F [x] / (f).

Looking back at Example 4.3.2, we see that [C : ]R] = 2 follows from Proposi-tion 4.3.4, since C = and the minimal polynomial of i over R is x2 + 1. Hereare some other examples of Proposition 4.3.4.

Example 4.3.5 Consider the extension Q C Since the minimal polynomialof is x2 2, the proposition implies that Q] 2 and that

(4.12) = I a,b E Q}.

Note also that this representation is unique.

Example 4.3.6 By Example 4.1.7, the minimal polynomial of over Q is

x4— lOx2+ 1. Thus =4, and every e can be

written

for unique a,b,c,d EQ.


Example 4.3.7 Let F (x) be the field of rational functions in the variable x withcoefficients in F. Then Proposition 4.3.4 implies that [F(x) : F] = oo, since x is notalgebraic over F.

B. The Tower Theorem. We can also determine how the degree behaves whenwe have successive extensions F C K C L. The following result is sometimes calledthe Tower Theorem.

Theorem 4.3.8 Suppose that we have fields F C K C L.(a) If [K: F] = Do or [L : K] oo, then [L: F] cx).

(b) If [K: F] <oo and [L: K] <00, then [L: F] = [L : K] [K : F].

Proof: We will prove the contrapositive of part (a): if [L: F] <00, then [K: F] <00and [L : K] <oc. Thus we may assume that L has finite dimension as a vector spaceover F. Let 'y1,... be a basis. Then:

• One easily sees that K C L is a subspace of L over the field F. Since L has finitedimension over F, so does any subspace. Hence [K : F] dimpK < 00.

• Take a EL. Since-y1,... "YN span Lover F, a = aj'yj, where a, E F. SinceF c K, we can consider this as a linear combination with coefficients in K. ThusL is spanned over K by a finite set, so that [L : K] = dimKL < Do.

Toprovepart(b),letm = [K:F] andn = [L:K],andpickbasesai,...,am ofKoverF and /3i,... , of L over K. We will prove that the mn products

a basis of L over F. This will prove the theorem.We first show that the a1/31 span L over F. Take -y E L. Since /3k,... , span L

over K, we can write -y = where b1,... , E K. Then, since aI,... ,amspan K over F, we have = where a11 E F. Combining these equations,we obtain

= E = EEajjajfij.j=I i=I i=I j=I

Since E F, this shows that the span L over F.To prove linear independence, suppose that we have a linear relation

m n

= 01=1 j=I

where E F. As above, we can write this as

E(Eaijai)/3j=0.j=1 i=1

92 EXTENSION FIELDS

The expressions in the large parentheses all lie in K, and since the are linearlyindependent over K, we conclude that

for 1<j<n.i= I

Since the are linearly independent over F and F, we must have = 0 forall i and j. This proves the desired linear independence. •

Here are two examples of the Tower Theorem.

Example 4.3.9 We will analyze Q C Q ( using

Q c

Proposition 4.3.4 shows that 1, form a basis of over Q, since x2 —2 is theminimal polynomial of over Q. Furthermore, part (a) of Exercise 8 of Section 4.1shows that x2 —3 is the minimal polynomial of over so that 1, forma basis of over Thus:

•• The proof of Theorem 4.3.8 shows that the products of the bases 1, and 1,

namely 1, = give a basis of over Q.

Example 4.1.16 showed that span over Q. We now seethat 1, form a basis that arises naturally from Theorem 4.3.8.

In Section 4.2, we used Maple and Mathematica to show that =(see (4.5)). We now give a different proof using

We just showed that : = 4, and Example 4.3.6 tells us that the sameThen

:Q] = :Q]

gives = 1.

Example 4.3.10 Let w = and L = Q(w, We will compute [L : Q] usingthe extension fields

To determine {Q : Q], first observe that x3 —2 is irreducible over Q. Since x3 —2has degree 3, one can prove this using Lemma A.1.19 and Proposition A.3.1, thoughit is quicker to use the Schönemann—Eisenstein criterion (Theorem 4.2.3) with p = 2.

By Proposition 4.3.4 we conclude that


We next compute Recall that x2 +x+ 1 has roots w and w2, neither ofwhich is real. Since c R, x2 +x + 1 has no root in this field, so that x2 +x + 1is the minimal polynomial of w over Hence

[L : =2,

since L = Then Theorem 4.3.8 implies that

[L:Q] = =2.3=6.

We will return to this example often.

Mathematical Notes

Here is one of the ideas used in this section.

• Algebras over a Field. The key idea of this section is that a field extension F C Lgives L the structure of a vector space over F, so that L is simultaneously a field and avector space. In general, there are many examples of rings that are also vector spacesover a field. Here is the general definition.

Definition 4.3.11 A (possibly noncommutative) ring R is an algebra over the fieldF is a vector space over F such that:(a) The vector space addition on R is the same as the ring addition on R.(b) The scalar multiplication on R is compatible with the ring multiplication:

(ab) r = (b . r) for alla,b E F andr ER,a.(rs)=(a.r)s=r(a.s) forallaEFandr,sER.

A field extension F C L makes L into an F-algebra. Other examples include thepolynomial ring F[xi,... , x n matrices (F).

Historical Notes

The idea of representing elements of a field as linear combinations has a longhistory. For example, in 1847 Cauchy took a polynomial f E F [x] of degree n andrepresented elements of F[x] / (f) as linear combinations of the cosets of!, x, . . . ,

Kronecker also represented elements of extension fields using linear combinations,and he was aware of the importance of linear independence. But in all of this work,the term "degree" applied only to degrees of polynomials.

In 1871 Dedekind developed a theory of field extensions that included the conceptof degree. He writes an extension as A C ci and gives the modem criterion forw1,... , E Ii to be linearly independent over A. Furthermore, if the w span ci, thenhe sets (ci,A) = n. He also knows Proposition 4.3.4, but only gives special cases ofTheorem 4.3.8.

94 EXTENSION FIELDS

The modern formulation of the results of this section is due to Emil Artin. Hedeveloped his approach to Galois theory in the 1920s. He turned Dedekind'sinto the degree [L: F] and made it the centerpiece of his theory of finite extensions.Artin profoundly transformed the way people think about Galois theory. We will saymore about this in Section 6.1.

For more details on the history of how these concepts developed, we refer thereader to [6] and [7].


Exercise 1. In (4.9) we represented elements of F(a) uniquely using remainders on divisionby the minimal polynomial of a. In this exercise you will adapt the proof of Proposition 4.3.4to the case of quotient rings. Suppose that f E F [x] has degree n > 0. Prove that every cosetof F [x] / (f) can be written as

where ao,al ,... E F are unique.

Exercise 2. Compute the degrees of the following extensions:(a)

(b)

(c)

(d)

Exercise 3. For each of the extensions in Exercise 2, find a basis over Q using the method ofExample 4.3.9.

Exercise 4. Suppose that F C L is a finite extension with [L: Fl prime.(a) Show that the only subfields of L containing F are F and L.(b) ShowthatL=F(a)foranyaEL\F.

Exercise 5. Consider the extension Q C L = We will compute [L : QI.(a) Show that x4 —2 and x3 —3 are irreducible over Q.

(b) UseQCQ('h)cLtoshowthat4l[L:Q]and[L:Q]< 12.(c) Use Q C C L to show that [L: Q] is also divisible by 3.(d) Explain why parts (b) and (c) imply that [L : Q] = 12. This works because 3 and 4 are

relatively prime. Do you see why?

Exercise 6. Suppose that a and are algebraic over F with minimal polynomials f and g,respectively. Prove the Reciprocity Theorem: f is irreducible over F(13) if and only if g isirreducible over F(a).

Exercise 7. Suppose we have extensions L.o C L1 C C Lm. Use induction to prove thefollowing generalization of Theorem 4.3.8:(a) forsome 1 <i�m,then[Lm :L0J 00.

(b) If[L :L1_i] <oo for all 1 <i < m,then

{Lm : Lo] = [Lm : Lm_ij[Lm_i : Lm_21 ... [L2 : L1][L1 : Lol.

ALGEBRAIC EXTENSIONS 95

4.4 ALGEBRAIC EXTENSIONS

Now that we know the basic properties of the degree of an extension, we can continueour study of algebraic elements. We begin with a definition.

Definition 4.4.1 Afield extension F C L is algebraic if every element of L is algebraicover F.

It turns out that finite extensions are always algebraic.

Lemma 4.4.2 Let F C L be a finite extension. Then:(a) F C L is algebraic.(b) Ifa E L, then the degree of the minimal polynomial of a over F divides [L: F].

Proof: An element a E L gives F C F (a) C L, and then the Tower Theorem impliesthat [F(a) F] is finite and divides [L: F]. Then (a) and (b) follow immediately fromProposition 4.3.4. U

Exercise I will show that the converse of this lemma is false—there are algebraicextensions that are not finite. So a finite extension is an especially nice algebraicextension.

We next explore the structure of finite extensions.

Theorem 4.4.3 Let F C L be afield extension. Then [L : F] <00 if and only therearea1,... ,am EL such that each a, is algebraic over F andL = F(ai, . . .

Proof: First suppose that [L: F] <oo. Let a1, . . . E L be a basis of L as a vectorspace over F (so that m dimFL). Then

L={aiai+...+amamlai,...,amEF}CF(ai,...,am)CLproves that L = F(ai, . . . Each a1 is algebraic over F by Lemma 4.4.2.

Going the other way, suppose that L = F(ai,... ,am) where each a, is algebraicoverF. LetL0 = F andL1 = F(ai,. . . for 1 <i < m. Thenwegetfieldextensions

(4.13) FL0CL1CCLmL,and Corollary 4.1.11 shows that

for 1 i < m. Since a, is algebraic over F, it is also algebraic over the larger fieldF. Then Proposition 4.3.4 implies that

[L, :L,_i] = :L,_i] < cx,

so that every successive extension in (4.13) has finite degree. Then the generalizationof the Tower Theorem given in Exercise 7 of Section 4.3 implies that

[L:F] [Lm :Lo] = ELm :Lm_i]...[Li :L.o] <00.

This completes the proof of the theorem. U

96 EXTENSION FIELDS

As an application of the theorem just proved, let us show that the sum and productof algebraic elements are algebraic.

Proposition 4.4.4 Let F C L be a field extension. If a, E L are algebraic over F,then so are a + and

Proof: Theorem 4.4.3 implies that F C F(aj3) is a finite extension and hence isalgebraic by Lemma 4.4.2. Thus every element of F(a, is algebraic over F. Sincea+$,a/3 E F(a,/3), the proposition is proved.

Corollary 4.4.5 Given any field extension F C L, the subset

M = {a E L I a is algebraic over F}

is a subfield of L containing F.

Proof: We have F CM since a E F is a root ofx—a E F[x], and Mis closedunder addition and multiplication by Proposition 4.4.4. Since —1 E F C M, we seethat a EM implies —a = —1 a OEM, then Exercise 1 ofSection 4.1 shows that 1/a E M. It follows that M is a subfield of L. •

Here is a classic example of this corollary.

Example 4.4.6 A complex number z E C is called an algebraic number if it isalgebraic over Q. By Corollary 4.4.5, we have the field of algebraic numbers

Q = {z E C z is an algebraic number}.

Later in the section we will prove that Q is algebraically closed.

We next show that being algebraic is transitive in the following sense.

Theorem 4.4.7 Let F C K C L. If a E L is algebraic over K and K is algebraic overF, then a is algebraic over F.

Proof: Let a be a root off = + +/3o E K[x], where . . . ,/3o E K are notall 0. By hypothesis, each is algebraic over F. Then M = . . . is a finiteextension of F by Theorem 4.4.3. Furthermore, M is constructed so that f E M[xI. Itfollows that a is algebraic over M, which implies that M C M(a) is a finite extension.By Theorem 4.3.8,

[M(a):F]=[M(a):M][M:F1 <oc.

Thus F C M(a) is finite and hence algebraic. This means that every element ofM(a), including a, is algebraic over F. U

Here is an example of this theorem.

ALGEBRAIC EXTENSIONS 97

Example 4.4.8 Theorem 4.4.7 implies that every complex solution of the equation

(4.14) x11 +(1 +3i)x+'?/Ti= 0

is an algebraic number. This follows because the coefficients are obviously algebraicover Q. (Do you see why there are no real solutions?) In Exercise 2 you will showthat the minimal polynomial of a solution of (4.14) has degree at most 1760. <1'

Theorem 4.4.7 also has the following immediate corollary.

Corollary 4.4.9 If we have field extensions F C K C L where L is algebraic over Kand K is algebraic over F, then L is algebraic over F. •

Mathematical Notes

Here are some of the ideas encountered in this section.

• The Field of Algebraic Numbers. In Example 4.4.6, we defined Q to be the setof all algebraic numbers in C. This field has the following nice property.

Theorem 4.4.10 The field Q ofalgebraic numbers is algebraically closed.

Proof: By Exercise 7 of Section 3.2 it suffices to show that every nonconstantpolynomial in has a root in Q. Given such a polynomial f, we can regard fas an element of C[xJ, since Q c C. Then f has a root a EC by the FundamentalTheorem of Algebra, and a is algebraic over Q because f e Q [x]. But Q is algebraicover Q by definition, so that a is algebraic over Q by Theorem 4.4.7. Thus f has theroot a E Q, and we are done.

One can also show that if Q C L is an extension such that L is algebraic over Qand L is algebraically closed, then L Q. More generally, if F is any field, thenthere is a field F, unique up to isomorphism, such that F is algebraic over F andalgebraically closed. We call F the algebraic closure of F. (Strictly speaking, Fis only unique up to a nonunique isomorphism. Hence we should say "an algebraicclosure" rather than "the algebraic closure.") A discussion of algebraic closures canbe found in [Jacobson, Vol. II, Sec. 8.1].

• Algebraic Integers. Finally, in addition to the notion of an algebraic number in C,one can also define an algebraic integer to be a complex number that is a root of amonic polynomial with integer coefficients. For example, and w = (—1 +are algebraic integers, since they are roots of x2 — 2 and x2 + x + 1, respectively; butone can show that w/2 is not an algebraic integer (see Exercise 3). Algebraic integersplay an important role in number theory. For example, Euler proved Fermat's LastTheorem for n = 3 by writing x3 + y3 = z3 as

x3 =z3 —y3 = (z—y)(z—wy)(z—w2y)

and using unique factorization in the ring of algebraic integers Z[wI. This subjectis called algebraic number theory. For an introduction to algebraic number theory,including the details of Euler's argument, see [10, Ch. 9].

98 EXTENSION FIELDS


Exercise 1. Lemma 4.4.2 shows that a finite extension is algebraic. Here we will give anexample to show that the converse is false. The field of algebraic numbers Q is by definitionalgebraic over Q. You will show that [Q: Q] = 00 as follows.(a) Given n � 2 in Z, use Example 4.2.4 from Section 4.2 to show that Q has a subfield L

such that [L: Q] = n.

(b) Explain why part (a) implies that Q] = 00.

Exercise 2. Let a e C be a solution of (4.14). We will show that the minimal polynomial ofa over Q has degree at most 1760. Let L = i,

(a) Show that [L : Q] < 1760.(b) Use Lemma 4.4.2 to show that the minimal polynomial of cs has degree at most 1760.

Exercise 3. In the Mathematical Notes, we defined an algebraic integer to be a complexnumber a e C that is a root of a monic polynomial in Z[x].(a) Prove that a E C is an algebraic integer if and only if a is an algebraic number whose

minimal polynomial over Q has integer coefficients.(b) Show that w/2 is not an algebraic integer, where w = (—1 +

Exercise 4. Use (4.10) and (4.11) to prove the following weak form of Lemma 4.4.2: ifn = [L: F] <oc, then every a E L is a root of a nonzero polynomial in F[x] of degree n.

Exercise 5. In 1873 Hermite proved that the number e is transcendental over Q, and in 1882Lindemann showed that is transcendental over Q. It is unknown whether ir + e and it e

are transcendental. Prove that at least one of these numbers is transcendental over Q.

Exercise 6. Let F be a field. Show that other than the elements of F itself, no elements ofF(x) are algebraic over F. Thus, even though [F(x) : F] = oc by Example 4.3.7, the fieldM = {a E F(x) a is algebraic over F} of Corollary 4.4.5 is as small as possible, namely F.

Exercise 7. Suppose that F is an algebraically closed field, and let F C L be an algebraicextension. Prove that F = L.

Exercise 8. In this exercise you will show that every algebraic extension of R is finite ofdegree at most 2. To prove this, consider an algebraic extension R C L.(a) Explain why we can find an extension L C K such that x2 + 1 has a root a E K.(b) Prove that L(a) is algebraic over TR(a) and that R(a) C.

(c) Now use the previous exercise to conclude that EL: R] 2 and that equality occurs if andonly if L C.

Exercise 9. Prove that a E Q is an algebraic integer if and only if a E ZZ.

REFERENCES

1. H. Cohen, A Course in Computational Algebraic Number Theory, Springer, New York,Berlin, Heidelberg, 1993.

2. D. Cox, Why Eisenstein proved the Eisenstein criterion and why Schönemann discovereditfirst, Amer. Math. Monthly 118 (2011), 3—21.

REFERENCES 99

3. R. Dedekind, Theory ofAlgebraic Integers, English translation by J. Stiliwell, CambridgeU. P., Cambridge, 1996. (Translation of 1877 French edition.)

4. H. L. Dorwart, Irreducibility of polynomials, Amer. Math. Monthly 42 (1935), 369—381.

5. C. F. Gauss, Disquisitiones Arithmeticae, Leipzig, 1801. Republished in 1863 as VolumeI of [Gauss]. French translation, Recherches Arithmétiques, Paris, 1807. Reprint byHermann, Paris, 1910. German translation, Untersuchungen über Höhere Arithmetik,Berlin, 1889. Reprint by Chelsea, New York, 1965. English translation, Yale U. P., NewHaven, 1966. Reprint by Springer, New York, Berlin, Heidelberg, 1986.

6. B. M. Kiernan, The development of Galois theory from Lagrange to Artin, Arch. Hist.Exact Sci. 8 (1971), 40—154.

7. I. Kleiner, Field theory: From equations to axiomatization, Parts I and II, Amer. Math.Monthly 106 (1999), 677—684, 859—863.

8. I. K]einer, The genesis of the abstract ring concept, Amer. Math. Monthly 103 (1996),417—424.

9. F. Lemmermeyer, Reciprocity Laws, Springer, New York, Berlin, Heidelberg, 2000.

10. I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers, Third Edition,Wiley, New York, 1972.

CHAPTER 5

NORMAL AND SEPARABLEEXTENSIONS

This chapter will study some important properties of field extensions. We will beginwith extensions obtained by adjoining all roots of a polynomial. These splitting fieldswill lead to the idea of normality. We will also consider the idea of separability forboth polynomials and field extensions. The chapter will end with the Theorem of thePrimitive Element.

5.1 SPLITTING FIELDS

Given a nonconstant polynomial f E F [xl, Theorem 3.1.4 shows that there is anextension F C L over which f splits completely. In this section we will consider thesmallest such extension.

A. Definition and Examples. We begin with a definition.

Definition 5.1.1 Let f E F[x] have degree n > 0. Then an extension F C L is asplitting field off over F if(a) f= c(x—csi)• where CE F and cs1 EL, and(b)


102 NORMAL AND SEPARABLE EXTENSIONS

Be sure you understand how this captures the idea of the smallest extension overwhich a polynomial splits completely. The existence of splitting fields follows fromTheorem 3.1.4, for if f E Fix] splits completely as! = c(x — cr') . . . (x— in L[x],then F (a!,... , a splitting field off over F. We will prove below thatall splitting fields off E F{X] are isomorphic.

In the subsequent text, whenever we say "L is a splitting field of f E F [x]," wewill tacitly assume that f is nonconstant.

Here are some examples of splitting fields.

Example 5.1.2 = is a splitting field of (x2 — 2)(x2 —3)overQ.

Example 5.1.3 In Example 4.1.10 we showed that

=

Thus Q(i, is a splitting field of x4 —2 over Q.

Example 5.1.4 In Exercise 1 you will prove that the field Q(w, considered inExample 4.3.10 is a splitting field of x3 —2 over Q.

Note that a splitting field off E F [x] depends on both the polynomial f and thefield F. For instance:

a splitting field of x2 + 1 over Q is Q(i);

a splitting field of x2 + 1 over R is C;

a splitting field of x2 + 1 over C is C.

Since the roots of a nonconstant polynomial f E F[x] are algebraic over F, itfollows from Theorem 4.4.3 that a splitting field of f over F is always a finiteextension of F. We can bound the degree of this extension as follows.

Theorem 5.1.5 Let f E F [x] be a polynomial of degree n > 0, and let L be a splittingfield of f over F. Then {L : F] <n!.

Proof: We will prove this by induction on n. When n = 1, f = ax + b has the root—b/a E F, since a ThusL= F in this case, and [L:F] 1! is clear.

Now suppose that f has degree n> 1, and let L = F(ai,... , be a splittingfield off over F. If we write f = (x — )g, then the division algorithm implies thatg E F(ai ) [x]. Furthermore, the roots of g are obviously a2,. . . , a splittingfield ofg over F(ai) is given by

=L,

where the first equality follows from Corollary 4.1.11. Since g E F(ai ) [x] has degreen 1, our inductive hypothesis implies that

[L:F(ai)] <(n—i)!.

SPLITTING FIELDS 103

To bound the degree of F C L, we use the extensions F C F(ai) C L. By theTower Theorem (Theorem 4.3.8), we have

[L:F] = [L:F(ai)][F(ai) :F] < (n— 1)![F(ai) :F].

However, we also know that : F] is the degree of the minimal polynomial ofa1 over F, by Proposition 4.3.4. Since f(ai) = 0, we obtain [F(ai):F] <n, andthen [L : F] <n! follows. •

Sometimes the bound in Theorem 5.1.5 is sharp, meaning that there are caseswhere equality occurs, though the inequality can also be strict. For instance:

• By Example 5.1.2, is a splitting field of (x2 —2) (x2 — 3) over Q and

has degree 4 <4! over Q.• By Example 5.1.4, Q(w, is a splitting field of x3 —2 over Q and has degree

6 = 3! overQ.We will see in the next chapter that the size of the splitting field is closely related tothe size of the Galois group of the extension.

B. Uniqueness. We next study the uniqueness of splitting fields. A givenpolynomial f e F[x] will have many distinct splitting fields. For example,and Q[t]/(t2 —2) are splitting fields of x2 —2 over Q. The key point is that whilethey are not the same, they are isomorphic.

In order to prove this result for all polynomials, we need to prove somethingmore general. Suppose that we have an isomorphism of fields F1 F2, and letfi E F1 [x] be a polynomial of degree n > 0. Applying to the coefficients of fi givesapolynomialf2 E F2k].

Now let L be a splitting field off1 over F1 for i = 1,2. This gives the picture

L1 L2

U U

F1

Although the splitting fields L1 and L2 may be constructed in quite different ways,the following theorem tells us that they are always isomorphic.

Theorem 5.1.6 Given ft E F1 [x] and : F1 F2 as above, there is an isomorphismL1 such that =

Proof: We will prove this by induction on n = deg(fi) = deg(f2). When n = 1, wesaw in the proof of Theorem 5.1.5 that L1 F1 and L2 = F2. The theorem follows inthis case by taking =

Now suppose that n> 1. We know that L1 = Fi(ai,... where a!,... arethe roots of fi. As in the proof of Theorem 5.1.5, we will use the extensions

(5.1) F1CF1(ai)CL1,

where Fi(ai) c is a splitting field of gt = f1 /(x ai). We now proceed in thefollowing five steps.


Step 1. We first create an abstract model for Let h1 Fi[x] be the minimalpolynomial of We know that h, is an irreducible factor of E F1 [xJ, since is

a root of fi. ThusFi(ai) =FI[cEI]

where we have used Proposition 4.1.14 (for the equality) and Lemma 4.1.13 (for theisomorphism). The resulting isomorphism takes to x+ (h1).

Step 2. We next find a root of f2 corresponding to The key point is that the fieldisomorphism F1 induces a ring isomorphism F1 [x] F2[x] that takes fito f2. This isomorphism takes factors to factors and irreducibles to irreducibles. Inparticular, h1 will map to an irreducible factor h2 of f2. Since f2 splits completelyover L2, so does h2 (do you see why?). This allows us to label the roots of 12 as

Step 3. The root of 12 gives the extensions

(5.2) F2 C F2(f31) CL2,

where c L2 is a splitting field of = f2/(x — fit). As in Step 1, we also have

F2(/31) =F2[fiuI

since h2 is the minimal polynomial of This isomorphism takes flu to x + (h2).

Step 4. Since Fi[x] F2[x] takes h3 to h2, it must take (h1) to (h2). This meansthat we get an isomorphism of quotient rings

Fi[x]/(hi)

that takes x + (h1) to x + (h2) and is on the coefficients. Combining this withSteps I and 3, we get an isomorphism

that takes to flu and satisfies

Step 5. Finally, since : F1 (c'i) F2(fl1) takes cu to flu andf' to 12, it also takesgu = fi/(x—cu) to = f2/(x—/3I). As noted above, L1 is a splitting field ofover F1 (ai), and in the same way, L2 is a splitting field of over F2(81).

We can now prove the existence of the desired isomorphism between L1 and L2.If we combine the extensions (5.1) and (5.2) together with the isomorphisms andp,, then we get the diagram

L1 L2

U U

(5.3) F2(fl1)

U U

F1 F2

SPLI1TING FIELDS 105

Since = f' / (x a1) has degree n — 1, Step 5 implies that we can apply the inductivehypothesistog1 E Fi(ai)[xI and : Fi(ai) F2(/31). This L1 L2, whoserestriction to F1 (a1) is But since IF, = it follows that the restriction ofto F1 is Thus is the desired isomorphism. U

When applied to the identity map 1F : F —÷ F and f F [x], Theorem 5.1.6 impliesthe following uniqueness result for splitting fields.

Corollary 5.1.7 If Li and L2 are splitting fields off e F [xJ, then there is an isomor-phism L1 that is the identity on F. •

Because of this corollary, we can now speak of the splitting field of f E F [x],

provided that we remember that splitting fields are unique up to isomorphism.One might wonder why we proved Theorem 5.1.6 if all we wanted was Corol-

lary 5.1.7. The answer lies in the inductive nature of the proof: if we begin withthe identity map = 1F : F —÷ F, then the inductive step (5.3) uses the isomorphism

F(ai) F(/31). So if we had stated Theorem 5.1.6 only for the identity, thenour inductive hypothesis would not apply, since need not be the identity.

We conclude this section with a further application of Theorem 5.1.6. The ideais that this theorem gives some interesting isomorphisms of a splitting field. Moreprecisely, the following result will play an important role in Chapter 6.

Proposition 5.1.8 Let L be a splitting field of a polynomial in F [x], and suppose thath E FIx] is irreducible and has roots a, E L. Then there is a field isomorphisma : L —* L that is the identity on F and takes a to /3.

Proof: Since h is the minimal polynomial of a, we have an isomorphism

F(a) = F[a] F[x]/(h)

that is the identity on F and sends a to x + (h). Similarly, using /3, we have

F(/3) = F[/3] F[x]/(h)

that is the identity on F and sends to x + As in Step 4 of the proof ofTheorem 5.1.6, we can put these together to get a field isomorphism

such that = /3 and is the identity on F.Now suppose that L is a splitting field off E F[xI. Then f E F(a)[x] and f E

F (/3) [x], which means that L is a splitting field off over both F (a) and F Thuswe have the following diagram of splitting fields

L L

U U

F(a) F(/3)


where takes f to f. Then Theorem 5.1.6 gives L L such that =Since p is the identity on F and maps to /3, a = is what we want. •

Here is an example of this proposition.

Example 5.1.9 L = is the splitting field of x2 —2 over Q. This polynomialis irreducible over Q and has roots ± L. Then Proposition 5.1.8 implies thatthere is an isomorphism o• : L —÷ L such that =

In the terminology of Chapter 6, an isomorphism a : L L that is the identity onF C L is an element of the Galois group Gal(L/F). We will use Proposition 5.1.8 toconstruct elements of Gal(L/F) when L is a splitting field over F.


Exercise 1. Show that a splitting field of x3 —2 over Q is Q(w, w =

Exercise 2. Prove that f F [x] splits completely over F if and only if F is the splitting fieldof f over F.

Exercise 3. Prove that an extension F C L of degree 2 is a splitting field.

Exercise 4. Find the splitting field of x6 — 1 E Q [x].

Exercise 5. We showed in Section 4.1 that f = x4 lOx2 + 1 is irreducible over Q. Show thatL = is the splitting field of f over Q.

Exercise 6. Letf E Q{x] be the minimal polynomial of = +(a) Show that f = x4 — 4x2 + 2. Thus [Q(a) Q] = 4.(b) Show that is the splitting field of f over Q.

Exercise 7. Letf=x3—x+1 EF3[x].(a) Show that f is irreducible over ]F3.(b) Let L be the splitting field off over F3. Prove that [L: F3] = 3.(c) Explain why L is a field with 27 elements.

Exercise 8. Let n be a positive integer. Then the polynomial f = f —2 is irreducible over Qby the Schönemann—Eisenstein criterion for the prime 2.(a) Determine the splitting field L off over Q.(b) Showthat[L:Q] =n(n—1)whennisprime.

Exercise 9. Letf e F[x] have degree n > 0, and let L be the splitting field of f over F.(a) Suppose that [L: F] = n!. Prove that f is irreducible over F.(b) Show that the converse of part (a) is false.

Exercise 10. Let F C L be the splitting field of f E F[x], and let K be a field such thatF C K C L. Prove that K C L is the splitting field of some polynomial in K[x].

Exercise 11. Suppose that f F[x] is irreducible of degree n > 0, and let L be the splittingfield of f over F.(a) Prove that F].(b) Give an example to show that n = [L F] can occur in part (a).

Exercise 12. In the situation of Theorem 5.1.6, explain why [L1 : Fi] = [L2 : F2].

Exercise 13. Let L = Q ( Use Proposition 5.1.8 to prove that there is an isomorphisma L L such that = and =

NORMAL EXTENSIONS 107

5.2 NORMAL EXTENSIONS

In this section, we will discover an important property of splitting fields. This willlead to the concept of a normal extension.

Being a splitting field is a very special property of a field extension. For example,we will see below that is not the splitting field of any f E Q[x]. The basicreason for this lies in the following proposition.

Proposition 5.2.1 Let L be the splitting field off E F[x], and let g E F[x] be irre-ducible. ifg has one root in L, then g splits completely over L.

Proof: We can assume that f and g are monic. Then L = , an), where

f = (x (x If/3 E Lisa root of g, then g is the minimal polynomial of/3 since g is irreducible and monic. We need to prove that all roots of g lie in L.

Proposition 4.1.15 implies that L = . . . , so that /3 is a polynomial in thei.e., = h(ai,... , for some h E F[xi,... Now consider the polynomial

(5.4) s(x)= fi EL[x].

This clearly has all of its roots in L. Furthermore, the factor corresponding to a e

is x , = x — /3, so that /3 is a root of s.If we could show that s E F[x], then gls would follow immediately, since g is the

minimal polynomial of /3. Since s splits completely over L, this would imply that galso splits completely over L.

Hence it suffices to prove that s E F [x]. We do this by going to the universalsituation, as we did for the polynomial in the proof of Theorem 3.2.4. Thepolynomial

S(x)= flhas coefficients in F[xi, . . . Furthermore, permuting x1, . . . permutes thefactors of S. It follows that if we multiply out 5, then we get an expression

S(x) = ..

where each pj(Xi , . . . E F[xi,... is symmetric. Since the are the roots off e F[xJ, Corollary 2.2.5 implies that

,E F. We conclude that

s(x) = E F[x].

As explained above, the proposition now follows. •

This proof of Theorem 5.2.1 uses the theory of symmetric polynomials fromChapter 2. See [Stewart, Ch. 101 for a proof that doesn't use symmetric polynomials.


Here is the example promised above.

Example 5.2.2 It is now easy to see why is not the splitting field of anypolynomial in Q[x], since x3 — 2 is irreducible over Q and obviously has a root in

If this field were a splitting field, then Proposition 5.2.1 would force x3 — 2

to split completely over But this is impossible, since C JR doesn'tcontain the complex roots of x3 —2.

In Exercise 1 you will prove similarly that is not the splitting field of anypolynomial in Q[x].

The property of Proposition 5.2.1 leads to the following definition.

Definition 5.2.3 An algebraic extension F C L is normal if every irreducible poiy-nomial in F [xJ that has a root in L splits completely over L.

In Exercise 2 you will show that F C L is normal if and only if the minimalpolynomial (relative to F) of every E L splits completely over L.

The following result reveals the strong link between normal extensions and split-ting fields.

Theorem 5.2.4 Suppose that F c L. Then L is the splitting field of some f E F{x] ifand only if the extension F C L is normal and finite.

Proof: First suppose that L is the splitting field off E F[x]. Then F C L is finite byTheorem 5.1.5 and is normal by Proposition 5.2.1.

For the converse, suppose that F c L is normal and finite. By Theorem 4.4.3, thefiniteness of this extension implies L F(ai,... , 'km), where each is algebraicover F. Let p1 E F[x] be the minimal polynomial of aj, and set f Wewill show that L is the splitting field off over F.

To prove this, first observe that every p, splits completely over L, since F C Lis normal and p, F[x] is irreducible with a root E L. It follows that f splitscompletely over L. Now let L' c L be the subfield of L generated by F and the rootsof f. Since the roots of f include . . . we have

This shows that L' = L, so that L is the splitting field off over F. •

We will see that normal extensions play an important role in Galois theory.

Historical Notes

Polynomials similar to s(x) in (5.4) appear in the work of Galois. For example, inhis first memoir on Galois theory, Galois says the following:

In fact, by multiplying together all of the factors of the form V — b, c,. . , d),where one operates on the tetters by all possible permutations, one will get anequation rational in V that is necessarily divisible by the equation in question.

(See [Galois, p. 51].) Here, a,b,c,...,d are roots of a polynomial f e F[xI, and,d) is an element of the splitting field F(a,b,c,...,d). Then we can

interpret Galois's statement as follows:

SEPARABLE EXTENSIONS 109

• By saying "an equation rational in V," Galois is asserting that the resulting poly-nomial in V has coefficients in F. This is exactly what we proved about thepolynomial s(x) in (5.4).

• When Galois says that this is "necessarily divisible by the equation in question,"he is referring to the minimal polynomial of ,d). This is what wecalled g in Proposition 5.2.1.

Thus, although normality does not appear explicitly in Galois's work, the abovequotation should make it clear that it is implicit in what he does. We will say moreabout Galois's results in Chapter 12.


Exercise 1. Prove that is not the splitting field of any polynomial in Q[x].

Exercise 2. Prove that an algebraic extension F C L is nonnal if and only if for every a E L,the minimal polynomial of a over F splits completely over L.

Exercise 3. Determine whether the following extensions are normal. Justify your answers.(a) Q C Q((,,), where (,, =(b)

(c) F = 1F3(t) C F(a), where t is a variable and a is a root of x3 — tin a splitting field.

Exercise 4. Give an example of a normal extension of Q that is not finite.

5.3 SEPARABLE EXTENSIONS

Given a nonconstant polynomial f E F [x] with splitting field F C L, we can write

(5.5)

It is important to realize that cVI,... are not always distinct. For example, f =— 2x + 1 E Q[x] has a1 = = 1. In this section, we will study those special

polynomials for which the roots are all different.We begin with some terminology. Given f as in (5.5), let . . . , 13r be the distinct

elements of L that appear among a1, . . . , and let m1 be the number of times x —appears in (5.5). Then we can write (5.5) as

f=ao(x_fii)ml ...(X_/3r)mr, a0EF, 131,...,I3rELdistinct, mI,...,mr� 1.

We call m, the muliplicity of and say that is a simple root if m, = 1 and a multipleroot if m, > 1.

Definition 5.3.1 A polynomial f E F [x] is separable if it is nonconstant and its motsin a splitting field are all simple.

In other words, f is separable if it has distinct roots. These definitions areindependent of splitting field used, since all splitting fields of f over F are isomorphic.


One tool used to study separability is the discriminant E F of a monicpolynomial f E F [x]. We defined (f) in Section 2.4 and showed in Proposition 2.4.3that if deg(f) > l,then

fJ (a a)2 when1<i<j<n

Another tool we will need is the formal derivative, which for a polynomial g =E F[x] is defined to be

g i—+ g' enjoys the usual properties from calculus, including

56(gh)' = g'h + gh'

for g,h E F[x] and a,b E F. See Exercise 1 for a proof of (5.6).Separability, the discriminant, and the formal derivative are related as follows.

Proposition 5.3.2 If f e F [xI is monic and nonconstant, then the following areequivalent:(a) f is separable.(b) 0.

(c) f andf' are relatively prime in F[x], i.e., gcd(f,f') = 1.

Proof: If deg(f) = 1, then = 1 by the definition of given in Section 2.4.It follows easily that (a), (b), and (c) are all true in this case. Hence we may assumethat deg(f) = n> 1.

For (a) (b), let ar,. . . , a,, be the roots off in some splitting field. The aboveformula for shows that 0 is equivalent to a, a3 for all i < j.

It remains to show (a) (c). Let L be a splitting field of f over F, so thatForagiveni,write

f(x) = h,(x) =

Differentiating, we obtain f'(x) = (x — + h'(x) by the product rule, and thenevaluating at gives

(5.7) f'(a1) = = a').

If (c) is false, then f and f' have a common factor g of positive degree. Sincewe must have g(aj) = 0 for some i, and then glf' implies that = 0. Hence0 = f'(a,) = a1), so that a, = a3 for some j i.

Conversely, if (c) is true, then 1 = Af + Bf' for some A, B E F [x]. Evaluatingthis at a gives 1 = so that f'(a1) 0. By (5.7), this implies that

is nonzero for all i. Hence a1,.. . , a,, are distinct. •


The definition of separable polynomial given in Definition 5.3.1 is nonstandard inthat it applies to arbitrary nonconstant polynomials with distinct roots, while mostbooks focus on irreducible polynomials with distinct roots. Fortunately, as long as werestrict to irreducible polynomials, Definition 5.3.1 is consistent with the literature.

We can also extend the concept of separability to algebraic extensions.

Definition 5.3.3 Let F C L be an algebraic extension.(a) a E L is separable over F if its minimal polynomial over F is separable.(b) F C L is a separable extension if every a E L is separable over F.

Since minimal polynomials are irreducible, this agrees with the definition ofseparable extension given in other texts.

We can interpret the separability of a polynomial in terms of its irreducible factorsas follows.

Lemma 5.3.4 A nonconstant polynomial f E F[x] is separable if and only if f is aproduct of irreducible polynomials, each of which is separable and no two of whichare multiples of each other

Proof: First assume that f is separable. If a factor off fails to have distinct rootsin a splitting field, then the same is true for f. Hence any irreducible factor off mustbe separable. Also, if the factorization of f into irreducibles includes two factors thatare multiples of each other, then the product of these factors would be a nonseparabledivisor of f. Hence the factorization of f must consist of separable, irreduciblepolynomials no two of which are multiples of each other.

Conversely, let f = g5, where . . . , are separable and irreducible, and notwo are multiples of each other. Then, in the splitting field off, each gj has distinctroots. Furthermore, suppose that g and gj share a root a for some i j. Since gjand gj are irreducible, this would imply that each was a constant times the minimalpolynomial of a, which is a contradiction. Hence f is separable. •

In order to make good use of Lemma 5.3.4, we need to understand when anirreducible polynomial is separable. Fortunately, many irreducible polynomials areautomatically separable.

Lemma 5.3.5 Let f E F [x] be an irreducible polynomial of degree n. Then f isseparable either of the following conditions is satisfied:(a) F has characteristic 0, or(b) F has characteristic p > 0 and pin.

Proof: Let f = a0f where n >0 and a0 0. Then f' =+ Condition (a) or (b) implies that n 0 in F, so that a0 0

implies na0 0. Hence f'is nonzero and has degree n — 1.

Since f is irreducible, its only divisors (up to constant multiples) are 1 and f.Hence g = gcd(f,f') must be 1 or f, up to constants. But and f' 0 implydeg(g) <deg(f') = n — 1. Thus g cannot be a multiple of f, so gcd(f,f') = g = 1..


One surprise of Lemma 5.3.5 is that separability is related to the characteristic.Here is another example of this phenomenon.

Example 5.3.6 Consider f = x" — 1 E F[x], where n > 0. By Proposition 5.3.2, f isseparable if and only if f is relatively prime to f' = However:

• If n 0 in F, then the only irreducible factor of f' is x, which clearly doesn'tdivide f. Thus f is relatively prime to f' in this case.

• If n = 0 in F, then f' is identically zero, in which case f divides f'. Hence f isnot relatively prime to f' in this case.

It follows that 1 E F [x] fails to be separable if and only if F has characteristic pand p divides n.

For the remainder of the section, we will consider fields of characteristic 0 andcharacteristic p separately. Since we encounter fields of characteristic 0 most often,we will begin with them.

A. Fields of Characteristic 0. Here is an application of Lemmas 5.3.4 and 5.3.5.

Proposition 5.3.7 1fF has characteristic 0, then:(a) Every irreducible polynomial in F {x] is separable.(b) Every algebraic extension ofF is separable.(c) A nonconstant polynomial f E F[x} is separable and only is a product of

irreducible polynomials, no two of which are multiples of each other

Pro of : Part (a) follows immediately from Lemma 5.3.5, and this implies part (b)by Definition 5.3.3. Finally, part (c) follows from part (a) and Lemma 5.3.4. .

In characteristic 0, we can get rid of multiple roots as follows.

Proposition 5.3.8 Let F have characteristic 0, and suppose that f E F [x] has thefactorization f = cg'1'1 . where c E F, gj E is monic and irreducible for

1 andgi,...,gj are distinct. Then

(5.8)gcd(f,f')

Further,nore, is separable and has the same roots as f in a splitting field.

Proof: Proposition 5.3.7 implies that gi is separable, and this polynomial andf clearly have the same roots in a splitting field. Hence it suffices to prove (5.8).

The factorization f = cg7' . implies that we can compute gcd(f,f') byfinding the highest power of g, that divides f' (do you see why?). If we write

f = h, =

then differentiating gives

f' = + = + g1h).


This shows that f'. If we had then gj + and thus gjSince gj is irreducible, this would force gj or gj h1. The latter is impossible bythe definition of h1, and the former is impossible because is nonzero of degreedeg(g) — 1 (this is where we use characteristic 0). Hence is the highest powerof gj dividing f', which implies that

gcd(f,f')

The desired formula (5.8) follows immediately. •

This proposition is more powerful than it seems. For example, suppose that wehave a polynomial f E F[x] that has multiple roots in a splitting field, say

(5.9) a0EF, /31,...,I3rdistinct, rn? 1.

If we ignore the multiplicities, then we get the separable polynomial

g=ao(x—di)...(x—(3r),

which has the same roots as f. There are three methods to find g:

• If we know the roots off, then we get g from the factorization (5.9). This requiresknowing the roots, which rarely happens.

• If we know the irreducible factorization f = . over F, then we getg = cgl by Proposition 5.3.8. This requires knowing the factorization, whichcan be time-consuming to compute.

• We get g from the gcd computation given in (5.8) of Proposition 5.3.8.

In practice, the third method is the most efficient. Here is an example.

Example5.3.9 —x'°+2x8—4x7+3x5—3x4+x3+3x2—x—1 E Q[xI.Using the gcd command in Maple or the PolynomialGCD command in Mathematica,one finds that

gcd(f,f')=x6—x5+x3—2x2+1.

It follows that

f x"—x10+2x8—4x7+3x5—3x4+x3+3x2—x—1(5.10) gcd(f,f') — x6—x5+x3—2x2+l

=x5+x2—x— i

is a separable polynomial with the same roots as f.

B. Fields of Characteristic p. We begin with an important property of suchfields.

Lemma 5.3.10 Let F be afield of characteristic p, and assurne that a, j3 E F. Then(a + = & + /3P and (a — = &


Proof: The binomial theorem implies that

= . . .+ . . . +

In the proof of Proposition 4.2.5, we showed that p — 1. Since Fhas characteristic p, the above identity reduces to (a + = & + In Exercise 2you will use this to prove that (a j3)" = & — /3P.

In Exercise 3 you will use Lemma 5.3.10 to show that 1 is the only pth root ofunity in a field of characteristic p.

Since (a/3)P = aP/3P, Lemma 5.3.10 implies that the map a —* a1' is a ring homo-morphism over any field F of characteristic p. This is the Frobenius homomorphismof F. We will use Frobenius when we discuss finite fields in Chapter 11.

Here is our first example of a nonseparable irreducible polynomial.

Example 5.3.11 Let F = k(t), where k has characteristic p and t is a variable. Weclaim that f = — t E F[x] is nonseparable and irreducible over F.

To prove this, note that f has no roots in F, by Exercise 9 in Section 4.2. Since pis prime, Proposition 4.2.6 implies that f is irreducible over F. Furthermore, if a E Lis a root of f in its splitting field L, then t. Using Lemma 5.3.10, it follows that

(5.11) (x—a)"=x1'—a1'=x"—t.

Thus f does not have distinct roots in its splitting field L and hence is not separable.The polynomial f also gives an example of a nonseparable finite extension.

Namely, a e L is a root of f, so that f is the minimal polynomial of a over F,since f is irreducible and monic. It follows that F c L is not separable.

Note also that by (5.11), a is the only root of f. Hence the splitting field isL = F(a). This implies [L : F] = p. since f is the minimal polynomial of a.

One caution is that over a field of characteristic p. not all irreducible polynomialsof degree p fail to be separable. Here is a simple example.

Example 5.3.12 ForthefieldF2oftwoelements,f=x2+x+ 1 E 1F2[x] isirreducible,since it has no roots in F2. It is also separable, since f' = 2x+ 1 = 1 is relativelyprime tof.

We will say more about characteristic p in the Mathematical Notes.

C. Computations. To determine whether a monic polynomial f F [x] is separable,one can use either or gcd(f,f') by Proposition 5.3.2. We will briefly discusshow to compute both using Maple and Mat hematica.

We begin with a gcd computation.

Example 5.3.13 Example 5.3.9 explained how Maple and Mathematica do this overQ. For example, if

(5.12) f=x6+lOx3+3x2+1 EQ[x],


one computes gcd(f,f') = 1, so that f is separable. However, since f has integercoefficients, we can reduce modulo p and obtain a polynomial E [x]. Then wecan ask whether is separable over IF,,.

For p = 2 or 3, we have = 0, since f' = 6x5 +30x2 +6x. Thus =f,, 1, so that f,, is not separable for these primes. For a larger prime

such as p = 557 (the reason for this choice will soon become clear), we compute thegcd over IF557 using the Maple command

Gcd(f6 + + + 1, + + 6*x) mod 557;

which gives the result x + 257. Thus is not separable. In Mathematica, thiscomputation is done using the command

+ + 3f2 + 1, + + 6x,

Modulus—> 557]

which gives the same answer x + 257.

The second approach to studying whether f is separable would be to compute thediscriminant Chapter 2 gave a cumbersome method for computing thatexpresses = x1)2 in terms of the elementary symmetric polynomials andthen evaluates the oj at the coefficients off (up to the usual sign). A more efficientapproach uses the resultant.

We will not discuss resultants in detail, for this would take us too far afield. Theidea is that forf, g E F [x], their resultant

Res(f,g,x) E F

is a polynomial in the coefficients off and g with the property that

Res(f,g,x) = 0 f and g have a common root in an extension ofF.

An introduction to resultants can be found in [1, Ch. 3, §5] and [3, pp. 97—104].For us, the most important property of resultants is that if f F[x] is monic of

degree n> 1, then

(5.13) =

(see [3, pp. 103—1041). In Maple and Mathematica, the resultant off, g is computedusingthecommandsresultant(f,g,x) andResultant[f,g,x].

Example 5.3.14 As an example, consider the polynomial f = x6 + lOx3 + 3x2 + 1given by (5.12). This leads to

= (— = —(—649684800) = 26.36.52.557.

As before, reducing f modulo p gives E In Exercise 4 you will show thate F,, is the congruence class of modulo p. Thus

f,, is separable over IF,, 0 mod p.


It follows that f,, is separable over IF,, if and only if p {2,3,5,557}. From here,one easily finds that

(5.14) gcd(f f')= x2+3," x+257, p=557,

1, otherwise.

You will compute a similar example in Exercise 5.

Mathematical Notes

Our treatment omits many interesting results about separability.

• Separable Extensions. Here are some conditions that imply separability.

Theorem 5.3.15(a) If L = , cx,,), where each cx, is separable over F, then F C L is separable.(b) 1fF C L is the splitting field of a separable polynomial, then F C L separable.(c) 1fF c K and K C L are separable extensions, then F C L is separable. •

We will defer our proof of part (a) until Chapter 7, since it uses some ideas fromGalois theory. (For a proof that doesn't use Galois theory, see Corollaries 1 and 3 of[Garling, Sec. 10.21.) In Exercise 6 you will show that part (b) follows from part (a).The proof of part (c) requires the concept of separable degree, which is discussed in[Grillet, Sec. 7.2].

• The Structure of Irreducible Polynomials. Although irreducible polynomials areseparable in characteristic 0, things are more complicated in characteristic p. In thiscase, irreducible polynomials are built from separable ones as follows.

Proposition 5.3.16 Let F have characteristic p. and let f E F [x] be irreducible.Then there is an integer e � 0 and a separable, irreducible polynomial g E F [x] suchthatf(x) = •

You will prove this in Exercise 7.

• Purely Inseparable Extensions. If an algebraic extension F C L is not separa-ble, then some (but not necessarily all) elements of L have nonseparable minimalpolynomials. Here is a simple example.

Example 5.3.17 Suppose that k has characteristic 3, and let t, u be variables. Con-sider F = k(t, u), and let F C L be the splitting field of f = (x2 t)(x3 u). Thus Lcontains elements cx, such that cx2 = t and = u. In Exercise 8 you will prove thefollowing:

• The minimal polynomial of cx over F is x2 — t, which is separable. Thus cx isseparable over F.


• The minimal polynomial of /3 over F is x3 — u, which is not separable (rememberthat F has characteristic 3). Hence /3 is not separable over F.

Thus some elements of L are separable over F while others are not. <IF>

However, some extensions have very few separable elements. Given F C L,it is clear that every a E F is separable over F. Thus we say that an algebraicextension F C L is purely inseparable if no element of L \ F is separable over F. Forexample, you will prove in Exercise 9 that the extension of Example 5.3.11 is purelyinseparable.

In general, if F C L is purely inseparable, then the minimal polynomial of a E Lis of the form xi" — a for some e 0 and a E F. This implies that the degree of afinite purely inseparable extension is a power of p (see Exercise 10).

Returning to the case of an arbitrary algebraic extension F C L in characteristic p,one can "separate" the separable elements from the inseparable ones. More precisely,one can prove the existence of a unique intermediate field F C K C L such that Kis separable over F and L is purely inseparable over K. A proof can be found inSection 8.7 of [Jacobson, Vol. III.

• The Squarefree Decomposition of a Polynomial. Proposition 5.3.8 shows thatif F has characteristic 0 and f F[x], then g = f/gcd(f,f') e F{x] is separable andhas the same roots as f. This means that

(5.15) f=gh,

where g is separable and every root of h has multiplicity at least 2 as a root of f. In thissituation, we call g the squarefree part off, and (5.15) its squarefree decomposition.Squarefree decompositions also exist when F has characteristic p (this is proved in[2, Tutorial 5, pp. 37—381). The difference is that in characteristic p, the squarefreepart g need not have the same roots as f (can you give an example?). See alsoExercise 11.



Exercise 2. Let F have characteristic p, and suppose that a, /3 E F. Lemma 5.3.10 shows that

(a + = + for all e � 0.

Exercise 3. Let F be a field of characteristic p. The nth roots of unity are defined to be theroots off — 1 in the splitting field F C L off — 1.

(a) If p n, show that there are n distinct nth roots of unity in L.(b) Show that there is only one pth root of unity, namely 1 E F.

Exercise 4. Let f Z[x] be monic and nonconstant and have discriminant Then letIF,,{x] be obtained from f by reducing modulo p. Prove that E F,, is the congruence

class of


Exercise 5. For f = x7 + x + 1, find all primes for which f,, is not separable, and computeas in (5.14).

Exercise 6. Use part (a) of Theorem 5.3.15 to show that the splitting field of a separablepolynomial gives a separable extension.

Exercise 7. Suppose that F is a field of characteristic p. The goal of this exercise is to proveProposition 5.3.16. To begin the proof, let f E F[xI be irreducible.(a) Assume that f' is not identically zero. Then use the argument of Lemma 5.3.5 to show

that f is separable.(b) Now assume that f' is identically zero. Show that there is a polynomial gi E F{x] such

thatf(x) =gi(x").(c) Show that the polynomial gi of part (b) is irreducible.(d) Now apply parts (a)—(c) to gi repeatedly until you get a separable polynomial g, and

conclude that f(x) = g(x" ) where e 0 and g E F[x] is irreducible and separable.

ExerciseS. Let F = k(t,u) and f = (x2 —r)(x3 —u) be as in Example 5.3.17. Then thesplitting field off contains elements a, such that a2 = t and /33 = u.

(a) Prove that x2 — t is the minimal polynomial of a over F. Also show that x2 t is separable.(b) Similarly, prove that x3 u is the minimal polynomial of /3 over F, and show that x3 — u

is not separable.

Exercise 9. Let F be a field of characteristic p. and consider f = xi" — a E F [xJ. We willassume that f has no roots in F, so that f is irreducible by Proposition 4.2.6. Let a be a rootoff in some extension of F.(a) Argue as in Example 5.3.11 that F(a) is the splitting field off and that [F(a) F] = p.

(b) Let /3 E F(a) \F. Use Lemma 5.3.10 to show that /3" E F.(c) Use parts (a) and (b) to show that the minimal polynomial of /3 over F is /3".

(d) Conclude that F C F(cs) is purely inseparable.

Exercise 10. Suppose that F has characteristic p and F C L is a finite extension.(a) Use Proposition 5.3.16 to prove that F C L is purely inseparable if and only if the minimal

polynomialofeverya ELisoftheformx" —aforsomee�OandaEF.(b) Now suppose that F C L is purely inseparable. Prove that [L: Fl is a power of p.

Exercise 11. Let f E F[x] be nonconstant. We say that f is squarefree if f is not divisible bythe square of a nonconstant polynomial in F [x].(a) Prove that f is squarefree if and only if f is a product of irreducible polynomials, no two

of which are multiples of each other.(b) Assume that F has characteristic 0. Prove that f is separable if and only if f is squarefree.

Exercise 12. Prove that f E F [x] is separable if and only if f is nonconstant and f and f' haveno common roots in any extension of F.

Exercise 13. Let F have characteristic p, and let F C L be a finite extension with p [L: Fl.Prove that F C L is separable.

Exercise 14. Let F C K C L be field extensions, and assume that L is separable over F. Provethat F C K and K C L are separable extensions. Note that this is the converse of part (c) ofTheorem 5.3.15

Exercise 15. Let f be the polynomial considered in Example 5.3.9. Use Maple or Mat hematicato factor f and to verify that the product of the distinct irreducible factors off is the polynomialgiven in (5.10).

THEOREM OF THE PRIMITIVE ELEMENT 119

Exercise 16. Let F have characteristic p and consider f = — x + a E F jxl.(a) Show that f is separable.(b) Let a be a root off in some extension of F. Show that a + 1 is also a root.(c) Use part (b) to show that f splits completely over F (a).(d) Use part (a) of Theorem 5.3.15 to show that F C F(a) is separable and normal.In ExerciseS of Section 6.2 you will show that if f is irreducible over F, then Gal(F(a)/F)Z/pZ, generated by the automorphism sending a to a + 1. This is related to a theorem ofArtin and Schreier, which states that in characteristic p. every separable, normal extension ofdegree p is the splitting field of an irreducible polynomial of the form x" — x + a.

Exercise 17. Let be a root of a polynomial f.(a) Assume that f(x) = (x for some polynomial h(x), and let f(m) denote the mth

derivative of f. Prove that f(m)(/3) = m!h(/3).(b) Assume that we are in characteristic 0. Prove that has multiplicity m as a root off if

and only if f(/3) = = ... = = 0 and(c) Assume that we are in characteristic p. How big does p need to be relative to m in order

for the equivalence of part (b) to be still valid?

5.4 THEOREM OF THE PRIMITIVE ELEMENT

Of the extension fields F C L studied so far, the nicest case is when L = F(a) forsome a E L. When this happens, we say that a is a primitive element of F C L.In this section, we will show that many but not all finite extensions have primitiveelements.

Here is the Theorem of the Primitive Element.

Theorem 5.4.1 Let F C L = F(a1,. . . , a finite extension, where each is

separable over F. Then there is a L separable over F such that L = F(a).Furthermore, F is infinite, then a can be chosen to be ofthefor,n

Proof: First assume that F is infinite and that L = F(ai,... , where each a1 isseparable over F. We will use induction on n to show that there are ti,... , E Fsuch that L = F(tiai + . + and ticti + + is separable over F.

We begin with the case n = 2. Given L = F(J3,-y), let f,g E F[x] be the minimalpolynomials of fi, -y, respectively, and set £ = deg(f), m = deg(g). In a splitting fieldof fg, the separability of j3,-y implies that

f has distinct roots = /3i ,132,.. . ,

g has distinct roots 'y = y1,y2, . . .

Since F is infinite, we can find A E F such that

z r

-ys —


This easily implies that

(5.16) for(r,s)

In particular, since = and -y = -y1, we have

(5.17) forl 2<j<m.

We first prove that F(/3 + A-y) = F(/3, -y). Since F(/3 + c F(/3, -y) is obvious,it suffices to show that fi,'y E + .X-y). We begin with -y. Observe that

• g(x) vanishes at 'y and lies in F[xJ C F(/3+ A'y)[x];

• f(/3 + — Ax) vanishes at 'y (check this!) and also lies in +Our strategy will be to study the greatest common divisor of the polynomials g(x)and + A-y — Ax). We first note that if the gcd were 1, then

A(x)g(x) + B(x)f(/3 + A-y — Ax) = 1

for some A,B E F(/3 + Ay)[x]. By the above bullets, evaluating this at x = -y wouldgive 0 1. Hence

h(x) = gcd(g(x),f(/3 + A-y — Ax)) E + A-y)[x]

has degree at least 1. If the degree were > 1, then h(x) Ig(x) implies that for some2 < j <m, would be a root of h(x) (do you see how this uses the separability ofg?). But since h(x) 1fC13 + A'y — Ax), must also be a root of f(/3 + A'y — Ax), i.e.,f(/3 + A'y — = 0. Since the roots off are /3 = . . . , 13t, this implies

/3 + Ay — = for some 1 <i < £,

which contradicts (5.17). Hence h has degree 1, and then h = x — -y follows, since'y is a root. But we also know hE F(/3+A-y)[x], so that -y F(/9+A-y). Then/3 = (/3 + A-y) — -y E F(/3 + A-y) follows immediately, since A E F. This completesthe proof that F(/3,-y) = F(/3 + A-y).

Next let p E F[x] be the minimal polynomial of /3 + A'y over F. We need to showthat p is separable. For this purpose, consider

(5.18) s(x)

Note that /3+ A'y is a root of s, since /3 = Furthermore, since f F[x], A E F,and -y1, . . . are the roots of g E F [xI, one can easily show that s e F [x] using thetechniques used in the proofs of Theorem 3.2.4 and Proposition 5.2.1. We leave thedetails as Exercise 1. It follows that p divides s in F [xI. However, we also havef = (x /3').. . (x — which when combined with (5.18) gives the formula

(5.19)i=1 j=1

THEOREM OF THE PRIMITIVE ELEMENT 121

Then (5.16) implies that s has distinct roots. Hence p is also separable (it divides s),which proves that + A'y is separable over F. Letting ti = 1 and t2 = A, we see thatthe theorem is true for n = 2.

Now suppose that n > 2 and that L = F(ai, . . . , a,1), where each a, is separa-ble over F. By our inductive hypothesis, we can find t1,... ,tn_I E F such thatF(ai,... = F(ao), where = + is separable over F.Then

L = F(aj,... ,a,,) = F(aj,... = =

By the proof for n = 2, we have F(ao + A E F, wherea0 + is separable over F. If we set = A, then + = t1a1 + + t,,a,, isthe desired separable primitive element. This completes the proof when F is infinite.

The proof of the theorem is very different in the case when F is a finite field. Wewill give the argument in Exercise 2. •

Here are two situations when the hypotheses of Theorem 5.4.1 are satisfied.

Corollary 5.4.2 Let F C L be a finite extension.(a) 1fF C L is separable, then there is a E L such that L = F(a).(b) 1fF has characteristic 0, then there is a E L such that L = F(a). Furthermore,

then a can be chosen to be of theform

Proof: We know that L = F(ai,...,an) since F CL is finite. In part (a), eacha, is separable since F C L is separable, so we are done by Theorem 5.4.1. Forpart (b), let F have characteristic 0. Then F is infinite and each a1 is separable byProposition 5.3.7. Again, we are done by Theorem 5.4.1. m

Every field of characteristic 0 contains a copy of 7L. In Exercise 3 you will usethis to show that in the equation a = + in part (b) of Corollary 5.4.2,we can assume that t1, . . . 7Z. This observation is due to Galois.

In some simple cases, one can explicitly find primitive elements.

Example 5.4.3 Consider Q C In the notation of the proof of Theo-rem 5.4.1, we have flu = = forf =x2 —2, and 'y1 = =for g = x2 —3. Then any A 0 in Q satisfies (5.16). Thus is a primitiveelement of Q C for all A E Q\{0}.

Not all finite extensions have primitive elements. By Corollary 5.4.2, such anextension cannot have characteristic 0. Here is an example in characteristic p.

Example 5.4.4 Let k be a field of characteristic p and let t, u be variables. Considerthe extension field

(5.20) F = k(t,u) CL,


where L is the splitting field of (xP — t) — u) E F[x]. Thus there are a, /3 E L witha1' = land /3P = u. By Exercise 4, we have L = F(a,/3) and [L:F1 =

Let us show that (5.20) has no primitive element. Given -y E L, we can useL=F(a,/3) =F[a,$] to write

-y = F,

where the sum is finite. Lemma 5.3.10 implies that

yP = =

and then = t and /31' = u give

yP = E F.

Hence -y is a root of xP — E Fix], so that [F(y) : F] p. Since [L: F] = p2. wehave L F(-y) for all -y E L. Thus F C L has no primitive element.

In Exercise 4 you will show that the extension (5.20) is purely inseparable.

Mathematical Notes

Theorem 5.4.1 leads to the following question about primitive elements.

• Existence of Primitive Elements. Corollary 5.4.2 tells us that all finite separableextensions have primitive elements. But this is not the full story, since the extensionF C L = F(a) discussed in Example 5.3.11 is not separable but has a primitiveelement. The following theorem of Steinitz characterizes all finite extensions thathave primitive elements.

Theorem 5.4.5 A finite extension F C L has a primitive element and only if thereare only finitely many intermediate fields F C K C L. •

A proof can be found in Section 4.14 of [Jacobson, Vol. I]. As an example of thisresult, consider the extension F C L from (5.20). Since this has no primitive element,there must be infinitely many fields K such that F C K C L. You will construct aninfinite collection of such fields in Exercise 5.

Historical Notes

We will see in Chapter 12 that Lagrange and Galois knew special cases of theTheorem of the Primitive Element.

REFERENCES 123


Exercise 1. Use the hints given in the text to prove that (5.18) has coefficients in F.

Exercise 2. Let F be a finite field, and let F C L be a finite extension. We claim that there isa E L such that L = F(a) and a is separable over F.(a) Show that L is a finite field.(b) The set L* = L \ {0} is a finite group under multiplication and hence is cyclic by Propo-

sition A.5.3. Let a E L be a generator. Prove that L = F(a).(c) Showthata'isarootofxm—l E F[x]forall0< i<m—l,andconclude

that

(d) Use part (c) to show that a is separable over F.

Exercise 3. In the equation a = tiai + in part (b) of Corollary 5.4.2, show that wecan assume that t,,. . . , E Z.

Exercise 4. In the extension F C L of Example 5.4.4, we have F = k(t, u), where k hascharacteristic p and L is the splitting field of (x" t) (x" u) E F [x]. We also have a, /3 E L

satisfying a" = t, /3" = u. Prove the following properties of F C L:(a) L=F(a,fi) and [L:F]=p2.(b) all yEL\F.(c) F C L is purely inseparable.

Exercise 5. Let F C L = F(a, /3) be as in Exercise 4, and consider the intermediate fieldsF C F(a + A/3) C L as A varies over all elements of F. Suppose that A p are two elementsofF such that F(a+A/3) = F(a+p13).(a) Show that a,/3 E F(a+A/3).(b) Conclude that F(a + Afi) = F(a, /3), and explain why this contradicts Example 5.4.4.It follows that the fields F(a + A/3), A E F, are all distinct. Since F is infinite, we see thatthere are infinitely many fields between F and L.

Exercise 6. Explain why the proof of Theorem 5.4.1 implies that F (/3 + Ay) = F (/3, when'y is separable over F, /3 is algebraic over F, and A satisfies (5.17).

Exercise 7. Let F C L = F(a,,... , a finite extension, and suppose that at,. . . , a,,_iare separable over F. Prove that F C L has a primitive element.

Exercise 8. Use Exercise 7 to find an explicit primitive element for F = k(t, u) C L, where khas characteristic 3 and L is the splitting field of (x2 t)(x3 u). Note that this extension isnot separable, by Exercise 8 of Section 5.3.

REFERENCES


2. M. Kreuzer and L. Robbiano, Computational Commutative Algebra 1, Springer, NewYork, Berlin, Heidelberg, 2000.

3. L. Weisner, Introduction to the Theory of Equations, Macmillan, New York, 1938.

CHAPTER 6

THE GALOIS GROUP

In this chapter we will define the Galois group of a finite extension F C L. We willthen study the Galois group of the splitting field of a separable polynomial and givesome examples of Galois groups.

6.1 DEFINITION OF THE GALOIS GROUP

If L is a field, then an automorphism of L is a field isomorphism a : L —+ L. We nowdefine one of the central objects in Galois theory.

Definition 6.1.1 Let F C L be a finite extension. Then Gal(L/F) is the set

{ a : L —+ L a is an automorphism, a(a) = afor all a E F}.

In other words, Gal(L/F) consists of all automorphisms of L that are the identityon F. The basic structure of Gal(L/F) is as follows.

Proposition 6.1.2 Gal(L/F) is a gmup under composition.

Proof: First suppose that a, 7- E Gal(L/F). Then ai- is the composition a whichis an automorphism because a, r are. Also, if a E F, then a o r(a) = a(r(a)) =


126 THE GALOIS GROUP

o-(a) = a, since a, i- are the identity on F. Hence composition gives an operation on

Gal(L/F), which is associative by standard properties of composition.The identity map 1L : L —+ L is an isomorphism that is the identity on F, so that

ii. E Gal(L/F). One easily checks that a o = o a = a for all a E Gal(L/F).Thus 1L is the identity element of Gal(L/F).

Finally, any a E Gal(L/F) is an automorphism, which means that its inverse

: L —÷ L is also an automorphism. Also, if a E F, then a = a(a), which implies

a'(a) = a1 (a(a)) = a. This shows that E Gal(L/F) and completes the proofthat Gal(L/F) is a group under composition. .

Because of this proposition, we call Gal(L/F) the Galois group ofF C L. In orderto compute Galois groups, we need to know how elements of Gal(L/F) behave. Webegin with the following simple observation.

Lemma 6.1.3 Let F C L be finite, andfix a E Gal(L/F). Given h E Fbi, . . . andthen

In particular; F[xl and /3 E L, then

a(h(/3)) =h(a(/3)).

Proof: This follows immediately because a preserves addition and multiplication

and is the identity on the coefficients of h. .

This lemma has some nice consequences concerning the Galois group.

Proposition 6.1.4 Let F C L be a finite extension and let a E Gal(L/F). Then:(a) If h E F[x] is a nonconstant polynomial with E L as a root, then

a motofh lying inL.

(b) If L = F(a1,... , then a is uniquely determined by its values on . . . ,

Proof: By Lemma 6.1.3, h E F{x] and 0 = imply that

0 = a(O) = =

which shows that a(a) E L is also a root of h. Part (a) follows.Turning to part (b), note that L = F[ai, . . . since L = is a finite

extension of F. Hence any /3 L can be written

for some polynomial h E F{x,,. . . By Lemma 6.1.3,

a(/3)

It follows that a : L —* L is uniquely determined by . . . .This proposition leads to our first result on the structure of Gal(L/F).

DEFINITION OF THE GALOIS GROUP 127

Corollary 6.1.5 Let F C L be a finite extension. Then its Galois group Gal(L/F) isfinite.

Proof: Since F CL is finite, L = where each is algebraic overF. Now suppose that a e Gal(L/F). By part (b) of Proposition 6.1.4, a is uniquelydetermined by a(a i), , a (a,,). Furthermore, if p, E F [x] is the minimal polynomialof aj, then part (a) shows that there are at most deg(p,) possibilities for a(aj). Thefiniteness of Gal(L/F) follows immediately. •

In Exercise I you will show that in the situation of Corollary 6.1.5, one has

Let us now use Proposition 6.1.4 to compute some Galois groups. We begin byobserving that Galois groups are sometimes unexpectedly small.

Example 6.1.6 Consider the extension

studied in Example 5.2.2. The minimal polynomial of over Q is x3 —2, which hasroots where w = The last two are not real and hence can'tlie in L. Hence every a E Gal(L/Q) must satisfy a is uniquelydetermined by it must be the identity. Thus Gal(L/Q) = { Do you seehow this argument uses both parts of Proposition 6.1.4?

Example 6.1.7 Let F = k(t), where k is a field of characteristic p, and let F C L bethe splitting field off = — t F[x]. If a E L is a root off, then L = F(a) andf = by Example 5.3.11. Thus a is the only root off. Arguing as in theprevious example, we see that Gal(L/F) = {1L}.

Here are some examples where the Galois group is nontrivial.

Example 6.1.8 Let 'r : C —÷ C be complex conjugation, i.e., r(z) = for z E C.By (A.5), we know that r is a homomorphism of fields, and it is an automorphismbecause To T is the identity. Furthermore, we have T(a) = a for all a R, so thatT E Gal(C/R). Thus Gal(C/R) has at least two elements, since 1c E

However, we also know that C = IR(i). Since the roots of x2 + 1 are ±i, Propo-sition 6.1.4 implies that a Gal(C/R) is determined uniquely by a(i) = ±i. Hence

has at most two elements. Combining this with the previous paragraph,we conclude that

Gal(C/R) = {lc,'r}.It follows that Z/2Z. <N

Example 6.1.9 Next consider the extension Q c L = Arguing as in theprevious example shows that a e Gal(L/Q) is determined uniquely by =

Thus Gal(L/Q) <2. There are two ways to see that equality occurs:

By explicit computation, one can show that a(a + = a is an auto-morphism of L.


L = is the splitting field of x2 —2 over Q. Since x2 —2 is irreducible over

Q and E L, Proposition 5.1.8 implies that there is an automorphism of L thattakes to and is the identity on Q.

Our last example will appear often in this chapter.

Example 6.1.10 For the extension Q C L = Proposition 6.1.4 impliesthat a Gal(L/Q) is determined uniquely by

(6.1) = =

This gives the inequality IGal(L/Q) <4. The natural question is whether all possiblesign combinations in (6.1) actually occur, i.e., whether IGal(L/Q) I =4. In Exercise 2you will prove this using Proposition 5.1.8 as in the previous example. We will learna much quicker method in Section 6.2.

Finally, we study what happens when we go to an isomorphic field.

Proposition 6.1.11 Suppose that F C Li and F C L2 are finite extensions, and let'p L1 —* be an isomorphism that is the identity on F. Then the map sending a topo a o defines a group isomorphism

Gal(Li/F) Gal(L2/F).

Proof: You will prove this in Exercise 3. •

Proposition 6.1.11 shows that isomorphic fields give isomorphic Galois groups.We use this as follows.

Definition 6.1.12 Letf E F[x]. The Galois group off over F is Gal(L/F), where Lis a splitting field off over F.

To check that Definition 6.1.12 makes sense, suppose that Li and L2 are splittingfields off E F[x}. Corollary 5.1.7 implies L1 via an isomorphism that is theidentity on F, and hence Gal(Li/F) by Proposition 6.1.11. Thus theGalois group off over F is well defined up to isomorphism.

Using this terminology, Example 6.1.8 tells us that the Galois group of x2 + 1 overR is Z/2Z.

Historical Notes

The definition of Galois group given here is very different from the one given byGalois. He only dealt with splitting fields, and for him, the Galois group consistedof certain permutations of the roots. We will give Galois's definition and explore itsrelation to Definition 6.1.1 in Chapter 12.

Isomorphisms of fields were first defined by Richard Dedekind in 1877 under thename "permutations." Here is his definition from [1, pp. 108—109]:

DEFINITION OF THE GALOIS GROUP 129

Now let Il be any field. By a permutation of we mean a substitution whichchanges each number

a, fi, a+fl, a—fl, afi, a/fl

of into a corresponding number

a', fi', (a+fl)', (a—fl)', (afi)', (a/fl)'

in such a way that

(a+fl)' =a'+fl'(a/3)' = a'fl'

are satisfied and the substitute numbers a', fl',.. are not all zero. We shall seethat the set of the latter numbers forms a new field,

In Exercise 4 you will show that this implies that the map —* given by a '-4 a'is an isomorphism of fields.

By 1894 Dedekind was also aware of the relevance of automorphisms to Galoistheory. Dedekind's influence can be seen in the work of Heinrich Weber, who gave acareful account of group theory and Galois theory in the first volume of his Lehrbuchder Algebra, which appeared in 1894. In this book Weber begins with Galois'sdefinition of the Galois group and shows how this leads to automorphisms of thesplitting field.

The final step in the evolution of the Galois group is due to Emil Artin, whoduring the 1920s made Definition 6.1.1 the starting point of Galois theory. Thefirst exposition of this approach appeared in the 1930 edition of [van der Waerden].Artin published his own account of Galois theory in 1938 and 1942. The latter wasenormously influential and is still in print as [Artin]. See [2] for more details.


Exercise 1. Let L = F(ai,. . . , ar), and let p1 E F[x] be a nonzero polynomial vanishing atExplain why the proof of Corollary 6.1.5 implies that Gal(L/F) I � deg(pi) . deg(p,,).

Exercise 2. Consider the extension Q C L = Vs). In Exercise 13 of Section 5.1, youused Proposition 5.1.8 to construct an automorphism of L that takes to — and is theidentity on By interchanging the roles of 2 and 3 in this construction, explain why allpossible signs in (6.1) can occur. This shows that IGal(L/Q)I = 4.

Exercise 3. This exercise will prove a generalized form of Proposition 6.1.11.(a) Let : L, L2 be an isomorphism of fields. Given a subfield F, C Li, set F2 = ço(Fi),

which is a subfield of L2. Prove that the map sending a E Gal(Li/Fi) toinduces an isomorphism Gal(Li/Fi) Gal(L2/F2).

(b) Explain why Proposition 6.1.11 follows from part (a).

Exercise 4. In the Historical Notes, we saw that Dedekind defined a "permutation" a '—+ a' tobe a map —+ 11' satisfying (a + fi)' = a' + /3' and (afl)' = a'fl' for all a, /3 E Dedekindalso assumes that = {a' a E and that the a' are not all zero.


(a) Show that 1 E maps to 1 E Il'. Once this is proved, it follows that a a' is a ringhomomorphism. (Recall that sending ito 1 is part of the definition of ring homomorphismgiven in Appendix A.)

(b) Show that the map a '—÷ a' is one-to-one.This shows that Dedekind's definition of field isomorphism is equivalent to ours.

Exercise 5. Prove the following inequalities:(a) <8.(b) . where p1,... ,pn are the first n primes.

In each case, one can show that these are actually equalities.

Exercise 6. If we apply Exercise I to the extension Q C L = V15), we get theinequality IGal(L/Q)l 8. Show that � 4.Exercise 7. Let F C L be a finite extension, and let a- : L —* L be a ring homomorphism thatis the identity on F. This exercise will show that a- is an automorphism.(a) Show that a is one-to-one.(b) Show that a- is onto.

6.2 GALOIS GROUPS OF SPLITTING FIELDS

In this section we will study the Galois group of the splitting field of a separablepolynomial. Recall from Section 5.3 that f is separable if it has distinct rootsin a splitting field. This is the situation considered by Galois.

We now prove the first main theorem of Galois theory.

Theorem 6.2.1 If L is the splitting field of a separable polynomial in F [x], then theGalois group ofF C L has order Gal(L/F)I = [L : F].

Proof: Our hypothesis implies that L = F(ai,.. where a1,...a separable polynomial f e F [x]. Then each a1 is separable over F (be sure

you can explain why). By the Theorem of the Primitive Element (Theorem 5.4.1),we can find fi L separable over F such that L = Let h E F[x] be the minimalpolynomial of Note that h is separable, since is.

Since L = F(i3), Proposition 4.3.4 implies that [L: Fl = m, where m deg(h). Toprove the theorem, we need to show that Gal(L/F) has m elements. We will use thefollowing ideas from Chapter 5:

• Normality (Section 5.2): If an irreducible polynomial has one root in a splittingfield, then all of its roots lie in the splitting field.

• Separability (Section 5.3): Separability means that a polynomial has distinct rootsin its splitting field.

• Isomorphisms (Proposition 5.1.8): If two elements in a splitting field L are rootsof the same irreducible polynomial over F, then there is an automorphism of Lthat is the identity on F and takes one root to the other.

As we will now explain, the theorem follows easily from these ideas.The above polynomial h E F [xl is separable and has a root e L. Since L is

a splitting field over F, the bullets for normality and separability imply that h has

GALOIS GROUPS OF SPLITTING FIELDS 131

distinct roots = /31,132,... m = deg(h), all of which lie in L. Now fix one ofthe roots, say /3g. Then /3 and are roots of the irreducible polynomial h. SinceL is a splitting field over F, the bullet for isomorphisms implies that there is anautomorphism of L such that = and cr1 is the identity on F.

It follows that . . . E Gal(L/F). Note that for i j, since oi(/3) == Thus Gal(L/F) has at least m distinct elements. But given any

0- E Gal(L/F), Proposition 6.1.4 and L = F(/3) imply that a is uniquely determinedby cr(13) E , 13m}. It follows that ci = cr, for some i. This completes the proofof the theorem. •

The following example illustrates the power of the theorem just proved.

Example 6.2.2 Consider Q C L = In Example 6.1.10, we saw that0- E Gal(L/Q) is uniquely determined by

which implies that Gal(L/Q) 4. We also asked whether equality occurs.This is now easy to decide, for [L : Q] = 4 by Example 4.3.9 and L is the splitting

field of the separable polynomial (x2 — 2)(x2 — 3). Hence all of the above signcombinations must occur. In particular, we can find ci, T E Gal(L/Q) such that

62(.)

In Exercise 1 you will show that Gal(L/Q) = {1L,0,T,ar} Z/2Z x Z/2Z (this isusually called the Klein four-group).

It is important to understand why the hypotheses splitting field and separable arenecessary in the proof of Theorem 6.2.1. We can see this in the first two examplesconsidered in Section 6.1:

• Consider Q c The Galois group is trivial by Example 6.1.6. Thisextension is not a splitting field, by Example 5.2.2.

• Consider F = k(t) C L, where k has characteristic p and L is the splitting field off = — t. The Galois group is trivial by Example 6.1.7. This polynomial is notseparable by Example 5.3.11.

In both of these examples, note that IGal(L/F)I < [L: F]. This is no accident, for inSection 7.1 we will prove that Gal (L/F) I < [L: Fl, with equality if and only if L isthe splitting field of a separable polynomial in F [x]. Such extensions will be calledGalois extensions in Chapter 7.


Exercise 1. Complete Example 6.2.2 by showing that Gal(L/Q) = { lL,0-, r, ar} and thatGal(L/Q) Z/27Z >< Z/2Z.


Exercise 2. Consider Q C L = Q(w, where w =(a) Explain why a E Gal(L/Q) is uniquely determined by a(w) E {w,w2} and E

(b) Explain why all possible combinations for a(w) and actually occur.In the next section we will show that Gal(L/Q) S3.

Exercise 3. Consider Q C L = where = By Proposition 4.2.5, theminimal polynomial of over Q is x4 +x3 +x2 +x+ 1.(a) Show that [L: = 20.

(b) Show that L is the splitting field of x5 —2 over Q, and conclude that Gal(L/Q) is a groupof order 20.

We will describe the structure of this Galois group in Section 6.4.

Exercise 4. Consider the nth root of unity (,, = We call Q C a cyclotomicextension of Q.(a) Show that Q C Q((,,) is a splitting field of a separable polynomial.(b) Given a E show that a(ç) = for some integer 1.(c) Show that the integer i in part (b) is relatively prime to n.(d) The set of congruence classes modulo n relatively prime to n form a group under mul-

tiplication, denoted (Z/nZ)*. Show that the map a i-+ [i], where aone-to-one group homomorphism

(e) The order of (Z/n7L)* is I(Z/nZYI = where /(n) is the Euler fromnumber theory. Prove that the homomorphism of part (d) is an isomorphism if and onlyif : Q] = (I)(n).

(f) Letp be prime. Use part (e) and Proposition 4.2.5 to show that Gal (Z/pZ)In Chapter 9 we will prove that : Q] = By part (e), this will imply that there is anisomorphism for all n.

Exercise 5. Let F have characteristic p, and assume that f = x" — x + a E F [x] is irreducibleover F. Then let L = F(a), where a is a root off in some splitting field. In Exercise 15 ofSection 5.3, you showed that F C L is a normal separable extension.(a) Show that IGal(L/F) = p. and use this to prove that Gal(L/F) 7L/pZ.(b) Exercise 15 of Section 5.3 showed that a + 1 is a root off. For i = 0,... ,p — 1, show

that there is a unique element of Gal(L/F) that takes a to a + i.(c) Use part (b) to describe an explicit isomorphism Gal(L/F) Z/pZ.

Exercise 6. Let f E F[x] be irreducible and separable of degree n, and let F C L be a splittingfield of f. Prove that n divides IGal(L/F)I.

6.3 PERMUTATIONS OF THE ROOTS

In Chapter 1 we saw that permutations of the roots of a cubic arise naturally fromCardan's formulas. In fact, the title of Section 1.2 was "Permutations of the Roots."We now explain more generally how Galois groups relate to permutations. As in theprevious section, we assume that L is the splitting field of a separable polynomialf E F[x]. Our goal is to interpret Gal(L/F) in terms of permutations of roots off.

Let n = deg(f). Then in L[xI we can writef as the product


where a0 0 and a i,... , E L are distinct. In this situation we get a map

(6.3) Gal(L/F) —*

as follows. Given a Gal(L/F), Proposition 6.1.4 implies that a(aj) is a root of f(since a, is), so that a(aj) = for some T(i) E {l,. . . ,n}. Note that T(i) isuniquely determined, since a1, . . .

a is (be sure you see why). It follows that r is a permutation, i.e.,r E S,,. This defines the map (6.3).

Proposition 6.3.1 The map Gal(L/F) —÷ described in (6.3) is a one-to-one gmuphomomorphism.

Proof: Suppose that a1, a2 E Gal(L/F) correspond to r1, E S,, via (6.3). Thismeans that = and similarly for a2 and T2. Then

a1 o = al (a2(aI)) = a1 = =

This shows that aI o a2 corresponds to T1T2, so that (6.3) is a group homomorphism.It remains to show that (6.3) is one-to-one. This follows immediately from

Proposition 6.1.4, since L = F(ai,... The proof is now complete.

Proposition 6.3.1 shows that for the splitting field of a separable polynomial ofdegree n, we can regard the Galois group as a subgroup of By Lagrange's Theorem,it follows that Gal(L/F) divides n!. Combining this with [L: F] = IGal(L/F) I fromTheorem 6.2.1, we get the following corollary.

Corollary 6.3.2 If L is the splitting field of a separable polynomial f E F [x], then[L: F] divides n!, where n = deg(f). U

Theorem 5.1.5 states that [L F] n! when L is the splitting field off E F[x] ofdegree n. Do you see how Corollary 6.3.2 refines this result when f is separable?

Here are some examples of Proposition 6.3.1.

Example 6.3.3 We know that the splitting field of f = (x2 —2) (x2 —3) over Q isL = Example 6.2.2 shows that Gal(L/Q) = { a, T, aT}, where a and'r satisfy

and

Let a! = = = and a4 = Then Gal(L/Q) is isomorphicto a subgroup of S4 by Proposition 6.3.1. The automorphism a fixes a!, a2 andinterchanges a3,a4. Hence a i-+ (34) e S4. One similarly shows that T '-+ (12), 50ar (34)(12) = (12)(34). ThusGal(L/Q) {e,(12),(34),(12)(34)} CS4.


Example 6.3.4 Consider the extension Q C L = Q(w, w = Since L isthe splitting field of x3 — 2 over Q (Exercise 1 of Section 5.1), we get a one-to-onegroup homomorphism Gal(L/Q) —* S3. However, we learned in Example 4.3.10 that[L:Q] = 6. Since [L:Q] = it follows that Gal(L/Q) S3. You willwork out the details of this isomorphism in Exercise 1.

When one thinks of Galois groups in terms of permutations, it makes sense toask how properties of the permutations relate to properties of the corresponding fieldextension. One nice example of this involves the following subgroups of

Definition 6.3.5 A subgroup H C is transitive every pair of elements i, j E

{l,...,n},thereisi-EHsuchthatr(i)=j.

For example, Sn is a transitive subgroup of itself, since the transposition (ii) takesito j. But not all subgroups of 5n are transitive.

Example 6.3.6 The subgroup {e, (12), (34), (12)(34)} C from Example 6.3.3 isnot transitive, since no element of the subgroup takes 1 to 3.

It is natural to ask if the subgroup of corresponding to Gal(L/F) is transitive.This question was answered by Camille Jordan in 1870 as follows.

Proposition 6.3.7 Let L be the splitting field of a separable polynomial f E F[x] ofdegree n. Then the subgmup of Sn corresponding to Gal(L/F) is transitive if andonly if f is irreducible over F.

Pmof: First suppose that f is irreducible with distinct roots cr1,. . . , E L. As in theproof of Theorem 6.2.1, we can use Proposition 5.1.8 to construct an automorphism

L L that takes to and is the identity on F. Then a E Gal(L/F), and thecorresponding permutation in 5n clearly takes ito j. Thus Gal(L/F) gives a transitivesubgroup of

Conversely, suppose that Gal(L/F) corresponds to a transitive subgroup of 5n.and let h be an irreducible factor off. We will show that deg(h) � n, which easilyimplies that f is irreducible (do you see why?).

For this purpose, let the roots of f be cvi,... ,a,, L. Since h is a nonconstantfactor of f, we can find i such that = 0. Now pick any j E {1,. . . ,n}. By ourtransitivity assumption, there is a E Gal(L/F) such that a(a1) = aj. Since h hascoefficients in F, part (a) of Proposition 6.1.4 implies that a rootof h. Since j was arbitrary and cu,... , a,, are distinct, it follows that h has at least nroots, which implies that deg(h) � n.

Mathematical Notes

Here are two topics for further discussion.

• The Galois Group of a Polynomial. In Section 6.1, the Galois group off E F [x]was defined to be Gal(L/F), where F C L is a splitting field of f over F. But when f


is separable of degree n, the Galois group off has extra structure given by its actionon the roots of f. Hence one can argue that the correct definition of "the Galois groupoff" is the homomorphism Gal(L/F) —÷ studied in this section.

• Transitive Group Actions. Definition 6.3.5 defines a transitive subgroup of Sn.This can be generalized to any group action: if a group G acts on a set X (as definedin Section A.4), then the action is transitive if for all x, y E X, there is g E G such thatg x = y. For example, if L is the splitting field of a separable polynomial f E F [x],

then Gal(L/F) acts on the roots of f. Hence Proposition 6.3.7 can be restated assaying that f is irreducible if and only if Gal (L/F) acts transitively on the roots off.

Historical Notes

Proposition 6.3.1 shows that for the splitting field of a separable polynomial ofdegree n, the Galois group Gal(L/F) is isomorphic to a subgroup of the symmetricgroup 5n• The permutations in this subgroup correspond to those permutations thatrespect the algebraic structure of the roots, i.e., those that come from automorphismsof the splitting field.

Galois defined his "group" to consist of certain arrangements of the roots of thegiven polynomial. In Chapter 12 we will show that this set of permutations in Sn is theimage of Gal(L/F) 5,,. Hence his group agrees with ours up to isomorphism. Whatis interesting is that Galois had no notion of automorphism, although automorphismsare implicit in his development of the theory.

In [Galois, p. 79j Galois defines transitive subgroups Of Sn and gives an example ofa nontransitive subgroup that in modern terms is written ((12), (345)) c S5. However,his terminology is different: He writes "irreducible" instead of "transitive." Thisshows that he also knew Proposition 6.3.7, though Jordan was the first to state theresult explicitly.


Exercise 1. Consider Gal(L/Q), where L = w = By Exercise 2 of Sec-tion 6.2, there are 0,T E Gal(L/Q) such that

31 31 31 2o(v2) =wv2, a(w) =w and r(v2) = v2, 'r(w) =w.

Find the permutations in S3 corresponding to and T.

Exercise 2. For each of the following Galois groups, find an explicit subgroup of S4 that isisomorphic to the group. Also, the Galois group is isomorphic to which known group? (By"known groups," we mean cyclic groups, dihedral groups, the quaternion group, symmetricgroups, alternating groups, products of these groups, etc. You may need to look up some ofthese in your abstract algebra text.)(a) Gal(Q(i,(b) Gal(Q(i,

Exercise 3. In the terminology of Exercise 2, Gal(Q(i, isomorphic to whichknown group? Explain your reasoning in detail.


Exercise 4. Consider the extension Q c L = Q(a), where a = In Exercise 6 ofSection 5.1, you showed that f = x4 4x2 + 2 is the minimal polynomial of a over Q and thatL is the splitting field off over Q. Show that Gal(L/Q) Z/47Z.

Exercise 5. Let f E F[x] be separable, where f = gi for g, E F[xJ of degree d > 0, andlet L be the splitting field off over F. Show that Gal(L/F) is isomorphic to a subgroup of theproduct group Sd1x X Sag.

Exercise 6. Let H be a transitive subgroup of Prove that HI is a multiple of n.

Exercise 7. Let f E F[x] be irreducible and separable of degree n and let F C L be a splittingfield of f. Use Exercise 6 and Proposition 6.3.7 to prove that n divides IGal(L/F)I. This givesan alternate proof of Exercise 6 of Section 6.2.

6.4 EXAMPLES OF GALOIS GROUPS

In this section we will give some interesting examples of Galois groups.

A. The pth Roots of 2. Let = be a pth root of unity, where p is prime.

By Section A.2 the roots —2 are for 0 j p — 1, so that

is the splitting field of xP — 2 over Q. Our goal is to describe Gal(L/Q).The minimal polynomial of over Q is x"1 + ... + 1 by Proposition 4.2.5 and

the roots of this polynomial are for 1 i p — 1 by Section A.2. Furthermore, the

minimal polynomial of over Q is —2 by the Schönemann—Eisenstein criterion,and its roots are listed above. Since p and p — 1 are relatively prime, the method usedin Exercise 5 of Section 4.3 implies that [L Q] = p(p — 1). (See also Exercise 8 ofSection 5.1.)

It follows from Theorem 6.2.1 that Gal(L/Q) is a group of order p(p 1). To

see what group this is, let a Gal(L/Q). Then Proposition 6.1.4 implies that a isuniquely determined by

In other words, there are integers 1 i p — 1 and 0 j p — I such that

(6.4) a(ç) =

a by The number of possible pairs (i,j) is (p — 1) . p =p(p 1). Since this is also the order of Gal(L/Q), it follows that all possible pairs

(6.5) (i,j)E{l,...,p—l}x{0,...,p—1}

must occur in (6.4).

EXAMPLES OF GALOIS GROUPS 137

To determine the group structure, we need to compute the composition of and

0r,s• This is done as follows:

ai,jo = = (ai,j(ç))T =iT

ai,j o = = =

=

This computation suggests that

ar,, = 0ir,is+j.

Unfortunately, the pair (ir, is + j) need not lie in (6.5). We can resolve this difficultyby realizing that for i E Z, depends only on the congruence class of i modulo p.In other words, for a = {iJ F,, = Z/pZ, the number = is well defined.

If we set F; = then for

(a,b) E

we can define 0a,b to be the element of Gal(L/Q) such that

= =

Then the above computation shows that 0'c,d aac,ad+b.

This composition formula leads to a geometric description of the Galois groupGal(L/Q). Given a,b E F,,, the function 'yab: —÷ F,, defined by "la,b(U) = au + bis an affine linear transformation. By Exercise 1, is one-to-one and onto if andonly if a 0, and all such form a group of order p(p — 1) under composition.This group is called AGL( 1, F,,), the one-dimensional affine linear group modulo p.To understand its structure, we take u E IF,, and compute

'Ya,b°'fc,d(t') = "Ya,b('Yc,d@')) =

= a(cu+d) +b = acu + (ad+b) =

Thus "lab° "lc,d = "Yac,ad+b' so that the map oa,b "tab gives an isomorphism

Gal(L/Q) AGL(l,IF,,).

Another way to understand the structure of AGL( 1, F,,) is via the subgroup

T = {"lIb I b

In Exercise 2 you will show that there is a group isomorphism T F,,. You will alsoprove that T is a normal subgroup of AGL( 1, IF,,) with quotient

(6.6) AGL(1,F,,)/T


As a group, F,, is cyclic of order p, and Proposition A.5.3 implies that 1F is cyclic oforder p — I. In the Mathematical and Historical Notes we will say more about howAGL( 1, is built from these cyclic groups.

B. The Universal Extension. In Chapter 2 we studied the elementary symmetricpolynomials di, . . . , a,, in variables Xi,... , Recall from Proposition 2.1.4 that

(X_Xi)...(XXn)=Xn_giXn_l++(_1)TUrXn_r++(_l)nUn.

This is the universal polynomial of degree n introduced in Section 2.2 and denotedby f. Note that f is a polynomial in x with coefficients in the field

K=F(o-i,...,o-,,).

Since the roots off are Xi,... ,x,,, it follows easily that

L=F(xi,...,x,,)

is the splitting field off over K. We call K C L the universal extension in degree n.Since f has distinct roots, Section 6.3 gives a one-to-one group homomorphism

Gal(L/K) —÷ We now prove that this map is an isomorphism.

Theorem 6.4.1 The universal extension K = F(cr1,... CL = F(xi,... indegree n is the splitting field of a separable polynomial. The action of the Galoisgroup on the roots of the universal polynomial of degree n gives an isomorphism

Gal(L/K)

Proof: We showed above that L is the splitting field of the universal polynomial f.Notice also that f is separable, since its roots x1, . . . are distinct.

To prove the final assertion of the theorem, we will use the action of Sn OflF[xi, . . . ,x,,] discussed in Section 2.4. Recall that for f F[xi,. . . and r e 5,,,

• f is the polynomial obtained by permuting the variables according to r. Thisaction has the properties

r(f+g) = r.f+r.g,(6.7) T.(fg) = (r.f)(T.g),

r• (yf) = (T'y) .f,

where 'i-,-y 5,, and f,g E F[xi,... You will prove this in Exercise 3.In Exercises 4 and 5 you will also show that f T f is a ring isomorphism from

F[xi, . . . to itself and hence extends to an isomorphism of its field of fractions.It follows that permuting the variables according to r gives an automorphism ofL = F(xi, . . . Since the elementary symmetric polynomials are fixed by theaction of r E this automorphism is the identity on F(ai, . . . ,ci,,).

We have thus shown that f i—÷ i- . f is an element of Gal(L/K). Under the mapGal(L/K) —* Sn of Proposition 6.3.1, this automorphism obviously maps to r. Since


r was an arbitrary element of we see that Gal(L/K) —* S,, is onto, which completesthe proof of the theorem. •

Chapter 7 will describe the Galois theory of the universal extension.

C. A Polynomial of Degree 5. Consider the polynomial f = x5 — 6x + 3, and let Lbe the splitting field off over Q. The Schönemann—Eisenstein criterion implies thatf is irreducible and hence separable. Thus Gal(L/Q) is isomorphic to a subgroupH C S5. We will show that H = S5, so that

(6.8) Gal(L/Q) S5.

We will sketch the proof and leave the details for Exercise 6.By Exercise 6 of Section 6.2, Gal(L/Q)l = HI is divisible by 5, since f is

irreducible. By Cauchy's Theorem (Theorem A. 1.5) from group theory, H must havean element g of order 5. Recall that g is a product of disjoint cycles whose order isthe least common multiple of the lengths of the cycles. Since g is in S5 and has order5, one easily sees that g is in fact a 5-cycle. Thus H contains a 5-cycle.

By the Fundamental Theorem of Algebra, we can assume that L C C, so that theroots of f can be regarded as complex numbers. Furthermore, using curve graphingtechniques from calculus, one also sees that f has exactly three real roots. It followsthat complex conjugation gives an element r E Gal(L/Q) that interchanges two ofthe roots and fixes the other three. Since i- maps to a transposition in S5, we concludethat H contains a transposition.

Hence, relabeling the roots appropriately, we may assume that H contains (12345)and (li) for some i E {2,3,4,5}. Since (12345)i_1 is a 5-cycle beginning (ii...),we can relabel the roots again so that H contains (12345) and (12). It is a classicresult in group theory that these two permutations generate S5. You probably studiedthis in your abstract algebra course (if not, you should do Exercise 7). This showsthat H = S5 and completes the proof of (6.8).

This example is taken from [Stewart, Chapter 14]. In Chapter 8 we will see thatf = — 6x + 3 is not solvable by radicals, since Gal(L/Q) Different proofsof (6.8) will be given in Examples 13.2.8 and 13.4.7 of Chapter 13.

Mathematical Notes

This section has several topics of interest to discuss.

• Specialization of Galois Groups. For the fifth roots of 2, we have

the splitting field of x5 —2 over Q has Galois group AGL( 1, F5),

while for the universal extension in degree 5 over F = Q, we have

the splitting field of x5 — cr1x4 + — J3X2 + —

overQ(ai,a2,u3,a4,as) has Galois group


Since the second polynomial is the universal polynomial of degree 5, the first can beregarded as the specialization of the second obtained by the mapping

oi —*0, U3 F-*O, a41-+O, a5 F—* 2.

However, the Galois groups are not the same, which implies that specialization doesnot always preserve the Galois gmup. This is part of what makes Galois theory sohard—polynomials of the same degree may have different Galois groups.

On the other hand, most specializations of the universal polynomial of degree nover Q have S. as their Galois group. This follows from the Hubert IrreducibilityTheorem, which is discussed in [Hadlock]. For example, one can prove that theGalois group off — x — 1 e Q[xI is Sn for all n � 2 (see [4, p. 42]).

• Semidirect Products. In the text we noted that the one-dimensional affine groupAGL(1,F,,) has a normal subgroup T F,, such that F,'. We willnow explain how AGL( 1, F,,) is the semidirect product of F,, and via the action of

In general, let G and H be groups, and assume that G acts on H (as defined inSection A.4) in the following special way: for any g E G, the map h i—* g . h givenby the action on g on H is a group homomorphism from H to itself. Then define abinary operation on the set H x G by

(6.9) (h,g)(h',g')=(h(g.h'),gg')

where h (g . h') is the product of h, g h' E H. The intuition behind this formula is thatwhen we multiply (h,g) and (h',g'), we "twist" by the action of g, since g is betweenh and h'. In Exercise 8 you will show that this defines a group, called the semidirectproduct H G.

For example, the action of on IF,, gives the semidirect product IF,'. In thisgroup, the product is given by

(b,a).(d,c) = (b+a.d,ac) = (ad+b,ac),

since the group operation is addition in F,, and multiplication in It follows that(b,a) gives an isomorphism

(6.10)

For any semidirect product H x G, the map (h, g) g is a group homomorphismthat is clearly onto. In Exercise 8 you will check that the kernel of this map is

Then the Fundamental Theorem of Group Homomorphisms implies that

In Exercise 9 you will explore how this relates to (6.6) and (6.10).


• The Extension Problem. Given groups G and H, a third group G1 is an extension ofH by G if G1 contains a normal subgroup H1 H such that G1 /H1 G. For example,when G acts on H by group homomorphisms as above, the semidirect product H G

is an extension of H by G.An important observation is that the same groups can have nonisomorphic exten-

sions. For example, Exercise 10 will show that the product IF,, x IF' and the semidirectproduct F,, are nonisomorphic extensions of IF,, by when p � 3.

The extension problem in group theory asks whether it is possible to classify allextensions of H by G. This is a difficult problem and is one of the reasons why groupsare hard to classify. The extension problem is also related to group cohomology.

Historical Notes

For the extension Q C L = Q (ç, one can describe Gal(L/Q) AGL( 1, F,,)in terms of permutations in Si,. We do this by replacing { 1, . . .

, p} with the congruenceclasses { [1], . . .

, [p] } = IF,,. Then the affine linear transformation 'Yal) defined in thetext becomes the element of 5,, represented by the permutation

(1 2 ... pa2+b ... ap+b

provided we think of the entries as congruence classes modulo p. This can beexpressed more succinctly in the form

(6.11)

We will see in Chapter 12 that the permutations (6.11) are implicit in the work ofLagrange.These permutations also appear explicitly in Galois's study of irreduciblepolynomials of prime degree that are solvable by radicals. We will have more to sayabout this in Chapter 14.

In the late nineteenth and early twentieth centuries the subgroup of S,, consistingof the permutations (6.11) was called the metacyclic group. These days the termmetacyclic is used more generally to mean any group G possessing a normal subgroupH such that both H and G/H are cyclic. In Exercise 11 you will show that the groupof permutations of the form (6.11) is metacyclic in this sense.

As for the Galois group of the universal extension computed in Theorem 6.4.1,Galois states this result as follows [Galois, p. 511:

In the case of algebraic equations, the group is nothing other than the collectionof 1.2.3. . . m possible permutations on the m letters...

Here, "algebraic equation" refers to the universal case where the "m letters" are theroots of the universal polynomial of degree m. However, we will see in Chapter 12that Galois's use of the word "permutation" is different from ours.



Exercise 1. Given a, b E IF,,, define F,, —* IF,, by 'Yab(U) = au + b.(a) Prove that Yab is one-to-one and onto if and only if a 0.(b) Suppose that a 0. Prove that the inverse function of is 'Ye-i —a

(c) Show thatAGL(l,F,,) = {'YabI (a,b) E F; x

is a group under composition.

Exercise 2. Consider the map AGL( I, F,,) —+ F; defined by 'Yab a.

(a) Show that this map is an onto group homomorphism with kernel T = {'ylb b E

Then use this to prove (6.6).(b) Show that

Exercise 3. This exercise is concerned with the proof of (6.7). Given r E S,,, observe thatf i—* -r f can be regarded as the evaluation map from F [Xl, . . . , to itself that evaluatesf(xi,. ..,Xn) at (Xr(l),...,Xr(n)).(a) Explain why Theorem 2.1.2 implies that f r f is a ring homomorphism. This proves

the first two bullets of (6.7).(b) Prove the third bullet of (6.7).

Exercise 4. Let -r E Prove that f r f is a ring isomorphism from F{xi,. . . ,x,] to itself.

Exercise 5. Let R be an integral domain, and let K be its field of fractions. Prove that everyring isomorphism : R —+ R extends uniquely to an automorphism : K —* K.

Exercise 6. As in the text, let f = — 6x+ 3.

(a) Use the hints given in the text to show that every element of S5 of order 5 is a 5-cycle.(b) Use curve graphing from calculus to show that f has exactly three real roots.

Exercise 7. Show that is generated by the transposition (12) and the n-cycle (12. . . n).

Exercise 8. Let G and H be groups where G acts on H by group homomorphisms. As in thetext, we let H G denote the set H x G with the binary operation given by (6.9).(a) Prove that H G is a group.(b) Prove that the map G —+ G defined by (h,g) g is an onto group homomorphism

with kernel H x {e}.(c) Prove that h (h,e) defines an isomorphism H H x {e} (where the group structure

on H x {e} comes from H x G).

Exercise 9. Explain how (6.6) and (6.10) relate to the last paragraph of the discussion ofsemidirect products in the Mathematical Notes.

Exercise 10. Let p � 3 be prime, and let F,, xi be the semidirect product described in theMathematical Notes.(a) Show that F,, is not Abelian.(b) Show that the product group IF,, x F; is Abelian.(c) Show that F,, x F; is an extension of F,, by

Since we already know that F,, is an extension of IF,, by we see that (a) and (b) givenonisomorphic extensions.

Exercise 11. The goal of this exercise is to show that the group G of permutations (6.11) ismetacyclic in the sense that G has a normal subgroup H such that H and G/H are cyclic. Showthat this follows from G AGL( I, F,,) together with (6.6) and Proposition A.5.3.

ABELIAN EQUATIONS (OPTIONAL) 143

Exercise 12. Let p be prime. Generalize part (a) of Exercise 6 by showing that every elementof S,, of order p is a p-cycle.

Exercise 13. Let L be the splitting field of 2x5 — lOx + 5 over Q. Prove that Gal(L/Q) Ss.

Exercise 14. let L = Q(ç, Prove that L = i.e., the splitting field of—2 over Q can be generated by two of its roots. Chapter 14 will show that this follows from

Galois's criterion for an irreducible polynomial of prime degree to be solvable by radicals.

Exercise 15. Let L = Q (ç, The description of Gal(L/Q) given in the text enables oneto construct some elements of Gal )). Use these automorphisms and Proposition 6.3.7to prove that —2 is irreducible over Q(ç).

6.5 ABELIAN EQUATIONS (OPTIONAL)

In this section we will discuss the following theorem of Abel:

If the roots of an equation of arbitrary degree are related among themselves insuch a way that all the roots can be expressed rationally by means of one ofthem, which we denote by x; if in addition whenever one denotes by Ox, OjX twoother arbitrary roots, one has

OOIX = OIOX,

then the equation to which they belong will always be solvable algebraically.

(See [Abel, p. 479].) Our goal is to interpret this theorem in terms of Galois theory.We begin by translating Abel's theorem into modern terminology. First observe

that Abel talks about an equation f = 0 rather than a polynomial f. This is typicalfor the early nineteenth century. We will assume that f is a nonconstant polynomialwhose coefficients lie in a field F. Since we prefer x to be a variable, we will replaceAbel's x with a. So a will be a root off in some extension field.

Now let a1 = a, a2,. . . , a splitting field L. Then, when Abelsays that the roots can be "expressed rationally" in terms of a, he means that thereare rational functions 61, with coefficients in F such that a, = 9,(a). In Exercise 1you will show that this is equivalent to

(6.12) L=F(a).

Here is an example of this from Chapter 4.

Example 6.5.1 If we let a = + v's, then (4.4) implies that f = x4 — lOx2 + 1factors as

(x—a)(x+a)(x— lOa+a3)(x+lOa—a3).

If we set Oi(x) =x, 02(x) = —x, 03(x) = lOx—x3, and 04(x) = —lOx+x3, then theroots off are O1(a), 02(a), 03(a), and 04(a).

The rational functions 0, in Abel's theorem give functions from L to L that usuallyfail to be automorphisms. For instance, the function 02(x) = —x in the above example


does not preserve multiplication, since 92(ab) = —ab differs from 92(a)02(b) =(—a)(—b) ab wheneverab 0.

In our notation, the displayed equation in the quote from Abel becomes

(6.13) = 1 <i,j <n.

In Exercise 2 you will show that the rational functions of Example 6.5.1 satisfy thiscondition. Following Kronecker and Jordan, we call f = 0 an Abelian equation if fis a nonconstant polynomial with a root satisfying (6.12) and (6.13).

The conclusion of Abel's theorem states that f is "solvable algebraically." Inmodem terms this means "solvable by radicals," which will be defined carefully inChapter 8. Thus Abel's theorem can be restated as follows.

Theorem 6.5.2 In characteristic 0, every Abelian equation is solvable by radicals.

The hypothesis about characteristic 0 is not in Abel's original statement but isneeded since the theory of solvability developed in Chapter 8 only applies to fieldsof characteristic 0. Abel always worked in characteristic 0.

Theorem 6.5.2 is a consequence of the following two theorems. Recall fromSection 6.1 that the Galois group off E F [x] is Gal(L/F), where L is a splitting fieldof f over L.

Theorem 6.5.3 The Galois group of an Abelian equation is an Abelian group.

Theorem 6.5.4 In characteristic 0, a polynomial with Abelian Galois group is solv-able by radicals.

We will prove Theorem 6.5.4 in Chapter 8. The proof will follow from Galois'scriterion for solvability by radicals together with the fact that every finite Abeliangroup is solvable. (All of these terms will be defined in Chapter 8.)

We now prove Theorem 6.5.3.

Proof of Theorem 6.5.3: If the Abelian equation is f = 0, where f E F [x], then fhas a root such that L = F (cr) is the splitting field of f. In particular, we haverational functions E F(x), 1 <i < n, such that the are the roots of f. Nowlet o, T E Gal(L/F). In Exercise 3 you will prove the following:

• a(a) = and = for some i and j.

• = ru in Gal(L/F) if and only if = r(a(a)) in L.

• = and =Since f = 0 is Abelian, the theorem follows easily from these bullets.

In the 1 880s, Weber applied the term "Abelian" to commutative groups becauseof this theorem.

ABELIAN EQUATIONS (OPTIONAL) 145

Historical Notes

The story of Abelian equations begins with Gauss, who showed in 1801 thatx" — 1 = 0 is solvable by radicals. We will study Gauss's work in Chapter 9, and inChapter 10 we will explore the surprising geometric consequences of his results. InExercise 4 you will show that f — 1 = 0 is an Abelian equation over Q.

In his 1829 paper Mémoire sur une classe particulière d 'equations résolublesalgébriquement [Abel, Vol. I, pp. 478—507], Abel states the theorem quoted at thebeginning of the section and goes on to say

After having explained this theory [the solvability of Abelian equations] ingeneral, I will apply it to circular and elliptic functions.

(See [Abel, Vol. I, p. 479].) In this passage, "circular functions" refer to the work ofGauss just mentioned, and "elliptic functions" refer to Abel's deep results on ellipticfunctions and complex multiplication. We will discuss a special case of this involvingthe lemniscate in Chapter 15. Abel died at age 26 before he could publish the fulldetails of his work.

Kronecker introduced the term "Abelian equation" in 1853 in the special casewhen the Galois group was cyclic. The general sense of the term, as defined here,is due to Jordan in 1870. Kronecker's interest in Abelian equations is related to hisamazing conjecture that the mots of an Abelian equation over Q can be expressedrationally in terms of a mot of unity. This was proved in 1886 by Weber and is nowcalled the Kronecker—Weber Theorem. The modem version of this theorem is statedas follows.

Theorem 6.5.5 Suppose that Q C L is a finite extension such that L C C. Then thefollowing conditions are equivalent:(a) Q C L is normal and Gal(L/Q) is Abelian.(b) There is a mot of unity = such that L C •

In the next chapter, you will prove (b) (a) in Exercise 14 of Section 7.3, anda proof of (a) (b) can be found in [3, pp. 125—129]. The proof of (a) (b) usesideas from algebraic number theory and is beyond the scope of this book.

The early history of group theory and Galois theory are closely related—after all,Galois was the person who introduced the term "group" into mathematics. So it isnot surprising that notions like Abelian equations from Galois theory influenced theterminology of group theory. We will see many more examples of this phenomenonin the next chapter.


Exercise 1. Assume that f E F[x] is nonconstant and has roots = a2,. . . , a,, in a splittingfield L. Prove that L = F (a) if and only if there are rational functions 0 E F (x) such thata, = 0,(a). Can we assume that the 0, are polynomials?

Exercise 2. Show that the equation x4 — lOx2 + 1 = 0 discussed in Example 6.5.1 is Abelian.

Exercise 3. Complete the proof of Theorem 6.5.3.


Exercise 4. Show that I = 0 is an Abelian equation over Q.

Exercise 5. Let f be the minimal polynomial of V'2 + over Q. Show that f = 0 is anAbelian equation.

Exercise 6. In this exercise, you will prove a partial converse to Theorem 6.5.3. Suppose thata finite extension F C L is normal and separable and has an Abelian Galois group.(a) Explain why F C L has a primitive element.(b) By part (a), we can find E L such that L = Let f be the minimal polynomial of

Prove that f = 0 is an Abelian equation over F.See Theorem 8.5.8 for the precise relation between Abelian equations and Abelian groups.

Exercise 7. Show that the implication (a) (b) of Theorem 6.5.5 is equivalent to Kronecker'sassertion that the roots of an Abelian equation over Q can be expressed rationally in terms ofa root of unity.

REFERENCES

1. R. Dedekind, Theory ofAlgebraic Integers, English translation by J. Stillwell, CambridgeU. P., Cambridge, 1996. (Translation of 1877 French edition.)

2. B. M. Kiernan, The development of Galois theory from Lagrange to Artin, Arch. Hist.Exact Sci. 8 (1971), 40—154.

3. D. A. Marcus, Number Fields, Springer, New York, Berlin, Heidelberg, 1977.

4. i-P. Serre, Topics in Galois Theory, Jones and Bartlett, Boston, 1992.

CHAPTER 7

THE GALOIS CORRESPONDENCE

This chapter will draw on the work we did in Chapters 4, 5, and 6 to state and provethe main theorems of Galois theory. We will also give some applications.

7.1 GALOIS EXTENSIONS

In Section 6.2 we learned that splitting fields of separable polynomials are especiallynice from the point of view of Galois theory. The main goal of this section is tocharacterize such extensions in terms of normality and separability. We will alsoapply this theory to study separable extensions.

A. Splitting Fields of Separable Polynomials. Before stating our main result,we introduce the idea of a fixed field. Suppose that we have a finite extension F C Lwith Galois group Gal(L/F). Given a subgroup H C Gal(L/F), we call

LH = =a for all a E H)

the fixed field of H. This terminology is justified by Exercise 1, where you will showthat LH is a subfield of L containing F.

Here is one of the important theorems of Galois theory.


148 THE GALOIS CORRESPONDENCE

Theorem 7.1.1 Let F C L be a finite extension. Then the following are equivalent:(a) L is the splitting field of a separable polynomial in F{x].

(b) F is the fixed field of Gal(L/F) acting on L.(c) F C L is a normal separable extension.

Proof: (a) (b): Let K be the fixed field of Gal(L/F). By Exercise 1 we haveF C K C L, and the goal is to show K = F. For this purpose, note that since L is thesplitting field of a separable polynomial f e F[x] over F, the same is true over thelarger field K, since f also lies in K[xI. By Theorem 6.2.1 it follows that

[L:F1 = and {L:K] = IGal(L/K)L

Next observe that Gal(L/K) C Gal(L/F), since if an automorphism ofLis the identityon K, then it is also the identity on the smaller field F. The reverse inclusion alsoholds, since every a e Gal(L/F) is the identity on K, for K is the fixed field ofGal (L/F). It follows that Gal (L/K) = Gal(L/F). Combining this with the aboveequations, we see that

[L:F] = {L:K].

Since [L:F] [L:K}[K:F],wehave[K:F] = 1,andK=Ffollows.(b) (c): Now suppose that F is the fixed field of Gal(L/F) and let a E L. We

will find the minimal polynomial of a over F using a construction due to Lagrange.Let = a, a2, . . . , a,. be the distinct elements of L obtained by applying the elementsof Gal(L/F) to a. Then consider the polynomial

(7.1) h(x) =fl(x_aj) EL[x].

We claim that h E F[x] and that h is irreducible over F.We first show that a E Gal(L/F) permutes the a,. By definition, a, = T(a) for

some r E Gal(L/F). Then a(a1) = a(i-(a)) = (ar)(a), which isa maps { a1,... , a permutation, since a is one-to-one.

Since a permutes the a1, it also permutes the factors x — a, of h. This shows thatthe coefficients of h are fixed by Gal(L/F) and hence lie in the fixed field, which is Fby assumption. Hence h E F[x], as claimed.

Next let g F[x] be the irreducible factor of h that vanishes at a. Then Proposi-tion 6.1.4 shows that a(a) is also a root of g for all a e Gal(L/F). Since the a1 arethe distinct elements of L obtained in this way, (7.1) shows that hIg. It follows that his irreducible over F, since g is an irreducible factor of h.

We conclude that h E is the minimal polynomial of a over F, since h isirreducible over F and has a as a root. The above formula for h also shows that h isseparable and splits completely over L. Hence:

• Normality: 1ff E F[x] is irreducible and has a root a L, then f is the polynomialh defined in (7.1) (up to a constant factor). Thus f splits completely over L, whichproves normality.

GALOIS EXTENSIONS 149

• Separability: If E L, then its minimal polynomial is the polynomial h. Thenis separable over F because h is, and separability follows.

This shows that F c L is normal and separable, as claimed.(c) (a): Finally, suppose that F c L is normal and separable. We can write

L = , where the minimal polynomial p, of a, over F is separable. Letqi,... ,q,. be the distinct elements of the set {pi,... and set

f is separable (the lemma applies because the q, are monic--—doyou see why?). Furthermore, the proof of Theorem 5.2.4 shows that L is the splittingfield off over F (you will check this in Exercise 2). Thus L is the splitting field overF of a separable polynomial in F [x], as claimed.

In light of this theorem, we make the following definition.

Definition 7.1.2 An extension F C L is called a Galois extension if it is a finiteextension satisfying any of the equivalent conditions of Theorem 7.1.1.

To see how Definition 7.1.2 works, consider the following extensions.

• The extension Q C is Galois, since is the splitting field of(x2 — 2)(x2 —3) over Q. This uses part (a) of Theorem 7.1.1.

• The extension Q c is not Galois, since x3 — 2 is irreducible over Q, has

a root in but does not split completely over This uses part (c) ofTheorem 7.1.1.

Here is one case where being a Galois extension is automatic.

Proposition 7.1.3 Suppose that F C L is a Galois extension and that we have anintermediate field F C K C L. Then K C L is a Galois extension.

Proof: We will use part (a) of Theorem 7.1.1. If F C L is Galois, then L is thesplitting field of a separable polynomial in f F [x]. By regarding f as an elementof it follows immediately that the same is true over the larger field K. (This isthe argument used in the proof of (a) (b) from Theorem 7.1.1.) .

While the proof of Proposition 7.1.3 seems easy, notice that it is much less obviousif we think in terms of parts (b) and (c) of Theorem 7.1.1.

The reader should also note that in the situation of Proposition 7.1.3, F C K neednot be Galois. Here is a simple example to illustrate this.

Example 7.1.4 By Example 4.1.10, Q c Q(i, is the splitting field of x4 —2 andhence is a Galois extension. Consider the intermediate fields Q(i) and ThenQ C Q(i) is Galois (it is the splitting field of x2 + 1), while Q C is not (x4 —2is the minimal polynomial of but doesn't split completely).

In Section 7.2 we will learn how to recognize exactly when F C K is Galois in thesituation of Proposition 7.1.3.


Definition 7.1.2 and Theorem 6.2.1 imply that IGal(L/F)I = [L:F] wheneverF C L is Galois. For an arbitrary finite extension, the relation between the order ofthe Galois group and the degree of the extension can be described as follows.

Theorem 7.1.5 Let F C L be a finite extension. Then:(a) IGal(L/F)I divides [L:F].(b) Gal(L/F)I < [L:F].(c) F C L is a Galois extension and only if = [L: F].

Proof: To prove part (a), let K be the fixed field of Gal(L/F). Then F C K C L,and the proof of (a) (b) from Theorem 7.1.1 implies that Gal(L/K) = Gal(L/F)(be sure you understand why). Thus K is the fixed field of Gal(L/K), so that K C Lis a Galois extension by Theorem 7.1.1. Hence

[L:F] [L:K][K:F] = IGal(L/K)I[K:FI = IGal(L/F)I[K:F],

where the first equality uses Theorem 4.3.8, the second uses Theorem 6.2.1 (K C Lis Galois), and the third uses Gal(L/K) = Gal(L/F). Hence the order of Gal(L/F)divides [L : F], as claimed.

Part (b) is an immediate consequence of part (a). As for part (c), note that onedirection follows from Theorem 6.2.1. For the converse, suppose that F C L is afinite extension with Gal(L/F)I = [L: F], and let K be the fixed field of Gal(L/F).If we can prove that K = F, then Theorem 7.1.1 will imply that F C L is a Galoisextension.

To show that K = F, first observe that the proof of part (a) given above impliesthat K C L is a Galois extension and that Gal(L/K) = Gal(L/F). Then

[L:F] = = IGal(L/K)I = [L:K],

where the first equality is by assumption, the second uses Gal(L/K) = Gal(L/F),and the third holds because K C L is a Galois extension. We conclude that K = Fjust as in the proof of (a) (b) from Theorem 7.1.1.

Theorem 7.1.1 gave three ways to characterize Galois extensions, and part (c) ofTheorem 7.1.5 gives a fourth. Putting these together, we see that a finite extensionF C L is Galois if and only if any of the following equivalent conditions is satisfied:

• L is the splitting field of a separable polynomial in F {x].• F is the fixed field of Gal(L/F) acting on L.• F C L is a normal separable extension.

• Gal(L/F)I = [L:F1.

B. Finite Separable Extensions. The theory of Galois extensions implies thefollowing characterization of finite separable extensions.

Proposition 7.1.6 Let F C L be afinite extension. Then L is separable over F andonly if L= where each is separable over F.


Proof: First assume that F C L is separable. Since it is also finite, Theorem 4.4.3implies that L has the desired form. For the converse, let L = . . , whereeach is separable over F. Our strategy will be to embed L in a larger field that isseparable over F.

Let p, be the minimal polynomial of over F, and let qi, . . . , q,- be the distinctelements of the set Then Lemma 5.3.4 implies that f = qi is

separable, since each qj is. Let M be the splitting field of f, regarded as a polynomialin L[x]. ThusM =L(81,... ,/3m), where 81,... are the roots off.

We claim that M = F(i31, . . . ,/3m). To see why this is true, note that we have theobvious inclusion

(7.2) F(13i,...,/3m) CL(/31,...,/3m) =M.

However, the roots . , I-3m include . . . , so that

L and /3k,. .. , which gives the inclusion

ML(/31,...,13m)CF(/31,...,/3m).

Combining this with (7.2), we see that M = F(/31 , . . . as claimed.This shows that M is the splitting field over F of the separable polynomial f. Then

F c M is Galois and hence separable by Theorem 7.1.1. Since L C M, every elementof L is separable over F, so that F C L is separable.

Proposition 7.1.6 has some nice consequences. For example, if F C Land /3 E Lare separable over F, then so are + /3, and (assuming /3 0). This inturn implies that in characteristic p, any finite extension can be written as a separableextension followed by a purely inseparable one. You will be asked to prove theseassertions in Exercises 3 and 4.

C. Galois Closures. The proof of Proposition 7.1.6 shows how to embed a finiteseparable extension F C L into a larger Galois extension. This leads to the idea ofGalois closure, which roughly speaking is the smallest extension of L that is Galoisover F. More precisely, we have the following result.

Proposition 7.1.7 Let F C L be a finite separable extension. Then there is anextension L C M such that:(a) M is Galois over F, i.e., F C M is a Galois extension.(b) Given any other extension L C M' such that M' is Galois over F, there is afield

homomorphism : M —* M' that is the identity on L.

Proof: Since F CL is finite and separable, we can write L = . . whereaj is separable over F. Following the proof of Proposition 7.1.6, we get an extensionL c M such that M is a splitting field over L of the separable polynomial f =qi . . .

where ,q,- are the distinct elements of {pi,...,pn} and p, is the minimal


polynomial of over F. As in the proof of Proposition 7.1.6, we see that F C M isa Galois extension.

To show that L C M satisfies part (b) of the proposition, let L C M' be an extensionwhere M' is Galois over F. By Theorem 7.1.1, F C M' is normal, so that each p, splitscompletely over M'. It follows that f splits completely in M'. Let M" C M' be thesubfield obtained by adjoining the roots off to F. Furthermore, since a1 E L C M'is a root off, we have L C M". Thus we can regard M" as a splitting field off overL. By the uniqueness of splitting fields (Corollary 5.1.7), there is an isomorphism

M —+ M" that is the identity on L. Since M" C M', we can regard p as a fieldhomomorphism : M —÷ M'. This completes the proof of the proposition. •

Be sure you understand why part (b) of the proposition implies that L C M can bethought of as the smallest extension of L that is Galois over F. The field constructedin Proposition 7.1.7 is called the Galois closure of L over F. In Exercise 5 you willshow that the Galois closure of F C L is unique up to an isomorphism that is theidentity on L.

Related to the idea of Galois closure is the normal closure of a finite extensionF C L. Roughly speaking, this is the smallest extension of L that is normal over F.The theory of normal closures is worked out in Exercises 6 and 7.

Historical Notes

Of the criteria for F C L to be a Galois extension given in Theorem 7.1.1, themost elegant is the one involving the fixed field of Gal(L/F). For Galois, this washis Proposition I, which was the first of his main results [Galois, p. 51]:

PROPOSITION I

THEOREM. For a given equation, let a,b,c,... be them roots. There isalways a group of permutations on the letters a, b, c,... that enjoys the followingproperty:

1° that every function of the roots that is invariant** under the substitutionsof the group, is rationally known;

2° conversely, that every function of the roots that is rationally detennined,is invariant under these substitutions*.

For Galois, "rationally known" and "rationally determined" refer to elements of afield F containing the coefficients of the given equation. Adjoining the roots of thisequation gives the splitting field L = F(a, b, c,...). Furthermore, Galois assumes thatthe given polynomial "does not have equal roots." Hence the polynomial is separable,so that L is a Galois extension of F. Since every element of L is a "function of theroots," parts 10 and 2° of Galois's Proposition I say that F is the fixed field of theGalois group acting on L. Thus we recover part (b) of Theorem 7.1.1.

In Galois's manuscript, Proposition I includes two notes, marked with ** and *

above. We will explain these notes when we discuss Galois's work in Chapter 12.We should also mention that Galois knew the formula (7.1) for the minimal

polynomial given in the proof of Theorem 7.1.1 (see [Galois, p. 85]). In Chapter 12,


we will see that this formula is a generalization of the resolvent polynomial definedby Lagrange in 1770.


Exercise 1. Given a finite extension F C L and a subgroup H C Gal(L/F), prove thatLH = {a E L o(cs) =a for all o E H} is a subfield of L containing F.

Exercise 2. In the proof of (c) (a) in Theorem 7.1.1, give the details of how the proof ofTheorem 5.2.4 shows that L is the splitting field of f over F.

Exercise 3. Suppose that F C Land that a, /3 E L are separable over F. Prove that a + /3, a/3,and a//3 (assuming /3 0) are also separable over F.

Exercise 4. Let F C L be a finite extension, and assume F has characteristic p. Then considerthe set K = {a E L a is separable over F}.(a) Use Proposition 7.1.6 to show that K is a subfield of L containing F. Thus F C K is a

separable extension.(b) Use part (c) of Theorem 5.3.15 to show that K C L is purely inseparable.

Exercise 5. Prove that the Galois closure of a finite separable extension F C L is unique up toan isomorphism that is the identity on L.

Exercise 6. In analogy with the Galois closure of a finite separable extension, every finiteextension F C L has a normal closure, which is essentially the smallest extension of L that isnormal over F. State and prove the analog of Proposition 7.1.7 for normal closures.

Exercise 7. Prove that the normal closure of a finite extension F C L is unique up to anisomorphism that is the identity on L.

Exercise 8. Let h be the polynomial (7.1) used in the proof of (b) (c) from Theorem 7.1.1.Show that there is an integer m such that

fJ

Exercise 9. For each of the following extensions, say whether it is a Galois extension. Be sureto say which of our four criteria (the three parts of Theorem 7.1.1 and part (c) of Theorem 7.1.5)you are using.(a)

Q C Q(a,/3), a,fi distinct roots ofx3 +x2 +2x+ 1.(c)

(d) C C(t), t a variable.(e) C C(t), t a variable, n a positive integer.

The ideas underlying the extensions given in parts (d) and (e) will be discussed in Section 7.5.

Exercise 10. Prove that Q(w, is the Galois closure of over Q.

Exercise 11. Construct the Galois closure of Q C

Exercise 12. Let F C L be an extension of degree 2, where F has characteristic 2.

(a) Show that L = F(a), where a is a root of an irreducible polynomial of degree 2.(b) Show that the minimal polynomial of a over F is separable.


(c) Conclude that F c L is a Galois extension with Gal(L/F) Z/2Z.(d) By completing the square, show that there is /3 E L such that L = F(/3) and /32 E F.For /3 as in part (d), let a = /32 E F. Then we can write /3 = This shows that if F hascharacteristic 2, then every degree 2 extension of F is obtained by taking a square root.

7.2 NORMAL SUBGROUPS AND NORMAL EXTENSIONS

In Chapter 5 we introduced normal extensions, and in abstract algebra you learnedabout normal subgroups. This section will explain why it is no accident that theseconcepts have the same name.

A. Conjugate Fields. In high school algebra one calls 2— the conjugate of2 + This terminology is used for subfields as follows.

Definition 7.2.1 Suppose that we have finite extensions F C K C L. Then, for anautomorphism a E Gal(L/F), we call

aK = E K}

a conjugate field of K.

We should write a(K) instead of aK, but we prefer the latter because it is lesscumbersome. Note that aK is a subfield of L, since a is a field isomorphism.

We can compute the degree of a conjugate field as follows.

Lemma 7.2.2 Let F C K C L and a Gal(L/F) be as in Definition 7.2.1. ThenFCaKCLand[K:F]=[aK:F}.

Proof: The inclusion F C aK is obvious, since F C K and a is the identity onF. Also, a E F and E K. It follows that

K —÷ aK is linear over F in the sense of linear algebra. Hence is anisomorphism of vector spaces over F, so that [K : F] = dimFK = dimraK = [aK : F]

by the definition of degree given in Section 4.3. •

Here is an example of conjugate fields.

Example 7.2.3 Consider the extension Q c Q(w, where w = Then wehave the following intermediate fields:

Q(w,

(7.3) Q(w)

NORMAL SUBGROUPS AND NORMAL EXTENSIONS 155

Recall that a E Gal(Q(w, is determined uniquely by

(7.4) a(w) E {w,w2} and E

In Exercise 2 of Section 6.2 we showed that all possible combinations of a(w) andactually occur. In Exercise 1 below you will check the following:

• has conjugate fields and• Q(w) equals all of its conjugates.Later in the section we will explain the second bullet using Galois theory.

We next relate intermediate fields to subgroups of the Galois group.

Lemma 7.2.4 Suppose that we have finite extensions F C K C L. Then:(a) Gal(L/K) is a subgroup of Gal(L/F).(b) If a E Gal(L/F), then Gal(L/aK) = aGal(L/K)a' in Gal(L/F).

Pmof: Each a e Gal(L/K) is an automorphism of L that is the identity on K. Thena e Gal(L/F) follows from F c K; hence Gal(L/K) C Gal(L/F). Since both aregroups under composition, we conclude that Gal(L/K) is a subgroup.

To prove part (b), let -y E aGal(L/K)a1 and aK. Then -y = for someT E Gal(L/K), and = a(a) for some a eK. Thus

y(/3) (a(a))= a(r(a))

a(a) =

where the third equality follows because 'r is the identity on K. Hence -y is the identityon aK, which implies that aGal(L/K)a' C Gal(L/aK). The opposite inclusion isequally straightforward (see Exercise 2), and the lemma follows. •

In group theory, a conjugate of a subgroup H C G is a subgroup of the formgHg' for some g E G. Thus part (b) of Lemma 7.2.4 tells us that conjugate fieldscorrespond to conjugate subgroups.

B. Normal Subgroups. The first main theorem of this section explains hownormal subgroups relate to normal extensions.

Theorem 7.2.5 Suppose that we have fields F C K C L, where F C L is a Galoisextension. Then the following conditions are equivalent:(a) K = aKfor all a E Gal(L/F), i.e., K equals all of its conjugates.(b) Gal(L/K) is a normal subgroup of Gal(L/F).(c) F C K is a Galois extension.(d) F c K is a normal extension.

Proof: We first show that (a) and (b) are equivalent. Proving (a) (b) is especiallyeasy, for K = aK and Lemma 7.2.4 imply that

Gal(L/K) = Gal(L/aK) = aGal(L/K)a'.


Thus Gal(L/K) is normal in Gal(L/F) since this holds for all a E Gal(L/F). Toprove (b) (a), first note that if Gal(L/K) is normal and a E Gal(L/F), then usingLemma 7.2.4 a second time implies that

Gal(L/K) aGal(L/K)a' = Gal(L/aK).

However, K C L and aK C L are Galois extensions by Proposition 7.1.3. Hence

K = fixed field of Gal(L/K) = fixed field of Gal(L/aK) = aK,

where the first and third equalities use Theorem 7.1.1.We next observe that (c) and (d) are equivalent. The implication (c) (d) is

trivial, since every Galois extension is normal and separable. For (d) (c), note thatsince F c L is Galois, it is also separable, and then any intermediate field F C K C Lis also separable over F. If in addition K is normal over F, then it is normal andseparable, and hence Galois.

Finally, we prove that (a) (d). For (a) (d), let f Fwith a root a E K. We need to show that f splits completely over K. In the proof ofTheorem 7.1.1 we showed that up to a constant, f is the polynomial

h(x) =fl(x—ai)

from (7.1), where al = a,a2, . . . ,a,- are the distinct elements of L obtained byapplying the elements of Gal(L/F) to a. Since a e K, each a, lies in a conjugatefield of K. Using (a), we conclude that a, E K for all i, so that h and hence f splitcompletely over K.

It remains to show that (d) (a). Take a E K and a e Gal(L/F), and let p bethe minimal polynomial of a over F. By Proposition 6.1.4, a(a) is also a root ofp. Since F c K is normal, p splits completely over K, which implies that a(a) EK.Thus aK C K, and then equality follows by Lemma 7.2.2. •

Here is an example of how this theorem works.

Example 7.2.6 Consider Q C L = Q(w, studied in (7.3). By the discussionfollowing (7.4), there are automorphisms a, i- E Gal(L/Q) such that

(7.5) a(w) = w, = and r(w) = w2, =

Label the roots of x3 —2 as a1 = = and a3 = and consider theisomorphism Gal(L/Q) S3 given by the action of the automorphisms on the rootsa1 ,a2, a3. Then it is easy to see that

r—+(23).

Since these permutations generate 53, it follows that a and i- generate Gal(L/Q).Now consider the fields in the diagram (7.3). Each such field K gives a subgroup

Gal(L/K) C Gal(L/Q). Furthermore, in Exercise 3 you will show that

(7.6) K1 C 1(2 C L Gal(L/Ki) D Gal(L/K2).


In other words, larger fields correspond to smaller Galois groups. Then we claimthat for the fields K of (7.3), the map K Gal(L/K) gives the following diagram ofsubgroups of Gal(L/Q):

{e}

(7.7) (a) (r) (o.2r) (ar)

Gal(L/Q)

In this diagram, (a) is the subgroup generated by a. Thus (a) = {e,a,a2}, since ahas order 3. Similarly, (r), (a2r), (aT) are subgroups of order 2.

To see how (7.3) gives (7.7), consider the case of Q(w). We know that

IGal(L/Q(w))I = [L : Q(w)] = : Q(w)] = = = 3.

Furthermore, (7.5) shows that a is the identity on Q(w), since a(w) = w. Thusa E Gal(L/Q(w)), and it follows easily that Gal(L/Q(w)) = (a). In Exercise 4you will give similar arguments for the other fields in (7.3) to verify that applyingK Gal(L/K) to (7.3) gives (7.7).

We can relate (7.3) and (7.7) to Lemma 7.2.4 and Theorem 7.2.5 as follows. Firstconsider Q(w). This is the splitting field of x2 +x+ 1 over Q, so that Q C Q(w) isGalois. By Theorem 7.2.5, this implies:

• Q(w) coincides with its conjugates in L, as we saw in Example 7.2.3.• Gal(L/Q(w)) = (a) is normal in Gal(L/Q).

We can also go the other way. Under the isomorphism Gal(L/Q) S3, (a) maps tothe normal subgroup A3. Thus (a) is normal in Gal(L/Q), so that Q C Q(w) is aGalois extension by Theorem 7.2.5.

We can do a similar analysis for Example 7.2.3 shows that this field hasthree conjugates in L. Hence

(7.8) = (T)

is not normal in Gal(L/Q), by Theorem 7.2.5. We can relate the conjugate fieldsof to the conjugates of (T) as follows. By Exercise 1, the conjugate fieldsof are itself, and Then Lemma 7.2.4 implies that theGalois groups of L over these fields are the conjugate subgroups

(T), a(T)a', a2(T)a2.

One easily checks that these are the subgroups (r), (a2r), (aT) from (7.7).


In group theory, normal subgroups are important because they lead to quotientgroups. Recall that if N is normal in G, then left cosets of N coincide with rightcosets, and the set GIN consisting of all cosets of N in G becomes a group undermultiplication, the quotient group. Theorem 7.2.5 shows that normal subgroups arisenaturally in Galois theory.

When Gal(L/K) c Gal(L/F) is normal, the second main theorem of this sectionexplains how to interpret the quotient group.

Theorem 7.2.7 Suppose that we have extension fields F C K C L, where F C K andF C L are Galois. Then Gal(L/K) is a normal subgroup of Gal(L/F), and there is anatural isomorphism of groups

Gal(L/F)/Gal(L/K) Gal(K/F).

Proof: 1fF C K is Galois, then Gal(L/K) is normal in Gal(L/F) by Theorem 7.2.5.It remains to relate Gal(L/F)/Gal(L/K) to Gal(K/F).

For a fixed a Gal(L/F), the restriction of a to K gives the isomorphism aIK:K aK. But Theorem 7.2.5 tells us that aK = K, since F C K is Galois. It followsthat is an automorphism of K. Since a is the identity on F, the same is true for

(do you see why?), so that aIK E Gal(K/F).It follows that a aIK defines a function

Gal(L/F) —+ Gal(K/F).

Furthermore, in Exercise 5 you will verify that for a, T E Gal(L/F), we have

(7.9)

where the first composition takes place in Gal(L/F) and the second in Gal(K/F).This shows that is a group homomorphism.

The kernel of is easy to determine, for if a Gal(L/F), then

a E Ker(c1) aIK = a is the identity on K a E Gal(L/K).

Thus = Gal(L/K), and then the Fundamental Theorem of Group Homomor-phisms implies that induces an isomorphism

Gal(L/F)/Gal(L/K) Im(F) C Gal(K/F).

The final step is to show that = Gal(K/F). The key point is that since allof the extensions involved are Galois extensions, their degrees equal the order of thecorresponding Galois groups. Thus

= IGal(L/F)/Gal(L/K)IGal(L/F)J [L:F] K F G I K FIGal(L/K)I [L:K] . ] a ( /

This shows that Im(c1) = Gal(K/F) and completes the proof of the theorem. .


Here is a simple example of Theorem 7.2.7.

Example 7.2.8 Consider Q c Q(w) C L = Q(w, Since Q C Q(w) is Galoisand Gal(L/Q(w)) where a is as in (7.5), the theorem implies that

Gal(Q(w)/Q) Gal(L/Q)/(a) S3/A3 Z/2Z.

Note that if r is as in (7.5), then Gal(Q(w)/Q) = {1Q(w),rIQ(W)}.

Mathematical Notes

There are two ideas in this section to comment on.

• The Galois Correspondence. In Section 7.3 we will see that (7.3) and (7.7) givean example of the Galois correspondence. It is easy to check that (7.7) gives allsubgroups of Gal(L/Q) for L = Q (w, (see Exercise 6). Then Theorem 7.3.2 willtell us that (7.3) gives all fields between Q and Q(w, This is not obvious—whilethe subfields in (7.3) are easy to find, how do we know that they are all the subfields?This is a good illustration of the power of the Galois correspondence.

We should also mention that (7.6) is also part of the Galois correspondence. Theidea behind (7.6) is that as the field K gets larger, the Galois group Gal(L/K) getssmaller. This explains why the arrows in (7.3) go up while those in (7.7) go down.

• Conjugate Fields. If F C K is not Galois, then K will have a certain number ofconjugate fields in L (assuming F C K C L and L is Galois over F). We claim thatthe number of such conjugate fields is related to the normalizer of a subgroup.

To see this, we first analyze when a conjugate equals the given field. Suppose thatF c K C L, where L is Galois over F, and let a E Gal(L/F). In Exercise 7 you willshow that

(7.10) K = aK Gal(L/K) = in Gal(L/F).

In group theory, the normalizer of a subgroup H C G is the set

=H}.

One can show that NG(H) is a subgroup of G, H is a normal subgroup of NG(H),and NG (H) is the largest subgroup of G in which H is normal. (You should doExercise 8 if you're not familiar with normalizers.) From (7.10), it follows that fora E Gal(L/F), we have

K = aK a is in the normalizer of Gal(L/K) in Gal(L/F).

Using this and standard facts about group actions, one can prove that if F C L isGalois and K is an intermediate field, then the number of conjugate fields of K in Lis given by the index

[Gal(L/F) :N] = IGal(L(F)I

where N is the normalizer of Gal(L/K) in Gal(L/F) (see Exercise 9 for the details).


Historical Notes

In a letter written the night before his fatal duel, Galois describes the concept ofnormal subgroup as follows [Galois, pp. 173—175]:

In other words, when a group G contains another group H, the group G can bedivided into groups that are obtained by performing the same substitution on thepermutations of H, so that G = H + HS + HS' + and it can also be dividedinto groups with the same substitutions so that G = H + TH + T'H + •••. Thesetwo decompositions do not ordinarily coincide. When they do coincide, thedecomposition is said to be proper.

In modern terms, equality of the decompositions

implies that the left cosets of H coincide with the right cosets, which is equivalentto the usual definition of normal subgroup. However, we will see in Chapter 12 thatGalois's "groups" are not quite what you might think.

Galois was also aware of Theorem 7.2.5, though again his terminology takes someexplanation. The details of what Galois knew can be found in [Edwards, pp. 47—66],[3, pp. 80—84], and [8].

The second main theorem of this section, Theorem 7.2.7, concerns quotient groups.Quotient groups weren't defined until much later in the nineteenth century, thoughhints can be found in the examples worked out in Galois's memoir (see [3, p. 82]).When discussing Galois's work in 1852, Betti made some further progress toward adefinition of quotient group, and by the 1 880s quotient groups were well established.For us, the key point is that both normality and quotient groups first arose in thecontext of Galois theory.


Exercise 1. In the diagram (7.3), verify the following.(a) has conjugate fields and

(b) Q(w) equals all of its conjugates.

Exercise 2. Complete the proof of Lemma 7.2.4 by showing that

Gal(L/crK) C oGal(L/K)ci,'.


Exercise 4. Verify that applying K Gal(L/K) to (7.3) gives (7.7). Don't forget to includethe extreme cases K = Q and K = L.

Exercise 5. Prove (7.9) in the proof of Theorem 7.2.7.

Exercise 6. For the extension Q C L = Q(w, we listed some subgroups of Gal(L/Q) indiagram (7.7). Prove that this gives all subgroups of Gal(L/Q).

Exercise 7. Suppose that F C K C L, where L is Galois over F, and let o E Gal(L/F). Showthat

K = Gal(L/K) = in Gal(L/F).

THE FUNDAMENTAL THEOREM OF GALOIS THEORY 161

Exercise 8. Let H be a subgroup of a group G, and let NG (H) = {g E G = H} be thenormalizer of H in G, as defined in the Mathematical Notes.(a) Prove that NG(H) is a subgroup of G containing H.(b) Prove that H is normal in NG(H).(c) Let N be a subgroup of G containing H. Prove that H is normal in N if and only if

NC NG(H). Do you see why this shows that NG(H) is the largest subgroup of G in whichH is normal?

(d) Prove that H is normal in G if and only if NG(H) = G.

Exercise 9. Let F c L be Galois, and suppose that F C K C L is an intermediate field. Thegoal of this exercise is to show that the number of conjugates of K in L is

[Gal(L/F) :N] = IGal(L/F)I

where N is the normalizer of Gal(L/K) in Gal(L/F). More precisely, suppose that the distinctconjugates of K are

K=alK,cr2K,...,orK,where = e. Then we need to show that r = [Gal(L/F) N].(a) Show that Gal(L/F) acts on the set of conjugates . . . ,orK}.(b) Show that the isotropy subgroup of K is the normalizer subgroup N.(c) Explain how r = [Gal(L/F) : N] follows from the Fundamental Theorem of Group Actions

(Theorem A.4.9 from Appendix A).

Exercise 10. In (7.5), explain why is complex conjugation restricted to Q(w,

Exercise 11. Consider the extension Q C L =(a) Show that Gal(L/Q) = {e,o-,r,ai-}, where

(b) Find all subgroups of Gal(L/Q), and use this to draw a picture similar to (7.7).(c) For each subgroup of part (b), determine the corresponding subfield of L and use this to

draw a picture similar to (7.3).(d) Explain why all of the subgroups in part (b) are normal. What does this imply about the

subfields in part (c)?In the next section, we will see that the Galois correspondence implies that the subfields youfound in part (c) give all subfields of L.

7.3 THE FUNDAMENTAL THEOREM OF GALOIS THEORY

We can now state the main result of this chapter, which describes precisely the relationbetween subgroups and subfields. Recall that if we are given a finite extension F C Land a subgroup H C Gal(L/F), then we have the fixed field

L11 = = for alla E H}.

In Exercise 1 of Section 7.1 you showed that LH is a subfield of L containing F. Thefirst part of the Fundamental Theorem of Galois Theory goes as follows.


Theorem 7.3.1 Let F C L be a Galois extension.(a) For an intermediate field F C K C L, its Galois group Gal(L/K) C Gal(L/F)

has fixed field= K.

Furthermore, Gal(L/K) I = [L : K] and [Gal(L/F) : Gal(L/K)1 [K: F].

(b) For a subgroup H C Gal(L/F), its fixed field F C LH C L has Galois group

Gal(L/LH) =H.

Furthermore, [L: LHI = HI and [LH : F] = [Gal(L/F) : H].

Proof: Part (a) follows easily from earlier results. We are assuming that F C L isGalois, so that K C L is also Galois by Proposition 7.1.3. Then K =LGaI(L/K) followsfrom Theorem 7.1.1 and the definition of Galois extension.

Since K C L and F C L are both Galois, we have IGal(L/K)I = [L : K] andIGal(L/F) I = [L: F] by Theorem 6.2.1. Using these equalities and the Tower Theo-rem (Theorem 4.3.8), we obtain

[Gal(L/F) :Gal(L/K)1 =

_______

== [K:F].

This completes the proof of part (a).To prove part (b), let H be a subgroup of Gal(L/F). This gives F C LH C L, and

since every a E H is the identity on LH, we have

(7.11) H C Gal(L/LH).

To prove that equality occurs, we will give a classic proof using the Theorem of thePrimitive Element. Observe that LH C L is a finite separable extension (since F C Lis), so that L = LH(a) for some a E L by Corollary 5.4.2. Then consider

h(x)= fl(x_a(a)).aEH

By standard arguments, the coefficients of h are fixed by H (be sure you can provethis carefully). Thus h E LH[XI satisfies h(a) = 0. It follows that if p E LH[x] is theminimal polynomial of a over LH, then This implies that

(7.12) ?deg(p)=[L11(a):L11]=[L:L11],

where the second equality follows because p is the minimal polynomial of a overLH. Combining (7.11) and (7.12), we obtain

[L:LH] IHI IGal(L/LH)I.

However, Proposition 7.1.3 implies that LH C L is Galois, so that we also haveGal (L/LH) I = [L : LII]. Then the above inequalities easily imply that

IHI = IGal(L/LH)I,


and H = Gal(L/LH) follows immediately. Similar to part (a), we conclude that[Gal(L/F) :H] = [LH : F]. We leave the details as Exercise 1. •

Here is the second part of the Fundamental Theorem of Galois Theory.

Theorem 7.3.2 Let F C L be a Galois extension. Then the maps between interme-diate fields F C K C L and subgmups H C Gal(L/F) given by

K '—* Gal(L/K),H LH

reverse inclusions and are inverses of each other Furthermore, if a subfield Kcorresponds to a subgroup H under these maps, then K is Galois over F if and onlyif H is normal in Gal(L/F), and when this happens, there is a natural isomorphism

Gal(L/F)/H Gal(K/F).

Proof: Composing the maps one way gives

K '—p Gal(L/K) K

by part (a) of Theorem 7.3.1, and going the other way gives

H LH '—p Gal(L/LH) = H

by part (b) of the theorem. This proves that the maps K '-+ Gal(L/K) and H '—p LHare inverses of each other. The map K '—+ Gal(L/K) is inclusion-reversing by (7.6),and H1 C H2 LH follows from the definition of fixed field.

The final assertions of the theorem follow from Theorems 7.2.5 and 7.2.7. •

We next give two examples of the Galois correspondence.

Example 7.3.3 Consider the extension Q C L = Q(w, w = Recall from(7.7) that Gal(L/Q) S3 has subgroups

{e}

(a) (T) (a2r) (ar)

Gal(L/Q)

Here, a, 'i- E Gal(L/Q) are as in (7.5), and Exercise 6 of Section 7.2 shows that theseare all subgroups of Gal(L/Q).


By (7.3), the corresponding fixed fields are

Q(w,

Q(w)

The key point is that according to Theorem 7.3.2, these are all subfields of L =Q(w, containing Q. Furthermore, note that the discussion of conjugate exten-sions, normal subgroups, etc., given in Example 7.2.6 verifies the fine details ofTheorem 7.3.2.

Here is a slightly more complicated example.

Example 7.3.4 We get a similar picture for the extension Q C L = Q(i, InExercise 2 you will describe Gal(L/Q) as follows.

• Gal(L/Q) is generated by elements r such that

a(i) = i, = and T(i) = —i, =

We also have o(a) = 4 and o(r) = 2.

• Gal(L/Q) D8, where D8 is the dihedral group of order 8.

We next work out the correspondence between subfields of L and subgroups ofGal(L/Q).

In Exercise 3 you will show that all subgroups of Gal(L/Q) are given by

{e}

(r) (v.2) (oar)

(7.13) NIZ(a2,r) (oP) (o2,o'r)

N1ZGal(L/Q)


and that the corresponding fixed fields are given by

(7.14) Nt/tNt VQ(i)

Again, Theorem 7.3.2 implies that this gives all subfields of L = Q(i, containing

Q. Exercise 3 will work out the details of the Galois correspondence.

Finally, let us give an interesting application of the Galois correspondence.

Proposition 7.3.5 Let F c L be a finite separable extension. Then there are onlyfinitely many intermediate fields F C K C L.

Proof: By Proposition 7.1.7 there is an extension L C M such that F C M is Galois.Then Theorem 7.3.2 implies that subfields ofMcontaining F correspond to subgroupsof Gal(M/F). Since Gal(M/F) is finite, it has only finitely many subgroups, so thatthere are only finitely many subfields of M containing F. Since F C L C M, it followsin particular that there are only finitely many intermediate fields between F and L..

In contrast, there are finite purely inseparable extensions that have infinitely manyintermediate fields. Here is a classic example.

Example 7.3.6 Let k be a field of characteristic p, and consider the extension

F=k(t,u)CL,

where L is the splitting field of (xP t) — u) E F [x]. This extension was studiedin Example 5.4.4, where we showed that it has no primitive element. Furthermore,Exercise 4 of Section 5.4 showed that F CL is purely inseparable and L = F(a,8),where & = t and /3P = u.

In Exercise 5 of Section 5.4 you proved that the intermediate fields

(7.15)

are all distinct as A ranges over the distinct elements of F. Since F is infinite, we seethat there are infinitely many intermediate fields.

In Exercise 4, you will show that Gal(L/F) = { 1L}. This means that in particular,Gal(L/F) has only one subgroup, namely {e}, yet F C L has the infinitely manyintermediate fields given by (7.15).


This example shows that the Galois correspondence can break down spectacularlyfor purely inseparable splitting fields.


Exercise 1. Complete the proof of Theorem 7.3.1 by showing that [Gal (L/F) H] = [LB F]for all subgroups H C Gal(L/F).

Exercise 2. Consider Q C L = Q(i,(a) Show that there are a, r E Gal(L/Q) such that

o(i) = i, = and T(i) = —i, =

(b) Prove that o(a) = 4, 0(T) = 2 and that r is complex conjugation restricted to L.(c) Prove that a and T generate Gal(L/Q).(d) Show that Gal(L/Q) D8, where D8 is the dihedral group of order 8.

Exercise 3. Let L = Q(i, and a,'r E Gal(L/Q) be as in Exercise 2 and Example 7.3.4.(a) Show that all subgroups of Gal(L/Q) are given by (7.13).(b) Show that the corresponding fixed fields are given by (7.14).(c) Determine which subgroups in part (a) are normal in Gal(L/Q), and for those that are

normal, construct a polynomial whose splitting field is the corresponding fixed field.(d) For the subfields in part (b) that are not Galois over Q, find all of their conjugate fields.

Also describe the conjugates of their corresponding groups.

Exercise 4. Prove that the extension F CL of Example 7.3.6 has Gal(L/F) = {1L}.

Exercise 5. Consider the extension F = C(t4) C L = C(t), where t is a variable.(a) Show that L is the splitting field of x4 — E F[x] over F.(b) Show that x4 t4 is irreducible over F.(c) Show that Gal(L/F) Z/47L.

(d) Similar to what you did in Exercise 3, determine all subgroups of Gal(L/F) and thecorresponding intermediate fields between F and L.

We will say more about this type of extension in Section 7.5.

Exercise 6. This exercise will work out the Galois correspondence for the splitting field L of

4x2 + 2 over Q. In Exercise 6 of Section 5.1 you showed that L = Q( + and thatGal(L/Q) Z/4Z. Now, similar to Example 7.3.4, determine all subgroups of Gal(L/Q) andthe corresponding intermediate fields of Q C L.

Exercise 7. Let = and consider the extension Q C L =(a) Show that L is the splitting field off =x6 +x5 +x4 +x3 +x2 +x+ 1 over Q and thatf

is the minimal polynomial of(b) Let (Z/7Z)* be the group of nonzero congruence classes modulo 7 under multiplication.

By Exercise 4 of Section 6.2 there is a group isomorphism Gal(L/Q)(Z/7Z) *

be the subgroup generated by the congruence class of —1. Prove that+ (a') is the fixed field of the subgroup of Gal(L/Q) corresponding to H.

Exercise 8. Let = + (f', where =(a) Show that the minimal polynomial of over Q is x3 + x2 2x — 1.

(b) Use Exercise 7 to show that the splitting field of x3 + x2 — 2x — 1 over Q is a Galoisextension of degree 3 with Galois group isomorphic to Z/37Z.

FIRST APPLICATIONS 167

Exercise 9. Let F be a field of characteristic different from 2, and let F C L be a finiteextension. Prove that the following are equivalent:(a) L is a Galois extension of F with Gal(L/F) Z/2Z x 7Z/2Z.(b) L is the splitting field of a polynomial of the form (x2 —a)(x2 — b), where a,b E F but

do not lie in F.

Exercise 10. Suppose that fi E C are algebraic of degree 2 over Q (i.e., they are both rootsof irreducible quadratic polynomials in Q[x]). Prove that the following are equivalent:(a) Q(a) =(b)(c) + is the root of a quadratic polynomial in Q [xi.

Exercise 11. Let F C L be a Galois extension, and let F C K C L be an intermediatefield. Then let N be the normalizer (as defined in the Mathematical Notes to Section 7.2) ofGaI(L/K) C Gal(L/F). Prove that the fixed field LN is the smallest subfield of K such that Kis Galois over the subfield.

Exercise 12. Let H be a subgroup of a group G, and let N = flgEG gHg.(a) Show that N is a normal subgroup of G.(b) Show that N is the largest normal subgroup of G contained in H.

Exercise 13. Let F C L be a Galois extension, and let F C K C L be an intermediate field. Ifwe apply the construction of Exercise 12 to Gal(L/K) C Gal(L/F), then we obtain a normalsubgroup N C Gal(L/F). Prove that the fixed field LN is the Galois closure of K.

Exercise 14. Prove the implication (b) (a) of Theorem 6.5.5.

Exercise 15. Let p be prime. Consider the extension Q C L = Q (ç, discussed inSection 6.4. There, we showed that Gal(L/Q) The group has twosubgroups defined as follows:

and

where = au + b, u e IF,,. Let T' and D' be the corresponding subgroups of Gal(L/Q).(a) Show that the fixed field ofT' is Q(ç).(b) What is the fixed field of D'? What are the conjugates of this fixed field?

7.4 FIRST APPLICATIONS

This section is devoted to three applications of the Galois correspondence.

A. The Discriminant. The discriminant E F of a nonconstant monic poly-nomial f F[x] was defined in Section 2.4. There, we showed that if f has degreen 2andf= in asplittingfieldLoff, then

= — E F.i<j

In Section 5.3, we saw that is separable if and only if 0. We define

EL.


Note that while is uniquely determined by f, the above square root depends onhow the roots are labeled.

1ff E F[xJ is separable, then by Section 6.3 the action of the Galois group on theroots . . . , off gives a one-to-one group homomorphism

Gal(L/F) —÷ Sn.

In we also have the alternating group An C Our first result shows thatcontrols the relation between An and Gal(L/F).

Theorem 7.4.1 Letf and F c L be as above, and assume that the characteristic ofF is different from 2.(a) If a E Gal(L/F) corresponds to r then

= sgn(r)

(b) The image of Gal(L/F) lies in An and only if E F (i.e., is thesquare of an element ofF).

Proof: The result is trivial if n = 1. Hence we may assume that n � 2. Recall fromProposition 2.4.1 that = —x3) F[xi, . . . has the property that

(7.16)

for all r E This gives the identity

—x,-(f)) = sgn(r) fJ(xj—xj)i<j i<j

in F[xi,... ,Xn]. Since the evaluation map F[xi,... —* L sending x, to is a ringhomomorphism (Theorem 2.1.2), it follows that

fl — ci,(f)) = sgn(r) — = sgn(r)i<j i<j

However, we also have = which implies that

i<j

This completes the proof of part (a). For part (b), observe that F C L is Galois, sothat F is the fixed field of Gal(L/F). Combining this with (a), we obtain

a E Gal(L/F),

where T is the image of a under the map Gal(L/F) —+ Since 0 and Fhas characteristic 2, the last condition is equivalent to sgn(r) = 1 for all r comingfrom Gal(L/F). Then we are done, since sgn(r) = 1 if and only if r E An.


This result allows us to compute the Galois group of an irreducible cubic.

Proposition 7.4.2 Let f F [x] be a monic irreducible separable cubic, where F hascharacteristic 2. If L is the splitting field off over F, then

Gal(L/F)fZ/3Z, if is a square in F,

— otherwise.

Proof: Exercise 6 of Section 6.2 implies that IGal(L/F) is a multiple of 3, sincef is irreducible and separable. We also have the one-to-one map Gal(L/F) —+ S3.

Since the only subgroups of S3 of order divisible by 3 are S3 and A3, the propositionfollows easily from Theorem 7.4.1. We leave the details as Exercise 1. .

Here is a simple example of this proposition.

Example 7.4.3 Consider f = x3 +x2 — 2x 1 E Q[xI. It is easy to see that f isirreducible over Q and hence is separable, since we are in characteristic 0. Using themethod discussed in Section 5.3 one computes that

By Proposition 7.4.2 the Galois group off over Q is cyclic of order 3.

In Exercise 2 you will compute the Galois groups of some other cubics, and inChapter 13 we will compute the Galois groups of quartics and quintics.

B. The Universal Extension. Consider the universal extension in degree n,

(7.17) K=F(o1,...,an)CL=F(xi,...,xn),

where as usual cxi,. . . , are the elementary symmetric polynomials. Recall fromSection 6.4 that this is the splitting field of the universal polynomial of degree n,

Theorem 6.4.1 implies that K C L is Galois with Galois group Gal(L/K) Sn.

Furthermore, if we identify Gal(L/K) with then o 5,, becomes the automorphismof L = F(xi, . . . ,X,,) that permutes the according to a.

Then the Fundamental Theorem of Galois Theory implies the following factsabout symmetric functions. As above, set

= [J(xj -xi).i<j

Theorem 7.4.4 Let R E F(xi,.. . be a rationalfunction.(a) R is invariant under S,, and only ifRE F(ai,...


(b) Assume that F has characteristic $ 2. Then R is invariant under An if and onlyif there areA,B E F(ai,. such that

Proof: Since (7.17) is a Galois extension, Theorem 7.3.1 implies that K is the fixedfield of Gal(L/K) = acting on L. This proves part (a).

To prove part (b), let M = LA,, be the fixed field of An acting on L. Since An hasindex 2 in Gal(L/K) = Theorem 7.3.1 implies that K C M is an extension ofdegree 2. However, (7.16) shows that EM, so that we have

K C C M.

By the Tower Theorem, it follows that

2 = [M: K] = {M: :Kj.

But (7.16) also shows that K (since F and hence K have characteristic 2).We conclude that M. Finally, since is a primitive element of thedegree 2 extension K C M, Proposition 4.3.4 implies that

This completes the proof of part (b). •

In Chapter 2 we proved part (a) of Theorem 7.4.4 by first considering the casewhen R is a polynomial in Xi,... ,Xn (Theorem 2.2.2) and then doing the case whenR is a rational function (Exercise 8 of Section 2.2). The proof given above is muchshorter. This illustrates nicely the power of Galois theory.

On the other hand, if R is a polynomial in x1,... ,Xn, then part (a) of Theorem 7.4.4does not assert that R is a polynomial in the at—the theorem only tells us that R is inthe field F (o',... , an). The point is that Galois theory deals with fields rather thanrings. In Exercise 3 you will study what happens in part (b) of Theorem 7.4.4 whenR is a polynomial in x1,... ,Xn.

C. The Inverse Galois Problem. The Galois group Gal(L/F) ofa finite extensionF C L is a finite group. But what finite groups can arise in this way? We will discusstwo aspects of this question.

We first note that a finite group G of order n is isomorphic to a subgroup of Sn.The easiest way to see this is to label the elements of G as gi,.

. . , Then the groupoperation on G can be represented by its Cayley table, where the entry in row i andcolumn j is g,g1. For example, the Cayley table of S3 is

______

e (123) (132) (12) (13) (23)e e (123) (132) (12) (13) (23)

(123) (123) (132) e (13) (23) (12)(7.18) (132) (132) e (123) (23) (12) (13)

(12) (12) (23) (13) e (132) (123)(13) (13) (12) (23) (123) e (132)(23) (23) (13) (12) (132) (123) e


In 1854 Cayley observed that every row of the Cayley table of G is a permutationof the elements of G (do Exercise 4 if you didn't prove this in your abstract algebracourse). It follows that for the ith row, there is a permutation E (where n =

I

such that

(7.19) gig1 =

In Exercise 5 you will compute the six elements of S6 given by the rows of (7.18),and in Exercise 6 you will show that in general, the map G —+ given by gj H—*

is a one-to-one group homomorphism. It follows that G is isomorphic to a subgroupof Combining this with the Galois correspondence for the universal extension indegree n gives the following nice result.

Theorem 7.4.5 Given a finite group G, there is a Galois extension whose Galoisgroup is isomorphic to G.

Proof: Let G be a finite group of order n, and let F be an arbitrary field. We knowthat the universal extension in degreee n,

K=F(t71,...,c7n)CL=F(xi,...,xn),

is a Galois extension with Galois group Gal(L/K) Sn.Since G is isomorphic to a subgroup of it follows that G is also isomorphic

to a subgroup H c Gal(L/K). Then the fixed field of H is an intermediate fieldK c L11 c L, and the Fundamental Theorem of Galois Theory tells us that LH C L isa Galois extension with Galois group

Gal(L/LH) = H G.

This shows that LH C L is the desired extension. •

However, this is not the end of the story, for in the extension LH C L constructedin Theorem 7.4.5, the smaller field LH depends on the group. In explicit examples,one is often interested in Galois groups of polynomials over Q. Thus the question is:Which finite groups can occur as the Galois group of a finite extension of Q? This is

called the inverse Galois problem for Q.There has been a lot of work on this problem, starting with Hilbert, who used his

irreducibility theorem (mentioned in the Mathematical Notes to Section 6.4) to showthat for every n 1, both and An can occur as Galois groups of Galois extensionsof Q. In Section 6.4 we also gave the example of x — 1, whose Galois groupover Q is Sn for n 2. Another example is the polynomial

Pn(X) = 1 +x+ + +

obtained by truncating the power series for eX. In 1930, Schur proved that the Galoisgroup of Pn over Q is An when n 0 mod 4 and otherwise (see [Chebotarev,p. 398] for references and further examples). In the case of a prime p. the paper [7]


uses elementary methods to construct a polynomial of degree p whose Galois groupover Q is The book Inverse Galois Theory [5] discusses some of the powerfulmethods used to study this unsolved problem in general. See also [1].

Historical Notes

The universal extension in degree n,

plays a central (though somewhat implicit) role in Lagrange's 1770 treatise on solvingequations by radicals (see the Historical Notes to Section 1.2). In his paper, Lagrangeproves many interesting results, including a theorem that (in modem terminology)says that if E L, then the intermediate field K C C L satisfies

= LGaI(LIK(W)).

For us, this is part of the Galois correspondence, yet Lagrange proved this result 60years before Galois. We will discuss Lagrange's work in more detail in Chapter 12.


Exercise 1. Give a detailed proof of Proposition 7.4.2.

Exercise 2. Compute the Galois groups of the following cubic polynomials:(a) x3—4x+2overQ.(b)

x3 —3x+ 1 overQ.(d) x3 t over C(t), t a variable.(e) x3 — t over Q(t), t a variable.

Exercise 3. This exercise will study part (b) of Theorem 7.4.4 when f is a polynomialin Xl,. . . that is invariant under An. The theorem implies that f = A +

B E F(ai,. . . , o,,). You will prove that A and B are polynomials in the 0. Recall that F is afield of characteristic 2.

(a) Show thatf+(12).f=2A.(b) In part (a), the left-hand side is a polynomial while the right-hand side is a symmetric

rational function. Use Theorem 2.2.2 to conclude that A is a polynomial in the 01.(c) Let P denote the product of f —A and (12). (f—A). Show that P =(d) Let B = u/v, where u, v E F[ai, . . . , c,,,] are relatively prime (recall that F[o1,. . .

,a,,] is

a UFD). In Exercise 8 of Section 2.4 you showed that is irreducible in F[ai, . . . ,an].Use this and the equation v2P = to show that v must be constant. This will provethatBEF[ai,...,a,,].

Exercise 4. Let G be a group of order n, and fix g E G.(a) Show that the map G G defined by h i—* gh is one-to-one and onto.(b) Explain why part (a) implies that each row of the Cayley table of G is a permutation of

the elements of G.

AUTOMORPHISMS AND GEOMETRY (OPTIONAL) 173

(c) Write G = {gi,... ,gn}, and fix g E G. Use part (a) to show the existence of E

satisfying gigj = as in (7.19).

Exercise5. LabeltheelementsofS3 asgi =e,g2 = (123),g3 = (132),g4 = (12),g5 = (13),and g6 = (23). Write down the six permutations a E S6 defined by the rows of the Cayleytable (7.18).

Exercise 6. In the situation of Exercise 4, let G = {gi,. . . , }, and assume that gigj = Let0k E Sn be the corresponding permutations determined by (7.19).

(a) Prove that = ak.(b) Prove that the map G —+ defined by g a1 is a one-to-one group homomorphism.

Exercise 7. Let f and F C L satisfy the hypothesis of Proposition 7.4.2, and assumethat F. Prove that Gal 75)) = 7L/3Z and that f is irreducible overF(

7.5 AUTOMORPHISMS AND GEOMETRY (OPTIONAL)

This optional section will explore some unexpected connections between geometryand Galois theory.

A. Groups of Automorphisms. The theory developed in Chapters 6 and 7 beginswith an extension F C Land then considers its Galois group Gal(L/F). We will nowchange our point of view and instead begin with a field L and a finite group G ofautomorphisms of L.

Here are two simple examples.

Example 7.5.1 Let L = Q((7), and consider the automorphism o• of L that mapsto Since 1 mod 7, we see that G = (a) = {1L,0-,a2} Z/3Z is agroupofautomorphisms of L.

Example 7.5.2 Let L = C(t), where t is a variable. It is easy to see that t i—* 1/tinduces an automorphism a of L. This gives the group G = (a) = { li., a} Z/2Zof automorphisms of L. <II>

Given a finite group G of automorphisms of a field L, we get the fixed field

LG C L.

Furthermore, the definition of LG easily implies that G C Gal(L/LG). However, muchmore is true, as we will now prove.

Theorem 7.5.3 Let G be a finite group of automorphisms of afield L. Then:(a)

(b) LG C L is a Galois extension.(c) Gal(L/LG) = G.

Proof: Let n = IGL We first claim that

(7.20) [L:LG]<n.


If (7.20) is false, then we can find as,. . . , a,,+i L that are linearly independentover LG. Also let the elements of G be = lL,a2, . . . ,o-,,. Then, given unknownsXi,.. consider the equations

=0,(7.21)

=0.

This is a system of n homogeneous equations in n + 1 unknowns with coefficients inthe field L.

Since the number of unknowns exceeds the number of equations, (7.21) must havea nontrivial solution (Xi,... = in Among all nontrivialsolutions in L1, pick one that has the fewest nonzero fit's. Relabeling, we canwrite this solution as (fit,... ,fir, 0,... , 0), where flu,... ,fir are nonzero. Then beinga solution of (7.21) means that

i1,...,n.Observe that r> 1 since 0 and 0. Furthermore, we may divide by flu andrelabel fi2, . . . ,fir to obtain

(7.22) aj(cu)+fi2Tj(a2)++/3raj(ar)=0, i=1,...,n.

Since cr1 = setting i = 1 in (7.22) gives

Hence fi2,. . . , fir cannot all lie in LG since a i,... , are linearly independent over

LG by assumption. Relabeling, we may assume that fir LG, so that a(/3r) fir forsome a E G. Now apply this a to (7.22) to obtain

aaj(au)+a(fl2)aaj(a2)+...+a(flr)aaj(ar) =0, i= l,...,n.

Since G = {a1,. . . is a group under composition, the product aa1 gives allelements of G as we vary cr,. Thus we obtain

(7.23) aj(ai)+a(fi2)aj(a2)+••.+a(fir)aj(ar) =0, i= 1,...,n.

Now multiply the equations of (7.22) by a(fir) and the equations of (7.23) by firand subtract. This choice of multipliers causes the coefficients of aj(ar) to cancel.Hence we are left with the equations

(a(fir) — fir) + (a(fir)fi2 fira(fi2))

fira(fir_i))ai(ar_i)0, i= 1,...,n.


Thus the (n + 1 )-tuple

(cx(fir) ,0)

is a solution of (7.21). It has at most r— 1 nonzero entries and is nontrivial, since13r. This contradicts our choice of r and completes the proof of (7.20).

It follows that LG C L is a finite extension. Furthermore, we have

LG C C LG,

where the first inclusion follows because elements of Gal(L/LG) are the identityon LG, and the second follows from G c Gal(L/LG). We conclude that LG is thefixed field of Gal(L/LG), and then Theorem 7.1.1 implies that LG C L is normal andseparable and hence is a Galois extension. This proves part (b) of the theorem.

SinceLG CL is Galois, Theorem 7.1.5 implies that [L:LGJ = IGal(L/LG)L Com-bining this with (7.20), we have

Gal(L/LG)l = [L:LGI <n= IGal(L/LG)I,

since G C Gal(L/LG). From here, parts (a) and (c) follow easily. •

Here is an example of how to use this theorem.

Example 7.5.4 In Example 7.5.2 we considered the group of automorphisms ofL=C(t)givenbyG= (o-), wherea(t) = l/t. Sincet+r' isobviouslyfixedbyG,we have

(7.24) C(t+t')CLGCL=C(t).

However, t is a root of (x—t)(x—t') =x2 (t+f')x+ 1 E C(t+t')[x]. Fur-thermore, the inclusions

C(t) C C(t+r')(t) C C(t)

show that = C(t). Thus C(t) is obtained by adjoiningt toSince t is a root of a quadratic equation with coefficients in C(t + t we have

[C(t) C(t < 2.

Theorem 7.5.3 implies that [L : LG] GI = 2. Using (7.24) and the Tower Theorem,it follows easily that LG = C(t + t').

B. Function Fields in One Variable. Example 7.5.4 is a Galois extensionconstructed from the field C(r) of rational functions in the variable t with coefficientsin C. More generally, the function field F(t), where F is an arbitrary field, has someinteresting subfields as follows.

Proposition 7.5.5 Assume that E F(t) is a rational function not in F, and writea = a(t)/b(t), where E F[t] are relatively prime. Then:


(a) a is transcendental over F.

(b) The polynomial a(x) — ab(x) E F(a){x] is irreducible over F(a).

(c) F(a) C F(t) is a finite extension of degree

[F(t) :F(a)] = max(deg(a),deg(b)).

Proof: If a is algebraic over F, then a satisfies an equation

+ = 0,

where n 1 and as,... e F. Substituting a = a(t)/b(t) into the above equation

and multiplying by b(t)" gives

(7.25) + + . . . + = 0

in the polynomial ring F[tI. This implies that

= b(t) (— . .

—1)

Since a(t) and b(t) are relatively prime, b(t) must be constant, say b0 E F. Then

substituting b(t) = b0 into (7.25) gives

a(t) is also constant (can you explain why?), say a0 E F.Then a = a(t)/b(t) = ao/bo e F, which is a contradiction. Part (a) follows.

For parts (b) and (c), first observe that since a is a rational function of t, we have

F(a) C F(a)(t) = F(a,t) = F(t).

In other words, F(t) is obtained by adjoining t to F(a). Thus [F(t) : F(a)] is the

degree of the minimal polynomial of t over F(a). To find the minimal polynomial, we

will use the relatively prime polynomials a (t), b(t) E F [t] appearing in a = a(t)/b(t).Consider the polynomial in x defined by a(x) — ab(x). Then:

• a(x) — ab(x) is a polynomial in x with coefficients in F(a).

• t is a root of a(x) — ab(x) since a(t) — ab(t) = a(t) — = 0.

• If a(x) = a0f +... and b(x) = +... have degrees n and m, then

a(x)—ab(x)=(aof+...)—a(boxm+...).

None of the coefficients can cancel because a F. Hence the degree of x ina(x) — ab(x) is max(n,m) = max(deg(a),deg(b)).


Now suppose that a(x) cth(x) is irreducible over F(a). Then the above bulletsimply that

[F(t):F(a)] =thedegreeofxina(x)—ab(x)= max(deg(a),deg(b)).

Thus part (c) of the proposition follows from part (b).To prove part (b), we begin in the polynomial ring F[x,y] for variables x,y. By

Theorem 2.1.1, F[x,yJ is a UFD. We first claim that a(x) —yb(x) is irreducible inF[x,y]. This is easy to see, for if

a(x) —yb(x) =AB, A,B E F[x,yI,

then A and B can't both have positive degree in y. We may assume that A E F[x].In Exercise 1 you will show that this implies that A divides a(x) and b(x). HenceA is constant, since a(x),b(x) are relatively prime. This proves that a(x) —yb(x) isirreducible in F[x,y].

Now consider a(x) —yb(x) as a polynomial in F(y)[xJ, i.e., as a polynomial inx with coefficients in F (y). We claim that it is irreducible over F (y) because it isirreducible in F[x,yj. This can be proved several ways—Exercise 2 uses Gauss'sLemma, and Exercise 3 gives a more elementary proof.

To apply this to our situation, recall that a is transcendental over F by part (a).This means that a can be regarded as a variable over F. Hence y a induces aring isomorphism F(y)[xJ F(a)[x] that takes a(x) —yb(x) to a(x) — ab(x). Thenthe previous paragraph implies that a(x) —a b(x) e F(a) [x] is irreducible over F (a).This completes the proof of the proposition. U

Here is an example that illustrates Theorem 7.5.3 and Proposition 7.5.5.

Example 7.5.6 Consider the automorphisms a and r of C(t) defined by

a(a(t)) = and T(a(t)) =

where = and a(t) is an arbitrary rational function in C(t). It is easy to seethat a has order n and i- has order 2. Furthermore, the computation

roaor(a(t)) = Toa(a(t')) =

shows that r o a o T = a a group G of automor-phisms of C(t) isomorphic to the dihedral group of order 2n. If we let L =then Theorem 7.5.3 implies that

LG c C(t)

is a Galois extension of degree 2n with Galois group isomorphic toTo describe this extension more explicitly, let

+1EC(t).


Proposition 7.5.5 implies that C C(t) also has degree 2n. Since + t'1is invariant under the action of o and T, we have extensions

+ c LG c C(t)

where C(t) has degree 2n over both smaller fields. Thus = LG, and

+ c C(t)

is a Galois extension with Galois group isomorphic to

C. Linear Fractional Transformations. Given a field F and a variable t,

F C F(t)

is an extension of infinite degree, so that the theory developed in previous sectionsdoesn't apply. But we can still define Gal(F(t)/F) to consist of all automorphismsof F(t) that are the identity on F. We will describe this group using matrices.

Let GL(2, F) be the group of 2 x 2 invertible matrices with entries in F. Then

(aEGL(2,F),

gives a rational functionat + b

EF(t).Ct + d

This relates to Gal(F(t)/F) as follows.

Theorem 7.5.7 Let F c F(t) be as defined above.(a) For 'y GL(2,F), the function F(t) —* F(t) defined by

is an automorphism that is the identity on F. Thus E Gal(F(t)/F).(b) The map -y F—* defines a group homomorphism GL(2,F) —+ Gal(F(t)/F).(c) The homomorphism of part (b) is onto, and its kernel consists of all nonzem

multiples of the identity matrix.

Pmof: For part (a), we first observe that the evaluation map sending t to E F(t)induces a ring homomorphism

F[t]—*F(t).

This map is one-to-one, since is transcendental over F by Proposition 7.5.5.Thus it extends to the field of fractions of F [tI, which gives the map

F(t) —* F(t)

described in the statement of the theorem. Hence is a one-to-one homomorphism(see Exercise 2 of Section 3.1). Furthermore, the image of is the subfield F

By Proposition 7.5.5, the extension C F(t) has degree

max(deg(at+b),deg(ct+d)) = 1,

since a or c is nonzero. Thus = F(t), so that is onto. It follows thatE Gal(F(t)/F), since it is obviously the identity on F.


This proves part (a) and shows that we have a map

GL(2,F) —* Gal(F(t)/F)

defined by 1ey) = The inverse in the definition of 1 is needed to make it agroup homomorphism, as you will prove in Exercise 4. Part (b) follows.

For part (c), let a E Gal(F(t)/F). Then = for all cE(t) E F(t),since a is the identity on F. This implies that F(a(t)) = a(F(t)) = F(t). Settinga(t) = A(t)/B(t), where A(t),B(t) E F[t] are relatively prime, we can write this asF(A(t)/B(t)) = F(t). By Proposition 7.5.5, it follows that

max(deg(A),deg(B)) = 1.

ThusA(t) = at+b and B(t) = ct+d, where a,b,c,d F. In Exercise 5 you willshow that

(7.26)(a GL(2,F).

Since a(t) = it follows that a == where is the matrix (7.26). Thisproves that 1' is onto.

Finally, take 'y E GL(2, F) in the kernel of (I). Then = so that

at + bt=a -(t)=ct+d

in F(t), where = Clearing denominators and collecting terms gives

Ct2 + (d a)t b = 0

in F [t]. Thus c = b = 0 and a = d, which shows that -y is a nonzero multiple of theidentity matrix. Since all nonzero multiples of the identity are in the kernel (checkthis), we get the desired description of the kernel. •

If '2 is the 2 x 2 identity matrix, then F*12 consists of all nonzero multiplesof the identity. Then the Fundamental Theorem of Group Homomorphisms andTheorem 7.5.7 imply that we have an isomorphism

GL(2,F)/F*12 Gal(F(t)/F).

The quotient group GL(2, F)/F*12 is denoted PGL(2, F). Thus

PGL(2,F) Gal(F(t)/F).

The group PGL(2,F) will play an important role in what follows. The "PGL" inPGL(2, F) stands for projective linear group. We will learn more about projectivelinear groups in Chapters 13 and 14.


0. Stereographic Projection. Let F be a field. Then define

F=FU{oo},where is a formal symbol that stands for the "point at infinity" of F. Given

/a b'\d) eGL(2,F) and aEF,

we get [-ri e PGL(2,F) = GL(2,F)/F*12 and set

aa+b(7.27)

ccx + d

In Exercise 6 you will show that this gives a well-defined action of PGL(2, F) onFurthermore, in Exercise 7 you will prove the following.

Proposition 7.5.8 Let (a! , a2, a3) and ,/32, be triples of distinct points ofF.Then there is a unique [y] E PGL(2, F) such that

[y] a1 =

fori=l,2,3.We will assume that F = C for the rest of the section. This will allow geometry

to give some interesting Galois groups. It is customary to call C = C U {oo} theRiemann sphere because we can map the unit sphere S2 C R3 to C by stereographicprojection.

The unit sphere S2 is defined by x2 +y2 + z2 = 1 in 1R3, and we identify a + bi E Cwith (a,b,0) so that C becomes the plane z = 0 in Then define

as follows. Given P E \ {(0,0, l)}, we draw the line connecting P to the northpole (0,0, 1) and define 7r(P) to be the point where the line intersects the xy-plane.Here is the picture:


In this picture, the xy-plane is shaded gray and the top half of S2 is shown as a wireframe. In Exercise 8 you will show that

fa b \ a b(7.28) ir(a,b,c)= (—,————,0) =—+i————,

1—c / 1—c 1—c

where the last equality uses the above identification of C with the xy-plane in R3.Under the map ir, the south pole (0,0, —1) maps to 0 C and the equator of thesphere maps to the unit circle {z e C I zI = 1}.

We then extend ii- to stereographic projection

*: —p C

bydefining*(0,0,l)=ooand*(P)=7r(P)forPES2\{(0,0,1)}. Notethat*isone-to-one and onto.

The key geometric property of * is the following. Consider a rotation r of S2about some axis through the origin. This gives a map

r: S2 —4 S2.

We then obtain the map

:C—+C

by composition. The remarkable fact is that this map is given by a linear fractionaltransformation. To state our result more precisely, let

Rot(S2)

be the group of all rotations of the sphere. A careful description of this group can befound in [9, Sec. 2.2], together with a proof of the following important result.

Theorem 7.5.9 Given r E Rot(S2), there is a unique [-y] PGL(2,C) such that

*oro*'(z) =

for all z E C. Furthermore, the map

Rot(S2) —4 PGL(2,C)

defined by r [-II is a one-to-one group homomorphism.

Here is an example of this result.

Example 7.5.10 Consider the octahedron with vertices at (± 1,0,0), (0, ± 1,0), and(0,0, ± 1) in R3. This is inscribed in the sphere S2 and when combined with stereo-graphic projection gives the picture


This shows the top half of the octahedron. Note that (1,0,0) becomes 1 E C and(0, 1,0) becomes i C, so that the six vertices of the octahedron map to the points

0,±1,±i,oo E C.

Now consider the rotation r1 of S2 by 180° about the x-axis. By Theorem 7.5.9,this gives an element of PGL(2, C) that takes 1 to itself and interchanges 0 and oo.However, note that

(0 1

0

satisfies

for z E C. This also takes 1 to itself and interchanges 0 and oc. Since [-y1] is uniquelydetermined by its values on 0, 1, oc (this is Proposition 7.5.8), it follows that r1corresponds to [-y1].

Similarly, consider the rotation r2 ofS2 by 90° counterclockwise about the z-axis.In Exercise 9 you will show that r2 corresponds to ['y2], where

(i 0

Note thatZ = iZ

forz E C.Finally, consider the rotation r3 ofS2 about the y-axis that takes (0,0, 1) to (1,0,0).

Under stereographic projection, this corresponds to an element E PGL(2, C) thatfixes ±i and maps

00 —+1 —+0 -+ —l —+00 —*..'.


Be sure that you can see this in the above picture. In Exercise 9 you will show that

(1 —l

1

Thusz— 1

. Z =

forz E C.In Exercise 10 you will show that the rotation group of the octahedron is isomorphic

to S4 and is generated by the rotations r1, r2, and r3. By the isomorphism ofTheorem 7.5.9, it follows that the subgroup

G = ['y3}) C PGL(2,C)

is isomorphic to S4. By Theorem 7.5.7, we can regard G as a group of automorphismsof L = C(t), and then Theorem 7.5.3 shows that

LGcL=C(t)

is a Galois extension with Galois group G S4.

Example 7.5.10 will have an unexpected application in Chapter 14.

Mathematical Notes

There are many ideas in this section to discuss.

• Finite Subgroups of Linear Fractional Transformations. Theorems 7.5.3and 7.5.7 show that any finite subgroup G C PGL(2, F) gives a Galois extension

LG cL=F(t)

with Galois group G. The remarkable fact is that for many fields F, the finitesubgroups G C PGL(2,F) have been classified.

For example, when F = C, G is isomorphic to one of the following groups:

A cyclic group C,, of order n, n � 1

A dihedral group D2,, of order 2n, n> 2

(7.29) The alternating group A4 of order 12

The symmetric group 54 of order 24

The alternating group A5 of order 60.

Furthermore, two finite subgroups of PGL(2, C) are conjugate in PGL(2, C) if andonly if they are isomorphic as abstract groups. Proofs of these assertions can befound in [2, Sec. 2.13] and [9, Secs. 2.2, 2.61.


Even more remarkable is the geometric origin of these subgroups of PGL(2, C).We saw in Example 7.5.10 how the octahedron gave a subgroup G C PGL(2,C)isomorphic to S4. In a similar way, the tetrahedron and icosahedron give subgroupsisomorphic to A4 and A5, respectively. Furthermore, you will show in Exercise 12how to realize and as the symmetry groups of polyhedra. Thus polyhedra giveGalois groups!

Chapter 14 will discuss the subgroups of PGL(2,F) when F is a finite field.

Invariant Theory. Examples 7.5.6 and 7.5.10 gave extensions with Galois groupsand S4. respectively. For the extension was given explicitly as

c C(t),

while for S4, we merely wrote

(7.30) LGcL=C(t),

In fact, one can show that for the group G = C PGL(2,C) of Exam-ple 7.5.10, the extension (7.30) is given by

c C(t).

While the invariance of + t" under is obvious, it is not at all clear that

(t8 + 14t4+a =—

is invariant under G ([-y1], ['y3]). But even if a is invariant, how does one finda in the first place? The answer involves invariant theory and is beyond the scopeof this book. But as a small hint of where a comes from, we note that t8 + 14t4 + 1has the following geometric description. The octahedron has eight faces that areequilateral triangles. If we project from the center of the sphere, the center of eachface gives a point on S2. Under stereographic projection, these give eight points of C.Then the roots oft8 + 14t4 + 1 are precisely these eight points.

Invariant theory gives similar formulas for the extensions coming from the tetra-hedron and icosahedron. Complete details can be found in Chapter 3 of [9].

• Lüroth's Theorem. The above paragraph suggests that for any finite subgroupG C PGL(2,C), there is a E C(t) such that if L = C(t), then

LGZC(a).

In fact, more is true: given any field F and any intermediate field F C K C F(t)with [F(t) : K] <00, there is a F(t) such that K = F(a). This result is known asLüroth's Theorem. Proofs may be found in [6, Sec. 9.5] and [9, Sec. 6.3].

• The Quintic and the Icosahedron. When we use the rotations of the icosahedron,we get a Galois extension

LGcL=C(t)


with Galois group G By [9, Sec. 4.8], there is W E C(t) such that

LG=C(W)

and C(W) C C(t) is the splitting field of the irreducible quintic

— lOWx3 +45W2x2 — W2 E C(W)[x].

This is called a Brioschi quintic.The results of Chapter 8 will imply that the Brioschi quintic is not solvable by

radicals over C(W). However, in the nineteenth century it was discovered that(roughly speaking) the solution of any quintic can be reduced to solving the Brioschiresolvent. This involves a rich collection of ideas described by Klein in [4] and morerecently in the book [91 and the poster [10].

The Galois group of an irreducible quintic will be computed in Chapter 13.


Exercise 1. Let P,Q E F[x,y) be polynomials such that PIQ and P E F[x], and write Q =ao(x)+aI(x)y+a2(x)y2+...+am(x)ym. ProvethatPlafori=0,...,m.

Exercise 2. In the proof of Proposition 7.5.5, we showed that a(x) — yb(x) is irreducible inF [x, y] and we want to conclude that it is also irreducible in F(y) [x]. Prove this using theversion of Gauss's Lemma stated in Theorem A.5.8.

Exercise 3. The proof of Proposition 7.5.5 shows that a(x) — yb(x) is irreducible in F{x,yI. Inthis exercise, you will give an elementary proof that a(x) — yb(x) is irreducible over F(y) [xi.Suppose that

a(x) —yb(x) =AB, A,B E F(y)[xI.

You need to prove that A or B is constant, which in this case means that A or B lies in F(y).(a) Show that there are nonzero polynomials g(y), h(y) E F [y] that clear the denominators of

A and B, i.e., g(y)A =A1 and h(y)B = B1 for someAi,B1 E F[x,y].(b) Show that g(y)h(y)(a(x) —yb(x)) = A1B1 in F[x,yJ and explain why a(x) —yb(x) must

divide either A1 or B1 in(c) Assume that A1 = (a(x) — yb(x))A2, where A2 E F[x,y]. Show that this implies that

g(y)h(y) = A2B1, and then conclude that B1 E F[y].(d) Show that B E F(y).

Exercise 4. Prove that the map 1 GL(2, F) —* Gal(F(t)/F) defined in the proof of Theo-rem 7.5.7 is a group homomorphism.


Exercise 6. In this exercise, you will prove that PGL(2,F) acts on = F Li {oo}.(a) First show that

aa+b (a bd

defines an action of GL(2, F) on Explain carefully what happens when a = oc.

(b) Show that nonzero multiples of the identity matrix act trivially on F, and use this to givea careful proof that (7.27) gives a well-defined action of PGL(2, F) on F.


Exercise 7. Proposition 7.5.8 asserts that we can map any triple of distinct points of F to anyother such triple via a unique element [-y] E PGL(2, F). We will defer the proof of existenceof [-y] until Exercise 24 in Section 14.3. In this exercise, we will prove the uniqueness part ofthe proposition, since this is what is used in Example 7.5.10.(a) First suppose that [-y] E PGL(2,F) fixes and also fixes two points a ofF. Prove

that is a nonzero multiple of the identity matrix.(b) Now suppose that {-yJ E PGL(2, F) fixes three distinct points of F, and let a be one of

these points. Show that there is [5J E PGL(2, F) such that [ö] a = 00. Then prove that'y is a nonzero multiple of the identity matrix by applying part (a) to 1.

(c) Show that the desired uniqueness follows from parts (a) and (b).

Exercise 8. Prove the formula (7.28) for stereographic projection.

Exercise 9. In Example 7.5.10, we considered rotations rI , T2, of the octahedron and definedmatrices y1, 'y2, y3 E GL(2, C). We also proved carefully that corresponds to [-y1] under thehomomorphism of Theorem 7.5.9. In a similar way, prove that r2 corresponds to [y2] and r3corresponds to [y3].

Exercise 10. The goal of this exercise is to prove that the symmetry group G of the octahedronis isomorphic to S4. By symmetry group, we mean the group of rotations that carry theoctahedron to itself. We think of G as acting on the octahedron.(a) Let v be a vertex of the octahedron. Use the action of G on v and the Fundamental

Theorem of Group Actions to prove that GI = 24.

(b) The eight face centers of the octahedron form the vertices of an inscribed cube. Explainwhy the octahedron and its inscribed cube have the same symmetry group.

(c) The cube has four long diagonals that connect a vertex to an opposite vertex. Explainwhy the action of G on these diagonals gives a group homomorphism G —+S4.

(d) Let , r2, r3 E G be the rotations described in Example 7.5.10. Explain how each rotationacts on the inscribed cube and describe its corresponding permutation in S4.

(e) Prove that the three permutations constructed in part (d) generate S4.(1) Use parts (a) and (e) to show that G Also prove that G is generated by rl , r2, r3.

See Section 14.4 for a different approach to proving that a group is isomorphic to S4.

Exercise 11. In Section 6.4, we defined the one-dimensional affine linear group AGL( 1, F,,)over the finite field F',,. More generally, if F is any field, then AGL( 1, F) consists of allfunctions : F —4 F defined by

aEF,

where a E F*, b E F, and the group structure is given by composition. In this exercise, youwill represent AGL(l,F) as a subgroup of PGL(2,F).(a) Show that the map 'Yab 'j'] defines a one-to-one group homomorphism

AGL(1,F) —÷ PGL(2,F).

(b) Consider the action of PGL(2, F) on P. Show that the isotropy subgroup of PGL(2, F)acting on 00 is the image of the homomorphism of part (a).

Exercise 12. In this exercise, you will construct polyhedra whose symmetry groups areisomorphic to and For consider the polyhedron whose vertices are the north andsouth poles of S2 together with the nth roots of unity along the equator. For n = 8, this givesthe following picture:


As usual, this shows only the top half of the polyhedron. Note that to obtain a three-dimensionalobject, we must assume n > 3.

(a) Show that the symmetry group of this polyhedron is isomorphic to when n 4 andS4 when n = 4.

(b) Now take the vertices on the equator and move them up in S2 so that they become thevertices of a regular n-gon lying in the plane z = c, where c> 0 is small. Prove that thesymmetry group of this polyhedron is isomorphic to

(c) Find polyhedra inscribed in S2 whose symmetry groups are C1 (the trivial group), C2, D4(the Klein four-group), and D8, respectively.

Notice that the symmetry groups of these polyhedra, together with those of the tetrahedron,octahedron, and icosahedron, give all of the groups listed in (7.29).

Exercise 13. Consider the automorphism ofL =C(t) defined by a(t) a(Q). This generatesa cyclic group G of automorphisms such that GI = n. Adapt the methods of Example 7.5.6to show that LG = C C(t) is a Galois extension whose Galoisgroup is cyclic of order n.

Exercise 14. Consider the automorphisms of L = F(t) defined by

a(a(t)) = and r(a(t)) = —t).

(a) Prove that a and r generate a group G of automorphisms of F(t) isomorphic to S3.(b) Show that G corresponds to the subgroup of PGL(2, F) consisting of all elements that

map the subset {0, 1,oo} CF to itself.(c) Prove that

LG =

______

and conclude thatC F(r)

is a Galois extension with Galois group G S3.


REFERENCES

I. C. U. Jensen, A. Ledet and N. Yui, Constructive Aspects of the Inverse Galois Problem,Cambridge U. P., Cambridge, 2002.

2. 0. A. Jones and D. Singerman, Complex Functions, Cambridge U. P., Cambridge, 1987.

3. B. M. Kiernan, The development of Galois theory fmm Lagrange to Artin, Arch. Hist.Exact Sci. 8 (1971), 40—154.

4. F. Klein, Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree,English translation by George Gavin Momce, Trübner & Co., London, 1888. Reprint byChelsea, New York, 1956.

5. G. Malle and B. H. Matzat, Inverse Galois Theory, Springer, New York, Berlin, Heidel-berg, 1999.

6. T. Mora, Solving Polynomial Equations Systems I, Cambridge U. P., Cambridge, 2003.

7. B. Osofsky, Nice polynomials for introductory Galois theory, Math. Mag. 72 (1999),2 18—222.

8. I. Radloff, Evariste Galois: principles and applications, Historia Math. 29 (2002), 114—137.

9. J. Shurman, Geometry of the Quintic, Wiley, New York, 1997.

10. Wolfram Research, Solving the Quintic, Poster, Wolfram Research, Champaign, IL, 1995.Available at http:/Ilibrary.wolfram.comlexamples/quintic/.

PART III

APPLICATIONS

The next four chapters give classic applications of Galois theory.Cardan's formulas lead to the notion of solvability by radicals. In Chapter 8, we

relate this to the notion of a solvable group and use Galois theory to show that ingeneral, polynomials of degree 5 cannot be solved by radicals.

Chapter 9 discusses the Galois theory of the cyclotomic extension Q Cwhere is a primitive nth root of unity. We also explain how Gauss analyzed thisextension when n is a prime.

Chapter 10, we study straightedge-and-compass constructions from the pointof view of Galois theory. This includes the classic Greek problems (trisecting theangle, duplicating the cube, squaring the circle) as well as the construction of regularpolygons. We also explore what happens when we go beyond straightedge andcompass to allow constructions using origami.

Finally, Chapter 11 explores the theory of finite fields. We consider the structureof finite fields and compute the Galois groups involved. We also describe irreduciblepolynomials and cyclotomic polynomials over a finite field.

CHAPTER 8

SOLVABILITY BY RADICALS

In this chapter, we will use the Galois theory developed in Chapter 7 to determinewhen a polynomial equation can be solved by radicals. The idea is to translate theproblem into group theory. Hence we begin with the group-theoretic concept ofsolvable group.

8.1 SOLVABLE GROUPS

Here is the basic definition of this section.

Definition 8.1.1 A finite group G is solvable there are subgroups

{e} = C .. C Gic Go =G

such that for i = 1, . . . , n we have:(a) G is normal in G1_1.(b) : G,] is prime.

Since G, is normal in part (b) of the definition can be replaced by theequivalent assertion that /G1 is a cyclic group of prime order.


192 SOLVABILITY BY RADICALS

We will show below that every finite Abelian group is solvable. Here is an exampleof a non-Abelian solvable group.

Example 8.1.2 The subgroups

{e} CA3 CS3

show that S3 is solvable, since each subgroup is normal in the next and the indices[A3: {e}] = 3, [S3 :A3] = 2 are prime. <11>

In Exercise 1 you will prove similarly that A4 and S4 are solvable. On the otherhand, we will see in Section 8.4 that and are nonsolvable for n � 5.

Here is our first result about solvability.

Proposition 8.1.3 Every subgroup of a solvable finite group is solvable.

Proof: Let G be finite and solvable with subgroups G, as in Definition 8.1.1. Givena subgroup H c G, set H, = fl H and note that

Ho=G0flH=GnH=H,=GnflH{e}flH={e}.

Then consider the group homomorphism

ir:H,_1

that sends h E H,_1 to the coset hG1 E G,_1 /G1. Observe that h E is in the kernelof if and only if hG, = G, which happens if and only if

hE H,_1n G, = (G,_1 nH) n G, = H n G,= H,,

where the second equality follows from G, C G_1. This shows that Ker(ir) = H,.Thus H, is normal in H1_1. By the Fundamental Theorem of Group Homomorphisms,we get an isomorphism

H1_1/H, =H,_i/Ker(ir) C G_1/G,.

Since is cyclic of prime order, it follows that H,_i/H, is either trivial orisomorphic to G,_1/G,. Thus either H11 = H1 or [H1_1 :H,] is prime. Hence, bydiscarding duplicates, the subgroups

{e} = ... C H,c H1_1C ... C H0 =H

show that H is solvable. .Here is one of the main theoretical tools for dealing with solvable groups.

Theorem 8.1.4 Let G be afinire group and H a normal subgroup. Then G is solvableand only H and G/H are.

Proof: First suppose that G is solvable. Then H is solvable by Proposition 8.1.3. Toshow that G/H is solvable, suppose that G, are subgroups of G as in Definition 8.1.1,and let ir: G —p G/H be the group homomorphism g gH. Then let G, = ir(G,). InExercise 2 you will show the following:

SOLVABLE GROUPS 193

(a) Go = G implies that G0 = G/H.(b) = {e} implies that = {eH} (where eH is the identity of G/H).

(c) G, normal in G_1 implies that G, is normal in G1_(d) The map —* G_1/G, given by gG, i—* ir(g)G1 is a well-defined onto

group homomorphism.

By assumption G,_1 /G, is a group of prime order. In Exercise 2 you will show thatthis fact and (d) imply that G,_1 /G, is either trivial or also has prime order. Thuseither G1_1 = G, or G] is prime. Then, as in the proof of Proposition 8.1.3,discarding duplicates among the subgroups

{eH} = ÔnC C ... C = G/H

shows that G/H is solvable.Conversely suppose that H and G/H are solvable. Let H,, i 0,... , £ be subgroups

of H satisfying Definition 8.1.1, and similarly let j = 0,... ,m be subgroups ofG/H satisfying the definition.

As above, we have ir : G —+ G/H. Given a subgroup K C G/H, set

(8.1) 7r'(K) {g E GI

ir(g) E K}.

In Exercise 3 you will verify that (K) is a subgroup of G. You will also checkthat the kernel of is

H ir' ({eH})

and thatG= ir1(G/H).

If we apply this to the subgroups

{eH} = Gm C .. C = GIN,

then we obtain the subgroups

H = C C = G.

However, we also have the subgroups

{e} = .. C H, C H,_1C ... C H0 = H.

We "glue together" these sequences of subgroups by defining G, C G to be

G — Jir'(Gj),'lHj_m,

Note that the sequences are "joined" at Gm = ir = H0 = H. It remains to showthat G, is normal in of prime index. For m < i < £ + m, this is obvious, since


G1 = Hi_rn in this range and the H1 satisfy Definition 8.1.1. For 0 i m, we leaveit as Exercise 4 to show that for this range of indices, G1 is normal in and that

(8.2)

Since the G, satisfy Definition 8.1.1, it follows that G is solvable. This completes theproof of the theorem. •

We next use Theorem 8.1.4 to show that Abelian groups are solvable.

Proposition 8.1.5 Every finite Abelian group G is solvable.

Proof: We will prove the proposition by complete induction on nI

The casen = 1 being trivial, assume that G is an Abelian group of order n> 1 and that theresult is true for all Abelian groups of order < n.

Let p be a prime divisor of IGI. If p = IGI, then G is cyclic of order p and hencesolvable. If p < IGI, then by Cauchy's Theorem (Theorem A.1.5), we can find g E G

of order p. Now let H = (g) be the subgroup generated by g. Then H is normal, sinceG is Abelian and = ii < IGI. It follows that the orders of H and G/H are strictlysmaller than GI = n. Hence H and G/H are solvable (by our inductive assumption),and then Theorem 8.1.4 implies that G is solvable. •

Here is an interesting non-Abelian solvable group.

Example 8.1.6 The one-dimensional affine linear group AGL( 1, 1Ff,) over wasintroduced in Section 6.4. There, the discussion leading up to (6.6) showed thatAGL( 1, lF,) has a normal subgroup T with quotient AGL( 1, lF'.

Since and are Abelian, they are solvable by Proposition 8.1.5, so thatAGL( 1, IF,,) is solvable by Theorem 8.1.4. We also know that AGL( 1, IF,,) is non-Abelian for p � 3 by part (a) of Exercise 10 of Section 6.4. This example will beimportant in Chapter 14.

Mathematical Notes

The definition of solvability is related to the ideas of simple groups, compositionseries, and the Jordan—Hölder Theorem. We will say more about these topics inSection 8.4. However, some standard results used to study solvable groups need tobe mentioned here.

Solvability and the Order of a Group. In some cases the solvability of a group isdetermined by its order. For example, in Exercise 5 you will prove the following.

Theorem 8.1.7 If p is prime, then every group of order pfl, n > 0, is solvable. •

In 1904, Bumside [4] generalized Theorem 8.1.7 as follows.

Theorem 8.1.8 If p and q are distinct primes, then every group of order pnqm,n,m > 0, is solvable. U

SOLVABLE GROUPS 195

In 1963, Feit and Thompson [5] proved the following surprising result.

Theorem 8.1.9 Every group of odd order is solvable. •

In spite of its simple statement, the proof of Theorem 8.1.9 uses some verysophisticated mathematics and takes 255 pages.

• Solvability and the Sylow Theorems. The Sylow Theorems imply some niceresults about solvability. As stated in Theorem A.5. 1, we have:

• (First Sylow Theorem) If p" is the highest power of a prime dividing the order ofa finite group G, then G has a subgroup of order pfl, called a p-Sylow subgroup.

• (Second Sylow Theorem) All p-Sylow subgroups of G are conjugate in G.• (Third Sylow Theorem) If G has N p-Sylow subgroups, then N 1 mod p and N

divides

Here are two examples of how the Third Sylow Theorem can be used to prove that agiven group is solvable.

Example 8.1.10 Let G be a group of order 14, and let N be the number of 7-Sylowsubgroups of G. Then N 1 mod 7 and Nj 14 by the Third Sylow Theorem. Itfollows easily that N = 1, so that G has a unique 7-Sylow subgroup H. Since anyconjugate of H is also a 7-Sylow subgroup, H coincides with its conjugates. Thus His normal, and then Theorem 8.1.4 easily implies that G is solvable.

Example 8.1.11 Let G have order 42. Arguing as in Example 8.1.10, one sees thatG has a normal 7-Sylow subgroup H. Then G/H has order 6, so that G/H Z/6Zor S3, both of which are solvable. Hence G is solvable by Theorem 8.1.4. 4>

In Exercises 6 and 7 you will combine similar arguments with Burnside's Theorem(Theorem 8.1.8) to show that all groups of order < 60 are solvable.


Exercise 1. Consider the groups A4 and S4.

(a) Show that {e, (l2)(34), (13)(24),(14)(23)} is a normal subgroup of S4.(b) Show that A4 and 54 are solvable.

Exercise 2. This exercise is concerned with the first part of the proof of Theorem 8.1.4.

(a) Prove assertions (a)—(d) made in the proof of the theorem.

(b) Suppose that : Mi M2 is an onto group homomorphism. If Mi = p, where p is

prime, then prove that 1M2I

= 1 or p.

(c) Explain how part (b) proves the assertion made in the text that G1_1 either is trivial or

has prime order.

Exercise 3. Consider the map ir: G —÷ G/H used in the proof of Theorem 8.1.4. Given asubgroup K C G/H, define as in (8.1).(a) Show that ir (K) is a subgroup of G containing H.(b) Show that H is the kernel of ir and that H =(c) Show that G =


Exercise 4. In the situation of (8.2), prove that G is normal in and that '—+

gives the isomorphism (8.2).

Exercise 5. In this exercise, you will prove Theorem 8.1.7. We begin with a classic resultfrom group theory about the center of a group of prime power order. Recall that the center ofa group G is the subset

Z(G) = {g E G I gh = hg for all h E G}.

Most courses in abstract algebra prove that Z(G) {e} when GI = p prime (see, forexample, [Herstein, Thm. 2.11.2]). You may assume this result.(a) In any group, show that (g) is normal for all g E Z(G).(b) Prove Theorem 8.1.7 using induction on n, where IGI = p is prime.

Exercise 6. In this exercise you will prove that groups of order 30 are solvable.(a) Use the method of Example 8.1.10 to prove that groups of order 10 or 15 are solvable.(b) Show that a group of order 30 is solvable if and only if it has a proper normal subgroup

different from {e}.(c) Let G be a group of order 30. Use the Third Sylow Theorem to show that G has one or

ten 3-Sylow subgroups and one or six 5-Sylow subgroups.(d) Show that the group G of part (c) can't simultaneously have ten 3-Sylow subgroups and

six 5-Sylow subgroups. Conclude that G must be solvable.See [Herstein, Sec. 2.12, Ex. 7] for further details on the structure of groups of order 30.

Exercise 7. Use Burnside's Theorem (Theorem 8.1.8) to show that groups of order < 60are solvable, with the possible exception of groups of order 30 or 42. When combined withthe previous exercise and Example 8.1.11, this implies that groups of order < 60 are solvable.In Section 8.4 we will prove that A5 is not solvable. Since A5 has order 60, it is the smallestnonsolvable group. (One can show that A5 is the only nonsolvable group of order 60 up toisomorphism.)

Exercise 8. Let G be a finite group, and suppose that we have subgroups

{e} = Gn C C G0 =G

such that G1 is normal in for i = 1,... ,n.(a) Prove that G is solvable if is Abelian for i = 1,... , n.

(b) Prove that G is solvable if /G1 is solvable for i = 1, . . . , n.

8.2 RADICAL AND SOLVABLE EXTENSIONS

The purpose of this section is to introduce the field theory needed to study solvabilityby radicals.

A. Definitions and Examples. The naive idea of solvability by radicals arisesfrom polynomials such as x3 + 3x + 1, whose unique real root is

by Example 1.1.1. This algebraic number is built by taking successive radicals.When we cast this in terms of fields, we are led to the following definition.

RADICAL AND SOLVABLE EXTENSIONS 197

Definition 8.2.1 Afield extension F C L is radtcal there are fields

F = F0 C F1c ...C

wherefori= l,...,n, there is'yj E F; with F; E m1 >0.

Notice that if we let b1 = E F;_i, then is an m1th root of b. This allows usto write = so that

b,EF,_1.

This shows that radical extensions are obtained by adjoining successive radicals.Here is our first example of a radical extension.

Example 8.2.2 For the field extension Q c + let = and 'y2 =

+ Then we have the extensions

Q

where=

= 2 eQ = = C Since

(be sure you can prove this), Q C Q (v"2 + is a radical extension.

An important observation is that some extensions are not radical but are containedin larger radical extensions. Here is an example.

Example 8.2.3 Let Q CL be a splitting field off=x3+x2—2x—-1 C Q[x]. InExample 7.4.3, we showed that f is irreducible over Q with discriminant

>0.

By Theorem 1.3.1, the roots off are real, which allows us to assume that L C R.Furthermore, since is a perfect square, Proposition 7.4.2 implies that Q C L isa Galois extension of degree 3. Cardan's formulas imply that Q C L is contained ina radical extension (see also Exercise 1).

However, the extension Q C L is not radical itself. We prove this as follows. IfQ C L were radical, then [L : Q] = 3 would imply that L = Q('y), where e Q forsome m > 3 (see Exercise 2 for the details). Then the minimal polynomial f of 'yover Q would divide xtm and have degree [L: QI = 3. Since Q C L is Galois, fwould split completely over so that three of 'y, . . would liein L. This is impossible, since L C R. Hence Q C L is not radical.

This example motivates the following definition.

Definition 8.2.4 A field extension F C L is solvable (sometimes called solvable byradwals) if there is afield extension L C M such that F C M is radical.


In this terminology, the extension Q C L considered in Example 8.2.3 is solvable,since it is contained in a radical extension.

B. Compositums and Galois Closures. In order to understand radical andsolvable extensions, we need to define the compositum of two or more subfields.

Definition 8.2.5 Suppose that we have afield Land two subfields K1 C Land K2 C L.

Then the compositum of K1 and K2 in L is the smallest subfield of L containing K1and K2. We denote the compositum by K1 K2.

In Exercise 3 you will show that the compositum always exists and that thecompositumofK1 CLandK2

(8.3)

For example, the compositum of and in R isWe next consider Galois closures. Proposition 7.1.7 proves that every finite

separable extension F C L has a Galois closure, which may be thought of as thesmallest Galois extension of F containing L. We can express the Galois closure as acompositum as follows.

Proposition 8.2.6 Suppose that F c L C M where F C M is Galois. Then thecompositum of all conjugate fields of L in M is the Galois closure ofF C L.

Pmof: The Theorem of the Primitive Element implies that L = F(a) for someE L. Since F C M is Galois, the minimal polynomial h of over F is separable and

splits completely over M, say h(x) = (x (x — ar), where

a Galois extension of F containing L. In Exercise 4 you will prove that F C Kis the Galois closure of F c L. In Exercise 4 you will also show that the conjugatefields of L in Mare F(a1) for i = 1,..., r. Then (8.3) implies that

F(ai)F(ar)=F(ci,...,ar)=K.

This proves that K is the compositum of the conjugate fields of L in M. •

C. Properties of Radical and Solvable Extensions. We begin with thefollowing useful lemma.

Lemma 8.2.7(a) 1fF CL andL CM are radical, then so is F CM.(b) Ifwe have F C K1 CL and F C K2 CL such that F C K1 is radical, then

K2 C K1 K2 is radical.

(c) Jfwe have F C C LandF C K2 C Lsuch thatF C K1 andF C K2 are radical,then F C K1 K2 is radical.

RADICAL AND SOLVABLE EXTENSIONS 199

Proof: Part (a) follows easily from the definition of radical extension by combiningthe sequences of fields used for F C L and L C M. We omit the details.

For part (b), we have fields F = F0 C F1 C C C F, =where e F1_1 for 1 <i < n. Then define fields

= K2,

(8.4)

F C K2 and induction to show that

(8.5)

for i = 0, . . . , n. This in turn implies that

E C

for i = 1,... ,n. It follows easily that K2 = F0' C C is radical. In Exercise 5 youwill show that is the compositum K1K2, which will prove part (b).

Finally, for part (c), note that K2 C K1K2 is radical by part (b). Then we are doneby part (a), since F C K2 is radical by assumption. U

We next use Proposition 8.2.6 to study the Galois closure of a radical extension.

Theorem 8.2.8 Ifan extension F C L is separable and radical, then its Galois closureis also radical.

Proof: Find an extension L C M such that F C M is Galois (such an extension existsby the existence of Galois closures). Given a E Gal(M/F), we get the conjugatefield F c aL c M. Exercise 6 shows that F c aL is radical because F C L is radical.

But once we that each conjugate field is radical over F, Lemma 8.2.7 tells usthat their compositum is also radical over F. Then we are done, since the compositumis the Galois closure by Proposition 8.2.6. •

The following corollary of Theorem 8.2.8 will be used in Section 8.5.

Corollary 8.2.9 If a finite extension F C L of characteristic 0 is solvable, then so isits Galois closure.

Proof: Since F c L is solvable, we have F C L C L' such that F C L' is radical.Furthermore, F c L' is separable (we are in characteristic 0) and hence has a Galoisclosure F c L' C M. Then F C M is radical by Theorem 8.2.8.

Now consider F C L C M. Since F C M is Galois, it contains the Galois closureof F C L by Proposition 8.2.6. Thus the Galois closure lies in the radical extensionF C M, so that the Galois closure is solvable by definition. U

In the next section, we will see how the solvable extensions defined here relate tothe solvable groups studied in Section 8.1.


Historical Notes

In 1824 Abel proved that the general quintic cannot be solved by radicals. Hepresented his proof in the privately printed Memoir on algebraic equations, in whichis demonstrated the impossibility of solving the general equation of the fifth degree[Abel, pp. 28—331 that he sent to the leading mathematicians of Europe. In thismemoir, Abel begins his proof as follows:

Let5 4 3 2

y —ay +by —cy +dy—e=Obe the general equation of the fifth degree and let us suppose that it is solvablealgebraically, that is, one can express y by a function formed by radicals of thequantities a, b, c, d, and e.

It is clear that in this case we can express y in the form

1

m being a prime number and R, p, P1, p2, etc. functions of the same form as y,and so on until we come to rational functions of the quantities a, b, c, d, and e.

(This is from the English translation in [9, pp. 155—169].) Abel's description of yin terms of radicals is a "top-down" version of the definition of radical extension.Definition 8.2.1 is a "bottom-up" approach that begins with the smallest field (herecontaining the coefficients a, b, c, d, e) and successively adds radicals. Abel insteadbegins with the largest field (here containing y) and successively strips away radicals.Another difference is that Abel focuses on individual elements rather than the fieldsin which they lie. Nevertheless, the above quotation contains a clear description of aradical extension. Be sure you understand this. I

The reader may wonder why Abel assumes that m is prime in We will see inLemma 8.6.2 that this is no restriction. Also note that Abel's "solvable algebraically"means "solvable by radicals" in modern terms. We will say more about Abel's proofin Sections 8.5 and 12.1.


Exercise 1. As in Example 8.2.3, let L be a splitting field of x3 +x2 — 2x 1 over Q. Also let= e.

(a) Show that the roots of x3 +x2 — 2x — I are 2cos(2jir/7) = + for j = 1,2,3.(b) Show that Q C L C Q((7), and explain why Q C Q((7) is radical.

Exercise 2. In the situation of Example 8.2.3, assume that Q C L is radical. Prove thatL = Q(-y) where -ym EQ for some m 3.

Exercise 3. Here you will prove two properties of compositums.(a) Prove that the compositum K1K2 exists.(b) Prove (8.3).

Exercise 4. This exercise is concerned with the proof of Propostion 8.2.6.(a) Show that K = F(aj,. . . , is the Galois closure of F C L.(b) Prove that the conjugates of L in M are the fields for i = 1,. . . , r.

SOLVABLE EXTENSIONS AND SOLVABLE GROUPS 201

Exercise 5. This exercise will complete the proof of part (b) of Lemma 8.2.7.(a) Prove (8.5).(b) Prove that the field defined in (8.4) is the compositum K1K2.

Exercise 6. Suppose we have finite extensions F C L C M and a E Gal(M/F), and assumethat F C L is radical. Prove that F C aL is also radical.

Exercise 7. Suppose that we have extensions F C K1 C L and F C K2 C L such that F C Kiand F C 1(2 are Galois. Prove that F C K1K2 is Galois. This will show that the compositumof two Galois extensions is again Galois.

8.3 SOLVABLE EXTENSIONS AND SOLVABLE GROUPS

The main question we will answer in this section is: When is a finite extension F C Lsolvable? Because of subtleties that can occur in characteristic p, we will make thefollowing simplifying assumption:

Alifields appearing in this section will have characteristic 0.

See Section 8.5 for an example to show what can go wrong in characteristic p.

A. Roots of Unity and Lagrange Resolvents. Section A.2 shows that givenone mth root of a complex number, we get the others by multiplying by the mth rootsof unity. Since radical extensions involve taking mth roots, it makes sense that rootsof unity will play an important role. However, the roots of unity in Section A.2 arecomplex numbers, while the fields considered here need not be subfields of C (eventhough they have characteristic 0). For this reason, we need to study roots of unityfor arbitrary fields of characteristic 0.

Given a positive integer m and a field L of characteristic 0, consider the splittingfield of xm — 1 over L. In Exercise 1 you will show that xm — 1 has m distinct roots inits splitting field. These roots form a group under multiplication, which is cyclic byProposition A.5.3. A generator ( of this group has the following two properties:

• Themdistinctrootsofxm— 1 are• The splitting field of xm — 1 over L is L( 1, (,••• , (rn—i) = L(().

We call ( a primitive mth root of unity in this situation. We claim that

(8.6) L C L(() is Galois and is Abelian.

To prove this, note that L C L(() is Galois, since L(() is the splitting field of theseparable polynomialxm — 1 E L[x]. Now suppose that a, r E Gal(L(()/L). Then o, rare determined by their values on (, and since the roots of xm — 1 are 1,(, . .

.

it follows that a(() = and r(() = for integers i, j.Thus

ar(() = = (o.(())f = ((i)J = (if

A similar computation shows that ra(() = = Then ar = ra, since ar andra are uniquely determined by their values on (. Hence is Abelian.


Given a Galois extension F C L and a primitive mth root of unity we get theextensions

L(()/ NL

F

We can relate the solvability of the various Galois groups as follows.

Lemma 8.3.1 Let F c L be a Galois extension, and be a primitive mth root ofunity. Then F c and c are also Galois, and

Gal(L/F) is solvable Gal(L(()/F) is solvable

is solvable.

Proof: In Exercise 2 you will prove that F C is Galois, which implies thatF(() C are also Galois. To prove the first equivalence, we use the extensions

F F C L are Galois, Theorem 7.2.7 implies that is anormal subgroup of Gal(L(<)/F) such that

Gal(L/F)

But is Abelian by (8.6) and hence solvable by Proposition 8.1.5. ThenTheorem 8.1.4 implies that is solvable if and only if Gal(L/F) is. Thisproves the first equivalence of the lemma.

For the second equivalence, consider the extensions

F C F in place of L), so that, arguing asabove, we get a group isomorphism

Gal(F is Abelian and hence solvable, and then Theorem 8.1.4implies that is solvable if and only if is. This provesthe second equivalence of the lemma. •

The following result will play a crucial role in our analysis of solvable extensions.The proof uses a clever construction due to Lagrange.

Lemma 8.3.2 Suppose that K c M is a Galois extension with Gal(M/K) 7Z/pZ,p prime. JfK contains a primitive pth root of unity then there is a E M such thatM=K(a)andaP EK.


Proof: By hypothesis, Gal(M/K) is cyclic of order p. Let a E Gal(M/K) be agenerator, and fix E M \ K. Then, for each i = 0, . . , p — 1, consider the Lagrangeresolvent defined by

(8.7) = + + + . . . + (a).

This easily implies that

= + ...

Since = 1 and is the identity, the final term on the right-hand side of the aboveequation simplifies to and then the equation becomes

=

so that

(8.8) =

Since K and 1, (8.8) easily implies that

a(4')

But a generates Gal(M/K), so that the above equation shows that is fixed bythe Galois group. Hence E K, since K C M is Galois. Also, when i = 0, (8.8)becomes a(ao) cEo, and then the argument just given shows that cEo E K.

Suppose for the moment there is some i between 1 and p — 1 such that 0. Forthese i's, we also have (' 1, and it follows that cEj. Combining this with(8.8), we conclude that a(cE1) so that K. This implies M = since[M : K] is prime. Then = has the desired properties, since E K.

It remains to consider what happens if = 0 for all i = 1, .. . , p — 1. In this case,we add up the equations (8.7) for i = 0,... ,p — ito obtain

aO

=

+

+ + + . . .

+ (-2(p-1) +

+ ($+ + C2(P_I) + . . . +

=

+ (1 + + + . . + +

+ (1 + . . .

In Exercise 3 you will show that

(8.9)


for i = 1,. . . , p — 1. It follows that the above formula for ao simplifies to

a0 = p13,

so that /3 = ao/p (remember that we are in characteristic 0). However, we provedabove that a0 E K, yet /3 K by assumption. This contradiction shows that at leastone of a1,. . . , is nonzero, which completes the proof of the lemma. •

B. Galois's Theorem. In Section 8.2, we showed that if F C L is solvable, thenwe can find an extension L C M such that F C M is Galois and solvable. For anarbitrary Galois extension, the wonderful fact is that the Galois group determineswhether or not the extension is solvable. The following theorem due to Galois is oneof the most important applications of Galois theory.

Theorem 8.3.3 Let F C L be a Galois extension. Then the following are equivalent:(a) F C L is a solvable extension.(b) Gal(L/F) is a solvable group.

Proof: We prove (a) (b) in three steps.

Reduction to the Radical Case. Since F C L is solvable, it lies in a radical extensionF c L'. By Theorem 8.2.8, the Galois closure F C M of F C L' is radical over F.Thus we have F c L c M where M is radical and Galois over F.

Suppose for the moment that Gal(M/F) is a solvable group. Since F C L isGalois, Theorem 7.2.7 implies that we have an isomorphism

Gal(L/F) Gal (M/F)/Gal(M/L).

Then Theorem 8.1.4 implies that Gal(L/F) is also solvable. Hence it suffices toprove that Gal(M/F) is solvable. In other words, we can assume that F C L is radicaland Galois.

Adjunction of Roots of Unity. Suppose that F C L is radical and Galois. If we adjoina primitive mth root of unity to both F and L, then part (b) of Lemma 8.2.7 impliesthat the resulting extension C L(() is radical, since L(() is the compositum of

and L. This extension is also Galois by Lemma 8.3.1. If we can show thatis solvable, then Lemma 8.3.1 will imply that Gal(L/F) is solvable.

Hence we can assume without loss of generality that F contains any mth root of unitywe want.

Proof of Solvability. Since F C L is radical, we have subfields

(8.10)

such that for i = 1,... ,n, we have F, = where E for some m, > 0.By the previous step, we can also assume that F contains a primitive m,th root ofunity for i = 1,. . . , n. In this situation, we claim that

(8.11) C F, is Galois with cyclic Galois group.


To prove this, note that 1, c,... , are the distinct m,th roots of unity, whichmeans that

are the distinct roots of — e [xl. Since E F C

F1 = is Galois. The proof that the Galois group iscyclic is fun and is left to the reader as Exercise 4. This completes the proof of (8.11).

We now prove solvability. Given the subfields (8.10), consider the subgroups

G = Gal(L/F1) C Gal(L/F).

Since the Galois correspondence is inclusion-reversing, (8.10) gives

{1L} = Gal(L/L) = = C C

C G1 C Go = Gal(L/Fo) = Gal(L/F).

Consider the extensions F1_1 C Fj C L. Then C L is Galois, since isan intermediate field of the Galois extension F C L. Furthermore, C Fj is alsoGalois by (8.11). Hence Theorem 7.2.7 implies that G1 is normal in G,_1 with

= Gal(L/F1_i)/Gal(L/F1) Gal(F,/F,_i).

By (8.11), we conclude that G,_1 /G, is Abelian. Since this is true for all i = 1,. . . , n,

part (a) of Exercise 8 from Section 8.1 implies that Gal(L/F) is solvable. Thiscompletes the proof of (a) (b).

It remains to prove (b) (a). We do this in two steps.

A Special Case. Let F C L be Galois with solvable Galois group. Assume in additionthat F satisfies the following special hypothesis:

(8.12) F has a primitive pth root of unity for every prime p dividing IGal(L/F) I.

We will prove that F C L is radical in this situation. Since Gal(L/F) is solvable,we have subgroups {1L} = C C Go = Gal(L/F) as in Definition 8.1.1. Thenconsider the fixed fields

F, = LG, C L.

Since the Galois correspondence is inclusion-reversing, this gives the fields

F = '-'GaI(L/F) = LG0 = F0 C Fi C

=L{1L} =L.

Furthermore, since G, is normal in the Galois correspondence together withTheorem 7.2.7 implies that

Gal(F,/F,_i).


Since : G1 is prime, 7L/pZ for a prime p. In Exercise 5 youwill prove that p divides Gal(L/F)I. By (8.12), F and hence contain a primitivepth root of unity.

It follows that C F, satisfies the conditions of Lemma 8.3.2. Thus F, isobtained from by adjunction of a pth root of an element of F,_ This provesthat F CL is a radical extension when F satisfies (8.12).

The General Case. Finally, we consider what happens when we only assumethat F C L is a Galois extension with solvable Galois group. In this situation,let be a primitive mth root of unity, where m = IGal(L/F)I. By Lemma 8.3.1,

is solvable since Gal(L/F) is.We relate the orders of these groups as follows. As in the proof of Lemma 8.3.2,

we have an isomorphism

Gal(L/F)

If you look back at the proof of Theorem 7.2.7, you will see that this isomorphismcomes from the homomorphism

—* Gal(L/F)

given by restricting an automorphism of to L. Since is asubgroup of we have a homomorphism

(8.13) —+ Gal(L/F)

also given by restriction to L. But the kernel of this map is the identity, since elementsof the kernel are the identity on both Land Thus (8.13) is one-to-one, whichby Lagrange's Theorem implies that

(8.14) m = IGal(L/F)I is a multiple of

Now let p be a prime dividing Then p divides m by (8.14).Since is a primitive mth root of unity, em/p is a primitive pth root of unity (seeExercise 6). Since em/p e we conclude thatF L replaced by F(() and respectively. It follows that C isradical by the Special Case. But F C is obviously radical (cv' = 1 E F), so thatF C is radical by part (a) of Proposition 8.2.7.

Since F C L in aradical extension of F. Hence F c L is solvable by definition. This completes theproof of the theorem. •

The proof of Theorem 8.3.3 implies that a solvable Galois extension becomesradical after adjoining a suitable root of unity. Here is the precise result.

Corollary 8.3.4 Let F C L be Galois and solvable, and let be a primitive mth rootof unity, where m = [L : F]. Then F C is radical.


Proof : If F C L is Galois and solvable, then Gal(L/F) is solvable. The GeneralCase of the proof of (b) (a) in Theorem 8.3.3 shows that F C L(c) is radical,where is a primitive mth root of unity form = IGal(L/F) I. Then we are done, sinceGal(L/F)I = [L:F] for Galois extensions. .

Exercise 7 will give a more refined version of this result.

C. Cardan's Formulas. We conclude this section with a surprising applicationof Lagrange resolvents. Let F = Q(w), where w = is our usual cube root ofunity. Note that w is primitive as defined at the beginning of the section.

We will study the universal cubic

J=x3—a,x2+a2x—o3=(x—xi)(x—x2)(x—x3).

If we regard this as a polynomial with coefficients in K = F(cTi , os), then thesplitting field off is the universal extension in degree 3,

K=F(cri,a2,o3) CL=F(xi,x2,x3),

with Galois group Gal(L/K) = S3 (we identify an automorphism with the permutationit induces on the roots). As noted in Example 8.1.2, the subgroups

{e} CA3 CS3

show that S3 is solvable. Hence K C L is solvable.A more interesting picture emerges when we apply the proof of Theorem 8.3.3

to this situation. Since (8.12) is satisfied, the Special Case tells us to take the fixedfields of the above subgroups. By Theorem 7.4.4, these fixed fields are

K C

K is the discriminant off and = (X2 —x2)(x3 —x1). (Thisdiffers by a sign from the formula for used in Theorem 7.4.4. However, it givesthe same field and leads to nicer formulas below.)

Since K C is clearly radical, we turn our attention to C L. Here,the Galois group is A3 Z/3Z (be sure you know why). Since K = Q(w, ai ,

contains the primitive cube root of unity w, Lemma 8.3.2 implies that there is a E Lsuch that

L = E

To get an explicit formula for a, we use the Lagrange resolvent a, defined in (8.7)for the generator a = (123) ofA3. Setting (= w, =x1, and i = 1 gives

a1 =x1 +w2a2.x1 =Xi +w2x2+wx3,

since = w2. This formula for relates nicely to Section 1.2:


• In (1.10) of Section 1.2 we showed that

zi = +w2x2+wx3)

is a root of the cubic resolvent (1.9). Up to the factor of this is precisely So

Galois theory explains where Zi in (1.10) comes from!• Furthermore, recall that the roots of the cubic resolvent listed in (1.10) were

obtained from z, by applying elements of S3. Thus (1.10) comes from Zi byapplying elements of Gal(L/K). It follows from (7.1) that the cubic resolvent isthe minimal polynomial of zi. Galois theory explains the cubic resolvent!

This is nice, but things get even better when we use our methods for computingsymmetric polynomials. Namely, e K( implies that = A + for someA,B E F(a,,a2,a3). Since ct1 is a polynomial in the Exercise 3 of Section 7.4implies that A, B E F [ai , a2, a3]. In Exercise 8 you will show that

=

q = + a,a2/3 — 03. This allows us to write

+W2X2+WX3 =

In Exercise 8 you will also show that if we set /3, = (23). then

/3i

and

= +ct, +/3,),

(8.15) X2 = +wct, +w2/3,),

X3 +w/3,).

If you compare this with (1.8), (1.18), and (1.19) in Section 1.2, you will see we havederived Cardan's formulas using Galois theory.

We will say more about solving polynomials by radicals in Section 8.5.

Historical Notes

Solvable groups first appeared in Galois's version of Theorem 8.3.3. Here is anextract from his statement of the theorem [Galois, pp. 57—59]:

I first observe that to solve an equation, it is necessary to reduce its groupuntil it contains only a single permutation

Given this, we will try to find the condition satisfied by the group of anequation for which it is possible to reduce the group [to a single permutation] byadjunction of radical quantities...


The second part of the quotation refers to a radical extension. Furthermore, since theradicals need not lie in the splitting field of the polynomial, Galois is describing asolvable extension.

The first part of the quotation explains the strategy used by Galois: As moreradicals are adjoined, the field gets bigger, so that under the Galois correspondencethe group gets smaller. Furthermore, if the splitting field is K C L, then the fixedfield of {e} c Gal(L/K) is L. This means that when the group is reduced to "a singlepermutation," we have found the splitting field.

Galois wants to know the "condition satisfied by the group" in this situation. Hismethod is to study how the Galois group changes under the adjunction of a pth rootfor some prime p. In [Galois, p. 59], he says the following:

We can always suppose ... that included among the quantities adjoinedearlier to the equation is a pth root of unity

Consequently, by theorems IL and III, the group of the equation shoulddecompose into p groups that have in relation to one another the following doubleproperty: 10 that one passes from one to the other by a single pennutation; 2°that they all contain the same substitutions.

The first part of this quotation refers to adjoining roots of unity, just as we did in theproof of Theorem 8.3.3. The second part seems more obscure until one realizes thatthe "double property" 10 and 2° is Galois's awkward way of saying normal subgroup.Then decomposing into "p groups" refers to cosets of the subgroup, so that we havea normal subgroup of prime index p. Since this happens for the radical adjunctionsthat reduce the Galois group to the identity, we see that the Galois group is solvable.This is the condition that Galois sought and is the first appearance of solvable groupsin mathematics.

Galois also asserts that the converse is true. The main point is Lemma 8.3.2, whichGalois states as follows [Galois, pp. 59—61]:

I say reciprocally that if the group of the equation can be partitioned into pgroups that have this double property, one can, by a simple extraction of a pthroot, and by adjunction of this pth root, reduce the group of the equation to oneof the partial groups.

The key step in his proof is a Lagrange resolvent, which Galois writes as

where O,Oi,... ,Op—i correspond to /3,a(/3),... in (8.7) and t is a pth rootof unity. Students usually find the proof of Theorem 8.3.3 to be straightforward,with the exception of the Lagrange resolvent (8.7)—this seems to come out of theblue. Yet here is Galois using essentially the same resolvent with no explanationwhatsoever. As we will learn in Chapter 12, Galois didn't need to say anything, forLagrange had worked out the theory of such resolvents in detail in 1770.

One observation is that when Galois says "partitioned into p groups," he seems tobe using the term "group" for both a subgroup and its cosets. In fact, the situation iseven more complicated, as we will see in Chapter 12. Given that we are at the birthof group theory, some confusion about terminology is understandable.



Exercise 1. Let m be a positive integer, and let L be a field of characteristic 0. Then let L C Mbe the splitting field of x" 1 E L[x].

(a) Prove that — I is separable.(b) Prove that the roots of xm — 1 lying in M form a group under multiplication.

Exercise 2. Assume that F c L is a Galois extension and that F has characteristic 0. Also,consider the extension L C obtained by adjoining a primitive rnth root of unity. Provethat F C is Galois.

Exercise 3. Prove (8.9), where is a primitive pth root of unity and 1 � p — 1.

Exercise 4. Consider the extension F1_1 C F of (8.11). In the discussion following (8.11), weshowed that this extension is Galois. We now describe its Galois group.(a) Let o- E Gal(Fi/Fi_i). Show that there is a unique integer 0 £ � — 1 such that

a(y1)

(b) Show that a- [f] defines a one-to-one homomorphism Gal(F/F1_i) —+ Z/rn1Z, whereis the congruence class of £ modulo rn.

(c) Conclude that Gal(Fi/Fi_i) is cyclic.

Exercise 5. Suppose that we have extensions F C F1_1 C F1 C L such that L is Galois over Fand F1 is Galois over F1_ . Prove that Gal(F1/F1_i) divides Gal(L/F)

Exercise 6. Let L be a field containing a primitive rnth root of unity and let n be a positivedivisor of rn. Prove that is a primitive nth root of unity.

Exercise 7. Let F C L be Galois and solvable (with F of characteristic 0). This exercise willconsider a variation of Corollary 8.3.4. Let pt,. . . , be the distinct primes dividing [L: F].(a) Show that F contains a primitive (pi . . . root of unity if and only if F contains a

primitive pth root of unity for i = 1,. . . , r.

(b) Prove that F C L is radical when F contains a primitive (pi . . pr)th root of unity.(c) Prove that F C is radical, where is a primitive (pi root of unity.

Exercise 8. This exercise concerns the details of our derivation of Cardan's formulas.(a) Use the computational methods of Section 2.3 to obtain the formulas for c4 and /3i stated

in the text.(b) Prove (8.15).

8.4 SIMPLE GROUPS

Here is the key definition of this section.

Definition 8.4.1 A gmup G is simple if its only normal subgmups are {e} and G.

Some simple groups are easy to find.

Example 8.4.2 If p is prime, then Lagrange's Theorem implies that the cyclic group

Z/pZ is simple. In Exercise 1 you will prove that these are the only nontrivial

Abelian finite simple groups.

SIMPLE GROUPS 211

Here are some more interesting simple groups.

Theorem 8.4.3 The alternating group is simple for all n � 5.

Proof: Our argument will use the following two properties of

• An I-cycle (i1 i2. . i,) lies in if and only if! is odd.• If n � 3, then is generated by 3-cycles.

The first property follows from the identity (A.2)

(1' . = (i1 i1)... (i1 i3)(i1 i2)

of Section A. 1, which shows that an I-cycle is a product of! — 1 transpositions. Thesecond property is less obvious and will be proved in Exercise 2.

Suppose that H {e} is a normal subgroup of It suffices to show thatH = To prove this, we first show that H contains a 3-cycle. By assumption Hcontains a nontrivial permutation a-. We will create a 3-cycle in H by considering thedecomposition of a into disjoint cycles.

Since contains the 3-cycle (II 1213) and H C is normal, it follows that

(8.16) a'(j1j2j3Y'a(f1J2J3) EH.

This will be useful because it will allow us to create some interesting elements of H.In Exercise 3 you will prove that the permutation (8.16) has the following property:

If neither j nor a(j) lies in {ji ,12,i3},(8.17) . .

. —1thena (3132J3) a(jij2.j3)fixesj.

This is important because the given permutation a E H might be very complicated,especially if n is large. But (8.17) shows that a' (Ii i2i3Y'a(ii 32)3) is a simplerpermutation, since it moves at most six elements of {1, . . . n}. Furthermore, thissimpler permutation lies in H by (8.16). We will exploit this by making carefulchoices of the 3-cycle (ji 32)3).

We now prove that H contains a 3-cycle by considering the following cases.

Case 1. First suppose that one of the cycles in a has length � 4, say

In this case, we claim that

(8.18) a1(i2i3i4)'a(i2i3i4) = (i1i3i4).

By (8.17), this permutation fixes all j {i1 i3, i4}, and from here it is easy to verify(8.18). We leave the details as part of Exercise 3. Since (8.18) and (8.16) imply that(i1 i3 i4) E H, we have the desired 3-cycle.

Case 2. Next suppose that a has a 3-cycle. If a is a 3-cycle, then we are done. Hencewe may assume that

a= (i1i2i3)(i4i5...)...


We claim that

(8.19) a'(i2i3i5)'a(i2i3i5)=(iti4i2i3i5).

The proof is similar to (8.18) and is part of Exercise 3. As in Case 1, it follows that(i1 i5) E H. This shows that H contains a 5-cycle. Then H contains a 3-cyclebyCase 1.

Case 3. Finally suppose that a is a product of disjoint 2-cycles. There must be atleast two since a E H C so that

a=This time, we have

(8.20) a'(i2i3i4)'a(i2i3i4) = (i1 i3)(i2i4)

(see Exercise 3). As usual, this shows that (i1 i3)(i2i4) e H. To turn this into a3-cycle, let i5 be distinct from i1,i2,i3,i4 (this is where we use n > 5). Then wecompute directly that

((i1 i3)(i2i4))'(i1 i3i5)' ((i1 i3)(i2i4))(i1 i3i5) = (i1 j5 i3).

Again we get a 3-cycle in H.

Since every a e in H must satisfy one of these three cases, we conclude that Hcontains some 3-cycle, say (jfk). We next claim that H contains all 3-cycles, since itis normal. To prove this, suppose that i', j', k' are distinct, and let 0 be a permutationthat satisfies

9(i)=i', 9(j)=j', 0(k)=k'.An important property of permutations is that for any cycle (i1 i2 . . i,), we have theidentity

(8.21) 0(ii i2 . . ' = (0(u) 9(i2) . . .9(u))

(see Exercise 3). This implies that

9 (i' f' k') E H, since H is normal in On the other hand, if 9then 0' = O(ij) E The above computation, performed using 9' instead of 0, showsthat (j'i'k') E H (you should verify this carefully). Then (i'j'k') (j'i'k')' e H,as claimed.

Thus H contains all 3-cycles. At the beginning of the proof, we noted that is

generated by 3-cycles. It follows immediately that H = and we are done. •

We next observe that non-Abelian finite simple groups are not solvable.

Lemma 8.4.4 Let G be a non-A belian finite simple group. Then G is not solvable.

SIMPLE GROUPS 213

Proof: Suppose that G is solvable. Then we can find a normal subgroup G1 C Go =G such that [G: G1] = [Go: G1] is prime. Since G is simple, we must have G1 = {e},since G1 G. Thus

= [G: Gj]IG1 = [G: Gi]I{e}I = [G: G1],

so that G has prime order. But this implies that G is cyclic and hence Abelian. Thelemma follows by contradiction. •

Combining Lemma 8.4.4 with earlier results gives us infinitely many nonsolvablegroups as follows.

Theorem 8.4.5 The alternating group and the symmetric gmup are solvableif and only if n 4.

Proof: The cases n = 1,2 are trivial, and we saw in Example 8.1.2 and Exercise 1of Section 8.1 that S3 and S4 are solvable. By Proposition 8.1.3 it follows that A3 andA4 are solvable (this is also easy to prove directly).

Now suppose that n � 5. Then is non-Abelian (the 3-cycles (123) and (124)don't commute) and simple (by Theorem 8.4.3). By Lemma 8.4.4 we conclude that

is not solvable for n � 5. Then Proposition 8.1.3 shows that is also not solvableforn�5. •

For later purposes, we determine the normal subgroups of

Proposition 8.4.6 If n> 5 and H C is a normal subgroup, then either H = {e},H = or H =

Proof: If H is normal in then H flAp is normal in (see Exercise 4). Sincen 5, Theorem 8.4.3 implies that H flAg is {e} or In the latter case, we have

C H, which easily implies that H = or since = 2.

Finally suppose that H flAg = {e}. If H {e}, then Exercise 5 will show thatH = {e,a}, where

is the product of an odd number of disjoint 2-cycles. Now pick k different from i and

j, and let 0 = (jk). Then (8.21) implies that

=0(ij)00(...)010...0'=(ik)(...)....

This is still a product of disjoint 2-cycles. Since one of the cycles is (ik), it can'tequal a = (ij)(... ).... Thus 0a0' H, which is impossible, since H is normal.This contradiction shows that H = {e} and completes the proof. .Mathematical Notes

The relation between simple and solvable groups is more interesting than indicatedby Lemma 8.4.4. The key observation is that all groups are "built" out of simplegroups by means of what are called composition series.


• Composition Series and the Jordan—Holder Theorem. Definition 8.1.1 says thata group G is solvable if we can find subgroups

(8.22)

such that G, is normal in G,_1 and G,] is prime for i = 1,... , n. This impliesin particular that the quotient /G1 is simple, since it has prime order.

More generally, if G is a finite group, then a composition series of G consists ofsubgroups (8.22) such that G, is normal in G,_1 and the quotient G_1/G1 is simplefor all i. We call the G1_1 /G, the composition factors of G.

Example 8.4.7 Let n � 5. Since An is simple, a composition series of Sn is

{e} CAn C Sn.

The composition factors are An/{e} and Sn/An Z/2Z.

It is straightforward to show that any finite group has a composition series (seeExercise 6). However, a given group may have more than one composition series.For example, the cyclic group Z/6Z = ([1]) has the composition series

{e} c ([2]) c Z/6Z and {e} c ([3]) c Z/6Z.

The factors of the first composition series are Z/2Z and Z/3Z, while the factors forthe second are Z/3Z and 7L/2Z. The Jordan—Holder Theorem asserts that any twocomposition series of a given group have the same length and that the correspondingcomposition factors can be permuted so that they become isomorphic. Hence thecomposition factors of a group are the simple groups from which the group is built.Here "built" refers to the extension problem discussed in the Mathematical Notes toSection 6.4. For more on composition series, see [Jacobson, Vol. I, Sec. 4.6].

In particular, a finite group is solvable if and only if its composition factors are the"simplest" simple groups, namely the Abelian ones. This shows that solvable groupsform a very special class of groups.

Historical Notes

The term "simple group" is due to Jordan. He was the first to prove that An issimple for n 5. However, concerning A5, Galois noted in 1832 that "the smallestnumber of permutations for which there is an indecomposable group is 5 .4.3 whenthe number is not prime" [Galois, p. 175]. The simplicity of A5 is also implicit in thework of Ruffini and Abel on the unsolvability of the quintic equation.

The idea of a composition series is due to Jordan. He proved that any twocomposition series have the same length and that the indices : G,J are unique upto a permutation. Later, once the concept of quotient group was better understood,Holder proved the Jordan—HOlder Theorem mentioned above.

SOLVING POLYNOMIALS BY RADICALS 215


Exercise 1. Let G be a nontrivial finite Abelian group. Prove that G is simple if and only ifG Z/pZ for some prime p.

Exercise 2. Prove that An is generated by 3-cycles when fl 3.

Exercise 3. This exercise is concerned with the proof of Theorem 8.4.3.(a) Prove (8.17).(b) Verify the identities (8.18), (8.19), and (8.20).(c) Verify the conjugation identity (8.21).

Exercise 4. Let H1 and H2 be subgroups of a group G and assume that H1 is normal in G.Prove that flH2 is normal in H2.

Exercise 5. Suppose that H C Sn is a subgroup such that H {e} and

H = {e, where a is a product of an odd number of disjoint 2-cycles.

Exercise 6. Let G be a finite group.(a) Among all normal subgroups of G different from G itself, pick one of maximal order and

call it H. Prove that G/H is a simple group.(b) Use part (a) and complete induction on GI to prove that G has a composition series.

Exercise 7. Show that the Feit—Thompson Theorem (Theorem 8.1.9) is equivalent to theassertion that every non-Abelian finite simple group has even order.

Exercise 8. Prove that Z/4Z and Z/2Z x 7Z/2Z are nonisomorphic groups with the samecomposition factors.

8.5 SOLVING POLYNOMIALS BY RADICALS

As in Section 8.3 we will assume the following:

Al/fields appearing in this section will have characteristic 0.

The goal of this section is to study the roots of polynomials using Galois theory.

A. Roots and Radicals. So far, our discussion of solvability by radicals hasfocused on field extensions. We now shift our attention to polynomials and theirroots.

Definition 8.5.1 Let f F[x] be nonconstant with splitting field F C L.(a) A root a L off is expressible by radicals over F if a lies in some radical

extension ofF.(b) The polynomial f is solvable by radicals over F if F C L is a solvable extension.

In Exercise 1 you will show that part (b) of this definition doesn't depend on whichsplitting field off over F we use.

Definition 8.5.1 implies that if a nonconstant polynomial in F[x] is solvable byradicals, then all of its roots are expressible by radicals. However, for an irreducible


polynomial, it turns out that solvability by radicals is satisfied as soon as one root isexpressible by radicals. Here is the precise result.

Proposition 8.5.2 Letf E F [xl be irreducible. Then f is solvable by radicals overF if and only has a root expressible by radicals over F.

Proof: One direction is obvious. Going the other way, suppose that f has a rootin some radical extension of F. This means that F C is solvable, so that by

Corollary 8.2.9, its Galois closure F C C M is also solvable. (Remember thatwe are in characteristic 0.)

Since a Galois extension is normal and f is irreducible over F with a root in M,we see that f splits completely over M. Thus M contains the splitting field off overF (in fact, M is the splitting field—see Exercise 2). The proposition follows, sinceF c M is solvable.

We can now apply the theory developed in Sections 8.3 and 8.4. Recall fromDefinition 6.1.12 that the Galois group off e F[x] is Gal(L/F), where L is a splittingfield off over F. Then Theorem 8.3.3 implies the following.

Theorem 8.5.3 A polynomial f e F [x] is solvable by radicals over F if and only ifthe Galois group off over F is solvable.

We can apply this to polynomials of low degree as follows.

Proposition 8.5.4 1ff E F[x] has degree n 4, then f is solvable by radicals.

Proof: 1ff is separable, then the Galois group off is isomorphic to a subgroup ofby Proposition 6.3.1, and we are done by Theorem 8.5.3, since is solvable for

n 4. See Exercise 3 for the case when f is not separable.

Once we get to degree 5, a different picture emerges.

Example 8.5.5 In Section 6.4, we showed that f = x5 — 6x + 3 has S5 as Galois groupover Q. But S5 is not solvable by Theorem 8.4.5, so that f is not solvable by radicalsover Q by Theorem 8.5.3. Furthermore, f is irreducible, so that by Proposition 8.5.2,no root of f is expressible by radicals over Q.

This example requires that we revise how we think about the roots of a polynomial.Most students come into a course on Galois theory thinking that the roots of apolynomial f e Q [xl are numbers like

etc.

The English word "root" comes from the Latin "radix," and the radical symbol isa modified version of the first letter "r" of "radix." Historically, "root" came to referto a solution of f(x) = 0 because of the intuition that roots are built from radicals.But the above example shows that this intuition is wrong. Roots of polynomials areintrinsically more complicated than just radicals.


B. The Universal Polynomial. The quadratic formula shows that the universalquadratic -

f=x2—aix+a2=(x—xi)(x—x2)

is solvable by radicals, and Cardan's formulas imply that the universal cubic

7=x3—aIx2+a2x—o3=(x—xI)(x—x2)(x—x3)

is solvable by radicals. Furthermore, once we know these formulas in the universalcase, then they apply to all polynomials of degree 2 and 3.

This discussion shows that asking if the quadratic formula and Cardan's formu-las generalize to polynomials of degree n is equivalent to asking if the universalpolynomial of degree n,

(8.23)

is solvable by radicals. By Section 6.4, the splitting field of f is the universalextension in degree n

whose Galois group Gal(L/K) is isomorphic to Then Theorem 8.5.3 impliesthat the existence of radical formulas generalizing the quadratic formula or Cardan'sformulas is equivalent to the solvability of

In particular, the solvability of implies the existence of radical formulas forpolynomials of degree 4. These are Ferrari 'Sformulas, to be discussed in Chapter 12.However, when n > 5, we have the following.

Theorem 8.5.6 If n > 5, then the universal polynomial f E K[x} of degree n is notsolvable by radicals over K, and no root off is expressible by radicals over K.

Proof: The first assertion follows from Theorem 8.5.3, since is not solvable whenn 5 by Theorem 8.4.5, and the second assertion follows from Proposition 8.5.2,since is irreducible over K. •

Thus, while we have the quadratic formula for polynomials of degree 2, Cardan'sformulas for degree 3, and Ferrari's formulas for degree 4, it is impossible to findradical formulas that apply to all polynomials of degree n when n 5.

It is important to keep in mind that for every n � 5, there are always somepolynomials of degree n, such as f — 2 E Q[x], that are solvable by radicals. It isonly when we try to solve all polynomials by radicals that we run into problems.

C. Abelian Equations. In 1829 Abel considered separable polynomials f E F[x}that have a root a such that the roots off are (a), . . , (a), where , arerational functions with coefficients in F satisfying

=


for all i, j. Following Kronecker and Jordan, we call f = 0 an Abelian equation inthis situation. Abel showed that Abelian equations are solvable by radicals. We canprove Abel's theorem as follows.

Theorem 8.5.7 Letf e F[xJ. 1ff = 0 is an Abelian equation, then f is solvable byradicals over F.

Proof: Theorem 6.5.3 states that the Galois group of an Abelian equation is Abelian.(As noted in Section 6.5, this result is the origin of the term "Abelian group.") SinceAbelian groups are solvable by Proposition 8.1.5, we are done by Theorem 8.5.3. .

This shows that Abel's theorem on the solvability of Abelian equations followsfrom Galois theory and basic facts about solvable groups.

We studied Abelian equations in the optional Section 6.5 of Chapter 6. Forthose who read that section, note that Theorem 6.5.2 is simply a restatement ofTheorem 8.5.7 and that Theorem 6.5.4 follows from Theorem 8.5.3 because Abeliangroups are solvable.

For irreducible polynomials, the relation between Abelian equations and Abeliangroups is especially nice.

Theorem 8.5.8 Letf E F[x] be irreducible and separable of degree n with splittingfield F C L. Then

f = 0 is an Abelian equation Gal(L/F) is an Abelian group.

Furthermore, when these conditions are satisfied, we have

IGaI(L/F)I = [L:F1 =n

andL = F(a)for any root EL ofF.

Proof: The implication was proved in Theorem 6.5.3. For the opposite im-plication, let E L be a root of F. Then F c c

c Gal(L/F), which is normal since Gal(L/F) is Abelian. ThusF c F(a) is Galois and hence normal by Theorem 7.3.2. Since f is irreducibleover F and has a root in F(a), it must split completely over F(a) by normality. Itfollows that L = This implies the final assertions of the theorem, and f = 0 is

Abelian by Exercise 6 of Section 6.5. •

D. The Fundamental Theorem of Algebra Revisited. In Chapter 3 we provedthe Fundamental Theorem of Algebra using the following two facts:

• Every polynomial of odd degree in R[x] has a root in IR (Proposition 3.2.2).• Every quadratic polynomial in C[xI splits completely over C (Lemma 3.2.3).

The proof given in Section 3.2 used induction on the power of 2 in the degree off EJR {xJ. Artin gave an elegant version of this argument using the Galois correspondence,the solvability of groups of prime power order (Theorem 8.1.7), and the First SylowTheorem (see the Mathematical Notes to Section 8.1). Here is Artin's proof.


Theorem 8.5.9 Every nonconstant polynomial in C[x] splits completely over C, i.e.,C is algebraically closed.

Proof: By Proposition 3.2.1, it suffices to prove that every nonconstant polynomialin JR [x] splits completely over C. Given such a polynomial f, let JR C L be its splittingfield. Since JR has characteristic 0, this extension is separable and hence Galois. LetG = Gal(L/IR), and define H C G as follows: If IGI is odd, then H = {e}, and if IGIis even, then H is a 2-Sylow subgroup of G. Hence H is a subgroup of G such thatHI is the highest power of 2 dividing IGI.

By the Galois correspondence, the fixed field JR C has degree [LH : JR] =[G : H] = This is odd by the definition of H, so that JR C LH has odd degree.It follows that if a E LH is a primitive element over R, then the minimal polynomialg E R{x] of a has odd degree. But by the first bullet above, g has a root in JR. Sinceminimal polynomials are irreducible, this means that g must have degree 1, whichimplies LH = JR.

Then the Galois correspondence implies H = G, so that is a power of 2, sayIGI = If n 0, then G is trivial, which implies that L = JR. Hence f splitscompletely over JR in this case. Now suppose that n 1. By Theorem 8.1.7, G issolvable, which by GI and Definition 8.1.1 means that we have subgroups

{e} = Ga_iC C G1C Go =G

such that is normal in G1_1 of index 2 for 1 i n. This gives the fixed fields

JR = C C C

such that LG,_ C has degree 2 for every i.Since n> 1, we have the degree 2 extension JR C LG1. The minimal polynomial

of a primitive element of this extension is a quadratic polynomial with no real roots.It follows easily that LG C.

Now suppose that n � 2. Since LG1 C LG2, we have a degree 2 extension of C. Bythe second bullet above, this is impossible, since every quadratic polynomial in C{x]splits completely over C. Hence we must haven = 1, which implies that IGI = 2 andL = LG C. It follows that! splits completely over C, as claimed. .

Notice that our proof of Theorem 8.5.9 translates the above two bullets into thefollowing field-theoretic facts about JR and C:

• JR has no extensions of odd degree> 1.• C has no extensions of degree 2.

The essence of Artin's argument is that these facts combine with Galois theory andresults from group theory (the First Sylow Theorem and the solvability of groups oforder to prove the Fundamental Theorem of Algebra.

Historical Notes

The universal polynomial f considered in (8.23) is sometimes called the "generalpolynomial" of degree n. We will see in Chapter 12 that Lagrange tried hard to solve


the general quintic in 1770. Using these methods, Ruffini proved the impossibilityof solving the general quintic by radicals in 1799, though his proof was difficult tofollow (see [2]). In 1824, Abel, also using the methods of Lagrange, found a proofthat came to be more generally accepted. The general quintic is discussed in [3], [8],and [11]. Also, one can prove the unsolvability of the general equation of degreen � 5 by radicals (Theorem 8.5.6) without using Galois theory (see [1]). See [9] foran account of Abel's proof and [10] for more on his life.

A tantalizing comment in [Galois, p. 72] suggests that Galois may have known theFirst Sylow Theorem. We don't know whether he had a proof or simply conjecturedthe result.

However, there is no doubt that Galois knew a lot about solvability by radicals.Chapter 14 will explore Galois's amazing insights about when an irreducible poly-nomial of degree p or p2, p prime, is solvable by radicals.


Exercise 1. Let F C L1 and F C L2 be splitting fields of f E F[xI. Prove that F C L1 issolvable if and only if F C L2 is solvable.

Exercise 2. Let f E F [x] be separable and irreducible, and assume that we have an extensionF C F(a) where a is a root off. Prove that the Galois closure of this extension (as definedin Section 7.1) is the splitting field off over F.

Exercise 3. Let F have characteristic 0 and suppose that f E F [x] has degree � 4 and is notseparable. Prove that f is solvable by radicals over F.

Exercise 4. Let f be the minimal polynomial of + over Q, where all of theindicated radicals are real. Prove that f is solvable by radicals over Q.

Exercise 5. Let F have characteristic 0, and assume that we have fields F C K C L. Alsosuppose that a E L is expressible by radicals over K and that the extension F C K is a solvableextension. Prove carefully that the minimal polynomial of a over F is solvable by radicalsover F.

Exercise 6. The proof of Theorem 8.5.9 used the Theorem of the Primitive Element to showthat R has no extensions of odd degree> 1. Prove this without using primitive elements.

8.6 THE CASUS IRREDUCIBILIS (OPTIONAL)

In this optional section we will complete our discussion of the casus irreducibilisbegun in Chapter 1. We will also give an example to show how solvability by radicalscan fail in characteristic p.

A. Real Radicals. By Section 1.3, a monic separable cubic polynomial f E R[x]with real roots has discriminant > 0. Then Cardan's fonnulas (8.15) imply thatthe complex number

—.V 27 1V 27

THE CASUS !RREDUC!BILIS (OPTIONAL) 221

appears in the formulas for the roots off, even though the roots are real.It is natural to ask whether it is possible to express the roots of f in terms of real

radicals in this situation. In some cases, such as f x3 + x2 — 5x — 5, the answer isyes, since f = (x+ 1)(x2 —5) has roots which are expressible using realradicals. However, we will show below that the answer is no whenever the cubic fis irreducible. This is the casus irreducibilis from Section 1.3.

We first give a careful definition of what it means for a real number to be expressibleby real radicals.

Definition 8.6.1 Let F be a subfield of lit Then:(a) F c L is a real radical extension if F C L is radical and L C R.(b) E R is expressible by real radicals over F if there is a real radical extension

F c L such that E L.

Before proving our main theorem, we need to study radical extensions. Our firstresult allows us to limit ourselves to prime radicals.

Lemma 8.6.2 1fF C K is a radical extension, then there are fields

where for i 1,. . . n, there is e F; such that F; = (ye) and E ifor someprime

Proof: We first show that the lemma is true for an extension F C F(y) with E Ffor some m> 1. If m is prime, then we are done, and if m is not prime, then let p bea prime dividing m and set 8 = -y'1. This gives extensions

F C F(8) C F(8)(y) = F(-y)

such that = 8 E F(8) and 8m/p = E F. If rn/p is prime, then we are done, andif not, pick a prime dividing rn/p and continue as above. Thus the lemma holds forF C F('y). Since any radical extension is a sequence of such extensions, the lemmafollows.

We next study extensions obtained by adjoining real prime radicals.

Lemma 8.6.3 Let E be a subfield of R, and suppose that -y E R satisfies -y E andm E E, where mis prime. Then g = xm — m is irreducible overE, and [E(y) : E] = m.

Proof: By Proposition 4.2.6, it suffices to show that g has no roots in E. If /3 E Eis a root of g, then /3 = for some mth root of unity It followsthat /3 = ±-y since /3 and -y are real and nonzero and the only real roots of unity are± 1. Thus -y = ±fi E E. This contradiction proves the lemma. •

The following result will be a key tool in our analysis of the casus irreducibilis.

Proposition 8.6.4 Suppose thatM C Lisa Galois extension with L C Rand [L M] =pfor an odd prime p. Then L cannot lie in a real radical extension of M.


Proof: Suppose that we have an extension

where y M, -y e R, and 'ym e M for some prime m. Then M C R implies that[M(-y) :M] =mbyLemma8.6.3. Wewilirelate [L(-y) :Me-y)] to {L:M] by consideringthe following diagram:

L('y)/ \(8.24) L M('y)\ /

M

If -y E L, then M(-y) = L, since 'y M and [L : M] is prime. It follows that

(8.25)

m that m is the minimalpolynomial of 'y over M. Since M C L is normal, xm m splits completely over L.The roots of this polynomial are -y for £ = 0,. . . , m — 1, where = Since

0, it follows that E L. This is impossible, since m is odd and L CHence -y L, so that {L('y) : L] m by Lemma 8.6.3. Using m = : and

the Tower Theorem, (8.24) easily implies that

(8.26) [L : M] = p.

Thus adjoining a real prime radical doesn't change the degree.By Lemma 8.6.2, a real radical extension M C K is obtained by adjoining suc-

cessive real prime radicals. Each time we do this, (8.26) shows that the degree isunchanged. In Exercise 1 you will use this to prove that

(8.27) [KL:K] = [L:M] =p,

where KL is the compositum of K and L. It follows that KL K, which in turn impliesthat L K (see Exercise 1). Since M C K is an arbitrary real radical extension of M,we conclude that L cannot lie in such an extension, as claimed. •

B. Irreducible Polynomials with Real Radical Roots. We can now statea generalized version of the casus irreducibilis proved by Holder in 1891 [61 andindependently by Isaacs in 1985 [7]. Their result shows that when an irreduciblepolynomial has all real roots, the roots are expressible by real radicals only in veryspecial cases.

Theorem 8.6.5 Let F be a subfield of R and let f F [x] be irreducible with splittingfield F C L C JR. Then the following conditions are equivalent:(a) Some root off is expressible by real radicals over F.

THE CASUS IRREDUCIBILIS (OPTIONAL) 223

(b) All roots off are expressible by real radicals over F in which only square rootsappear

(c) F C L is a radical extension.

(d) [L:F] isapowerof2.

Proof: Some implications of the proof are easy. For example, (b) (a) is trivialand (c) (a) follows from L C R.

Now suppose that (d) holds. This implies that IGal(L/F) is a power of 2. As inthe proof of Theorem 8.5.9, this leads to subfields

F = LG0 C LG, C C L

where each field has degree 2 over the previous. Since the characteristic is differentfrom 2, each field is obtained from the previous by adjoining a square root. Thisshows that F C L is radical, so that (d) (c) follows. We also obtain (d) (b) sinceL CR.

It remains to prove (a) (d). We have f E F[x} with splitting field F C L C R.Now assume that some root a of f lies in a real radical extension F C K and that[L: Fl is not a power of 2. Our goal is to derive a contradiction.

We will use a clever idea from [7] to reduce to the situation of Proposition 8.6.4.Let p be an odd prime dividing [L : Fl. We claim the following:

(8.28) There is a e Gal(L/F) of order p such that a(a) a.

Let us first use (8.28) to get our desired contradiction. Given a as in (8.28), letM = L(a) be the fixed field of the cyclic group generated by a. Then the Galoiscorrespondence implies that M C L is a Galois extension such that

[L:M] = = = I(a)I =p.

By Proposition 8.6.4, L lies in no real radical extension of M.On the other hand, a E L because L is the splitting field off, yet a M because

a(a) a. Since [L : M] is prime, L = M(a) follows. Furthermore, we are assumingthat a E K, where F C K is real radical. Hence:

• L=M(a)CMK.• M C MK is real radical, since F C K is (see Exercise 2).

These bullets imply that L lies in a real radical extension of M, which contradicts theprevious paragraph.

It remains to prove (8.28). Since p divides [L : FJ = Gal(L/F)I, Cauchy's The-orem implies that Gal(L/F) has an element r of order p. Let the roots of f be

= a,... ,am, m = deg(f). Using L = F(ai,...,am) and T we see thata, for some i. However, f is irreducible, so that by Proposition 5.1.8, there

is a, E Gal(L/F) such that = Then has order p and

= = a.


It follows easily that o- = satisfies the conditions of (8.28). The proof oftheorem is complete. •

Theorem 8.6.5 has the following useful corollary.

Corollary 8.6.6 Let F be a subfield of JR. and assume that f E F [x] is irreducibleand deg(f) is not a power of 2. 1ff splits completely over R, then no root off isexpressible by real radicals over F.

Proof: Let F C L C R be the splitting field off over F, and let a E L be a root off. Then the extensions

F C F(a) C L

and the Tower Theorem imply that [L: F] is a multiple of deg(f). Since deg(f) isnot a power of 2, the same is true for [L : Fl. Then the corollary follows from theequivalence (a) (d) of Theorem 8.6.5. •

In concrete terms, Corollary 8.6.6 says that if a polynomial with real roots isirreducible over a subfield F C Rand has degree not a power of 2, then it is impossibleto express any of its roots using real radicals over F. In particular, this is true for anyirreducible cubic with real roots, which is the casus irreducibilis.

Here is an example of Theorem 8.6.5.


f = x4 — 4x2 + x + 1.

By Exercise 3, f is irreducible over Q and all of its roots are real. In Chapter 13 wewill show that the Galois group off over Q is isomorphic to S4. If L is the splittingfield of f over Q, it follows that [L: Q} = 24. This is not a power of 2, so that byTheorem 8.6.5, no root off is expressible in terms of real radicals. Yet f is solvableby radicals since S4 is solvable.

We can use Corollary 8.6.6 to construct solvable extensions that are not radical.

Example 8.6.8 Consider f = x3 + x2 — 2x — 1. In Example 8.2.3, we showed thatthe splitting field Q C L off is solvable but not radical. This follows immediatelyfrom Corollary 8.6.6 since f is irreducible of degree 3 with all real roots.

We will also see that Theorem 8.6.5 has applications to the geometric constructionsconsidered in the Mathematical Notes to Section 10.1.

C. The Failure of Solvability in Characteristic p. One surprise is that themethods used to study real radicals can also shed light on solvability by radicals forfields of characteristic p. When we considered solvable extensions in Sections 8.3and 8.5, we explicitly assumed that we were working in characteristic 0. As we willsee, this is necessary because of the lack of pth roots of unity in characteristic p.

We begin with a result on adjoining prime radicals. We say that a field F containsall roots of unity if xm — 1 splits completely over F for all integers m � 1.

THE CASUS !RREDUCIBILIS (OPTIONAL) 225

Lemma 8.6.9 Let F be a field that contains all roots of unity. Also assume that Fhas an extension containing an element y such that y F but m E Ffor some primem. Then g = xm — m is irreducible over F and [F('y) : F] = m.

Pmof: If g has a root F, then /3Pfl = m which implies that = for somemth root of unity Then F imply that 'y e F, which is a contradiction. Henceg has no roots in F and the lemma follows from Proposition 4.2.6. •

We now state our main result.

Proposition 8.6.10 Let M be a field of characteristic p that contains all roots ofunity. Then any Galois extension M C L of degree p is not solvable.

Proof: We need to prove that L cannot lie in a radical extension of M. The argumentis remarkably similar to the proof of Proposition 8.6.4. Consider an extension

M C M(y)

suchthat'y Mand-ytm E Mforsomeprimem. Lemma8.6.9impliesthatg =xm _yIflis irreducible over M and [M(-y) M] = m.

Following the proof of Proposition 8.6.4, we have the diagram (8.24). If -y E L,then = L since [L : MI is prime. Using (8.25), we see that m = p. Then theminimal polynomial of -y over M is g = x" — However, M C L is Galois andhence separable, so that -y would be separable over M. Yet g = x' = (x y)P isclearly inseparable.

This contradiction shows that -y L, so that [L(-y) : L] = m by Lemma 8.6.9. Since[M(-y) MI = m, the Tower Theorem and (8.24) imply that

[L(y):M(y)i = [L:MJ p.

Thus adjoining a prime radical doesn't change the degree. Also, the extensionM(y) C L(y) is Galois (you will prove this in Exercise 4), and contains allroots of unity because M does.

From here, it is straightforward to show that L lies in no radical extension of M.We leave the details as Exercise 4. •

We construct an extension M c L that satisfies Proposition 8.6.10 as follows.

Example 8.6.11 Let k be an algebraically closed field of characteristic p (the ex-istence of such a field is proved in [Jacobson, Vol. II, Sec. 8.1]), and let M =where t is a variable. Since xtm — 1 k[x] splits completely over k for any m, it followsthat M contains all roots of unity.

Following Artin and Schreier, we consider the polynomial

f =x1' —x+t E M[x].

Let L be a splitting field off over M. We know that M C L is a Galois extension byExercise 15 of Section 5.3. In Exercise 5 you will show that there is a one-to-onegroup homomorphism

(8.29) Gal(L/M) —+ 7L/pZ.


Since [L M] = IGal(L/M) , it follows that [L MI = 1 or p. The former would implythat L = M, which would mean that f splits completely over M. However, Exercise 6will show that f has no roots in M. Hence [L M] p.

It follows that M C L is a Galois extension of degree p. Since M contains all rootsof unity, Proposition 8.6.10 implies that M c L is not solvable.

In Example 8.6.11, note that Gal(L/M) Z/pZ since M C L is a Galois extensionof degree p. So the Galois group is Abelian and hence solvable, yet the extension isnot solvable. This shows that the relation between solvable extensions and solvableGalois groups breaks down in characteristic p.

To see where the problem is, note that a key step in the proof of Proposition 8.6.10is the observation that in characteristic p, the polynomial g = — = (x — is

not separable and hence 'y can't lie in a nontrivial Galois extension. The inseparabilityof g can be explained by the small number of pth roots of unity in characteristic p,for any two roots of g = x' — differ by a pth root of unity, but — 1 = (x 1

)1)

implies that the only pth root of unity is 1.It turns out that if one avoids extensions whose degree is divisible by p, then the

relation between solvable extensions and solvable Galois groups works out nicely incharacteristic p. See Exercise 7 for a proof.

Historical Notes

Although the casus irreducibilis for the cubic equation dates back to the sixteenthcentury, rigorous proofs didn't appear until late in the nineteenth century. Mollamegave a proof in 1890, followed a year later by the more general result of Holderproved in the text. A quick proof for cubics can be found in [van der Waerden].

We should also mention the following result of Loewy from the 1920s.

Theorem 8.6.12 If F C and f E F [xl is irreducible of degree 2mn, n odd, then fhas at most 2m roots expressible by real radicals over F. U

When f is of odd degree, Theorem 8.6.12 implies that at most oneroot can be expressible by real radicals. For cubics, this is consistent with Cardan'sformulas (see Example 1.1.1). References and a proof of Theorem 8.6.12 when thedegree is odd can be found in [Chebotarev, p. 350].


Exercise 1. Here are some details from the proof of Proposition 8.6.4.(a) Prove (8.27).(b) Prove that KL = K if and only if L C K.

Exercise 2. Let F C K be a real radical extension and that F C M C R. Prove that M C MKis a real radical extension.

Exercise 3. Show that the polynomial f = x4 — 4x2 + x + 1 of Example 8.6.7 is irreducibleover Q and has four real roots.

REFERENCES 227

Exercise 4. Complete the proof of Proposition 8.6.10.

Exercise 5. This exercise will consider the polynomial f = x" x + t from Example 8.6.11.Let a E L be a root off.(a) Show that the roots off are a, a + 1,. . . , a + p 1.

(b) Let o- E Gal(L/M). By part (a), o-(cs) = a + i for some i. Prove that a F—* [i] gives thedesired one-to-one homomorphism (8.29).

Exercise 6. Let k be a field and let M = k(t), where t is a variable. The goal of this exerciseis to prove that if n> 1, then there is no element E M such that /3" — /3 + t = 0.

(a) Write = A/B, where A, B E k[t] are relatively prime polynomials. Prove that /3" — /3 += 0 implies that BIA and hence that B is constant.

(b) Show that A" —A +t 0 for all polynomials A E k[t].

Exercise 7. Suppose the F is a field of characteristic p and that F C L is a Galois extension.Also assume that Gal(L/F) is solvable and that EL: F]. Prove that F C L is solvable.

REFERENCES

1. R. 0. Ayoub, On the nonsolvability of the general polynomial, Amer. Math. Monthly 89(1982), 397—401.

2. R. 0. Ayoub, Paolo Ruffini's contributions to the quintic, Arch. Hist. Exact Sci. 23 (1980),253—277.

3. I. G. Bashmakova and G. S. Smirnova, The Beginnings and Evolution of Algebra, Englishtranslation by A. Shenitzer, MAA, Washington, DC, 1999.

4. W. Burnside, On groups of order Proc. London Math. Soc. 2 (1904), 388—392.

5. W. Feit and J. Thompson, Solvability of groups of odd order, Pacific J. Math. 13 (1963),775—1029.

6. 0. L. Holder, Ueber den Casus Irreducibilis bei der Gleichung dritten Grades, Math.Annalen 38(1891), 307—3 12.

7. I. M. Isaacs, Solution of polynomials by real radicals, Amer. Math. Monthly 92 (1985),571—575.

8. B. M. Kiernan, The development of Galois theory from Lagrange to Artin, Arch. Hist.Exact Sci. 8(1971), 40—154.

9. P. Pesic, Abel's Proof, MIT Press, Cambridge, MA, 2003.

10. A. Stubhaug, Niels Henrik Abel and His Times, Springer, New York, Berlin, Heidelberg,2000.

11. H. Wussing, The Genesis of the Abstract Group Concept, English translation by A.Shenitzer, MIT Press, Cambridge, MA, 1984.

CHAPTER 9

CYCLOTOMIC EXTENSIONS

In this chapter we will explore the Galois theory of cyclotomic extensions, which areextensions of the form Q C Q = This will involve a study of cyclotomicpolynomials and Gauss's theory of periods. In the next chapter we will apply theseresults to determine which regular polygons are constructible by straightedge andcompass.

9.1 CYCLOTOMIC POLYNOMIALS

In Section 4.2 we showed that if p is prime, then

=x"1

is the minimal polynomial of = over Q. In this section, we will describe theminimal polynomial of

=

over Q, where n is now an arbitrary integer 1. We will also compute the Galoisgroup But first, we need two facts from elementary number theory.


230 CYCLOTOMIC EXTENSIONS

A. Some Number Theory. We begin with the Euler Given a positiveinteger n, we define to be the number of integers i such that 0 i < n andgcd(i, n) = 1. We can interpret in terms of the ring Z/nZ as follows. Theinvertible elements of this ring form the set

(Z/nZ)* = {[iI e Z/nZ [i][j] = [1] for some [j] Z/nZ}.

One easily sees that (Z/nZ)* is a group under multiplication. In Exercise 1 you willshow that (Z/n7Z)* has order 4(n). In other words,

(9.1) =

Our first lemma gives the basic properties of the 4-function.

Lemma 9.1.1 Let be defined as above.(a) If n and mare relatively prime positive integers, then =(b) If n> 1 is an integer; then

cb(n)=nfJ (1_I),pin

where the product is over all primes p dividing n.

Proof: Since gcd(n,m) = 1, Lemma A.5.2 implies that there is a ring isomorphisma Z/nmZ Z/nZ x 7L/mZ. In Exercise 2 you will show that a induces a groupisomorphism

(Z/nmZ)* x (Z/mZ)*.

Then follows immediately from (9.1).Next observe that if p is prime and a � 1, then counts the number of integers

i such that 0 i < and In other words, if

S = {j E Z I 0 j <pa and pLi},

then = pa— However, for some 0 j <pa if and only if j = pt for

some 0 £ < Thus ISI= pa_i, so that =pa _pai•

For arbitrary n> 1, write n = where the p, are distinct primes and a, � 1for all i. Using part (a) and the formula = pa — pa_i, we obtain

cb(n) =. . = . .

.

=nfl(i_i).

This completes the proof.

Our second lemma is sometimes called Fennat's Little Theorem.

CYCLOTOMIC POLYNOMIALS 231

Lemma 9.1.2 Ifp is prime, then a mod pfor all integers a.

Proof: Since the congruence is true when p a, we may assume that a. Then[a] e (Z/pZ)*, so that = [11, since (Z/pZ)* is a group of order p — 1 undermultiplication. In congruence notation, this means that 1 mod p. The desiredcongruence follows by multiplying each side by a. •

B. Definition of Cyclotomic Polynomials. Our next task is to define thecyclotomic polynomial (x) for n � 1 and show that it has integer coefficients. Webegin with the factorization

(9.2) f—1= fiO<i<n

Then define the nth cyclotomic polynomial (x) to be the product

(9.3) [JO<i<n

gcd(i,n)=1

Thus the roots of are for those 0 i < n relatively prime to n. It followsthat (x) has degree Combining this with (9.1), we see that

q5(n) = I(Z/nZ)*I.

This link between and (Z/nZ)* will be used to determineIn Section 8.3 we defined a root off — ito be aprimitive nth root of unity if its

powers give all roots off 1. In Exercise 3 you will prove that in our situation, theprimitive nth roots of unity are for 0 < i < n and gcd(i, n) = 1. Thus the roots of

(x) are the primitive nth roots of unity in C.Here are some examples of cyclotomic polynomials.

Example 9.1.3 When n = 2, the only primitive square root of unity is —1, so that(1)2(x) = x+ 1. When n = 4, the primitive fourth roots of unity are i and i3 = —i, sothat

Since c1) 1(x) = x— 1, we get the factorization

x4 1 = (x— l)(x+ l)(x2+ 1) =

Proposition 9.1.5 will show that — 1 has a similar factorization.

Example 9.1.4 Let p be prime. Since 1, . . . p — 1 are relatively prime to p, it followsthat

xp —

Using —1 = (x— i)(xP' + +x+ 1), we obtain + +x+ 1,which agrees with the definition of given in Section 4.2.


In the following discussion we will write din to indicate that d is a positive divisorof n. We now state some elementary properties of cyclotomic polynomials.

Proposition 9.1.5 (x) is a monic polynomial with integer coefficients and hasdegree Furthermore, these polynomials satisfy the identity

(9.4) — 1 =din

Proof: (x) is monic by definition and has degree as shown above. Nextwe prove the factorization (9.4). The basic idea is that every number i in the rangeo i < n gives a divisor d = gcd(i, n) of n. Since different values of i can give thesame d, we can organize the factorization (9.2) according to d. This gives

f_i=fl Hdin O<i<n

gcd(i,n)=d

For a fixed positive divisor d of n, the corresponding part of this factorization is

(9.5) [fO<i<n

gcd(i,n) =d

But gcd(i,n) = d implies that i = dj and n = where gcd(j, = 1. Also:

•

•

It follows that (9.5) can be written

fJ0<1<1

which by (9.3) is the cyclotomic polynomial 1 (x). Thus the above factorization ofx" — 1 becomes

din

Then (9.4) follows since d is a positive divisor of n if and only if is.It remains to show that (x) has integer coefficients. We prove this by complete

induction on n. The base case n = 1 is trivial, since = x— 1. Furthermore, ifn> 1, then (9.4) and our inductive hypothesis imply that

— 1 = fJdln,d<n

= (x) a monic polynomial g(x) with integer coefficients.


Hence is the quotient of x" 1 by g(x). Since f — 1 and g(x) lie in Z[xIand g(x) is monic, the refinement of the division algorithm presented in Exercise 4implies that (x) E Z[xI. This completes the proof. •

Here are some examples of how to use the identity (9.4).

Example 9.1.6 Let p be prime. Proposition 9.1.5 implies that

— 1 = and — 1 =

Thus— 1 = (x" —

It follows that

=

X l=xp_1+xP_2+...+x2+x+1x— 1

by replacing x with

Example 9.1.7 In the examples of cyclotomic polynomials given so far, the coeffi-cients are always 0 or ±1. This is true for all n < 105. You will show in Exercise Sthat is the polynomial

1 +x+x2—x5—x6—2x7 —x8—x9+x'2+x'3+x'4+x'5

+x35 +x36 —x39 —x40 2x4' —x42 —x43 +x47 +x48.

As n increases, the coefficients of (x) can get arbitrarily large (see [11 and [4]).<1I>

C. The Galois Group of a Cyclotomic Extension. The first step in computingis to prove that is irreducible. For this, we need the following

application of symmetric polynomials and Lemma 9.1.2.

Lemma 9.1.8 Letf E 7L[x] be monic of positive degree, and let p be prime. 1ff,, isthe monic polynomial whose roots are the pth powers of the roots off, then:(a)

(b) The coefficients off and are congruent modulo p.

Proof: 1ff has roots r = deg(f), then


Similarly, f(x) + + In these for-mulas, al,.. , are the elementary symmetric polynomials from Chapter 2.

Observe that a, (xc,... , xfl is a symmetric polynomial. In Exercise 6 you willshow that the algorithm of Theorem 2.2.2 implies that

(9.6)

where S(ai,... , a polynomial in a1,... , with integer coefficients. However,if we reduce modulo p, then Lemma 5.3.10 implies that

as polynomials with coefficients in IF,. (see Exercise 6 for details). Combining thiswith (9.6), we see that the coefficients of S(ai, . . . , a,.) are all divisible by p.

Now substitute 'yi,••• , 'Yr for x1, . . . in (9.6). Since a ('y',... , Z for alli and S has integer coefficients, we conclude that a1 (-yf,... , 'yj') e Z. Since thecoefficients of S are all divisible by p, we also have

=aj(7l,...,'yr)" mai(y1,...,'yr) modp,

where the second congruence follows from Lemma 9.1.2. Thus the coefficients offand are congruent modulo p.

We now show that (x) is the minimal polynomial of over Q.

Theorem 9.1.9 The cyclotomic polynomial (x) is irreducible over Q.

Proof: Let f Q[x] be an irreducible factor of Then Gauss's Lemma, in theform of Corollary 4.2.1, allows us to assume thatf E Z[xJ and that

(9.7) =f(x)g(x),

for some g Z[x]. We can also assume that f and g are monic, since is.Let p be a prime not dividing n. The first step in the proof is to show that

(9.8) If is a root off, then so is

We will prove (9.8) by contradiction, so suppose that = 0 and 0.As in Lemma 9.1.8, let f,, E Z[x] be the monic polynomial whose roots are the

pth powers of the roots of f. In Exercise 7 you will show that the roots of f,, aredistinct primitive nth roots of unity, which implies that f,, divides (x). 1ff and

f would divide f,,, since f is irreducible. This would force

f = since they are monic of the same degree. But f = f,, is impossible, since0 and = 0 (the latter follows from = 0 by the definition of fr).

Thus they have no common roots, so that (9.7) can be written

=

Since (x), f(x), and (x) are monic with integer coefficients, the refined divisionalgorithm of Exercise 4 implies that the same is true for h(x).


Consider the map sending q(x) E Z[x] to the polynomial E IF1, [x] obtainedby reducing the coefficients of q(x) modulo p. Since f(x) = by Lemma 9.1.8,the above factorization implies that f2 (x) divides (x) in IF,, [xl. Thus f2 (x) divides

1, so that — 1 is not separable in IFp[xl. But — 1 is separable, since pin.This contradiction completes the proof of (9.8).

Now let be a fixed root of f and let be any primitive nth root of unity. InExercise 7 you will show that = (i for some j relatively prime ton. Let j = p1be the prime factorization of j, and note that each p, is relatively prime to n. Thensuccessive application of (9.8) shows that

are roots off. Hence every primitive nth root of unity is a root off. Since f divideswe conclude that f = Thus is irreducible, since f is. •

Theorem 9.1.9 implies that (x) is the minimal polynomial of over Q. Thus

[Q : Q] deg (x)) = which proves the following corollary.

Corollary 9.1.10 Q]

This makes it easy to compute the Galois group of a cyclotomic extension.

Theorem 9.1.11 There is an isomorphism such that a Emaps to [t] E (ZL/nZ)* and only =

Proof: We know from (8.6) that Q C is a Galois extension. Furthermore,an element a e is uniquely determined by which is a root of

because is. Thus = for some £ relatively prime ton. By Exercise 4of Section 6.2, the map a i—* [tJ is a well-defined one-to-one group homomorphism

(Z/nZ)'. Then Corollary 9.1.10 implies that

= = = (Z/nZ)*I

It follows that is an isomorphism.

In the next chapter we will use Corollary 9.1.10 to characterize those n for whicha regular polygon with n sides is constructible by straightedge and compass.

Historical Notes

While both Lagrange and Vandermonde made significant use of roots of unity,the first systematic study of cyclotomic extensions is due to Gauss. Most of Gauss'sresults appear in Section VII of Disquisitiones Arithmeticae [5], published in 1801.This amazing book covers a wide range of topics in number theory. In particular,Gauss introduces the congruence notation a b mod n and proves a version ofGauss's Lemma (Theorem A.3.2).

In Section VII Gauss studies the extension Q C where p is prime. As wewill see in the next section, Gauss constructs primitive elements for all intermediate


fields and essentially describes the Galois correspondence. In Article 365 of [5]

he applies his results to the constructibility of regular polygons by straightedge andcompass. We will discuss this in the next chapter.

To study Q C Gauss needed to know that = + + 1 is irre-ducible over Q. Not surprisingly, he proves this using Gauss's Lemma. For generaln> 1, the entry dated June 12, 1808 of Gauss's mathematical diary (see [6]) reads asfollows:

The equation... that contains all primitive roots of the equation x" — 1 =0 cannotbe decomposed into factors with rational coefficients, proved for compositevalues of n.

Unfortunately, Gauss's proof has been lost. The first published proof that (x) isirreducible (Theorem 9.1.9) appeared in 1854 and is due to Kronecker. Our proof isbased on arguments of Dedekind, as presented by Jordan in 1870. The key step is(9.8), which we proved using Lemma 9.1.8. Schönemann's proof of this lemma datesfrom 1846, though Gauss proved it much earlier in an unpublished continuation of[5]. A modem proof of (9.8) is sketched in Exercise 8.


Exercise 1. Prove that a congruence class [i] E 7L/nZ has a multiplicative inverse if and onlyif gcd(i,n) = 1. Conclude that (Z/n7L)* has order Be sure that you understand whathappens when n = 1.

Exercise 2. Assume that gcd(n,m) = 1. By Lemma A.5.2, we have a ring isomorphisma: 7L/nmZ Z/nZ x Z/mZ that sends [a]nm to ([a],,, Prove that a induces a groupisomorphism (Z/nmZ)* (Z/nZ)* x (7L/m7Z)*.

Exercise 3. Let = E C. Prove that for 0 < i < n and gcd(i, n) = 1 are the primitiventh roots of unity in C.

Exercise 4. Let R be an integral domain, and let f, g E R[X], where f 0. If K is thefield of fractions of R, then we can divide g by f in K[x] using the division algorithm ofTheorem A. 1.14. This gives g = qf + r, though q, r E K[x] need not lie in R[x].(a) Show that dividing x2 by 2x+ 1 in Q{x] gives x2 = (2x+ 1) + r, where q,r E Q{x] are

not in Z[x], even though x2 and 2x+ 1 lie in Z[x].(b) Show that if f is monic, then the division algorithm gives g = qf + r, where q, r E R[x].Hence the division algorithm works over R provided we divide by monic polynomials.

Exercise 5. Verify the formula for (x) given in Example 9.1.7.

Exercise 6. This exercise is concerned with the proof of Lemma 9.1.8.(a) Letf E Z[xi,. . . , x,,] be symmetric. Prove that lisa polynomial in ar,.. . , oP,, with integer

coefficients.(b) Letpbeprime and

Exercise 7. This exercise is concerned with the proof of Theorem 9.1.9.(a) Let ( be a primitive nth root of unity, and let i be relatively prime to n. Prove that ('is a

primitive nth root of unity and that every primitive nth root of unity is of this form.(b) Let y1,. be distinct primitive nth roots of unity and let i be relatively prime to n.

Prove that . . . , are distinct.


Exercise 8. This exercise will present an alternate proof of (9.8) that doesn't use symmetricpolynomials. Assume that ( is a root of f such that f((") 0. As in the text, q(x) E Z[x]maps to the polynomial E iF,, [x]. Let g(x) be as in (9.7).

(a) Prove that (is a root of g(x"), and conclude that f(x) g(x").

(b) Use Gauss's Lemma to explain why f(x) divides g(x") in Z[x], and conclude thatf(x)divides in Fp{x].

(c) Use Exercise 7 to prove that = and conclude thatf(x) divides

(d) Now let h(x) e be an irreducible factor off(x). Show that h(x) divides so that

h(x)2 dividesf(x)

(e) Conclude that h(x)2 divides x' 1 E Fp[xI.

(f) Use separability to obtain a contradiction.

Exercise 9. In proving Fermat's Little Theorem a mod p. recall from the proof of

Lemma 9.1.2 that we first proved a"1 1 mod p when a is relatively prime top. For general

n> 1, Euler showed that a is relatively prime to n. Prove this. What

basic fact from group theory do you use?

Exercise 10. Prove that a cyclic group of order n has generators.

Exercise 11. Prove that n =

Exercise 12. Here are some further properties of cyclotomic polynomials.

(a) Given n, let m = p. Prove that (x) = This shows that we can reduce

computing (x) to the case when n is squarefree.

(b) Let n> I be an odd integer. Prove that c12fl (x) = ( —x).

(c) Let p be a prime not dividing an integer n> 1. Prove that =

Exercise 13. We know when p is prime. Use this and Exercise 12 to compute

and

Exercise 14. The Möbius function is defined for integers n � 1 by

11, ifn=l,

10, otherwise.

Prove that = 0 when n> 1.

Exercise 15. Let be the Möbius function defined in Exercise 14. Prove that

= —

din

This representation of (x) is useful when studying the size of its coefficients.

Exercise 16. Let n and m be relatively prime positive integers.(a) Prove that Q((nm)

(b) Prove that 4,, (x) is irreducible over Q ((m)•


9.2 GAUSS AND ROOTS OF UNITY (OPTIONAL)

In this section we will explore how Gauss studied Q C Q(ç), where p is an oddprime. Working 30 years before Galois, Gauss described the intermediate fields ofthis extension and used his results to show that 1 = 0 is solvable by radicals.

A. The Galois Correspondence. If p is an odd prime, then Proposition 9.1.11implies that

Let us recall what we know about this group:

• (7L/pZ)* is cyclic of order p — 1 by Proposition A.5.3.• For every positive divisorf of p — 1, (7L/pZ) * has a unique subgroup H1 of order

f by Theorem A. 1.4.

Following Gauss, we let e = Thus

ef = p — 1,

and H1 has index e in (Z/pZ)*. We will use this notation throughout the section.One further fact not mentioned earlier is the following:

• 1ff and f' are positive divisors of p — 1, then H1 C H1' if and only if f If.You will prove this in Exercise 1. Hence we can easily check when one subgroup iscontained in another.

By the isomorphism Gal(Q(ç)/Q) (Z/pZ)" and the Galois correspondence,the intermediate fields of Q C Q(ç) are the fixed fields

[i]EH1}

as f ranges over all positive divisors of p — 1. These fixed fields have the followingnice properties.

Proposition 9.2.1 The intermediate fields Q C Lj- C satisfy:(a) Lj is a Galois extension of Q of degree e.(b) 1ff and f' are positive divisors of p 1, then L1 1 Lf' if and only If'.(c) 1ff and f' are positive divisors of p — 1 such that then is

cyclic of order fl/f.

Proof: You will supply the straightforward proof in Exercise 2. U

In particular, if p — 1 = q,- is the prime factorization of p — 1, then we getsubfields

(9.9) Q = Lq1...q, C Lq2...qr C C Lq,_1q, C Lqr CL1 =

where : = qi. Thus every element of is the root of apolynomial of degree qj over

All of this is a simple application of Galois theory. The surprise is that Gaussunderstood most of this, including (9.9). Before discussing Gauss's results, let us doan example.

GAUSS AND ROOTS OF UNITY (OPTIONAL) 239

Example 9.2.2 Let p = 7. Then (9.9) with p — 1 = 6 = 3 •2 becomes

Q=L6CL2cL1

where L2 is the fixed field of the unique subgroup of order 2 of Gai(Q((7)/Q).To make this more explicit, consider m = + = + = 2cos(2ir/7).

In Exercise 3 you will show that corresponds to the subgroup {e,'r} ofGal(Q((7)/Q), where r is complex conjugation. This subgroup has order 2, whichimplies that

L2 =Q(m).

In Exercise 3 you will also show that the conjugates of m over Q are

= + = 2cos(4ir/7) and = + = 2cos(6ir/7),

and that , are roots of the cubic equation

y3 +y2—2y— 1=0.

It is easy to check that is a root of x2 — mx + 1 E L2[xI. From here we canexpress (7 in terms of radicals as follows. Applying Cardan's formulas to the abovecubic, one sees that

(9.10)

provided that the cube roots are chosen correctly (see Exercise 3). Then applying the

quadratic formula to x2 — + I = 0 gives

(9.11)

where we use the same cube roots as in (9.10).

Notice how (9.11) is similar to the formula

i

4 2

from Exercise 8 of Section A.2. These formulas were known to Lagrange andVandermonde in the 1 770s. Vandermonde also worked out a similar formula forwhich is more surprising in that it required solving an equation of degree 5 by radicals(see [Tignol, Ch. lii).

B. Periods. In Section VII of Disquisitiones, Gauss proves the existence of radicalformulas for for any odd prime p. His proof uses periods, which for positivedivisors f of p 1 are carefully chosen primitive elements of L1 over Q.


Let ef = p 1, and let H1 c (7Z/pZ)* be the unique subgroup of order f. Givenan element a = [i] E (Z/pZ) set = This is well defined, since = 1. Hencewe can make the following definition.

Definition 9.2.3 Let A e Z be relatively prime to p. This gives [A] e (Z/pZ)* andthe coset [A]H1- of Hf in (7Z/pZ)*. Then we define an f-period to be the sum

(f,A)=aE[AIHj

Here are some simple properties of f-periods.

Lemma 9.2.4 Let ef = p 1, and let (f, A) be defined as above. Then:(a) Two f-periods either are identical or have no terms in common.(b) There are e distinct f-periods.(c) The f-periods are linearly independent over Q.(d) Let a E Gal(Q(ç)/Q) satisfy o-(ç) = Then, for any f-period (f, A),

a((f,A)) = (f,iA).

Proof: Recall that i,ç, ., e Q(ç) are linearly independent over Q, since

[Q(ç) = p — 1. Multiplying by shows that the same is true for ç, . . .

This implies that two f-periods coincide if and only if the corresponding cosets ofH1 are equal. Then part (a) follows because cosets are either identical or disjoint, andpart (b) because the number of cosets is the index of Hf in (Z/pZ) which is e =Then part (c) is a consequence of part (a) together with the linear independence of

• ,(P overQ.For part (d), observe that = Thus = c implies that

aE[A]H1 aE[A]Hf bE[iAIHj

where the third equality follows via the substitution b = [i]a.

Here are some particularly simple periods.

Example 9.2.5 Since p is odd, the unique subgroup of (Z/pZ)* of order 2 is H2 ={[l], [—l]}. The cosets of this subgroup are [A]H2 = {[A], [—A]}, so that the 2-periodsare

The number of 2-periods is e =In particular, when p = 7, the distinct 2-periods are (2, 1), (2,2), and (2,3). These

were denoted and in Example 9.2.2.

We now prove that f-periods give the desired primitive elements.


Proposition 9.2.6 Let Lf be the fixed field of H1. Then:(a) Let (f,A1),...

, (f, Ae) be the distinct f-periods. Then

g(x)=(x—(f,Ai))...(x—(f,Ae))

is in Q [xJ and is the minimal polynomial of any f-period over Q.

(b) Any f-period is a primitive element of L1 over Q.

Proof: An f-period ij = (f,A) corresponds to a coset [A}H1. If [iJ (Z/pZ)*,then the f-period (f, iA) corresponding to [iA]H1 is a conjugate of ij over Q, byLemma 9.2.4. Since [iA]Hj gives all cosets of H1 as we vary [ii, the conjugates of

over Q are the e distinct f-periods (f, A1),. . . , (f, As). Then part (a) follows fromthe formula for the minimal polynomial given in equation (7.1) of Chapter 7.

Hence Q C Q CL1 are extensions of degree e. Since(Z/pZ)* has a unique subgroup of index e, the Galois correspondence implies that

= Lf. This proves part (b). U

As a corollary, we get the following interesting basis of Lf over Q.

Corollary 9.2.7 The f-periods fonn a basis of L1 over Q.

Proof: The f-periods lie in L1 by Proposition 9.2.6. Furthermore, Lemma 9.2.4tells us that thee such periods are linearly independent over Q. The corollary follows,since [L1 : Q] = e by Proposition 9.2.1. •

Our next task is to describe the extension L1 in terms of periods, where f andf' are positive divisors of p — 1 satisfying f If'. Set d = f'/f, so that [L1 : L1'I = d.Any f-period (f, A) is a primitive element of Lj- over We need to describe itsminimal polynomial over

This is done as follows. Observe that H1 is a subgroup of index d =f'/f in Hp.Hence every coset of in (Z/pZ)* is a disjoint union of d cosets of H1. (DoExercise 4 if you are unsure of this.) In particular, is a disjoint union

(9.12) = .

where we may assume Ai = A, since [A]H1 C This leads to the followingdescription of the desired minimal polynomial.

Proposition 9.2.8 Let f and f' be positive divisors of p — 1 such that and setd = f'/f. Given an f-period (f, A), let A1 = A, A2,... ,Ad be as in (9.12). Then

h(x) = (x— (f,A1)) (x— (f,Ad))

is in [xI and is the minimal polynomial of (f, A) over L1'.


Proof: The proof is similar to what we did in part (b) of Proposition 9.2.6. Setting

= (f, A), we need to show that as a varies over the elements a(ij)give the f-periods (f, A1),..., (f, Ad).

To prove this, let a E so that a(ç) = for [i] E Hf'. Then

=a((f,A)) = (f,iA).

This f-period corresponds to the coset However,

[iA]Hf C = [A] =

where the final equality uses [i] By (9.12), it follows that [iA]Hf [AJ]Hf forsome j, so that a(ii) = (f,iA) = Since every (f,A1) arises in this way (seeExercise 5), the proposition is proved. •

We will give an example of Proposition 9.2.8 below.

C. Explicit Calculations. The above results are pretty but somewhat abstract. Tocompute specific examples, we need a concrete way to work with periods. The keyidea, due to Gauss, is to pick a generator [g] of the cyclic group (Z/pZ) Since thisgroup has order p — 1, it follows that

(Z/pZ)* = {[l],[g],[g2],... ,[gP_2]}.

In other words, the p — 1 numbers 1, g, g2,... , represent the nonzero congruenceclasses modulo p. We call g a primitive root modulo p.

Given a primitive root g and ef = p — 1 as usual, Exercise 1 implies that isgenerated by ge, i.e.,

Hf=

It follows that the coset [A]Hj gives the f-period

f—I

(9.13) (f,A) = + + + + =

1=0

So far, we have assumed that [A] E (Z/pZ)*, i.e., pTA. However, (9.13) makes sensefor any integer A. Since = 1, one easily sees that

(f,A)=f when

For an arbitrary A E Z, we call (f, A) a generalized period. Thus a generalized periodis an ordinary period if ptA and is equal to f if

In order to compute the minimal polynomials appearing in Propositions 9.2.6and 9.2.8, we need to know how to multiply f-periods. Gauss expressed the productof two f-periods in terms of generalized periods as follows.


Proposition 9.2.9 If (f, and (f, it) are f-periods with p and p then

f—I

=[X'IE[AIHf j0

Proof : Following [5, Art. 345], we set h = ge, so that

f—I=

since [h] generates Hf. We also have = [Aht]Hf for any £, which implies that(f,A) = (f,Aht). Thus

f—I f—I

it) =

f—I f—I f—I f—I

= =

t=0 j=0 j=0 £=0

f—I

1=0

This gives the desired formula, since h = •

Here is an example from [5, Art. 3461.

Example 9.2.10 In this example and three that follow, we will consider the 6-periodsfor p = 19. In Exercise 7 you will show that g = 2 is a primitive root modulo 19.Since f = 6 implies e = 3, the unique subgroup of order 6 in (7L/197L)* is generatedby [8]. Thus

H6 = {[1], [8], [812, [8], [7], [18], [11], [12]} c (Z/19Z)*.

For simplicity, we will write [n] as n, so that

H6= {l,7,8,1l,12,18}.

The e = 3 cosets of H6 in (Z/ I 9Z) * are H6 together with

2116 = {2, 14,16,22,24, 36} = {2,3,5, 14,16, 17},

4116 = {4,28,32,44,48,72} = {4,6,9, 10,13, 15}.

(Remember that we are working modulo 19.)According to Proposition 9.2.9,

(6,1)2 = (6,1 +1)+(6,7+1)+(6,8+1)+ (6,1 1+l)+(6, 12+1)+(6, 18+1)=(6,2)+(6,8)+(6,9)+(6,12)+(6,13)+6,


where the second equality uses (6, 19) = 6. This shows that generalized periods canarise when we multiply ordinary periods. However,

(6,8) = (6,1)

since 8 and 1 lie in the same coset of H6. Using similar simplifications, we get

(6,1)2 = 2(6,1) + (6,2) + 2(6,4) +6.

By Exercise 6 we also have

(6,l)+(6,2)+(6,4)=—1.

Then the formula for (6, 1)2 simplifies to

(6,1)2 =4—(6,2).

You will work out similar formulas in Exercise 7.

Example 9.2.11 Still assuming p = 19, our next task is to compute the minimalpolynomial of the 6-periods over Q. We will use the notation of the previousexample. By Proposition 9.2.6, the minimal polynomial is

(9.14) (x— (6,1)) (x— (6,2)) (x— (6,4)).

In Exercise 7 you will use the methods of Example 9.2.10 to show that

(6,l)(6,2) = (6,1)+2(6,2)+3(6,4),(9.15) (6, l)(6,4) = 2(6,1) +3(6,2) + (6,4),

(6,2)(6,4) =3(6,1)+(6,2)+2(6,4).

Note that these sum to 6(6, 1) +6(6,2) +6(6,4) = —6, since (as noted above) (6, 1) +(6,2)+(6,4) = —1.

Using (9.15) and (6, 1)2 = 4 — (6,2) (from Example 9.2.10), we have

(6,l)(6,2)(6,4) = (6,l)(3(6,1)+(6,2)+2(6,4))= 3(6,1)2+ (6, l)(6,2) +2(6, l)(6,4)

= 12—i-5(6,1)+5(6,2)+5(6,4)=7

(see Exercise 7 for the details). It follows that multiplying out (9.14) gives

(9.16) x3+x2—6x—7.

This is the minimal polynomial of the 6-periods over Q. Its splitting field is Q C L6,

the extension generated by the 6-periods.


Example 9.2.12 Now consider the 3-periods for p = 19. Since 6/3 = 2, we seethat L6 C L3 is an extension of degree 2. Hence 3-periods have quadratic minimalpolynomials over L6.

Since 2 is a primitive root modulo 19, the subgroup H3 C (Z/ 1 9Z)* is generatedby [216 = [812 = Using the notation of Example 9.2.10, we have

H6={1,7,8,ll,12,18}={1,7,11}U{8,12,18}=H3U8H3.

This shows that(6,1)= (3,1)+(3,8),

and in a similar way, one obtains

(6,2) = (3,2) + (3,16),(6,4) = (3,4)+(3,13).

However, Proposition 9.2.9 implies that

(3,l)(3,8) = (3,1+8)+(3,7+8)+(3,11+8)= (3,9)+(3,15)+3,

and since (3,9) = (3,4) and (3,15) = (3, 13) (do you see why?), we get

(3, 1) (3, 8) = (3, 4) + (3, 13) +3= (6, 4) +

By Proposition 9.2.8, the minimal polynomial of (3,1) and (3,8) overL6 is

(9.17) (x—(3,1))(x—(3,8))=x2—(6,1)x+(6,4)+3.

Exercise 7 will consider the minimal polynomials of the other 3-periods.

Example 9.2.13 The 1-periods for p 19 are the primitive 19th roots of unity(l,A) = for A = 1,...,18. In Example 9.2.12, we noted that H3 = {l,7,ll},which means that

(3,1) =By Exercise 7 the minimal polynomial of and over L3 is

(9.18) 1.

Combining this with (9.17) and (9.16), one can write an explicit formula for thatinvolves only square and cube roots.

In Exercises 8 and 9 you will use similar methods to derive the formula

cos(27r/17) = +(9.19)

due to Gauss. In Chapter 10, we will see that this leads immediately to a straightedge-and-compass construction of a regular 17-gon.


One reason these methods work so well is that the f-periods are linearly inde-pendent over Q by Lemma 9.2.4. Hence any linear combination of f-periods withcoefficients in Z or Q is unique. However, we've seen cases where generalizedperiods (f, A), also occur. But this is no problem, since (f, A) = f in such acase, and we also know that the distinct f-periods sum to —1 (see Exercise 6). Thusa generalized f-period can be expressed in terms of ordinary f-periods. Hence wecan always reduce to an expression involving only f-periods, where we know thatthe representation is unique.

D. Solvability by Radicals. When studying Q C we saw in (9.9) that aprime factorization p — I = qi gives intermediate fields

such that •q, : .q,] = qj. If we focus on one of these fields and the nextlarger one, then we get an extension of the form

(9.20) Ljq C L1

where fq divides p — 1 and q is prime. The theory of periods shows that (f, 1) isa primitive element of Lf and the examples given above make it clear that in anyparticular case we can compute the minimal polynomial of (f, 1) over Lfq.

When p = 19, the minimal polynomials found in Examples 9.2.10—9.2.13 havedegrees 2 or 3. Hence their roots can be found by known formulas. But when p = 11,

the period (2,1) = 2cos(2ir/ll) has minimal polynomial

y5+y4—4y3—3y2+3y+ 1

(see Exercise 10). Is this polynomial solvable by radicals? More generally, are theminimal polynomials of periods solvable by radicals?

For a theoretical point of view, this question is trivial, since Q C Q(ç) is a radicalextension = 1 E Q). It follows by definition that any f-period (f, A) is expressibleby radicals over Q, since (f, A) E Q(ç). Things become even more trivial if yourecall that when we studied solvability by radicals in Chapter 8, we felt free to adjoinany roots of unity we needed, including ç.

Hence it appears that solving the minimal polynomials of periods by radicals iscompletely uninteresting. The problem is that this ignores the inductive nature ofwhat's going on. The real goal, which goes back to Lagrange's strategy for solvingequations, is to construct pth roots of unity using only radicals and roots of unity oflower degree (we will discuss Lagrange's strategy in Chapter 12). This is what Gaussdoes in Disquisitiones.

Thus, when studying Q C we may assume inductively that we know allmth roots of unity for m <p. Furthermore, as explained in the discussion preceding(9.20), it suffices to consider the extension Lfq C L1, where we may assume that thefq-periods are known. The idea is to express an f-period in terms of radicals thatare qth roots involving fq-periods and qth roots of unity. These roots of unity areknown, since q < p.


To do this in practice, we will use Lagrange resolvents. Let w be a primitive qthroot of unity. In Exercise 11 you will prove that

Gal(Lf(w)/Lfq(w)) Gal(Lf/Lfq) Z/q7L.

Since Lfq(W) contains a primitive qth root of unity, Lemma 8.3.2 implies that Lf(w) isobtained from Lfq (w) by adjoining a qth root. Furthermore, the proof of Lemma 8.3.2shows that the element adjoined is a Lagrange resolvent. Recall from (8.7) that ifa is a generator of Gal(Lf(w)/Lfq(w)) and /3 E Lf(W), then we get the Lagrangeresolvents

for i = 0,... ,q — 1. We will use /3 = (f, 1) ELf C Lj(w). In Exercise 11 you willshow that we can pick the generator a so that for any f-period (f, A),

(9.21) a((f,A)) = (fge/qA)

(note that e, since fq 1). Thus the above Lagrange resolvents can be written

(9.22) = (f, 1) + . ..

If we set A1 = a7, then we can define = Then the f-periods in (9.22) can beexpressed in terms of radicals as follows.

Theorem 9.2.14 Let and A, = be defined as above.(a)

(b)

(fgte/q) = . . .

Before beginning the proof, let us explain the f-periods appearing in the theorem.The extension Lfq C Lf corresponds to the subgroups Hf C Hfq of (Z/pZ) *• Since

e = Exercise 1 shows that these subgroups are generated by [ge] andrespectively. In Exercise 11 you will use this to prove that

(9.23) Hfq = Hf U U u . . . u

By Proposition 9.2.8, the f-periods (f, are the conjugates of (f, 1) over Lfq.

Proof of Theorem 9.2.14: Part (a) follows easily from the properties of Lagrangeresolvents presented in the proof of Lemma 8.3.2. For part (b), let = Thenfor any integer m we have

=

q—1 q—I

= ( (f,At) = q(f,Am),


where the last equality follows from Exercise 9 of Section A.2. This gives the desiredformula for (f, )'m), since = for i > 0.

From a computational point of view, the results of this section give a systematicmethod for expressing A = c4 in terms of fq-periods and qth roots of unity. Thisworks because f-periods and fq-periods are linearly independent not only over Qbut also over Q(w), where w is a primitive qth root of unity (you will prove thisin Exercise 12). Thus the radical formula for given in Theorem 9.2.14 isexplicitly computable.

Mathematical Notes

Here are comments on two topics relevant to what we did in this section.

• Primitive Roots Modulo p. The formulas presented in this section illustrate theusefulness of knowing primitive roots modulo p. Gauss explains a method for findingprimitive roots in [5, Art. 73—74]. See also [10, p. 163].

Let gp denote the smallest positive primitive root modulo p. For example, 2 is aprimitive root modulo 19, which implies that gj9 = 2. In 1962 Burgess [3] provedthat for any r > 0 there is a positive constant C(r) such that

gp

for all odd primes p. This says that gp can't be too big relative to p. On the otherhand, Kearnes [9] proved in 1984 that given any integer m > 0 there are infinitelymany primes p > m such that gp > m. So gp can still get large.

If we fix a primitive root g modulo p, then the discrete log problem asks thefollowing: Given an integer a not divisible by p. find i such that a g' mod p.We write this as i = logga. It is easy to describe an algorithm for finding log5a(divide a — g' by p for i = 0, 1,2 and stop when the remainder is zero). Butfinding an efficient algorithm for logga is much more difficult. Several modemencryption schemes, including the Pohig—Hellman symmetric key exponentiationcipher (described in [10, Sec. 3.11) and the Diffie—Heilman key exchange protocol(described in [2, Sec. 7.4] and [10, Sec. 3.1]), would be easy to break if discrete logswere easy to compute.

Primitive roots modulo p are also used in the Digital Signature Algorithm sug-gested by the National Institute of Standards and Technology. A description can befound in [2, Sec. 11.5]. As above, one could forge digital signatures if discrete logswere easy to compute.

There are also purely mathematical questions about primitive roots modulo p. Alist of unsolved problems can be found in [7, Sec. F.9].

• Periods and Gauss Sums. Let p = 17. By Exercise 9 we have

(8,3) = I


which easily implies(8,1) (8,3) =

In Exercise 13 you will show that this can be written

(9.24)

where, for an odd prime p. the Legendre symbol is defined by

('0,

if px2 modphasno solution.

More generally, for an odd prime p. a quadratic Gauss sum is defined to be

Gauss used these sums to prove quadratic reciprocity. He also proved the remarkableformula

ifpmlmod4,— ifpm3mod4.

Notice how this generalizes (9.24). A careful discussion of quadratic Gauss sumscan be found in [8, Ch. 6].

Historical Notes

Most results of this section are implicit in Section VII of Disquisitiones. Themain difference is that we have stated things in terms of the Galois correspondence,which to each divisor f of p 1 associates the subgroup Hf of (Z/pZ) * and thesubfield Lf of Q(ç). For Gauss, on the other hand, each divisor f gets associated tothe collection of f-periods (f, A). In general, he considers elements rather than thefields in which they lie. For example, consider [5, Art. 3461, which asserts that given(f, A), any other f-period (f, can be expressed as

(f, A) + a2(f, A)2 + ... + ae_i(f,

for some uniquely determined integers a0, . . . , For us, this gives the uniquerepresentation of (f, tt) as an element of Lf = Q ((f, A)).

Another difference is our use of cosets. For example, if g is a primitive rootmodulo p and f divides p — 1, then Gauss notes that the distinct f-periods are


where e = For us, this follows from Lemma 9.2.4, since H1 C (Z/pZ)* isgenerated by [ge], so that its cosets in (Z/pZ)* are

[1]H1, [g]H1, [g2]Hf,... ,

Cosets give a conceptual basis for what Gauss is doing, and the same is true for theminimal polynomials computed in Proposition 9.2.8.

It is also interesting to note that Gauss makes implicit use of the Galois groupWe saw in Lemma 9.2.4 that a(ç) = implies that a((f,A)) =

(f, k.\). Now consider the following quote from [5, Art. 345]:

IV. It follows that if in any rational integral algebraic function F = u, v,...)we substitute for the unknowns t, u, v, etc. respectively the similar periods(f, X), (f, ii), (f, ii), etc., its value will be reducible to the form

A +B(f, l)+B'(f ,g)+B"(f ,g2)... +BE(f,ge_l)

and the coefficients A, B, B', etc. will all be integers if all the coefficients in F are

integers. But if afterward we substitute (f,kA), (f,ku), (f,kv), etc. for t,u,v,etc. respectively, the value of F will be reduced to A + B(f, k) + B'(f, kg) + etc.

A "rational integral algebraic function" is a polynomial with coefficients in Q. Hereis an example of what this means.

Example 9.2.15 In Example 9.2.10, we showed that

(6,1)2=4_(6,2)

when p = 19. Using k = 2, the above quotation from Gauss tells us that

(6,2)2 = 4— (6,4).

In modem terms, this follows by applying the automorphism cr Ethat takes to So the Galois action is implicit in Gauss's theory!

Gauss's result thatxP — 1 is solvable by radicals is less compelling from the modemperspective, though it is still interesting when one thinks inductively. But historically,being able to solve special but nontrivial equations of high degree was important.Here is what Gauss says in [5, Art. 3591:

Everyone knows that the most eminent geometers have been unsuccessful inthe search for a general solution of equations higher than the fourth degree,or (to define the search more accurately) for the THE REDUCTION OF MIXEDEQUATIONS TO PURE EQUATIONS. ... Nevertheless, it is certain that thereare innumerable mixed equations of every degree that admit a reduction to pureequations, and we trust that geometers will find it gratifying if we show that ourequations are always of this kind.

For Gauss, an equation is "pure" if it is of the formxm —A = 0 and "mixed" otherwise.Thus, reducing "mixed equations to pure equations" is what we call solvability by


radicals. Of course, in saying "our equations," Gauss is referring to the minimalpolynomials satisfied by the periods, as constructed in Proposition 9.2.8.

Gauss's study of the pth roots of unity is an important midpoint in the developmentleading from Lagrange to the emergence of Galois theory. Gauss uses Lagrange'sinductive strategy to work out the Galois correspondence for Q C and histheory of periods makes everything explicit and computable. He also shows thatLagrange resolvents are the correct tool for studying solvability by radicals, pavingthe way for Galois's analysis of the general case.

In spite of its beauty, what Gauss does in Section VII of [5] is not perfect. Someproofs are omitted and others have gaps. For example, Gauss does not prove theassertion about the Galois action made in the quotation before Example 9.2.15. Also,as noted in [Tignol, p. 1951, Gauss's study of solvability assumes without proof thatwhen fq divides p — 1, the f-periods are linearly independent over the field generatedby the qth roots of unity. (You will prove this in Exercise 12.)

Galois was very aware of Section VII of Disquisitiones. For example, Galoisdescribes the "group" of Q c n prime, as follows [Galois, pp.51—53]:

In the case of the equation = 0, if one supposes a = r, b = r5, c =g being a primitive root, the group of permutations will simply be as follows:

a b c d k

b c d ... k a

c d ... k a b

k a b c ... i

in this particular case, the number of permutations is equal to the degree ofthe equation, and the same will be true for equations where all of the roots arerational functions of each other.

Here, r is a primitive nth root of unity. Each line is a cyclic permutation of the oneabove it, which leads to a cyclic group of order n — 1. This quotation also reveals thatfor Galois, a "permutation" was an arrangement of the roots and that the permutations(in the modern sense) are obtained by mapping the first arrangement in the table tothe others. You will work out the details of this in Exercise 14. We will say moreabout how Galois thought about Galois groups in Chapter 12.


Exercise 1. Let G be a cyclic group of order n and let g be a generator of G.(a) Let f be a positive divisor of n and set e = n/f. Prove that H1 = (ge) has order f and

hence is the unique subgroup of order f.(b) Let f and f' be positive divisors of p — 1. Prove that Hf C if and only if f If'.

Exercise 2. Prove Proposition 9.2.1.

Exercise 3. Let m be as in Example 9.2.2.(a) We know that is a root ofx6 +x5 +x4 +x3 +x2 +x+ 1 = 0. Dividing byx3 gives

x3 i = 0.


Use this to show that are roots ofy3 +y2 — 2y— 1.

(b) Prove that : Q] = 3, and conclude that is the fixed field of the subgroup{e,r} C Gal(Q((7)/Q), where r is complex conjugation.

(c) Prove (9.10).

Exercise 4. Let A C B be subgroups of a group G, and assume that A has index d in B. Provethat every left coset of B in G is a disjoint union of d left cosets of A in G.

Exercise 5. Complete the proof of Proposition 9.2.8.

Exercise 6. Prove that the sum of the distinct f-periods equals —1.

Exercise 7. This exercise is concerned with the details of Examples 9.2.10, 9.2.11, 9.2.12, and9.2.13.(a) Show that 2 is a primitive root modulo 19.(b) Use the methods of Example 9.2.10 to obtain formulas for (6,2)2 and (6,4)2.(c) Show that the formulas of part (b) follow from (6,1)2 = 4— (6,2) and part (d) of

Lemma 9.2.4.

(d) Prove (9.15) and use this and Exercise 6 to show that (6, l)(6,2)(6,4) = 7.

(e) Find the minimal polynomials of (3,2) and (3,4) over the field considered in Exam-pie 9.2.12.

(f) Show that (9.18) is the minimal polynomial of (19 over the field L3 considered in Exam-ple 9.2.13.

Exercise 8. In this exercise and the next, you will derive Gauss's radical formula (9.19) forcos(2ir/l7).(a) Show that 3 is a primitive root modulo 17.(b) Show that

H8 = {1,2,4,8,9, 13,15, 16},

H4 = {l,4, 13, 16},

H2 = {l, 16},

where we write the congruence class [n] modulo 17 as n.

(c) Use Propositions 9.2.8 and 9.2.9 to compute the following minimal polynomials:

Extension Primitive Elements Minimal Polynomial

QCL8 (8,l),(8,3) x2+x—4

L8CL4 (4,l),(4,2)(4,3),(4,6)

x2—(8,1)x—l

x2—(8,3)x—l

L4 CL2 (2,1),(2,4) x2—(4,l)x+(4,3)

The resulting quadratic equations are easy to solve using the quadratic formula. But how dothe roots correspond to the periods? For example, the roots (8, 1), (8,3) of x2 + x — 4 are(—1 ± .,/17)/2. How do these match up? The answer will be given in the next exercise.

Exercise 9. In this exercise, you will use numerical computations and the previous exercise tofind radical expressions for various f-periods when p = 17.


(a) Show that

(8,1) =2cos(2ir/17)+2cos(4ir/17)+2cos(8ir/17)+2cos(l6ir/17),(4,1) = 2cos(2ir/17) +2cos(8ir/17),(4,3) = 2cos(6ir/17)+2cos(lOir/17),

(2, 1) = 2cos(2ir/17).

Then compute each of these periods to five decimal places.(b) Use the numerical computations of part (a) and the quadratic polynomials of Exercise 8

to show that

(8,1) =

________

(4,1) =

(4,2)= l+V'i7-

(4,3)=

(c) Use the quadratic polynomial x2 —(4, l)x+ (4,3) and part (b) to derive (9.19).

Exercise 10. Let p 11. Prove that y5 + y4 — 4y3 3y2 + 3y + 1 is the minimal polynomialof the 2-period (2,1) = 2cos(2ir/ll).

Exercise 11. Let C be the extension studied in Theorem 9.2.14. Thus f and fq dividep — 1, and q is prime. As usual, ef = p I and g is a primitive root modulo p. Finally, let wbe a primitive qth root of unity.

(a) Let T e Gal(Q(ç)/Q) satisfy=

and let a' = TIL be the restriction of r to

L1. Prove that a' generates Gal(L1/L1q).(b) Prove that Gal(Lf(w)/Lfq(w)) where the isomorphism is defined by

restriction to Lf.(c) Let a E Gal(Lf(w)/L14(w)) map to the element a' E Gal(Lf/Lfq) constructed in part (a).

Prove that a satisfies (9.21).(d) Prove the coset decomposition of Hfq given in (9.23).

Exercise 12. Let p be an odd prime, and let m be a positive integer relatively prime to p.(a) Prove that i,ç,. . , are linearly independent over Q(ç).

(b) Explain why part (a) implies that. . ,

are linearly independent over(c) Let — 1. Prove that the f-periods are linearly independent over


Exercise 14. Consider the quotation from Galois given at the end of the Historical Notes.(a) Show that the permutations obtained by mapping the first line in the displayed table to

the other lines give a cyclic group of order n 1. Also explain how these permutationsrelate to the Galois group.

(b) Explain what Galois is saying in the last sentence of the quotation.

Exercise 15. What are the 1-periods?


Exercise 16. Redo Exercise 3 using periods.

Exercise 17. Let f be an even divisor of p — 1, where p is an odd prime. Prove that everyf-period (f,A) lies in R.

REFERENCES

1. G. Bachman, On the coefficients of cyclotomic polynomials, Mem. Amer. Math. Soc. 106(1993).

2. J. A. Buchmann, Introduction to Cryptography, Springer, New York, Berlin, Heidelberg,2001.

3. D. A. Burgess, On character sums and L-series, Proc. London Math. Soc. 12 (1962),193—206.

4. G. P. Dresden, On the middle coefficient of a cyci otomic polynomial, Amer. Math. Monthly111 (2004), 53 1—533.

5. C. F. Gauss, Disquisitiones Arithmeticae, Leipzig, 1801. Republished in 1863 as VolumeI of [Gauss]. French translation, Recherches Arithmétiques, Paris, 1807. Reprint byHermann, Paris, 1910. German translation, Untersuchungen über Höhere Arirhmetik,Berlin, 1889. Reprint by Chelsea, New York, 1965. English translation, Yale U. P., NewHaven, 1966. Reprint by Springer, New York, Berlin, Heidelberg, 1986.

6. J. J. Gray, A commentary on Gauss's mathematical diary, 1796—1814, with an Englishtranslation, Expo. Math. 2 (1984), 97—130. (The Latin original of Gauss's diary isreprinted in [Gauss, Vol. X.1J.)

7. R. K. Guy, Unsolved Problems in Number Theory, Springer, New York, Berlin, Heidel-berg, 1994.

8. K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer,New York, Berlin, Heidelberg, 1982.

9. K. Kearnes, Solution of Problem 6420, Amer. Math. Monthly 91 (1984), 521.

10. R. A. Mollin, Introduction to Cryptography, Chapman & Hall/CRC, Boca Raton, FL,2001.

CHAPTER 10

GEOMETRIC CONSTRUCTIONS

The idea of geometric constructions using straightedge and compass goes back to theancient Greeks. A straightedge is an unmarked ruler. This chapter will explore thesurprising connection between geometric constructions and Galois theory. Topicscovered include classic problems from Greek geometry, the work of Gauss describedin Chapter 9, and the use of origami to solve cubic and quartic equations.

10.1 CONSTRUCTIBLE NUMBERS

We assume that you remember how to do standard straightedge-and-compass con-structions such as bisecting a given angle, erecting a perpendicular to a given line ata given point, and dropping a perpendicular from a given point to a given line.

To prove theorems about geometric constructions, we need a careful descriptionof what a construction is. The basic idea is that we begin with some known points.From these points, we use straightedge and compass to construct lines and circles:

Cl. From we can draw the line £ that goes through and /3.

C2. From /3 and -y, we can draw the circle C with center -y whose radius is thedistance from to /3.


256 GEOMETRIC CONSTRUCTIONS

In this labeling, "C" stands for "Construct." From these lines and circles, their pointsof intersection (assuming they are nonempty) give new points:

P1. The point of intersection of distinct lines Li and £2 constructed as above.The points of intersection of a line £ and circle C constructed as above.

P3. The points of intersection of distinct circles C1 and C2 constructed as above.

Here, "F" stands for "Point." We regard these newly constructed points as known.Then we can apply Cl, C2, P1, P2, and P3 to our enlarged collection of known points.We continue this process until the construction is completed.

For us, the plane will be the field of complex numbers C, so that constructing apoint means constructing a complex number. Our constructions will all start fromthe same two numbers, 0 and 1. This leads to the following definition.

Definition 10.1.1 A complex number a is constructible there is afinite sequenceof straightedge-and-compass constructions using Cl, C2, P1, P2, and P3 that beginswith 0 and 1 and ends with a.

Here are some examples of constructible numbers.

Example 10.1.2

(a) From 0 and 1, we get the x-axis using Cl and the circle of radius 1 centered at1 using C2. These intersect in the numbers 0 and 2. By P2, 2 is constructible.Iterating this shows that every n E Z is constructible.

(b) In Exercise 1 you will use standard methods from high school geometry toconstruct the line perpendicular to the x-axis at 0. This will show that the y-axisis constructible. Then use C2 to construct the circle of radius 1 centered at 0.These intersect in ±i. By P2, i E C is constructible.

These constructions will be useful later.

Example 10.1.3 Suppose that we can construct a regular polygon with n sides some-where in the plane (we call this a regular n-gon). Using two consecutive vertices andthe center of the n-gon, we get the triangle shown here:

(We may have constructed the center in the process of constructing the n-gon. Ifnot, then the center is constructible, since it is the intersection of the bisectors of the

CONSTRUCTIBLE NUMBERS 257

interior angles.) One easily sees that the angle at the center is 0 = 2ir/n. In Exercise 2you will show how to copy this angle to the origin:

Intersecting this with the unit circle shows that the nth root of unity = is

constructible. In Exercise 2 you will show that this process can be reversed. Henceis constructible if and only if a regular n-gon can be constructed by straightedge

and compass. Section 10.2 will determine those n's for which this is possible.

The set of constructible numbers has the following properties.

Theorem 10.1.4 The set = {a E C a is constructible} is a subfield of C. Fur-the rmore:

(a) wherea,bEllta E implies that

Proof: We first show that 'it' is a subgroup of C under addition. Given a <it' \ {0},construct the line connecting 0 and a by Cl and the circle of radius al centered atthe origin by C2. These intersect in ±a, so that —a is constructible by P2.

Now suppose that a and 13 are constructible. If a, and 0 are not collinear, thenuse C2 twice to construct the circle of radius a with center 13 and the circle of radius

with center a. One of the points of intersection is a + 13:

By P3, we conclude that a + 13 is constructible. In Exercise 3 you will show that thisis also true when a, 13, and 0 are collinear. Since 0 <it' by definition, it follows that'it' is a subgroup of C under addition.

We next prove part (a). Given a = a + ib <it', we can drop perpendiculars froma to the x-axis and y-axis constructed in Example 10.1.2. This shows that a, ib 'it'.

Since the circle of radius libi centered at 0 intersects the x-axis at b, C2 and P2 implythat b 'It'. Conversely, given a, b <it' fl R, applying C2 and P2 to the circle of radiusbl centered at 0 shows that ib is constructible. By the previous paragraph, a + ib is

constructible and part (a) follows.


Now let a, b E n {x E R x > 0} and consider the following two figures:

ia

0 tid= 1/a

Figure 1 Figure 2

Recall that i was constructed in Example 10.1.2. In Figure 1 we construct ib as aboveand then use Cl to draw the line 1 containing i and a. By high school geometry, wecan draw the line 1' through ib that is parallel to 1. Then P1 shows that 1' and thex-axis intersect at a constructible real number c. But c = ab follows easily by similartriangles, so that ab is constructible. In a similar way Figure 2 shows that 1/a isconstructible. We leave the details as Exercise 3. This exercise will also show that

fl R is a subfield of JR.To show that is closed under multiplication and taking reciprocals of nonzero

elements, let a = a + ib and fi = c + id be constructible numbers. Then

= (a+ib)(c+id) ac— bd+i(ad+bc).

However, a,b,c,d E by part (a), so that ac—bd,ad+bc E since thelatter is a subfield of JR. Using part (a) again, we conclude that

aa —b

— a+iba—ib — a2+b2

Using part (a) and the fact that is a subfield of JR, we easily see that 1/a EThus is a subfield of C.

Finally, we show that is constructible when a is. We can assume that a 0.If we write a re'9, r = a > 0, then it suffices to show that is constructible.To prove this, note that the constructibility of a implies the following:

• First, using the x-axis and the line containing 0 and a (by Cl), we can constructthe angle 9, which we can then bisect by the usual straightedge-and-compassconstruction. Thus the angle 9/2 is constructible.

• Second, the circle of radius r = al centered at 0 (by C2) intersects the x-axis at±r. By P2, we see that r is constructible.

• Third, if we can construct then we can construct the circle of radiuscentered at the origin by C2. Then P2, applied to this circle and the angle 9/2constructed above, implies that is constructible.

ib

1'

0 ac = ab


To study ..,[r, let r> 0 be constructible and define the point /3 by the diagram

/3 = the intersection

(10.1) of the linex= 1with the semicircle

In Exercise 3 you will show that /3 is constructible. By Euclidean geometry, thetriangle with vertices 1, /3, and 1 + r is a right triangle. The two smaller triangles thatshare the side determined by 1 and /3 are similar, so that

1 dd = r'

where d is the distance from 1 to Thus d2 = rand hence d = Since d is easilyseen to be constructible, we conclude that is constructible. •

Here is an example of how to use Theorem 10.1.4.

Example 10.1.5 By Exercise 8 of Section A.2, = is given by the formula

i

4 2

Since the field is closed under the operation of taking square roots, it follows easilythat is constructible. By Example 10.1.3, we conclude that a regular pentagon canbe constructed by straightedge and compass.

We call the field of constructible numbers. We next study the structure of

Theorem 10.1.6 Let be a complex Then E if and only if there aresubfields

Q = F0 c Fic C F,,

F,, and : Fj_1] = 2for 1 <i < n.

Proof: First suppose that we have Q = F0 C C F,, C C where [F,: = 2. ByExercise 12 of Section 7.1, F, = for some E F,_1. We will prove thatF, C i < n. The case F0 = Q C follows because 'if' is asubfield of C. Now suppose that c W. Then E is constructible, whichimplies E by Theorem 10.1.4. Thus F, C W, as claimed. Thisshows that F,, c so that in particular, any E F,, is constructible.

Conversely, given E 'a', we need to create successive quadratic extensions thatstart from Q and eventually contain We will prove that there are extensions

1 l+r


Q = F0 c C c C where [F; : = 2 such that contains the real andimaginary parts of all numbers constructed during the course of constructing a. Thetheorem will follow, since a = a + ib will imply that a,b E so that a E

We will prove this by induction on the number N of times we use P1, P2, orP3 in the construction of a. When N 0, we must have a = 0 or 1, in whichcase we let = F0 = Q. Now suppose that a is constructed in N> 1 steps,where the last step uses P1, the intersection of distinct lines £1 and £2. But then

was constructed from distinct points a1 and /3i using Cl, and similarly £2 wasconstructed from distinct points a2 and /32. By our inductive assumption, there areextensions Q = Fo C C C C where [F;: Fj_1] = 2 such that contains the realand imaginary parts of al,/31,a2,/32. We will prove that contains the real andimaginary parts of a.

The line £i has an equation of the form alx+ b1y = Cl and goes through aiSince the real and imaginary parts of a1, 13' lie in Exercise 4 implies that we canassume that the coefficients a1, b1 lie in Similarly, £2 has an equation of theform a2x + b2y = C2 where a2, b2, C2 E Hence the real and imaginary parts of agive the unique solution of the equations

a3x + b1y =a2x + b2y = C2.

In this situation, linear algebra tells us that the matrix

(ai b1

b2

is invertible (be sure you can explain why), so that the unique solution is

(x'\ — (ai (c1— b2)

It follows immediately that the real and imaginary parts of a lie inNext suppose that the last step in the construction of a uses P2, the intersection of

a line £ and a circle C. Thus £ is the line through a1 /3i (from Cl), and C is the circlewith center 'Y2 and radius a2 — /321 (from C2). The five points a1, /3i , a2, /32, 'Y2 comefrom earlier steps in the construction, so that by our inductive assumption, there areextensions Q = Fo C C C C where [F;: = 2 such that contains the realand imaginary parts of these five points. We will show that the real and imaginaryparts of a lie in or in a quadratic extension of

As above, £ is given by an equation

(10.2) a1x+biy=c,,

where a1, b1 ,Ci E In Exercise 4 you will show that C is given by an equation

(10.3) x2+y2+a2x+b2y+c2 = 0,


where a2,b2,c2 Now suppose that 0. Then dividing (10.2) by a1 andrelabeling, we can assume that the line £ is given by x + bjy = Substitutingx = —biy+ci into (10.3) gives the quadratic equation

(—biy+ci)2+y2+a2(—biy+ci)+b2y+c2=0.

By the quadratic formula, the values of y involve the square root of an expression inIf this lies in then so do y and x = —bjy + and it follows that the real and

imaginary parts of a lie in On the other hand, if this square root does not lie inthen it lies in a quadratic extension C Then y and x = —biy + ci also lie in

which shows that the real and imaginary parts of a lie in a quadratic extensionof When a1 = 0, the argument is similar and is left as part of Exercise 4.

Finally, suppose that the last step in the construction of a uses P3, the intersectionof distinct circles Ci and C2. As above, we can find Q F0 C C C C where

Fj_1J = 2 such that the circles are given by equations

(10.4)x2+y2+aix+biy+ci=O,x +y +a2x+b2y+c2=0,

where all of the coefficients lie in Furthermore, we know that the real andimaginary parts of a give a solution of (10.4).

If we subtract these equations, we get the equation

(10.5) (ai—a2)x+(bi—b2)y+(ci—c2)=0.

Since the circles (10.4) are distinct but not disjoint, one easily sees that the coeffi-cients of x and y in (10.5) don't vanish simultaneously. Thus (10.5) defines a line.Furthermore, if we combine this equation with the first equation of (10.4), then weare in the previous case of the intersection of a circle and a line. We conclude thatthe real and imaginary parts of a lie in or in a quadratic extension of Thiscompletes the proof.

Corollary 10.1.7 is the smallest subfield of C that is closed under the operationof taking square roots.

Proof: By Theorem 10.1.4, we know that a E implies that E Now let Fbe any subfield of C closed under taking square roots, and suppose that a E ByTheorem Thefirst paragraph of the proof of Theorem 10.1.6 shows that C F. Thus a E

F follows as desired. U

Theorem 10.1.6 also has the following useful consequence.

Corollary 10.1.8 If a then [Q(a) : QI = for some m � 0. Thus everyconstructible number is algebraic over Q, and the degree of its minimal polynomialover Q is a power of 2.


Proof: If st', then Theorem 10.1.6 gives extensions Q = Fo C C C Cwhere :F,_i] = 2 and E Hence

by the Tower Theorem. However, we also have Q c C Using the TowerTheorem again, we conclude that Q] divides : Q] = •

Some of the most famous problems in Greek geometry are trisection of the angle,duplication of the cube, and squaring the circle. Using Corollary 10.1.8, we can solvethese as follows.

Example 10.1.9 Trisection of the Angle. We know that every angle can be bisectedusing straightedge and compass. We will prove that this is not true for trisections, i.e.,there exist angles that cannot be trisected by straightedge and compass. Suppose, forinstance, that we could trisect a 120° angle in this way. Since we can construct a 120°angle from 0 and 1 by straightedge and compass (see Exercise 5), a trisection of thisangle would imply that we could construct a 40° angle from 0 and 1 by straightedgeand compass. Intersecting this with the unit circle centered at the origin, it wouldfollow that the 9th root of unity = would be a constructible number (since40° 2ir/9 radians).

However, Theorem 9.1.9 implies that the minimal polynomial of is the cyclo-tomic polynomial and the factorization

x9— 1 = = (x— 1)(x2+x+ 1)(x6+x3+ 1)

from Proposition 9.1.5 shows that x6 +x3 + 1 is the minimal polynomial of ByCorollary 10.1.8, is not constructible. This contradiction proves that we cannottrisect 120° using straightedge and compass. In Exercise 6 you will show that it is

also impossible to trisect 60° by straightedge and compass.

In Section 10.2, we will use the results of Section 9.1 to determine all n for whichis constructible.

Example 10.1.10 Duplication of the Cube. Here, the problem is to take a givencube and construct one with exactly twice the volume. We can pick our units ofmeasurement so that the given cube has edges of length 1. In these units, the volumeis also 1, which means that we need to construct a cube of volume 2. Since volumeis edge length cubed, it follows that if we could duplicate the cube, then we couldconstruct a number s such that s3 = 2, i.e., s = Furthermore, since the cubehas edge length 1, we can assume that the construction begins with 0 and 1. Itfollows that duplicating the cube by straightedge and compass implies that s =is constructible. But x3 — 2 is the minimal polynomial of over Q, so thatis not constructible, by Corollary 10.1.8. This contradiction proves that we cannotduplicate the cube by straightedge and compass.

Example 10.1.11 Squaring the Circle. This is the problem of constructing a squarewhose area is equal to that of a given circle. If we pick our units of measurement so


that the given circle has radius 1, then the circle has area Since a square of areahas side it follows that if we could square the circle, then we could construct

Furthermore, since the circle has radius 1, we can assume that the constructionbegins with 0 and 1. It follows that squaring the circle by straightedge and compasswould imply that is constructible.

Since is a field, the constructibility of would imply that ir = isalso constructible. Then Corollary 10.1.8 would imply that is algebraic overQ. However, in 1882 Lindemann proved that ir is transcendental over Q. A self-contained proof can be found in [Hadlock, Sec. 1.7]. This contradiction shows thatwe cannot square the circle by straightedge and compass.

One could also ask whether the converse of Corollary 10.1.8 is true. In otherwords, if a C is algebraic over Q and the degree of its minimal polynomial is apower of 2, is a constructible? The following result will answer this question.

Theorem 10.1.12 Let a E C be algebraic over Q, and let Q C L be the splitting fieldof the minimal polynomial of a over Q. Then a is constructible and only if [L : QIis a power of 2.

Proof: First suppose that [L : Q] is a power of 2. We will follow the proof ofthe Fundamental Theorem of Algebra given in Section 8.5. Since Q C L is Galois,it follows that Gal(L/Q)I = [L:Q] is a power of 2, say IGal(L/Q)I = ByTheorem 8.1.7, Gal(L/Q) is solvable, which by Definition 8.1.1 means that we havesubgroups

{e} = Gm C Gm_i C C G1 C Go = Gal(L/Q)

such that is normal in of index 2 (since IGal(L/Q) I = This gives

Q = C L,

where [LG : LGI] = 2 for all i. By Theorem 10.1.6, a EL is constructible.Turning to the converse, we first show that Q c is a normal extension. For this,

let a E and let f be the minimal polynomial of a over Q. We need to prove that fsplits completely over Since a is constructible, Theorem 10.1.6 gives extensions

ThenletQcMbetheGalois closure of Q C as constructed in Proposition 7.1.7. In Exercise 7 you willshow that we may assume that M C C.

Note that f splits completely in M, since M is normal over is irreducibleoverQ, Nowlet/3EMbeanyrootoff. ByProposition 5.1.8, there is a- E Gal(M/Q) such that cr(a) = /3. Applying ci to thefields Q = C C C M gives

such that [a(F) : a(F1_1)] = : fl_i] = 2 for all i. By Theorem 10.1.6, /3 = ci(a) E

is constructible. This shows that f splits completely overIt follows that contains a splitting field L off over Q. By the Theorem of the

Primitive Element, we have L = for some E L. Since 'y E Corollary 10.1.8


implies that [Q (-y) : Q] = [L: Q] is a power of 2, as claimed. This completes the proofof the theorem. .

We can use Theorem 10.1.12 to show that the converse of Corollary 10.1.8 is false.Here is an example.

Example 10.1.13 Let a be a root of the polynomial

f=x4—4x2+x+l.One easily checks that f is irreducible over Q, so that [Q(a) : QI = 4. However, inChapter 13 we will show that the splitting field L off over Q satisfies [L : Q] = 24.

By Theorem 10.1.12 we conclude that a is not constructible.

Mathematical Notes

There are several issues that are worthy of further comment.

• Starting Configurations. According to Definition 10.1.1, a constructible numberis constructed by a sequence of constructions that always begins with 0 and 1. It ispossible to begin constructions with different starting configurations. For example,

three noncollinear points a, /3, -y determine an angle with vertex a and rays a/3 andThis angle can be bisected by straightedge and compass, even though a,/3,'y

need not be constructible.We will not develop the theory of such constructions beyond the comment that

the trisection of the angle is most naturally stated in this context: given an angledetermined by a, /3, -y as above, one seeks a construction that trisects this angle bystraightedge and compass. In Example 10.1.9 we showed that this is impossible byfinding a particular case of a, /3, -y to which we could apply Corollary 10.1.8.

• Compasses. In C2, the compass uses points a /3 to give the radius, with thecenter given by a third point -y. This is slightly different from what Euclid does, forhe uses the compass with points a /3 where the center is either a or /3. One canprove that this more restrictive notion of compass (called the Euclidean compass inMartin's book [151) gives the same set of constructible points.

More surprising is the fact that we can dispense with the straightedge entirely.The Mohr—Mascheroni Theorem states that a E C is constructible if and only if thereis a sequence of Euclidean compass constructions that starts with 0 and 1 and endswith a. A proof can be found in [15, Ch. 31.

• Straightedge and Dividers. A set of dividers is a tool that can copy line segments.In other words, given points a /3 and a point -y on a line 1, dividers allow us toconstruct points , e I such that the distance from or 52 to 'y equals the distancefrom a to /3, as in the following picture:

52


Let .9 denote the set of real numbers that can be obtained from 0, 1, and iby a sequence of straightedge-and-dividers constructions. By [15, Thm. 5.6], .9is a subfield of R. A more interesting property of .9 is that if a, b e .9, then

+ b2 .9. To prove this, note that the y-axis is constructible using 0, i, and thestraightedge, so that given b E .9, we can construct ib using our dividers. Combiningthis with a E .9, we get the diagram

Now use the dividers to transfer the line segment from a and ib to the positive x-axis,starting from 0. The Pythagorean Theorem implies that v'a2 + b2 E .9, as claimed.

In general, a subfield of R that contains + b2 whenever it contains a and b iscalled Pythagorean. Thus .9 is Pythagorean, and in Exercise 8 you will show that.9 is the smallest Pythagorean subfield of R. This is an analog of Corollary 10.1.7.We call .9 the field of Pythagorean numbers.

The most interesting result about .9 is the following analog of Theorem 10.1.12.

Theorem 10.1.14 Let a E R be algebraic over Q, and letf be the minimal polyno-mial of a over Q with splitting field L. Then the following are equivalent:(a) a(b) All roots off are real, and a is constructible.(c) All roots off are real, and {L Q] is a power of 2.

Proof: The equivalence (a) (b) is proved by Auckly and Cleveland in [3, p. 225].Then (b) (c) follows from Theorem 10.1.12. •

For those who read the discussion of the casus irreducibilis in Section 8.5, wenote the following corollary of Theorems 10.1.14 and 8.6.5.

Theorem 10.1.15 Let a E be algebraic over Q. Then a E .9 if and only if a isexpressible by real radicals and all conjugates of a over Q are real. •

This shows an unexpected relation between geometric constructions and solvabil-ity by radicals.

Numbers in .9 are constructed using straightedge and dividers. Since a compasscan be used as a pair of dividers, one sees easily that

Using Theorem 10.1.14, we prove that these fields are not equal as follows.


Example 10.1.16 Consider

_______

a=Then a E fl since is closed under taking square roots. However, the minimalpolynomial of a is

f = — 4x2 —4,

which has roots ± '/2 + Two of these roots are not real, so that a byTheorem 10.1.14.

• Marked Rulers. The straightedge we've been using is an unmarked ruler. Butsuppose instead that we have a marked ruler, which is a straightedge with two markson it one unit apart. Such a ruler allows the construction of some interesting linesand points. Provided one starts from the points 0, 1, i in C, one can prove that allstraightedge-and-compass constructions can be done using only a marked ruler. Wewill see in Section 10.3 that there are also marked-ruler constructions for trisectingangles and taking cube roots.

Historical Notes

There are two versions of where the problem of duplicating the cube first arose.In one version, King Minos was unhappy with the cubical tomb of his son Glaucusand ordered its size doubled. In the other, a delegation from Athens asked the oracleat Delos for advice about a plague in Athens. The Athenians were told to double thesize of the cubical altar of Apollo. In both versions, the first attempt was to doublethe sides, which multiplies the volume by a factor of eight. The point, of course, isthat what was required was to double the volume, which means that the sides mustbe multiplied by Because of the popularity of the second version of the story,the duplication of the cube is sometimes called the Delian problem.

Less is known about the origin of the other two Greek problems. It may havebeen something like the following. A line segment is easily bisected or trisected bystraightedge and compass, and bisecting angles is equally easy. Hence it is naturalto ask about trisecting angles. Similarly, a rectangle or triangle is easily squared bystraightedge and compass. Since the next most basic geometric figure is a circle, it isnatural to ask whether it is also squarable.

Greek geometers never solved these problems, though the search for solutions ledto some wonderful mathematics. For example, consider the lune indicated by theshaded region:


Around 440 B.C., Hippocrates of Chios discovered this lune in the course of his workon squaring the circle. The above figure has a quarter circle with radius OA OBand a semicircle with diameter AB. The lune is the region outside the quarter circlebut inside the semicircle. Hippocrates showed that the shaded region has the samearea as the triangle AOB (you will verify this in Exercise 9). It follows that this luneis squarable, so that at least part of a circle is squarable. Hippocrates found othersquarable lunes, and in the twentieth century, Chebotarev and Dorodnov showed thatthere are only five squarable lunes. A proof can be found in [18]. See also [20].

Somewhat later, Hippias of Elis (ca. 425 B.C.) discovered a curve called thequadratrix, which he used to trisect angles and square the circle. In modern terms,this curve is given by

(10.6)

In Exercise 10 you will use this curve to trisect angles and square the circle. Asfor duplication of the cube, Menaechmus introduced parabolas around 350 B.C. as a

by-product of his work on duplication. In Exercise 11 you will show that duplicationof the cube can be solved by intersecting two parabolas. We will see in Section 10.3that many other constructions can be done by intersecting conic sections.

There is a lot more to say about Greek work on these problems. For example,the spiral of Archimedes r = 0 has some nice applications that you will study inExercise 12. Numerous other examples can be found in Chapter 4 of [9]. Greekgeometry is more interesting than what you learned in high school, in part because theGreeks had these great problems to inspire them. In modern mathematics, unsolvedproblems play a similar role of inspiring research. For example, the inverse Galoisproblem mentioned in Section 7.4 is still unsolved and is being actively studied bymany researchers.

The three Greek problems were solved in the nineteenth century. In 1837 Wantzelshowed that duplication of the cube and trisection of the angle cannot be done bystraightedge and compass. His argument used the irreducibility of certain cubicpolynomials. This is similar to what we did in Exercise 6 (for trisection of the angle)and Example 10.1.10 (for duplication of the cube). The first page of Wantzel's paperis reproduced on page 84 of [Escofier]. Finally, as noted in Example 10.1.11, theproblem of squaring the circle was solved in 1882 when Lindemann showed that iris transcendental over Q.

We should also note that the variants on straightedge-and-compass constructionsmentioned in the Mathematical Notes have a long and interesting history. This isdiscussed in [15]. See also the Historical Notes to Section 10.3.


Exercise 1. In part (a) of Example 10.1.2 we constructed the x-axis. In a similar way showthat the y-axis is constructible. For each step in your construction be sure to say which of Cl,C2, P1, P2, and P3 you are using.


Exercise 2. Suppose that a, /3, -y are noncollinear and consider the rays and emanatingfrom a that go through /3 and 'y respectively. We call this the angle formed by a, Alsoassume that a, /3, -y are constructible.(a) Prove that there is a constructible number S with positive y-coordinate such that the angle

formed by a, /3, -y is congruent to the angle formed by 0, 1,5. As in Exercise 1, each stepin the construction should be justified by Cl, C2, P1, P2, or P3.

(b) Prove the claim made in Example 10.1.3 that = e2"1" is constructible if and only if aregular n-gon can be constructed by straightedge and compass.

Exercise 3. This exercise covers the details omitted in the proof of Theorem 10.1.4.(a) Let a, /3 be constructible numbers such that 0, a, /3 are collinear. Prove that a + /3 iS

constructible.(b) Let a E n {x E JR x> 0}. Use Figure 2 in the proof of Theorem 10.1.4 to show that

1/a is constructible.(c) In the proof of Theorem 10.1.4, we showed that '6'fl {x E JR Ix> 0} is closed under

addition, multiplication, and multiplicative inverses. Use this to prove that fl JR is asubfield of JR.

(d) Prove that the number /3 pictured in (10.1) is constructible (assuming that r is con-structible).

Exercise 4. This exercise covers the details omitted in the proof of Theorem 10.1.6.(a) Suppose that a line goes through distinct points al = UI + iv1 and /3! = u2 + iv2, where

u!,vI,u2,v2 lie in a subfield F CR. Prove that LI is defined by an equation of the formaix+biy=c1 where a1,b1,ci EF.

(b) Suppose that a2 and are complex numbers whose real and imaginary parts liein a subfield F C JR. Prove that the circle C with center 'Y2 and radius a2 — /321 has anequation of the form (10.3) with a2,b2,c2 E F.

(c) In the proof of Theorem 10.1.6, we considered the equations (10.2) and (10.3) whenai 0. Explain what happens when a! = 0 in (10.2).

Exercise 5. In this exercise you will give two proofs that = is constructible.(a) Give a direct geometric construction of with each step justified by citing Cl, C2, P1,

P2,orP3.(b) Use Theorem 10.1.6 to show that is constructible.

Exercise 6. Show that it is impossible to trisect a 60° angle by straightedge and compass.

Exercise 7. Suppose we have extensions Q C F C C where [F Q] is finite. Prove that thereis a field M such that F C M C C and Mis a Galois closure of Fover Q.

Exercise 8. In the Mathematical Notes we defined the field C JR and what it means for asubfield F C JR to be Pythagorean.(a) Let a be a real number. Prove that a E if and only if there is a sequence of fields

Q=F0c such

the smallest Pythagorean subfield of JR.

Exercise 9. Show that the lune illustrated in the Historical Notes has the same area as thetriangle AOB in the illustration.

Exercise 10. The quadratrix is the curve y = xcot(7rx/2) for 0 <x 1. In this problem, youwill use this curve to square the circle and trisect the angle.


(a) Show that 2/it = xcot(irx/2), i.e., the quadratrix meets the y axis at y = 2/it.We will follow Hippias and include this point in the curve.

(b) Show that we can square the circle starting from 0 and 1 and constructing new pointsusing Cl, C2, P1, P2, or P3, together with the intersections of constructible lines with thequadratrix.

(c) A point (a, b) on the quadratrix determines an angle 9 as pictured below. Prove that0 = ira/2.

(d) Suppose that we are given an angle 0 < 0 <ir/2. Prove that we can trisect 9 startingfrom 0, 1, and 0 and constructing new points using Cl, C2, P1, P2, or P3, together withthe intersections of constructible lines with the quadratrix.

Explain how the method of part (d) can be adapted to trisect arbitrary angles.

Using the quadratrix, what else can you do to angles besides trisecting them?

Exercise 11. Explain how the points of intersection of the parabolas y = x2 and x = 2y2 enableone to duplicate the cube. Your explanation should include a picture.

(a) Explain how the spiral and 0 = ir/2 enable one to square the circle.

(b) Given an angle Oo, explain how the spiral enables one to trisect 9o.

2/it

(a,b)

(e)

(f)

Exercise 12. The spiral of Archimedes is the curve whose polar equation is r = 9:


10.2 REGULAR POLYGONS AND ROOTS OF UNITY

Our next task is to apply the theory developed in Section 10.1 to the question ofwhich regular polygons can be constructed by straightedge and compass. Our maintool will be the cyclotomic extension Q C studied in Chapter 9.

Before stating our main result, we need some terminology: An odd prime p is aFe rmat prime if it can be written in the form

p = + 1

for some integer m � 0. Following Gauss, we can now characterize constructibleregular polygons as follows.

Theorem 10.2.1 Let n > 2 be an integer Then a regular n-gon can be constructedby straightedge and compass and only if

where s 0 is an integer and. ,Pr are r 0 distinct Fe rmat primes.

Proof: In Example 10.1.3 we saw that a regular n-gon is constructible by straight-edge and compass if and only if is constructible. Using results proved earlier, wecan determine when is constructible as follows:

• By (8.6), Q C is a Galois extension.• By Theorem 10.1.12, it follows that is constructible if and only if : QI is

a power of 2.

• By Corollary 9.1.10, = where is the Euler de-fined in Section 9.1.

We conclude that is constructible if and only if is a power of 2.First suppose that n = ' 'Pr, where P1,... ,Pr are distinct Fermat primes.

Then part (b) of Lemma 9.1.1 gives the formula

l\ J2 (p1l)"'(prl), s>O,—

pin — 1), s—O.

It follows that is a power of 2, since each p, is a Fermat prime.

Conversely, suppose that is a power of 2, and let the factorization of n be

n = '' where qi,... , are distinct primes and the exponents ai, . . . , are all

� 1. Then part (b) of Lemma 9.1.1 gives the formula

If qj is odd, then we must have a, = 1, since is a power of 2, and we alsoconclude that q, — I must be a power of 2. However, in Exercise 1 you will show


Because of Euler's negative result, there was little interest in Fermat primes untilGauss discovered their relation to the constructibility of regular polygons. The firstentry in his famous mathematical diary, dated March 30, 1796, reads as follows:

The principles upon which the division of the circle depend, and geometricaldivisibility of the same into seventeen parts, etc.

(See [11].) Gauss wrote this one month before his 19th birthday.The details of what Gauss proved about regular polygons appear in Section VII

of Disquisitiones Arithmeticae [101. As explained in Section 9.2, Gauss studied theequations satisfied by periods (special primitive elements of intermediate fields) ofthe extension Q c Q(ç), where p is prime. Then, in Article 365 of [101, he applieshis results to show that is constructible when p is a Fermat prime. Though heasserts that the converse is true, the first published proof of this is due to Wantzel in1837. In Article 366 Gauss describes which are constructible when n is arbitrary(Theorem 10.2.1), though his proof is again incomplete.

Gauss knew that a straightedge-and-compass construction of a regular 17-gon wasa big deal. As he says in Article 365 of [101:

It is certainly astonishing that although the geometric divisibility of the circleinto three and five parts was already known in Euclid's time, nothing was addedto this discovery for 2000 years.

But rather than give an explicit construction, Gauss shows that

cos(27r/17)=

Exercises 8 and 9 of Section 9.2 explain how this formula follows from Gauss'stheory of periods. From here one can design a construction for the regular 17-gon,though it is not very efficient. A more elegant construction can be found in [Stewart,Ch. 171, along with a reference for Richelot's construction of the regular 257-gonin 1832. There is also the story of Professor Hermes of Lingen, who late in thenineteenth century worked 10 years on the construction of the regular 65537-gon.

We conclude with some remarks about arc length. This was an important topicin the seventeenth and eighteenth centuries. For example, by inscribing a regularn-gon in the unit circle, one easily sees that constructing the n-gon by straightedgeand compass is equivalent to dividing a circle into n equal arcs by straightedge andcompass. Another example involves the lemniscate, which is the curve in the planedefined by the polar equation r2 = cos2O:

REGULAR POLYGONS AND ROOTS OF UNITY 273

In 1716 Fagnano discovered a method for doubling and halving_an arc of the lem-

niscate. In particular, he showed that the circle of radius \/\ñ — 1 (drawn in dashesin the illustration at the bottom of the previous page) divides each quadrant of thelemniscate into arcs of equal length. Hence the lemniscate can be divided into eightequal arcs by straightedge and compass. In Chapter 15 we will explore a remarkablegeneralization of this discovered by Abel.


Exercise 1. Suppose that 2/c + 1 is an odd prime. Prove that k is a power of 2.

Exercise 2. Let p be prime. In Example 9.1.6, we showed that

1.

The goal of this exercise is to prove that (x) is irreducible over Q using only theSchönemann—Eisenstein criterion.(a) Explain how the formulas of Example 9.1.6 imply that

(x+ i)P2 —1 = ((x+ i)P_ 1).

(b) Let (x + 1) be the reduction of (x + 1) modulo p. Show that

x"2 1).

(c) Show that (x + 1) is irreducible over Q by the Schönemann—Eisenstein criterion. Asin the proof of Proposition 4.2.5, this will imply that the same is true for (x).

Exercise 3. Using only Proposition 4.2.5, Theorem 10.1.12, and Exercise I, show that ç is

constructible if and only if p is a Fermat prime.

Exercise 4. Prove that

((n) = (m

when mm, m > 0, and use this to conclude that if is constructible and mm, m > 0, then (mis constructible.

Exercise 5. Suppose that n = 2sp1 . Pr, where pi,. . . ,m- are distinct Fermat primes. Thenis constructible by Exercise 3.

(a) Show that (2s is constructible.(b) Assume that (a' (b are constructible and gcd(a, b) = 1. Prove that (ab is constructible.(c) Conclude that (,, is constructible, since (2c, (P' . . , (i,, are.

Exercise 6. Now suppose that is constructible for some n > 2. The goal of this exerciseis to prove that if p is an odd prime dividing n, then p is a Fermat prime and p2 jn. Thisand Exercise 5 will give a proof of Theorem 10.2.1 that doesn't require knowing that (x) isirreducible for arbitrary n.(a) Let p be an odd prime dividing n. Use Exercises 3 and 4 to show that p is a Fermat prime.(b) Now assume that p is an odd prime and Use Exercise 4 to show that is

constructible. Then use Theorem 10.1.12 and Exercise 2 to obtain a contradiction.In Chapter 15 we will use a similar strategy to prove Abel's theorem about straightedge-and-compass constructions on the lemniscate.


Exercise 7. Prove that 3, 5, 17, 257, and 65537 are Fermat primes.

Exercise 8. Use 1og10(F33) 2331og10(2) to estimate the number of digits in the decimalexpansion of F33. Then do the same for F2478782.

10.3 ORIGAMI (OPTIONAL)

In this optional section, we will use origami—the art of Japanese paper folding—todo some constructions not possible by straightedge and compass. We will also givea careful description of origami numbers and explain what they mean from the pointof view of Galois theory.

A. Origami Constructions. We begin with a classic origami construction thatshows how to trisect an angle. Take an arbitrary angle 0 between ir/4 and ir/2, andput it in the bottom left corner of a square sheet of paper. This gives the picture onthe left:

(10.7)

P2

P1

Thus 0 is the angle between the line 12 and the bottom of the sheet. Then, as indicatedon the left, fold the sheet twice to obtain two lines parallel to the bottom such thatthe line is equidistant to the parallel lines through the points Pi and P2.

Now, turning to the picture on the right, do a classic origami move that folds thesheet so that P1 moves to a point on and P2 moves to a point Q2 on 12. Youshould try this on a sheet of paper (a rectangular sheet will work fine). In Exercise 1you will prove that the angle made by the bottom and the segment P1 Qi is 0/3. Thuswe have trisected an arbitrary angle ir/4 0 ir/2 using origami!

Origami also makes it easy to double or halve angles. From this and the aboveconstruction, it follows that one can trisect any angle using origami (see Exercise 2).

We can also solve cubic equations using origami. But before explaining this, weneed to think about the underlying geometry of the trisection given in (10.7). Thesurprise is that we are dealing with simultaneous tangents to parabolas. To see howthis works, consider the geometric description of a parabola, which is defined as the

ORIGAMI (OPTIONAL) 275

locus of all points P equidistant from a fixed point P1 (the focus) and a fixed line(the directrix). This gives the following picture:

(10.8)

In this picture, the segments and PQ1 have equal length, and PQ1 is perpendicularto the directrix 11. The key point, which you will prove in Exercise 3, is that Qi is thereflection of P1 about the tangent line at P (the dashed line in the picture). You willalso prove the converse. Thus we have the following result.

Lemma 10.3.1 In the plane, let P1 be a point not on a line 11. Then, given anotherline £, the reflection of Pi about £ lies on li and only is tangent to the parabolawith focus Pj and directrix li.

To see how this relates to origami, look back at (10.7). The origami move we usedtook P1 to E and P2 to Q2 E 12 and was done by folding along the dashed line.This means that the reflection of Pi about the dashed line lies on so that the dashedline is tangent to the parabola with focus P1 and directrix 11 by Lemma 10.3.1. Thesame argument shows that the dashed line is also tangent to the parabola with focusP2 and directrix l2. We conclude that using origami, one can find the simultaneoustangents to two given parabolas.

Here is an example of how to use this.

Example 10.3.2 Let us find the real roots of the cubic equation x3 + ax + b = 0,

where a,b E and b 0. Following the paper [1], we consider the parabolas

(10.9) (y 2bx and y =

Let £ be a line with slope m that is simultaneously tangent to these parabolas, sayat points (xi,yi) on the first and (x2,y2) on the second. In Exercise 4 you will usecalculus to show that the slope of the tangent line to the first parabola at (x1 is

b

y'

P


This implies that m 0 andy' — = from which we easily conclude that

— (b)2 b

(10.10)X1

2b — 2b 2m2'b a= — + -.m 2

Computing the slope of the tangent line to the second parabola at (X2 , y2) gives

(10.11) x2=m,

which easily implies thatm2

Y2 =

If we substitute these values into m = (Y2 —yi)/(x2 —xi), then we obtain

2m3—b

Since m 0, it follows without difficulty that m satisfies the equation

(10.12) m3+am+b=0.

Hence the slopes of the simultaneous tangents to the parabolas (10.9) are roots of thecubic m3 + am + b. In Exercise 5 you will do this using origami. 4>

B. Origami Numbers. Our next task is to give a careful description of the numberswe get when we add the origami move used in (10.7) to the constructions Cl and C2defined in Section 10.1. More precisely, consider the following origami construction:

From not lying on lines £2, we can draw a line £ that reflects a1 toa point on and a2 to a point on £2.

The dashed line in (10.7) is an example of C3. There are situations where no line £satisfies the conditions of C3 (see Exercise 6). Hence what C3 really says is that weare allowed to use such a line £ whenever it exists.

By Lemma 10.3.1, C3 enables us to draw a simultaneous tangent to two givenparabolas (assuming there is such a tangent). Notice that C3 constructs only the lineL. This is because in origami, a line is a fold and a point is an intersection of folds.Of course, once we have £, we can construct the reflections of a! and a2 about £ byfurther straightedge-and-compass constructions.

The constructions Cl, C2, and C3 create circles and lines, and intersecting themusing P1, P2, and P3 from Section 10.1 gives new points that can be used for furtherconstructions. We define origami numbers as follows.

Definition 10.3.3 A complex number a is an origami number if there is a finitesequence of constructions using Cl, C2, C3, P1, P2, and P3 that begins with 0 and 1and ends with a.


This definition appears to involve compass, straightedge, and origami. However,in Chapter 10 of Martin's book [15], it is shown that all straightedge-and-compassconstructions can be done using origami (called "paperfolding" by Martin). Inparticular, one can replace Cl, C2, C3, P1, P2, and P3 with constructions that involveonly origami and give the same set of origami numbers.

The set of all origami numbers has the following structure.

Theorem 10.3.4 The set C = {a E C a is an origami number} is a subfield of C.Furthermore:(a) Leta =a+ib, where a,b ER. Thena E C and only E 0.

(b) a E C implies that e C.(c) A complex number a lies in C and only there are subfields

Q = F0 C F1c C C

such that a E and [F,: = 2 or 3for 1 <i

Proof: We refer to [15, Ch. 101 for the proof that C is a subfield of C. The proofof part (a) is similar to what we did in Theorem 10.1.4 and is omitted.

To prove part (b), write a in polar form as a = re'0. We may assume r> 0. Usingthe compass, we can transfer r to the x-axis, and then the straightedge-and-compassconstruction given in (10.1) shows that E C. Since we can also bisect 0 bystraightedge and compass, it follows that

E C.

For the cube root, we can trisect 0 using (10.7) and Exercise 2. To constructconsider the parabolas (10.9) with a 0 and b = —r. By Exercise 7 the foci al,a2and directrices li ,12 of these parabolas are defined over any subfield of R containingr and hence can be constructed from r by straightedge and compass. Applying C3to ai , a2 and ,l2, we can construct a simultaneous tangent £ to these parabolas. By(10.12), £ has slope m = This easily implies that E C. Since w = E C(do you see why?), it follows that

i=0,1,2.

Intheproof of part (c) we will say thatfieldsQ=Foc cCforma2-3tower if [F, : = 2 or 3 for 1 i < n (this differs slightly from the terminologyused by Videla in [21]). Now suppose that Q = F0 C C is a 2-3 tower. Wewill prove that C C by induction on n. Since the case n = 0 is obvious, we mayassume that C C. Given a E we know that a is a root of a polynomialf E C[xI of degree at most 3, since : 2 or 3. 1ff has degree 1, then a E Cis immediate, and if f has degree 2 or 3, then, by the quadratic formula or Cardan'sformula, a can be expressed in terms of square roots, cube roots, and elements of C.By part (b), it follows that a E C.


Going the other way, let a be an origami number. We will show that there is a 2-3tower Q = F0 c ... c c C such that contains the real and imaginary parts of allnumbers constructed in the course of constructing a. The theorem will follow, sincea = a+ib will imply that a,b E so that a (We used the same strategy inthe proof of Theorem 10.1.6.)

We will prove this by induction on the number N of times we use P1, P2, or P3 inthe construction of a. First suppose that a is constructed in N> 1 steps and that thelast step uses P1. Thus a is the intersection of distinct lines Li and £2 created earlierin the construction. If both lines come from Cl, then we are done, as in the proof ofTheorem 10.1.6. However, if we used C3 to construct either of the lines, then morework is needed.

If Li was created using C3, then Li is simultaneously tangent to two parabolaswhose foci and directrices were created earlier in the construction. We claim thathas an equation whose coefficients lie in a 2-3 tower. To prove this, first consider thespecial case when the parabolas are of the form (10.9) for some a,b E R. Here, ourinductive assumption and Exercise 7 imply that a, b lie in a 2-3 tower. Then the slopem of L1 satisfies the cubic equation (10.12), so that we can extend the 2-3 tower toget one that contains a, b, m. By (10.10), the point (xi ,yi) E has coordinates in the2-3 tower. It follows that Li has an equation Ax + By = C whose coefficients lie in thesame 2-3 tower. In the general case when L1 was created using C3 for two arbitraryparabolas, one can argue similarly that has an equation whose coefficients lie in a2-3 tower. We omit the details.

It follows that if Li or L2 (or both) were created using C3, then there is a 2-3tower containing the coefficients of their defining equations. As in the proof ofTheorem 10.1.6, we conclude that the coordinates of the intersection of Li and L2 liein the same tower.

Next suppose that we use P2 to create a, so that a comes from the intersection ofa circle and a line. By our inductive assumption and the above argument, we mayassume that the circle and line are defined by equations whose coefficients lie in a2-3 tower, and then we are done by the argument of Theorem 10.1.6.

Finally, when we use P3, the argument is identical to what we did in Theo-rem 10.1.6. This completes the proof of the theorem. •

Here is a nice example of Theorem 10.3.4.

Example 10.3.5 The 2-3 tower

Q C Q(2cos(2ir/7)) C

shows that is an origami number. It follows that a regular heptagon (7-gon) can beconstructed by origami. <II>

We can also characterize origami numbers using Galois theory.

Theorem 10.3.6 Let a E C be algebraic over Q and let Q C L be the splitting fieldof the minimal polynomial of a over Q. Then a is an origami number and only if[L : Q] = 2a3b for some integers a, b � 0.


Proof: The argument is similar to the proof of Theorem 10.1.12. If a E C, onefirst proves that Q C C is a normal extension, so that L C C. Then the formula forEL: Q] follows by applying Theorem 10.3.4 to a primitive element of Q C L. For theconverse one uses Burnside's pnqm Theorem (Theorem 8.1.8) to show that Gal(L/Q)is solvable because IGal(L/Q) I = [L: QI = The desired 2-3 tower is then easilyconstructed using the Galois correspondence and the definition of solvable group.We leave the details as Exercise 8.

Here are two examples of Theorem 10.3.6.

Example 10.3.7 The results of Section 9.1 imply that Q C Q is a Galois exten-sion of degree 10. This is not of the form 2a31', so that cannot be constructed byorigami.

Example 10.3.8 Let a E C be a root of f = x6 +x +1. Using Maple or Mathematica,one easily checks that f is irreducible over Q, so that Q C Q(a) is an extension ofdegree 6. However, even though 6 = 2.3, a is not an origami number. This followsfrom the galois command in Maple, which shows that the splitting field Q C L off has Galois group Gal(L/Q) S6. Then Theorem 10.3.6 implies that a C, since[L:Q] = =6! = 24.32.5.

You will prove the following corollary of Theorem 10.3.6 in Exercise 9.

Corollary 10.3.9 Let f(x) E Q[x] be a polynomial of degree 4. Then the mots off(x) are origami numbers, i.e., we can solve f(x) = 0 by origami. •

The papers [7] and [14] give explicit descriptions of how to solve cubics byorigami. The paper [7] also treats cubics.

C. Marked Rulers and Intersections of Conics. Origami numbers can bealso constructed with a marked ruler or intersections ofconics. We will discuss thesemethods briefly (without proofs), beginning with the marked ruler.

A marked ruler is a straightedge with two marks on it one unit apart. This issometimes called a twice-notched straightedge. A marked ruler can construct a linein two ways: first, by connecting two known points, and second, by verging, whichgiven a known point P and known lines and 12 draws a line through P that meetsat and 12 at Q2 such that the segment QI Q2 has length 1:

Qi


A marked-ruler construction begins with the points 0, 1, and i. At each step, oneconstructs a new line by applying either of the two operations just described to thealready constructed lines and points and then intersecting the new line with otheralready constructed lines to get new points in C.

Here is a quartic equation that comes from verging with a marked ruler.

Example 10.3.10 Let l1 be the line y = x and 12 be the liney = — and let P = 0).Then let £ be a line with slope m through P. In Exercise 10 you will show that £meets 11 at the point

m(10.13) QI = (xi,xi) e wherexi = 2m-2

and meets 12 at the point

(10.14) Q2 = (x2, E 12, wherex2= 2m±

If we think of £ as the marked ruler, then verging from P with Ii and l2 means thatthe distance from Qi to Q2 is 1. This gives the equation

( m m \2 ,i —m m+½(2m+1) =1,

which simplifies to the quartic equation

(10.15) 7m4—16m3—21m2+8m-l-4=0.

The roots of this equation are all real and represent the slopes of the four lines throughP that are constructed with a marked ruler by verging with the lines l1 and 12.

In Exercise 9 of Section 13.1, we will see that the Galois group of (10.15) is S4(this can also be done using the galois command in Maple). Hence the splittingfield is an extension of Q of degree 24. This is not a power of 2, so that these linesare not constructible with straightedge and compass.

See Exercise 11 for an example where verging leads to a cubic equation. Giventhat origami also solves cubic and quartic equations, the following result proved in[15, Ch. 101 is not surprising.

Theorem 10.3.11 Let a E C. Then a can be constructed using a marked ruler if andonly if a is an origami i.e., a E 0. •

We next consider conics. These can be defined geometrically in terms of foci,directrices, and eccentricities, or one can work algebraically, giving separate treat-ments for ellipses, hyperbolas, and parabolas. We will use a third approach, whichdefines a conic to be a curve in the plane defined by an equation of the form

(10.16) F(x,y) = ax2 +bxy+cy2+dx+ey+f= 0


Wealsoassumethat(10.16)hasatleast one solution with x,y real. This excludes equations like x2 +y2 + 1 = 0.

We write the equation (10.16) in matrix form as follows. Let

(ac

f)and let

fx

Then one easily checks that

(10.17) F(x,y) =x'Ax,

where is the transpose of x. Then the conic C defined by (10.16) is nondegenerateif det(A) 0. As shown in [21, if C is nondegenerate, then

b2—4ac<0 Cisanellipse,

b2—4ac=O Cisaparabola,

b2 — 4ac> 0 C is a hyperbola.

To do constructions by intersecting conics, start with 0 and 1, and construct eithera line connecting two already constructed points or a conic whose coefficients arepreviously constructed real numbers. Then we get new points by intersecting theselines and conics. This gives the following set of complex numbers.

Theorem 10.3.12 Let a e C. Then a can be constructible by intersecting conics ifand only if a is an origami i.e., a E U

Proof: This is proved by Alperin in [1]. Alternatively, Videla shows in [211 that ais constructible by conics if and only if a lies in a 2-3 tower. When we combine thiswith part (c) of Theorem 10.3.4, the theorem follows immediately. •

Putting together the results from this section, we have the following equivalencesfor a complex number a:

a is an origami number a is constructible by marked ruler

a is constructible by intersecting conics

a is algebraic over Q, and the Galois groupof its minimal polynomial has order


Mathematical Notes

Here are two topics for further discussion.

• Marked Ruler and Compass. By using a marked ruler and a compass, one can doconstructions beyond what is possible by marked ruler alone. A marked ruler allowsus to verge using a point and two lines, but with a compass to draw circles, we canalso verge using a point and two circles or a point, a circle, and a line. An exampleof the latter is Archimedes' angle trisection:

(10.18)

Here, we have a unit circle centered at 0 and a point P on the circle that makes theindicated angle 9 with the line 1. Then verging from P with I and the circle gives thedashed line containing points Q on the circle and R on I that are one unit apart. InExercise 12 you will prove that ZPRO = 0/3. (Note that we can trisect angles usingonly the marked ruler. Exercise 13 gives such a trisection due to Pappus.)

Using a marked ruler and compass also enables us to construct points not possibleby marked ruler alone. An example in Baragar's paper [4] shows that the real rootsof the polynomial

x5—4x4+2x3+4x2+2x—6

can be constructed using a marked ruler and compass (what Baragar calls a "compassand twice-notched ruler construction"). This polynomial is irreducible over Q, and

the methods used to analyze x5 — 6x + 3 in Section 6.4 imply that its splitting fieldQ c L has Galois group Gal(L/Q) S5. It follows that the roots of this polynomialare not expressible in terms of radicals and are not origami numbers, but can beconstructed using marked ruler and compass. However, it is not known exactlywhich numbers can be constructed in this way.

In the exercises you will show that marked-ruler-and-compass constructions canbe interpreted in terms of intersecting conchoids and limaçons with lines and circles.Further details may be found in [41.

• Origami and Dual Conics. Origami and intersections of conics lead to the sameset of complex numbers. Since origami involves simultaneous tangents to parabolas,it is reasonable to ask if origami has an intrinsic connection to intersections of conics.The answer involves the dual conic of a parabola, whose points correspond to tangentlines of the parabola. Then simultaneous tangents to two parabolas correspond tointersections of their dual conics. To make these ideas precise, one needs to work in


the projective plane, which is beyond the scope of this book. See [11 for a discussionof the ideas involved.

Historical Notes

What we call "conics" are more properly called "conic sections," for they weredefined by the Greeks as the intersections of a cone with a plane. One of the firstGreek geometers to consider conic sections was Menaechmus (ca. 350 B.c.). He wasa student of Plato and Eudoxus. He showed how to duplicate the cube by intersectingtwo parabolas (Exercise 11 of Section 10.1). Thus the idea of solving cubic equationsby intersecting conics goes back to the very beginning of the study of conic sections.

In his book On the Heptagon in the Circle, Archimedes (287—212 B.c.) may haveconstructed a regular heptagon using the intersections of conics. Although this bookno longer exists, works by Islamic geometers such as Thãbit ibn Qurra (826—90 1) onthe same problem mention Archimedes' book and use these methods to construct theregular heptagon.

One of the major works of Greek geometry is the Conic Sections by Apollo-nius (ca. 262—190 B.c.). This treatise introduced the terms ellipse, parabola, andhyperbola. A description of the Conic Sections can be found in [9, Ch. 6].

A later writer was Pappus (ca. 300), who wrote extensive commentaries on variousaspects of Greek geometry. His work contains the first known description of a conicsection in terms of focus, directrix, and eccentricity, though his description probablyappeared in earlier but now lost works. Pappus gave a nice angle trisection usingintersections of conics (see Exercise 14).

There is also a large Islamic literature on constructions using conic sections. Asnoted by Martin [15, p. 135], over a dozen conic constructions of the regular heptagonwere found by Islamic geometers during the Middle Ages. Besides Thabit ibn Qurramentioned earlier, another prominent geometer is Abu Ali Hasan ibn al-Haytham(ca. 965—1039), known in the West as Alhazen. He is best known for the problem ofdescribing reflections in a circular mirror, which he solved by intersecting a hyperbolaand a circle.

The emergence of equations for conics, such as (10.16), took a while. The standardequations for the ellipse, hyperbola, and parabola are implicit in many of the resultsproved by the Greeks, but it wasn't until the work of L'Hôspital in 1707 that theywere written down in their modem form. Much more on the history of the conicsections can be found in [5] and [6].

The marked ruler was first used by Nicomedes (ca. 240 B.c.) to construct cuberoots using a marked ruler (see Exercise 15). Pappus, in one of his commentaries onGreek geometry, described verging as moving a ruler "about a fixed point until bytrial the intercept [the portion of the ruler lying between the given lines] was found tobe equal to the given length." We've already seen that Archimedes and Pappus useda marked ruler to trisect angles. Nicomedes also introduced the conchoid, which is acurve created by verging with a line (see Exercise 16). In 1593, Viète proposed thatverging with a marked ruler be allowed for geometric constructions. See [4] and [15]for more details on the history of the marked ruler.


The connection with paperfolding or origami seems to be more recent. One of theearliest references is Geometric Exercises in Paper Folding by T. Sundara Row [19],published in Madras in 1893. The origami trisection given at the beginning of thesection was discovered in the 1970s by Hisashi Abe and is taken from [12]. Morereferences on origami can be found in [1], [3], [121, [14], and [16], and Hull's book[131 contains a wealth of activities related to origami and mathematics. We also notethat origami is equivalent to a construction called "mira" described in [8] and [15].


Exercise 1. This exercise will use the diagram

P2

P

P1

to prove that the origami construction described at the beginning of the section trisects theangle 0 formed by the line 12 and the bottom of the square.(a) Let Q be the intersection of the line segments and Prove that Q lies on the

dashed line 1.(b) Prove that 0 is congruent to a + fi.(c) Use triangles PQ1 and AP2PQI to prove that /3 and -y are congruent.(d) Use triangle to prove that a is congruent to + -y.

(e) Conclude that a is congruent to 20/3 and that the angle formed by P1Q1 and the bottomof the square is 0/3.

Exercise 2. In the text we showed how to trisect an angle between ir/4 and 7r/2 by origami.(a) Explain how to bisect and double angles by origami.(b) Explain how to trisect an arbitrary angle by origanii.

Exercise 3. Let P1 be a point not lying on a line Ii in the plane. Drop a perpendicular fromPi to 11 that meets Ii at a point S. Then choose rectangular coordinates such that P1 lies onthe positive y-axis and the x-axis is the perpendicular bisector of the segment P1S. In thiscoordinate system, P1 = (0,a) and is defined by y = —a, where a >0.(a) The parabola with focus P1 and directrix 11 is defined to be the set of all points Q that are

equidistant from P1 and Prove that it is defined by the equation 4ay = x2.

(b) Let Q = (xo,yo) be a point on the parabola. Prove that the y-intercept of its tangent lineis —yo.


(c) Let Q = (xo,yo) be a point on the parabola, and let Qi E be obtained by dropping aperpendicular from Q. Prove that is the reflection of Pi about the tangent line to theparabola at Q.

(d) Past (c) proves one direction of Lemma 10.3.1. Prove the other direction to complete theproof of the lemma.

Exercise 4. Show that the tangent line at a point (Xl,yi) on the first parabola in (10.9) hasslope given by

bm=yi —

Exercise 5. In the text we showed that the slopes of the simultaneous tangents to the parabolasin (10.9) are roots of (10.12). In this exercise, you will give an origami version of this in thespecial case when a = 2 and b = 1. Begin with a square sheet of paper folded so that thebottom edge touches the top. This fold will be the positive x-axis, and the left edge of the sheetwill be the directnx for the first parabola in (10.9).(a) Describe the origami moves one would use to construct the foci and directrices of the

parabolas in (10.9) when a = 2 and b = 1. Also construct the y-axis. Exercise 7 will behelpful.

(b) Now perform an origami move that takes the focus of each parabola to a point on thecorresponding directrix. Explain why there is only one way to do this.

(c) Part (b) gives a line whose slope in is the real root of x3 + 2x + 1. Explain what origamimoves you would use to find the point on the x-axis whose coordinates are (m, 0).

Exercise 6. Suppose that in the situation of C3, we have points not lying on lines£2. Also assume that and £2 are parallel and that there is a line £ satisfying C3 (i.e., £

reflects cs, to a point on £, for i = 1,2). Prove that the distance between Li and £2 is at most thedistance between and Cs2. This makes it easy to find examples where the line described inC3 does not exist.

Exercise 7. Consider the parabolas (y — = 2bx andy = from (10.9).(a) Show that the first parabola has focus b, and directrix x = —

(b) Show that the second parabola has focus (0, and directrix y = —

Hence the focus and directrix of the first parabola are defined over any subfield of R containinga and b. For the second, this is true over any subfield of R.

Exercise 8. Complete the proof of Theorem 10.3.6 sketched in the text.

Exercise 9. Prove Corollary 10.3.9.

Exercise 10. In Example 10.3.10, prove that £ meets Ii and 12 at the points Qi and Q2 given in

(10.13) and (10.14). Also draw the four lines whose slopes are the roots of (10.15).

Exercise 11. This exercise will give an example of a cubic equation that arises from verging.Consider the lines Ii defined by y = 0 and 12 defined by y = x and verge from P = (1, usinga marked ruler. Show that this gives the vertical line x = 1 together with three nonvertical lineswhose slopes m satisfy the cubic equation

4m3+m2-4m+1 =0.

Also show that the nonvertical lines cannot be constructed by straightedge and compass.

Exercise 12. Prove that LPRO = 0/3 in the construction (10.18).


Exercise 13. According to [15], Pappus used a marked ruler to trisect angles as follows. Givenan angle 0 <9 < ir/2, write it as 9 = LPOA, where:

• The distance between P and 0 is 1/2.• The line determined by P and A is perpendicular to the line determined by 0 and A.

Any angle 0 < 9 < can be put in this form by a marked-ruler construction. Finally, let l2be the line through P that is perpendicular to Then verging with 0 and the lines and 12

gives points Q and R E 12 such that Q and R are one unit apart:

Prove that LQOA = 9/3.

12

Exercise 14. As explained in [211, Pappus used intersections of conics to trisect angles asfollows. Consider the unit circle centered at the origin, and let 9 satisfy 0 < 9 < ir/2. ThenP = (cos9, sin 9) is the corresponding point on the unit circle. We assume that P is known.Also let 0 = (0,0) be the origin, and setA = (1,0). Thus 9 = LPOA.(a) Consider the curve C consisting of all points Q = (x,y) such that the distance from P to

Q is twice the distance from Q to the x-axis. The curve C intersects the unit circle at apoint R lying in the interior of LPOA. Prove that LROA = 9/3.

(b) Show that the curve C is a hyperbola. It follows that we have trisected an angle using theintersection of a hyperbola and a circle, i.e., an intersection of conies.

Exercise 15. In this exercise, we discuss a marked-ruler construction of cube roots due toNicomedes and taken from [15]. Let k be a real number such that 0 < k < 8, and consideran isosceles triangle such that AC and BC have length 1 and AB has length k/4. Thenextend AC and AB as indicated in the picture below, and choose D on the extension of AC sothat AD also has length 1. Finally, draw the line through D and B.

Verging from C with the lines and 12 indicated above gives points Q E 11 and R E 12 that areone unit apart. Assume that Q D.(a) Explain why the restriction 0 < k < 8 is necessary.(b) Prove that the distance between B and R is(c) Explain how to give a marked-ruler construction of for any k > 0.

P

1/2

R

0 A

A

1

B 12 R

Ii

D


Exercise 16. Let P be a point distance b > 0 from a line 1. Put a marked ruler though P withone mark at R E 1. When R moves along 1, the other mark or Q2 (depending on which sideof lit is on) traces out the conchoid of Nicomedes. When b < 1 we get the picture

We can relate the conchoid to construction problems as follows.(a) Suppose we are given a point P and lines li, 12, and assume that P Ii. Prove that a

point Q is obtained by verging with P and 12 if and only if Q is one of the points ofintersection of 12 with the conchoid determined by P and

(b) Prove that the angle trisection of (10.18) can be interpreted as the intersection of the unitcircle with the conchoid determined by P and 1.

(c) Suppose that P = (0,0) and I is the horizontal line y = —b. Prove that the polar equationof the conchoid is

r= bcsc6± 1,

where the minus sign gives the portion of the curve above I and the plus sign gives theportion below.

(d) Under the assumptions of part (c), show that the Cartesian equation of the conchoid is

(x2+y2)(y—b)2 =y2.

By part (a), verging is the same as intersecting the conchoid with a line. Since the aboveequation has degree 4, this explains why verging leads to an equation of degree 4.

Exercise 17. Let P be a point on a circle, and consider a marked ruler that goes through P.If we place one mark on a point Q on the circle, then the other mark R1 or R2 (depending onwhether it is inside or outside the circle) traces out a curve called the limaçon of Pascal:

This curve was known to Jordanus Nemorarius (1225—1260) and Albrecht Dürer (147 1—1528)and possibly the ancient Greeks. It was rediscovered by Etienne Pascal (father of Blaise Pascal)


about a century after Dürer. In 1650 Roberval, unaware of the earlier work, named the curvein Pascal's honor.(a) Show that the angle trisection (10.18) can be interpreted as the intersection of the line I

with the limaçon determined by the circle and the point P.(b) Let P = (0,0) and let C be the circle of radius a and center (a,0). Show that the

corresponding limaçon has polar equation

r= 1+2acosO.

(c) In the situation of part (b), show that the Cartesian equation of the limaçon is

(x2+y2—2ax)2 =x2+y2.

Exercise 18. A Pierpont prime is a prime p > 3 of the form p = + 1. Prove that a regularn-gon can be constructed by origami (or by marked ruler or by intersections of conics) if andonly if = 2a3bp1

Ps, where a, b � 0 and pi,..., are distinct Pierpont primes. This wasfirst proved by Pierpont in [17].

REFERENCES

1. R. C. Alperin, A mathematical theory of origami constructions and numbers, New YorkJ. Math. 6 (2000), 119—133.

2. J. W. Archbold, Introduction to the Algebraic Geometry of a Plane, Edward Arnold,London, 1948.

3. D. Aucldy and J. Cleveland, Totally real origami and impossible paper folding, Amer.Math. Monthly 102 (1995), 215—226.

4. A. Baragar, Constructions using a compass and twice-notched straightedge, Amer. Math.Monthly 109 (2002), 151—164.

5. J. L. Coolidge, A History of the Conic Sections and Quadric Surfaces, Oxford U. P., 1945.Reprint by Dover, New York, 1968.

6. J. L. Coolidge, A History of Geometrical Methods, Oxford U. P., 1940. Reprint by Dover,New York, 1963.

7. B. Carter Edwards and J. Shurman, Folding quartic roots, Math. Mag. 74 (2001), 19—25.

8. J. W. Emert, K. I. Meeks and R. B. Nelson, Reflections on mira, Amer. Math. Monthly101 (1994), 54.4—549

9. H. Eves, An Introduction to the History of Mathematics, Sixth Edition, Brooks/Cole,Pacific Grove, CA, 1990.


11. J. J. Gray, A commentary on Gauss's mathematical diary, 1796—1814, with an Englishtranslation, Expo. Math. 2 (1984), 97—130. (The Latin original of Gauss's diary isreprinted in [Gauss, Vol. X.l].)

REFERENCES 289

12. 1. Hull, A note on "impossible" paper folding, Amer. Math. Monthly 103 (1996), 240—241.

13. T. Hull, Project Origami: Activities for Exploring Mathematics, A K Peters, Ltd., Welles-ley, MA, 2006.

14. T. Hull, Solving cubics with creases: The work of Beloch and Lill, Amer. Math. Monthly118 (2011), 307—3 15.

15. G. E. Martin, Geometric Constructions, Springer, New York, Berlin, Heidelberg, 1998.

16. V. Pambuccian, Axiomatizing geometric constructions, J. Appl. Log. 6 (2008), 24-46.

17. J. Pierpont, On an undemonstrated theorem of the Disquisitiones A rithmeticce, Bull. Amer.Math. Soc. 2 (1895—1896), 77—83.

18. M. M. Postnikov, The problem of squarable lunes, Amer. Math. Monthly 107 (2000),645—651.

19. T. S. Row, Geometric Exercises in Paper Folding, Addison & Co., Madras, 1893. Alsopublished by Open Court, Chicago 1901. Reprint by Dover, New York, 1966.

20. P. Stevenhagen and H. W. Lenstra, Jr., Chebotarëv and his density theorem, Math. Intel-ligencer 18, no. 2 (1996), 26—37.

21. C. R. Videla, On points constructible from conics, Math. Intelligencer 19, no. 2 (1997),53—57.

CHAPTER 11

FINITE FIELDS

The main topic of this chapter is the theory of finite fields. We will study theirexistence and uniqueness and compute their Galois groups. We will also considerirreducible polynomials over finite fields.

11.1 THE STRUCTURE OF FINITE FIELDS

In this section we discuss the basic properties of finite fields.

A. Existence and Uniqueness. The simplest examples of finite fields are F,,,the integers modulo a prime p. These relate to arbitrary finite fields as follows.

Proposition 11.1.1 Let F be a finite field. Then:(a) There is a unique prime p such that F contains a subfield isomorphic to F,,.(b) F is a finite extension of F,,, and

Fl =pfl, wheren=

Proof: Every field of characteristic 0 contains a subfield isomorphic to Q and

hence is infinite. Thus F has characteristic p for some prime p. Furthermore, the


292 FINITE FIELDS

discussion of characteristic in Section A. 1 shows that p7L C Z is the kernel of the ringhomomorphism that sends m E Z to 1 E F. By the Fundamental Theorem of RingHomomorphisms, F contains a subfield isomorphic to Z/pZ = F,,.

The map F,, —* F makes F an extension field of F,,. Following our usual practice(see Definition 3.1.2), we identify IF,, with its image and writeF as a vector space over F,,. The elements ofF give finitely many vectors in F, whosespan over F,, is obviously F. It follows that F is a finite-dimensional vector spaceover IF,,. As in Section 4.3, this means that F is a finite extension ofF,,. Furthermore,if n = [F : F,,], then we can find a basis a1,... ,a,, of F over F,,. Hence every elementof /3 F can be written uniquely as

ajEIF,,.

Since the a, can be any of the p elements of IF,,, there are possibilities for ThusFl = This completes the proof. •

For the rest of this chapter, we will assume as in the above proof that a finitefield F contains F,, as a subfield. Our first major result is that F is the splitting fieldover F,, of a particularly simple polynomial.

Theorem 11.1.2 Let F be a finite field with q = p" elements. Then:(a)

a x

F has q elements, its multiplicative group F* = F \ {O} is a group withq 1 elements. It follows that a for all a E F.This proves part (a) and shows that the q elements of F are roots of — x. Thenpart (b) follows since — x is monic of degree q. Hence — x splits completelyover F. Since every element of F is a root, — x can't split completely over anystrictly smaller field. Thus F is a splitting field —XE •

Using this theorem, we obtain the following uniqueness result for finite fields.

Corollary 11.1.3 Two finite fields with the same number of elements are isomorphic.

Proof: Corollary 5.1.7 implies that any two splitting fields of x E F,, [x] areisomorphic. Then Corollary 11.1.3 follows immediately from Theorem 11.1.2. •

We next show that a finite field of order p" exists for any p and n.

Theorem 11.1.4 Given any prime p and any positive integer n, there is a finite fieldwith pfl elements.

Proof: Let q = p", and let L be an extension of F,, such that — x splits completelyover L. Since we are in characteristic p, the derivative of — x is —1, so that

— x, — x)') = 1. Thus — x is separable and hence has distinct roots in L.This means that F = {a e L

I= a} is a subset of L consisting of q elements. In

THE STRUCTURE OF FINITE FIELDS 293

Exercise 1 you will show that F is a subfield of L. It follows that F is a finite fieldwith q = pfl elements. .

Given any q p" as in Theorem 11.1.4, the finite field of order q constructed inthe theorem is unique up to isomorphism by Corollary 11.1.3. Hence we can speakof "the" finite field with q elements. We will denote this field as Fq. Since these fieldswere first described by Galois (see the Historical Notes), lFq is sometimes denoted asGF(q), where "GF' stands for "Galois Field."

One can use Theorem 11.1.2 to count the number of roots of a polynomial in afinite field as follows.

Proposition 11.1.5 1ff E is nonconstant and n 1, then the number of rootsoff in is the degree of the polynomial gcd(f,xP' — x).

Proof: Let g = gcd(f,xP' — x), where the gcd is computed in A usefulobservation is that if one replaces with any larger field, then one gets the samepolynomial g (you will prove this in Exercise 2). Thus we may compute the gcd in

[xJ. If we denote the elements of this field by for i = 1,... , then

by part (b) of Theorem 11.1.2. This is the irreducible factorization of x in[x]. Hence g is the product of those x — that divide f. Since x — divides f if

and only if = 0, we obtain the product formula

g= fif(at)=O

The proposition now follows immediately. U

Here is an example to illustrate Proposition 11.1.5.


f=x" +x5+2x+1 EIF7[X].

To compute the number of roots in we need to compute gcd(f, — x). In Maple,we do this using the command

mod 7;

In Mathematica, we would type

Modulus —> 7]

In both cases, the output is the polynomial

294 FINITE FIELDS

By Proposition 11.1.5,1 has three roots in F73. Furthermore, replacing V3 within the above computation gives a gcd of 1, so that f has no roots in F74.

One drawback of this method is that the degree of x7" — x increases rapidly. Forexample, if we replace with in the above computation, then Maple gives

x8+3x7+x6+x5+5x4+x3+4x+6

(thus f has eight roots in F78), whereas Mathematica gives an error message becausethe degree of x = xSlMSOl x is too large for PolynomialGCD.

B. Galois Groups. We next compute the Galois group of the extension F,, C Fq.

Theorem 11.1.7 If q = p's, then:(a) F,, C is a Galois extension of degree n.(b) The map : Fq —+ Fq defined by Frob,,(cx) is an automorphism of Fq

that is the identity on F,,.(c) generates Gal(Fq/Fp). Thus there is a group isomorphism

Gal(Fq/Fp) Z/nZ

that sends E Gal(Fq/Fp) to [1] E 7L/nZ.

Proof: In the proof of Theorem 11.1.4, we noted that — x is separable. ThenTheorem 11.1.2 implies that Fq is the splitting field of a separable polynomial. Hence

C Fq is Galois. Proposition 11.1.1 implies that = n since q = p".Turning to part (b), observe that by Lemma 5.3.10,

Since we also have Frob,,(1) = = 1 and

Frob,,(a$) = = =

Frob,, is a ring homomorphism. By Exercise 2 of Section 3.1, is

also one-to-one and hence onto, since it maps the finite set Fq to itself. Thus Frob,, isan automorphism of Fq. Since it is the identity on by Lemma 9.1.2, we concludethat Frob,, E Gal(Fq/Fp).

For part (c), we first note that since F,, C Fq is Galois, we have

Gal(Fq/F,,)I = [Fq:FpJ fl,

where the second equality uses q = pfl and Proposition 11.1.1. It follows that theorder of Frob,, divides n. Suppose that (Frob,,)r is the identity, where 0 < r < n.

Here, (Frob,,)' denotes the r-fold composition of Frob,, with itself, so that

times r times

= (...(csP)P...)P =


Thus, if (Frobp)' is the identity element of Gal(lFq/Fp), then

& =

for all lFq. Since 0 < r < n, this implies that the polynomial — x of degreepr <p's = q has q roots, which is clearly impossible. Hence has order n, whicheasily gives the desired isomorphism Gal(lFq/]Fp) Z/n7L. •

We call the Fmbenius automorphism of lFq. We next use Theorem 11.1.7to determine when one finite field is contained in another.

Corollary 11.1.8 Let and befinitefields. Then is isomorphic to a subfieldof andonly ifmln.

Proof: First suppose that is isomorphic to a subfield of Writing this as aninclusion, we obtain

c cProposition 11.1.1 and the Tower Theorem imply that

n = :lFp] = =

This shows that m divides n.Conversely, suppose that mn. Since is cyclic of order n by Theo-

rem 11.1.7, we know that has a subgroup H of order By the Galoiscorrespondence of Section 7.3, the fixed field F of H is an extension

F c

satisfying

=m.

Using Proposition 11.1.1, we see that F has order ptm. By Corollary 11.1.3, it followsthat F is a subfield of isomorphic to

In Exercise 3 you will prove Corollary 11.1.8 using neither Theorem 11.1.7 northe Galois correspondence.

When Corollary 11.1.8 gives F C As usual, we identify withF, which gives the inclusion

Then we can generalize Theorem 11.1.7 as follows.

Theorem 11.1.9 Let mm and C Then there is a group isomorphism

G 1/It? lit? \ '7//fl—

that sends E to [1] E

296 FINITE FIELDS

Proof: You will prove this in Exercise 4. •This result makes it easy to work out the Galois correspondence of F,, C The

key point is that subfields of correspond to subgroups of Z/nZ, yet subgroups ofZ/nZ correspond to positive divisors of n. Here is an example inspired by the classicbook [131 by Lidl and Niederreiter.

Example 11.1.10 For 1F230, the above remarks show that subfields of IF230 correspondto positive divisors of 30. This gives the following Galois correspondence:

F230 {0}/INZ/5Z Z/3Z Z/2Z

tN/ N7/\F22 IF23 F25 Z/157L Z/1OZ Z/6Z/

7L/30Z

To understand the diagram, recall that an intermediate field IF2 C F C F230 corre-sponds to the subgroup Gal(IFyo/F) C Gal(1F2,o/F2) Z/30Z. Combining this withTheorem 11.1.9, we see that F = corresponds to Gal(lF23o/lF2m) Z/ Whenwe do this for each divisor of 30, we obtain the above diagram.

Mathematical Notes

Finite fields are used in many different areas of mathematics. Here are three ofparticular importance.

• Finite Groups. If F is any field, then the set GL(n, F) of invertible n x n matriceswith entries in F is a group under matrix multiplication. In particular, if F is afinite field, then GL(n, F) is a finite group. These groups play an important role inboth Galois theory and the theory of finite groups. We will say more about this inSection 14.3.

• Equations over Finite Fields. Given a nonzero polynomial f e R[x,y], the so-lutions of f(x,y) = 0 lying in R2 form a curve in the plane. For instance, the unitcircle is defined by x2 +y2 — 1 = 0. Similarly, given a polynomial f E IFp{x,yI, wecan consider solutions of f(x,y) = 0 lying in Such equations were of interest toGauss. For instance, the equation x2 + y2 + x2y2 1 mod p appears in the last entryin his mathematical diary [101.

Things get more interesting when one realizes that for f E [x, y], we can alsoconsider solutions of f(x, y) = 0 lying in for any n> 1. Since is finite, thereare only finitely many such solutions, though as n gets larger, the number of solutions


increases. This is all controlled by the zeta function of the equation. See [9, Ch. 5],[11, §11.5], and [12, PP. 158—1601 for more about zeta functions.

Coding Theory. Information in a computer is often represented as a string of 0'sand 1 's. A good example is the ASCII code, which uses numbers between 0 and127 to represent letters, digits, punctuation marks, and common special characterson computers. The ASCII value of "g" is 103 (decimal) = 1100111 (binary), while"G" is "071" (decimal) 1000111 (binary). Thus the ASCII code uses which isa vector space of dimension 7 over F2.

In algebraic coding theory, a code consists of a subset of (the set of codewords), where q is usually a power of 2. In the study of error-correcting codes,one wants the code words to be widely spaced so that errors in transmission can bedetected, yet if they are too widely spaced, the code is not very efficient. Findinggood codes is an active area of research. An introduction to coding theory can befound in Lidl and Niederreiter [13, Sec. 9.11. One surprise is that an important classof codes (the so-called Goppa codes) involves equations f(x, y) =0 over finite fields.See Moreno's book [14] for an introduction.

Historical Notes

The finite field IF,, first arose in the study of congruences modulo the prime p.For an arbitrary modulus n, congruences are implicit in the work of Fermat, Euler,Lagrange, and Legendre, though Gauss was the first to give an explicit definition.All of these people knew that congruences modulo a prime have special properties.For example, in Disquisitiones, Gauss states the following result:

A congruence of the mth degree

+cxm_2+etc.+Mx+N_=O

whose modulus is a prime number p that does not divide A, cannot be solved inmore than m different ways, that is, it cannot have more than m noncongruentroots relative to p...

(See [8, Art. 43].) In modern terms, this says that a polynomial in of degree mhas at most m roots in F,,, which is true because IF,, is a field. Exercise 5 will give anexample of how this can fail when the modulus is not a prime.

In 1830, Galois published Sur Ia théorie des nombres in the Bulletin des sciencesmathématiques de Ferussac (see [Galois, pp. 113—127]). Galois begins the paper bynoting that for congruences F (x) 0 mod p as above, "one customarily considersonly integer solutions." If instead one considers "incommensurable" solutions, Galoisstates that "I have arrived at certain results that I think are new." Essentially everythingin this section can be traced back to these results of Galois.

His construction goes as follows. Consider a polynomial F E Z[x] of degree ii> Iwhose reduction modulo p in IF,, [x] is irreducible of degree v (Exercise 6 explainshow Galois stated this). Since v> 1, it follows that the congruence

(11.1)

298 FINITE FIELDS

has no integer solutions. Thus, as stated in [Galois, p. 113], "One must regard theroots of this congruence as a kind of imaginary symbol." In analogy with the usualsymbol for Galois uses ito denote a root of the congruence (11.1). Then heconsiders expressions of the form

where a,a1,... are integers modulo p. There are p" different choices for ct,which give the finite field with pV elements. Galois proves the following facts aboutthis field:

•• The elements of this field are the roots of x (Theorem 11.1.2).• All irreducible polynomials of degree ii lead to the same field (Corollary 11.1.3).• Primitive roots exist, i.e., the nonzero elements of the field form a cyclic group

under multiplication (Proposition A.5 .3).Galois also knew Proposition 11.1.5, which he states as follows:

Next, to get the integer solutions, it suffices, as M. Libn appears to have beenthe first to remark, to find the greatest factor common to Fx = 0 and = 1.

If now one wants to have imaginary solutions of the second degree, onefinds the greatest factor common to Fx = 0 and = 1, and in general, thesolutions of order v will be given by the greatest factor common to Fx = 0 andx/,,,_I = 1.

(See [Galois, pp. 123—125].) Note that Galois uses an equal sign = to denotecongruence modulo p. Do you see how his version of Proposition 11.1.5 counts onlynonzero solutions?

Galois's arguments are sketchy and assume the existence of a root i of (11.1).Nevertheless, his account is remarkably complete. For instance, given an arbitraryii> 1, he uses a splitting field of — x to prove the existence of a polynomial F ofdegree ii that is irreducible modulo p. This is his way of proving Theorem 11.1.4.As for the Galois group Galois doesn't state Theorem 11.1.7 directly.However, if i is a root of (11.1), then Galois knew that the other roots are given by

jP2, , which is the usual way we use the Galois group to find the otherroots of an irreducible polynomial (see Exercise 7). Also, in Section 14.3 we will seehow Galois used to construct some interesting matrix groups.

However, Galois was not the only person to discover finite fields. Gauss describeda theory of "higher congruences" in an unpublished manuscript from around 1800,and Schönemann described a theory of finite fields based on congruences in 1846.Here is Schönemann's approach. Let f e ZIxI be monic of degree n whose reductionmodulo p is irreducible in [xl. Also let ct e C be a root off. Then Schönemannconsiders expressions of the form where E Z[xl, and he writes

(11.2)

to mean = + pR(cx), where R E 7L[xI. Using this, he shows that everyis congruent modulo (p, to an expression of the form


where 0< a <p — 1 for i = 0,... ,n 1. There are p" such expressions.From a modern perspective, lies in the ring Z[a], and the congruence classes

of (11.2) give the quotient ring

(11.3)

where is the ideal of Z[a] generated by p. It follows that Schönemann'sconstruction leads to a ring with elements.

We prove that this ring is a field as follows. In Exercises 8 and 9 you will showthat since f is monic, the evaluation map q(x) '—* q(&) induces a ring isomorphism

Z[x]/fZ[x]

Furthermore, you will also show that this isomorphism induces an isomorphism

Z{xI/(p,f)

where

(11.4) (p,f) =pZ[xJ+fZ[x] = {pR(x)+f(x)S(x) IR(x),S(x) EZ[x]}.

Thus we can interpret Schönemann's construction as taking the quotient of Z[x] by(p, f) in two steps: first quotient out by f to get and then quotient out by p toget (11.3). However, we can reverse the order of the steps: first quotient out by pto get Z[x] /pZ[x] F,, [x], which sends f to f F,, [xl, and then quotient out by f toobtain the isomorphism -

(see Exercise 9). Since f E F,, [x] is irreducible, the results of Chapter 3 show thatF,, [xl / (f) is an extension field of F,, in which f has a root (namely, the coset of x).

Combining these isomorphisms gives

(11.5) Z[a]/pZ[cx]

which proves that Schönemann's construction gives a finite field with pfl elements.Also, this isomorphism sends the coset of to the coset of x. If we let i denote thiscoset, then we recover Galois's construction of the finite field. So everything fitstogether nicely. See [5] for more on Schönemann's work on finite fields.

Besides the constructions of Schönemann and Galois, the isomorphisms (11.5)show that we can represent a finite field with elements as

(11.6) Z[xl/(p,f)

whenever f is monic of degree n and irreducible modulo p. This construction isimplicit in Schönemann's work and was made explicit by Dedekind in 1857. UnlikeGalois (who assumed the existence of i) and Schönemann (who used the FundamentalTheorem of Algebra to find a), Dedekind's construction is purely algebraic andbecame the standard way to define finite fields (see, for example, Dickson's book [6]

300 FINITE FIELDS

from 1901). In 1893 E. H. Moore showed that any finite field is isomorphic to oneof the form (11.6). He also introduced the notation GF(p") for the finite field withp" elements constructed in (11.6), although these days the notations GF(p") andare often used interchangeably.

For more details on the history of finite fields, we refer the reader to [7, Vol. I, pp.233—252] and [13, pp. 73—78].


Exercise 1. Let F,, C L be an extension such that — x splits completely over L, where q = pfl,

and let F be the set of roots of this polynomial. Prove that F is a subfield of L.

Exercise 2. Suppose that f, g E F{x] are polynomials, not both zero, and let h be their greatestcommon divisor as computed in F [x]. Now let L be an extension field of F. Prove that h is thegreatest common divisor off, g when considered as polynomials in L[x].

Exercise 3. Give a proof of Corollary 11.1.8 that uses neither Theorem 11.1.7 nor the Galoiscorrespondence.

Exercise 4. Prove Theorem 11.1.9.

Exercise 5. As noted in the text, if f E 7L[x] has degree n and its leading coefficient is notdivisible by a prime p. then f(x) 0 mod p has at most n solutions modulo p. Here are twoquestions that explore what happens when n = 2 and the modulus is arbitrary.(a) How many solutions does the congruence — 1 0 mod 8 have modulo 8?(b) Fix an integer m> 1, and assume that every polynomial of degree 2 in ZL/m7L[x] has at

most two roots in Z/mZ. Is m prime?

Exercise 6. Let F E 7Z[x] have degree n, and assume that the leading coefficient of F is notdivisible by p. Prove that the reduction of F modulo p is irreducible over F,, if and only if itis not possible to find polynomials Z[xJ, where <n, such that

= F(x) +px(x).

This is how Galois defines irreducibility modulo p in [Galois, p. 113].

Exercise 7. Let f E F,, [xj be irreducible of degree ii. Use (7.1) and Theorem 11.1.7 to proveGalois's observation that if i is one root off in a splitting field, then the other roots are given

ExerciseS. LetlandfbeidealsinaringR,andletl+J={r+sI rE I,sEJ}betheirsum.Also let 7 = {r + J r E !}. This is a subset of the quotient ring R/J.(a) Prove that I + J is an ideal of R and that I is an ideal of R/J.(b) Show that the map r + (I + J) '—* (r + J) + 7 defines a well-defined ring isomorphism

R/(I+J) (R/J)/7.

Exercise 9. Let f E Z[x] be monic and irreducible, and let a E C be a root of f. Then letf E F,, [x] be the reduction off modulo the prime p. and let (p, f) be as in (11.4).

(a) Prove that the map q(x)+fZ[x] '—+ q(a) is a well-defined ring isomorphism Z{x]/fZ[x]Z[a].

(b) Use Exercise 8 to prove that Z[x]/(p,f) Z[a]/pZ[cs].(c) Similarly prove that ?'L[x]/(p,f) Fp[x]/(f).

IRREDUCIBLE POLYNOMIALS OVER FINITE FIELDS (OPTIONAL) 301

Exercise 10. Letf=2+2.r+2x2+2x3+2x4+2x5+2x6+2x7+x8+x9+x'° E ]F3[x].

(a) Use the method of Example 11.1.6 to determine the number of roots off in IF33 and IF37.(b) Explain why the splitting field off over IF3 is IF321.

Exercise 11. Let f E be an irreducible polynomial of degree n. Prove that f splitscompletely in

11.2 IRREDUCIBLE POLYNOMIALS OVER FINITE FIELDS (OPTIONAL)

Our presentation of finite fields in Section 11.1 was very abstract—it wasn't until theHistorical Notes that we explicitly constructed as IF,, [x] / (f), where f E IF,, [x] isirreducible of degree n. Yet whenever finite fields are implemented on a computer,such a representation is essential. It follows that we need a good understanding ofirreducible polynomials in IF,, [xi.

A. Irreducible Polynomials of Fixed Degree. We begin with the followingeasy result concerning irreducible polynomials in IF,, [x].

Proposition 11.2.1 Letf E IF,, [x] be irreducible of degree m. Then:(a) fdividesxP" —x.(b) f is separable.(c) Given an integer n 1, f divides — x f has a root in

Proof: We begin with part (c). Let be a root of f in a splitting field over IF,,. Sincef is irreducible, c IF,, (cr) has degree m. Then F,, (cr) by Proposition 11.1.1and Corollary 11.1.3. From here, the second equivalence of part (c) follows directlyfrom Corollary 11.1.8. Since f is irreducible over IF,,, we also have f Igcd(f, — x)if and only if deg(gcd(f,xP' — x)) > 0. Then the first equivalence follows fromProposition 11.1.5.

We get part (a) by taking n = m in part (c), and part (b) follows immediately, sincexv'" — x is separable by the proof of Theorem 11.1.4.

More generally, one can show that if F is any finite field and f E F [x] is irreducible,then f is separable (see Exercise 1). Since irreducible polynomials in characteristico are always separable, it follows that nonseparable irreducible polynomials canoccur only over infinite fields of characteristic p. Do you see how this relates to theexamples of such polynomials presented in Example 5.3.11?

We next want to count the number of irreducible polynomials of fixed degree in[xJ. This number is finite, since we are working over a finite field. Furthermore,

since any irreducible polynomial becomes monic after multiplying by a suitableconstant, it suffices to compute the number

Nm {f IF,, [x] I f is monic irreducible of degree m}

These numbers are related as follows.

302 FINITE FIELDS

Theorem 11.2.2 Let Nm be as defined above. Then, for any n � 1, we have

=

mm

where the sum is over all positive divisors of n.

Proof: Since x is separable, we know that it factors as a product of distinctirreducible polynomials in [x]. Furthermore, since it is monic, we can assumethat the polynomials in the factorization are also monic. Finally, part (c) of Proposi-tion 11.2.1 shows that the polynomials in the factorization are all monic irreduciblepolynomials of [x] whose degree m divides n.

This allows us to write xP" — x as follows. Let

= {f E [xJI f is monic irreducible of degree m},

so that Nm = I. Then the previous paragraph implies that

(11.7) x1"_xz=fl Hfmm

(be sure you understand why). Since every f E has degree m, taking the degreeof each side of (11.7) gives the desired formula. •


Example 11.2.3 The monic irreducible polynomials of degree 1 in are of theform x a for a e Thus N1 = p. Then the theorem implies that

p2 =2N2+Ni =2N2+p,

so that N2 = — p). In Exercise 2 you will use this to prove that

(11.8)

N4 that we can findirreducible polynomials of degrees 2 and 4. In particular, this proves the existence offinite fields of orders p2 and p4.

To generalize the formulas in this example, we will use the Möbius function p.(n)from Exercise 14 of Section 9.1. This function is defined by

11, ifn=1,= ifn =p' fordistinctprimesp1,.

.

0, otherwise.

Then we have the following formula for


Theorem 11.2.4 The number of monic irreducible polynomials of degree n inis given by

mm

Proof: Let F be a complex-valued function defined on the set of positive integers.Then we get another such function G defined by

G(n) =mmn

where as usual the sum is over all positive divisors of n. The Möbius inversionformula asserts that in this situation, we can express F in terms of G as follows:

F(n) =mm

A proof of the Möbius inversion formula can be found in most books on numbertheory. See [11, Sec. 2.21 or [15, Sec. 4.3].

In particular, if F(n) = then Theorem 11.2.2 implies that

G(n) = >F(m) = = p's.mm mm

By the inversion formula, we obtain

= F(n) = =mm mm

The desired formula follows immediately.

Here are some examples of how this theorem works.

Example 11.2.5 When n = 4, Theorem 11.2.4 implies that

N4 =

= .p4+(—l)•p2+O.p)

=

which agrees with (11.8). Similarly, when n = 6, you will show that

N6=

in Exercise 3.

In Exercise 4 you will use Theorem 11.2.4 to prove that > 0 for all n � 1.

304 FINITE FIELDS

B. Cyclotomic Polynomials Modulo p. Theorem 11.1.2 tells us that = afor all a E Fq. It follows that = 1 when a $ 0, so that every nonzero elementOf Fq is a root of unity. In characteristic 0, the minimal polynomial of a primitive dthroot of unity in C is the cyclotomic polynomial We will now explore whathappens when we reduce these polynomials modulo p.

By Section 9.1, (x) is the monic polynomial whose roots are the primitive dthroots of unity in C. Furthermore, has integer coefficients, is irreducible ofdegree and has the factorization

(11.9)din

in Z [x]. The reduction of (x) modulo p should be denoted (x), but for simplicitywe will denote it by 1'd(X). Thus (11.9) becomes an identity over in Wewill concentrate on the case when gcd(d,p) = 1. This restriction is explained inExercise 5.

We begin by describing the roots of (x). Recall that a dth root of unity a isprimitive if d is the smallest positive integer such that = 1.

Proposition 11.2.6 If gcd(d, p) = 1 and q = pfl, then the following are equivalent:(a)

(b) (x) splits completely in lFq.

(c) has a root in Fq.Furthermore, when these conditions are satisfied, the roots of '1d (x) in lFq consist ofthe primitive dth roots of unity.

Proof: We first study the primitive dth roots of unity in characteristic p. Observethat x" 1 is separable, since gcd(d,p) = 1. Hence x" — 1 has d roots in a splittingfield. These roots form a group under multiplication, which is cyclic of order d byProposition A.5.3. Such a group has generators (Exercise 10 of Section 9.1),so that there are primitive dth roots of unity. By Exercise 11 of Section 9.1,

Lid

and by (11.9) forn =d,

Lid

From these facts, it is straightforward to prove by complete induction on d that theroots of (x) are the primitive dth roots of unity in characteristic p.

We now prove (a) (b). Applying (11.9) with n = q — 1, we obtain

—1 = fiLIq—I

Since — 1 splits completely in lFq and dlq — 1, we see that (x) splits completelyFq. The implication (b) (c) is trivial. Finally, to prove (c) (a), note that a root


E lFq of is a primitive dth root of unity by the above analysis. Thend in F), and (a) is proved.

The final assertion of the proposition now follows immediately. •

We next compute the irreducible factors of 4d (x). Since gcd(d, p) = 1, we have[p] E (Z/dZ)*. Let m denote the order of [pJ in this group. Writing [p]m = [1] as a

congruence, we see that m is the smallest positive integer such that dlpm 1. Thisnumber determines the degree of the irreducible factors of as follows.

Theorem 11.2.7 Given d, let m be as above. Then 1d (x) is the product ofirreducible polynomials in IF,, [x] of degree m.

Proof: Let f be an irreducible factor of (x). To show that deg(f) = m, it sufficesto show that the smallest positive integer £ such that d 1i' — 1 is £ = deg(f). We provethis as follows. For an integer £ � 1, observe that

d — 1 (x) splits completely over

f splits completely over

f has a root in

The first equivalence is by (a) (b) of Proposition 11.2.6, the second follows from(b) (c) of the same proposition because fI (x), the third follows because F,, Cis Galois and hence normal, and the fourth is by part (c) of Proposition 11.2.1. Theseequivalences show that deg(f) has the desired property.


Example 11.2.8 Let p = 2. Ford =5, one easily sees that [2] has order 4 in (Z/5Z)By Theorem 11.2.7, is the product of = 1 irreducible polynomials ofdegree 4 in IF2 [x]. Thus (x) = x4 + x3 + x2 + x + 1 is irreducible in F2 [x]. Its roots

are the primitive 5th roots of unity in IF16.When d = 15, [2] also has order 4 in (Z/15Z)*. Thus 4'I5(x) is the product ofI 5)/4 = 8/4=2 irreducible polynomials of degree 4. In Exercise 6 you will verify

that the factorization in F2 [x] is

= x8 +x7 +x5 +x4 +x3 +x+ 1 = (x4 +x3 + 1)(x4 +x+ 1).

The roots of these polynomials are the primitive 15th roots of unity in 1F16.

C. Berlekamp's Algorithm. We conclude this section by explaining how todetermine whether a given nonconstant polynomial f E IF,, [x] is irreducible. Onemethod for doing this would be to list the finitely many nonconstant polynomialsof degree <deg(f) and divide each into f. We will give a much more efficientmethod based on Berlekamp's factoring algorithm that uses a nice combination oflinear algebra and the division algorithm.

306 FINITE FIELDS

Suppose that we want to test whether a given polynomial f E IF,, [x] is irreducible.We may assume that f has degree n> 1. Furthermore, since irreducible polynomialsover F,, are separable by Proposition 11.2.1, we may also assume that f is separable.We will use the quotient ring

R IF,, [x] / (f).

By the division algorithm, every element of the ring R can be written uniquely in theform

a vector space over IF,,, R has dimension n. Now consider the map

T : R —* R

definedby T(g+ (f)) = g" + (f). This is well defined, for ifg+ (f) = h+ theng h+fB for some BE F,,{x]. Since we are in characteristic p. we have

gP = (h+fB)" =h"+fB" =h"+f•f"'B",

which implies gP + (f) = + (f). Furthermore, it is easy to see that T is linear over(you will prove this in Exercise 7). The identity map 'R R —* R is also linear.

Then we get the following unexpected result.

Theorem 11.2.9 Letf E F,, [x] be separable of degree n> 1, and let R = F,, [x] / (f).Then f is irreducible and only if the linear map T — R -4 R has rank n — 1.

Proof: If f is irreducible, then R is a field and T is the Frobenius automorphism.Hence the kernel of T — 1R consists of the solutions of = This gives IF,,, so thatthe kernel has dimension 1. By the dimension formula from linear algebra, it followsthat T 1R has rank n—i.

On the other hand, if f is reducible, then f gh where g, h E IF,, [x] have degree<deg(f). Furthermore, g and h must be relatively prime, since f is separable. Hencewe can such thatAg+Bh= 1.

Observe that Ag + (f) is in the kernel of T 1R if and only if (Ag)P — Ag is amultiple off. Using the binomial theorem and f = gh, we have

(Ag)" =Ag(i —Bh)"1 r=Ag(l (p— 1)Bh+ . . .+ (—l)P1(Bh)P1)=Ag—gh.(p— l)AB+...+gh•(—l)"'AB"'h"2

Ag mod f.

It follows that Ag + is in the kernel. Interchanging the roles of Ag and Bh, we seethat Bh + (f) is also in the kernel.

If we can show that Ag + (f) and Bh + (f) are linearly independent elements ofR, then T — 1R will have rank at most n —2 and the theorem will follow. So supposethat some linear combination of these cosets is zero, i.e., there are a, b E IF,, such thataAg + bBh is a multiple off = gh. Then

aAg+bBh=ghC


for some C e [x]. Since g and h are relatively prime, it follows easily that gIbB andBut Ag +Bh = 1 implies that gcd(g,B) = gcd(h,A) = 1, so that we must have

a = b = 0. This proves the desired linear independence. .One useful observation is that from the vector Ag + (f) in the kernel of T —

we can recover the factorization off, since g = gcd(f, Ag) follows from f =gh andAg + Bh = 1. In general, elements of the kernel can be more complicated, but it is stillpossible to use them to find the irreducible factorization off. This is Berlekamp'salgorithm, which is described in [13, Sec. 4.1].

Here is an example to show how Theorem 11.2.9 can be used.

Example 11.2.10 Let f = x5 + x4 + 1 IF2 [x]. Note that f is separable, sincegcd(f, f') = 1. Then R = F2 [x] / (f) is a vector space over F2 of dimension 5 withbasis 1 + (f),x+ (f),x2 + (f),x3 + (f),x4 + (f), which for simplicity we write asl,x,x2,x3,x4.

Note that T : R —* R is the squaring map, since p = 2. To compute the matrix ofT,

we apply T to each basis element and represent the result in terms of the basis:

1 +x+x4,

x4 = 1 +x+x2+x3 +x4.

Here, x6 = 1 + x + x4 means that 1 + x + x4 is the remainder of x6 on division byf x5 +x4 + 1, and similarly for the last line.

It follows that the matrix ofT — 1R with respect to the basis l,x,x2,x3,x4 is

10011 10000 0001100011 01000 0101101001—00100=0110100001 00010 0001100111 00001 00110

(remember that we are in characteristic 2). One easily sees that this matrix has rankat most 3, since the first column is zero and the sum of the last three columns is zero.Since 3 <4 = deg(f) 1, Theorem 11.2.9 implies that f is reducible.

Historical Notes

In the early 1 800s Gauss showed that the number of monic irreducible polynomialsin [x] of degree n is given by the formula

(11.10)

308 FINITE FIELDS

where >a is the sum over all distinct primes dividing n, >ab is the sum over allproducts of distinct primes dividing n, and so on. In Exercise 8 you will showthat (11.10) is equivalent to the formula given in Theorem 11.2.4. Schönemanndiscovered (11.10) independently in 1846. He and Gauss also knew the factorization

—xgivenin (11.7).It is also possible to count the number of monic irreducible polynomials in ]Fq [x]

of degree n. In Exercise 9 you will prove the analogs of Theorems 11.2.2 and 11.2.4for arbitrary finite fields.

The Berlekamp algorithm is much more recent. The relation between irreducibilityand the rank ofT — 'R given in Theorem 11.2.9 is due to Butler [4] in 1954. He provedmore generally that if f E [x] is separable of degree n > 0, then the rank of T —

is n — k, where k is the number of irreducible factors of f in IF,, [xl (see Exercises 10and 11). This theorem can be generalized to any finite field. In 1967 Berlekamp[2] rediscovered Butler's result and used it as the basis for his factoring algorithm.Some of the beginning steps of his method will be discussed in Exercise 12, and thedetails can be found in [13, Sec. 4.1]. Berlekamp's algorithm works best for smallfinite fields; other methods are used for larger ones.

We refer the reader to Chapters 3 and 4 of [13] for much more on the mathematicsand history of polynomials over finite fields. See also [5].


Exercise 1. Let f E F[xJ be irreducible, where F is a finite field. Prove that f is separable.

Exercise 2. This exercise concerns Theorem 11.2.2 and the factorization (11.7).(a) Compute N3 and N4 using only Theorem 11.2.2.(b) Write down the factorization (11.7) explicitly when pfl = 4 and 8.

Exercise 3. Use Theorem 11.2.4 to compute N6 and N36.

Exercise 4. In Theorem 11.1.4 we used splitting fields to show that a field of order pfl existsfor any prime p and integer n 1. When Galois and others considered this question in thenineteenth century, their approach was to prove the existence of an irreducible polynomial inF,, [x] of degree n. In other words, they needed to prove that > 0.(a) Prove that > 0 using Theorem 11.1.4.(b) Suppose that we have proved Theorem 11.2.4 but not Theorem 11.1.4. Use this to prove

that > 0.

Exercise 5. Let F be a field of characteristic p, and let a E F be a root of unity. Prove thatthere is some d � 1 relatively prime to p such that a is a dth root of unity.

Exercise 6. This exercise is concerned with Example 11.2.8.(a) Show that N4 = 3 when p = 2. Then write down these three irreducible polynomials

explicitly.(b) Verify the factorization of 4 (x) given in the example.(c) Show that the roots of x4 + x3 + 1 and x4 + x + 1 are the reciprocals of each other.

Exercise 7. As in the discussion of Berlekamp's algorithm, let R = IF,, [x] / (f) and considerthe pth-power map T : R —* R. Prove that T is a linear map when R is regarded as a vectorspace over


Exercise 8. Prove that Gauss's formula (11.10) is equivalent to the formula given in Theo-rem 11.2.4.

Exercise 9. State and prove analogs of Theorems 11.2.2 and 11.2.4 that count monic irreduciblepolynomials of degree n in lFq [x], where q is now a power of the prime p.

Exercise 10. Suppose that a monic polynomial f E IF,, [x] has a factorization f = fi where

f',. . . aredistinctmonic irreduciblepolynomials. LetR = andletR, = IF,,{xl/(fi)for i = 1,. . . , k. Then consider the map

co:R—+Rix"xRk

defined by

+ (f)) = (g + (fi),.. . ,g+ (fk) ).

The goal of this exercise is to prove that cp is a ring isomorphism when we make R1 x x Rkinto a ring using coordinatewise addition and multiplication.(a) Prove that is a well-defined ring homomorphism.

(b) Prove that is one-to-one.

(c) Show that Rand R1 x x Rk have the same dimension when considered as vector spacesover

(d) Use the dimension theorem from linear algebra to conclude that is a ring isomorphism.

Exercise 11. In the situation of Theorem 11.2.9, let T R —* R be the pth-power map, whereR = IF,, [x] / (f) and f is separable of degree n. The goal of this exercise is to prove that therank ofT 1R is n — k, where k is the number of irreducible factors of fin F,, [xl. We will usethe isomorphism R R' = R1 x x R,, constructed in Exercise 10.(a) Let T' : R' —+ R' be the map that is the pth power on each coordinate. Prove that p

induces an isomorphism between the kernel ofT — 1,, and the kernel ofT' —(b) Prove that the kernel ofT' — 1,?' has dimension k as a vector space over IF,,.

(c) Prove that T — 1R has rank n — k, and use this to give another proof of Theorem 11.2.9.

Exercise 12. Let f E F,, [xJ be monic and separable of degree n> 1, and assume that T — lRhas rank n — 1. By Theorem 11.2.9, f is reducible. In this exercise, you will use the kernelof T — 'R to produce a nontrivial factorization off.(a) Show that the constant polynomials in F,, [xl give a one-dimensional subspace of the kernel

ofT— 1R.(b) Prove that there is a nonconstant polynomial h E F,, [xl of degree <n such that flh" — h.

Parts (c), (d), and (e) will use h to produce a nontrivial factorization off.(c) Explain why h = HaEF (h — a) in F,,[x].

(d) Use parts (b) and (c) to show that f = IIIOEF gcd(f,h — a).

(e) Use deg(h) <n to show that h — a) when a E F,,. Conclude that the factorizationof part (d) is nontrivial, i.e., gcd(f, h — a) is a nonconstant factor off of degree <n forat least two a E IF,,.

The basic idea of Berlekamp's algorithm is that one can factor f into irreducibles by takingthe gcd's of the nontrivial factors gcd(f, h — a) produced by part (e) as we vary h and a.

Exercise 13. Consider the polynomial f = x6 + x4 + x + 1 E F2 [x].(a) Use Exercise 11 and the method of Example 11.2.10 to show that f is the product of three

irreducible polynomials in F2 [x]. Also find a basis of the kernel ofT — lR.

310 FINITE FIELDS

(b) One element of the kernel is (0,0, 1, 1,0, 1). This corresponds to h = x2 + x3 + x5, sincewe're using the basis of R given by the cosets of 1,x, . . . ,x5. Show that gcd(f,h) andgcd(f,h+ 1) give a nontrivial factorization f = as in Exercise 12.

(c) Pick an element h' of the kernel not in the span of 1 and h. Compute h') andgcd(g,h'+l)fori= 1,2.

(d) Part (c) should show that f is a product of three nonconstant polynomials. Why is thisthe irreducible factorization off?

Exercise 14. In this exercise we will count the number of primitive elements of the extensionF,, C This is the number

(a) Use Corollary 11.1.8 to prove that P" = Pm.

(b) Use the Möbius inversion formula to conclude that = This formulawas first proved by Dedekind in 1857.

(c) Explain how the formula of part (b) relates to Theorem 11.2.4.

Exercise 15. This exercise will illustrate how the word "primitive" is sometimes overused

in mathematics. In the previous problem, we computed the number of primitive elements ofF,, C In this problem, we consider the primitive roots of which are generators ofthe cyclic group The minimal polynomial over F,, of a primitive root of is called aprimitive polynomial for These are the minimal polynomials of the primitive (p" — 1)st

roots of unity in characteristic p.(a) Prove that has — 1) primitive roots, where / is the Euler(b) Prove that every primitive polynomial for has degree n.(c) Prove that the product of the primitive polynomials for is —1(x).

Exercise 16. Consider the trinomial f = xr +x' + 1 E F2[x], where r > s > 0 and r is prime.Prove that f is irreducible over F2 if and only if fix2 — x. If in addition r is large and f isprimitive in the sense of Exercise 15, then one can use f to make a pseudo-random numbergenerator that takes a long time to repeat itself. For example, x43112609 + x21078848 + 1 is aprimitive trinomial of large degree. See [3] for more details.

Exercise 17. In Section 4.2, we used the Schönemann—Eisenstein criterion to prove thatCI)p(x) = + ... +x+ 1 is irreducible over Q, where p is prime. Here is a very differentproof. We know that primitive roots modulo p exist. By Dirichlet's theorem on primes in

arithmetic progressions, it follows that there is a prime £ such that E (Z/pZ) *has order

p — 1. Prove that c1,, (x) is irreducible modulo £ and conclude that it is irreducible over Q.This argument is due in Schönemann in 1845 (see [5]).

REFERENCES

1. E. Berlekamp, Algebraic Coding Theory, McGraw-Hill, New York, 1968.

2. E. Berlekamp, Factoring polynomials over finite fields, Bell System Tech. J. 46 (1967),1853—1859. See also Chapter 6 of [1].

3. R. P. Brent and P. Zimmerman, The great trinomial hunt, Notices Amer. Math. Soc. 58(2011), 233—239.

REFERENCES 311

4. M. C. R. Butler, On the reducibility of polynomials over a finite field, Quart. J. Math.Oxford Ser. (2) 5 (1954), 102—107.

5. D. Cox, Why Eisenstein proved the Eisenstein criterion and why Schönemann discoveredit first, Amer. Math. Monthly 118 (2011), 3—21.

6. L. E. Dickson, Linear Groups with an Exposition of the Galois Field Theory, B. G.Teubner, Leipzig, 1901. Reprint by Dover, New York, 1958.

7. L. E. Dickson, History of the Theory of Numbers, Carnegie Institute, Washington, DC,1919—1923. Reprint by Chelsea, New York, 1971.


9. D. M. Goldschmidt, Algebraic Functions and Projective Curves, Springer, New York,Berlin, Heidelberg, 2003.

10. J. J. Gray, A commentary on Gauss's mathematical diary, 1796—1814, with an Englishtranslation, Expo. Math. 2 (1984), 97—130. (The Latin original of Gauss's diary isreprinted in [Gauss, Vol. Xl].)


12. N. Koblitz, A Course in Number Theory and Cryptography, Springer, New York, Berlin,Heidelberg, 1987.

13. R. Lidl and H. Niederreiter, Finite Fields, Encyclopedia of Mathematics and Its Applica-tions 20, Addison-Wesley, Reading, MA, 1983.

14. C. J. Moreno, Algebraic Curves over Finite Fields, Cambridge U. P., Cambridge, 1991.

15. I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers, Third Edition,Wiley, New York, 1972.

PART IV

FURTHER TOPICS

The final four chapters of the book sample the further riches of Galois theory.Chapter 12 explores the history of Galois theory. We begin with Lagrange, who

studied an important special case. We then explain how Galois thought about histheory and discuss Kronecker's approach to this subject.

Given an arbitrary polynomial, how do we find its Galois group? In Chapter 13we show that in principle this can always be done. We also explore various moreefficient methods for dealing with polynomials of small degree.

Chapter 14 continues our study of solvability by radicals. We give Galois'swonderful criterion for when an irreducible polynomial of prime degree is solvableby radicals. Then, following Galois, we consider irreducible polynomials of prime-squared degree. This requires a careful study of the theory of permutation groups.

Finally, Chapter 15 discusses Abel's theorem on straightedge-and-compass con-structions on the lemniscate. This involves some truly wonderful mathematics. Inparticular, we use certain elliptic functions constructed from the lemniscate to createextensions of Q(i) with Abelian Galois groups.

CHAPTER 12

LAGRANGE, GALOIS, ANDKRON ECKER

This chapter will explore the contributions to Galois theory made by Lagrange,Galois, and Kronecker. Our account of these great mathematicians will touch on thehigh points of their work on the roots of polynomials.

As you read this chapter, you will be asked to look back at numerous argumentsin previous chapters. The goal is to gain a better understanding of where thesearguments came from and to see how the basic concepts of Galois theory evolved.

12.1 LAGRANGE

As we noted in the Historical Notes to Section 1.2, Lagrange's 1770 treatise Reflexionssur Ia resolution algébrique des equations studied the known methods for solvingequations of degree 4 and analyzed these methods using permutations. Lagrange'shope was that these methods could be adapted to equations of degree � 5. Thissection will discuss some of Lagrange's ideas and explain why his approach wasdoomed to failure for degree � 5.

Lagrange's goal in 1770 was to understand the roots of an arbitrary polynomial.However, when dealing with expressions in the roots, Lagrange was concerned"only with the form" of such expressions and not "with their numerical quantity"


316 LAGRANGE, GALOIS, AND KRONECKER

[Lagrange, p. 385]. In modem terms, this means the following. Given a field Fand roots . , ce,, of a polynomial f E F[xj of degree n, then for Lagrange, an

"expression" in the roots is a quotient of the form

. .

where A,B e F[xi,... are polynomials in variables x1,... with coefficients inF. Hence the "form" of this expression is the rational function

A(xi,. .. ,x,,)E F(xi, . . .

By using only the "form," Lagrange is dealing with the case when the roots are

variables Xi,... These are the roots of the universal polynomial of degree n,

1=

where a1, . . . , are the elementary symmetric polynomials from Section 2.1.The coefficients off lie in K = F(ai, . . . , an), and the splitting field off over K

is L = F(xi,... We will use the universal extension in degree n,

K CL,

throughout this section. We will also assume that F has characteristic 0 so that wecan use the results on solvability by radicals proved in Chapter 8. Theorem 6.4.1

shows that K C L is a Galois extension with Galois group

(12.1)

where a E gives the automorphism that sends a rational function E L to therational function a obtained from by replacing x, with X,(j) for each i. We arethus in a rich mathematical context: We know explicitly how the Galois group acts,we have the Galois correspondence, and we understand solvability by radicals.

Lagrange, on the other hand, was working 60 years before Galois and 150 yearsbefore Artin formalized the Galois correspondence. Lagrange's main tools were thefollowing results from the theory of symmetric functions:

• A polynomial in F[xi,. . . that is unchanged by all permutations in lies in

• A rational function in L = F(xi,... that is unchanged by all permutations inlies in K = F(o-i,... (i.e., K is the fixed field of (12.1) acting on L).

We proved the first bullet in Theorem 2.2.2 and the second in Exercises 7 and 8 fromSection 2.2. Using these facts, Lagrange discovered several important parts of Galoistheory, even though the concept of "group" didn't exist in 1770. Lagrange didn'teven have Cauchy's efficient notation for expressing permutations.

We will now explore some of what Lagrange did in his Reflexions.

LAGRANGE 317

A. Resolvent Polynomials. Fix a rational function L = F(xi,... andconsider the rational functions a E S,,. Let

(12.2)

be the distinct rational functions we get in this way. The polynomial

(12.3) 0(x) =

with roots is a special case of the polynomial (7.1) used in the proof ofTheorem 7.1.1. Reread the proof of Theorem 7.1.1, especially the part where weshow that (7.1) is the minimal polynomial of an element in a Galois extension. Besure you understand how (7.1), applied to e L, gives the polynomial 0 definedabove. It follows that 0 has coefficients in K = F(o-i, . . . is separable, and is theminimal polynomial of over K. Hence we have proved the following.

Proposition 12.1.1 The polynomial 0 from (12.3) lies in K[xI and is separable andirreducible. U

We call 0 the resolvent polynomial of Hence the "Galois" construction ofminimal polynomials given in (7.1) is actually due to Lagrange. In Exercise 1 youwill follow Lagrange by proving that 0 E K[xJ using the second of the above bullets.

Here is an example of a resolvent polynomial from Chapter 1.

Example 12.1.2 Let n = 3 and Zi = +w2x2 +wx3), w = You will checkin Exercise 2 that S3 acting on gives the six elements Zi, Z2 = (23) WZ1, WZ2,

w2Z2 listed in (1.10) and that the resulting resolvent is

O(z) (z — zi)(z —Z2)(Z— —wz2)(z— w2z1)(Z— w2z2)

=z6+qz3—p3/27,

where

ala2

These formulas become identical to those derived in (1.2) and (1.4) of Section 1.1 ifwe usex3 +bx2+cx+d in place of x3 — aix2+a2x— a3.

One of Lagrange's main goals was to replace the clever substitutions used to deriveCardan's formulas in Section 1.1 with the following systematic process:

• Pick zi L = F(xi ,x2,x3) whose resolvent 0(z) is easy to solve.• Express the roots x1 ,x2,x3 in terms of the roots of the resolvent.

Example 12.1.2 does the first step of this process, since O(z) = z6 + qz3 — p3 /27 = 0

can be solved by the quadratic formula. The second step is equally easy, since

Zi +w2x2+wx3),

Z2 = +wx2+w2x3),


together with 1 +w+w2 = 0 and a1 = Xi +X2 +x3, imply that

= +Zi +Z2,

= +WZ2,

X3 = +W2Z1 +WZ2.

Comparing this with (8.15) from Section 8.3 reveals that our "Galois" approach toCardan's formulas is virtually identical to the "Lagrange" approach just described.The major difference is that in (8.15), we used the Lagrange resolvent

X1 +W X2+W2X3 =Xi +WX2+WX3

rather than Zi = +W2X2 +WX3) as above. The name "Lagrange resolvent" is noaccident, as we will see later in the section.

Here is another example of a resolvent polynomial.

Example 12.1.3 Consider YL = x1x2 +X3X4 E L = F(xi ,x2,x3 ,X4). In Exercise 3 youwill show that the action of S4 on y' gives the three polynomials

Yi = XIX2 +X3X4, Y2 = +X2X4, = +X2X3,

and that the corresponding resolvent, as a polynomial in y, is

0(y) = (y—(xix2+x3x4))(y—(xIx3+x2x4))(y—(xlx4+x2x3))

= y3 — a2Y2 + (aia3 — 404)y — — +4a2a4.

This will be useful later in the section when we solve the quartic equation.

In Example 12.1.3, note that when the 24 permutations a E S4 are substitutedinto a the result is always one of the three polynomials Lagrangemade the crucial observation that this happens because many permutations leaveyi = x1x2 +x3x4 unchanged. From the modem point of view this is best stated usingthe language of group actions. Group actions are discussed in Section A.4 and havebeen used in several places in the text, most prominently in the proof of Theorem 6.4.1in Section 6.4.

In terms of group actions, we can describe what Lagrange did as follows. In theproof of Theorem 6.4.1, we showed that acts on L = F (Xi, . . . , xn). For our chosenrational function E L, we have the orbit

where the last equality uses the notation of (12.2). We also have the isotmpy subgroup

Since every a E satisfies a = we can write this symbolically as

LAGRANGE 319

Now consider çojin the orbit of This implies that = cr E Sn.

One easily sees that a• = çoj for every a E (be sure you can show this). Asabove, we write this symbolically as

(12.4)

In this way, we partition into r cosets of Since each coset haselements, we conclude that

=

Although he didn't use this terminology, the above partition is implicit in whatLagrange did. Hence Lagrange in essence proved the following:

• divides the order of Sn. Thus the index

=

is an integer. This is a special case of Lagrange's Theorem.• r = SnI/IH((p)I [Sn: Thus thenumber of elements in the orbit

of is the FundamentalTheorem of Group Actions from Appendix A.4.

These results are more than just special cases: They represent the first time these issueswere considered in mathematics. The name "Lagrange's Theorem" was chosen inhonor of Lagrange's analysis of this situation.

On the other hand, the details of Lagrange's arguments are quite different fromours. To see how he approached these matters, we need to think in terms of resolventpolynomials. For this purpose, consider the polynomial of degree n! defined by

e(x)= [J(x_cr.cp).

To compare this with the resolvent

6(x) =

we organize the product formula for tB according to cosets of in Sn. The keyobservation is that (12.4) and imply that

fl (x— a• = (x— = (x—

Since this holds for i = 1, . . . , r, we obtain the following theorem of Lagrange.

Theorem 12.1.4 Given E L F(x1,... ,Xn), the polynomials 0 and 8 are relatedby the equation

0(x) =


In the degree of the resolvent polynomial 8 is the index

= .Here is an example to show how this result can be used.

Example 12.1.5 Example 12.1.3 shows that the resolvent of yi = x1x2 +x3x4 hasdegree 3. Thus, by Theorem 12.1.4, the isotropy group H(y1) has = 8 elements.Also note that H(yi) contains (12) and (1324). In Exercise 3 you will show that

H(yi) = ((12),(1324)) CS4

and that H(yi) is isomorphic to the dihedral group D8 of order 8.

It is fun to read Lagrange's statement of Theorem 12.1.4:

One can show in the same manner that, if the function

is by its own nature such that it conserves the same value when two, or three, ora greater number of different permutations are made among the roots x', x", x",x1" the roots of the equation e =0 will be equal three by three, or four byfour, or etc.; so that the quantity e will be equal to a cube or a square-square94, or etc., and consequently the equation e = 0 will reduce to that of 0 = 0,

whose degree will be equal to or to or etc.

(See [Lagrange, pp. 370—371].) Here, is Lagrange's notation for n!, and his f isour In this statement, Lagrange says that if f is fixed by 2, 3, etc. permutations,then the resolvent has degree etc. At first glance, this seems wrong, forthe denominator is one more than the number of permutations. The reason for thediscrepancy is that Lagrange didn't count the identity permutation.

In most courses on group theory, students usually study cosets and Lagrange'sTheorem in one part of the course and group actions in another. Pedagogically,this makes sense, but it is also important to remember that historically, things areoften more complicated. In considering resolvent polynomials, Lagrange had to dealwith many issues all at once. It is a testament to his power as a mathematician thatLagrange could see what was important and thereby enable his successors to sort outthe details of what he did.

B. Similar Functions. In [Lagrange, pp. 358—359] Lagrange says that "one callsfunctions similar those that vary at the same time or remain the same when one makesthe same permutations among the quantities of which they are composed." In modernterms this means that E L = F(xi ,... are similar if for all o E we have

Thus rational functions and are similar if and only if they have the same isotropysubgroup, i.e., =

LAGRANGE 321

Lagrange makes a careful study of similar functions, though his most amazingresult concerns the more general situation where we have L with the propertythat is fixed by every permutation that fixes In terms of isotropy groups, thiscondition can be written

C

(be sure you understand this). Assuming we know can we determine whichsatisfy the above condition? Here is Lagrange's remarkable answer.

Theorem 12.1.6 Suppose that rational functions L = F(xi, . . . ,x,,) have theproperty that is fixed by every permutation fixing Then is a rationalfunctionin with coefficients in K = F(cri,... ,

Pmof: Our first proof of the theorem will use the Galois correspondence for K C L.Using e L, we get the intermediate field K C K(co) C L. Then:

• Under the group isomorphism (12.1), Gal(L/K(ço)) C Gal(L/K) corresponds toH(ço) c Be sure you understand how this follows from Proposition 6.1.4.

• By hypothesis, is fixed by H(ço), so that is in the fixed field• By the Galois correspondence, =Thus and hence is a rational function in with coefficients in K.

Our second proof is taken from [31 and is much more in the spirit of Lagrange.As above, let = , ço,. be the different rational functions obtained by letting

act on Fix i between 1 and r. The proof of Theorem 12.1.4 shows that cpjcorresponds to some left coset of say Since doesn't affectit follows that every element of takes to = . In this way, we getelements V-'i = , which need not be distinct (do you see why?).

Using the çoj and consider the function

(12.5))

0(x) 0(x)

X—(pl

where 0(x) is the resolvent of In spite of the denominators, W(x) is actuallya polynomial in x, since 9(x) is by definition divisible by x ,x — Anelement a permutes the so that — + + cor) isunaffected by Since 0(x) e K[x], the coefficients of W (x) must be symmetric.Hence 1J1(x) E K[x].

Next observe that if we evaluate the polynomial

Ho

at then we get — çoj) when i = 1 and 0 when 2 i r. Looking at theformula (12.5) for W(x), we conclude that

(12.6)


However, 0(x) = (x — (x — and (5.7) imply that

(12.7) O'(cpl) = — — Pr) = — (pj).

j=2

Since = and = equations (12.6) and (12.7) give the equation

(12.8)

This expression lies in K(ço) since '11(x) and 0'(x) have coefficients in K. •

One advantage of the second proof of Theorem 12.1.6 is that it gives an explicitformula (12.8) for expressing in terms of though in practice computing thisformula can be unpleasant. While this proof differs from Lagrange's, it uses çoj and

in the same way, and Lagrange knew the formula (12.5), which is closely relatedto the Lagrange interpolation formula stated in Exercise 1 of Section 4.2.

Here are some simple applications of Theorem 12.1.6.

Example 12.1.7 Let L = F(xi ,x2,x3,x4), where F has characteristic 2. Theisotropysubgroupofti =xl+x2—x3—x4is((12),(34)) CS4. Sinceyi =xjx2+x3x4is fixed by these permutations, we conclude that Yl E K(t1).

Example 12.1.8 Since fields of characteristic 0 are infinite, we can pick distinctelements t1,... , E F. Now consider

= tixi+ ... + EL = F(xi,. ..

Ifa E then a.fi Since are distinct, it followsthat a = /3 if and only if a is the identity. Thus {e}. This means thatany E L is fixed by H(/3) = {e}, so that K(S) by Theorem 12.1.6. Sincewas an arbitrary element of L, we see that L K(/3), i.e., is a primitive element ofL. Furthermore, = {e} and Theorem 12.1.4 imply that the resolvent of /3 hasdegreen!. Thus

0(y) fJ(y_a./3)= fJ

is the resolvent of /3. This will be useful in the next section. <N

Lagrange was aware of Example 12.1.8, so that the idea of a primitive elementdates back to the very beginnings of Galois theory.

From the point of view of Galois theory, Theorem 12.1.6 is exciting in that itreveals a strong connection with the Galois correspondence. It gets even better whenwe bring in Lagrange's similar functions. Here is the precise result.

LAGRANGE 323

Theorem 12.1.9 Let E L = F(xi,... Then and are similar functionsand only =

Proof: We have = = by the Galoiscorrespondence. Using (12.1), this gives K(ço) = H(ço) = (be sureyou understand why). Since and are similar if and only if H(ço) thetheorem follows.

This theorem shows that the intrinsic object corresponding to similar functionsis the field they generate when adjoined to K = F(ai,... For us, it is naturalto think in terms of fields. Lagrange, on the other hand, was writing before settheory was developed, so that he and his contemporaries tended to think of individualelements rather than the sets in which they lie. At the same time, Lagrange knewthat individual functions e L weren't intrinsic, which is why he introduced similarfunctions. Taken together, similar functions and Theorem 12.1.6 show that Lagrangehad an implicit understanding of the Galois correspondence for the extension K C L.

C. The Quartic. After analyzing the solutions of cubic and quartic equations,Lagrange states his strategy for solving equations as follows [Lagrange, p. 355]:

As should be clear from this analysis that we have just given of the main knownmethods for the solution of equations, all these methods reduce to the samegeneral principle, namely to find functions of the roots of the proposed equationsuch that: 10 the equation or equations by which they are given, i.e., of which theyare the roots (equations that are usually called reduced equations), happen to beof a degree smaller than that of the proposed equation, or at least decomposableinto other equations of a smaller degree than this one; 2° the values of the desiredroots can be easily deduced from them.

Here, "functions of the roots of the proposed equation" are elements E L, and"reduced equations" are the corresponding resolvent polynomials. So Lagrange'sidea is to look for resolvent polynomials that either have smaller degree or factor intopolynomials of smaller degree.

To see what this means in practice, let us apply Lagrange's methods to Ferrari'ssolution of the universal quartic equation

x4 — + a2x2 — a3x+ = 0.

We first describe what Ferrari did (with some of the algebra left to the exercises).Write the above equation as

X4—cJIX3 = —a2x2+a3x—a4.

Since F has characteristic 0, we can add the quantity

2 2 2 ai y2yx

to each side, where y is yet to be chosen. In Exercise 4 you will show that this leadsto the equation

2 °1 Y2 2 Y2(12.9) (x


We next choose y so that the right-hand side of (12.9) is also a perfect square. Theright-hand side is quadratic in x. In general, if A $ 0, then

Applying this to the right-hand side of (12.9), you will show in Exercise 4 thatB2 = 4AC leads to the cubic equation

(12.10) y3 — a2y2+(aIa3 — +4o2a4 = 0.

This is the resolvent from Example 12.1.3 and is called the Ferrari resolvent.If Yi is a root of this resolvent, then the above formula for Ax2 + Bx + C shows

that the right-hand side of (12.9) becomes

/ \/ _fLy+o.3 \2(y+——o2)(x+ 2' I"

It follows that (12.9) can be written as

(x2_ = (Yi _a2)(X+)2

2 2 4

2 2

x gives the four roots x1 , X2 , X3 X4 of ourquartic equation. This is Ferrari's solution of the quartic.

One of Lagrange's main observations is that the auxiliary polynomials and radicalsused in solving cubics or quartics come from expressions built from the roots andhence can be explained in terms of resolvent polynomials. For example, the Ferrariresolvent (12.10) is the resolvent polynomial ofy1 = xlx2 +x3x4 from Example 12.1.3.Exercise 5 will show how yi = X1X2 +X3X4 follows from (12.11).

We can also explain the square root in (12.11) in the same way. Using Yi =X1 X2 + x3x4 and setting

t1 X1+X2X3X4,

one checks thata2

=

This allows us to define

(12.12)

LAGRANGE 325

The isotropy subgroup ofti is easily seen to be ((12), (34)), which means thatits resolvent polynomial 0(t) has degree 6. By (12.12), t1 is a root of the quadraticpolynomial

which has coefficients in K(yi). To get a polynomial with coefficients in K, we usethe other roots Y2 , y3 of the quartic resolvent (12.10). This gives

(12.13) (t2—4y1 (t2 —4y2—0?+472) (t2 —a1+4t72).

In Exercise 6 you will show that this polynomial lies in K[t] and hence is the resolvent0(t) since it has degree 6.

The passage from Lagrange quoted at the beginning of our discussion of thequartic states that we need resolvent polynomials "of a degree smaller than that of theproposed equation, or at least decomposable into other equations of a smaller degreethan this one." In terms of what we just did for the quartic, this means the following:

• For Yl = xix2 +x3x4, the resolvent has degree 3, which is smaller than 4. So wecan find Yi ,Y2,Y3 by Cardan's formulas.

• For t1 = x1 + x2 — — X4, the resolvent has degree 6, but since we already knowwe can decompose the resolvent into quadratics as in (12.13). Then we

get ti in terms of yi by extracting a square root.

Hence Ferrari's solution can be seen as a special case of Lagrange's strategy.We can also describe the above derivation in terms of fields and Galois groups.

Example 12.1.7 shows that we have fields

(12.14) KCK(yi)CK(ti)CL.

Since K c K(yi) has degree 3, we can use Cardan's formulas to express the splittingfield, and (12.12) shows that K(y1) C K(ti) is obtained by adjoining a square root. Ifwe take the Galois groups of (12.14), then Gal(L/K) S4 gives the subgroups

H(t1) = ((12),(34)) C H(y1) = ((l324),(12)) CS4.

This differs from what we did in Chapter 8. There, we wanted a chain of subgroupswhere each was normal in the next larger. In contrast, H(yi) C S4 is not normal (seeExercise 3). Hence Lagrange did not follow a strictly "Galois-theoretic" approachto solving the quartic. However, the way he built up the solution using extensions ofsmaller degree shows that he had the beginnings of a theory of solvability.

There is still more to say about the quartic, since the fields in (12.14) only givet1 =x1 +x2 —X3 —x4. We needto explainhow to gofromhereto therootsx1 ,x2,x3,x4.Rather than pursuing Ferrari's solution (12.11), we will switch to Euler's solution,which follows naturally from what we have done so far.

The key idea is to work simultaneously with the extensions (12.14) and theirconjugates. For K(y1), this means using K(y2) and K(y3). As for K(t1), the sixconjugates ofti are ±t1 , ±t2, ±t3, where t2 = (23) . ti = Xi — x2 + X3 — X4 and t3 =(24).t2=xI —x2—x3+x4. By (12.12), weknow that

=


and similarly one can show that

= and t3 =

where y2 = (23) and = (24) Yi. We can thus express t1,t2,t3 in terms ofradicals once we find Yi ,Y2,Y3 using Cardan's formulas.

In Exercise 7 you will show that the equations

U1 =XI+X2+X3+X4,

(12.15)tI =X1+X2—X3—X4,

t2 X1 X2+X3—X4,

t3 X1 —X2—X3+X4

imply that

x1 =

X2 +t1 —t2—t3),

X3 = —t1 +t2—t3),

= —t1 —t2+t3).

Thus the x1 can be expressed as sums involving three square roots. However, wecan't make independent choices of signs, since this would lead to eight values for theroots. The point is that t1 , t2, t3 satisfy the identity

(12.16) t1t2t3 = — + 8cr3

(see Exercise 8), so that knowing two of the square roots determines the third. Hence,if Yi , Y2, y3 are the roots of the quartic resolvent (12.10), then the four roots of thequartic are

(12.17)

where the ± signs are chosen so that the product of the radicals is the right-hand sideof (12.16). This is Euler's solution of the quartic.

Lagrange discusses other solutions of the quartic and interprets them in terms ofresolvents. In general, his approach to solving equations anticipates many featuresof Galois theory, though there are important differences, especially in the appearanceof nonnormal subgroups and the use of conjugate fields.

D. Higher Degrees. Although Lagrange's methods work wonderfully for equa-tions of degree 4, they fail for degrees 5 and greater. One way to see this is by thetheory of Chapter 8, which tells us that K C L is not solvable by radicals for n � 5,since Gal(L/K) is not a solvable group for n � 5.

It is also possible to describe this failure in terms of Lagrange's strategy. Sincethe degree of a resolvent polynomial is the index of the isotropy subgroup, findingresolvents of small degree is equivalent to finding subgroups of Sn of small index.However, as soon as n > 5, such subgroups are hard to find, as we will now prove inthe following theorem.

LAGRANGE 327

Theorem 12.1.10 Let n > 5.(a) ff11 C is a subgroup of index [Sn : H]> 1, then either H = An or [Sn : H] � n.

(b) ff11 C An is a subgroup of index [An : H] > 1, then [An :H] � n.

Proof: To prove part (a), we first note that there is L whose isotropy group isprecisely the subgroup H, i.e., H = H(p). You will prove this in Exercise 9. Thenwrite the distinct rational functions of the form a• for a E 5n as

By Theorem 12.1.4 we know that r = [Sn : H]. Now consider the set

In Exercise 10 you will show that N is a subgroup of Note also that every a E Nfixes = which implies that N C H.

The key point of the proof is that N is a normal subgroup of Sn. To prove this,we must show that T 1aT E N for all a E N and 'r E Fix i between 1 and r. If

çoj = for some j, then . çof = çoj. Using a E N, this implies that

This is true for all i, so that y1ar E N. Thus N is normal in Sn.Since N C H Sn, Proposition 8.4.6 implies that either N = {e} or N = H = An.

To complete the proof, we must show that N {e} implies [Sn H] � n.We will show that N = {e} and r = [Sn H] <n lead to a contradiction. First

observe that every r e 5n permutes the pi. The number of is r, so that they canbe permuted in r! ways. Yet the number of r's is n!. Since r < n implies r! <n!,there must be r1 T2 in 5n that give the same permutation of the ço,. Thus

forall i=l,...,r,

which easily implies that

forall i=1,...,r.

Thus EN, so that N {e}, since r2. This contradicts N = {e}.You will prove part (b) in Exercise 11. •

To see how this messes up Lagrange's strategy, suppose that n � 5 and thatthe resolvent 8(x) of E L has degree > 1. Since deg(8) = [Sn :H(ço)], part (a)of Theorem 12.1.10 tells us that either deg(O) � n or = An, in which case

E by Theorem 7.4.4. Hence the only reasonable way to begin Lagrange'sstrategy is to pick = But then continuing his strategy would entail findinga proper subgroup of An of index <n (you will verify this in Exercise 12). Suchsubgroups don't exist (this is part (b) of Theorem 12.1.10), so that Lagrange's strategyfails forn >5.


E. Lagrange Resolvents. To see where Lagrange resolvents come from, recallthat the solution of the cubic used

= +WX3 +W2X2), w

and the above solution of the quartic used

tl=xl+x2—x3—x4z=xl+(—1)x3+(—l)2x2+(—l)3x4.

Aside from the factor of both expressions involve the roots multiplied by roots ofunity to increasing powers. Here's how Lagrange says this [Lagrange, p. 356]:

As to equations that do not exceed the fourth degree, the simplest functions thatyield their solution can be represented by the general formula

+y2x,,, + . . .

x',x",x", . . . being the roots of the proposed equation, which is assumedto have degree andy being a root different from 1 of the equation

— 1 = 0

In Lemma 8.3.2 of Chapter 8, we used the name "Lagrange resolvent" for suchexpressions. There, we wanted to show that Galois extensions of prime degree p areobtained by adjoining a pth root when the smaller field contains a primitive pth rootof unity. Our main tool was the Lagrange resolvent (8.7):

aj

You should reread the proof of Lemma 8.3.2, especially (8.7) and (8.8).We can apply the formula for a1 to the extension K C L as follows. Let a =

(12.. .n) S,, Gal(L/K), let = Xi, and let be an nth root of unity. Thenreplacing p with n in the above formula gives the Lagrange resolvent

(12 18)=x1 +(axi •x1

= Xi + C'x2 + +

This agrees with Lagrange's "general formula." One can prove the identity

a = (12.. .n) fixes

= = (x1 + C'x2 + + . .. +

The proofs are identical to the arguments used in the proof of Lemma 8.3.2. Onecan also show that if is a primitive nth root of unity, then (12. . . n) generates theisotropy subgroup of £1, = a?, i.e., H(01) = ((12.. .n)) (see Exercise 13). It followsthat the resolvent polynomial of has degree n! /n = (n — 1)!. Lagrange states thisas follows [Lagrange, pp. 332—333]:

LAGRANGE 329

one gets an equation in 9 of degree 1.2.3... — 1), whose roots are thevalues of 9 that come from the permutations of the — 1 rootsx" ,x",... ignoringthe root x'.

(As above, Lagrange uses instead of n for the degree.) In Exercise 14 you willwork out how the final part of this statement relates to the proof of Theorem 12.1.4.

We conclude with an unexpected property of Lagrange resolvents. Let n = p beprime and = Then (12.18) with = gives the Lagrange resolvent

(12.19) +

where i = 0, 1, . . . , p — 1. Ignoring = a1, we set

Lagrange forms the polynomial

(12.20) —

so that T, U, X,... are the elementary symmetric polynomials of 0k,... La-grange then asserts that "the coefficients T, U,X,... are each given by an equationof degree 1.2.3. .

.— 2)," i.e., degree (p —2)! [Lagrange, p. 333].

By claiming that these resolvents have degree (p — 2)!, Lagrange is in effectsaying that their isotropy subgroups have order p! / (p — 2)! = p(p — 1). In fact, onecan prove that the isotropy group of T is the subgroup C consisting of thepermutations

1 _<a_<p—1,

where everything is interpreted modulo p. In the Historical Notes to Section 6.4, weshowed that is isomorphic to the affine linear group AGL( 1, F,,). Furthermore,we will prove in Proposition 14.1.4 of Section 14.1 that is a maximal solvablesubgroup of S,,. So Lagrange essentially found a maximal solvable subgroup of

Lagrange concludes that his strategy fails in degree 5 and that

if the algebraic resolution of equations of degrees greater than four is not im-possible, then it must depend on some functions of the roots, different from thepreceding.

(See [Lagrange, p. 357].) In spite of this failure, it is impressive to see how farLagrange got. His 1770 treatise is one of the great works of algebra.

Historical Notes

In the sixteenth century del Ferro, Tartaglia, and Cardan solved the cubic, andFerrari (a student of Cardan) solved the quartic. This was followed by mathematicianssuch as Viète, Hudde, Descartes, Tschimhaus, Euler, and Bézout, who simplified andimproved these solutions and found some entirely new solutions. Many of thesemethods are analyzed by Lagrange in his Reflexions.


The late eighteenth century was a time of active research on the roots of poly-nomials. Besides the work of Lagrange just discussed, we have Euler's solutionof the quartic, which appeared in his 1770 algebra text [10, PP. 282—2881 (a nicediscussion of Euler's method can be found in [4, pp. 104—107]). Euler also foundmany examples of quintics that are solvable by radicals and gave an incomplete proofof the Fundamental Theorem of Algebra. In 1772 Lagrange used the methods ofhis Reflexions to fill most of the gaps in Euler's proof—though, as mentioned in theHistorical Notes to Section 3.2, his argument is still incomplete.

Another important development in the early l770s was Vandermonde's Mémoiresur Ia resolution des equations. This paper covers much of the same material asLagrange's Reflexions, though Vandermonde's approach is different from Lagrange's.In particular, he considered permutations in more detail than Lagrange and understoodhow resolvents relate the action of permutations. He also used these methods totreat cubic and quartic equations and independently discovered Lagrange resolvents,though he didn't pursue the general theory to the same depth as Lagrange. One wayin which he went significantly beyond Lagrange was his treatment of the equation

— 1 = 0. For example, Lagrange notes that x11 — 1 reduces to solving a quintic,but Vandermonde actually solved the resulting quintic by radicals. This may havebeen part of what inspired Gauss to investigate x1' — 1 = 0, p prime. His results arediscussed in Section 9.2.

Lagrange hoped to solve equations by finding functions of the roots that gave aresolvent of small degree. In this section, we learned that

the degree of the the number of distinct — the index of theresolvent of L values a E isotropy subgroup of

In 1845, Cauchy studied "the problem of the number of values that can be assumed byfunctions," which by the last equality means the study of the index of subgroups ofThis was one of the important problems in the early history of group theory. The keyresult is Theorem 12.1.10, which we used to show the failure of Lagrange's strategy.In modern terminology, here are some highlights of how we got from Lagrange toTheorem 12.1.10:

• In 1799 Ruffini published a proof that the quintic is not solvable by radicals. Hisproof was hard to follow, but he did show that S5 had no subgroups of index 3 or 4,which is part (a) of Theorem 12.1.10 for n = 5. He also proved the irreducibilityof resolvent polynomials.

• In 1815 Cauchy generalized Ruffini's result by showing that the index of a sub-group H is either 2 or at least the largest prime <n. Cauchy used the word"index" to denote the number of values of a function, which is where the modernterm "index" comes from. The same paper also proved that is generated by3-cycles. Cauchy also emphasized the importance of the identity permutation andintroduced the two-row notation for permutations.

• In 1824 Abel gave the first generally accepted proof that the general quintic isnot solvable by radicals. He used Cauchy's (and Ruffini's) results on the index ofsubgroups of S5.

LAGRANGE 331

• In 1832 Galois defined a normal subgroup of a group and asserted that a noncyclicsimple group has order at least 60. Note that A5 I = 60.

• In 1845 first Bertrand and then Cauchy proved part (a) of Theorem 12.1.10, thoughtheir proofs axe quite different from the one given here. Cauchy also introducedthe cycle notation now taught in introductory courses in abstract algebra.

• In 1869 Jordan defined the concept of simple group, and in 1870 he showed thatAn is simple for n � 5.

• In 1879 Kronecker proved Theorem 12.1.10 using the simplicity of An for n � 5.This is the proof used in the text.

Further results on the index of subgroups in can be found in [15, pp. 138—139] or[16, p. 274, Note 1201.

The Historical Notes to Section 8.5 mentioned the work of Ruffini and Abel onthe unsolvability of the quintic. Now that we know Lagrange's Reflexions, we canget a better idea of what they did. Very roughly, both Ruffini and Abel tried to provethe unsolvability of the quintic by showing the nonexistence of the required resolventpolynomials, which in terms of group theory reduces to Theorem 12.1.10 for n = 5.

But this alone is not enough, for Lagrange's theory only deals with rational functionsof the roots. But suppose that there were formulas for the roots x1, . . . that involvedexpressions like + ... + Lagrange's methods would no longer apply. Sothe first thing Abel had to do was prove that if the quintic were solvable by radicals,then one could write the solution entirely within L = F(xi, . . . assuming that Fhad suitable roots of unity. This is discussed in [1]. See also [9] and [10] in thereferences to Chapter 8.

Then comes Galois, whose work in the early 1 830s is the main topic of the nextsection. The important thing to say here is that Galois's analysis of solvability byradicals led to the concept of solvable group and gave a dramatically simpler approachto all of these questions. Namely, once one proves that is not solvable for n > 5,then one immediately concludes that the general polynomial of degree n � 5 isn'tsolvable by radicals and that Lagrange's strategy for the quintic must fail. Resultslike Theorem 12.1.10 are simply not needed. This shows the power of good ideas.And the fact that Cauchy was still pursuing Lagrange's approach in 1845 shows howlong it took to understand these ideas.

Further comments on eighteenth-century algebra can be found in Chapter 6 of [2].The book [Tignol] has chapters on the work of Lagrange, Vandermonde, Ruffini, andAbel. A good description of Lagrange's Reflexions can be found in [11]. This paperalso discusses the subsequent history of Galois theory.


Exercise 1. Let 0(x) be the resolvent polynomial defined in (12.3). Use the second bulletfollowing (12.1) to show that 0(x) E K[x].

Exercise 2. Work out the details of Example 12.1.2.

Exercise 3. This exercise concerns Examples 12.1.3 and 12.1.5.


(a) Compute the resolvent 8(y) of Example 12.1.3. This can be done using the methods ofSection 2.3.

(b) Letyi =xlx2+x3x4. ShowthatH(yi)=((12),(1324))cS4.

(c) Show that H(yi) is not normal in S4.(d) Show that H(yi) is isomorphic to D8, the dihedral group of order 8.

Exercise 4. Verify (12.9) and (12.10).

Exercise 5. This exercise will study the quadratic equations (12.11). Each quadratic has tworoots, which together make up the four roots xi ,x2,x3,x4 of our quartic.(a) For the moment, forget all of the theory developed so far, and let y be some root of the

Ferrari resolvent (12.10). Given only this, can we determine how y relates to the x? Thisis surprisingly easy to do. Suppose are the roots of (12.11) for one choice of sign,and xk , are the roots for the other. Thus i, j, k, I are the numbers 1,2,3,4 in some order.Prove that y is given by y = +XkXl.

(b) Now let yl = XIX2 +x3x4, and define the square root in (12.11) using (12.12). Show thatthe roots of (12.11) are xI for the plus sign and X3 ,X4 for the minus sign.

Historically, the Ferrari resolvent was just a tool for solving the quartic. Lagrange was the firstto observe that the roots of (12.10) can be expressed in terms of the roots of the quartic. Hisargument [Lagrange, p. 2621 is similar to what we did here.

Exercise 6. Explain why the polynomial (12.13) has coefficients in K = F(a,,a2,a3,o4).

Exercise 7. Show that (12.15) implies the equations for xI ,X2,X3,X4 given in the text.

Exercise 8. Let ti , t3 be defined as in (12.15).(a) Lagrange noted that any transposition fixes exactly one of t1, t2, t3 and interchanges the

other two, possibly changing the sign of both. Prove this and use it to show that t1t2t3 is

fixed by all elements of S4.(b) Use the methods of Chapter 2 to express t1 t2t3 in terms of the The result should be the

identity (12.16).

Exercise 9. Let H be a subgroup of S,,. In this exercise you will give two proofs that there isEL such that H =

(a) (First Proof.) The fixed field LH gives an extension K C LH. Explain why the Theoremof the Primitive Element applies to give E LH such that L11 = Show that this phas the desired property.

(b) (Second Proof.) Let m = be a monomial in Xl, . . . ,x,, with distinct exponentsal,...,an. Then define

Prove that = H.

Exercise 10. Prove that the subset N C Sn defined in the proof of Theorem 12.1.10 is asubgroup of Sn.

Exercise 11. Let H be a proper subgroup of An with n 5. Prove that H] � n.

Exercise 12. The discussion following Theorem 12.1.10 shows that if we are going to useLagrange's strategy when n � 5, then we need to begin with = V's, which has isotropysubgroup An. Suppose that E L is our next choice, and let 8(x) be the resolvent of Since

LAGRANGE 333

we regard as known, we may assume that The idea is to factor 8(x) oversay 0 = R1 . . . where E [x] is irreducible. This is similar to how (12.13)

factors the resolvent of ti over K(yi). Suppose that enables us to continue Lagrange'sinductive strategy. This means that some factor of 0, say has degree <n. Your goal is toprove that this implies the existence of a proper subgroup of of index <n.(a) Prove that � 2.(b) Since 0 splits completely over L, the same is true for Let E L be a root of R3 and

consider the fieldsK C CM = c L.

Let C be the subgroup corresponding to Gal(L/M) C Gal(L/K) under (12.1). Provethat C A. and that : is the degree of R3.

(c) Conclude that <n implies that is a proper subgroup of of index <n.With more work, one can show that deg(R1) = : for all i and that

2

It follows that s = 1 or 2.

Exercise 13. Let be a primitive nth root of unity, and let a = (xi + @2 + +Prove that ((12...n)) CSn.

Exercise 14. Let a, be as in (12.18). The quotation given in the discussion following(12.18) can be paraphrased as saying that the roots of the resolvent of 0 = come from thepermutations of the n — I roots X2, . . . , that ignore the root xi. What does this mean?(a) Show that each left coset of ((12.. .n)) in can be written uniquely as o((12. .

where a fixes 1.(b) Explain how Lagrange's statement follows from part (a).

In general, we say that gi,. . . , G are coset representatives of a subgroup H C G ifgi H,. . . , g,,,H are the distinct left cosets of H in G (so m = [G HI). Thus Lagrange's quotationgives an explicit set of coset representatives for ((12. . . n)) C S,,.

Exercise 15. Given the Lagrange resolvents a!,. . . , defined in (12.19), the goal of thisexercise is to prove that

(a) Write = for 1 <1< p, so that a, = Then show that

=j=i j,I=I

(b) Given an integer m, use Exercise 9 of Section A.2 to prove that

Jp, ifm 0 mod p,

— otherwise.

(c) Use parts (a) and (b) to prove the desired formula for

Exercise 16. Prove that Theorem 7.4.4 follows from Theorem 12.1.6 and Proposition 2.4.1.


Exercise 17. In Theorem 12.1.9, we used the Galois correspondence to show that rationalfunctions and are similar if and only if K(p) = Give another proof of this resultthat uses only Theorem 12.1.6.

Exercise 18. Consider the quartic polynomial f =x4 + 2x2 — 4x + 2 E Q[x].(a) Show that the Ferrari resolvent (12.10) is y3 — 2y2 8y.

(b) Using the root y' = 0 of the cubic of part (a), show that (12.11) becomes

x2 = 1)

and conclude that the four roots off are

and

(c) Use Euler's solution (12.17) to find the roots of f. The formulas are surprisingly different.We will see in Chapter 13 that this quartic is especially simple. For most quartics, the formulasfor the roots are much more complicated.

Exercise 19. This exercise will prove a version of Theorem 12.1.10 for a subgroup H of anarbitrary finite group G. When G = Theorem 12.1.10 used the action of on L and wroteH = for some E L. In general, we use the action of G on the left cosets of H definedbyg.hH =ghHforg,h E G.(a) Prove that g hH = ghH is well defined, i.e., hH = h'H implies that ghH = gh'H.(b) Prove that H is the isotropy subgroup of the identity coset eH.(c) Let m = [G : H], so that the left cosets of H can be labeled giH, . .. ,gmH. Then, for

g E G, let a E Sm be the permutation such that g . g,H = Prove that the mapg a defines a group homomorphism G —+ 5m•

(d) Let N be the kernel of the map of part (c). Thus N is a normal subgroup of G. Prove thatN C H.

(e) Prove that [G : N] divides m!.(f) Explain why you have proved the following result: If H is a subgroup of a finite group G,

then H contains a normal subgroup of G whose index divides {G: H]!.(g) Use part (f) and Proposition 8.4.6 to give a quick proof of Theorem 12.1.10.

Exercise 20. Let G be a finite group and let p be the smallest prime dividing IGI. Prove thatevery subgroup of index p in G is normal.

Exercise 21. Part (a) of Theorem 12.1.10 implies that when n 5, the index of a propersubgroup of Sn is either 2 or � n.(a) Prove that always has a subgroup H of index n. This means that equality can occur in

the bound [Sn : H] � n.(b) Give an example to show that Theorem 12.1.10 is false when n = 4.

12.2 GALOIS

In this section we will explore several aspects of Galois's work. Our discussion willbe based on his 1831 memoir on Galois theory, entitled Mémoire sur les conditionsde résolubilité des equations par radicaux. See [Galois, pp. 42—71] for the Frenchoriginal and [Edwards, pp. 101—1131 for an English translation.

GALOIS 335

A. Beyond Lagrange. In Section 12.1 we saw that Lagrange studied the universalcase where the roots are variables Xi,... In contrast, Galois created a theory thatapplies to arbitrary polynomials. To see the difference, recall the quotation fromGalois given in the Historical Notes to Section 7.1 [Galois, p. 51]:

PROPOSITION I

THEOREM. For a given equation, let a,b,c,... be the m roots. There isalways a group of permutations on the letters a, b, c,.. that enjoys the followingproperty:

10 that every function of the roots that is invariant* * under the substitutionsof the group, is rationally known;

2° conversely, that every function of the roots that is rationally determined,is invariant under these substitutions*.

As noted in Section 7.1, this asserts that in a Galois extension F C L, the field F is thefixed field of the Galois group Gal(L/F). For our purposes here, the most interestingpart of the proposition is the double asterisk **, which refers to the following marginalnote in Galois's manuscript [Galois, p. 50]:

Here we call invariant not only a function whose form is invariant under thesubstitutions of the roots among themselves, but also those [functions] for whichthe numerical value does not vary under the substitutions.

In Section 12.1 we saw that Lagrange's concern was "only with the form" of ex-pressions and not "with their numerical quantity." In this marginal note, we see thatGalois is consciously going beyond Lagrange. We finally have a theory that appliesto all polynomials, not just the universal one.

The single asterisk * in the above quote will be discussed in the Historical Notes.

B. Galois Resolvents. To understand the splitting field of a separable polynomial,Galois used a variation of Lagrange's notion of resolvent polynomial. Suppose thatf E F[x] can be written f = ao(x co)... (x — in a splitting field L, wherea1,... ,a,, are distinct. We also assume that F is infinite. Given t1,... ,t,, E F,consider the polynomial of degree n! defined by

(12.21) s(y) = fi (y— + . . .

The discussion following (5.4) in the proof of Proposition 5.2.1 in Section 5.2 showsthat s(y) E F [y]. You should reread this argument, which uses symmetric functionsand is similar to Galois's.

In this situation, Galois asserts that since . . . , a,, are distinct, one can findt1,...,t,, EFsothatthen!elements

tlaa(I) + a ES,,,

are all distinct. In other words, t1, . . . , F may be chosen so that s(y) is separable.When this happens, we call s(y) a Galois resolvent of f. Exercises 1 and 2 will


prove that such t1,... , exist. Galois uses the letter V to denote t1 ct1 + +Following Lagrange, he refers to V as a "function of the roots."

Here is an example of a Galois resolvent.

Example 12.2.1 We will compute a Galois resolvent of f = (x2 2)(x2 3) =— 5x2 + 6 E Q[xI, which has roots and Let (t1 ,t2,t3,t4) =

(0, 1,2,4). In the notation used by Galois,

V 1

Using Maple or Mathematica, one can compute that (12.21) gives

s(y) = 731025000000— 5765769000000y2+ 13335274350000y4

— 12343809230400y6+5171341381036y8— 1110939359380y'°

+ 12973035 1909y'2— 8413645990y'4+ 30839421 1y16

— 6392440y18 + 73339y2° — 430y22 +y24

A computer calculation also shows that gcd(s(y),s'(y)) = 1, which implies that s(y)

is a Galois resolvent off. Factoring s(y) into irreducibles gives

s(y) = (900— 132y2+y4)(25— 118y2+y4)(361 —70y2+y4)

(36— 60y2 +y4)(100 — 28y2 +y4)(25 22y2 +y4).

Hence the Galois resolvent is reducible in this case.

As stated in [Galois, p. 49], the key property of V is the following:LEMMA III. If the function V is chosen as indicated in the preceding article,then it has the property that every root of the given equation [our f] can be

expressed rationally as a function of V.

This lemma says the roots ct1,... , lie in F(V). It follows easily that

since V = + E F(ct1,. . . = L. Thus Lemma III implies that V is aprimitive element of the splitting field off over F.

Galois's proof of Lemma III is so terse that when Galois submitted his memoirto the French Academy in 1831, Poisson complained that the proof was insufficientbut could be completed using Lagrange's methods [Galois, p. 50]. A discussion ofGalois's proof can be found in [Edwards, 3 you will useLagrange's methods to prove Lemma III.

Let us compare Galois's Lemma III with Example 12.1.8, where we considered

fortheextension K = F(o-i,... CL = F(xi,. . . In Example 12.1.8, we usedTheorems 12.1.4 and 12.1.6 to show that L = K(fi) and that

O(y)= fi

GALOIS 337

is the resolvent of /3. Thus 0(y) is irreducible over K by Proposition 12.1.1. Sowe have a primitive element /3 and an irreducible polynomial 0(y) of degree n!. InGalois's situation, we have V and s(y), and although V is a primitive element (byLemma III), Example 12.2.1 shows that s(y) need not be irreducible. Thus, whilesome of Lagrange's results apply to arbitrary polynomials (such as the constructionof primitive elements), others do not (such as the irreducibility of resolvents).

As we have defined things, V is a root of the Galois resolvent s = s(y) e FEY].

Since s can be reducible over F, we let h h(y) E FEY] be the minimal polynomialof V over F. Then h is an irreducible factor of s. We will let m denote the degree ofh. Note that h is separable, since s is.

Galois makes the crucial observation that the roots of h interact with the roots ofthe original polynomial f as follows [Galois, pp. 49—51]:

LEMMA IV. Suppose that one forms the equation of V [our s], and that one takesone of its irreducible factors, such that V is a root of an irreducible equation [ourhi. Let V, V', V", ... be the roots of this irreducible equation. If a = isone of the roots of the given equation [our then çb(V') will also be a root ofthe given equation.

(In the original, Galois wrote a = f(V). We have changed f to because we use ffor the given polynomial.)

Pro of of Lemma IV: Since L = F(V) contains the root a of f, we can write a =where E F [x]. Also note that by normality, h splits completely over L. This showsthat V, V', V",••• E L. In particular, L contains the roots V and V' of the irreduciblepolynomialh. By Proposition 5.1.8, we can find a E Gal(L/F) such that a(V) = V'.(This proposition played a crucial role in our development of Galois theory, especiallyin the proof of Theorem 6.2.1.) Then Lemma IV follows immediately from

0= a(0) = a(f(a)) = f(4(a(V))) =

where the fourth equality uses f,4 E F[x] and a E Gal(L/F). •

Galois's argument is different from ours and doesn't mention automorphismsexplicitly. But it should be clear that automorphisms and how they act on roots areimplicit in the statement of Lemma IV. Galois's proof of Lemma IV is described in[Edwards, pp. 51—521.

C. Galois's Group. We next explore how Galois defined the Galois group. Heconsidered only splitting fields of separable polynomials. We will show that in thissituation, Galois's definition is equivalent to the one given in Section 6.1.

Consider the splitting field L of a separable polynomial f E F k]. As above, weassume that F is infinite. For us, the Galois group Gal(L/F) consists of automor-phisms of L that are the identity on F. However, Proposition 6.3.1 shows that we caninterpret Gal(L/F) in terms of permutations of the roots of f. Thus we can thinkof Gal(L/F) as consisting of all permutations of the roots that come from automor-phisms. In other words, we consider only those permutations that preserve the fieldoperations. Here is an example of what this means.


Example 12.2.2 Let L = Q(+ \/2 ± be the splitting field of f = — 4x2 +2over Q, and consider the permutation of the roots defined by

This is not consistent with the field_operations, since + i—* — should

imply that — + F-* — \/2 — Hence this permutation doesn't come froman automorphism in Gal(L/Q).

Galois did not use the notion of field automorphism. So how did he decide whichpermutations to use? His approach is based on the primitive element V and minimalpolynomial h constructed above. We will use the following notation. Let

V,V',V",...

denote the roots of h. Furthermore, since L = F(V) by Lemma III, the roots i,... ,off can be written

(12.22) p1(v),

where have coefficients in F. Then Galois describes his groupas follows [Galois, p. 53]:

No matter what the given equation [our fi is, one can find a rational function Vof the roots such that all of the roots are rational functions of V. Given this V,let us consider the irreducible equation of which V is a root (lemmas III and IV)[our h]. Let V, V', V" be the roots of this equation.

Let ço2V be roots of the proposed equation.

Write down the following m permutations of the roots:

(V), ...,(V'), piV', .

(V"), coy", p1V", .

coV(m1), o2V(ml), . .

I say that this group of permutations has the desired property.

(In the original, n and m are interchanged. We have switched them in order to beconsistent with the notation used here.) In this table, the first entry of a row is a labelfor the row, and the remaining n entries of the row are roots off by Lemma IV.

One complication is that for Galois, the word "permutation" has a different mean-ing than it has for us. We will discuss this in the Mathematical Notes below. Fornow, we will understand the above quote as saying that Galois's "group" consists ofthem permutations obtained by mapping co(V), cpi(V), ço2(V),. . . to them rows displayed in the quote. These permutations are related to the Galois groupGal(L/F) as follows.

GALOIS 339

Theorem 12.2.3 Let L be the splitting field of the separable polynomial f in F[x]and let the roots off be denoted as in (12.22). Proposition 6.3.1 gives a one-to-onegroup homomorphism Gal(L/F) —* Then the image of this map consists of the mpermutations described by Galois.

Proof: As above, V is a primitive element of F C L, and h is the minimal polynomialof V over F. The m roots of h will be denoted = v, = v', v(2) =

Note also that h is separable. Then the proof of Theorem 6.2.1 implies that

Gal(L/F) = {cri,...,0m},

where is the automorphism of L that takes the primitive element V to the rootof h. As in the proof of Lemma IV, it follows that

(12.23) = =

for any polynomial with coefficients in F.In the homomorphism Gal(L/F) —+ Sn from Proposition 6.3.1, ci, maps to the

permutation that takes to

a1(p(V)), (V)), . . . (V)).

Using (12.23), this can be rewritten as

(12.24) 'p' . . .

One easily sees that (12.24) is the ith row displayed in the above quote since V, V',v" V(m_l) are now ..., V(m_l). Hence the images of the a,are the m permutations described by Galois. U

This theorem shows that for the splitting field of a separable polynomial, thedefinition used by Galois is equivalent to Definition 6.1.1. The Historical Notes toSection 6.1 give a brief description of how we got from Galois's group to the modernGalois group Gal (L/F). A more detailed explanation appears in [11].

D. Natural and Accessory Irrationalities. Beforeexplaining Galois's strategyfor solving equations, we need to discuss some classical terminology. Let F C L bethe splitting field of a separable polynomial f E F[x]. Then adjoin a quantity to F,where /3 is a root of an auxiliary equation that we assume to be known. For example,/3 could be a radical or a root of a resolvent equation. If /3 F, then we call /3 anatural irrationality when /3 E L and an accessory irrationality when L.

Example 12.2.4 Letf e F [x] be solvable by radicals with splitting field F C L.

• If F c L is radical in the sense of Section 8.2, then we can obtain the roots offby adjoining natural irrationalities.

• If F c L is solvable but not radical, then at least one of the radicals adjoined mustbe an accessory irrationality.


The quantity /3 F in the above discussion gives an extension F C K = F(/3).Then it is easy to see that

/3 is a natural irrationality K C L,

/3 is an accessory irrationality K L.

When K L, we can assume that K and L are both contained in some larger field(see Exercise 4). Then we get a diagram

KL/ \(12.25) K L\ /

F

where KL is the compositum of K and Las in Definition 8.2.5. The relation betweenGal(KL/K) and Gal(L/F) is described by the following result.

Theorem 12.2.5 Suppose that we have a diagram (12.25) where F C L is a Galoisextension and F C K is finite. Then K C KL is a Galois extension and the restrictionmap a '—* alL defines an isomorphism

Gal(KL/K) Gal(L/KflL) C Gal(L/F).

Proof: In Exercise 5 you will show that K C KL is Galois whenever F C L is. Bythe results of Chapter 7, Gal(KL/L) C Gal(KL/F) is normal since F C L is Galois,and thus aL L for all a E Gal(KL/F). Thus a gives a map

alL : L —* L.

Since a IL is the inverse of alL (see Exercise 6), alL is an automorphism of L.Furthermore, a is the identity on F, which implies that the same is true for alL.

When restricted to Gal(KL/K) C Gal(KL/F), this gives a map

(12.26) Gal(KL/K) —+ Gal(L/F).

You will show that this is a group homomorphism in Exercise 6.To see that (12.26) is one-to-one, suppose that a E Gal(KL/K) and alL is the

identity on L. Then a is the identity on both K and L, which easily implies that a isthe identity on KL (see Exercise 6). Thus (12.26) is one-to-one.

Finally, we need to show that the image is Gal(L/K fl L). For this purpose, letH C Gal(L/F) be the image of (12.26). By the Galois correspondence, it sufficesto show that LH = KflL. First suppose E LH. Then is fixed by all alL e H,which means that a is fixed by all a E Ga1(KL/K) and hence is in the fixed fieldof Gal(KL/K). Thus a E K since K C KL is Galois, and a e KflL follows fromaELHCL. ThisprovesthatLHCKflL.

GALOIS 341

For the other inclusion, let a E K fl L. Then in particular, a is in K and hence isfixed by all a E Gal(KL/K). Thus alL(a) a for all alL E H, which implies a E LH,as claimed. This completes the proof. •

Theorem 12.2.5 is sometimes called the Theorem on Natural Irrationalities. Tosee why, suppose that K L, i.e., K is obtained from F by adjoining accessoryirrationalities. Then the isomorphism

Gal(KL/K) Gal(L/KflL)

of Theorem 12.2.5 means that K C KL and KflL CL have the same Galois group. ButKnL lies inside Land hence is obtained from F by adjoining natural irrationalities.Thus, from the point of view of Galois theory, Theorem 12.2.5 implies that we don'tneed accessory irrationalities.

E. Galois's Strategy. In Section 12.1 we saw that Lagrange formulated hisstrategy for solving equations in terms of resolvents. However, there are groupslurking in the background. For instance, our discussion of the quartic used

Yi XIX2 +x3x4 and ti = Xi +X2 X3 X4,

whose isotropy subgroups are

H(y1) = ((12),(1324)) DH(t1) = ((12),(34)).

So the idea of reducing to smaller groups is implicit in what Lagrange was doing.Getting smaller groups is the main goal of Galois's strategy. In the Historical Notes

to Section 8.3, we gave the following quote where Galois discusses his approach tosolvability by radicals [Galois, pp. 57—59]:

I first observe that to solve an equation, it is necessary to reduce its groupuntil it contains only a single permutation...

Given this, we will try to find the condition satisfied by the group of anequation for which it is possible to reduce the group [to a single permutation] byadjunction of radical quantities

In the first sentence, Galois states the goal of reducing the Galois group to the identity,and in the second, he says that in the case of solvability by radicals, the goal is toreduce the Galois group by adjoining radicals.

In its most general form, Galois's strategy is to successively adjoin known quanti-ties (radicals or roots of resolvents) in order to reduce the Galois group to the identity.This adjunction process gives an extension F C K that we regard as known. Thesplitting field off over K is easily seen to be K C KL, which is one of the extensionsin the diagram (12.25). By Theorem 12.2.5, we have

(12.27) Gal(KL/K) a subgroup of Gal(L/F).

Thus going from F to K gives a subgroup of the original Galois group. Furthermore,if the new Galois group is the identity, then KL = K, which implies that L C K. SinceK is known, it follows that the roots of f are also known.


Here is an example of how this works.

Example 12.2.6 One easily checks that! = x3 + 9x —2 E Q [x} is irreducible over Qwith real root given by

The other roots off are a complex conjugate pair = since = —3024 isnegative. If L is the splitting field of f over Q, then Proposition 7.4.2 implies thatGal(L/Q) S3, since Q.

To make the Galois group smaller, we adjoin /3 = + 2\fl to Q, which gives

K = Q(/3). In Exercise 7 you will show that — and hence lie in K. Thismeans that f factors as (x — i)g, where g E K[x] has roots Thus KL isobtained from K by adjoining cr3. Since K C R and cr2, are not real, it followsthat {KL: K1 = 2. This shows that Gal(KL/K) Z/2Z.

Furthermore, if we think of KL as the splitting field off over K, then we still havethe map Gal (KL/K) —÷ S3. Given how we've labeled the roots, the image of this mapis clearly ((23)) c S3. So adjoining /3 reduces the group of permutations from S3 tothe smaller group ((23)).

In Exercise 7 you will show that if we adjoin w = to K, then K' = K(w)contains all roots off, so that K'L = K'. Hence the Galois group has been reducedto Gal(K'L/K') = {e}, which completes Galois's strategy.

To fully understand Galois's strategy, we need to think in terms of permutations.If our separable polynomial f has degree n, then the action of the Galois group onthe roots gives a map

(12.28) Gal(L/F) —*

whose image is Galois's group by Theorem 12.2.3. Now let F C K be a finiteextension. Since K C KL is the splitting field off over K, we get a similar map

(12.29) Gal(KL/K) H' Sn.

In Exercise 8 you will show that these maps are compatible with the isomorphismgiven in (12.27). Hence, when we regard Gal(KL/K) and Gal(L/F) as subgroupsof the former is contained in the latter.

This makes Galois's strategy easy to understand. He works with a fixed separablepolynomial f of degree n. For him, the group off lies in Sn, but the field he worksover keeps changing. Furthermore, each time he enlarges the field by adjoiningsomething known (a radical or a root of a resolvent), he passes from the group to asubgroup (which may be the whole group). This leads to extensions

F c K1 C K2 C

where each K, is regarded as consisting of things that are known. If at some point,say for Km, the group reduces to the identity, then Gal(KmL/Km) = {e}, which as

GALOIS 343

noted above implies that L C Km. This allows us to express the roots off in terms ofknown quantities.

Historical Notes

When reading Galois, one must keep in mind the distinction between arrangementsof roots and permutations of roots. If you look back at the quotation giving Galois'sdefinition of his group, you will see that he lists m arrangements of the roots. Thecorresponding permutations come from mapping the first arrangement to the others.To complicate matters, Galois uses different terminology from us:

Us GaloisArrangement PermutationPermutation Substitution

So when Galois says "group of permutations," he really means "group of arrange-ments." But later in the memoir, we find the following [Galois, pp. 53—55]:

It is evident that in the group of permutations considered here, the order of theletters is not of importance, but rather only the SUBSTITUTIONS of the letters bywhich one passes from one permutation to another.

By Theorem 12.2.3, these substitutions form a subgroup of isomorphic to theGalois group.

Galois knew the difference between arrangements and substitutions, and wasaware that the latter formed a group in the modem sense [Galois, p. 47]:

if one has substitutions S and T within such a group, one is sure to have thesubstitution ST.

From the modem point of view, substitutions are more important. But this was notclear to Galois, especially given the vivid visual image provided by groups of ar-rangements. This is evident from Galois's definition of his group, and other examplescan be found in [12]. Galois's memoir is written in terms of arrangements, althoughchanges made shortly before his death in 1832 indicate that Galois was thinkingabout switching to substitutions. For example, we quoted Galois's Proposition I atthe beginning of the section. This quotation includes an asterisk * that refers to amarginal note where Galois says "Put everywhere in place of the word permutationthe word substitution" [Galois, p. 50]. But then Galois crosses this out!

It took a while for the mathematical community to understand Galois's ideas.In 1866 the third edition of Serret's Cours d'algèbre supérieure included a partialaccount of Galois theory. As quoted in [11, p. 110], Serret comments that "Galoisused the notion of groups of permutations [our arrangements] ..., but it seems betterfor us to keep to substitutions." This quote also shows that "substitution" was thecommon name for elements of S,. in the nineteenth century. Another example of thisis Jordan's 1870 text Traité des substitutions et des equations algébriques [Jordan 1],which gave the first complete account of Galois theory.

Our discussion of Galois's strategy did not state his version of Theorem 12.2.5 (theTheorem on Natural Irrationalities). The reason is that one needs to understand the


distinction between arrangements and permutations before reading Galois's version,which goes as follows [Galois, p. 55]:

PROPOSITION II

THEOREM. If one adjoins to a given equation [our f] the root r of anauxiliary irreducible equation ",

1° one of two things will occur: either the group of the equation will not change;or it will be partitioned into p groups each belonging to the original equationwhen one adjoins each of the roots of the auxiliary equation;

2° these groups have the remarkable property, that one passes from one to anotherby applying to all of the permutations of the first the same substitution of theletters.

The asterisk * indicates that r was a root of an auxiliary equation "of prime degreep" in an earlier version of Proposition II [Galois, p. 54]. This is the "p" that appearsin 10. In Exercise 9 you will use the Galois correspondence and Theorem 12.2.5to show that if [K : Fl = p, then Gal(KL/K) is isomorphic either to Gal(L/F) or toa subgroup of index p in Gal(L/F). The latter corresponds to "partitioned into pgroups" in the above quotation.

It appears that Galois first proved Proposition II in the prime-degree case. Thenight before his fatal duel, he realized that his proof applied in greater generality.Writing in haste, he changed part but not all of the statement of Proposition II. Healso knew that his proof was incomplete—this is where he writes "Je n'ai pas letemps" ("I don't have time") [Galois, p. 54].

This explains Proposition II up to the appearance of p. But what about theremainder of 10? The idea is that instead of adjoining one root r of the auxiliaryequation, one could adjoin a different root r' of the same equation. This gives adifferent extension F C K' = F(r'). Then going from F to K' will reduce the group,but possibly in a different way. In modem terms, F C K and F C K' are conjugateextensions. By Theorem 12.2.5, Gal(KL/K) and Gal(K'L/K') are isomorphic tosubgroups of Gal(L/F). Then Galois's observation in 10 is that these subgroups areconjugate in Gal(L/F). You will prove this in Exercise 10.

The precise meaning of 2° of Proposition II will be explored in Exercises 11and 12. Galois's proof of Proposition II can be found in [5] and [12].

This concludes our discussion of Galois. However, to fully appreciate whatGalois did, the reader should keep in mind Galois's other contributions to Galoistheory, many of which were discussed earlier in the book:

• Extension fields (Historical Notes to Section 4.1).• The Galois correspondence (Historical Notes to Section 7.1).• Normal subgroups (Historical Notes to Section 7.2).• Solvable groups and solvability by radicals (Historical Notes to Section 8.3).• Finite fields (Historical Notes to Section 11.1).

This is an impressive list for someone who died at age 20. There is also Galois'samazing work on irreducible polynomials of degree p and p2, where p prime. Thiswill be described in Chapter 14.

GALOIS 345

For a fuller account of Galois's mathematical work, the reader should consult [5],[11, PP. 80—84], [12], [Edwards], or [Tignol, Ch. 14]. The biography of Galois [13]describes his short but intense life.


Exercise 1. Let F be an infinite field and let V be a finite-dimensional vector space over F.The goal of this exercise is to prove that V cannot be the union of a finite number of propersubspaces. This will be used in Exercise 2 to prove the existence of Galois resolvents.

Let W1,. . . , Wm be proper subspaces of V such that V = W1 U. U Wm, where m> 1 is thesmallest positive integer for which this is true. We derive a contradiction as follows.(a) Explain why there isv E W1 \(W2 U... U Wm).(b) There is w E V \ W1, since Wi is a proper subspace. Using v from part (a), we have

AV+ wE V = W1 U for all A E F. Explain why this implies that there are A1 A2

inFsuchthatAjv+w,A2v+wE forsomei.(c) Now derive the desired contradiction.

Exercise 2. Suppose that we have an extension F C L, where F is infinite. The goal of thisexercise is to show that if . . , a,, E L are distinct, then t1, . . . , t,, E F can be chosen so thatthe polynomial s(y) defined in (12.21) has distinct roots. Given a Tin let

Wo,r = {(ti,. . . ,tn) E F" — a1-(I))t1 = 0 in L}.

(a) Prove that W,r is a subspace of and that W,y,r F".(b) Show that part (a) and Exercise limply that there are i,... , t, E F such that the polynomial

s(y) from (12.21) has distinct roots.

Exercise 3. This exercise will prove Galois's Lemma III using the methods of Lagrange.Let V = (i + + tnan, where t,, are chosen so that the Galois resolvent s(y) from(12.21) is separable. Also let V,, = tiQa(i) + . . for a E Sn. Prove that each

a rational function in V with coefficients in F by adapting the second proof ofTheorem 12.1.6.

Exercise 4. In the discussion preceding (12.25), we have extensions F C L, which is a splittingfield of f F[x], and F C K = where fi is a root of an irreducible polynomial in F[x].Given the many ways in which extension fields can be constructed, these extensions mightnot have much to do with each other. Prove that there is an extension F C M that containssubfields F c ci M and F c K1 ci M such that L,

are the identity on F. Thus, by replacing L, K with the isomorphicfields L1, K a larger field, as claimed in the text.

Exercise 5. Suppose that F C L is the splitting field of a separable polynomial f E Fix]. Alsosuppose that we have another finite extension F C K such that the compositum KL is defined.Prove that K C KL is the splitting field of f over K.

Exercise 6. This exercise will complete the proof of Theorem 12.2.5. Given a E Gal(KL/K),we showed in the text that alL maps L to L.(a) Show that (aT)IL = aILTIL.(b) Use part (a) to show that is the inverse function of alL.(c) Use part (a) to show that (12.26) is a group homomorphism.


(d) Let a be an automorphism of KL that is the identity on both K and L. Prove that a is theidentity on KL.

Exercise 7. This exercise is concerned with the details of Example 12.2.6. As in the example,

letL be the splitting field of f =x3 +9x— 2 overQ and set K = Q(/3), where =(a) Show that E K.(b) Show that K' = K(w), w = contains all roots of f.

Exercise 8. In Theorem 12.2.5, we have the map (12.26) defined by a '-+ alL. However, ifF C L is the splitting field of a separable polynomial f E F[x] of degree n, then we also havemaps (12.28) and (12.29). Prove that these maps are compatible, i.e., that a E Gal(KL/K) andalL E Gal(L/F) map to the same element of S,, under (12.28) and (12.29).

Exercise 9. In the situation of Theorem 12.2.5, suppose that F C K is an extension of primedegree p. Prove that Gal(KL/K) is isomorphic to either Gal(L/F) or a subgroup of index pin Gal(L/F).

Exercise 10. Suppose that we have a diagram (12.25) as in Theorem 12.2.5. Also assumethat K = and let K' = F(fl'), where and have the same minimal polynomial overF. You will show that Gal(KL/K) and Gal(K'L/K') give conjugate subgroups of Gal(L/F).This is the modern version of what Galois says in 1° of Proposition II.(a) Let F C M' be the Galois closure of the extension F C M constructed in Exercise 4.

Explain why we can regard L, K, and K' as subfields of M'.(b) Explain why we can find r E Gal(M'/F) such that T(K) = K'.(c) Show that TIL E Gal(L/F) maps K fl L to K' fl L. Thus K fl L and K' fl L are conjugate

subfields of L.(d) Use Lemma 7.2.4 to show that in Theorem 12.2.5, GaI(KL/K) and Gal(K'L/K') map to

conjugate subgroups of Gal(L/F).

Exercise 11. Let A denote the set of arrangements described by Galois. This is Galois's"group." For simplicity, we write the first arrangement on Galois's list as a! a,,. Then letG be the set of permutations that take the first element of A to the others. Theorem 12.2.3implies that G is a subgroup of S,, isomorphic to Gal(L/F).

We also have the action of on the set of all n! arrangements of roots by

as,, =

This induces an action of G on the set of arrangements.(a) Explain why A is the orbit of ai a,, under the G action.(b) Show that the map G A defined by a a is one-to-one and onto.

Exercise 12. In the situation of Theorem 12.2.5, let G C S,, correspond to Gal(L/F), andH C Sn correspond to Gal(KL/K). By Exercise 8, we know that H C G. Also let A bethe set of arrangements studied in Exercise 11. Then a left coset aH C G gives a subsetaH . cx,, C A, and since the map a i—* a at a,, is one-to-one and onto, the setsaH . a,, partition A into disjoint subsets. We claim that these are the "groups" that appearin 1° and 2° of Galois's Proposition II.(a) Given any two such "groups" aH a,, and rH . an, prove that there is -y E G

such that (as Galois says in 2°) one passes from one to the other by applying -y to allarrangements in the first.

(b) So far, it seems like Galois is describing cosets. However, as pointed out in [12], Galoisthought of these "groups" differently. This is seen by explaining how they relate to 10 of

KRONECKER 347

Galois's proposition. Let M' be the field used in Exercise 10, and let r E Gal(M'/F).Then K' = i-(K) is a conjugate of K. Let a E G be the permutation corresponding toTIL E Gal(L/F). Show that is the subgroup of corresponding to Gal(K'L/K').

(c) Using the setup of part (b), consider the "group" aH CA. Prove that aHa' Cis the set of all permutations of S, that map the first element of this "group," namely

o,,, to another element of the "group." (Remember that this is the process forturning a "group" of arrangements into a subgroup of Sn.)

Combining parts (b) and (c), we see that what Galois says in 10 of Proposition II is fullyconsistent with what we did in Exercise 10.

Exercise 13. This exercise will show that not all choices of the t' in (12.21) give Galoisresolvents. As in Example 12.2.1, f = — 2)(x2 —3) has roots and

This time we will use (t1 ,t2, t3, = (0, 1,2,3). Show that (12.21) gives the polynomial

s(y) = 1679616— 45722880y2 +445417056y4 — 1935550800y6

+4169468065y8 —4504515400y'° +2268233020y12 —432170200y'4

+ 3678l990y16 — 1483000y18 + 29596y2° — 280y22 + y24

= (81 —90y2+y4)2(16—40y2+y4)2(l — 10y2+y4)2.

This does not have distinct roots, so that s(y) is not a Galois resolvent.

Exercise 14. Use Theorem 12.2.5 and standard results about Galois extensions to prove that= [L : KflL]. Then explain why this implies that Gal(KL/K)I <IGal(L/F)I if

and only if F is a proper subfield of K fl L.

Exercise 15. Let F c L and F C K be Galois extensions such that KL is defined. We will alsoassume that K fl L = F. The goal of this exercise is to prove that F C KL is a Galois extensionwith Galois group

Gal(KL/F) Gal(L/F) x Gal(K/F).

(a) Prove that F C KL is Galois and that a E Gal(KL/F) implies that alL E Gal(L/F) andalK E Gal(L/K).

(b) Use part (d) of Exercise 6 to show that there is a one-to-one group homomorphism

Gal(KL/F) —+ Gal(L/F) x Gal(K/F).

(c) Use Exercise 14 and the Tower Theorem to show that [KL: F] = [K : F] [L: F].(d) Conclude that the map of part (b) is an isomorphism.

12.3 KRONECKER

In this section we will explore how Kronecker combined ideas of Lagrange, Gauss,and Galois to give a powerful construction of the splitting field of a separable poly-nomial over a field of characteristic 0.

A. Algebraic Quantities. In 1882 Kronecker published the important paperGrundzüge einer arithmetischen Theorie der algebraischen Grössen [Kronecker,Vol. II, pp. 237—3871. In English, the title is "Foundations of an arithmetic theory


of algebraic quantities," which signals Kronecker's intention to create a generalfoundation for dealing with algebraic objects.

Kronecker begins his Grundzuge by describing the fields that he will work over,although he deliberately avoids using Dedekind's terminology of "fields." The quota-tion from Dedekind given in the Historical Notes to Section 4.1 shows that Dedekind'sdefinition is very abstract: anything that satisfies the field axioms is afield. Kronecker,on the other hand, wants to emphasize that the objects he deals with are very con-crete. We will use the term "field" when discussing what Kronecker does, thoughKronecker would not be entirely comfortable with this practice.

Kronecker's basic objects of study are elements of a Rationalitäts-Bereich (domainof rationality). Such a domain is built out of finitely many algebraischen Grössen(algebraic quantities) 9k', 9k", ER",. .., which can be variables or roots of polynomials(we will say more about this below). Then an element of the Rationalitäts-Bereich isa rational function with integer coefficients in these quantities. In modern terms, thisis the field

(12.30)

since by clearing denominators, every element of L can be written as a quotient ofpolynomials with integer coefficients. For Kronecker, however, the emphasis is moreon the elements than on the field.

The basic operation on such fields is adjunction. Given L as in (12.30), consideran irreducible polynomial with coefficients in L. Then a root of this polynomial isan algebraic quantity that gives a new Rationalitäts-Bereich when adjoined to L.In §2 of the Grundzüge, Kronecker assumes without comment that exists. As wewill see, he eventually explains why this assumption is valid.

Here is a simple example of Kronecker's adjunction process.

Example 12.3.1 Given a variable x, the field Q(x) is an example of a Rationalitäts-Bereich. Furthermore, in Exercise 1 you will verify that y2 — 4x3 — x is irreducibleas a polynomial in y with coefficients in Q(x). Thus, by adjoining a root of thispolynomial to Q(x), we get a new Rationalitäts-Bereich

Q(x,y1)

In particular, = + x is an algebraic quantity.

Kronecker then studies the structure of the fields (12.30). His main result is thatany such field can be written as an extension

(12.31) Q c c

where , can be regarded as variables over Q, and 15 is algebraic overIn Exercise 2 you will show that this follows from the result

of Steinitz discussed in the Mathematical Notes to Section 4.1, together with theTheorem of the Primitive Element.

KRONECKER 349

In contrast to modern presentations, the fields considered by Kronecker are con-structed explicitly. This constructive attitude runs very deep. For example, ratherthan just defining what it means for a polynomial to be irreducible, Kronecker givesa method for deciding whether or not a polynomial with coefficients in a field ofthe form (12.31) is irreducible. To do this, he first discusses polynomials inand describes the algorithm presented in Proposition 4.2.2 for factoring polynomialsover Q. He then gives a terse explanation of how to factor in the general case. Themissing details can be found in Edwards's book [7].

B. Module Systems. Besides developing a theory of fields, the Grundzügealso considers rings, ideals, and quotient rings. This begins in §5 of the Grundzüge,where Kronecker introduces the Integritäts-Bereich (domain of integrality) built from

, In modern terms, this is the integral domain

R =.

consisting of all polynomials in 91', 91", 9l",... with integer coefficients. The fieldof fractions of R is the Rationalitäts-Bereich , 91", 91", ...).

The next step is to define certain ideals of R. In §21 of the Grundzuge, Kro-necker takes finitely many elements M1, M2 ,M3,... E R and defines the module system(M1 , M2, M3,...) to consist of all linear combinations with coefficients in R

A1M1+A2M2+A3M3+•, A1,A2,A3,... ER.

Given M, M' E R, Kronecker then defines

(12.32) MmM' (modd.Mi,M2,M3,...)

to mean that M — M' is contained in (M1, M2 ,M3,...). These days, we say "ideal"rather than "module system" and we write the ideal as

(In Exercise 3 you will prove that this is an ideal of R.) Then (12.32) means thatM — M' E I, which is equivalent to the equality M + I = M' + I of cosets in R/I.

It follows that Kronecker is developing the basic language of ideals and quotientrings. However, Kronecker didn't use Dedekind's term "ideal," because Dedekindallowed his ideals to be very abstract, while Kronecker was only interested in theexplicitly constructed ideals described above.

For us, the most important application of these ideas came in Section 3.1. Recallthat in the proof of Proposition 3.1.3, we showed that if f E F is an irreduciblepolynomial, then L = F[x] / (f) is an extension field of F such that a = x+ (f) E L isa root off. This is how we proved the existence of roots.

This is equally important for Kronecker, for he uses a similar construction to givea precise meaning to the term "algebraic quantity." The idea is as follows: In (12.31),let G(x) be the minimal polynomial of 15 over Q(911,912,913,...). Then one canreplace the extension field Q(Q5,911 ,912,913,...) with the quotient ring

(12.33) Q(911,912,913,. . . )[xI/(G),


where the coset x + (G) plays the role of the root Q5. Since , arevariables in (12.31), we now have a rigorous construction of the algebraic quantity15 in terms of polynomials and ideals.

In this construction, Kronecker preferred to work with rings rather than fieldssince he wanted to avoid denominators as much as possible. So Kronecker wouldreplace (12.33) with

where one has now suitably cleared denominators so that G is an irreducible polyno-mial in , ,xI. This quotient ring is an integral domain whose field offractions is the corresponding field (12.33).

Here is a simple example of what these presentations look like.

Example 12.3.2 Consider the field Q (x, + x) constructed in Example 12.3.1.This is Q c Q(x) C Q (x, v"4x3 + x). As a polynomial in y, the minimal polynomialof +x is y2 —4x3 —x, so that (12.33) becomes

Q(x)[y]/(y2 —x).

Kronecker's presentation, which uses Z rather than Q, would be to take the field offractions of the integral domain

—4x3

Notice how polynomials in several variables appear naturally in this example.

As noted in [7], Kronecker was aware that this construction allows one to dispensewith "algebraic quantities." Kronecker states this as follows:

the whole arithmetic theory of algebraic quantities can be reduced to a theoryof entire functions of variables and unknowns with integer coefficients

(see [Kronecker, Vol. II, p. 377]). In the nineteenth century, "entire function" meantpolynomial. Thus Kronecker is saying that we can construct all algebraic quantitiesusing congruences of polynomials in several variables with coefficients in Z.

C. Splitting Fields. One of the points made in [8] is that Kronecker's conceptionof algebraic quantity evolved during the writing of the Grundzüge. The early sectionsof the Grundzüge don't give a precise definition of algebraic quantity, yet the latersections provide the language needed for this purpose (as noted in the above quota-tion). But to rewrite the Grundzüge from this new point of view would have been anoverwhelming task. Hence one needs to look at Kronecker's subsequent papers tosee how he worked out these ideas.

For us, Kronecker's 1887 paper Em Fundamentalsatz der ailgemeinen Arithmetik[Kronecker, Vol. III, pp. 209—240] is the most relevant. Here, he uses his new methodsto give an explicit construction of splitting fields. His presentation and proof involvea deep understanding of the ideas of Lagrange, Gauss, and Galois.

Let F be a field of characteristic 0, and let

(12.34) f(x) =f EF[x]

KRONECKER 351

be a separable polynomial. In his Fundamentalsatz paper, Kronecker uses F =where n> 1 and are variables. So for

Kronecker, the coefficients are explicitly known objects. His goal was to describethe roots off using module systems and congruences.

Kronecker was inspired by Galois's approach to the Galois group. Galois assumedthat the roots co,... , a,, off lie in some extension of F. Let us recall some of theideas developed in Section 12.2. The Galois resolvent (12.21) is the polynomial

s(y) = fi (y— + +tnaa(n))),

where t1,... , t,, E F are chosen so that s(y) is separable. Then the key player forGalois is the irreducible factor h(y) of s(y) that vanishes at V = t1a1 + ... + t,1cx,,.

Recall from Section 12.2 that by Galois's Lemma III, V is a primitive element ofthe splitting field of f over F. Thus each root a is a polynomial in V withcoefficients in F.

This implies that in the splitting field F(V), we can write

f(x)=fJ(x—ai) =fl(x—co1(V)).

Furthermore, since h(y) is the minimal polynomial of V over F, we can replace F(V)with the quotient ring F[y]/(h), where the coset y+ (h) plays the role of V. If wesubstitute this into the above equation and use congruence notation m, then we canwrite the above factorization as

(12.35) modh(y).

This is close to what Kronecker states in his Fundamentalsatz paper. However, ourderivation of (12.35) assumed knowledge of the roots , a,, off. Kronecker'sgoal is to compute this factorization without knowing the roots of the polynomial inadvance.

How does Kronecker accomplish this? In reading Gauss's 1815 proof of theFundamental Theorem of Algebra (discussed in the Historical Notes to Section 3.2),Kronecker learned the strategy of applying Lagrange's methods in the universal caseand then specializing to the specific polynomial at hand. This works as follows.

In the universal situation, the variables Xi,... are roots of the universal poly-nomial of degree n,

J = (x — x1) (x — x,,) = — + . + (— l)'1a,,,

which has coefficients in F (a i,. . . , a,,). Then the substitution a, '—+ c, takes f to

ISince F has characteristic 0, we can pick distinct integers t1, . . ,t,, E Z C F. This

gives the universal Galois resolvent

(12.36) S(y)= [f


where = t3x1 + + The theory of symmetric polynomials from Chapter 2shows that S(y) lies in Q[ai,...

Now let s(y) E F[y} be the polynomial obtained from S(y) by the substitutioni—+ c,. Kronecker then claims that t1, .. . , e 7L can be chosen so that s(y) is

separable. Once this is done, he will have constructed a Galois resolvent of f withoutknowing the roots of f. Exercises 1 and 2 of Section 12.2 proved the existenceoft1,... in F (rather than in Z) using the roots . Of course, Kroneckerneeds to use a different argument. Exercises 4 and 5 will show how to find the desired

t,, E 7Z without using the roots.In Section 3.2, we observed that Gauss used a similar method to compute the

polynomial Z(x, u) = fl1 (x — + + from (3.13) without knowingthe roots off. Do you see how this could have influenced Kronecker?

Given the Galois resolvent s(y) E F[y}, Kronecker then factors s(y) into a productof irreducible polynomials using the methods mentioned earlier in the section. Leth(y) be one of the irreducible factors. This gives the polynomial h(y) we need for(12.35). The important thing for Kronecker is that h(y) was constructed withoutknowing the roots off.

It remains to find explicit formulas for the polynomials in (12.35). For thispurpose, we turn to the methods of Lagrange discussed in Section 12.1. Since S(y) isthe resolvent polynomial of = t1x1 + + one can find explicitly computablepolynomials in F[cr1,... such that

(12.37)

where is the discriminant of S. You will prove this in Exercises 6—8.The substitution c, maps S(y) to s(y) and to i(s). Furthermore,

0, since s(y) is separable. Thus maps to some polynomial(y) E F [y]. We can now state the construction of the splitting field of f given in

Kronecker's Fundamentalsatz paper [Kronecker, Vol. II, p. 216].

Theorem 12.3.3 Let F have characteristic 0. Let f E be monic and separableof degree n > 0, and let s(y) and (y), i = 1,... , n, be constructed as above. Then,for any irreducible factor h(y) E F[x] of s(y), we have

modh(y).

This congruence means that each side is a polynomial in x whose coefficients areequal in the quotient ring F[y]/(h(y)). Furthermore, is a splitting fieldoff overF. •

We will not prove this here. The reader should consult [7] for a modern (but fullyconstructive) proof of Theorem 12.3.3.

Kronecker remarks that because of Theorem 12.3.3, "one is then relieved of theintroduction of algebraic quantities in many... algebraic investigations" [Kronecker,Vol. III, p. 216]. This is his clearest statement of how to avoid algebraic quantities.

KRONECKER 353

Kronecker's choice of the word Fundamentalsatz ("Fundamental Theorem") in thetitle of his paper indicates the importance he attaches to this result.

Theorems 12.3.3 and 3.1.4 both prove that a splitting field of f E F[x] exists,though Kronecker's theorem differs from Theorem 3.1.4 in two ways:

• Rather than construct the splitting field L using a sequence of quotient rings (asin Theorem 3.1.4), Kronecker constructs all roots off simultaneously using justone quotient ring L = F{yJ/(h(y)).

• We will see in Chapter 13 that Kronecker's construction leads directly to analgorithm for computing the Galois group off.

Hence Kronecker's construction of the splitting field contains a lot of informationabout the roots of the polynomial. It is harder than what we did in Theorem 3.1.4,but for a good reason.

Theorem 12.3.3 uses a lot of mathematics, including ideas of Lagrange (the uni-versal case), Gauss (relating the universal to the specific), and Galois (the Galoisresolvent). It is impressive how Kronecker was able to synthesize all of this mathe-matics into one theorem. One irony is that while Kronecker is given credit as the firstto prove the existence of the roots of a polynomial, his version of this result is rarelymentioned, since most books use the proof of Theorem 3.1.4 given in Chapter 3.While the modern proof illustrates the power of abstract algebra, it does not reflectthe richness of the historical context that led to Kronecker's proof of the existence ofsplitting fields.

Historical Notes

Congruences modulo a polynomial or module system play a central role in Kro-necker's construction of algebraic quantities. When Kronecker wrote his Grundzügein 1882, there were many known examples of congruences, including:

• am b mod n, fora,b E 7Z(Gauss, 1801).

• modx2+1, for ER[x] (Cauchy, 1845).

• mod pZ[&], for E (Schönemann, 1846).

• P(x) Q(x) mod (p,f(x)), for P(x),Q(x) E 7L[x] (Dedekind, 1857).

While Kronecker did not originate the use of congruences in the polynomial setting,he was clearly the first to realize the full power of this construction.

Our discussion of Kronecker's work omitted several important topics covered inthe Grundziige. For example, Kronecker created a theory of divisors to generalizeKummer's ideal numbers. This differs from Dedekind's theory of ideals, which isanother way to generalize ideal numbers. An exposition of divisor theory can befound in [6]. Kronecker also considered discriminants in detail.

We should also note that our discussion made liberal use of set theory. Forexample, when we gave Kronecker's definition of module system (M1 , M2,...) in aring R, we immediately translated it into the ideal

(M1,M2,...)={A1M1+A2M2-I-... IA1,A2,...ER}.


Kronecker, in contrast, says that A1M1 + A2M2 "contains the module system(M1 ,M2,...)" [Kronecker, Vol.11, p.335]. To us, this seems backwards. As explainedto me by Edwards, Kronecker's use of "contains" is similar to saying that 6 "contains"2 as a divisor. This makes sense when one realizes the importance of divisors inKronecker's mathematical thought. An introduction to Kronecker's views of settheory and the foundations of mathematics can be found in [9].

An important development in the late nineteenth century was the realization thatone needs to study the foundations of mathematics. It was no longer sufficient tosimply assume the existence of mathematical objects such as algebraic quantities.Rather, one had to give a rigorous proof of their existence. But instead of Kronecker'sconstructive vision of what this meant, the set-theoretic approach of Dedekind andCantor came to dominate. This explains why modem mathematics is firmly basedon sets and why abstract algebra is so different from high school algebra.

This chapter began with Lagrange's attempts to understand the roots of polynomi-als and ended with Kronecker's impressive construction of splitting fields. Along theway, we were able to witness the brilliance of Galois and the beginnings of modemalgebra. It has been a remarkable odyssey.


Exercise 1. Prove that y2 4x3 — x is irreducible when considered as an element of Q (x) [y].

Exercise 2. Show that (12.31) follows from the Theorem of the Primitive Element and thetheorem of Steinitz mentioned in the Mathematical Notes to Section 4.1.

Exercise 3. Let R be a commutative ring and let M1,. . . , M5 be elements of R. Prove that theset (M1,... = { A R} is an ideal ofR.

Exercise 4. In the discussion leading up to Theorem 12.3.3, we have the polynomial S(y)FEat,.. defined in (12.36). Then s(y) F[y] is obtained by '—+ where c1 is asin (12.34). Both of these polynomials depend on ti, . . . The goal of this exercise is toshow that if f is separable, then s(s) is a nonzero polynomial when ti,. . . are regarded asvariables. Since F has characteristic 0, part (a) of Exercise 5 implies that 0 for someti,. . . ,tn Z.

To prove that s(s) is a nonzero polynomial in ti,. . let F C L be the splitting field off constructed in Theorem 3.1.4. Thus f = (x— Qi) . . .(x— cr,,) in L[x].(a) If we regard the t as variables, explain why S(y) becomes a polynomial in y with

coefficients in . . , a,,, ti,... , t,,]. Conclude that s(y) F[ti,. . . ,t,,,y] and hence thatA(s)

(b) Explain why s(y) = (' — + . . . in L[ti,. . . ,t,,,y].

(c) Use part (b) and the separability off to show that s(y) has distinct roots, all of which liein L[ti,. . . Conclude that s(s) is a nonzero element of F[r1 ,. . .

Exercise 5. Let F be a field, and let g F [t1, . . . , t,,] be nonzero.(a) Suppose that F has characteristic 0, so that Q C F. For each i, pick a nonnegative integer

N, such that the highest power oft appearing in g is at most N,, and let

A ={(al,...,a,,) a, Z, 0 �a, <NJ.

Prove that there is (at,... ,a,,) A such that g(ai,. . . ,a,) 0.

KRONECKEA 355

(b) Now suppose that F has characteristic p and is infinite. Modify the argument of part (a)

to illustrate why the hypothesis "F is infinite" is needed in part (b).

Exercise 6. In F[xi,. . . consider the polynomial

f= (x—xi)..(x—x,,) +...+(—l)"an.

As noted in Section 2.2, we can regard f F[ai,. . . , as the universal polynomial of degreen. The goal of this exercise is to show that if f' denotes the derivative of f, then there arepolynomialsA,B E F[ai,. . .,a,,,xl such that deg(A) <n—2, deg(B) <n— 1, and

=

Here, is the discriminant defined in Section 2.4. The proof given here is taken from Gauss's1815 proof of the Fundamental Theorem of Algebra (see [14, pp. 293—295]).(a) Show that

+— (Al

+...+

is a polynomial in x of degree at most n — 1 whose coefficients are symmetric polynomials

(b) Prove that Bf' vanishes when x = x.(c) Conclude that — Bf' is divisible by f, and set

A= fShow that A and B have the desired properties.

Exercise 7. Let f F[XI be monic of degree n > 0 with discriminant F. UseExercise 6 to show that there are A,B E F[x] with deg(A) <n —2, deg(B) � n — 1 such thatthe coefficients of A and B are polynomials in the coefficients off and A f + Bf' =

Exercise 8. This exercise is concerned with from (12.37). Let S(y) be as in (12.36).(a) Show that applying (12.5) and (12.8) from the proof of Theorem 12.1.6 with f = /3 =

tixi+ + and g = gives

xi—

where

=Es,,

Also prove that 4,(y) E F[cri,. . . ,o-,,,y].

(b) Use Exercise 7 to show that there are polynomials A, B F [ai, . . . , a,,,y] such thatA(y)S(y)+B(y)S'(y) = Also show that s(S).

(c) Use part (b) to show that (12.37) holds with = B(y)4t(y).


REFERENCES

1. R. G. Ayoub, Paolo Ruffini 's contributions to the quintic, Arch. Hist. Exact Sd. 23(1980),253—277.

2. I. G. Bashmakova and 0. S. Smirnova, The Beginnings and Evolution of Algebra, Englishtranslation by A. Shenitzer, MAA, Washington, DC, 1999.

3. L. E. Dickson, Introduction to the Theory ofAlgebraic Equations, Wiley, New York, 1903.Reprinted in Congruence of Sets and Other Monographs, Chelsea, New York, 1967.

4. W. Dunham, Euler: The Master of Us All, MAA, Washington, DC, 1999.

5. H. M. Edwards, A note on Galois theory, Arch. Hist. Exact Sci. 41 (1990), 163—169.

6. H. M. Edwards, Divisor Theory, Birkhäuser, Boston, Basel, Berlin, 1990.

7. H. M. Edwards, Essays in Constructive Mathematics, Springer, New York, Berlin, Hei-delberg, 2004.

8. H. M. Edwards, Kronecker's arithmetical theory of algebraic quantities, Jber. d. Dt.Math.-Verein. 94(1992), 130—139.

9. H. M. Edwards, Kmnecker's views of the foundations of mathematics, in The History ofModem Mathematics, Volume I, edited by D. E. Rowe and J. McCleary, Academic Press,San Diego, CA, 1989.

10. L. Euler, Elements of Algebra, English translation by J. Hewlett, Springer, New York,Berlin, Heidelberg, 1984. Reprint of 1840 English edition.

11. B. M. Kiernan, The development of Galois theory from Lagrange to Artin, Arch. Hist.Exact Sci. 8(1971), 40—154.

12. I. Radloff, Evariste Galois: principles and applications, Historia Math. 29 (2002), 114—137.

13. L. Rigatelli, Evariste Galois 1811—1832, Springer, New York, Berlin, Heidelberg, 2000.

14. D. E. Smith, A Source Book in Mathematics, Volume One, Ginn, Boston, New York, 1925.Reprint by Dover, New York, 1959.

15. B. L. van der Waerden, History ofAlgebra, Springer, New York, Berlin, Heidelberg, 1985.

16. H. Wussing, The Genesis of the Abstract Group Concept, English translation by A.Shenitzer, MIT Press, Cambridge, MA, 1984.

CHAPTER 13

COMPUTING GALOIS GROUPS

Galois groups are not easy to compute. As Galois says in the "Discours préliminaire"to his first memoir on Galois theory [Galois, p. 39]:

If now you give me an equation that you have chosen at will, and about whichyou want to know if it is or is not solvable by radicals, I cannot do any morethan indicate the means for answering your question, without wanting to chargeeither myself or any other person with doing it. In a word, the calculations areimpractical.

Even with the aid of modern computers, it is not easy to compute the Galois groupof a polynomial of large degree (currently 50 or higher) unless the polynomial hassome special structure.

This chapter will explore some (but not all) ways of computing Galois groups ofarbitrary polynomials; Chapters 14 and 15 will describe special classes of polynomialsfor which it is possible to say more about the Galois group.

13.1 QUARTIC POLYNOMIALS

In Section 7.4 we explained how to compute the Galois group of a monic irreducibleseparable cubic polynomial f over a field F of characteristic different from 2. Recall


358 COMPUTING GALOIS GROUPS

that up to isomorphism the only possibilities for the Galois group are Z/3Z and S3,and that these cases are distinguished by whether or not the discnminant (f) is thesquare of an element of F.

In this section we will prove a similar result for a monic irreducible quarticpolynomial f E F [x], where F has characteristic 2. Note that f is necessarilyseparable by Lemma 5.3.5. We will write f in the form

(13.1) f=x4—c1x3+c2x2—c3x+c4, c1,c2,c3,c4EF.

In computing the Galois group off, the key players will be the discriminant off

A(f) = 144 + — —

(132)—

+ —

computed by the methods of Section 5.3, and the Ferrari resolvent of f

(13.3) Oj(y)

Below we explain how (13.3) relates to the theory developed in Section 12.1.The Galois group of f is Gal(L/F), where L is a splitting field of f over F.

Proposition 6.3.1 implies that there is a subgroup G C 54 such that

(13.4) Gal(L/F) G C S4.

Since we only need Gal(L/F) up to isomorphism, we can focus on the subgroup G.But G depends on how we label the roots. In Exercise I you will show that if wechange the labels, then G gets replaced by a conjugate subgroup in S4. Since theroots are intrinsically unlabeled, our goal is to compute G up to conjugacy.

Theorem 13.1.1 Let F have characteristic 2, and f E F[x] be monic and irre-ducible of degree 4. Write f as in (13.1), and let and Of(y) be defined as in(13.2) and (13.3). Then the subgroup G from (13.4) is determined as follows:(a) If Of(y) is irreducible over F, then

G_JS4,— 1A4, if E F2.

(b) If Of(y) splits completely over F, then

Furthermore, Of(y) splits completely over F and only if it is reducible over Fand E F2.

QUARTIC POLYNOMIALS 359

(c) If 01(y) has a unique root in F, then

and

G is conjugate to or 4/3 + — 4c2 = 0 and

((1324)) 7L/4Z, otherwise,

where D8 is the dihedral group of order 8. Furthermore, Oj- (y) has a unique rootin F if and only if it is reducible over F and F2.

Proof: In Section 12.1, we defined the universal Ferrari resolvent to be

9(y) = (y (X1X2 +X3X4)) (y — (x1x3 +x2x4)) (y — (xix4 +x2x3))

1ff has roots a a2, a3, a4 E L, then the evaluation map x1 '—* a, takes '—+ c1 andhence takes 0(y) to the Ferrari resolvent Oj(y) of f defined in (13.3). It follows thatthe roots of Oj(y) are

(13.5) a1a2+a3a4, a1a3+a2a4, ala4+a2a3.

In particular, Of(y) splits completely in L.Using this, we can now prove part (a). Since f and are monic and irreducible

over F, they are the minimal polynomials over F of a1 and at a2 + a3a4, respectively.By the Tower Theorem, we see that [L F} is divisible by 12, so that GI = IGal(L/F)

I

is also divisible by 12, since F C L is Galois. In Exercise 2 you will show that A4is the only subgroup of S4 with 12 elements. Thus the hypothesis of part (a) impliesthat G = A4 or S4. Then we are done by Theorem 7.4.1.

Before proving parts (b) and (c), we first observe that and f have the samediscriminant, i.e.,

(13.6) =

In the universal case, this was proved in Exercise 9 of Section 2.4. In Exercise 3 youwill explain why this implies (13.6). Since f is separable, we conclude that is aseparable cubic.

Now suppose that is reducible over F. Since is a cubic, this implies that ithas a root /3 e F. By (13.5), we may relabel the roots of f so that

(13.7) /3=aia2+a3a4EF.

As explained earlier, relabeling the roots of f replaces G with a conjugate subgroup.We will analyze how (13.7) affects the Galois group. What follows is a special

case of a general phenomenon that will play a central role in Sections 13.2 and 13.3.We claim that (13.7) implies that

Gc ((1324),(12))


The rough idea is that the Galois group shrinks when a resolvent has a root in F.To prove our claim, suppose that e Gal(L/F) corresponds to r E G C S4. Since

E F, one easily computes that

= o-(8) =

ifrE((1324),(12)),= ifrE (23)((1324),(12)),

ifTE (24)((1324),(l2)),

where

S4 ((1324),(12))u(23)((l324),(l2))u(24)((l324),(12))

is the decomposition of S4 into left cosets of ((1324), (12)). Since is separable,this implies that G C ((1324), (12)) as claimed.

As in part (a), we know that 4 divides IGI, since f is irreducible. It follows thatI

= 4 or 8. Furthermore, we found all subgroups of D8 when we worked out theGalois correspondence for Q c Q(i, in Section 7.3. In Exercise 4 you will usethis to show that G is one of the four groups

(13.8) ((l2),(34)), ((12)(34),(13)(24)), ((1324)), ((1324),(l2)).

Since f is irreducible, Proposition 6.3.7 implies that G is a transitive subgroup of S4(reread Section 6.3 if you've forgotten what transitive means). The first group listedin (13.8) is not transitive, so that G is one of the remaining three groups. Parts (b)and (c) of the theorem describe how we distinguish among these possibilities.

We begin with part (b). Since F has characteristic 2, Exercise 5 implies that amonic reducible cubic g E F[x] splits completely over F if and only if E F2.By (13.6), we conclude that if is reducible over F, then it splits completely over Fif and only if E F2. This proves the final assertion of part (b). Also, whensplits completely over F, Theorem 7.4.1 and E F2 imply that G C A4. Of thegroups in (13.8), only ((12)(34), (l3)(24)) Z/2Z x Z/2Z lies in A4. This provespart (b). No conjugacy is needed, since ((12)(34), (13)(24)) is normal in S4.

The final assertion of part (c) follows from the final assertion of part (b). Nowsuppose that /3 e F is a root of and that The last condition impliesthat G A4, so G must be one of the last two groups of (13.8). Our method fordistinguishing these begins with Euler's formula (12.17) for the roots of the universalquartic f. This formula involves the square root

Yi =xlx2+x3x4,

which is related to the roots XI ,x2,x3,x4 of f via

4Y1 —402 = (xi +X2 X3 X4)


(see the discussion leading up to (12.12)). If we apply the evaluation map x '—÷ a,and use (13.7), then we obtain

(13.9) 4/3-I-cf —4C2 = (at +a2—a3 —a4)

It follows that

(13.10) —4c2) = +Q2 —a4) EL.

Now suppose that —4C2 0, i.e., —4C2 E F*. If G = ((1324)),then Gal(L/F) has a generator a that maps to (1324). One easily computes that

a fixes (13.10), and since a generates the Galois group, we conclude

that + — 4C2) E F. Thus + — 4c2) E (F*)2.

On the other hand, if G = ((1324),(12)), then some a e Gal(L/F) maps to (12).For this a, we have

= and a(ai +az—a3—a4) =aI +a2—a3—a4.

Hence a takes (13.10) to its negative. Since F has characteristic 2 and (13.10) is

nonzero, we have + — 4c2) F. Thus + — 4c2) (F*)2.

The above argument fails when 4fl + — 4c2 = 0 (be sure you see why). In thiscase, we will use

(13.11) /32—4c4=(aIa2+a3a4)2—4a1a2a3a4=(ala2—a3a4)2.

In Exercise 6 you will show that + — 4c2 =0 implies that — 4c4 e F*. Then,arguing as above, one easily sees that

— 4c4) = — a3a4)

if and only if G = ((1324), (12)) (see Exercise 6). This completes the proof.

We now give some examples of Theorem 13.1.1.

Example 13.1.2 Consider f = x4 — 4x2 + x + 1 E One can show that f isirreducible of discriminant = 1957 = 19. 103 and that its resolvent

Or(y) =y3+4y2—4y— 17

is irreducible over Q. By Theorem 13.1.1, the Galois group of f is S4, so that thesplitting field off has degree 24 over Q. This has the following consequences:

• In Example 8.6.7 of Section 8.5, we used f as an example of an irreduciblepolynomial of degree 4 whose roots are all real yet cannot be expressed by real


radicals. This follows from Theorem 8.6.5, since the degree of the splitting fieldover Q is not a power of 2.In Example 10.1.13 of Section 10.1, we used f as an example of a polynomial ofdegree 4 whose roots are not constructible. This follows from Theorem 10.1.12,since the degree of the splitting field over Q is not a power of 2.

In Exercises 7 and 8 you will apply Theorem 13.1.1 to other quartic polynomialsfrom earlier in the text.

Our next example is taken from [Chebotarev, p. 253].

Example 13.1.3 Suppose that f = x4 + ax3 + bx2 + ax + 1 E F[x] is irreducible,where F has characteristic 2. In this case, the resolvent is

Of()?) =y3 —by2+(a2—4)y—2a2+4b

= (y—2)(y2+(2—b)y+a2—2b),

and the discriminant is

= (4b—a2—8)2(b—2a+2)(b+2a-l-2).

By Theorem 13.1.1, it follows that the Galois group off is Z/2Z x 7L/2Z if and onlyif(b—2a+2)(b+2a+2)—— (b+2)2 —4a2 F2.

The above factorization of is easy to find using Maple or Mathematica.In Exercise 9 you will show that the factor y — 2 of (y) arises naturally from thesymmetry of f. <11>

Here is an example taken from [18] that illustrates part (c) of Theorem 13.1.1.

Example 13.1.4 Assume that f = x4 + bx2 + d E F hascharacteristic 2. Also assume that d F2. In Exercise 10 you will show that! hasdiscriminant

= l6d(b2—4d)2

and resolvent

Oj-(y) =y3 —by2 —4dy+4bd= (y—b)(y2—4d).

Since d F2 and is reducible, part (c) of Theorem 13.1.1 applies. Using+ — 4c2 = 4b + 02 — 4b = 0 and /32 — 4C4 = b2 — 4d, we see that the Galois

group off is D8 if d(b2 — 4d) and 7L/4Z otherwise.See [18] for an analysis of what happens when d E F2.

In Section 13.3, we will give a version of Theorem 13.1.1 that works for all fields,not just those of characteristic 2. To prepare for this, we need the criterion from[18] for distinguishing between Z/4Z and D8 in part (c) of Theorem 13.1.1.


Proposition 13.1.5 As in Theorem 13.1.1, let! = x4 — Cl X3 +C2X2 —c3x+c4 E F[x]and assume that F has characteristic 2. Also assume that the Ferrari resolventGj(y) has a unique root /3 e F. Then the Galois group off is isomorphic to eitherZ/4Z or D8, and the former occurs and only jf(y2 — cly+ C2 /3)(y2 — fly+splits completely over F

Proof: We assume the same setup as the proof of part (c) of Theorem 13.1.1. Webegin with an observation about quadratic polynomials. Let g = y2 + Ay + B E F [x]

be such that either = 0 or g is irreducible over F. Also let F C be aquadratic extension where a e F. In part (a) of Exercise 12, you will show that

(13.12) gsplits completely

We next observe that h1 = y2 — c1y + c2 — has discriminant

where the last equality is (13.9). Since has a unique root /3 = ala2 +in F, the proof of part (c) of Theorem 13.1.1 gives a- E Gal(L/F) that maps to(1324) E S4. Then + a2 — a3 — cr4) = —(a1 + a2 — a3 — a4), which impliesthat either 0 or h1 is irreducible over F. Similarly, h2 = y2 — fly + c4 hasdiscriminant = /32

— and using a- and(13.1l), we see that either = 0

or h2 is irreducible over F. Then (13.12) implies that

(13 13)h1 h2 splits completely over F

If h1h2 splits completely over then (13.13) and part (c) of Theo-rem 13.1.1 imply that Gal(L/F) 7L/47Z. Conversely, if Gal(L/F) Z/4Z, thenL contains a unique quadratic extension of F, which must be F since

F2 by part (c) of the theorem. Since h1 and h2 split completely over L bypart (b) of Exercise 12, they split over quadratic extensions of F contained in L.Hence h1 h2 splits completely over F •

The proof of Proposition 13.1.5 is based on [221. Another method for handlingpart(c) of Theorem 13.1.1 is described in Exercise 13.

Mathematical Notes

The text contains several ideas that will be developed in subsequent sections. Hereare some remarks to help us see what is involved.

Transitive Permutation Groups. We noted in the proof of Theorem 13.1.1 thatG is a transitive subgroup of S4, since f is irreducible. Transitive subgroups ofwill play a prominent role in this chapter and the next. For example, implicit inTheorem 13.1.1 is the following classification of the transitive subgroups of S4.


Theorem 13.1.6 Up to conjugacy, the transitive subgroups of are

S4, A4, ((1324),(12)), ((1324)), ((12)(34),(l3)(24)).

Proof: Let G C 54 be a transitive subgroup. If we can prove that G arises from theGalois action on the roots of some monic irreducible quartic over a field of charac-teristic 2, then G is conjugate to one of the above five groups by Theorem 13.1.1.

We will find the desired quartic polynomial using the methods of Section 7.4.When F Q, the universal extension in degree 4,

K=Q(ai,a-2,cr3,o-4) CL=Q(xI,x2,x3,x4),

is the splitting field of the universal quartic f = a1X3 + cr2x2 — a3x + Givena transitive subgroup G C 54 Gal(L/K), the corresponding fixed field

KcMcL

satisfies Gal(L/M) G. Also observe that L is the splitting field over M of f andthat f is irreducible over M since G is transitive (this is Proposition 6.3.7). As notedabove, we are now done by Theorem 13.1.1. •

In practice, a standard strategy for computing Galois groups is the reverse of whatwe did in this section. When considering irreducible polynomials f E F[X} of degreen, one first finds all transitive subgroups of up to conjugacy and then, for eachsuch subgroup, determines criteria for the Galois group of f to be conjugate to thatsubgroup. This is the approach we will use for the quintic in Section 13.2.

• Resolvents. The general theory of resolvents is based on the ideas of Lagrangediscussed in Section 12.1. For 91(y), recall from the proof of Theorem 13.1.1 that webegan in the universal case with xlx2 +X3X4 and constructed the Ferrari resolvent O(y)of the universal quartic f. Then specializing to f gave Of(y). All of the resolventsconsidered in this chapter will be constructed similarly.

Besides Of(y), Theorem 13.1.1 needs to know whether or not E F2. Thiscan be stated in terms of the resolvent polynomial y2 (f), since E F2 if andonly if y2 — has a root in F.

In general, if a resolvent has a root in F, then this puts strong restrictions on theGalois group. For example, if y2 — has a root in F, then the group G liesin A4 (be sure you understand why), and if 91(y) has a root in F, then the proof ofTheorem 13.1.1 shows that some conjugate of G lies in ((1324), (12)). By combininginformation from different resolvents, we can obtain precise information about theGalois group. We will pursue these ideas in Section 13.3.

We can also use resolvents to explain part (c) of Theorem 13.1.1. This part of thetheorem says that if /3 E F is a root of and then the Galois group isD8 when /3 satisfies the condition that either

0 and —4c2)


or

and

The first part of the condition implies that y2 + — 4C2) has no roots inF. Because of the appearance of /3, we will call y2 — (4/3+ — 4C2) a relativeresolvent in Section 13.3. Similarly, the second part of the condition can be stated interms of the relative resolventy2 — —4c4.

Furthermore, (f) (4/3+ — 4c2) and (/32 — 4c4) are nonzero if and onlyif the corresponding relative resolvents have simple roots (as defined in Section 5.3).We will see in Section 13.3 that simple roots of resolvents or relative resolvents areneeded in order to get useful information about the Galois group.

• Diophantine Equations. So far, we have always had a fixed polynomial whoseGalois group we wanted to determine. But if we let the coefficients of the polynomialvary, then the criteria of Theorem 13.1.1 lead to some interesting equations. Webegin by revisiting an earlier example.

Example 13.1.7 Let f = x4 + ax3 + bx2 + ax + 1 E Q[xI and assume that f is irre-ducible over Q. By Example 13.1.3, the Galois group of f over Q is Z/2Z x Z/2Zif and only if (b + 2)2 — 4a2 E Q2. The latter is equivalent to saying that

(13.14) (b+2)2—4a2=c2

for some c E Q. If we write this as

4a2+c2 = (b+2)2,

then f = x4 + ax3 + bx2 + ax + 1 E Q[x] has Z/2Z x Z/2Z as Galois group if andonly if there is c Q such that (x,y,z) = (2a,c,b+2) lies on the cone

x2 +y2 = z2.

Hen we have an equation where we only want solutions whose coordinates all liein ç This is an example of a Diophantine equation. Such equations are an important

of number theory.In Exercise 11 you will show that if f = x4 + ax3 + bx2 + ax + 1 is irreducible

with positive integer coefficients, then the Galois group is Z/2Z x Z/2Z if and onlyif there is c > 0 in Z such that (2a, c, b +2) is a Pythagorean triple, i.e, the integers2a, c, b +2 are the sides of a right triangle with b + 2 as hypotenuse.

Here is a polynomial that leads to a more sophisticated equation.

Example 13.1.8 Assume that f = x4 + x + b E 7L[xI is irreducible over Q. Thispolynomial has discnminant

= 256b3 —27


and resolvent01(y) =y3 —4by— 1.

It is easy to see that 0J(y) is irreducible over Q, since its only possible rational rootsare ± 1 (be sure you can explain why). Thus the Galois group over Q is either 54 orA4, and the latter happens if and only if 256b3 —27 E Q2. In fact, we can replace Q2with Z2 since b E 7L. Thus the Galois group is A4 if and only if there is c E Z suchthat the point (x,y) = (b,c) lies on the curve

y2 =256x3—27.

This is an example of an elliptic curve. A famous theorem of Siegel asserts that suchan equation has at most finitely many integer solutions (see [31]). So there are atmost finitely many integers b such that f = x4 + x + b has Galois group A4 over Q.

This example can be extended in several ways. First, one could allow b to be arational number. Then one seeks rational points on the above elliptic curve. Someof the deepest conjectures in number theory involve rational points on elliptic curves(see [31] for an introduction). Another direction would be to consider polynomialsx4 + ax+ b E Z[xI with Galois groupA4. This problem is solved in [35] using methodsfrom algebraic number theory.

Historical Notes

The first person to give a systematic method for finding the Galois group of aquartic was F. Hack, in his unpublished 1895 dissertation. Many books and papershave addressed this problem—see the references in [18], to which one can add[Escofier] and [Garling]. Our version of Theorem 13.1.1 is based on [17].


Exercise 1. Let f E F{x] be separable of degree n, and let cXI,. , a,, be the roots off in asplitting field F c L off. In Section 6.3 we used the action of the Galois group on the roots toconstruct a one-to-one group homomorphism 4)1 : Gal(L/F) —* S,,. Now let . . , f3, be thesame roots, possibly written in a different order. This gives 4)2 Gal(L/F) —+ Sn. To relate 4)1

and 4)2, note that there is 'y E Sn such that j3, = for I <i < n. Now define the conjugationmap : Sn -4 Sn by =(a) Prove that 4)2(b) Let G C be the image of 4)1. Explain why part (a) justifies the assertion made in the

text that "if we change the labels, then G gets replaced with a conjugate subgroup."

Exercise 2. Prove that A4 is the only subgroup of S4 with 12 elements.

Exercise 3. Explain carefully why (13.6) follows from Exercise 9 of Section 2.4.

Exercise 4. Use Example 7.3.4 from Chapter 7 to show that (13.8) gives all subgroups ofK(1324),(12)) oforder4or8.

Exercise S. Let F be a field of charactenstic 2, and let g E F[x] be a monic cubic polynomialthat has a root in F. Prove that g splits completely over F if and only if E F2.


Exercise 6. This exercise is concerned with the proof of part (c) of Theorem 13.1.1. Letf(x) = X4 CiX3 + C2X — C3X + C4 as in the theorem.(a) Suppose that f has roots ai such that a! +a2 — a3 — a4 = a!a2 — a3a4 = 0.

Prove that f is not separable.

(b) Let be a root of the resolvent Oj(y). Use part (a) to prove that + — 4c2 and /32 — 4c4

can't both vanish when f is separable.(c) Suppose that 4/3 + — 4c2 = 0 in part (c) of Theorem 13.1.1. Prove carefully that G is

conjugate to if and only if

Exercise 7. In Exercise 18 of Section 12.1 you found the roots of f =x4 + 2x2 — 4x + 2 E Q[x]using the formulas developed in that section. At the end of the exercise, we said that "thisquartic is especially simple." Justify this assertion using Theorem 13.1.1.

Exercise 8. In Example 10.3.10, we showed that the roots off = 7m4 — 16m3 — 2 1m2 + 8m +4 E Q[m] can be constructed using origami. Show that the splitting field of f is an extension ofQ of degree 24. By the results of Section 10.1, it follows that the roots of f are not constructiblewith straightedge and compass, since 24 is not a power of 2.

Exercise 9. As in Example 13.1.3, let! = x4 +ax3 +bx2 +ax+ 1 E F[x], and let a be a rootoff in some splitting field off over F. Show that a root off, and then use (13.5)to conclude that 2 is a root of the resolvent Oj(y).

Exercise 10. As in Example 13.1.4, let f = x4 + bx2 + d E F [xl, where d F2. Computeand Oj(y).

Exercise 11. In Example 13.1.7 we showed that if f = x4 +ax3 +bx2 +ax+ I E 7L[x] isirreducible over Q, then its Galois group is Z/27Z x Z/2Z if and only if there is c E Q suchthat 4a2 +c2 =(b+2)2.(a) Show that c e Z, and use the irreducibility of f to prove that c 0. Hence we may

assume that c> 0, so that (2a, c, b + 2) is a Pythagorean triple.(b) Show that 32 + 42 = + 122 = 132, 72 + 242 = 252, and 82 + 152 = 172 give two

examples of polynomials with 7Z/2Z x Z/2Z as Galois group (two of the triples givereducible polynomials).

Exercise 12. This exercise is concerned with the proof of Proposition 13.1.5.(a) Prove(13.12).(b) Prove that the two polynomials h1 and h2 defined in the proof of the proposition factor as

hI=(y—(cI+c2))(y—(a3+a4))andh2=(y—ala2)(y—c53a4).

Exercise 13. Suppose that f E F[x] satisfies the hypothesis of part (c) of Theorem 13.1.1,and let a be a root off. Prove that G 7Z/4Z if f splits completely over F(a), and G D8

otherwise. This gives a version of part (c) that doesn't use resolvents. Since we can factorover extension fields by Section 4.2, this method is useful in practice.

Exercise 14. Use Theorem 13.1.1 to compute the Galois groups of the following polynomialsin Qixi:(a) x4+4x+2.(b) x4+8x+12.(c) x4+l.

(d) x4+x3+x2+x+l.(e) x4—2.


Exercise 15. In the situation of Theorem 13.1.1, assume that has a root in F. In theproof of the theorem, we used (13.5) and (13.7) to show that G is conjugate to a subgroup ofD8. Show that the weaker assertion that IGI = 4 or 8 can be proved directly from (12.17).

Exercise 16. Consider the subgroups ((12),(34)) and of S4.

(a) Prove that these subgroups are isomorphic but not conjugate. This shows that whenclassifying subgroups of a given group, it can happen that nonconjugate subgroups canbe isomorphic as abstract groups.

(b) Explain why the subgroup isn't mentioned in Theorems 13.1.1 and 13.1.6.

13.2 QUINTIC POLYNOMIALS

Polynomials of degree 5 have a richer Galois theory than those of degree 4. Thereare some obvious reasons for this: The computations are more complicated becausethe degree is higher, and the groups are more complicated because they need not besolvable. The surprise is that quintic equations have strong relations to many otherareas of mathematics, including:

• Geometry. The rotational symmetry group of the icosahedron is A5, and geomet-rically defined invariants of this group action have consequences for the quintic.

• Iteration. The "Galois theory" of Newton's method for solving polynomialequations due to Doyle and McMullen [9] uses A5 in a crucial way.

• Elliptic Functions. These functions arise in complex analysis and number theory,yet can also be used to find roots of quintics that can't be solved by radicals.

Because of such connections, quintic equations are the subject of entire books, inparticular those by King [19], Klein [20], and Shurman [30].

The aims of this section are more modest. We will focus on computing the Galoisgroup of a quintic and in particular on detenrnning when a quintic is solvable byradicals. As we will see, this will involve some substantial mathematics.

A. Transitive Subgroups of S5. If a quintic polynomial f e F[x] is separableand irreducible, then its Galois group Gal(L/F) (where F C L is a splitting field)is isomorphic to a transitive subgroup of S5. So we begin by determining thesesubgroups up to conjugacy. Our first result is elementary.

Lemma 13.2.1 Let G c be a subgroup. Then the following are equivalent:(a) G is transitive.(b) GI is divisible by 5.

(c) G contains a 5-cycle.

Proof: For (a) (b), recall that the order of an orbit divides the order of the group,by the Fundamental Theorem of Group Actions (Theorem A.4.9). Then we are done,since transitivity implies that { 1,2,3,4,5 } is an orbit of the action of G.

The implication (b) (c) is proved using the argument given in the discussionfollowing (6.8) in Section 6.4. Finally, for (c) (a), note that repeatedly applying

OUINTIC POLYNOMIALS 369

the five cycle (i1 i4 i5) to i1 gives

Transitivity follows immediately, since {i1,i2,i3,i4,i5} = {l,2,3,4,5}. •

It turns out that we already know most of the transitive subgroups of 55 up toconjugacy. More precisely, we have the following subgroups:

• The full symmetric group 55 of order 120.• The alternating group A5 of order 60.• The cyclic group ((12345)) of order 5.• By Section 6.4, the one-dimensional affine linear group AGL( 1, IF1,) is the group

of order p(p 1) consisting of maps i '—÷ ai + b where i, a, b E IF,, and a 0. Ifwe set p 5 and regard { 1,2,3,4,5 } as congruence classes modulo 5, then

AGL(l,1F5) CS5

is a subgroup of order 20. In particular, translation by 1 (i i + 1) is the 5-cycle(12345) and multiplication by 2 (i 2i) is the 4-cycle (1243). Be sure youunderstand this.

In Exercise 1 you will show that AGL(1,1F5) is generated by (12345) and (1243).Furthermore, (12345) is an even permutation while (1243) is odd. Hence

(13.15) AGL(l,1F5)flA5

is a proper subgroup of AGL(1,1F5) containing ((12345)). The group (13.15) alsocontains (1243)2 = (14) (23) (multiplication by 4—do you see why?). In Exercise 1you will show that

AGL(l,1F5)nA5 = ((12345),(14)(23))

where D10 is the dihedral group of order 10.The subgroup (13.15) and the four subgroups described in the bullets give five

subgroups of S5, all transitive by Lemma 13.2.1 since their orders are divisible by 5.These groups fit together in the diagram

S5

AGL(1,F5) A5

(13.16) t /AGL(1,IF5)flA5

t((12345))

We now classify transitive subgroups of S5 up to conjugacy.


Theorem 13.2.2 Every transitive subgroup G C S5 is conjugate to one of the groupsin the diagram(l3.16).

Proof: By Lemma 13.2.1, G contains a 5-cycle. Hence, replacing G by a conjugateif necessary, we may assume that (12345) e G. The key idea of the proof is toconsider the number of cyclic subgroups of order 5 contained in G.

First suppose that ((12345)) is the only such subgroup of G. Then we haveg e G. In the language of the Mathematical

Notes to Section 7.2, this means that G is contained in the normalizer

(((12345))) = {g S5I

g((12345))g' = ((12345))}.

In Section 14.1 we will more generally consider the normalizer

..p))) = {g E Si,, I g((12...p))g1 =

where p is now any prime, and we will prove in Lemma 14.1.2 that

This is part of Galois's brilliant analysis of which irreducible polynomials of primedegree are solvable by radicals. Rather than repeat the argument here, we will simplyassume this result from Chapter 14.

It follows that if ((12345)) is the only subgroup of G of order 5, then G is asubgroup of AGL( 1, IF'5). In Exercise 1 you will show that this implies that G is oneof the groups

((12345)), AGL(1,1F5) flA5, or AGL(1,]F5).

It remains to consider what happens when G contains more than one subgroupof order 5. In Exercise 2 you will prove that ((12345)) is a 5-Sylow subgroup ofG. By the Third Sylow Theorem (see Theorem A.5. 1 in Appendix A), the numberof 5-Sylow subgroups of G is congruent to 1 modulo 5. Since we have more thanone, we must have at least 6. Furthermore, each 5-Sylow subgroup has four 5-cycles,and any two such subgroups intersect in the identity. Since we have at least sixsuch subgroups, G must contain at least twenty-four 5-cycles. Yet has exactlytwenty-four 5-cycles by Exercise 2. It follows that G contains all 5-cycles.

We are almost done. The easily verified identity

(ijklm)(ijmlk) = (ikj), {i,j,k,l,m} = {l,2,3,4,5},

shows that G contains all 3-cycles. We know from Section 8.4 that A5 is generatedby 3-cycles. Hence G contains A5, so that G is A5 or S5. This completes the proof..

In Exercise 3 we will give a more elementary version of the above argument thatdoesn't use the Third Sylow Theorem.

As a corollary of Theorem 13.2.2, we get the following criterion for an irreduciblequintic to be solvable by radicals.

QUINTIC POLYNOMIALS 371

Corollary 13.2.3 Assume that f E F [x] is irreducible of degree 5 and that F hascharacteristic 0. Then f is solvable by radicals over F and only if its Galois groupover F is isomorphic to a subgroup of AGL( 1, F5).

Proof: By Theorem 8.5.3, f is solvable by radicals if and only if its Galois groupis solvable. The Galois group is isomorphic to a subgroup of S5, but A5 and S5 aren'tsolvable, by Theorem 8.4.5. So one direction of the corollary follows immediatelyfrom Theorem 13.2.2. For the other direction, we note that AGL( 1 ,Fs) and hence allof its subgroups are solvable by Example 8.1.6 from Section 8.1. •

Section 14.1 will discuss Galois's generalization of Corollary 13.2.3 in which 5 isreplaced by an arbitrary prime p.

B. Galois Groups of Quintics. Letf E F[x] be monic, separable, and irreducibleof degree 5, where F is a field of characteristic 2. We will determine the Galoisgroup off over F using a discriminant, a sextic resolvent, and a factorization.

The discrin-iinant is the usual discnminant and the factorization will bedescribed in the statement of Theorem 13.2.6. So let us turn our attention to thesextic resolvent. The idea is to find a polynomial

hEFki,...,xsl

with the property that

AGL(l,F5) = {o E a.h = h}.

Thus h should have AGL( 1 ,F5) as its symmetry group. We will use the polynomial

h = u2

where

(13 17)x1x2+x2x3+x3x4+x4x5+x5x1

—XIX3—X3X5—X5X2—X2X4—X4XI.

The signs are best explained using the diagram

2(13.18)

In the formula (13.17) for u, the coefficient of (where i f) is —l if i and fareconnected by a line segment in (13.18) and +1 otherwise. Since (12345) rotates thestar by 2ir/5 radians, it follows that (12345) . u = u.

4 3


On the other hand, (1243) takes (13.18) to the diagram

5

2

Here, i and j are connected by a line segment if and only if they are not connected in(13.18). Hence (1243) interchanges all signs, so that (1243) .u = —u.

It follows that h u2 is fixed by (12345) and (1243). Since these generateAGL(l,F5), we see that h is invariant under AGL(1,1F5). This is the full symmetrygroup of h, as we now prove.

Lemma 13.2.4 Leth = u2, where u is defined in (13.17). Then

AGL(1,1F5) = {aES5 Iah=h}

Pmof: LetG = {a eSs I a•h h}. ThenAGL(1,]Fs) C Gbytheaboveargument.If G were strictly bigger than AGL( 1 then (13.16) would show that G = S5, thatis, h would be symmetric. However, observe that

h=(xjx2+x2x3+•..—xix3—...)2

= + — — +

where terms involving x4 or x5 are not shown. This makes it easy to see that h is notsymmetric. Thus G which implies that G = AGL(1,1F5). .

By Exercise 4, left coset representatives of AGL( 1, F5) in Ss are

e, (123),(234), (345), (145),(125).(13.19)

Thus the orbit of Ss acting on h consists of

h1 =e•h=h, h2 = (123).h, h3 = (234).h,

h4=(345).h, h5=(145).h, h6=(125).h

(be sure you can explain why). This enables us to form the universal sextic resolvent

(13.20) O(y) = — h,).

The methods of Section 12.1 imply that 0(y) has coefficients in F [a', a2, a3, a4, as].

31'


Our given monic separable irreducible quintic can be written as

f = — Ci X4 + C2X3 — C3X2 + C4X — C5 E F[x].

Let its splitting field be F CL, and let be the roots of f in L. Underthe evaluation map x, '—* we know that '—* c, E F. Thus (13.20) maps to thesextic resolvent of f

(13.21) = [J(y_j3j) e

where= EL.

The structure of 9j-(y) is described by the following proposition, whose proof wedefer until later.

Proposition 13.2.5 Given f E F [x] as above, its sextic resolvent can be written

0j(y) = (y3 +b2y2 +b4y+b6)2

where b2,b4,b6 E F.

The Galois group off is Gal(L/F) for a splitting field L off over F. Also recallthat Gal(L/F) G C S5. where G is transitive. Here is our main result.

Theorem 13.2.6 Assume that f F [xJ is monic, separable, and irreducible of degree5 and that F is afield of characteristic 2. Then the subgroup G C S5 defined abovehas the following properties:(a) G CA5 and only E F2.

(b) G is conjugate to a subgroup of AGL( 1, IF5) if and only if the sextic resolventOJ-(y) defined in (13.21) has a root in F.

(c) G is conjugate to ((12345)) if and only if f splits completely over F(a), whereis a root off.

Proof: Part (a) follows from Theorem 7.4.1 since F has characteristic 2. Also,part (c) is relatively straightforward and is left as Exercise 5.

It remains to prove part (b). If G is conjugate to a subgroup of AGL( 1, IF5),then relabeling the roots if necessary, we may assume that G C AGL( 1, IF5). Let anarbitrary a E Gal(L/F) correspond to r E G. Then

=== (T h)(ai, a2, a3 , a4, a5) = h(a1,a2, a3,a4, a5) = flu,


where the last line follows from r E G C AGL(l,1F5) and Lemma 13.2.4. SinceF C L is a Galois extension, we must have i3i E F. Thus 61(y) has a root in F.

Conversely, suppose that Of(y) has a root in F. If G is conjugate to a subgroup ofAGL(l,F5), then we are done. So assume that this is not true. Since G is transitive,Theorem 13.2.2 implies that A5 C G. Let TI,...,T6 denote the 3-cycles in (13.19)that satisfy h = and let e Gal(L/F) map to Ti. The existence of a, followsfrom A5 C G. Then arguing as above shows that

i=l,...,6

(be sure you can supply the details). By assumption, some E F, and then the aboveequations easily imply that = /36. Hence the sextic resolvent is

Oj(y) = (y—/3i)6.

Comparing this with Proposition 13.2.5, we obtain the identity

(13.22) (y—fii)6

where /91,b2,b4,b6 E F. Multiplying this out and comparing the coefficients of y5,y4, andy3 gives the equations

= 2b2,

2b2b4 + 2b6.

In Exercise 5 you will verify these equations and use them to show that

b2=—3/3,, b4=3/3?,

since F has characteristic 2. Then (13.22) becomes

= (y3 — 3/3iy2 + — /33)2—

=

Hence 2wA(f) = 0. Yet F has characteristic 2, and 0, since f is separable.This contradiction completes the proof of the theorem. .

The structure of the sextic resolvent plays a crucial role in the above proof. So itremains to prove Proposition 13.2.5.

Proof of Propositi on 13.2.5: We will prove the proposition in the universal case andthen specialize. As in the proof of Theorem 13.2.6, let be the 3-cycle in (13.19)with Ti . h = h1. Then set

U1 = U,


where u is defined in (13.17). This gives the polynomial

9(y) = fJ(y — u,).

Using Maple or Mathematica and the methods of Section 2.3, one computes that

(13.23) 9(y) = y6 +B2y4 + B4y2 +B6 —

where = [L <i<j<5 (x, — is the square root of the discriminant, and

2B2= 8a1a3—3a2—20o4,

B4 = — + 160103 + + —

— + — +240a1a2a5 —

B6 8a4cr3 — — + —

(13.24) — + —

— 1 + 224

— — +

— — + +6400 1a2a4a5 — — +

You will do this computation in Exercise 6. The lovely ideas that underlie theformulas in (13.23) and (13.24) are explained in Exercises 7 and 8.

Since h, = 9(y) relates to the universal sextic resolvent 9(y) as follows:

= [J(y2 — = fl(y2

=fl(y_uj)(y+uj)=(_l)6fl(y_uj)(_y_uj)

= 9(y)9(-y).

Combining this with (13.23), we see that 0(y2) is the product

(y6 + B2y4 + B4y2 + B6 — (y6 + B2y4 + B4y2 + B6 +

which easily implies that

0(y2) = (y6 + B2y4 + B4y2 + B6)2 —

Replacing y2 with y, we obtain the universal formula

0(y) =

Then the evaluation ci, '—* c, gives the desired formula for Oj(y).


The proofjust given shows that Oj(y) = (y3 + b2y2 + b4y + b6)2 — y, whereb2, b4, b6 are obtained from (13.24) by replacing with c,. This will be useful in theexamples computed below.

We can also describe the irreducible factorization of 91(y) as follows.

Proposition 13.2.7 Letf E F[x] and G C S5 be as in Theorem 13.2.6. Then:

A5 C G 91(y) is irreducible over F,

G is conjugate to a Oj(y) = (y— /3)g(y), where E Fsubgroup of AGL(l,1F5) andg(y) E F[y] is irreducible over F.

Proof: You will prove this in Exercises 9 and 10. U

C. Examples. We first note that Theorem 13.2.6 and diagram (13.16) lead to thefollowing table for determining the Galois group Gal(L/F) G C S5:

Is Does Of(y) have Does f(x) split G up toin F2? a root in F? completely over conjugacy

No No S5

Yes No A5

No Yes AGL(l,F5)

Yes Yes No AGL(1,1F5)flA5

Yes Yes Yes ((12345))

In this table, denotes a root of f, and the dashes in the third column indicate

cases when the first two columns determine the Galois group. You will prove the

correctness of the table in Exercise 11.Here are three examples of how to use this table.

Example 13.2.8 In Section 6.4 we showed that the Galois group off = — 6x+3over Q is S5. We can verify this as follows. In Exercise 12 you will show thatf isirreducible with discriminant

= —1737531

and sextic resolvent

Oj-(y) = (y3+ 120y2+8640y—69120)2+2'°l737531y.

You will also show that is irreducible over Q. Since is not in Q2 thetable implies that the Galois group is

We will return to this example in Section 13.4. We next give an example that uses

the third column of the table.


Example 13.2.9 Consider f = —2 E In Exercise 13 you will showthat! is irreducible over with discriminant

= 50000 =


Of(y) = y6 — 2'°50000y = y6 — 2'455y.

This obviously has a root in Since is a square in the tabletells us that the Galois group is either cyclic of order 5 or dihedral of order 10. Todistinguish these, we need to check if f splits completely when we adjoin a root to

But f can't split completely over since its roots aren't all real.Hence the Galois group off over is AGL( 1, IF5) nA5

In Exercise 13 you will redo this example using results from Chapters 6 and 7.

Example 13.2.10 A quintic polynomial studied by De Moivre and Euler is

f = + px3 + + q e Q[x].

We will assume that! is irreducible over Q. In Exercise 14 you will show that f hasdiscriminant

— (4p5+3125q2)2 — (4p5+3125q2)23125 — 55


0j(y) = (y3 — + 1 lp4y + + 4000pq2)2 _2b0

You will also verify that Oj(y) has a root y = 5p2 E Q. Since Q2 (do you seewhy?), the table implies that the Galois group is AGL( 1, IF5).

In Exercise 14 you will give an elementary proof that the polynomial given inExample 13.2.10 is solvable by radicals.

D. Solvable Quintics. We first point out the following immediate consequenceof Corollary 13.2.3 and Theorems 13.2.6.

Corollary 13.2.11 Assume that f e is monic and irreducible of degree 5 andthat F has characteristic 0. Then f is solvable by radicals over F if and only if itssextic resolvent 01(y) has a root in F. •

As an application of this corollary, we will determine when an irreducible quinticof the form

f=x5-Fax+bEF[x]is solvable by radicals. We will assume that F has characteristic 0. Here is thesomewhat surprising result.


Theorem 13.2.12 Assume that f = x5 + ax + b E F[x] is irreducible, where a 0and F has characteristic 0. Then f is solvable by radicals over F and only if thereare A,/1 E F such that

—b—a— (Al)4(A26A+25)' —

Proof: In Exercise 15 you will show that f has discriminant

= 256a5 + 3125b4


Oj(y) = (y3 20ay2 + 240a2y+ 320a3)2 — 2'°(256a5 +3125b4)y.

By Corollary 13.2.11, f is solvable by radicals if and only if 9f(y) has a root in F.Since a 0, a root E F of Of(y) can be written /3 = aA for A E F. We can also

write b = for some F. With these substitutions, it follows that Gf(y) has aroot in F if and only if there is A E F such that

0= (aA)

((aA)3 20a(aA)2 + 240a2(aA) + — 2'°(256a5 + 31 (aA),

which (after some algebra) simplifies to

0=2'2a5((A6— iOA5+55A4— 140A3+l75A2—

Since a 0, this is equivalent to

—a

A6—10A5+55A4—140A3+175A2—106A+25— 3125Ap4

— (A—1)4(A2—6A-+-25)'

where the factorization of the denominator is easily done in Maple or Mathematica.Using this and b = a and b.

We will say more about this result in the Historical Notes.

Mathematical Notes

There are three topics we need to discuss further.

• Resolvents. The sextic resolvent appearing in Theorem 13.2.6 used h = u2, whereu is given in (13.17). This resolvent appears in [Chebotarev] and [36], for example.The paper [1] contains an especially nice discussion of how h relates to the stardiagram (13.18).


However, h = u2 is not the only possibility. One can instead use

01 = +X?X3X4 +X32X1X5

+X42X3X5

This leads to the sextic resolvent found in [5] and [10].The table used in Examples 13.2.8—13.2.10 distinguishes between the groups

((12345)) and AGL(l,lF'5)nA5 by factoring! over F(a), where a is a root of f.A natural question is whether this can be done with resolvents. The answer is yes,with a small complication. For example, if G C AGL( 1, F5) flA5, then the algorithmgiven in [5] computes

d=(aia2(a2—ai)+a2a3(ai—a2)+a3a4(a4—a3)2

+a4a5(a5 —a4)+asaj(ai —as))

One can prove thatd E F and that ifd 0, then G = ((12345)) if and only ifd E F2.

The latter condition is equivalent to the resolvent y2 — d having a root in F. Alsonote that d 0 guarantees that the resolvent is separable.

The problem is that d =0 can occur. When this happens, one performs a Tschirn-haus transformation to change f into a polynomial for which d 0. We will saymore about Tschirnhaus transformations in the Historical Notes. This complicationis why we used the factonzation of f over F(a) in part(c) of Theorem 13.2.6.

• Radical Solutions. When a quintic is solvable by radicals, it is natural to wantexplicit formulas for the roots. These can be complicated.

Example 13.2.13 Using our methods, it is straightforward to see that the Galoisgroup overQ of f =x5 + 15x+ 12 E Q[x] is isomorphic to AGL(1,F5). In [1] and[33], it is shown that

f—75+21V1O\'/5a125 ) 125 )

125 ) 125 )

is a root of f.

For an arbitrary solvable quintic, an algorithm for writing down the roots explicitlyis described in [10] and [211. For the special case of solvable quintics of the formx5 +ax+b, the solutions are described in [1] and [33]. See also [23].

• Normal Forms. A quintic of the form x5 + ax + b is said to be in Bring—Jerrardnormal form. When F has characteristic 0, it can be shown that an arbitrary quintic inF can be transformed into Bring—Jerrard form using a Tschirnhaus transformation,though in order to do so, one might need to replace F with a solvable extension. Aprocedure for doing this is described in [Dehn], [Postnikov], and [38]. In Exercise 16


YOU will show that in characteristic 5, not all quintic polynomials can be put intoBring—Jerrard form.

Two Bring—Jerrard quintics x5 + ax + b and x5 + a'x + b' in F[x] are equivalent ifthere is e F* such that

x5 +a'x+b' = A5((\x)5 +a17%x)+b).

Hence the roots of x5 +a'x+b' are the roots of x5 +ax+b multiplied by )C'. Usinga form of Theorem 13.2.12, [34] shows that when F = Q, there are infinitely manyinequivalent Bring—Jerrard quintics over Q.

However, if we switch to quintics of the form x5 + ax2 + b, there is a similar notionof equivalence, but here, [341 shows that up to equivalence, there are only five suchquintics. The argument reduces to finding x,y E Q such that

This is an elliptic curve (see the Mathematical Notes to Section 13.1) with onlyfinitely many solutions over Q.

One can show that an arbitrary quintic in characteristic 0 can be transformed intoa Brioschi quintic

— lOWx3+45W2x—W2.

The procedure for obtaining this normal form is described in [30, Ch. 5]. As for theBring—Jerrard form, a solvable extension of F may be required to obtain the Bnoschiform (see [30, Fig. 5.9.1]). The surprise is that the Brioschi quintic is deeply relatedto Section 7.5. This is because the rotational symmetries of the icosahedron give anextension K c C(t) (t a variable) with Galois group A5. The books [20] and [30]explain how a complete understanding of this extension enables one to find the rootsof any quintic polynomial.

Historical Notes

The first serious attempt to find the roots of polynomials of degree n 5 is due toTschirnhaus in 1683. His idea was to simplify

by a substitution of the form

(13.25)

now called a Tschirnhaus transformation. Eliminating x from these two equationsgives an equation in y of degree n that can be significantly simpler if b0,. . . , arechosen carefully.

Example 13.2.14 Given x3 + 3x + 1 e Q [xJ, consider the Tschirnhaus transformationy = a + bx + x2. In Exercise 17 you will show that eliminating x leads to the equation

(13.26) y3 + (6— 3a)y2+ (9 + 3b+ 3b2— 12a+ 3a2)y+P(a,b) = 0,


where P(a, b) is a polynomial in a and b. You will also show that the coefficients ofy2 andy vanish if and only if a and b satisfy

a=2 and b2+b—l=0.

If we pick b = l)/2, then the above cubic becomes

=2

1

In Exercise 17 you will use this to solve the original cubic. Note that the resultingTschimhaus transformation is defined over a degree 2 extension of Q.

Tschimhaus transformations can be used to solve all cubic and quartic equations.They fail in degree 5, though Bring in 1786 and Jerrard in 1834 showed that an arbi-trary quintic can be put into Bring—Jerrard form using Tschirnhaus transformationsdefined over suitable solvable extensions of the original field.

The quintic polynomial x5 + px3 + + q from Example 13.2.10 was solved byDe Moivre in 1706. The polynomial reappears in a 1764 paper of Euler devoted toalgebraic equations. In this paper, Euler writes an equation of degree 5 as

x5 =Ax3+Bx2+Cx+D.

(Can you explain why he omitted the x4 term?) Let the roots be a, /3, 6, in somesplitting field. Euler was seeking a formula of the form

(13.27)

where v is a root of an equation of degree < 4, and similar formulas for '7,6, f withthe radicals multiplied by suitable fifth roots of unity. Euler shows that this strategyworks for polynomials of degrees 2, 3, and 4, but for degree 5 he succeeds only insome special cases.

More precisely, he shows that if certain of the coefficients d, are zero in(13.27), then the original quintic reduces to one of the three special forms:

= D,

(13.28) x5 =5Px2+5Qx+Q2/P+P3/Q,x5 = 5Px3 — 5P2x + D.

We can analyze the resulting Galois groups as follows. Assume that the polynomialsof (13.28) have coefficients in Q and are irreducible. Then the Galois group of thefirst polynomial is isomorphic to AGL( 1, IF5) by Section 6.4, and the same is truefor the third polynomial by Example 13.2.10. You will show in Exercise 18 that theGalois group is also AGL(1,F5) for the second polynomial in (13.28).

However, if we adjoin the fifth roots of unity (standard procedure in the eighteenthcentury), then you will show in Exercise 18 that the first two equations of (13.28)


have cyclic Galois group over Q((5) while the third has Galois group isomorphic toAGL(1,1F5) flA5. Details of what Euler did can be found in [25].

The first sextic resolvents for the quintic are due to Lagrange, Malfatti, andVandermonde around 1770. Lagrange used

z = 2(4(x2xs +x3x5) +x4x5)+4(x2x4 +x,x5)

+4(x3xs +X1X2) +x2x3)) +

, 2 2 2 2 2 2 2+X1X2) +X2X3

Lagrange was led to this using the ideas discussed at the end of Section 12.1. Thepolynomial z is invariant under AGL( 1, IF5) and is the root of a sextic resolvent

z6 Az5 + Bz4 — Cz3 + Dz2 — Ez + F.

Lagrange computes A explicitly in terms of The formulas aresimilar to (13.24), except that Lagrange computed them by hand (no computers backthen!). Rather than continue with B, C, D, E, F, Lagrange comments that they can becomputed by similar methods and goes on to say:

But we shall not insert here such details which, besides that they would requirevery long calculations, would moreover not cast any light on the resolution ofequations of the fifth degree.

(See [Lagrange, p. 342].) For Lagrange, getting a resolvent of degree 6 was nothelpful, since he wanted to reduce to equations of degree smaller than 5 (which wenow know to be impossible). The irony is that one can use Lagrange's sextic to obtaina criterion similar to Corollary 13.2.11 for deciding which quintics are solvable byradicals. (This was done by Galois—see below.)

Malfatti's sextic resolvent is closely related to Lagrange's. He computed all ofits coefficients in terms of the elementary symmetric polynomials and knew that thequintic was solvable by radicals when the sextic had a rational root. The conversewas proved by Luther in 1847. The resolvent h = u2 used in the text is due to Jacobiin 1835. He was the first to prove (13.23). In 1861, Cayley independently discoveredthis resolvent and related it to the star diagram (13.18) (see also [1]).

Galois was naturally the first to think about this in terms of the Galois group.He showed that Lagrange's sextic resolvent has a rational root if and only if thecorresponding quintic is solvable by radicals. We will see in Section 14.1 that Galoisalso generalized this to irreducible polynomials of prime degree.

Theorem 13.2.12 about when a Bring—Jerrard quintic is solvable by radicals wasfirst proved by Runge in 1885. In the same year, Glashan and Young publisheddifferent versions of the same result. A modern proof appears in [33]. See also [23].

More on the history of the quintic equation can be found in [27], [39], and [40].


Exercise 1. As explained in the text, we can regard AGL( 1 ,1F5) as a subgroup of S5.(a) Prove that AGL(l,lF5) is generated by (12345) and (1243).(b) Prove that AGL(l,1F5)flA5 is generated by (12345) and (14)(23).


(c) Prove that the group of part (b) is isomorphic to the dihedral group D10 of order 10.(d) Prove that ((12345)), AGL(1,lF5) flAs, and AGL(l,1F5) are the only subgroups of

AGL( 1, IF5) containing ((12345))

Exercise 2. This exercise will consider some simple properties of S5.(a) Prove that ((12345)) is a 5-Sylow subgroup of S5 and more generally is a 5-Sylow

subgroup of any subgroup G C S5 containing ((12345)).(b) Prove that has twenty-four 5-cycles.

Exercise 3. Let G c S5 be transitive, and let N be the number of subgroups of G of order 5.In this exercise, you will use an argument from [Postnikov] to prove that N = I or 6 withoutusing the Sylow Theorems. Let C = {T E 55 \ G r is a 5-cycle}.(a) Prove that = defines an action of G on C.(b) Let T E Ss be a 5-cycle. Prove that a E satisfies = i- if and only if a E (T).(c) Use parts (a) and (b) to prove that Gl divides Cl.(d) Prove that 4N + Cl = 24.(e) Use parts (c) and (d) to prove that N = 1 or 6.

Exercise 4. Prove that (13.19) gives coset representatives of AGL( 1 ,IF5) in S5.

Exercise 5. Complete the proof of part (b) of Theorem 13.2.6. Then prove part (c).

Exercise 6. In this exercise, you will use Maple or Mathematica to prove (13.23) and (13.24).(a) The first step is to enter (13.17) and call it, for example, ul. Then use substitution

commands and (13.19) to create u2 u6. For example, u2 is obtained by applying(123) to ul. In Maple, this is done via the command

u2 subs({xl = x2,x2 = x3,x3 = xl}, Ui);

whereas in Mathematica one uses

u2 ul /. {xl—> x2,x2—> x3,x3—> xl}

(b) Now multiply out e(y) = (y —1.11) . . . (y — u6) and use the methods of Section 2.3 toexpress the coefficients of e(y) in terms of the elementary symmetric polynomials.

(c) Show that your results imply (13.23) and (13.24).

Exercise 7. Consider AGL( 1, IF5) flAs C Ss, and let u be defined as in (13.17).(a) Prove that the symmetry group of u is AGL( 1, IFs) fl A5.(b) Prove that (13.19) gives coset representatives of AGL(l,IF5) flA5 mA5.

Exercise 8. Let ul,...,u6 be as in the proof of Proposition 13.2.5, and let T ES5 be atransposition.(a) For each i, prove that T = —uj for some j.(b) Let 0(y) = (y — u,) and write this polynomial as

0(y)=y6+B1y5+B2y4+B3y3+B4y2+Bsy+B6.

(c) Explain how part (b) and the results of Chapter 2 imply that the coefficients B2, B4, B6 arepolynomials in al,a2,a3,a4,a5. This explains why the formulas (13.24) exist.

(d) Use Exercise 3 of Section 7.4 to show that the coefficients B1, B3, B5 must be of the formwhere B is a polynomial in aI,a2,as,a4,a5.


(e) Note that has degree 10 as a polynomial in xl,x2,x3,x4,x5. By considering thedegrees of B,,B3,Bs as polynomials in xI,x2,x3,x4,x5, show that part (d) implies thatB1 = B3 = 0 and that B5 is a constant multiple of This explains (13.23).

Exercise 9. This exercise will prove the first equivalence of Proposition 13.2.7.(a) First suppose that is irreducible. Prove that IGI is divisible by 6, and explain why

this implies that A5 C G.(b) Now suppose that A5 C G. Prove that Gal(L/F) acts transitively on . . . However,

we don't know that /3,,. are distinct.(c) Let p(y) be the minimal polynomial of flu over F. By part (b), it is also the minimal

polynomial of /32,... ,/36. Prove that 01(y) = p(y)m, where m = 1,2,3, or 6. The proof ofTheorem 13.2.6 shows that m = 6 cannotoccur, and m = I implies that Oj(y) is irreducibleover F. It remains to consider what happens when m = 2 or 3.

(d) Show that (y3 + ay2 + by + c)2 = Of(y) implies that = 0. Hence this case can'toccur.

(e) Show that (y2 + ay + b)3 = Oj(y) implies that 4b = a2, and then use this to show that

Exercise 10. This exercise will prove the second equivalence of Proposition 13.2.7. Note thatone direction follows trivially from Theorem 13.2.6. So we can assume that G C AGL(l,F5)and that = (y — )g(y) where E F.(a) Use (12345) E G to prove that Gal(L/F) acts transitively on $2,. .. As in the previous

exercise, we don't know if $2,... ,/36 are distinct.(b) Let p(y) be the minimal polynomial of $2 over F. By part (a), it is also the minimal

polynomial of/33, . . . Prove that Of(y) = (y— /3,)p(y)m, where m = 1 or 5. If m = I,then we are done. So we need to rule out m = 5.

(c) Show that (y — /3i)(y — = Of(y) implies that = $2. and then use this to show that

Exercise 11. Show that the table preceding Example 13.2.8 follows from the diagram (13.16)and Theorem 13.2.6.

Exercise 12. Let f = x5 — 6x + 3 E Q[x] be as in Example 13.2.8. Compute and 01(y)and show that Of(y) is irreducible over Q.

Exercise 13. Let f = —2 E be as in Example 13.2.9.(a) Compute and(b) In Section 6.4 we showed that the Galois group off over Q is isomorphic to AGL( 1, F5).

Use this and the Galois correspondence to show that the Galois group over is

isomorphic to AGL(1,F5)flA5.

Exercise 14. Let f = x5 + px3 + q E Q[xI be as in Example 13.2.10, and assume thatf is irreducible over Q.(a) Compute and 91(y).(b) Factor Of(y) e Q[x], and conclude that 5p2 E Q is a root of Oj4y).(c) Show that the substitution x = z — transforms f into z5 — + q.(d) Use past (c) to give an elementary proof that f is solvable by radicals over Q.

Exercise 15. As in Theorem 13.2.12, let f = x5 + ax + b. Compute and

Exercise 16. Let f = x5 + ax + b E F[x], where f is separable and irreducible and F hascharacteristic 5. The goal of this exercise is to prove the observation of [28] that the Galoisgroup of f over F is solvable.


(a) Prove that a 0.

(b) Use Exercise 5 from Section 6.2 to show that the Galois group of f over F is cyclic whena = —1.

(c) Show that there is a Galois extension F C L with solvable Galois group such that f isequivalent (as defined in the Mathematical Notes) to a polynomial of the form x5 — x + b'for some b' E L.

(d) Conclude that the Galois group off over F is solvable.

(e) Show that there is a field F of characteristic 5 and a monic, separable, irreducible quinticg e FkI that cannot be transformed to one in Bring—Jerrard form defined over any Galoisextension F C L with solvable Galois group.

In [28] Ruppert explores the geometric reasons why things go wrong in characteristic 5.

Exercise 17. Following Example 13.2.14, consider the equations x3 + 3x + I = 0 and y =a + bx + x2.(a) Use Maple or Mathemarica and Section 2.3 to eliminate x and obtain (13.26).

(b) Show that coefficients of y2 and y in (13.26) both vanish if and only if a = 2 andb2+b— 1=0.

(c) The equation for y becomes trivial to solve when a = 2 and b = — I )/2. We couldthen solve for x using y = a + bx + x2, but there is a better way to proceed. Note that

= —bx2 —ax+yx

follows from y = a + bx + x2. Furthermore, we can use y = a + bx + x2 to eliminate the x2in the above equation. Then use x3 + 3x + 1 =0 to obtain an equation in which x appearsonly to the first power. Solving this gives a formula for x in terms of y. The generalversion of this argument can be found in [Lagrange, p. 223].

Exercise 18. This exercise is concerned with the polynomials (13.28). As in the HistoricalNotes, we will assume that they lie in Q [x] and are irreducible.

(a) Show that + (P/Q) is a root of x5 — 5Px2 — 5Qx — Q2/P — P3/Q.

(b) Prove that the Galois group of x5 — 5Px2 — 5Qx — Q2/P — P3/Q over Q is isomorphic toAGL(1,F5).

(c) Prove that over Q the first two polynomials of (13.28) have cyclic Galois groupwhile the third has Galois group isomorphic to AGL( I, IFs) flAs.

Exercise 19. Use the methods of this section to compute the Galois group over Q of eachof the following polynomials. Be sure to check that they are irreducible. Remember that inSection 4.2 we learned how to factor polynomials over a finite extension of Q.(a) x5+x+l.(b) x5+20x+l6.(c) x5+2.(d) x5—5x+12.(e) x5 +x4 — 4x3 — 3x2 + 3x + 1.

Exercise 20. In the Mathematical Notes to Section 10.3, we noted that the roots of thepolynomial x5 — + 2x3 + 4x2 + 2x —6 E Q[x] can be constructed using a marked ruler andcompass. Show that this polynomial is not solvable by radicals over Q.


13.3 RESOLVENTS

So far, we have explained how to compute Galois groups of polynomials of degree 4or 5. It is time to turn our attention to polynomials of higher degree. We will see thatgeneralizations of the resolvents used in Sections 13.1 and 13.2 lead to a systematicstrategy for computing Galois groups.

A. Jordan's Strategy. In Theorem 13.1.1 we used the Ferrari resolvent to helpdetermine the Galois group of an irreducible polynomial of degree 4, and the sexticresolvent played a similar role in Theorem 13.2.6 for polynomials of degree 5. Thisstrategy for computing Galois groups was first described by Jordan in 1870:

The path to follow to treat this question [computing the Galois group] will bethe following: 1° one will form the various groups of the possible substitutionsG, G',... among the roots of the equation; 2° let G be one of these groups, chosenat will: one will affirm for oneself whether or not it contains the group of theequation by forming a function of the roots, invariant under the substitutionsof G and variable for other substitutions, calculating by the method of symmetricfunctions the equation [the resolvent] that has for roots the various values ofand looking for a rational root. Among the groups of the series G, G',... that inthis way contain the group of the equation, the smallest will be the group itself.

(See [Jordan 1, p. 2761.) The sextic resolvent Oj(y) used in Theorem 13.2.6 followsthis model nicely: for G = AGL(1,F5), we have = h = u2, whose symmetry groupis precisely G, and the universal sextic 8(y) is the polynomial "that has for roots thevarious values of p," which when evaluated at the coefficients of an irreducible quinticf E F[X] gives the resolvent Of(y). By Theorem 13.2.6, the question of whether theGalois group off is conjugate to a subgroup of AGL( 1, F5) is equivalent to "lookingfor a rational root' i.e., a root of 01(y) in F.

As the discussion of the sextic 81(y) reveals, Jordan's description is not perfect.In fact, it omits three important things:

• First, one needs to distinguish between the resolvent in the universal case and itsspecialization to the given polynomial.

• Second, having a rational root only implies that G contains the Galois group up toconjugacy.

• Third, this can fail if the rational root is not simple.

We will say more about these items below. Nevertheless, Jordan's description isremarkably close to some of the modem methods used to compute the Galois groupof an irreducible separable polynomial f e F {xI of degree n. Let Gj C S,, correspondto the Galois group off over F. Thus

Gal(L/F) Gj C Sn,

where F c L is a splitting field. In earlier sections, Gf was called G, but in thediscussion that follows, G will instead denote an arbitrary transitive subgroup of

Here is the step-by-step process for determining G1.

Step 1: Classify Groups. Transitive subgroups G C have been classified up toconjugacy for n 32 (see [4] and [15]). Published tables [7] go up to n = 15.

RESOLVENTS 387

Step 2: Find Polynomials. For each G C from Step 1, we need to find a polynomialin x1,... whose symmetry group is H(co) = G. Stauduhar's 1973 paper [36],

which pioneered the modem approach to Jordan's strategy, lists a polynomial foreach transitive subgroup of S4, S5, and S7 (and S6 with some errors noted in [13]).We will follow the standard convention that has coefficients in 7L.

Step 3: Compute Resolvents. Take G and from Step 2. Following the method ofSections 13.1 and 13.2, we compute the resolvent e (y) in the universal case, write itscoefficients in terms of the elementary symmetric polynomials, and then specializeto the coefficients of f. In terms of the roots i,... , a,, off, the resolvent is

0f(Y)

where p2,... , is the orbit of under the action (so m is the index of Gin 5,,). The problem is that the universal resolvent might be huge.

When F = Q, we can avoid this difficulty as follows. Suppose thatf e Z[xJ is irre-ducible of degree n. (Exercise 1 will explain why we can restrict to polynomials withinteger coefficients.) Then compute accurate numerical approximations ,

of the roots off, and multiply out the approximate resolvent

(13.29)

However, since the true resolvent has integer coefficients in this case (you will provethis in Exercise 2), it follows that if we have computed the approximate roots a,accurately enough, then the true resolvent is obtained from the approximate oneby rounding its coefficients to the nearest integer. Doing this rigorously requires acareful understanding of the numerical issues involved. See [36] for the details.

Example 13.3.1 Consider f = x5 — 6x+ 3 E Q[x] from Example 13.2.8. To 16decimal places, the approximate roots off are

= —1.6709352644808655592,

= —0.1181039225949867235 — l.587459l621207593640i,

= —0.1181039225949867235 + 1.587459l621207593640i,

= 0.50550 12304055246668,

= 1.4016418792653143394.

Evaluating the polynomials h• = from Section 13.2 at these numbers gives

= —43.4376362799772861 +28.6930156587206645i,

/32 = —7 1.5507381341784308 —94.8067689529853707i,

/33 = —71.5507381341784308 +94.8067689529853707i,

/34 = —5.0116255858442831 + 9.9920056672183422i,

/35 = —5.0116255858442831 — 9.9920056672183422i,

/36 —43.4376362799772861 —28.6930156587206645i


as approximate roots of the sextic resolvent (y). It follows that (13.29) becomes

Of(y) y6 +240.0000001y5+31680.00001y4+ 1935360.001y3

+ 58060800.02y2 + (584838144.3 — 0.07i)y

+ 4777574402 + 0.1 1095 86026i.

However, multiplying out the formula for (y) given in Example 13.2.8 shows that

Of(Y) y6+240y5+31680y4+1935360y3

+ 58060800y2 + 584838 l44y + 4777574400.

Looking at the constant term, we see that our approximation is not good enough.Hence we need to increase the accuracy of the roots of f.

The above calculation was done in Maple; Mathematica gives a similar result.The moral is that you need to know what you are doing when working numerically.Other methods for computing resolvents are discussed in [16].

Step 4: Use Resolvents. Suppose that! e F [xl is irreducible and separable of degreen and that G C and with = G give the resolvent e1(y) E F[y]. We can usethis to determine the location of the Galois group G1 C as follows.

Proposition 13.3.2 Letf E F[x] be separable and irreducible of degree n.(a) IfGf is conjugate to a subgroup of G, then Bf(y) has a root in F.(b) If 0f(Y) has a simple root in F, then G1 is conjugate to a subgroup of G.

Proof: Recall from Section 5.3 that a root of e1(y) is simple if the correspondinglinear factor appears exactly once in the factorization of e1(y) over a splitting field.

If G1 is conjugate to a subgroup of G, then G1 C G follows by suitably relabelingthe roots , off. Then G1 C G easily implies that . . , is invariantunder Gal(L/F) and hence lies in F (be sure you can supply the details).

Conversely, let /3 E F be a simple root of e1(y). By relabeling the roots off, wemay assume that /3 = ,an). If Gf G, then there is r E G1 such that T G.Then r . so that the resolvent may be written

= (y— (y — . . .

You will prove in Exercise 3 that = , E F and T E G1 imply that

which is impossible because = . . is a simple root of 01(y). Thiscontradiction proves that G1 C G.

For irreducible quartics, the Ferrari resolvent used in Theorem 13.1.1 is separablewhen f is separable, and for irreducible quintics, the same is true for the sextic

RESOLVENTS 389

resolvent used in Theorem 13.2.6. Hence all roots of these resolvents are simple.But in general, resolvents can have multiple roots. Here is an example.

Example 13.33 Let n = 4 and = +x2 —X3 —X4). In Exercise 4 you willverify that the symmetry group of is G = ((1324)) C when F has characteristic

2. Thus the corresponding resolvent e1(y) has degree 6. For the polynomialf = x4 + + d, d F2, from Example 13.1.4, you will show in Exercise 4 that

0f(Y) = + —

This has the rational root 0 E F, yet we showed in Example 13.1.4 that the Galoisgroup is not contained in ((1324)) when d(b — 4d2) F2. So ef(Y) fails to giveaccurate information about the Galois group, because 0 is not a simple root.

This example shows the importance of the word "simple" in Proposition 13.3.2.

Step 5: Repair Resolvents. Resolvents computed by the above process can failif their rational roots are not simple. To fix this, the standard method is to use aTschimhaus transformation (see the Historical Notes to Section 13.2) to change f toa different polynomial g. In [14], it is shown that this can always be done in such away that f and g have the same Galois group and the corresponding resolvent eg (y)is separable. See also [1, Algorithm 6.3.4]. Then redo Step 4 with g and êg(y).

Aside from some clever tricks, this method for computing Galois groups is thebasis of the algorithm used in [5] for polynomials of degrees 4, 5, 6, and 7. However,we will see below that the galois command in Maple computes Galois groups usinga slightly different approach.

B. Relative Resolvents. The idea of a relative resolvent was introduced in Sec-tion 12.1 in the universal case. Relative resolvents also are implicit in Theorem 13.1.1,as we will now explain.

Example 13.3.4 According to Example 13.3.3, ((1324)) C S4 is the symmetry groupof = (xi + x2 x3 — x4) in characteristic 2. In Exercise 4 you showed thatin the universal case, leads to the universal resolvent

e(y)=

wherey1 = X1X2 +x3x4,y2 = x1x3 +x2x4,y3 = XIX4 +x2x3 are the roots of the universalFerrari resolvent 9(y). 1ff = — CIX3 + — c3x + C4 E F[xI is irreducible andseparable and has roots a a2, a3, a4, then as usual x, i—* a, gives the resolvent

(13.30) e1(y) =

where /3i ,/32, /33 are the roots of the Ferrari resolvent 91(y).


Suppose that we have already computed Oj(y) and found that it has a root in F. Asusual, we may assume that = ala2 + a3a4 E F, so that Gf c ((1324), (12)). Todecide if Gf lies in the subgroup ((1324)), we could use the above resolvent e1(y).But since we already know we could instead use the factor

(13.31) +c?—4c2) EF[y],

which is an example of a relative resolvent. If (13.31) has a simple root in F, thenGf = ((1324)) follows by the relative version of Proposition 13.3.2.

Note that(13.31)has a simple root in F when + —4C2 0 and +— 4c2) E F2, just as described in part (c) Theorem 13.1.1. Do you see how part (c)

handles the case when the relative resolvent (13.31) has a multiple root?

The general theory of relative resolvents is described in [161 and [36]. Their mainadvantage is that they have smaller degree and hence are easier to compute. You willprove a version of Proposition 13.3.2 for relative resolvents in Exercise 5.

C. Quartics in All Characteristics. The treatment of quartics in Section 13.1assumed characteristic 2. Here, inspired by Keith Conrad [6], we use the ideasof this section to compute the Galois group of a monic irreducible separable quarticpolynomialf =x4 — c1 x3 +c2x2 — c3x+c4 E F[x] forany fieldF. As in Section 13.1,f has roots a! , a2, a3, a4, discriminant (f), and Ferrari resolvent

The main problem concerns the discriminant. In characteristic 2, Theorem 7.4.1tells us that Gf C if and only if E F. This fails in characteristic 2 since

= (ai — a2)(aI — a3)(aI— a4)(a2 a3)(a2 — a4)(a3 — a4)

= (a1 +a2)(ai +a3)(aI +a4)(a2+a3)(a2+a4)(a3+a4)

is clearly invariant under the Galois action and hence lies in F.In terms of symmetric polynomials, the problem is that the symmetry group of

= — depends on the characteristic. We will replace with

D = a = +cEA4

+X2X3X4 +XIX2X4

In Exercise 6 you will show that in characteristic 2,

D=

and that in all characteristics, = D — D', where D' (12) D. Thus D is "half"of v"3 (the positive terms in characteristic 2). By part (b) of Exercise 9 ofSection 12.1, the symmetry group of D is A4 in all characteristics.

Now let D(f) = D(ai,a2,a3,a4) and D'(f) = D'(ai,a2,a3,a4). The S4-orbitof D consists of D and D', so that we get the quadratic resolvent polynomial

(13.32) Dj(y) = (y D(f))(y — D'(f)) = y2 —Ay+B,

RESOLVENTS 391

where

A = cic2c3 — +4c2c4,

B = + + + — — +

+ — + + +

The formula forA follows from Example 2.2.4, and the formula for B can be computed

using the methods of Section 2.3.Using D1(y) and Of(y), we have the following preliminary result about the Galois

group of f. Recall that Gal(L/F) c S4. where L is a splitting field of f over F.

Theorem 13.3.5 Letf e F[x] be monic, separable and irreducible of degree 4. Thenthe subgroup Gf C S4 is determined as follows:(a) If is irreducible over F, then

f S4, (y) is irreducible over F,

if D1(y) splits completely over F.

(b) If Of(y) splits completely over F, then G1 Z/27L x 7Z/2Z. Furthermore, Oj(y)splits completely over F if and only if 01(y) and D1(y) are reducible over F.

(c) If has a unique root in F, then Gf is isomorphic to either 7L/47L or D8 (thedihedral group of order 8). Furthermore, Of(y) has a unique mot in F if andonly if 01(y) is reducible over F and D1 (y) is irreducible over F.

Proof: We first study the case when G1 CA4. The resolvent D1 (y) from (13.32) hasdiscriminant (D(f) D'(f))2, which is since (as noted above) = D — D'.It follows that D1(y) is separable since f is. Hence the roots of Df(y) are simple,and then Proposition 13.3.2 implies that

(13.33) G1cA4 Df(y)hasarootinF.

The proof of part (a) is now identical to what we did in Theorem 13.1.1, exceptthat we use (13.33) rather than Theorem 7.4.1 to determine whether Gf = A4 or S4.

For the rest of the proof, suppose that Oj(y) is reducible over F. Recall fromthe proof of Theorem 13.1.1 that f and have the same discriminant. Sincef is separable, it follows that the same is true for Of(y). We can assume that

= a root of (y) in F. As in the earlier proof, this implies that Gfis one of the three groups

If Df(y) is reducible over F, then Gf C by (13.33), which implies that G1 =((12) (34), (13) (24)) by the earlier proof. Since these permutations fix the other roots

a3 + a2a4, a1 a4 + a2a3 of Oj(y), it follows that 01(y) splits completely over F.

On the other hand, ifD1 (y) is irreducible over F, then G1 A4 by (13.33), which

forces G1 to be ((1324)) or ((1324), (12)). Both of these groups contain (1324),


which takes U1a3 + a2a4 to ala4 + a2a3. Hence = aIa2 + is the only rootof Oj(y) contained in F since Of(y) is separable.

From here, parts (b) and (c) of the theorem follow easily. •

To complete Theorem 13.3.5, we need to distinguish between 7L/47L and D8 inpart (c) of the theorem. The criterion given in part (c) of Theorem 13.1.1 fails in

characteristic 2. The reason is twofold: First, is always a square in F in thiscase, and second, characteristic 2 implies that

4/3 + — 4c2 =

is also a square in F. However, by replacing with Dj(y), we get the followingversion of Proposition 13.1.5 that works in all characteristics.

Proposition 13.3.6 As in Theorem 13.3.5, letf = x3 + c2x2 — c3x+ c4 E F[x].Also assume that Oj'(y) has a unique mot /3 E F. Then Gf is isomorphic to eitherZ/4Z or D8, and the former occurs and only if (y2 — + c2 — fl)(y2 — fly + c4)splits completely over F (D(f)).

Proof: Let L = F (at, a2, a3, a4) be the splitting field off over F. If G1then L contains a unique quadratic extension of F, which must be F (D(f)) sinceDf(y) = (y D(f))(y — D'(f)) is irreducible over F by part (c) of Theorem 13.3.5.However, y2 dy + c2 /3 andy2 — fly + c4 splitcompletely over L since

(13y2 —cly+c2—fl = (y— (ai +a2))(y— (a3+a4)),

y2—/3y+c4=(y—ala2)(y—a3a4)

by Exercise 12 of Section 13.1. Hence they split completely over the unique quadratic

extension of F contained in L. It follows that (y2 — + c2 — /3) (y2 — fly + C4) splitscompletely over F (D(f)).

On the other hand, suppose that Gf D8. By the proof of Theorem 13.1.1, we

may assume /3 = a1a2+a3a4 and Gf = ((l324),(12)). Let a EGal(L/F) map to

(l3)(24) E G1. Then a fixes D(f) since (l3)(24) is even. Hence

(13.35) a is the identity on F(D(f)).

Now suppose that (y2 ciy + c2 — /3) (y2 — fly + c4) splitscompletely over F (D(f)).Then (13.34) implies that ai + a2, ala2 E F(D(f)). Since a(ai + a2) a3 + a4and a(aIa2) = a3a4, we conclude from (13.35) that

al+a2=a3+a4 and aIa2=a3a4.

Part (a) of Exercise 6 ofSection 13.1 implies that f is not separable. This contradictionimplies that (y2 — c iy + c2 /3) (y2 — fly + C4) does not split completely over F (D(f))

U

Proposition 13.3.6 appears in [6]. Here is an example from the same paper.

RESOLVENTS 393

Example 13.3.7 Let F = k(u), where u is a variable and k has characteristic 2. Let

f = + (u + l)x2 + ux +1E F[xI.

In Exercise 7, you will verify the following:• f is irreducible and separable over F.• D1(y)=y2+u2y+u5+u3+u2.• Oj(y)=y3+(u+l)y2+u2=(y+u)(y2+y+u).Hence fi = u, so that the quadratic polynomials of Proposition 13.3.6 are given

andy2—uy+1=y2+uy+l. InExercise 7 you will show that y2 + uy + 1 has no roots in the splitting field of Dj(y).Then G1 D8 by Proposition 13.3.6.

One can also distinguish between Z/4Z and D8 using relative resolvents. Incharacteristic 2, +x2 —X3 —X4) has symmetry group ((1324)) and givesthe relative resolvent y2 — + c? — 4c2) described in Example 13.3.4. But

(13.36) =

has symmetry group ((1324)) in all characteristics. In the situation of part (c) ofTheorem 13.3.5, we have G1 C ((1324), (12)) D8, and to decide whether or notGf = ((1324)) Z/4Z, we can use the relative resolvent coming from (13.36),namely

where F is a root of 01(y) and

A = /3(clc3 — 2c4) — — +2c2c4,

B = + — + + /3(4c1c2c3c4 — — —

+ — + — + +

By Proposition 13.3.2, we get G1 = ((1324)) when y2 — Ay + B has a simple rootin F. This is the approach taken in [36]. The problem is that y2 —Ay+B may failto have simple roots, which as mentioned earlier in the section requires the use ofTschirnhaus transformations. Hence Proposition 13.3.6 is better for our purposes.

0. Factoring Resolvents. So far, we have asked whether resolvents have arational root. But there are situations where the irreducible factorization of a resolventcan be useful, even if none of the factors have degree 1. Here is an example.

Example 13.3.8 Let f be an irreducible separable quartic, and consider the sexticresolvent 01(y) given in (13.30). Assume that e1(y) is separable. In Exercise 8you will prove that G1 is conjugate to ((1324), (12)) if and only if = g(y)h(y),where g(y), h(y) E F[y] are irreducible of degrees 2 and 4, respectively. <V


A much more interesting example involves the group GL(3, IF2) of invertible 3 x 3matrices with entries in IF2.In Section 14.3 we will see that GL(3,IF2) is a simplegroup of order 168. The smallest non-Abelian simple group is A5 of order 60, andone can prove that GL(3, IF2) of order 168 is the next smallest.

Following [11] and [32] we will show that

g=x7— l54x+99EQ[xI

is an irreducible polynomial whose Galois group over Q is isomorphic to GL(3,1F2).Our tool will be the factorization of a resolvent of degree 35.

First observe that GL(3, IF2) acts on the eight-element vector space by matrixmultiplication. The origin 0 = (0,0,0) is fixed, but the seven nonzero vectors getpermuted. In Exercise 9 you will show that labeling these vectors Vi,. . . , V-i induces

a one-to-one group homomorphism

(13.37) GL(3,F2) —* 57.

For simplicity we will identify GL(3,1F2) with its image under this map and henceregard GL(3,1F2) as a subgroup of S7.

Now consider = x1 +x2 +x3 E Q[xi,...,x7]. If we are given a polynomialf E Q of degree 7 with roots i,... , then we get the resolvent

ef(Y)= fiI<i<j<k<7

There is one factor for each three-element subset of { 1,... , 7}, so that Oj(y) hasdegree = 35. Then we have the following interesting result of [32].

Proposition 13.3.9 Let f be irreducible of degree 7, and let ef(Y) be theabove resolvent of degree 35, which we assume to be separable. Then the Galoisgroup off over Q is isomorphic to GL(3, IF2) if and only if e1(y) g(y)h(y), whereg(y), h (y) e Q [y] are irreducible of degrees 7 and 28, respectively.

Proof: First suppose that the Galois group off over Q is isomorphic to GL(3,IF2).The transitive subgroups of S7 are known (see [3, pp. 206—209]) and in particular,any subgroup of S-i isomorphic to GL(3,IF2) is conjugate to the subgroup comingfrom (13.37). By relabeling the roots, we may assume that

Gj = GL(3,F2) c 57.

Since 01(y) is separable, its irreducible factorization is governed by the action ofthe Galois group on its roots, which is equivalent to the action of GL(3,]F2) onthree-element subsets of { 1, . . . , 7} (be sure you understand this). Hence we need tounderstand the action of GL(3, F2) on unordered triples of nonzero vectors of

A one-dimensional subspace of F? is a line through the origin, which consists of 0and a nonzero vector, since we are over IF'2. Hence there are seven such subspaces. InExercise 10 you will show that IF? also has seven two-dimensional subspaces, eachof which consists of 0 together with three linearly dependent nonzero vectors.

RESOLVENTS 395

It follows that of the 35 possible triples of nonzero vectors in 1Ff, seven consistof linearly dependent vectors while the remaining 28 consist of linearly independentvectors. In Exercise 10 you will show that GL(3,F2) acts transitively on each ofthese sets of triples. As explained above, this describes the Galois action on the rootsof ej(y), and the desired factonzation follows.

The converse is proved in [32]. Let the Galois group of f be isomorphic toGf C S7. Since e1(y) is separable, the Galois action on its roots is equivalent tothe action of G1 on three-element subsets of { 1,.. . , 7}. The conjugacy classes ofsubgroups S7 are known, and for each conjugacy class, one can compute the orbits ofits action on unordered triples. These are listed in Table I in [32]. Inspection of thislist shows that GL(3, IF2) c S7 is the only subgroup (up to conjugacy) such that theorbits have lengths 7 and 28. Thus G1 must be conjugate to GL(3,1F2) when e1(y)has irreducible factors of degrees 7 and 28. U

Here is the example mentioned earlier.

Example 13.3.10 For f = — 154x + 99, [11] computes that

ej(y) = y35 — 6160y29 +29898y28 — 38277624y23 —41255676y22

+37518228y2' + 18524283008y'7+ 6522421752y'6

+27295157736y'5+ 35173338750y'4— 2894923232432y"

+489571380144y10— 4925879415072y9+ 3933790086996y8

— 702099623709y7+ 149674336745472y5 — 96219216479232y4

— 2577300441 4080y3 + 21 354775085952y2 + 946763427456y

— 1217267263872.

Using Maple or Mathematica, the irreducible factorization of e1(y) over Q is easilycomputed to be

êf(y) = g(y)h(y),

whereg(y) =y7 —23ly3—462y2+77y+66

and h(y) is the polynomial

y28 + 23 1y24 + 462y23 — 6237y22 + 29832y2' + 5336 1y2°

1245090y'8+3958878y'7— 11719092y'6+30817248y'5

— 157564l43y'4+[lpt]319312224y'3—796323990y'2+14819061l8y"

— 299438l313y'°+5889443406y9+ 965064177y8 — 4595839182y7

+ 33180883659y6— 84492l27566y5+ 181691003340y4

— 382065796728y3 + l52613801648y2+ 3S862251O4Oy — 18443443392.

By Proposition 13.3.9, the Galois group of f over Q is isomorphic to GL(3,1F2).


The galois command in Maple computes the Galois group over Q of an irre-ducible polynomial of degree 9 in Q [xl. The algorithm used by Maple involvesfactoring resolvents of the above type. See [32] for more details.

The computer algebra programs GAP [12] and Magma [24] can also compute theGalois groups over Q. GAP can handle polynomials of degree 15. Magma, onthe other hand, has no a priori limitation on the degree of the polynomial, thoughcomputations for degrees � 50 are rarely successful.

Mathematical Notes

We will discuss two topics from this section.

Trinomials of Degree 7. In the text, we showed that the Galois group over Q of154x+99 is isomorphic to GL(3,1F2). In the late 1960s,x7 —7x+3 was shown

to have the same property. Bruin and Elkies [2] consider the problem of findingall trinomials f = ax7 + bx + c E Q{x] with GL(3, IF2) as Galois group. We say thatanother trinomial g is equivalent to f if

g = )'.(a(jix)7

for some Q'. By [2], equivalence classes of tnnomials over Q whose Galoisgroup is contained in GL(3, IF2) correspond to solutions (x,y) E Q2 of the equation

(13.38) y2 =x(81x5+396x4+738x3+660x2+269x+48).

By finding all solutions (x,y) e Q2 (including points at infinity), one gets the resultthat, up to equivalence, the only trinomials ax7 + bx+ c e Q[xl with GL(3,1F2) asGalois group are

x7 — 7x+3, x7 — 154x+99, 372x7 —28x+9, 4992x7 —23956x+34 113.

Details and references can be found in [2]. Also, [111 and [32] include references toother papers on polynomials of degree 7 with GL(3, IF2) as Galois group,

The equation (13.38) is another example of a Diophantine equation. In contrastto Example 13.1.8, this is not an elliptic curve. Instead, it has genus 2 (while ellipticcurves have genus 1). By Faltings's proof of the Mordell Conjecture, it is known thatequations of genus � 2 have at most finitely many rational solutions, i.e., solutionswith (x,y) Q2. But the proof is not constructive, so that in practice it can be difficultto prove that one has found all rational solutions.

• Groups and Geometry. The group GL(3, IF2) is important for reasons related togroup theory and geometry. Let us begin with the group theory. In the MathematicalNotes to Section 11.1, we defined GL(n, F) to be the group of invertible n x n matriceswith entries in the field F. This group contains the subgroup SL(n, F) of matricesof determinant 1. Furthermore, taking the quotient of each of these groups by thesubgroup consisting of multiples of the identity matrix gives groups

PGL(n,F) and PSL(n,F).

RESOLVENTS 397

We will say more about these groups in the Mathematical Notes to Section 14.3.There, we will see that aside from PSL(2,F2) S3 and PSL(2,F3) the groupPSL(n, Fq) is simple whenever n � 2.

In particular, PSL(3,F2) is simple. However, by Exercise 11 we have

GL(3,1F2) = SL(3,F2) PGL(3,1F2) = PSL(3,1F2).

This explains why GL(3,1F2) is simple. In Section 14.3 we will also see that

PSL(3,F2) PSL(2,Fi).

For this reason, some papers, such as [11], use PSL(2,F7) instead of GL(3,1F2).For any field F, GL(3, F) and PGL(3, F) have interesting geometric properties.

For GL(3, F), the geometric object it acts on is the vector space F3 of dimension 3over F. For PGL(3, F), the corresponding geometric object is the proj ective plane P2over F. Although this is beyond the scope of the book, we will make one commentrelated to the proof of Proposition 13.3.9. There, we observed that has seven one-dimensional subspaces and seven two-dimensional subspaces. Once you understandthe geometry of P2 over F = IF2, this follows immediately from pmjective duality.More on projective geometry can be found in [29].


Exercise 1. Let f(x) e Q [xl.(a) Prove that there are A, E Q such that g(x) = Af(px) E 7L[x] is monic.(b) Prove that f and g have isomorphic Galois groups over Q.

Exercise 2. Let f(x) = x" — + + E Z[x], and let ef(y) be the resolventbuilt from Prove that e1(y) E Z[xl.

Exercise 3. In the proof of Proposition 13.3.2, we asserted that

= . . ,

follows from /3 = . , cx,,) E F and r E G1. Prove this.

Exercise 4. As in Examples 13.3.3 and 13.3.4, let = +x2 —X3 —X4).

(a) Show that the symmetry group of is G = ((1324)) C S4 in characteristic 2.

(b) Show that in the universal case, leads to the resolvent

e(y)=fl(y2

where yi = XIX2 +x3x4,y2 = x1x3 +x2x4,y3 = XIX4 +X2X3 are the roots of the universalFerrari resolvent

(c) Let ê1(y) be obtained by specializing the resolvent e(y) of part (b) to I = x4 + bx2 + d.Show that

e1(y) =y2((y2 —

Exercise 5. This problem will state and prove a relative version of Proposition 13.3.2. Fix asubgroup H C S,, and suppose that f E F[x] is separable of degree n and that G1 C H. Now


let G C H be a subgroup. We want to know whether or not G1 lies in the smaller subgroup G.Let E F[xi , . . . have Gas its symmetry group and let = . be the orbit ofH acting on Then set

e"(y) = E F[x1,. ..,xn][y}.

Finally, if ,. . . , are the roots off in a splitting field L, let

eL[y]

be the polynomial obtained by x, '—+ z.(a) Explain why the degree of (y) is the index of G in H.(b) Prove that e7(y) E F[y].(c) Assume that G1 is conjugate within H to a subgroup of G (this means that

r E H). Prove that has a root in F.(d) Assume that (y) has a simple root in F. Prove that Gj is conjugate within H to a

subgroup of G.We call (y) a relative resolvent. You will verify in Exercise 12 that (13.31) from Exam-pie 13.3.4 is an example of a relative resolvent.

Exercise 6. LetD= X1X2X3 E Fixi ,x2,x3,x4].

(a) Prove that D = — — + in characteristic 2.

(b) Prove that = D (12) 'Din all characteristics.

Exercise 7. As in Example 13.3.7, let f = x4 + (u + l)x2 + ux+ I e Fjx), where F = F2(u).(a) Use Gauss's Lemma and the Schönemann—Eisenstein criterion to showthat f is irreducible

over F. (These results apply since 1F2[u] is a PID.)(b) Verify the formulas for D1 (y) and (y) given in Example 13.3.7.(c) Show that y2 + uy + I is irreducible over the splitting field of Df (y).

Exercise 8. Let f E F [x] be an irreducible quartic, where F has characteristic 2. Also letbe the sextic resolvent defined in Example 13.3.4. The goal of this exercise is to show

that G1 C S4 determines the irreducible factorization of over F. We will assume thatis separable.

(a) First suppose that G1 = A4 or S4. Prove that ef(y) is irreducible over F.(b) Now suppose that Gf = ((1324), (12)>. Prove that ej(y) = g(y)h(y), where g(y), h(y) E

F [x] are irreducible of degrees 2 and 4, respectively.(c) Suppose that Gf = ((l2)(34),(13)(24)>. Prove that e1(y) = gl(y)g2(y)g3(y), where

g(y) E F[x] is irreducible of degree 2.

(d) Finally, suppose that G1 = ((1324)>. Prove that e1(y) = gi (y)g3 (y), where gi (y),g2 (y), g3 (y) E F [x] are irreducible of degrees 1, 1, and 4, respectively.

(e) Explain why parts (a) through (d) enable one to determine Gf up to conjugacy using onlyand

Notice that the claim made in Example 13.3.8 now follows immediately.

Exercise 9. The action of GL(3, F2) on the nonzero vectors of gives a group homomorphismGL(3, F2) —+ S7. Prove that this map is one-to-one.

RESOLVENTS 399

Exercise 10. Consider the vector space(a) Prove that has exactly seven two-dimensional subspaces.(b) Forafield F,IetB = {{vj , v3} CF3 I , V3 are linearly independent overF}. Prove

that GL(3, F) acts transitively on B.(c) Let F be as in part (b). Prove that GL(3, F) acts transitively on the set of two-dimensional

subspaces ofF3.Be sure you understand how parts (b) and (c) apply to the proof of Proposition 13.3.9.

Exercise 11. Prove that GL(3,F2) = SL(3,1F2) PGL(3,1F2) = PSL(3,]F2).

Exercise 12. Prove that (13.31) from Example 13.3.4 is an example of a relative resolvent inthe sense of Exercise 5.

Exercise 13. In the proof of Proposition 13.3.9, we showed that when GL(3, IF2) C acts onthree-element subsets of { 1,. . . , 7}, the orbits have lengths 7 and 28. We also asserted that upto conjugacy, GL(3, IF2) is the only subgroup of S7 with this property. In this exercise, youwill study the action of some other subgroups of S7.(a) Prove that A7 and act transitively on three-element subsets of { 1, . . . , 7}. Thus there is

one orbit of length 35 for these groups.(b) In Section 13.2, the group AGL( 1 ,1F5) C played an important role in understanding

the Galois group of a quintic. In a similar way, we have AGL(l,1F7) C S7 provided wethink of the indices as congruences classes modulo 7. Prove that the orbits of AGL( 1, IF7)acting on the triples {O, l,2} and {O, l,3} have 21 and 14 elements, respectively.

Exercise 14. The quadratic resolvent D1(y) used in Theorem 13.3.5 to compute the Galoisgroup of a quartic in all characteristics was defined for a polynomial f of degree 4. Here youwill study what happens when f is monic of degree n. We begin with the polynomial

D=

where F is a field of any characteristic.(a) Prove that is the symmetry group of D.(b) Prove that = —xi) satisfies = D — D', where D' = (12) .D(c) Let f E F[x] be n and let . . be the roots of f in some splitting

field L. Then define D(f) = D(cxj,. . . and D'(f) = . . and set

D1(y) = (y—D(f))(y—D'(f)).

Prove that D1 (y) E F [y] and that the discriminant of Df (y) is = fl1 —)2•

Note that D(f) and D'(f) depend on how we order the roots while the polynomial D1(y)depends only on f.

(d) Assume that f is separable and let Gal(L/F) C Prove that G1 C if and onlyif Dj(y) splits over F.

This gives a version of Theorem 7.4.1 that works in all characteristics.

Exercise 15. Let f = x3 — CIX2 + C2x C3 E F[x] and let D1(y) E be as in the previousexercise.(a)

(b) Assume in addition that f is separable and irreducible. Explain how Dj(y) determinesthe Galois group off up to isomorphism.

This gives a version of Proposition 7.4.2 that works in all characteristics. See [6] for someexamples.


13.4 OTHER METHODS

This section will explore further tools for computing Galois groups. We begin witha result of Kronecker that works in complete generality but is not very efficient.However, his method also leads to a quick proof of a result of Dedekind that usesreduction modulo p to obtain useful information about Galois groups over Q.

A. Kronecker's Analysis. In Section 12.3 we studied Kronecker's constructionof the splitting field of a separable polynomial f E F[x] of degree n. Let us recallhow this works.

Assume that F is infinite, and let . . be the roots of f in a splitting fieldF c L. We saw in Section 12.2 that there are t1,. . . E F such that then! elements

are distinct. Thus

(13.39) s(y) = [J (n))) eL[y]

is separable of degree n!. We showed in Section 12.3 that s(y) E F[y] and that ifh(y) E F is any irreducible factor of s(y), then the quotient

is a splitting field of f over F. It follows that the degree of h(y) is the order of theGalois group off over F.

This construction seems to require the roots ai,... , However, the universalversion of s(y) is given by

(13.40) S(y)= fi

The theory of symmetric polynomials tells us how to write S(y) explicitly as apolynomial in F[cr1,... , Then specializing to the coefficients off gives s(y)as in (13.39). Furthermore, Exercises 4 and 5 of Section 12.3 show how to pick

E F (without knowing the roots) so that s(y) is separable.Here is an example of this process.

Example 13.4.1 Consider f = x3 +x2 — 2x — 1 e Q[xl. In Exercise 1 you will showthat if we set t1 = 1, t2 = 1, and t3 = 2, then the universal polynomial (13.40) becomes

S(y) y6 — + + 1402)y4 + — + 20a3)y3

+ + 1 + — )y2 + +

— — + 140c72o3)y+4t4 — +

+ + 76a?a3 — 322a1a2a3 +

OTHER METHODS 401

Using '—* '—÷ —2,a3 '—+ 1, we obtain

s(y)=y6+4y5—26y4—76y3+193y2+240y—377

= (y3+2y2— l5y+ l3)(y3+2y2— 15y—29),

where the second line is the irreducible factorization in Q[yI. This shows that s(y)

is separable (do you see why?). Thus the Galois group of f over Q has order 3.(This also follows from the theory of Chapter 9, since f is the minimal polynomialof 2cos(2ir/7) =

Besides giving the order of the Galois group, Kronecker observed that by modify-ing the above construction, one can extract the entire Galois group from an irreduciblefactor of s(y). The idea is that instead of letting t1,... , be elements of F, we letthem be variables. To prevent confusion, we will label these variables u1,. . . , andwrite (13.39) as

(13.41) Su(Y)= JJ EL[ui,...,un,y].

The subscript u is a reminder that (y) is a polynomial in the n + 1 variablesU1,. . . , In Exercise 2 you will show that the coefficients lie in F, so that

su(y)EF[uj,...,un,y].

Furthermore, we can compute (y) by first working in the universal situation andthen specializing to f. Thus we can find s,, (y) without knowing the roots off.

The polynomial ring F[ui,... , has two key structures:

• F[uj,. . . , ,y] has an Sn-action given by permutations of u1,... ,•For the splitting field L of f, L[ui, . . . , has the same two structures plus a third:

• L[ui,... , has a Gal(L/F)-action given by the Galois action on L.

As in the previous section, we write the Galois group off over F as

Gal(L/F) C Sn.

The above structures, when applied to Su(y), give the following description of Gf.

Theorem 13.4.2 Assume that f E F [x] is monic and separable of degree n, whereF is an arbitrary field. Also let h E F [ui,. . . , ,y] be an irreducible factor of thepolynomial (y) E F [u1, . . . , , y] constructed above. Then G1 C is conjugate tothe subgroup

Proof: In Exercise 3 you will show that (13.41) is the irreducible factorization ofsu(y) in L[u,,... ,Un,Y]. Thus we can pick a E such that

(13.42)


is a factor of h in L[ui,... , The permutation a will be fixed for the remainderof the proof. Our goal will be to prove that G = a 1Gf a.

Consider the polynomial

= fJ (y + +

(13.43)'yEGaI(L/F)

= fiStandard arguments imply that h is invariant under the action of Gal(L/F), so thath E F[ut,... Un,YJ since F C L is Galois. We can relate h to h as follows. Pick any

E Gal(L/F). Since h has coefficients in F and (13.42) divides h in L[ui,...it follows that

y— + .divides -y.h = h in Hence h divides h in L[ui,...,un,yI, which byExercise 3 implies that h divides h in F [ui, . . . ,

, y]. Since h is irreducible, it followsthat h = h after multiplying h by a suitable constant.

Now suppose that r E satisfies r h = h. This implies that r applied to (13.42)is a factor of h in L[ui,. . . ,

, y]. Since (13.43) is the irreducible factorization of h inL[Ui,...,Un,yI, we must have

y — + + = y + . . . +

for some e Gj. Since u1,... , u,, are distinct variables, this implies that

T(i) = f == oij)

It follows that = since , . . . are distinct. Thus

r = E

This shows that G c a1 Gf a. You will prove the opposite inclusion in Exercise 3,which implies G = Gf a. This completes the proof of the theorem.

Theorem 13.4.2 gives an algorithm for computing the Galois group off:

• EF[ui,...,un,y].• Factor (y) into irreducibles, and let h be an irreducible factor.• For each T E compute y . h and compare it with h.• Then the Galois group of f over F is isomorphic to {r E

Ir h = h}.

For n large, this algorithm is extremely inefficient. For example, (y) has degree10! = 3628800 in y when deg(f) = 10. Finding an irreducible factor h of couldtake a long time. And even if we could find h, then we would need to compute 'r . h

for all 3628800 permutations r E Thus this algorithm is not useful in practice,although it is a completely general method for computing Galois groups.

OTHER METHODS 403

Here is an example of how to use Theorem 13.4.2 when n is small.

Example 13.4.3 Considerf x3 + x2 — 2x — 1 E Q [xi. In Exercise 4 you will showthat (y) has the irreducible factorization

Su(Y)=

— + + — 3u,u2y+ — — 3U2u3y +

— U1U2U3 — — + + — 3ulu2y

in Q [UI , U2, u3, (This calculation was done in Mathematica.) Let h be the firstfactor multiplied by —1. You will show in Exercise 4 that

h = +u2+u3)y2+ (7(uIu2+ulu3+u2u3)—2(uI +u2+u3)2)y

+ 7UIU2U3 — (UI + u2 + u3)3 + + +

In this formula for h, everything is symmetric in ul,u2,u3 except for the last set ofparentheses. Thus T E S3 fixes h if and only if fixes

+ u?u3

It follows easily that the group G of Theorem 13.4.2 is ((123)) C S3.

Note that f is not required to be irreducible over F. Here is an example.

Example 13.4.4 For f = — 1 E you will show in Exercise 5 that

x (y2+(UI

x —U2U3).

In each factor, the terms of degree 0 in y are symmetric in U2, U3. So the coefficientof y is the crucial term. It follows that the first factor gives G = ((12)), the secondgives G = ((13)), and the third gives G = ((23)).

Although Example 13.4.4 is trivial from the point of view of Galois theory, it doesshow that Theorem 13.4.2 determines the Galois group only up to conjugacy.

In the earlier part of this section we assumed that F was infinite so that we could findE F such that then! elements t1 I) were distinct. In contrast,

Theorem 13.4.2 applies to all fields, even finite ones. This works because UI,... ,

are variables, so that the expressions Ui I) + + are automatically distinctby the separability off.

We will soon see that applying Theorem 13.4.2 over a finite field has some niceconsequences.


B. Dedekind's Theorem. Given a polynomial f e Z[x] and a prime p, we letf E [x] be the polynomial obtained by reducing the_coefficients of f modulo p.Then the following theorem of Dedekind shows how f can give information aboutthe Galois group off over Q.

Theorem 13.4.5 Letf e Z [x] be monic and separable of degree n. Given a prime psuch that let

_ _

wheref1 . E lFp[x] are monic and irreducible. Also set d, deg(j). Then:(a) The Galois group off over IF,, is cyclic of order lcm(di ,d2, . . . ,dr).(b) The Galois group off over Q contains an element that acts on the roots off

according to a product of disjoint cycles of the form

.

di-cycle d2-cycle d,-cycle

Hence the Galois group off contains an element of order lcm(di ,d2, . . .

Proof: First observe that f is separable, since p{ and is the reduction ofmodulo p by Exercise 4 of Section 5.3.

Part (a) is an easy application of Chapter 11. Since

x"—x= fi

a separable polynomial in IF,,[x] splits completely over F,,m if and only if it divides—x. Thus:

f splits completely over F,,m f, splits completely over IF,,m for all i

f, divides x for all i

d = divides m

lcm(di,d2,...,dr)dividesm,

where the second equivalence uses our above observation and the third equivalenceuses part (c) of Proposition 11.2.1. This easily implies that the splitting field of fover IF,, is d = lcm(di,d2,... ,dr). Since is cyclic of order d byTheorem 11.1.7, part (a) follows.

For later purposes, let us describe the action of on the roots of f.By Theorem 11.1.7, the Galois group is generated by the Frobenius automorphisma as'. Since f, is irreducible, Exercise 7 of Section 11.1 implies that if a is a rootoff1, then all roots are given by

d1=deg(j).

Hence the action of a i—* a1' on the roots of is given by a d-cycle. Since f isseparable, it follows that a '—+ acts on the roots of f according to a product ofdisjoint cycles of lengths d1, . . . , dr, just as in part (b) of the theorem.

OTHER METHODS 405

We turn to part (b). Consider the universal version of (y) defined by

Su(y)= fi E7L[Xi,...,Xn,Ui,...,Un,y].

aES,

This is symmetric in x1, . . . so that

Su(y)

by the Fundamental Theorem of Symmetric Polynomials over 7L (see Exercise 6 ofSection 9.1).

Write the polynomials f and f as

E

J=f—eix"'This gives

Su(y) e . . . obtained from via '—*

E obtained from via i-+ëj.

Thus (y) is the reduction of s,, (y) modulo p, since ãj is the reduction of c, modulo p.We relate Su(y) and to the Galois groups of f andf as follows. As usual, the

Galois group off over Q maps to a subgroup G1 C that records the Galois actionon the roots. Given an irreducible factor h of (y) over Q, Theorem 13.4.2 impliesthat Gf is conjugate to

By Exercise 6 we may assume that h is an irreducible factor of (y) in the ringun,y]. Reducing this modulo p gives E . . , If is an irre-

ducible factor of h, then it is also an irreducible factor of (y), so that by Theo-rem 13.4.2, the Galois group off over IF,, gives a subgroup of conjugate to

We G. Toprovethis,Then a su(y) = Sa(y) implies that h1 is also an irreducible factor of Su(y). SinceZ[ui,.. . , is a UFD, we must have

for some polynomial q e . . . ,Un,YI. Reducing this modulo p gives

Su(Y) /Z12IqE lFp[ui,...,un,yI.

Furthermore, the Sn-action is compatible with reduction modulo p. so that h1 = a . himplies h1 = a .h. Since g divides h, we see that = a divides a = h1. By theabove equation, this implies that divides Yet over the splitting field L of


f, (13.41) implies that Su(y) is a product of distinct irreducible factors. This easilyimplies that the same is true over Hence we have a contradiction, which provesthat a . h = h whenever o• = Thus G C G.

By part (a) the Galois group of f over IF,, contains an element whose action onthe roots of f is given by a product of disjoint cycles of lengths d1,... ,dr. Since theconjugate of a product of disjoint cycles of lengths d1,... , d,. is a permutation of thesame form, we see that G and hence G contain a permutation of the desired form.Since G is conjugate to Gf, the Galois group of f must contain an automorphismwhose action on the roots is as described in part (b) of the theorem. •

Here is an example to illustrate part (b) of Theorem 13.4.5.

Example 13.4.6 Consider f = x5 + 20x + 16 e Q [x]. In Exercise 7 you will verifythat f is irreducible with discriminant = 21656. This shows that the Galoisgroup off over Q is isomorphic to a subgroup of A5.

Working modulo 7, we have the irreducible factorization

f—(x+2)(x+3)(x3+2x2+5x+5) inIFi[x].

Since part (b) of Theorem 13.4.5 implies that the order of the Galois groupis divisible by 3. The classification of transitive subgroups of S5 given in (13.16)shows that A5 has no proper transitive subgroup of order divisible by 3. Hence theGalois group of f over Q is isomorphic to A5.

Our next example uses the cycle decomposition of part (b) of Theorem 13.4.5.

Example 13.4.7 In Section 6.4 we showed that f = x5 — 6x +3 E Q has S5 as itsGalois group over Q. If you look carefully at the argument given in Section 6.4,you'll see that we showed first that the image of the Galois group in Ss contains a5-cycle and a 2-cycle, and second that any 5-cycle and 2-cycle generate S5.

Using part (b) of Theorem 13.4.5, it is easy to get the required cycles. Considerthe irreducible factonzations

f=x5+4x+3 inlF5[xJ,

f=(x+2)(x+7)(x+13)(x2+12x+13) inlFii[x].

The first gives a 5-cycle and the second gives a 2-cycle. The theorem applies to theseprimes, since = —1737531 = _34. 19.1129.

In Exercise 8 you will give a different proof that the Galois group is S5 by reducingmodulo 11 and using the method of Example 13.4.6.

The paper [26] discusses an approach to computing Galois groups that uses The-orem 13.4.5 more systematically.

Mathematical Notes

We will discuss three topics related to Theorem 13.4.5.

OTHER METHODS 407

• Reduction Modulop. Given a monic polynomial f E Z[xI, Theorem 13.4.5 showsthat its factorization modulo p gives interesting information about the Galois groupof f over Q. The reduction is interesting for other reasons connected to what isknown as class field theory. This is a large topic, so we will confine ourselves to twoexamples.

The first concerns the case when f is the quadratic polynomial f — a, wherea E Z. Since = 4a, we know that j e IF,, is separable when For such aprime p, f splits completely modulo p if and only if the congruence

x2 a mod p

has an integer solution. When the latter holds, we say that a is a quadratic residuemodulo p. Quadratic residues play an important role in number theory and are relatedto the Legendre symbol defined in the Mathematical Notes to Section 9.2.

A deeper example is the following observation of Kronecker. The polynomial

fzr (x3 lOx)2+31(x2— 1)2 E Z[x]

has discriminant _26 .38. Now consider the following question: For

which primes p does the factorization off modulo p include a linear factor? In otherwords, when does f have a root modulo p? The amazing answer, due to Kronecker,is that if p> 3 is a prime different from 31, then

f(x)mOmodpforsomexE7Z p=x2+3ly2forsomex,yeZ.

So our question characterizes primes of the form x2 + 3 ly2. Kronecker never pub-lished a proof of this result, which today is regarded as part of class field theory andcomplex multiplication. See [8] for an introduction to this rich subject.

The Chebotarev Density Theorem. Let f E Z{x) be monic and separable ofdegree n with splitting field Q C L. Given a prime p that does not divideTheorem 13.4.5 implies that Gal(L/Q) contains an element that corresponds to theFrobenius automorphism & in the Galois group over IF,,. This element ofGal(L/Q) is called the Artin symbol of p. denoted ok,. Since the proof of Theo-rem 13.4.5 involves choices related to the ordering of the roots, the Artin symbolis well defined only up to conjugacy in Gal(L/Q). The Chebotarev Density Theoremdescribes the behavior of a1, as we vary the prime p:

• Up to conjugacy, every element of Gal(L/Q) equals a1, for some prime p.

• If we fix a conjugacy class C of Gal (L/Q), then the percentage of primes p whoseArtin symbols lie in C is proportional to Cl.

In the second bullet, the "percentage of primes" needs to be defined carefully. Thisand the Chebotarev Density Theorem are discussed in §8.B of [8]. See also [37].

We can reformulate this in terms of Gal(L/Q) G c as follows. A permutationhas cycle typed1,...,dr, whered1

product of disjoint cycles (including 1-cycles) of lengths d1,. . . , dr. In Exercise 9


you will prove that two elements of are conjugate if and only if they have the samecycle type. For a fixed cycle type d1,... , dr, the set

(13.44) {aEG o-hascycletypedi,...,dr}

is a union of conjugacy classes in G (see Exercise 10). Hence, if we fix the cycletyped1,. . . , dr of an element of G, then the Chebotarev Density Theorem implies thefollowing:

• There is some prime p for which the irreducible factors of f modulo p havedegrees

• The percentage of primes for which the irreducible factors of f modulo p havethese degrees is proportional to the number of elements of G with this cycle type.

Here is an example of what this looks like in practice.

Example 13.4.8 Consider! = x4 — 7x3 + 1 9x2 — 23x + 11 E 7L[x], which has =53• For the 200 primes 7 p 1237, it is straightforward to compute the degreesof the irreducible factors of f modulo p using Mathematica or Maple. When wetabulate the resulting degree patterns and the percentage of primes corresponding toeach pattern, we obtain:

(13 45)irreducible factors 4 of degree 1 2 of degree 2 1 of degree 4

percentage of primes 25% 23% 52%

The last column shows that f remains irreducible for some primes, so that the Galoisgroup off contains a 4-cycle by Theorem 13.4.5. In 54, a 4-cycle (abcd) generatesthe subgroup

((abcd)) = {e = (a)(b)(c)(d), (ac)(bd), (abcd), (dcba)}.

For such a subgroup, the distribution of cycle types is:

(1346) cycle type 1,1,1,1 2,2 4

percentage of elements 25% 25% 50%

By the Chebotarev Density Theorem, the close match with (13.45) strongly suggeststhat the Galois group off is cyclic of order 4. However, there could be a large primewhose degree pattern is not in (13.45). Hence this is not a rigorous computation ofthe Galois group.

The papers [16] and [26] discuss this method for computing Galois groups.

• Bad Reduction Modulop. When! e Z[x] is monic and separable, Theorem 13.4.5gives information about the Galois group off by reducing f modulo primes p suchthat p If instead thenf is not separable and our arguments fail.

When this happens, more advanced methods using the decomposition group andthe inertia group can still provide useful information about the Galois group. Forexample, in Section 6.4 we mentioned that f = x — 1 has Galois group over Q

OTHER METHODS 409

when n � 2. As explained on page 42 of reference [4] to Chapter 6, this is proved byreducing modulo the primes dividing the discriminant.


Exercise 1. Verify the computations given in Example 13.4.1.

Exercise 2. Prove that the polynomial (y) defined in (13.41) lies in K[ui,. .. , u,, , y].

Exercise 3. This exercise is concerned with the proof of Theorem 13.4.2.(a) Let . . . , E L. Prove that y + is irreducible in L[ui,. . . , ,y]. (This implies

that (13.41) is the irreducible factorization of (y) in . , u,, ,y].)(b) Letg,h E F[uj,. . andassumethatinthelargerringL[ui,. . . wehaveh=gq

forsomeqEL[uj,...,un,y].(c) In the final part of the proof of Theorem 13.4.2, we showed that G C a. Prove the

opposite inclusion.

Exercise 4. Consider the polynomial (y) when f = x3 + x2 — 2x 1 from Example 13.4.3.(a) Compute Su(y) E Q[ui, u2, U3,y], and derive the factorization given in Example 13.4.3.(b) Let h be the first factor of Su(y) given in Example 13.4.3, multiplied by —1 so that it is

monic in y. Using SymmetricReduction in Mathematica or NormalForm in Maple asin Section 2.3, write h as a polynomial in y so that its coefficients are of the form

a symmetric polynomial in UI, U2, U3 + a remainder in , U2, U3.

This should give the formula for h given in Example 13.4.3.

Exercise 5. Use the method of part (a) of Exercise 4 to derive the factorization of (y) givenin Example 13.4.4.

Exercise 6. As in the proof of Theorem 13.4.5, suppose that we have Su(y) E Z[ui, . . . , un,y]

and h E Q [ul,. . . , u,, , y] is an irreducible factor of su(y) when (y) is regarded as an elementof Q[ui,. . . , In this exercise we will study how close h is to being an irreducible factor

(a) We know that the rings Z[xi, . . . and Q[xi, . . . ,xn] are both UFDs. Prove that iff . . is irreducible and nonconstant, then it is also irreducible when regardedas an element of Q[xi,. . . ,x,].

(b) Prove that if (y) and h are as above, then h is a Q-multiple of an irreducible factor ofsu(y)

Exercise 7. Let f = x5 + 20x + 16 E Q[x] be the polynomial of Example 13.4.6. Show that Iis irreducible over Q, and compute its discriminant and irreducible factonzation modulo 7.

Exercise 8. Compute the Galois group of f = x5 — fix +3 over Q using reduction modulo 11and the method of Example 13.4.6.

Exercise 9. Prove that two permutations in are conjugate if and only if they have the samecycle type.

Exercise 10. Let G be a subgroup of Sn. For a fixed cycle type di,. . . , consider the set(13.44) of all elements of G with this cycle type.(a) Prove that this set is either empty or a union of conjugacy classes of G.


(b) Give an example where the set is empty, and give another example where it is a union oftwo conjugacy classes of G.

Exercise 11. This exercise will explore the ideas introduced in Example 13.4.8.(a) For each transitive subgroup of S4, make a table similar to (13.46) that lists the number

of elements of each possible cycle type for that subgroup.(b) For each polynomial in Exercise 14 of Section 13.1, compute its factorization modulo

200 primes, and record your results in a table similar to (13.45). Use this to guess theGalois group of each polynomial.

REFERENCES

I. B. C. Berndt, B. K. Spearman, and K. S. Williams, Commentary on an unpublished lectureby G. N. Watson on solving the quintic, Math. Intelligencer 24, no. 4 (2002), 15—33.

2. N. Bruin and N. D. Elkies, Trinomials ax7 + bx + c and ax8 + bx + c with Galois groups oforder 168 and 8 168, in Algorithmic Number Theory (Sydney, 2002), edited by C. Fiekerand D. R. Kohel, Lecture Notes in Comput. Sci. 2369, Springer, Berlin, 2002, 172—188

3. W. Bumside, Theory of Gmups of Finite Order, Cambridge U. P., Cambridge, 1897.

4. J. J. Cannon and D. F. Holt, The transitive permutation groups of degree 32, Experiment.Math. 17(2008), 307—3 14.

5. H. Cohen, A Course in Computational Algebraic Number Theory, Springer, New York,Berlin, Heidelberg, 1993.

6. K. Conrad, Galois groups of cubics and quartics in all characteristics, preprint. Avail-able at http: Ilwww.math.uconn. edu/-kconrad/blurbs/galoistheory/cubic—quarticalichar . pdf. (Note that the hyphen is not part of the URL.)

7. J. Conway, A. Hulpke, and J. McKay, On transitive permutation groups, LMS J. Comput.Math. 1 (1998), 1—8.

8. D. Cox, Primes of the Form x2 + fly2, Wiley, New York, 1989.

9. P. Doyle and C. McMullen, Solving the quintic by iteration, Acta Math. 163 (1989),15 1—180.

10. D. Dummit, Solving solvable quintics, Math. Comp. 57 (1991), 387—401.

11. D. W. Erbach, J. Fischer, and J. McKay, Polynomials with PSL(2, 7) as Galois group, J.Number Theory 11(1979), 69—75.

12. GAP: Groups, Algorithms and Programming. Available at http://www.gap—system.org/.

13. K. Girstmair, On invariant polynomials and their applications in field theory, Math.Comp. 48(1987), 78 1—797.

14. K. Girstmair, On the computation of resolvents of Galois groups, Manuscripta Math. 43(1983), 289—307.

15. A. Hulpke, Constructing transitive permutation groups, J. Symbolic Comput. 39 (2005),1—30.

REFERENCES 411

16. A. Hulpke, Techniques for computation of Galois groups, in Algorithmic Algebra andNumber Theory (Heidelberg, 1997), edited by B. Matzat, G.-M. Greuel, and G. Hiss,Springer, Berlin, 1999, 65—77.

17. Y. Ishibashi and T. Nakamizo, A note on the Galois group of a quartic polynomial, Bull.Fac. School Ed. Hiroshima Univ. 19 (1997), 39—41.

18. L.-C. Kappe and B. Warren, An elementary test for the Galois group of a quartic polyno-mial, Amer. Math. Monthly 96 (1989), 133—1 37.

19. R. B. King, Beyond the Quartic Equation, Birkhäuser, Boston, Base!, Berlin, 1996.

20. E Klein, Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree,English translation by George Gavin Mornce, Trübner & Co., London, 1888. Reprint byChelsea, New York, 1956.

21. S. Kobayashi and H. Nakagawa, Resolution of solvable quintic equation, Math. Japon. 37(1992), 883—886.

22. H. F. Kreimer, Review of[17], Math. Reviews, 98a: 12005.

23. D. Lazard, Solving quintics by radicals, in The Legacy of Niels Henrik Abel, edited by 0.Laudal and R. Piene, Springer, Berlin, 2004, 207—225.

24. Magma Computational Algebra System. Available at http: I/magma, maths. usyd.edu. au/magma!.

25. A. L. Maistrova, PemeBMe B pa6oTax .T1.[Solution of algebraic equations in the works of L. Euler], Istor.-Mat. Issled. No. 29(1985), 189—199. Unpublished English translation by A. Shentizer.

26. J. MCKay, Some remarks on computing Galois groups, SIAM J. Comput. 8 (1979), 344—347.

27. J. Pierpont, Zur Geschichte der Gleichung des V. Grades (bis 1858), Monatsh. Math.Phys. 6(1895), 15—68.

28. W. M. Ruppert, On the Bring normal form of a quintic equation in characteristic 5, Arch.Math. 58(1992), 44—46.

29. P. Samuel, Projective Geometry, translated by S. Levy, Springer, New York, Berlin,Heidelberg, 1988.

30. J. Shurman, Geometry of the Quintic, Wiley, New York, 1997.

31. J. H. Silverman and J. Tate, Rational Points on Elliptic Curves, Springer, New York,Berlin, Heidelberg 1992.

32. L. Soicher and J. MCKay, Computing Galois groups over the rationals, J. Number Theory20(1985), 273—281.

33. B. K. Spearman and K. S. Williams, Characterization of solvable quintics x5 +ax+b,Amer. Math. Monthly 101 (1994), 986—992.

34. B. K. Spearman and K. S. Williams, On solvable quintics x5 + ax + b and x5 + ax2 + b,Rocky Mountain J. Math. 26 (1996), 753—772.

35. B. K. Spearman and K. S. Williams, Quartic trinomials with Galois groups A4 and V4.Far East J. Math. Sci. (FJMS) 2 (2000), 665—672.

36. R. P. Stauduhar, The determination of Galois groups, Math. Comp. 27 (1973), 981—996.


37. P. Stevenhagen and H. W. Lenstra, Jr., Chebotarëv and his density theorem, Math. Intel-ligencer 18, no. 2 (1996), 26—37.

38. J. Stiliwell, Eisenstein 'sfootnote, Math. Intelligencer 17, no. 2 (1995), 58—62.

39. E. Weisstein, CRC Concise Encyclopeadia of Mathematics, Second Edition, Chapman &Hall/CRC, Boca Raton, FL, 2003.

40. Wolfram Research, Solving the Quintic, Poster, Wolfram Research, Champaign, IL, 1995.Available at http://library.wolfrain.com/examples/quintic/.

CHAPTER 14

SOLVABLE PERMUTATION GROUPS

This chapter will study solvability by radicals for irreducible polynomials of degreep or p2. where p is prime. These results go back to Galois and illustrate his amazinginsight into group theory. We will also discover why Galois invented finite fields.

While Galois's result for degree p is relatively easy to prove, understanding thecase of degree p2 requires the theory of permutation groups (subgroups of Sn).We will see that in degree p2. irreducible polynomials can be either primitive orimprimitive. The case of solvable imprimitive subgroups of S,,2 will be consideredfirst, followed by the more complicated case of solvable primitive subgroups. Theproofs will involve surprising amounts of group theory.

14.1 POLYNOMIALS OF PRIME DEGREE

The goal of this section is to prove the following wonderful theorem of Galois.

Theorem 14.1.1 Let F be afield of characteristic 0, and let f be irreducibleof prime degree p. Then the following are equivalent:(a) f is solvable by radicals over F.(b) For every pair of roots cs /3 off, F (cr, /3) is the splitting field off over F.


414 SOLVABLE PERMUTATION GROUPS

(c) For some pair of roots /3 off, F(a, /3) is the splitting field off over F.(d) The Galois group off over F is isomorphic to a subgroup of AGL( 1, F,,).

The proof will be given later in the section. Recall from Section 6.4 that the groupAGL( 1, IF,,) consists of all functions

'7a,b(u)_a11+1), aElF,', belF,,.

If we identify the congruence classes [1], [2J,.. , [p] E IF,, with the numbers 1,2,. . , p.then AGL( 1, IF',,) becomes a subgroup of S,,. Thus

AGL(l,F,,) c S,,

is a subgroup of order p(p — 1), and an element 'Yab E AGL( 1, F,,) is the permutation

2a+b pa+b) —

where we interpret everything modulo p. In particular, let 8 = Then we havethe p-cycle

0= =(12...p)eAGL(l,F,,).

Here are two useful facts about 9 and AGL( 1 ,IF,,).

Lenuna 14.1.2(a) AGL(1,IF,,) is the normalizer of (8) in(b) If 'r E S,, satisfies e then r E AGL(l,F,,).

Proof: The normalizer of (8) in S1., consists of all 'r e S,, such that T(0)T' = (9).In Exercise 1 you will show that r lies in the normalizer if and only if

Since (i) = i + £, the above equation is equivalent to the identity

(14.1) i=l,...,p,

where as usual we interpret everything modulo p. This implies that

and more generally, one easily proves that for any positive j,

i=l,...,p(see Exercise 1). Then setting i = p gives

'r(j) = -i-(p + = T(p) + =

POLYNOMIALS OF PRIME DEGREE 415

This shows that r = E Conversely, it is easy to see that any

Ya,b E satisfies (14.1) with a. This proves part (a) of the lemma.For part (b), first observe that (0) is a p-Sylow subgroup of since

= p(p — 1). Furthermore, it is unique by the Second Sylow Theorem(see Theorem A.5.l), since (0) is normal in by part (a).

Now assume that r and r9r1 E Then (rOr') is a also p-Sylow subgroup of By uniqueness, (0) = = T(0)r1. Thus rnormalizes (9) and hence lies in by part (a). •

We will use the following lemma several times. See Exercise 2 for a proof.

Lemma 14.1.3 Suppose that H is a normal subgroup of a finite group G and letg E G. If the order of g is relatively prime to [G: HI, then g E H.

We can now characterize solvable transitive subgroups of S,,.

Proposition 14.1.4 Every solvable transitive subgroup G C S,, is conjugate to asubgroup of AGL( 1, containing 0.

Proof: Since G is transitive, the orbit of any i E { 1,... ,p} is all of {1,. . . ,p}. Thusp divides the order of G by the Fundamental Theorem of Group Actions. HenceG contains an element of order p, by Cauchy's Theorem. This element must be ap-cycle, since we are in (the order of a permutation is the least common multipleof its cycle lengths). By Exercise 9 of Section 13.4, any p-cycle in is conjugateto 0. Replacing G with a suitable conjugate, we may assume that 0 E G.

Since G is solvable, we can find subgroups

{e} = G0 C G1 C C C G

suchthatGt_i isnormalinGt and [Gt:Gt_i1 l,...,n.Let i be the smallest index such that 0 E G1. Note that i>We first claim that G_1 I = p. To see why, suppose that [G,: I = q, where

q p is prime. Since 9 E G1 has order p, Lemma 14.1.3 implies that 0 e G_1, whichcontradicts the definition of i. Hence = p.

We next claim that i = 1. If i> 1, then there is r E G1_1 such that r(j) = k forsome j k mod p. Then maps k to j, so that p = e fixes k. Thus pis a product of disjoint cycles of lengths <p. Since p is prime, it follows that theorder of p is relatively prime to p = [Gi: G,_1]. Hence p E G1_1 by Lemma 14.1.3,and then of E G1_1 follows from i- e G,_1. Since j k mod p, this implies that9 which contradicts the definition of i. Hence i = 1.

Since G0 = {e}, these claims imply that G1 has order p and contains 0. Since 0has orderp, we conclude that G1 = (9). It follows that G1 C

Now let 1 <j < n be the largest index such that C Suppose thatj <n, and take E Then 0 E G1 C G3 implies that rOr' E since is

normal in This gives Y8T1 E sothatT E by part(b)ofLemma 14.1.2. SinceT E G1+, was arbitrary, we concludethat G1+i CThis contradicts the definition of j. Hence we must have j = n, which gives thedesired inclusion G = C •


We now have the tools needed to prove Galois's theorem.

Proof of Theorem 14.1.1: Let f have roots i, .. . , a splitting field L. ThenGal(L/F) is isomorphic to G C Si,, where a E Gal(L/F) maps to r E G such that

= By Proposition 6.3.7, G is transitive, sincef is irreducible.First consider (a) (d). 1ff is solvable by radicals over F, then G is transitive (by

the above) and solvable (by Theorem 8.5.3). Using Proposition 14.1.4, we concludethat G is conjugate to a subgroup of AGL( 1, Fr,). This proves (a) (d). For theconverse, note that AGL( 1, IF,,) is solvable by Example 8.1.6, and then any subgroupis also solvable by Proposition 8.1.3. Thus the Galois group off over F is solvable,so that f is solvable by radicals over F by Theorem 8.5.3.

We next prove (b) (c) (d) (b). The first implication is obvious. For thesecond, observe that in the proof of Proposition 14.1.4, the first paragraph appliesto any transitive subgroup of S,,. Thus we may assume that (0) C G. Now supposethat L = F i the

F

the first has degree p, since f is irreducible over F, and the second has degreeat most p — 1, since is a root of f/ (x — a,) E F (a,) [xl. By the Tower Theorem,

(14.2) GI = IGal(L/F)I = [L:F] =pm, 1 cm

p is prime, it follows that (0) is a p-Sylow subgroup of G. According to theThird Sylow Theorem (see Theorem A.5.1), the number of p-Sylow subgroups of Gdivides I GJ and is congruent to 1 modulo p. In Exercise 3 you will use this and (14.2)to show that (9) is the unique p-Sylow subgroup of G and hence is normal in G. Itfollows that G is contained in the normalizer of (0) in Si,. By Lemma 14.1.2, thenormalizer is AGL(1 ,]F,,), so that G C AGL(1 Thus Gal(L/F) is isomorphic toa subgroup of

For (d) (b), relabel the roots off so that G C AGL( I, Fr,). We need to show thatF(a,, is the splitting field of f over F for any i j. By the Galois correspondence,it suffices to show that the only element a Gal(L/F) fixing F(a,, aj) is the identity.Since a corresponds to T G and G C we see that r =a and b E IF1,. Then

a aai+b =

aaj+b=cEj.

This gives the equations ai + b = i and aj + b = j, which modulo p have the uniquesolution a = 1,b = 0, since i f modulo p. Thus T = is the identity, so that a isthe identity in Gal(L/F). Hence F(a,, aj) is the splitting field. •

We first encountered the affine linear group AGL( 1, IFP) as the Galois group of

Q c Q(ç, in Section 6.4. This extension is the splitting field of —2 Q[x],which is obviously solvable by radicals. Hence what we did in Section 6.4 is a perfectillustration of Theorem 14.1.1.

POLYNOMIALS OF PRIME DEGREE 417

Mathematical Notes

The proof of Theorem 14.1.1 uses the following concept from group theory.

Frobenius Groups. We showed above that F(cx1, aj) is the splitting field by arguingthat the identity is the only element of AGL( 1, IF,,) that fixes i and j. This generalizesas follows. If a finite group G acts transitively on a set X such that 1 < lxi <and for every x y in X the identity is the only element of G fixing x and y, then wesay that G is a Frobenius group. When this happens, the isotropy subgroup of anyx E X is called a Frobenius complement. A discussion of Frobenius groups can befound in [3, Sec. 3.4] and [14, p. 90]. See also Exercise 4.

Historical Notes

Galois considered Theorem 14.1.1 to be one the best applications of his theory.His version of the theorem is as follows [Galois, p. 69]:

PROPOSITION VIII

THEOREM. In order that an irreducible equation of prime degree be solv-able by radicals, it is necessary and sufficient that when any two of the roots areknown, the others can be deduced from them rationally.

If we are working over a field F, then "deduced from them rationally" means thatthe other roots are rational functions with coefficients in F in the known rootsThis implies that F(cs,/3) is the splitting field. Thus Galois's theorem is (a) (b)of Theorem 14.1.1. Galois especially liked this result because its statement doesn'tmention Galois theory, yet the Galois group is crucial to the proof.

As for part (d) of Theorem 14.1.1, Galois says the following [Galois, p. 67]:

Therefore, "if an irreducible equation of prime degree is solvable by radicals,then the group of the equation contains only substitutions of the form

Xk Xak+b

a and b being constants."

Reciprocally, if this holds then I say that the equation will be solvable byradicals.

This is (a) (d) of Theorem 14.1.1. Galois denotes the roots of the polynomial asxo,... where n is prime (his n is our p). Furthermore, on page 65, Galois says"We set in general x,, = = x1,..." Thus Galois treats the indices modulo njust as we treat them modulo p.

Galois published his Proposition VIII separately in 1830, before he had workedout his general theory of solvability. It is possible that thinking about this case(and the group AGL( 1, IF,,) in particular) led Galois to the general idea of normalsubgroups and solvability. The irony is that by focusing attention on the special caseof Theorem 14.1.1, Galois distracted his contemporaries from the real depth of hisinnovations.


Galois also formulated solvability by radicals in terms of the resolvents discussedin Chapters 12 and 13. Let f be irreducible of degree p. Since AGL( 1 ,IF,,) has index(p —2)! in S,,, the theory of Section 13.3 constructs a resolvent E F[y] suchthat if the Galois group off is isomorphic to a subgroup of AGL( 1, IF',,), then ef(y)has a root in F, and the converse is true provided that the root is simple. Because ofthis, Galois asserts that to check solvability by radicals,

it suffices to know whether or not this auxiliary equation [our e1(y)] of degree1.2.3. . (n — 2) has a rational root.

(See [Galois, p. 69].) Here, Galois's n is our p, and "rational root" means root in F.Although Galois (like Jordan) missed the importance of simple roots, this resultcan be regarded as the generalization of Corollary 13.2.11 for an arbitrary prime.Galois also knew how to build the resolvent e1(y) using the methods discussed inSection 12.1. For example, the polynomial (12.20) appears in Galois's memoir.

In a letter written to Crelle in 1828, Abel stated a version of (a) (b) (c) fromTheorem 14.1.1 (see [Abel, Vol. II, p. 270]). Also, an earlier letter to Crelle gaveformulas for the roots of a solvable polynomial of degree 5 and claimed that similarresults apply in degrees 7, 11, 13, etc. Unfortunately, no details of his proofs areknown. Kronecker fleshed out Abel's ideas in an 1853 paper that is discussed in [5].


Exercise 1. This exercise is concerned with the proof of part (a) of Lemma 14.1.2. Let

(a) Prove that E S,, lies in the normalizer of (0) if and only if r9 = for some 1 �p—i.

(b) Prove that (14.1) implies that r(i+j) = r(i) for all positive integers j.

Exercise 2. Let H be a normal subgroup of a finite group G and let g E G. The goal of thisexercise is to prove Lemma 14.1.3.(a) Explain why = H in the quotient group a/H.(b) Now assume that gcd(o(g), [G: H]) = 1. Prove that g E H.

Exercise 3. Let G satisfy (14.2). Use (14.2) and the Third Sylow Theorem to prove that G hasa unique p-Sylow subgroup H of order p. Then conclude that H is normal in G.

Exercise 4. The definition of Frobenius group given in the Mathematical Notes involves agroup G acting transitively on a set X. Prove that a group G is a Frobenius group if and onlyifGhasasubgroupHsuchthat 1< HI <

Exercise 5. Let F be a subfield of the real numbers, and let f E F[x] be irreducible of primedegree p> 2. Assume that f is solvable by radicals. Prove that f has either a single real rootor p real roots. This was proved by Kronecker in 1856 using methods due to Abel (see [15]).

Exercise 6. By Example 8.5.5, f = — 6x+ 3 is not solvable by radicals over Q. Give a newproof of this fact using the previous exercise together with the irreducibility off and part (b)of Exercise 6 from Section 6.4.

Exercise 7. Use Lemma 14.1.3 and part (a) of Lemma 14.1.2 to give a proof of part (b) ofLemma 14.1.2 that doesn't use the Sylow Theorems.

IMPRIMITIVE POLYNOMIALS OF PRIME-SQUARED DEGREE 419

Exercise 8. Let f E Fix] be irreducible of prime degree p � 5, where F has characteristic 0,and let a be roots off in some splitting field. If F(a, contains all other roots of

f is solvable by radicals by Theorem 14.1.1. But suppose that there is some third root 'ysuch that -y E F(a, Is this enough to force f to be solvable by radicals?(a) Use the classification of transitive subgroups of S5 from Section 13.2 to show that the

answer is "yes" when p = 5.

(b) Use the polynomial x7 — I 54x + 99 from Example 13.3.10 to show that the answer is "no"when p = 7.

14.2 IMPRIMITIVE POLYNOMIALS OF PRIME-SQUARED DEGREE

Having studied polynomials of prime degree p. we turn our attention to polynomialsof degree p2. In this section, we will see that such polynomials can be either primitiveor imprimitive. Our main result (Theorem 14.2.15) will describe the Galois group ofan irreducible imprimitive polynomial of degree p2 that is solvable by radicals. Theprimitive case will be considered in Sections 14.3 and 14.4.

The proof of Theorem 14.2.15 will require that we study permutation groups,i.e., subgroups of After defining primitive and imprimitive permutation groups,we will concentrate on the imprimitive case and use wreath products to classify allsolvable transitive imprimitive subgroups of Sp2. Primitive permutation groups willbe considered in Section 14.3.

A. Primitive and Imprimitive Groups. By Section 6.3, the Galois group of aseparable polynomial f of degree n gives a permutation group G C that records theGalois action on the roots off. An important idea in Galois theory is that propertiesoff should be reflected in the properties of G. For example, Proposition 6.3.7 saysthat f is irreducible over F if and only if G is transitive. Thus "transitive" is thepermutation group analog of "irreducible" for polynomials.

We next consider the concepts of imprimitive and primitive, which apply to bothpolynomials and permutation groups. We will begin with the former, where the ideais that separable polynomials come in two flavors, according to whether or not theroots break up into "blocks" under the action of the Galois group.

Before giving the general definition, let us consider an example.

Example 14.2.1 The polynomial f = —2 E Q[x] is separable with roots

and

We have written the roots in two blocks that have the following nice property: Ifwe apply a E Gal(Q(i, .ñ)/Q) to the first block of roots, then the result is eitherthe first block or the second block, and the same is true if we apply a to the secondblock. This follows because a(—ct) = —a(a). Hence the action of the Galois grouprespects the block structure of the roots.

This leads to the following general definition.


Definition 14.2.2 Letf E F[x] be a separable polynomial with splitting field L.(a) f is imprimitive the set of roots off can be written as a disjoint union

such that for every o- e GaI(L/F) and 1 <i k, we have a(R,) = some1 j k. We also require thatk> 1 and lfor some i.

(b) f is primitive if it is not imp rimitive.

In the definition of imprimitive, the R are the blocks, and a(R,) = R means thatthe Galois group preserves the block structure of the roots. The requirements thatk> 1 and some 1R11> 1 exclude the trivial block structures where there is only oneblock or where every block consists of a single root.

When the polynomial is also irreducible, we get some useful information aboutthe size of the blocks in the imprimitive case.

Lemma 14.2.3 Letf E F [xl be irreducible and separable of degree n. Assume thatf is imprimitive with roots R1 U as in Definition 14.2.2. Then every R, hasthe same number of elements, say 1. Thus n = kl, where k> 1 and 1> 1.

Proof: Given blocks R1 and R3 with i j, pick E R, and /3 E Since f isirreducible and L is its splitting field over F, Gal(L/F) acts transitively on the roots.Thus there is a E Gal(L/F) such that a(a) = Since f is imprimitive, we havea(R1) = so that = 1R11, since a is one-to-one. If 1 = 1R11, then

n=IRI[+••.+IRkI=kl,

since f is separable. Then k> 1 and 1> 1 follow from Definition 14.2.2. •

Here are some easy examples.

Example 14.2.4 If f is irreducible and separable of prime degree p, then f cannotbe imprimitive, since it is impossible to write p = ki with k> 1 and 1> 1. Henceirreducible separable polynomials of prime degree are automatically primitive.

However, if f is irreducible and separable of degree p2. then f can be eitherprimitive or imprimitive. In the latter case, we must have p blocks, each consistingof p roots. When p = 2, we saw an instance of this in Example 14.2.1.

We translate these concepts into group theory as follows.

Definition 14.2.5 Let G be a subgroup of Then:(a) G is imprimitive there is a disjoint union

{l,...,fl}=RIU...URk,

such that for every r E G and every 1 <i < k, we have 'r(R1) = R3for some1 j k. We also require thatk> I and lfor some i.

(b) G is primitive if it is not imprimitive.


Here is an example of an imprimitive permutation group.

Example 14.2.6 The subgroup G = ((1324), (34)) C S4 is imprimitive via the blocksR1 = {l,2} and R2 = {3,4}. This follows because (1324) maps each block to theother while (34) takes each block to itself.

If we label the roots of x4 —2 as = a2 = = = thenGal(Q(i, G C S4. Do you see how the above blocks relate to those usedin Example 14.2.1?

Lemma 14.2.3 has the following group-theoretic analog.

Lemma 14.2.7 Let G be a transitive subgroup of Assume that G is imprimitivewith blocks R1,... ,Rk as in Definition 14.2.5. Then every R has the same number ofelements, say 1. Thus n = ki, where k> I and 1> 1. •

We omit the proof because it is identical to the proof of Lemma 14.2.3. Be sureyou understand this.

B. Wreath Products. Let f e F[x] be separable, irreducible, and imprimitive ofdegree n. How does being imprimitive restrict the Galois group of f? As we willsee, the answer involves the concept of wreath product.

By what we've done so far, our question reduces to the study of transitive imprim-itive subgroups G c By Lemma 14.2.3, we have k> 1 blocks, each consistingof 1> 1 elements, where n = ki. To begin our analysis, we will regard S,, = Ski aspermutations of the product { 1,. .. , k} x { 1,. . . , l} and use the blocks

(14.3) {l,...,k}X{l,...,l}=RIU...URk, R,={i}x{l,...,l}.

Then consider

(14.4)

Sk is an imprimitive subgroup of Ski = with respect to theblocks R1, . . . , Rk. We call Sk 1 Si the wreath product of Sk with

We can describe an element o• E Sk 1 Si as follows. Since R, = {i} x { 1,. . . , l} anda(R,) there is a unique e such that for all (i,j) ER1, we have

a(i,j) = e

Thus describes how a maps R1 to If we write a = (r;p1,. . . ,ILk), then

(14.5) SklSi = {(r;/21,...,/Lk) Sk, l-t1,...,/-tk E Sd.

In more concrete terms, think of a dresser with k drawers and I items in each drawer.Then elements of the wreath product (14.5) permute the items in the dresser bypermuting the drawers via r and permuting the items in each drawer via the pj.

Using (14.5), we can describe some interesting subgroups of Sk Given sub-groups A C Sk and B C Si, define the set

A lB = {(r; /11,. .. ,,Uk) Ir E A,p1,... ,/1k E B}.


This is a subgroup with the following properties.

Lemma 14.2.8 Let A C Sk and B C S1 be subgroups. Then:(a) A lB is a subgroup of Sk S1.(b) The map (r; Pk) defines a group homomorphism A B A that is

onto and whose kernel is isomorphic to Bk = B x ... x B (k times).

Proof: Giveno=(T;pi,...,pk)EA1B, wefirstshowthata' EA1B. Sinceamaps R to via it is clear maps RT(j) to R via If we set j =then i = r (j), so that maps R3 to RT-l(j) via It follows that

(14.6) =

This obviously lies in A B. In Exercise 1 you will show that

(14.7) (r;pI,...,pk)(T';pc,... '14) =

Hence A B is closed under multiplication, and part (a) follows.It remains to prove part (b) of the lemma. The multiplication fonnula (14.7) shows

that (r; Pk) r is a homomorphism, which is clearly onto by the definitionof A B. Furthermore, its kernel is clearly the set

{(e;pI,...,pk)

Then (14.7) shows that the obvious map to is a group isomorphism. .The subgroups Sk 1 S1 C Ski = S. have the following important property.

Proposition 14.2.9 Every transitive imprimitive subgroup of is conjugate to asubgroup of Sk 1 S1 for some nontrivialfactorization n = ki.

Proof: Let G c be transitive and imprimitive. By Lemma 14.2.7, we have blocks

where each has I elements and every o E G maps to some To compare thisto (14.3), pick T E Ski with the property that = for 1 i k. Such a r existsbecause IR, = One easily checks that if a E G maps to then ar maps

Itfollowsthatr'GTCSklSI. •

Proposition 14.2.9 implies the following result about the Galois group of animprimitive polynomial.

Corollary 14.2.10 Let f E F[xJ be separable, irreducible, and imp rimitive of degreen. Then n has a nontrivial factorization n = kI such that the Galois group off overF is isomorphic to a subgroup of Sk lSi.

Here is an example of this result in degree 6.


Example 14.2.11 Suppose that f = x6 + bx4 + cx2 + d e F [x] is irreducible andseparable over a field of characteristic 2. It is easy to see that this polynomial isimprimitive, for if is a root, then so is —ce. Hence the roots can be partitioned intothe three blocks

R1 = R2 = R3 = {-y,—'y}

that are obviously permuted by the Galois group. Thus the Galois group off overF is isomorphic to a subgroup of S3 S2 C S6. This group has order 48. For

4x2 + 1 E Q[x], the galois command of Maple shows that its Galois group overQ has order 48. Hence the Galois group is isomorphic to

For x6 — 4x2 1 E Q[xI, the discriminant is 262292. Thus its Galois group over Qis isomorphic to a subgroup of (S3 iS2) nA6, which has order 24 (you will verify thisin Exercise 2). The galois command shows that the Galois group has order 24, sothat the Galois group is isomorphic to (S3 152) nA6 in this case. See Exercise 3 formore on the structure of these groups.

When n = p2 and p is prime, the only nontrivial factorization of n is n = p p.This gives the following corollary of Proposition 14.2.9 that will be useful later inthe section.

Corollary 14.2.12 Every transitive imprimitive subgroup of S,,2 is conjugate to asubgroup of Sp

For an irreducible imprimitive polynomial of degree p2. it follows that the Galoisgroup is isomorphic to a subgroup of S,, iSo. By (14.5), the order of this group is

=

This may seem like a large number, but it is actually quite small in comparison withSp21 = (p2)!. Here is an example.

Example 14.2.13 When p = 17, we have

ISI7H = 289! 2.1 x 10587, while 151715171 = (l7!)18 8.3 x 10261.

So IS17 1S171.

We conclude by determining the structure of S2 S2.

Example 14.2.14 The order of S2 S2 is = 8. To figure out which group oforder 8 this is, recall from Example 14.2.6 that ((1324), C 54 is imprimitive.This has order 8 and by Corollary 14.2.12 is conjugate to a subgroup of 521 S2. Itfollows that

((l324),(34)).

In particular, S2 152 is a dihedral group of order 8.


C. The Solvable Case. We now have all of the tools needed to classify solvableimprimitive subgroups of The key player is the solvable permutation groupAGL(1,IF,,) c S,,. Using Lemma 14.2.8, we obtain the wreath product

c S,,iS,, C S,,2.

This allows us to describe all transitive imprimitive solvable groups of

Theorem 14.2.15 Let G be a transitive subgroup of S,,2. Then the following areequivalent:(a) G is solvable and imprimitive.(b) G is conjugate to a subgroup of

Proof: Since is solvable, is also solvableby Exercise 4. Then (b) (a) follows easily, since every subgroup of S,, S,, isimprimitive and every subgroup of AGL( 1 ,IF,,) AGL( 1, is solvable.

We now consider (a) (b). Let G C S,,2 be transitive, solvable, and imprimitive.

By Corollary 14.2.12, we may assume that G C S,, 2Sf. Let G' be the image of

G under the homomorphism S,, S,, —÷ S,, of part (b) of Lemma 14.2.8. We claim

that G' is transitive. To prove this, take any i and j, and pick u E R, and v E R3.Since G is transitive, there is a = ,p,,) E G such that a(u) = v. Thena(R,) = which implies that -r(i) = j. Hence G' C S,, is transitive. Since G' issolvable by Theorem 8.1.4, Proposition 14.1.4 implies that 5G' C AGL( 1, forsome 8 S,,. It follows that after conjugating G by (8; e,. . . , e) E 5p iS,,, we haveG' C Thus

(14.8)

and we are halfway done with the proof.Now fix i between 1 and p, and consider the group

In Exercise 5 you will show that the map G1 —+ 5,, defined by

(14.9)

is a group homomorphism. By Exercise 5 the image G C of this map is transitiveand solvable. Then Proposition 14.1.4 implies that there is 8, e such that =(12. ..p) E 81G81' C Hence, after we conjugate G by (e;61,...,8,,),we may assume that

(14.10) and OEG

for all i. Notice that (14.8) continues to hold.Now let a = G be arbitrary, and fix j between 1 and p. We

will prove that E as follows. By (14.10) with i = -r(j), we can find


(p;v1,... e G such that p(i) i and v, = 9. Using (14.6) and (14.7), we obtainthe element

(14.11)

where = (see Exercise 5). Since

T'pT(j) =r'(i) =j,we see that -y E so that by (14.9),

= E C

AGL( 1, IF,,) by part (b) of Lemma 14.1.2. Since j was arbitrary

a polynomial is solvable by radicals if and only if its Galoisgroup is solvable. Hence we have the following corollary of Theorem 14.2.15.

Corollary 14.2.16 Let f E F[x] be irreducible and imprimitive of degree p2, andassume that F has characteristic 0. Then f is solvable by radicals over F and onlyif the Galois group off over F is isomorphic to a subgroup of the wreath product

.This corollary shows that the size of the Galois group of an irreducible solvable

imprimitive polynomial of degree p2 is bounded by

pP+l(p_l)P+I.

As p gets larger, this becomes very small in comparison with the size of S,2.

Example 14.2.17 When p = 17, Si72 I 2.1 x 10587 and Si7 8.3 x 10261 byExample 14.2.13. In contrast,

IAGL(l,1F17) = 17181618 6.6 x

Hence, while a random polynomial of degree 172 can have a Galois group as largeas an irreducible solvable imprimitive polynomial of this degree has a muchsmaller Galois group.

Mathematical Notes

Here are some further remarks on wreath products.

• Wreath Products. The wreath product defined in the text can be generalized asfollows. Let G be any group and let A C be a permutation group. Then set

A = {(r;gi,... ,gn) E A, E G}.


Following (14.7), we define a group operation on this set via

(r;gj,... =

In Exercise 6 you will show that this makes A G into a group that satisfies part (b)of Lemma 14.2.8. You will also show that if G is finite, then

IA1GI =

One surprise is that we can represent a wreath product as a semidirect product. SeeExercise 7 for the details. Further information about wreath products can be foundin [7, p. 811 and [14, pp. 219—228].

Historical Notes

The term "primitive" is due to Galois, though he said "not primitive" instead of"imprimitive." Like us, he began with polynomials. Here is his definition of "notprimitive" [Galois, p. 163]:

One calls equations not primitive the equations that are, for example, of degreemn, which decompose into m factors of degree n by means of a single equationof degree m. Such are the equations of Gauss. Primitive equations are those thatdo not possess such a simplification.

To understand this, suppose that f E F]x] is separable of degree mn with splittingfield F c L. Having "a single equation of degree m" means that we adjoin all rootsof such a polynomial of degree m. This gives a subfield

FcKCL

such that F c K is Galois, and having f decompose "into m factors of degree n bymeans of" this subfield means that there is a factonzation

(14.12) fEK[xI.

We can relate this to our definition of imprimitive as follows. Let f E F [x] be monic,separable, and imprimitive, and assume also that f is irreducible (as is implicit inGalois's definition). By Lemma 14.2.3, we can assume that the roots of f fall intom blocks R1, . . . , Rm, each consisting of n roots (so that f has degree mn). Then let

be the monic polynomial whose roots are the elements of R,. In Exercise 8 youwill show that J E K[xJ, where F c K is a Galois extension determined by the blockstructure of the roots. Hence we recover (14.12).

In the above quotation, Galois claims that the cyclotomic equations considered byGauss are imprimitive. You will prove this in Exercises 9 and 10.

We also note that Galois applied the terms "primitive" and "not primitive" to bothpolynomials and groups. See, for example, [Galois, p. 79].




Exercise 2. The wreath product S3 S2 C So can be thought of as the subgroup of all per-mutations that preserve the blocks R1 = {l,2},R2 = {3,4},R3 = {5,6}. As noted in Exam-ple 14.2.l1,S31S2 hasorder6•23 =48.(a) Show that (S3 S2) fl A6 has order 24.(b) Show that is the centralizer of (12)(34)(56) in S6 (meaning that S3 iS2 consists of

all permutations in S6 that commute with (12) (34) (56)).(c) Use part (b) to show that S3 152 is isomorphic to ((S31S2)nA6) x S2.

See the next exercise for more on S3 152 and (S3 iS2) flA6.

Exercise 3. One of the challenges of group theory is that the same group can have radicallydifferent descriptions. For instance, S4 and the group G = (S3 S2) flAo appearing in Exam-ple 14.2.11 both have order 24. In this exercise, you will prove that they are isomorphic. Wewill use the notation of Exercise 2.(a) There is a natural homomorphism G —* S3 given by how elements of G permute the blocks

R1 , R2, R3. Show that this map is onto, and express the elements of the kernel as productsof disjoint cycles.

(b) Use the Sylow Theorems to show that G has one or four 3-Sylow subgroups.(c) Show that A6 has no element of order 6.(d) Use part (c) and the kernel of the map G —+ S3 from part (a) to show that G has four

3-Sylow subgroups.(e) G acts by conjugation on its four 3-Sylow subgroups. Use this to prove that G S4.

(1) Using Exercise 2, conclude that iS2 S4 x S2.

We note without proof that S3 152 54 x S2 is also isomorphic to the full symmetry group(rotations and reflections) of the octahedron.

Exercise 4. Let A and B be solvable permutation groups. Prove that their wreath product Al Bis also solvable.

Exercise 5. This exercise will complete the proof of Theorem 14.2.15.(a) Let G1 —* 5,, be the map defined in (14.9). Prove that it is a group homomorphism and

that its image C 5,, is transitive and solvable.(b) Let cy = . .

and (p; . ,v,,) beas in the proof of Theorem 14.2.15. Thus wehave a fixed j such that i = r(j), ii = 0, and p(i) = i. Now let = (r'pr;Ai, . . . , A,,)

be as in (14.11). Prove carefully that A, =

Exercise 6. Let A be a subgroup of 5,,, and let G be any group. Then define A 1 G as in theMathematical Notes.(a) Prove that Al G is a group under the multiplication defined in the Mathematical Notes.(b) State and prove a version of part (b) of Lemma 14.2.8 for A 1G.(c) Prove that A1GI=IAIIGI" when Gis finite.

Exercise 7. Let A 1 G be as in Exercise 6, and let H be the set of all functions

(a) Given E H, define e H by = Prove that this makes H into agroup isomorphic to the product group


(b) Elements of A 1G can be written (r, where 4 H. Prove that in this notation, (14.7)becomes

=

(c) Ac acts on {1,...,n}. Show that this induces an action ofA on H via ='(i)) Be sure you understand why the inverse is necessary.

(d) The action of part (c) enable us to define the semidirect product H A. Using thedescription of A G given in part (b), prove that the map

'—* (T.çb,r)

defines a group isomorphism A G H x A. This shows that wreath products can berepresented as semidirect products.

Exercise 8. The goal of this exercise is to relate Definition 14.2.2 to Galois's definition of notprimitive. Let f E F[x] be monic, separable, and irreducible with splitting field F c L. Alsoassume that f is imprimitive with blocks of roots given by R1,. . . , where each block has nelements (thus deg(f) = mn). Let f be the monic polynomial whose roots are the elements ofR, and let K ci L be the fixed field of E Gal(L/F) I = R1 for all i}.(a) Show thatf = f f E for all i.(b) In Galois's definition, K is obtained by adjoining the roots of a separable polynomial

of degree m. In modern terms, Galois wants F ci K to be a Galois extension such thatGal(K/L) is isomorphic to a subgroup of Sm. Prove that the field K defined in part (a)has these properties.

See Exercise 14 for some examples.

Exercise 9. Assume that G ci Sn is transitive and Abelian.(a) Prove that = n by considering the isotropy subgroups of G.(b) Prove that G is primitive if and only if GI is prime.Thus a transitive Abelian permutation group is imprimitive unless it is cyclic of prime order.

Exercise 10. Let 4,, (x) be the cyclotomic polynomial whose roots are the primitive pth rootsof unity, where p is prime. We know that is irreducible of degree p — 1. In the quotationgiven in the Historical Notes, Galois asserts that is imprimitive.(a) Prove Galois's claim for p> 3 using Exercise 9.(b) Explain why we need to assume that p > 3 in part (a).

Exercise 11. Given a prime p, let C,, ci 5,, be the cyclic subgroup generated by the p-cycle(12. . . p). As explained in the text, this gives the wreath product C,, C,, ci Prove thatC,, i C,, is a p-Sylow subgroup of

Exercise 12. Let f be an irreducible imprimitive polynomial of degree 6, 8, or 9 over a fieldof characteristic 0. Prove that f is solvable by radicals over F.

Exercise 13. Let f = x6 + bx3 + c E F [x] be irreducible, where F has characteristic differentfrom 2 or 3. We will study the size of the Galois group off over F.(a) Show that f is separable. Thus we can think of the Galois group as a subgroup of S6.(b) Show that x6 + bx3 + c is imprimitive and that its Galois group lies in S2 S3. Also show

that IS2 1S3 = 72. Thus the Galois group has order � 72.

(c) Let F C L be the splitting field of f over F. Use the Tower Theorem to show that[L : F] <36. Hence the Galois group has order at most 36.

Using Maple, one can show that the Galois group of x6 + 2x3 —2 over Q has order 36 andhence is as large as possible.

PRIMITIVE PERMUTATION GROUPS 429

Exercise 14. Here are some examples to illustrate Galois's definition of imprimitive. We willuse the notation of Exercise 8. Let F be a field of characteristic different from 2 or 3.(a) Letf = x6 + bx4 + cx2 + d E F [xl be irreducible with splitting field F C L. Show that the

splitting field of x3 + bx2 + cx + d gives an intermediate field F C K C L such that F C Kis Galois andf = flf2f3, where f, E K[x] has degree 2 for i = 1,2,3. Also explain howK relates to the field K constructed in Exercise 8.

(b) Work out the analogous theory when f = x6 + bx3 + c E F[x] is irreducible.

Exercise 15. Let G C be transitive. Prove that G is primitive if and only if the isotropysubgroups of G are maximal with respect to inclusion.

Exercise 16. Let p be prime. The ring 7L/p2Z is not a field, but one can still define the groupAGL(l,Z/p2Z). Its action on Z/p2Z allows us to write AGL(1,Z/p2Z) C(a) Prove that AGL( 1, Z/p27Z) is solvable and transitive of order p3 (p — 1).

(b) Prove that AGL( 1, Z/p2Z) C S,,2 is imprimitive.

14.3 PRIMITIVE PERMUTATION GROUPS

We now consider primitive permutation groups. Our main result is a powerful theoremof Galois on the structure of solvable primitive permutation groups. In order to provethis, we will define doubly transitive groups and use finite fields to construct someinteresting permutation groups. We will also study the minimal normal subgroupsof a solvable group. As an added bonus, we will learn why Galois was interested infinite fields.

The theory developed in this section will also be used in Section 14.4 when weclassify solvable primitive subgroups of S,,2.

A. Doubly Transitive Permutation Groups. For permutation groups, doubletransitivity is defined as follows.

Definition 14.3.1 A subgroup G C is doubly transitive if whenever we havei,i',j,j' E {l,...,n}suchthat

and

there is a E G such that

a(i)=j and a(i')=j'.

We already know an example of a doubly transitive group.

Example 14.3.2 In Section 14.1, we considered AGL( 1, IF,,) as a subgroup of 5,,. Toprove that this is doubly transitive, consider i i' and j j', where we now regardthese as elements ofF,,. Since i i', there are a,b e F,, such that

ai+b=j and ai'+b=j',

and j j' implies that a 0. Thus the condition of Definition 14.3.1 is satisfied byE AGL( 1, IF,,). We will generalize this example later in the section.


The concepts of doubly transitive, primitive, and transitive are related as follows.

Proposition 14.3.3 Let G C S,, be a subgroup. Then:

G doubly transitive G primitive G transitive.

Proof: First suppose that G is doubly transitive and imprimitive. Then we haveblocks R1,... ,Rk, where k> 1 and 1R11 > 1 for some i. For this i, pick i1 i2 inand also pick i3 for some j i. Then we have pairs i1 i2 and i1 i3, so thatby double transitivity we can find a E G such that

a(ii)=it and a(i2)r=i3.

Now consider a (R,), which by assumption is one of the blocks R1,. . . , Thena(ii) = i1 R, implies that o-(R1) = R1, while a(i2) = i3 e implies that a(R,)This contradiction proves the first implication.

The second implication will be proved in Exercise 1. •

Doubly transitive permutation groups also have the following property.

Proposition 14.3.4 if G C Sn is doubly transitive, then IGI is divisible by n(n — 1).

Proof: Let P = { (i, J) I

1 <i, f <n, i j} be the set of pairs of distinct elements of{l,. . . ,n}. This sethas n(n l)elements, and Gacts onPviaa.(i,j) = (a(i),a(j)).The crucial observation is that G acts transitively on P because G is doubly transitiveon {l, . . . ,n}. Thus the G-orbit of any (i,j) e P has n(n — 1) elements. Using theFundamental Theorem of Group Actions, we conclude that n(n — 1) divides the orderofG. •

B. Affine Linear and Semilinear Groups. The finite fields introduced inChapter 11 lead to some important permutation groups. Let Fq be a finite field withq = pm elements, p prime, and let be the standard n-dimensional vector spaceover As in the Mathematical Notes to Section 11.1, GL(n,Fq) is the group ofinvertible n x n matrices with entries in Fq. This acts on by matrix multiplicationwhen elements of 1F are regarded as column vectors.

Using GL(n, lFq), we construct the larger group AGL(n, lFq) of affine linear trans-formations, which are maps 'yAv: —* defined by

AEGL(n,Fq),

Thus AGL(n, Fq) combines linear maps with translations. Note that GL(l, lFq) lFq*,

so that when q = p. AGL( 1, IF,,) is the one-dimensional affine linear group studied inSections 6.4 and 14.1.

The group AGL(n, lFq) contains the subgroups

(14 13)q I V E C AGL(n,IFq),

GL(n,Fq) E GL(n,IFq)} C AGL(n,IFq),


where E GL(n,]Fq) is the identity matrix and 0 E is the zero vector. Forsimplicity, we will write (14.13) as

C AGL(n,Fq) and C AGL(n,Fq).

In Exercise 2 you will show that is a normal subgroup of AGL(n, Fq) with quotientisomorphic to GL(n, IFq). You will also express AGL(n, Fq) as a semidirect product

via the action of GL(n,Fq) OflBy using the Galois group Gal(lFq/Fp), we can enlarge AGL(n,lFq) as follows.

An affine semilinear transformation is a map —* defined by

=Ao(u)+v, A E & E Gal(IFq/Fp), V E

These maps form the affine semilineargmup AFL(n,Fq).When q = p, one sees that AFL(n,lFq) = AGL(n,lFq), and when q = ptm, m> 1,

you will prove in Exercise 3 that AGL(n,lFq) is normal of index m in AFL(n,Fq).

Furthermore, we have inclusions

C AGL(n,Fq) C AI'L(n,Fq),

and is a normal subgroup of AFL(n, Fq) (see Exercise 3).These groups act on which means that they can be regarded as subgroups of

Sqn. As permutation groups, they have the following important properties.

Proposition 14.3.5 The groups AGL(n,Fq) and AFL(n,Fq) are doubly transitivesubgroups of They are also primitive.

Proof: In Exercise 4 you will prove that AGL(n, Fq) is doubly transitive whenacting on Hence the same is true for the larger group AFL(n,]Fq). Both groupsare then primitive by Proposition 14.3.3. U

In Section 14.4, we will study solvable subgroups of Applying the abovetheory, we get subgroups

(1414)q=p2 andn=1q=p andn=2 F

However, is a vector space over IF,, of dimension 2. In Exercise 3 you will show

that elements of AI'L( 1, are linear when considered as maps between vectorspaces over IF,,. It follows that if we use a basis to identify with then (14.14)gives the inclusions

(14.15) C C C AGL(2,IF,,) c sf2.

C. Minimal Normal Subgroups. Before proving Galois's theorem on solvableprimitive permutation groups, we need to take a detour into pure group theory.


Definition 14.3.6 A normal subgroup N of a group G {e} is minimal if N {e}and all nontrivial subgroups of N (i.e., subgroups of N different from {e} and N) arenot normal in G.

Here are some examples of minimal normal subgroups.

Example 14.3.7 Let n � 5. Then is clearly a minimal normal subgroup of Sn,since is simple.

Example 14.3.8 The translation subgroup IF,' is a normal subgroup of the affinelinear group AGL(n, Fe), where as above we identify v IF,' with the translation

E Since is Abelian, any subgroup of F,' is normal in Butwhen is such a subgroup normal in AGL(n, To answer this, note that

,w° ,v = ,Av

by part (b) of Exercise 2. Since GL(n, F,,) acts transitively on JFJ' \ {O} by part (c)of Exercise 4, it follows easily that if {O} H C IF,' is normal in thenH = Thus F' C is a minimal normal subgroup. <31>

Example 14.3.9 Consider the wreath product G = S2 C S21. where I � 5. Part (b)of Lemma 14.2.8 shows that the subgroup

A, xA, CS21A, =G

is normal in G. We will regard N = A, x A, as a subgroup of G. In Exercise 5 youwill prove that N has the following properties:

• The nontrivial normal subgroups of N are {e} x A, and A, x {e}.• The factors of A, x A, get permuted under conjugation by elements of

IrES2} CS2?A, =G.

Now suppose that a nontrivial subgroup H C N = A, x A, is normal in G. ThenH is normal in N, so that H = {e} x A, or A, x {e} by the first bullet. But thesesubgroups can't be normal in G, by the second bullet. We conclude that N = A, x A,is a minimal normal subgroup of G. <1>

The minimal normal subgroups in these examples are simple (Example 14.3.7) orproducts of simple groups (Examples 14.3.8 and 14.3.9). The following result showsthat this is no accident.

Proposition 14.3.10 Let N be a minimal normal subgroup of a finite group G. Thenthere is a simple group A such that we have an isomorphism

n times

forsomen� 1.


Pmof: Let A be a minimal normal subgroup of N. Given g E G, set Ag = gAg'.Exercise 6 shows that Ag is a minimal normal subgroup of N isomorphic to A.

We will first prove that N A'1 for some n � 1. If A = N, then we are done.So suppose that A N. By the minimality of N, we know that Ag1 A for somegi E G. Since the intersection of normal subgroups of N is normal in N and since Ais minimal in N, we must have flA {e}. Then Exercise 7 implies that

= {aai a EA, a1 EAg1} C N

is a normal subgroup of N isomorphic to the product group A x Ag1. If AAg1 = N,then we are done, since

Ag C AAg1 for all g E G, then it is easy to show that AAg1is normal in G (see Exercise 6). This is impossible by the minimality of N. Hencethere is e G such that Ag2 AAg1. Then (AAgi) flAg2 = {e}, since the left-handside is normal in N and lies in the minimal normal subgroup Ag2. Arguing as in theprevious paragraph, N contains the subgroup

XAg2

If AAg1Ag2 = N, then we are done, and if not, we continue as above. In Exercise 6you will show that this eventually leads to the desired isomorphism N

A is simple. The isomorphism N A'1 takes A C N toA x {e} x x {e} C If B CA is a nontrivial normal subgroup, then N A'1

takes B to the subgroup

Bx{e}x...x{e}cAxAx..•xA=A'1,which is easily seen to be normal, since B is normal in A. It follows that B is normalin N. This is impossible, since A is a minimal normal subgroup of N. Hence A mustbe simple, and the proposition is proved. .

When N is a minimal normal subgroup of a solvable group G, the simple groupA appearing in N A'1 must also be solvable. The only solvable simple groups arecyclic of prime order, so that A as groups. Thus we have proved the followingcorollary of Proposition 14.3.10.

Corollary 14.3.11 Let N be a minimal normal subgroup of a finite solvable group.Then there is a prime p such that N for some n 1.

D. The Solvable Case. Before proving Galois's structure theorem for solvableprimitive permutation groups, we need some preliminary definitions and results.

We first want to say more about inclusions such as AGL(n, lF'q) C which weobtain by identifying with { 1, . . . , q'1}. This is done carefully as follows. Givena set T, let S(T) = T —* T is one-to-one and onto}. This is a group undercomposition, called the symmetry group of T. Here are some examples.

Example 14.3.12 Since every affine linear or semilinear transformation of isone-to-one and onto, we have natural inclusions

(14.16) AGL(n,lFq) C AFL(n,lFq) C S(F).


Example 14.3.13 For a more basic example, note that S({ 1,. . . , £}) is the symmetricgroup Sg.

If T has £ elements, then there is a one-to-one onto map y: T —* {l, . ..

It is

easy to check that = -y o p o 'y' defines a group isomorphism

Under a subgroup of G C S(T) maps to a subgroup of St. In Exercise 8 you willshow that if we use a different map ': T —+ {1,. .

.then G maps to a second

subgroup of St conjugate to the first.In particular, a one-to-one onto map -y : —* { 1, . . .

,

} gives a group isomor-phism : Applying to (14.16) gives subgroups of also calledAGL(n,Fq) and AFL(n,Fq). Since -y is not unique, these subgroups are only definedup to conjugacy in

We next define what it means for a permutation group to be regular. Given a groupG and g e G, define : G —* G by (pg(h) = gh. One easily shows that (pg E S(G).Since = the mapping g gives an isomorphism

In general, if T is any set, then a subgroup G C S(T) is regular if there is a one-to-oneonto map -y : G —f T such that the isomorphism : S(G) S(T) takes {(pg g E G} CS(G) to G C S(T). When T is finite, it follows that every regular subgroup of S(T)has T elements.

Make sure you understand how this definition captures the idea that G C S(T) isregular when the action of G on T looks like the action of G on itself given by thegroup operation of G. Here are some examples.

Example 14.3.14 Let G be a group with n elements. In Section 7.4 we used theCayley table of G to show that G is isomorphic to a subgroup of In Exercise 9you will show that this subgroup is regular.

Example 14.3.15 Consider AGL(n,lFq) C If v e F, then is translationby v, so that when we identify v with translation by v, we see that the translationsubgroup

C AGL(n,Fq)

is a regular subgroup of Furthermore, if we use-y : {1, . .. to regard

C AGL(n,lFq) as subgroups of then is regular in Sq'.

The following lemma will be useful in our proof of Galois's structure theorem.

Lemma 14.3.16 Suppose that G C Sg is a subgroup. Then:(a) If G is primitive and N {e} is normal in G, then N is transitive.(b) If G is transitive and Abelian, then G is regular.


Proof: For part (a), consider the orbits of N acting on { 1, , £}. Fix an orbit N

j E {l,. ..

and take a E G. Since N is normal in G, we have

a(N.j) .a(j) z=N.cr(j).

This shows that G preserves the block structure given by the orbits of N. Since G isprimitive, the block structure is trivial, so that either there is only one orbit or everyorbit has only one element. The latter is impossible (since N {e}), and the formerimplies that N is transitive. This proves part (a).

Turning to part (b), consider the isotropy subgroup G1 off e { 1, . . . , £}. We claimthat G1 = {e}. To prove this, let T E G and observe that

= rG1r1 = G1,

where we use (A. 19) and the fact that G is Abelian. Since G is transitive, we concludethat the isotropy subgroups of G are equal. Thus e G1 fixes not only j but also

Hencer =e, sothatG1 = {e}.In Exercise 10 you will show that a subgroup of St is regular if and only if it is

transitive with trivial isotropy subgroups. It follows that G is regular. •

We can now prove the following great theorem of Galois.

Theorem 14.3.17 Let G C St be a solvable primitive permutation group. Then£ = p prime, and (up to conjugacy)

C G C C

Proof: Let N be a minimal normal subgroup of G. Since G is primitive, part (a) ofLemma 14.3.16 implies that N is transitive, and since G is solvable, Corollary 14.3.11implies that N 1F.

In particular, N is transitive and Abelian, so that N is regular by part (b) ofLemma 14.3.16. It follows immediately that £ = NI = p's, as claimed in the theorem.Furthermore, since N being regular means that N C is the image of

]FCS(F)

under the isomorphism : coming from some one-to-one onto map-y : lF —+ {l,. . Hence, to study NC G C we will consider

C G' C S(F),

where G' maps to G under An element g E G' C gives a bijection

(14.17)

We will show that (14.17) is affine linear, which will imply that G' CTo describe how G' acts on let C G' be the isotropy subgroup of 0 E 1F.

Now consider (14.17) for g E G'0 and write translation by v as Since IFJ' C G' is


normal, we have g .0= 0, we compute theaction of g on v as follows:

(g.0) = = w.

This shows that the map v g •v corresponds to conjugation by g on the normalsubgroup 1F C G'. Since conjugation is a group homomorphism, v '—f g V mustalso be a group homomorphism. Such a map is automatically linear over byExercise 11. Thus (14.17) gives a element of GL(n, F,,) when the latter is regardedas consisting of permutations of In other words, any element of is of the form'YA,O for some A E

We now prove that G' C As above, translation by V E isGiven g E G', let v g .0. Then E G' maps 0 to 0 and hence lies in Thus

= which implies that g = v'YAO = "fA,v E AGL(n, IF,,). This shows thatG' C AGL(n, and completes the proof of the theorem. •

When applied to polynomials, Theorems 8.5.3 and 14.3.17 imply the followingstructure theorem for the Galois group of a primitive solvable polynomial.

Corollary 14.3.18 Let F be afield of characteristic 0, and letf E F[x] be primitive.1ff is solvable by radicals over F, then f has degree p"for some prime p and integern � 1, and the Galois group off over F is isomorphic to a subgroup ofcontaining the translation subgroup •

Theorem 14.3.17 shows that a solvable primitive permutation group G satisfies

Furthermore, the final part of the proof shows that the isotropy subgroup G0 of 0 Ecan be regarded as a subgroup of GL(n, such that

G is uniquely determined by Go. So it makes sense to ask if there is anythingspecial that we can say about Go. As we will see, the answer involves the followingdefinition.

Definition 14.3.19 Go C GL(n, is irreducible if there is no nontrivial subspaceV C (i.e., no subspace V {0} and such that g(V) C Vfor all g E Go.

Using this, we get the following useful result.

Proposition 14.3.20 Assume that G is a permutation group satisfying

1F C G C C

and let G0 C GL(n, IF,,) be the isotropy subgroup of 0. Then:


(a) G is primitive if and only if G0 is irreducible.(b) G is solvable if and only if G0 is solvable.

Proof: For part (a), we will prove that G is imprimitive if and only if G0 is reducible.First assume that G is imprimitive with blocks R1, . . . , Rk. Since 1F C G and 1F' actstransitively on itself, we know that G is transitive. By Lemma 14.2.7, it follows that

Suppose for simplicity that 0 E R1. We claim that R1 is a subspace of To

prove this, take v E R1 and observe that v 0 = v, since acts by translation. SinceG preserves the blocks, we must have v R1 = R1, which means v + w E R1 for allw e R1. Since v E R1 was arbitrary, R1 is closed under addition and hence is asubgroup, because R1 is finite. Exercise 11 then implies that R1 is a subspace.

However, every g E Go maps 0 to 0 and hence R1 to R1, since G preserves theblocks. This shows that R1 is a nontrivial subspace of such that g(Rj) = R1 forall g E Go. Hence Go is reducible.

Conversely, suppose that there is a nontrivial subspace V such that g(V) C V forall g EGo. Then 1 <p's, and g(V) = V for all g. Now let Rl,...,Rk be thecosets of V in In Exercise 12 you will show that G is imprimitive with respect tothe blocks R1,... ,Rk. This completes the proof of part (a).

The proof of part (b) is a straightforward application of the results of Section 8.1.See Exercise 12 for the details. •

Theorem 14.3.17 and Proposition 14.3.20 imply that classifying solvable primitivesubgroups of reduces to the study of solvable irreducible subgroups of GL(n,We will use then = 2 case of this strategy in Section 14.4 when we consider solvableprimitive subgroups of

Mathematical Notes

Some important ideas from group theory appear in this section.

Multiply Transitive Groups. Besides transitive and doubly transitive groups, onecan define k-transitive subgroups of for 1 k n as follows. A subgroup G Cacts on the set Pk of ordered k-tuples of distinct elements of { 1, . . . , n} by

ci (iI,...,ik) (a(iI),...,o(ik)), a E G, (iI,...,ik) E Pk.

Then G is k-transitive if G acts transitively on Pk. In Exercise 13 you will show thatis n-transitive and is (n 2)-transitive, and in Proposition 14.3.5 we showed

that AGL(n, is 2-transitive (i.e., doubly transitive).An example of a 4-transitive group is the Mathieu group

M11=((210)(411)(57)(89),(1438)(2569))c511.

This is a simple group of order 7920 and is the smallest sporadic group in theclassification of finite simple groups. Some of the many interesting aspects ofmultiply transitive groups are discussed in [3, Ch. 7] and [7, Secs. 5.7, 5.8].


• Finite Simple Groups. The group GL(n, Fq) is finite whenever Fq is a finitefield. This leads to an interesting finite simple group as follows. First observe thatGL(n, Fq) contains the normal subgroups

SL(n,lFq) = {A E GL(n,Fq)I

det(A) = l},

where is the n x n identity matrix. The group SL(n,lFq) is normal because it is thekernel of the homomorphism det: GL(n,Fq) —* 1F, and 1F1,, is normal because itselements commute with all n x n matrices.

The pmjective linear group is the quotient group

PGL(n,IFq) =

which is also finite. Furthermore, inside this group we have the subgroup

PSL(n,Fq) C PGL(n,lFq)

consisting of all elements of PGL(n, lFq) represented by an element of SL(n, Fq). Theremarkable fact is that PSL(n, JFq) is almost always simple.

Theorem 14.3.21 LetFq be afinite field and n> 1 be an Then PSL(n,Fq)is a simple group except when n = 2 and q = 2 or 3. U

A proof can be found in [2] or [8]. In Exercises 14 and 15 you will show thatPSL(2, F2) S3 and PSL(2, IF3) which are not simple. You will also show that

= IPSL(2,IF5)I =60 and IPSL(2,Fi)I = IPSL(3,1F2)I = 168.

One can prove that PSL(2, F4) PSL(2, IF5) A5 and that every non-Abelian simplegroup of order < 200 is isomorphic to either A5 or PSL(2,F7) (see [8,Satz 6.151 and [14, pp. 106—107]). In Example l3.3.lOandExercise 9 of Section 13.3we showed that the Galois group of x7 — 1 54x + 99 over Q is

GL(3,1F2) PSL(3,F2).

The paper [1] gives a nice description of the isomorphism PSL(2,1F7)Finally, we should mention that other finite simple groups can be constructed using

matrices over finite fields. These groups play an important role in the classificationof finite simple groups. See [6] for an introduction.

• The O'Nan—Scott Theorem. Theorem 14.3.17 describes the structure of solvableprimitive permutation groups. This is a special case of the 0 'Nan—Scott Theorem,which describes the structure of arbitrary primitive permutation groups. The 0' Nan—Scott Theorem is a basic tool in the study of primitive permutation groups. The fullstatement of the theorem (see [3, Ch. 4]) is beyond the scope of this book.

However, it is possible to give a brief glimpse into what this theorem says. Weneed the following concept.


Definition 14.3.22 The socle of a finite group G is the subgroup H generated by theminimal normal subgroups of G.

In Exercise 16 you will show that the socle is a product of finite simple groups.It is also obviously normal in G. For a primitive permutation group G C S1. onecan prove the stronger result that the socle H c G is a transitive subgroup such thatH Am for some finite simple group A.

The O'Nan—Scott Theorem classifies a primitive permutation group G C St ac-cording to its socle H There are two cases, each with several subcases:

Regular Socle. If H is regular, then G falls into one of two classes:• Abe/ian Socle. A = and H = C G C• Non-Abe/ian Socle. H where m � 6 and A is non-Abelian, and G is a

"twisted wreath product" with restricted isotropy subgroups (see [3, Sec. 4.7]).

Nonregular Socle. Here, H is non-Abelian and G falls into one of three classes:• Almost Simple. H = A, where A is non-Abelian, and A C G C Aut(A), where

Aut(A) is the group of all automorphisms of A, and G/A is solvable.• Diagonal. H where m> 2, and G is a "subgroup of a wreath product with

diagonal action" (see [3, Sec. 4.7]).• Pmduct. H where m > 2 and G is a "subgroup of a wreath product with

product action" (see [3, Sec. 4.7]).One way to think of Theorem 14.3.17 is that it explains how solvable primitive

permutation groups relate to the larger class of all primitive permutation groups: Theyfit into the "regular Abelian socle" class of the O'Nan—Scott Theorem.

The O'Nan—Scott Theorem has many applications in group theory. For example,we know from Section 14.2 that doubly transitive groups are primitive. One can showthat doubly transitive groups belong to the Abe/ian Socle or Almost Simple cases ofthe 0' Nan—Scott Theorem. This and the classification of finite simple groups leadto a classification of all doubly transitive permutation groups. See [3, Sec. 7.7] for adiscussion of this result.

Historical Notes

Why did Galois invent finite fields? After all, the main focus of his research wason the roots of polynomials. This question is now easy to answer using Galois'sown words. Before giving the quotation, we recall from the Historical Notes toSection 11.1 that Galois considered elements of finite fields as "imaginary solutions"of congruences. In this language, here is what Galois had to say about the importanceof finite fields [Galois, p. 125]:

It is mainly in the theory of permutations ... that the consideration ofimaginary roots of congruences appears to be indispensable. This gives a simpleand easy method to recognize in which case a primitive equation is solvable byradicals, as I will now try to give the idea in a few words.

Given an algebraic equation fx = 0 of degree p", suppose that theroots are denoted by Xk, where the index k has the pU values determined by thecongruence b (mod. p).


Take any arbitrary rational function of V of the roots Xk. One transformsthis function by substituting everywhere the index k with the index (ak +a, b, r being arbitrary constants satisfying a" = 1 b" b (mod. p) and ran integer.

This is taken from Galois's article on finite fields. In the second paragraph of thequotation, Galois explains how elements of can be regarded as permutations ofthe finite field The function V in the third paragraph is an element of the splittingfield off, and the substitutions described by Galois form the affine semilinear groupAFL( 1, The formula (ak + b)P differs from the definition of semilinear givenin the text, but later in the article Galois explains that when using this group,

the value substituted fork in every index can be put in the three forms

(ak+b)pr = (a{k+ b'})" = a'k" +b" =

(See [Galois, p. 125].) The formula + b" is the one we used to define1, This group uses both the field and the Galois group /IF,,).

These quotes show that Galois's reason for introducing AFL( 1, is that hewants to "recognize in which case a primitive equation is solvable by radicals."Galois knew that AFL( 1, is solvable and plays an important role in determiningwhen a primitive polynomial is solvable by radicals. We will have more to say aboutthis in the next section.

Theorem 14.3.17 is the major result of this section and is due to Galois, thoughhe stated only the polynomial version given in Corollary 14.3.18. In his letter toChevalier written the night before his fatal duel, Galois describes his theorem asfollows [Galois, p. 177]:

1° In order that a primitive equation be solvable by radicals, it must be ofdegree pLy, p being prime.

2° All of the permutations of such an equation are of the form

Xk,i,m,... / Xak+bI+cm+ ..+f,aik+bil+cim+.

k, I,m being LI indices that take the p values indicating all of the roots. Theindices are taken modulo p. that is to say, the roots are the same when one addsa multiple of p to one of the indices.

The group obtained by using all substitutions of this linear form containsall together pfl(pfl — — p) . .

. (p's — permutations.

Notice how item 2° describes Also observe that the final sentencereplaces i' with n. In Exercise 17 you will prove Galois's assertion that

(14.18)

We should also mention the observation of [12, p. 133] that Abel knew Galois's

assertion 10 about the degree of a primitive polynomial solvable by radicals. Here ishow Abel stated the result [Abel, Vol. II, p. 2221:

If an irreducible equation of degree divisible by prime numbers distinct fromeach other, is solvable algebraically, then one can always decompose intotwo factors /ii and such that the given equation is decomposable into


equations, each of degree and whose coefficients depend on equations ofdegree p1.

When we compare this with Galois's definition of primitive given in the HistoricalNotes to Section 14.2, we see that Abel is saying that if an irreducible polynomialf issolvable by radicals, then f is imprimitive whenever its degree is not a prime power.The above passage appears in an unfinished manuscript that Abel wrote shortly beforehis death. It shows how Abel was also struggling to understand what it means for apolynomial to be solvable by radicals.

Finally, the simple groups coming from finite fields were first studied by Jordan.In 1870, Jordan gave an incomplete proof that PSL(n, F,,) is simple except for n = 2and p = 2 or 3. In his proof, Jordan used what we now call Jordan canonical formto study matrices in GL(n, IF,,). This canonical form uses the eigenvalues of thematrix, which are roots of the characteristic polynomial. Hence the eigenvalues liein finite extensions of IF,,. This shows that more general finite fields arise naturallywhen analyzing GL(n, IF,,). Jordan went on to consider GL(n, though the firstcomplete proof of Theorem 14.3.21 is due to Dickson in 1897.


Exercise 1. The goal of this exercise is to prove that primitive permutation groups are transitive.Assume that G C Sn is primitive but not transitive, and derive a contradiction as follows.(a) Explain why n> 1.(b) Let the orbits of G acting on { 1,. . . , n} be R1, . . . , R, (see Section A.4 if you have forgotten

about orbits). Explain why k> 1 and why elements of G map every orbit to itself.(c) Conclude that G is imprimitive. Be sure to take into account the case when every orbit

consists of a single element.

Exercise 2. Let E AGL(n,IFq) be translation by v E F', and let E AGL(n,IFq) bearbitrary.(a) Prove that'y

(b) Prove that =(c) Part (b) shows that the translation subgroup ]F' C AGL(n, Fq) is normal. Prove that the

quotient group AGL(n, Fq)/1F is isomorphic to GL(n, Fq).(d) Prove that AGL(n, IF,) is isomorphic to the semidirect product ni GL(n, IFq), where

GL(n, lFq) acts on by matrix multiplication.

Exercise 3. Consider the affine semilinear group AFL(n, IFq) for q = ptm.(a) Prove that AGL(n, Fq) is a normal subgroup of AFL(n, F,,) of index m.(b) Prove that is a normal subgroup of ArL(n, ]Fq).(c) Prove that elements of AFL(n, IF,,) give maps IF,' that are affine linear over F,,.

Exercise 4. Let F be any field. The definition of AGL(n, F,,) given in the text extends toAGL(n, F). The goal of this exercise is to prove that AGL(n, F) is doubly transitive when weregard elements of AGL(n, F) as permutations of the vector space(a) Use C AGL(n,F) to show that AGL(n,F) acts transitively on F".(b) Inside AGL(n, F), we have the isotropy subgroup of 0 E Prove that this isotropy

subgroup is GL(n,F).(c) Prove that GL(n, F) acts transitively on F' \ {0}.


(d) Use Exercise 19 below to conclude that AGL(n, F) is doubly transitive.

Exercise 5. Let A and B be non-Abelian simple groups. You will show that A x {eB} and{eA} x B are the only nontrivial normal subgroups of A x B. Let N C A x B be a normalsubgroup different from {(eA,eB)}, A x {en}, and {eA} X B.

(a) Prove that A x {eB} and {eA} x B are normal in A x B. Hence, if we can show thatN = A x B, then we will be done.

(b)

(a,b) EN be as in part (b). Show that ,ea) EN for any at E A.(d) Given eA a E A, prove that there is at E A such that aat ala. Then combine this with

parts (b) and (c) to show that Nfl (A x {eB}) = A x {ea}.(e) Part (d) implies that Ax {eB} C N, and the inclusion {eA} x BC N is proved similarly.

Use this to prove that N = A x B.Exercise 18 will explore various aspects of this argument.

Exercise 6. Let A C N be a minimal normal subgroup, where N is normal in a larger group G.Given g E G, we set A9 =(a) Prove that A9 is isomorphic to A and is a minimal normal subgroup of N.(b) Fix gt E G and consider M9. By Exercise 7, we know that AAg1 is a subgroup of N.

Assume that Ag C M91 for all g E G. Prove that M91 is normal in G.(c) Use the following idea to complete the proof of Proposition 14.3.10. Let d be the set

of all subgroups of N of the form A91 . . such that the map (at,... ,an) '—+ a!defines an isomorphism

Ag1 x xAgn

A = Ae E d. Then pick an element of d of maximal order.

Exercise 7. Let H and K be normal subgroups of a group G. Let HK = {hk h E H, k E K}.(a) Prove that HK is a normal subgroup of G.(b) Assume that H fl K = {e}. Prove that hk = k/i for all h E H, k E K.(c) As in part (b), assume that H fl K = {e}. Prove that the map H x K HK defined by

(h,k) '—* hk is a group isomorphism.

Exercise 8. Suppose that -y, -y' T —* { 1,... , are one-to-one and onto. As explained in thetext, these give isomorphisms : S(T) St.(a) Explain why a = ('y') — is an element of St.(b) Let a E St be as in part (a), and let & : —+ St be conjugation by a. Thus &(r) = ara

forTESt. Prove thatThis proves that and 7 differ by conjugation by an element of St.

Exercise 9. Let G be a group of order n. In Section 7.4 we constructed a subgroup H C Snisomorphic to G. Prove that H is regular in Sn.

Exercise 10. A permutation group G C St is regular if there is a one-to-one onto map-y: gE G}CS(G)tOGCSt. Recallthat

E S(G) is defined by p9(h) = gh for/i E G. The goal of this exercise is to show that G isregular if and only if it is transitive with trivial isotropy subgroups.(a) Let G C St be regular. Prove that G is transitive and that the isotropy subgroups of G are

trivial.(b) For the rest of the exercise, assume that G is transitive with trivial isotropy subgroups.

Define -y: G —÷ {l, . ..

by -y(r) = T(1) forT E G. Prove that this map is one-to-oneand onto.


(c) The map -y of part (b) gives S(G) Show that = g, and conclude that G isregular.

Exercise 11. We can regard F as both a group (under addition) and a vector space over(under addition and scalar multiplication). However, since we are over 1Ff, scalar multiplicationcan be built out of addition. Use this observation to prove the following:(a) Any subgroup of F is a subspace.(b) Any group homomorphism -y : 1F —+ F; is linear.

Exercise 12. This exercise will use the notation of the proof of Proposition 14.3.20.(a) Suppose that V C 1F is a nontrivial subspace such that g(V) C V for all g E Go. Use the

cosets of V in 1F to prove that G is impriniitive.(b) Explain why 1F is normal in G, and prove that G/1F Go. Use this to prove part (b) of

Proposition 14.3.20.

Exercise 13. Consider the definition of k-transitive given in the Mathematical Notes.(a) Prove that is n-transitive.(b) Prove that An is (n — 2)-transitive when n > 3.

Exercise 14. Consider the groups GL(2,lFq), SL(2,Fq), PGL(2,Fq), and PSL(2,lFq) definedin the Mathematical Notes.(a) Prove that 1)(q2— 1).

(b) Prove that = = q(q2 — 1).

(c) Prove that PSL(2, ]Fq) = SL(2, lFq)/{±12}, and conclude that

IPSL(2

(d) Compute PSL(2,Fq)jforq=2,3,4,5,7.

(e) Show that GL(3,F2)l = = 168.

Exercise 15. Prove that GL(2,F2) = SL(2,F2) PSL(2, IF2) S3 and PSL(2, IF3)

Exercise 16. Let G be a finite group with socle H. Prove that H is isomorphic to a product offinite simple groups.

Exercise 17. Prove Galois's formula (14.18) for

Exercise 18. Here are some observations related to Exercise 5.(a) Give an example to show that Exercise 5 is false if we drop the assumption that A and B

are non-Abelian.(b) Let A1,... ,Ar be non-Abelian simple groups. Determine all nontrivial normal subgroups

OfAiXXAr.

Exercise 19. Let G C Sn be transitive, and let G be the isotropy subgroup of i E {l,... ,n}.Thus G = {a E G C(i) = i}.(a) Prove that G is doubly transitive if and only if G1 acts transitively on { 1, . . . , n} \ {i}.(b) More generally, let k � 2. Prove that G is k-transitive if and only if G acts (k — 1)-

transitively on {1,...,n}\{i}.

Exercise 20. Let G C S,, be doubly transitive. Proposition 14.3.3 implies that G is transitive.Prove that G is transitive directly from the definition of doubly transitive.


Exercise 21. Generalize (14.15) by showing that we have inclusions

= C C C AGL(nm,]Fp) C

Exercise 22. Show that AGL(n, Fq) is isomorphic to the subgroup

{AEGL.(n,IFq),

where is the (n+l) x (n+1) matrix such that the upper leftn x ncomer isA, the firstn entries of the last column are v, and the first n entries of the last row are all zero.

Exercise 23. Use Theorem 14.3.21 to show that AGL(2, F,,) is not solvable for p> 3.

Exercise 24. The action of PGL(2, F) on F = F U {oo} was introduced in Section 7.5. Inparticular, Exercise 11 of that section implies that the isotropy subgroup of PGL(2, F) at thepoint can be identified with AGL( 1, F). Use part (c) of Exercise 4 and Exercise 19 to provethat the action of PGL(2, F) on F is 3-transitive (also called triply transitive).

Exercise 25. Prove that AGL( 1, F4) and AFL( 1, S4.

Exercise 26. Compute the orders of the groups in (14.15).

14.4 PRIMITIVE POLYNOMIALS OF PRIME-SQUARED DEGREE

Let f e F [x] be a primitive polynomial of degree p2, where p is prime and F hascharacteristic 0. The main goal of this section is to understand which Galois groupscan occur when f is solvable by radicals over F.

By Chapter 8 and Section 14.2, this is equivalent to classifying the solvableprimitive subgroups of up to conjugacy. The answer is more complicated than inthe imprimitive case. Instead of the single subgroup AGL( 1, F,,) AGL( 1, IF,,) C S,2used in Theorem 14.2.15, the primitive case will require three subgroups, denotedM1, M2, and M3.

Our strategy will be to first describe the M, and then show that every primitivesolvable subgroup of is conjugate to a subgroup of one of them. The resultsof Section 14.3 imply that most of the proofs will take place in AGL(2, F,,) and

You will see a lot of 2 x 2 matrices in this section.

A. The First Two Subgroups. The subgroups M1 and M2 are relatively easy todescribe. The first subgroup is the affine semilinear group

(14.19)

from (14.15). This subgroup has the following properties.

Proposition 14.4.1 The subgroup M1 = AFL( 1, C is solvable, doubly tran-sitive, and primitive. Furthermore, Mi I = 2p2(p2 — 1).

Proof: In Exercise 1 you will prove that M1 is solvable and compute its order. Thenwe are done, since M1 is doubly transitive and primitive by Proposition 14.3.5. •

PRIMITIVE POLYNOMIALS OF PRIME-SQUARED DEGREE 445

The second subgroup is constructed as follows. A pair of affine linear transfor-mations-y,-y' E give

defined by

(14.20) =

In Exercise 2 you will show that 8 is an affine linear transformation of Thus wehave an inclusion

AGL( 1, IF,,) acts on the first coordinate of a point in and the secondAGL( 1, F,,) acts on the second coordinate. To get a more interesting group, we addthe matrix that switches the coordinates. This gives the group

(14.21)

where the last inclusion is from (14.15). This subgroup has the following properties.

Proposition 14.4.2 The subgroup M2 C S,,2 described in (14.21) is solvable and,when p> 2, primitive. Furthermore, M21 = 2p2(p — 1)2.

Proof: In Exercise 2 you will verify that has order 2 and satisfies

x AGL(1,IF,,) x

It follows that x CM2 is a subgroup of index 2. From here,it is easy to compute 1M2 and show that M2 is solvable (see Exercise 2).

It remains to prove that M2 is primitive. First note that M2 contains the translationsubgroup since F,, C AGL( 1, F,,). Hence, by Proposition 14.3.20, M2 is primitiveif and only if the isotropy subgroup (M2)o C GL(2, F,,) is irreducible. In Exercise 2you will verify that (M2)o is generated by the matrices

(14.22)

Let {0} V c be a subspace such that C V for all matrices -y in (14.22). Ifwe can show that V = then (M2)o will be irreducible and we will be done.

Take (a,b) (0,0) in V. Using from (14.22), we see that E V forall E When a and b are both nonzero, this gives (p — 1)2 elements of V.Since p> 2 implies that (p 1)2 > p. we conclude that V = in this case. On theother hand, if a = 0, then b 0, and using from (14.22) shows that (b, 0) E V.Adding this to (0, b) E V, we obtain (b, b) E V with both coordinates nonzero. Hencewe are reduced to the previous case, so that V = The case when b = 0 is handledsimilarly. •


Notice how the proof of primitivity uses E M2. In fact, it is easy to see thatthe smaller group AGL( 1, IF,,) x AGL( 1, F,,) is imprimitive (see Exercise 2).

It is also interesting to compare the subgroups M1 and M2. By Propositions 14.4.1and 14.4.2, we have

— 2p2(p2—1) p+l— 2p2(p—l)2 —

Thus Mi > In Exercise 3 you will show that when p> 3, M2 is not doubly

transitive and is not isomorphic to a subgroup of Mi. So M1 and M2 are quite distinctas subgroups of S,,2.

B. The Third Subgroup. The third subgroupM3 is harder to describe than the firsttwo. We begin with a lemma about 2 x 2 matrices that will prove to be surprisinglyuseful. Recall that in any group G, the centralizer CG(g) of g E G is the subgroupconsisting of all elements of G that commute with g. Also let '2 E GL(2, F,,) denotethe identity matrix.

Lemma 14.4.3 Ifg E GL(2,F,,)\F12, then

= {m E GL(2,IF,,) m = a!2 + bgfor some a,b E

Proof: Every a12 + bg E GL(2, F,,) obviously commutes with g. Now take m in(g). Since g you will prove in Exercise 4 that there is v E such

that v and gv form a basis of Hence there exist a, b e F,, such that

mv=av+bgv= (a!2 +bg)(v).

Using mg = gm, we obtain

mgv = g(mv) = g(av + bgv) = agv + bg2v = (a!2 + bg)(gv).

This implies that m = a!2 + bg, since their corresponding linear maps agree on a basis

U

The subgroup M3 is constructed using the projective linear group PGL(2, IF,,)discussed in the Mathematical Notes to Section 14.3. The normal subgroup

= {A12I

A e F;} cgives the quotient group

=

where the image of m = E GL(2, IF,,) in the quotient will be denoted

[in] = e PGL(2,IF,,).

To define M3, we will construct a subgroup of PGL(2, isomorphic to S4 when

p > 2. By Exercise 1 of Section 8.1, S4 has the normal subgroup

((12)(34),(13)(24)) (Z/2Z)2.


Our strategy for finding a subgroup of PGL(2, isomorphic to S4 uses a carefullychosen subgroup isomorphic to (Z/2Z)2. Here is the precise result.

Proposition 14.4.4 Assume that p> 2. Then:(a) There exist g, h E GL(2, F,,) such that gh = —hg and det(g) = det(h) 1.

(b) Let g, h be as in part (a). Then g2 h2 —'2 and [g], [hi E PGL(2, generatea subgmup H such that

H = ([g], [hi) (Z/2Z)2.

Furthermore, the centralizer C(H) = (H) (consisting of all elementsof PGL(2, IF,,) that commute with every element of H) satisfies

and the normalizer N(H) (H) satisfies

N(H) defined in part (b) are unique up to conjugacy byelements of

Proof: In Exercise 5 you will prove that there are s, t E such that s2 + t2 = —1.

Then let(0 —i\ Is tand h=

—s

One easily computes that g, h have the desired properties. This proves part (a).For part (b), we have g, h with gh = —hg and det(g) = det(h) = 1. We first show

that g2 = —'2. Since det(g) = 1, the characteristic polynomial P(x) = det(g —x12)

can be written P(x) = x2 + ax + 1. Then the Cayley—Hamilton Theorem implies that

g2+ag+12=0.

(Do Exercise 6 if you didn't study this in your linear algebra course.) Conjugatingby hand using = —g easily implies that

g2 —ag+!2 = 0.

Adding these equations and dividing by 2 (p> 2) implies that

g = 12,

and reversing the roles of g and h gives h2 = Note also that neither g norh lies in since gh = —hg hg (p> 2). It follows easily that the subgroupH = KEg], [h]) C is isomorphic to (7L/2Z)2.

To study C(H), first note that if m1,m2 e then

(14.23) m1m2 = ±m2m1 [mii[m2] = [m2][mii.


One direction is obvious, since [±m21 = [m2]. For the other direction, observethat {mi][m2} = [m2][mi] implies that m1m2 = Am2m1 for some A E Taking thedeterminant of each side shows that A2 = 1, so that A = ± 1.

Now let [m] e C(H). Then mg = ±gm and mh = ±hm by (14.23). If both signsare +, then mE Lemma 14.4.3 implies that

m=a12+bg=c12+dh,

If b 0, then g would be a linear combination of 12 and h and hence would commutewith h. This is impossible, since gh = —hg and p > 2. Thus b = 0, which shows thatmis a multiple of!2. It follows that [m] = ['21 E H.

On the other hand, if mg = —gm and mh = hm, then

(mh)g = m(hg) = m(—gh) = (—mg)h = (gm)h = g(mh),

(mh)h = (hm)h = h(mh),

so that mh E By the above paragraph, [mh] = [121, hence[m] = [h] E H since h2 = —'2. The remaining possibilities for the signs are handledsimilarly and imply that [m] = [g] or See Exercise 7 for the details. This showsthat C(H) C H. The other inclusion is trivial, since H is Abelian. Thus C(H) = H.

To describe N(H), first observe that N(H) acts on H by conjugation. The iden-tity element is fixed, so that conjugation permutes the three nonidentity elements[g], [h], [gh]. It follows that we have a group homomorphism

N(H) —* S3,

where an element m E N(H) maps to the permutation of [g], [hi, [ghi given by conju-gation by m.

The kernel of consists of those m E N(H) that conjugate every element of H toitself. In other words, = C(H), which is H by the above. Since HI = 4, itfollows immediately that IN(H) 24, with equality if and only if p is onto.

To prove that is onto, note that (12 +g)(12 — g) = 212 since g2 = —'2. Thus

(!2+g)' =

since p > 2. Then conjugating h by '2+ g gives

=

= =gh,

where we use gh = —hg and g2 = —'2. It follows easily that [12 + gi conjugates [g]to itself and interchanges [h] and [gh]. Thus ['2 + is an element of N(H) that mapsto a 2-cycle in S3. Similarly, [12 + h] conjugates [h] to itself and interchanges [g] and[gh], so that [12 + h] e N(H) maps to a different 2-cycle. Since S3 is generated by anytwo distinct 2-cycles, we see that is onto and IN(H) I = 24.

The final step is to show that N(H) S4. In Exercise 8 you will show that N(H)has four 3-Sylow subgroups. Then the action of N(H) on its 3-Sylow subgroups

PRIMITIVE POLYNOMIALS OF PRIME-SOUARED DEGREE 449

gives a group homomorphism N(H) —* S4. In Exercise 8 you will prove that this mapis an isomorphism. The proof of part (b) is now complete.

For part (c), first suppose that g F12 satisfies g2 = —'2. By Exercise 4, there isV E such that v,gv form a basis of Since g takes v to gv and gv to g2v = —v, itfollows that

Q is the matrix whose columns are v and gv. This easily implies that allelements g Fy12 satisfying g2 12 are conjugate.

Now suppose that we have g, h and g', h' as in the statement of the proposition.Then the above paragraph shows that g and g' are conjugate, say g = Qg'Q'.Replacing g', h' with their conjugates by Q, we may assume that g = g'. We need toconjugate h to h' in a way that preserves g.

Since gh = —hg and gh' —h'g, it is easy to see that h'h' commutes with gand hence lies in C(g) = Note also that det(h'h') = 1, since det(h) =det(h') = 1. In Exercise 9, you will show that this implies that

(14.24) h'h'=det(m)m2, mEC(g).

Since g Lemma 14.4.3 implies that m is a linear combination of '2 and g.Using this together with gh = —hg and g2 = one easily computes that

mhm=ch,

c = +det(m). Combining this with (14.24), we obtain

mhm' = (mhm)m2 (±det(m)h)m2

= ±h(det(m)m2) ±h(h1h') = ±h'.

Since m E C(g), it follows immediately that [m] conjugates H = to H' =([gj, [h']). This easily implies the corresponding statement for N(H) and N(H'). Theproposition is now proved.

We can give explicit generators for the subgroup N(H) described in Proposi-tion 14.4.4. When p 1 mod 4, we can find an element i E of order 4. InExercise 10 you will show that N(H) is generated by the images of the matrices

14(0 1'\ Ii O"\ (1—1

.25)0,)' 1)'

i with E C in (14.25) gives the matrices from Example 7.5.10. There,we showed that the images of these matrices in PGL(2, C) generate the rotationalsymmetry group of the octahedron. Furthermore, this symmetry group is isomorphicto 54 by Exercise 10 of Section 7.5. So it is nice to see that the same matrices work inIF,, when p 1 mod 4. Explicit generators forN(H) when p 3 mod 4 are describedin Exercise 10.


We can finally construct M3. Assume that p > 2 and consider the homomorphism

ir : —+ —*

where the first map takes to A and the second is the quotient map that takes A to[A]. By Proposition 14.4.4, we have S4 N(H) c Then M3 is definedto be the inverse image of this subgroup under ir. Thus

(14.26) M3

The subgroup M3 has the following properties.

Proposition 14.4.5 Let p > 2. The subgmup M3 C described in (14.26) issolvable and primitive. Furthermore, M3 = 24p2(p — 1).

Proof: It is straightforward to show that M3 is solvable because S4 is, and the orderof M3 is also easy to compute. We leave this as Exercise 11.

It remains to prove that M3 is primitive. By Proposition 14.3.20, it suffices toshow that (M3 )o is irreducible. First observe that (M3 )o is the inverse image ofN(H) C under the quotient map —+ Let V Cbe a one-dimensional subspace mapped to itself by (M3)o. Thus V is a simultaneouseigenspace for all elements of (M3 )o. We derive a contradiction as follows.

Proposition 14.4.4 and the definition of M3 imply that (M3 )o contains elementsg,hsuch that gh= —hg. Now letvE Vbe nonzero. Thengh(v) = —hg(v). However,since v is an eigenvector for g and h, one easily sees that gh(v) = hg(v). This gives acontradiction since p > 2, and the irreducibility of (M3)0 follows.

It is possible to describe M3 more explicitly. First, one can show that M3 isisomorphic to the semidirect product (M3)o. Furthermore, a careful descriptionof the structure of (M3)o can be found in [16, Ch. 5], where (M3)o is denoted by M4when p 1 mod 4 and by M3 when p 3 mod 4.

One important observation is that except for certain small primes p, the subgroupsM1, M2, and M3 of AGL(2, IF,,) satisfy

(14.27)

Hence we really need three subgroups. We showed above that (14.27) holds forM1andM2 whenp>3. Thencomparing M31 = 24p2(p— 1) with Mu = 2p2(p2 —1) and

2p2(p — 1)2 shows that Mi M3 and M2 M3 when p> 13. Furthermore,(M1 )o and (M2)o have Abelian subgroups of index 2, which easily implies thatM3 and M3 M2. See Exercise 12 for more details, including a precise list ofthe exceptions to (14.27).

C. The Solvable Case. We can now state our main result concerning solvableprimitive subgroups of Sp2. Since every subgroup of S22 = S4 is solvable, we willassume that p> 2.


Theorem 14.4.6 Let G C be primitive, where p> 2 is prime. Then the followingare equivalent:(a) G is solvable.(b) G is conjugate to a subgroup of one of the gmups M1, M2, M3 defined in (14.19),

(14.21), (14.26), respectively.

Proof: The proof of (b) (a) is easy, since we know that M1, M2, and M3 aresolvable by Propositions 14.4.1, 14.4.2, and 14.4.5.

To prove (a) (b), first note that G is conjugate to a subgroup of AGL(2, IF,,) con-taining by Theorem 14.3.17. Furthermore, Proposition 14.3.20 implies that theisotropy subgroup G0 of 0 E is irreducible and solvable. It is also straightforwardto show that G is uniquely determined by G0 C GL(2, F,,). Thus it suffices to provethatin GL(2,IF,,), G0 isconjugatetoasubgroupof i E {1,2,3}.

We also note that we can assume that C Go. To see why, note that matricesin commute with all elements of GL(2, F,,). This makes it easy to see that thesubgroup of GL(2, F,,) generated by Go and 1F12 is solvable (you will prove thiscarefully in Exercise 13). If this larger group lies in some (M,)o up to conjugacy, thenso does Go. Hence we may assume that C Go. In particular, F,,*12 is an Abeliannormal subgroup of Go.

Let A C Go be an Abelian normal subgroup containing F12 of maximal order.The proof now breaks up into two cases, depending on A.

Case 1: First suppose that A $ and pick g e A \ lFl2. Then consider thecentralizer

C(g)

and its normalizerN(C(g)) =

Our strategy will be to first prove that

(14.28) Go C N(C(g))

and then show that N(C(g)) is conjugate to either (M1 )o or (M2)o.To prove (14.28), take m e Go. Then E A, since g E A and A is normal

in Go. We also know that A c C(g), since A is Abelian. Hence mgm' E C(g). ByLemma 14.4.3, this implies that

=a12+bg, a,beF,,.

Now take an arbitrary element h E C(g). Using Lemma 14.4.3 again, we can writeh = c12 +dg, where c,d E IF,,. Then

= m(c12 + dg)m' = c12 + dmgm'=c12+d(a12+bg)= (c+da)12+dbg.

This lies in C(g) by Lemma 14.4.3. Thus m normalizes C(g), so that m E N(C(g)).This completes the proof of (14.28).


The next step is to study N(C(g)). Here, our main tool will be the characteristicpolynomial P(x) = det(g X12) of g. This is a quadratic polynomial with coefficientsin F,,. There are three possible behaviors for P(x):

• P(x) is reducible and separable.• P(x) is reducible and nonseparable.• P(x) is irreducible.We will consider each possibility separately.

Reducible and Separable. We will show that N(C(g)) is conjugate to (M2)o. Byhypothesis, the eigenvalues of g are in F,,, which means that g is diagonalizable(be sure you can explain why). Hence there is Q E GL(2, F,,) such that

If we replace G0 with its conjugate then we may assume that

(a 0

We will show that N(C(g)) = (M2)o in this situation.Using Lemma 14.4.3, it is easy to see that

(14.29) C(g) = { °),i,v E

(see Exercise 14). Now let m = E N(C(g)). Then mgm1 e C(g), which bythe above description of C(g) implies that

(a O'\ i(p/3,Jm v)'

where v because a If we multiply on the right by m and compare entries,then it is straightforward to show that b = c = 0 or a = d = 0. Hence

(a o\ (0 b'\ (0 1\(c 0m= d) or m= = o) d

Since (M2)0 is generated by the matrices (14.22), it follows that m E (M2)0. Thus

N(C(g)) C (M2)0.

The opposite inclusion is straightforward to prove (see Exercise 14). We concludethat N(C(g)) = (M2)o.

Reducible and Nonseparable. We will show that this case can't occur, since G0is irreducible. By hypothesis, the only eigenvalue of g is a E IF,,. Hence there isQ E such that

=


where /3 0 because g Now replace Go with and note that Goremains irreducible. Hence we may assume that

/3

In Exercise 15, you will use Lemma 14.4.3 to show that

(14.30) C(g)= { ")

e

and you will prove that the normalizer of C(g) is

(14.31) N(C(g))= {

Also recall from (14.28) that G0 C N(C(g)).We obtain a contradiction as follows. Let V C be the subspace spanned by

the vector Since every element of (14.31) takes V to itself, we see thatGo C N(C(g)) cannot be irreducible. This gives the desired contradiction.

Irreducible. We will show that N(C(g)) is conjugate to the subgroup (M1 )o. SinceM1 = it is easy to see that (M1)o is the group consisting ofsemilinear maps : lFp2 —+ E a E defined by

(14.32) aa(u), u e

While need not be linear over IF,,2, it is always linear over F,, (do you see why?).To represent as an element of GL(2, F,,), we will use an isomorphism

(14.33) T :

of vector spaces over IF,,. Let (IF,,2) be the group of vector space isomorphisms—* that are linear over IF,,. Then we have a group isomorphism

(14.34) GL(2,IF,,)

where an IF,,-linear isomorphism 4: IF,,2 lFp2 maps to the matrix representing

(you will verify this in Exercise 16). Under (14.34), the subgroup

FL(1,IF,,2) c

maps to(M1)o C

Different isomorphisms T in (14.33) give different isomorphisms (14.34) that arerelated by conjugation in GL(2,IF,,) (see Exercise 16).


By assumption, g E GL(2, F,,) has irreducible characteristic polynomial P(x). Toanalyze N(C(g)), we will make a special choice of T in (14.33). Consider thefollowing bases of and

• Since P(x) has degree 2, it splits completely in Let a E be a root, andnote that a F,,, since P(x) is irreducible over IF,,. Then 1, a form a basis of IF,,2as a vector space over IF,,.

• Write g = and observe that c 0, since otherwise P(x) would have rootsa,d E IF,,. Then (h), form a basis of as a vector space over IF,,.

Using these bases, define T IF,,2 by

T(1) = and T(a) =

We claim that for this choice of T, the element of (IF,,2) corresponding tog E GL(2,IF,,) via (14.34) is multiplication by a.

More precisely, define F,,2 —÷ IFp2 by = aJ3 for e IF,,2. We must showthat

=g, i.e.,

where we now think of g as the linear map given by matrix multiplication. To provethis, first note that

(14.35) =T(a) = = =goT(l).

If the characteristic polynomial of g is P(x) = x2 + ax + b, then

g2 + ag + b!2 = 0

by the Cayley—Hamilton Theorem. Using this and T(a) = go T(1) from (14.35), weobtain

goT(a) =g2 oT(1) = (—ag—b12)oT(1) = —agoT(1) —bT(1)

= —aT(a) bT(1).

Since a2 + aa + b = 0, we also have

= T(a2) = T(—aa—b) = —aT(a)—bT(1).

Thus T = go T(a). This and (14.35) imply that T o-yQ = goT. We concludethat g corresponds to 'Ya under (14.34), as claimed.

It follows that N(C(g)) c GL(2, IF,,) corresponds to C under(14.34), where in the latter inclusion, the centralizer and normalizer are now computedrelative to (IFp2). Thus, if we can prove that

(14.36) = ['L(1,IF,,2)

when a E \ IF,,, then it will follow that N(C(g)) = (M1 )o. Be sure you understandthis.


We now prove (14.36). In Exercise 17 you will show that if a E then

(14.37)

/3 This implies that

m(a) = = =

m(a2) = moya(a) = = /3m(ct) =

where the last equality of the second line uses the first line. Thus

0 = m(0) = m(a2 +aa + b) = m(a2) +am(a) + bm(1)

= /32m(l)+a/3m(l)+bm(l) = (/32 +a/3+b)m(l).

Since m(1) 0 and F,,2 is a field, we must have

(14.38) /32+a/3+b=0.

To relate this to I'L( 1, write the Galois group of IF,,2 over IF,, as

= {e,a} Z/2Z,

where e is the identity and a has order 2. Then the roots of P(x) x2 + ax + b are aand a(a). Hence (14.38) implies that

/3=a or

In Exercise 17 you will show that if we set 5 = m(1), then

(1439)/3=a

/3=a(a)

in the notation of (14.32). This proves that C The oppositeinclusion is straightforward (see Exercise 17), and (14.36) follows.

Case 2: Next assume that is the maximal Abelian normal subgroup of Gocontaining We will show that G0 is conjugate to a subgroup of (M3)o. Firstnote that (M3 )o c GL(2, is the inverse image of the subgroup N(H) C PGL(2,

from Proposition 14.4.4. c be the image of Go C SinceC Go, it suffices to prove that C N(H) after a suitable conjugation.

Fix a minimal normal subgroup B' c G'0 as defined in Section 14.3, and let B bethe inverse image of B' in Since C Go, we have

lFpJ2 C B C Go.

Note also that B is normal in G0. We now prove some basic facts about B' and B.


For B', first note that is nontrivial, since Go is irreducible (be sure you can fillin the details). Then B' is a minimal normal subgroup of the nontrivial groupThis has two useful consequences:

• B' is generated by the conjugates (with respect to of any of its nonidentityelements (Exercise 18).

• B' is Abelian, since is solvable (Corollary 14.3.11).

(The solvability of G is used twice in the proof of Theorem 14.4.6: at the beginningof the proof to reduce to Go C GL(2, Fr), and here to imply that B' is Abelian.)

For B, recall that its center Z(B) consists of all elements of B commuting withevery element of B. We claim that Z(B) is as small as possible, i.e.,

(14.40) Z(B)

To see why, observe that Z(B) is normal in Go because B is (you will prove this inExercise 18). Note also that Z(B) is Abelian and contains F,12. But the hypothesisof Case 2 is that A = F12, which means that is the maximal Abelian normalsubgroup of Go containing 1F12. The equality (14.40) follows immediately.

The next step is to find some interesting elements of B. More precisely, we claimthat there are g, h E B such that

(14.41) gh = —hg, det(g) = det(h) = 1.

To prove this, take [mi] E B', [mi I [12]. The conjugates of [mi] generate B', so that Bis generated by and the conjugates of Since F,,!2, (14.40) implies thatm1 doesn't commute with at least one of its conjugates, say m2. Then g = E B

has det(g) = 1, since det(mi) = det(m2). It is also easy to see that g doesn't commutewith so that g Hence the conjugates of [g] generate B', which means thatB is generated by Fy12 and the conjugates of g. Arguing as above, g has a conjugateh such that gh hg. Also note that det(h) = 1. Since [g] [h] = [h] [g] (B' is Abelian),(14.23) implies the desired equation gh = —hg.

Let g,h B satisfy (14.41). Then [g] and [hi generate the subgroup H defined inProposition 14.4.4. Thus

H C B' C

Since B' is Abelian, we have B' c C(H) = H, where the last equality is by Proposi-tion 14.4.4. Thus H = B'. Since B' is normal in we also have G'0 C N(B') = N(H).As noted at the beginning of Case 2, this completes the proof of the theorem. •

A much more sophisticated proof of Theorem 14.4.6 can be found in [17, §21].This reference studies solvable subgroups of GL(n, for arbitrary n and £.

When p is large, Theorem 14.4.6 implies that solvable primitive subgroups ofare relatively small in size. Here is an example.


Example 14.4.7 When p = 17, Propositions 14.4.1, 14.4.2, and 14.4.5 imply thatthe orders of M1, M2, M3 are

105,

105,

1M31 =24. 172(l7_ l)= 1.1 x i05.

By Theorem 14.4.6, solvable primitive subgroups of S172 are extremely small whencompared to S1721 2.1 x 10587. In contrast, recall from Example 14.2.17 that thelargest solvable imprimitive subgroup of S172 has order

AGL(l,1F17) = 17181618 6.6 x

Thus being solvable and primitive is much more restrictive than being solvable andimprimitive.

Combining Corollary 14.2.16 and Theorem 14.4.6, we get the following criterionfor when an irreducible polynomial of degree p2 is solvable by radicals.

Corollary 14.4.8 Let f F be irreducible of degree p2, where F is a field ofcharacteristic 0. Then f is solvable by radicals over F

f and the Galois group off over F is isomorphic to a subgroupof the wreath product or

(b) f is primitive and the Galois group off over F is isomorphic to a subgroup ofthe groups M1, M2, andM3 defined in (14.19), (14.21), and (14.26). •

Mathematical Notes

This section includes some interesting ideas from group theory.

• Solvable Linear Groups. For most of the proof of Theorem 14.4.6, we workedwith the group G0 c GL(2, From this point of view, the argument showed thatevery solvable irreducible subgroup of GL(2, IF,,) is conjugate to a subgroup of (M1(M2)o, or (M3)o. A systematic approach to the study of solvable linear groups can befound in [9] and [17].

• Doubly Transitive Solvable Permutation Groups. In Proposition 14.4.1, weshowed that M1 = AFL( 1, is solvable and doubly transitive. What is moresurprising is that, with some exceptions for small primes, this group contains allsolvable doubly transitive subgroups of Sf2.

Theorem 14.4.9 Let p > 23 be prime. Then every solvable doubly transitive sub-group G C is conjugate to a subgroup of Mi =

Proof: Since G is solvable, Theorem 14.4.6 implies that G is conjugate to a subgroupof M1, M2, or M3. Furthermore, since G is doubly transitive, Proposition 14.3.4implies that I is divisible by p2 (p2 1). However,

IM2I=2p2(p—1)2 and 1M3I=24p2(p—l)

are not divisible by p2 (p2 — 1) when p > 23. This proves the theorem.


The following much stronger result was proved by Huppert in 1957.

Theorem 14.4.10 Let G c S, be solvable and doubly transitive. Then £ = ptm forsome prime p. Furthermore, if pm {32,52,72, 112,232,34}, then G is conjugate toa subgmup of

Proof: Our hypothesis implies that G is solvable and primitive, and then £ = ptm byTheorem 14.3.17. This is the easy part of the proof. For the rest of the argument, see[11, §7 of Ch. XII]. U

In fact, one can prove that up to conjugacy, all solvable doubly transitive permu-tation groups lie in AFL( 1, except for the 13 groups described in [101.

• Classifying Permutation Groups. Besides the two classes of groups just discussed(solvable linear groups and solvable doubly transitive groups), there has been a lot ofwork on classifying other sorts of interesting groups. Here is a brief sample of whathas been done:

• Solvable primitive subgroups of for n <256. See [16].• Primitive subgroups of Sn for n 1000. See [13].• Transitive subgroups of S,,2 for all primes p. See [4}.• All subgroups of See [2, Ch. XII] or [8, §8 of Ch. II].

The last bullet has some unexpected relations with Section 7.5 and Lemma 14.4.4.See Exercise 19 for an interesting subgroup of when p 1 mod 8.

Historical Notes

Galois worked very hard to understand solvable primitive subgroups, thoughhis research was incomplete at the time of his death. In the Historical Notes toSection 14.3, we gave quotations from Galois's paper on finite fields describingAFL( 1, and his version of Theorem 14.3.17, which asserts that up to conjugacy,a solvable primitive group G satisfies

F

In this paper, Galois notes that AFL( 1, is solvable and that any polynomial whoseGalois group is a subgroup of this group is solvable by radicals. He also makes thefollowing intriguing statement [Galois, p. 125]:

This remark would be of little importance if I had not already demonstratedthat reciprocally, a primitive equation would not be known to be solvable byradicals, without satisfying the conditions that I have just stated. (I excludeequations of the 9th and 25th degree.)

Galois seems to be saying that, with a few exceptions, a solvable primitive permutationgroup satisfies

(14.42) GCAFL(1,FA,4.

As we know from Theorem 14.4.6, this is not correct, for the groups M2 and M3are counterexamples when n = 2. On the other hand, if we replace "primitive" with


"doubly transitive," then we get a statement close to Huppert's theorem about doublytransitive solvable groups (Theorem 14.4.10). Furthermore, in his letter to Chevalier,Galois indicates that the above assertion is "too restricted. There are few exceptions,but there are some" [Galois, p. 177]. So it is hard to know exactly what Galoiswas thinking. Nevertheless, the results of this chapter make it abundantly clear thatGalois's insight into permutation groups was nothing short of astonishing. The readermay wish to consult [12] for further discussion of these issues.

The proof of Theorem 14.4.6 given in the text is based on suggestions of WaltParry and Jordan's 1868 paper Sur Ia resolution algébrique des equations primitivesdes degre p2 (p étant premier impair) [Jordan2, pp. 17 1—195]. Jordan was awarethat his results provide counterexamples to some of Galois's assertions.

Readers interested in learning more about the history of transitive permutationgroups should consult the introduction to [13] and Appendix A of [16].


Exercise 1. Prove that M1 = AFL( 1, is solvable, and compute its order.

Exercise 2. This exercise will study the subgroup M2 C AGL(2, defined in (14.21).(a) Prove that the map 5 defined in (14.20) gives an element of AGL(2,FP).(b) Prove that has order 2 and normalizes AGL( 1, F,,) x AGL( 1, F,,) C AGL(2, F,,).(c) Prove that M2 is solvable, and compute its order.(d) Prove that (M2)o is generated by the matrices in (14.22).(e) Prove that x AGL(1,F,,) C AGL(2,F,,) is imprimitive in Sf2.

Exercise 3. Let M1 and M2 be the groups defined in the text, and assume that p > 3. Provethat M2 is not doubly transitive and not isomorphic to a subgroup of M1.

Exercise 4. Let V be a vector space of dimension 2 over a field F, and let T : V —+ V be alinear map that is not a multiple of the identity. Also assume that T is an isomorphism. Provethat there is v E V such that v and T(v) form a basis of V over F.

Exercise 5. Fix a E F,,, p > 2. The goal of this exercise is to find 5, t E F,, with s2 + t2 = a.(a) Let S = {s2 SE Prove that SI = (p+ 1)/2.(b) LetS' = {a s E Show that SflS' 0, and use this to prove the existence of

s,t E F,, such that s2 +t2 = a.

Exercise 6. LetA = be a 2 x 2 matrix with entries in a field F.(a) Prove that the characteristic polynomial of A is P(x) = — tr(A)x + det(A), where

tr(A) = a + d and det(A) = ad — bc are the trace and determinant of A.(b) Prove that P(A) = A2— tr(A)A +det(A)12 is the zero matrix.

The Cayley—Hamilton Theorem generalizes part (b) by showing that P(A) is the zero matrixwhen P(x) is the characteristic polynomial of an n x n matrix A.

Exercise 7. Complete the proof of C(H) = H from Proposition 14.4.4 begun in the text.

Exercise 8. Let G be a group with a normal subgroup H (Z/27Z)2 such that CG(H) = Hand the map G —* Aut(H) given by conjugation is onto. The goal of this exercise is to provethat G S4. Note that IGI = 24 by the proof of Proposition 14.4.4.


(a) Use the Sylow Theorems to show that G has one or four 3-Sylow subgroups. Then useCG(H) = H to show that the number is four.

(b) Let H1 be a 3-Sylow subgroup of G. Use part (a) and the Sylow Theorems to show thatits normalizer has order 6.

(c) Now consider the homomorphism G S4 given by the action of G by conjugation onthe 3-Sylow subgroups. Use part (b) to prove that cannot contain an element oforder 3.

(d) Conclude that the image of contains A4. It follows that if is not an isomorphism, thenG contains a normal subgroup of order 2.

(e) Prove that G cannot contain a normal subgroup of order 2. Thus G

This exercise is closely related to Exercise 3 of Section 14.2.

Exercise 9. Let g and C(g) be as in the proof of part (c) of Proposition 14.4.4.(a) Show that C(g) is Abelian and contains(b) If m E C(g), then it is easy to see that det(m)m2 E C(g). By part (a), it follows

that = det(m) m2 defines a group homomorphism : C(g) —* C(g). Prove that= F,,!2 and = C(g)J/(p— 1).

(c) Prove thatI

det(w) = 1}.(d) Explain why we may assume that g = Then use Lemma 14.4.3 and Exercise 5 to

show that det : C(g) —* is onto. Conclude that

Idet(w) = l}.

The equality proved in part (d) shows that every element of C(g) of determinant 1 is of the formdet(m)m2 for some m E C(g). This will be used in the proof of part (c) of Proposition 14.4.4.

Exercise 10. Consider the subgroup N(H) c PGL(2, IF,,) defined in Proposition 14.4.4.(a) Prove that the images of the matrices (14.25) generate N(H) when p 1 mod 4.(b) Prove that generators of H and the images of the matrices

/1 —i\ / s t—1and

—s

from [17, p. 1631 generate N(H) when p 3 mod 4.

Exercise!!. LetM3 be as in Proposition 14.4.5. Show thatM3 is solvable of order 24p2(p —1).

Exercise 12. Consider the subgroups Mi, M2, and M3 defined in the text.(a) Show that (Mi )o and (M2)o have Abelian subgroups of index 2, and use this to prove that

neither can contain (M3)o. This proves that M3 Mi and fy13 M2.

(b) Explain why M3 = AGL(2,F3) when p = 3.

(c) Show that has an element of order p+ 1, and use this to prove that M1when p > 3.

(d) Show that A'!2 M3 when p > 5.(e) Show that M2 C A'!3 when p = 5.

It follows that the only exceptions to (14.27) are C M3 and M2 C M3 when p = 3 andM2 C M3 when p = 5. This result is due to Jordan.

Exercise 13. Let G0 C GL(2, be solvable. Prove that the subgroup generated by Go andIF;!2 is also solvable.

Exercise 14. Letg = where a,/3 E ]F,' and a /3.


(a) Prove (14.29).

(b) Let m = E N(C(g)). In the argument following (14.29), we claimed that b = c = 0

or a = d = 0. Supply the missing details.

(c) Prove that (M2)o C N(C(g)).

Exercise 15. Prove (14.30) and (14.31).

Exercise 16. Let V, W be vector spaces over a field F, and let AutF(V) be the group of vectorspace isomorphisms V V. Also let T : V —+ W be a vector space isomorphism.(a) Prove that To o T' induces a group isomorphism : AutF (V) AUtF (W).

(b) Let T' : V —* W be another isomorphism. Prove that there is P E Autp(W) such thatT' = o T. In the notation of part (a), : AutF (W) AutF (W) is conjugation by 1.

(c) In the situation of part (b), prove that 'YT' = °yT.

Exercise 17. Fix a E and let be as defined just before (14.35).(a) Prove (14.37) and (14.39). For (14.37), you should use the argument from the proof of

Lemma 14.4.3.

(b) Prove that FL(l,F,,2)

Exercise 18. Let M be a finite group.(a) Let A C M be a minimal normal subgroup, and let g e be in A. Prove that A is generated

by the elements ash varies over all elements of M.

(b) Let A C M be a normal subgroup. Prove that the center Z(A) of A is normal in M.

Exercise 19. In the Mathematical Notes, we mentioned that all subgroups of PSL(2, lFq) areknown up to conjugacy. We will do a small part of this classification by proving that PSL(2, IF,,)contains a subgroup isomorphic to S4 when p 1 mod 8. To begin, note that by Exercise 10,the images of the matrices (14.25) generate a subgroup of PGL(2, IF,,) isomorphic to S4.(a) Explain why has an element of order 8. Then i = has order 4.

(b) Compute (1 + i)2 and use this to prove that there is a E F,, such that a2 = 2.

(c) Show that the matrices (14.25) lie in after multiplication by suitable elementsof Hence their images generate a subgroup of PSL(2, F,,) isomorphic to S4.

(d) Over C, = cos(2ir/8) + isin(2ir/8) = (1 + i)/v'i How does this relate to part (b)?More generally, one can prove that if q = ptm and p > 2, then PSL(2, Fq) always contains acopy of A4, and it contains a copy of if and only if q ±1 mod 8 (see [2, Ch. XII] or [8, §8of Ch. II]). You should compare the list of groups given in these references with (7.29), whichasserts that the finite subgroups of PSL(2, C) = PGL(2, C) are cyclic, dihedral, or isomorphicto A4, S4, orA5.

Exercise 20. Assume that g,h E satisfy gh = —hg and det(g) = det(h) = 1, as inpart (a) of Proposition 14.4.4. Also assume that p> 2.(a) Prove that the subgroup (g, h) c GL(2, F,,) is isomorphic to the quaternion group Q =

{±1,±i,±j,+k},wherei2 =j2 =k2 = —1,ij= —ji=k, and —l EZ(Q).(b) Prove that (M3)0 is the normalizer of (g,h) in GL(2,F,,).

The quaternion group is an example of an extraspecial 2-group. The normalizer of an extraspe-cial 2-group in GL(2, Fq) is part of Aschbacher's classification of subgroups of GL(n, IFq). Thisis explained (briefly) in [131.


REFERENCES

1. E. Brown and N. Loehr, Why is PSL(2, 7) GL(3, 2)?, Amer. Math. Monthly 116(2009),727—732.

2. L. E. Dickson, Linear Groups with an Exposition of the Galois Field Theory, B. G.Teubner, Leipzig, 1901. Reprint by Dover, New York, 1958.

3. J. D. Dixon and B. Mortimer, Permutation Groups, Springer, New York, Berlin, Heidel-berg, 1996.

4. E. Dobson and D. Witte, Transitive permutation groups of prime-squared degree, J.Algebraic Combin. 16 (2002), 43—69.

5. H. Edwards, The construction of solvable polynomials, Bull. Amer. Math. Soc. 46(2009),397—411, 703—704 (Errata).

6. D. Gorenstein, Classifying the finite simple groups, Bull. Amer. Math. Soc. 14 (1986),1—98.

7. M. Hall, The Theory of Groups, Macmillan, New York, 1959. Reprint by Chelsea, NewYork, 1972.

8. B. Huppert, Endliche Gruppen I, Springer, New York, Berlin, Heidelberg, 1967.

9. B. Huppert, Lineare auflösbare Gruppen, Math. Z. 67 (1957), 479—518.

10. B. Huppert, Zwe(fach transitive, auflosbare Permutationsgruppen, Math. Z. 68 (1957),126—150.

11. B. Huppert and N. Blackburn, Finite Groups III, Springer, New York, Berlin, Heidelberg,1982.

12. I. Radloff, Evariste Galois: Principles and Applications, Historia Math. 29 (2002), 114—137.

13. C. M. Roney-Dougal and W. R. Unger, The affine primitive permutation groups of degreeless than 1000, J. Symbolic Comput. 35 (2003), 421—439.

14. J. S. Rose, A Course on Group Theory, Cambridge U. P., Cambridge, 1978.

15. M. Rosen, Niels Hendrik Abel and equations of the fifth degree, Amer. Math. Monthly102 (1995), 495—505.

16. M. W. Short, The Primitive Soluble Permutation Groups of Degree less than 256, Springer,New York, Berlin, Heidelberg, 1992.

17. D. A. Suprunenko, Matrix Groups, Translations of Mathematical Monographs, Volume45, AMS, Providence, RI, 1976.

CHAPTER 15

THE LEMNISCATE

The lemniscate is the plane curve defined by the equation (x2 + y2)2 = — y2. Hereis a picture:

We will consider the Galois groups of polynomials arising from division of thelemniscate into arcs of equal length. This will allow us to prove the followingwonderful theorem of Abel [Abel, Vol. I, p. 314]:

One can divide the entire circumference of the lemniscate into m equal parts byruler and compass alone, if m is of the form T' or 2" + 1, the last number beingat the same time prime; or as well if m is a product of several numbers of thesetwo forms.

Galois Theory. Second Edition. By David A. Cox 463Copyright © 2012 John Wiley & Sons, Inc.

464 THE LEMNISCATE

Abel goes on to say that this theorem is "precisely the same as that of M. Gauss,relative to the circle." You will verify this in Exercise 1.

To prove Abel's theorem, we will study doubly periodic functions of a complexvariable and the theory of complex multiplication. We will also learn why Eisensteinproved his irreducibility criterion.

15.1 DIVISION POINTS AND ARC LENGTH

To formulate Abel's theorem on the lemniscate carefully, we need to define then-division points of the lemniscate and study the arc length of this curve.

A. Division Points of the Lemniscate. In Section 10.2 we used the nth rootsof unity to determine when a regular n-gon can be constructed by straightedge andcompass. In terms of the unit circle centered at the origin, the nth roots of unitydivide the circle into n segments of equal length, starting from (1,0). For n = 5, thefifth roots of unity 1, give the picture:

(5

In general, the nth roots of unity i = 0, . . . , n 1, are the n-division points of theunit circle. Then Gauss's theorem of Section 10.2 can be restated as the assertion thatthe n-division points of the unit circle are constructible by straightedge and compassif and only if n is a power of 2 times a product of distinct Fermat primes.

Abel, following hints of Gauss, asked the same question for the lemniscate. Here,the n-division points of the lemniscate are obtained as follows. Begin at the originand follow the curve into the first quadrant, down into the fourth quadrant, backthrough the origin into the second quadrant, down into the third quadrant, and finallyback to the origin. As we do this, we mark those points that give one-nth of the totalarc length, two-nths of the arc length, etc. For n = 5, this gives the picture:

DIVISION POINTS AND ARC LENGTH 465

The n-division points divide the lemniscate into n segments of equal length. Whenn is odd, as in the above picture, the middle segment straddles the origin. When nis even, the n-division points are symmetric about the x- and y-axes, with the middledivision point at the origin. For n = 6, the 6-division points give the picture:

The n-division points on the lemniscate will lead to some remarkable polynomialsanalogous to the cyclotomic polynomials. The Galois theory of these polynomi-als will enable us to understand when the n-division points can be constructed bystraightedge and compass.

At the beginning of the chapter, we defined the lemniscate using the Cartesianequation (x2 +y2)2 — y2. In Exercise 2 you will show that in polar coordinates,the lemniscate is given by the equation

(15.1) r2=cos(29).

The polar coordinate r will play a central role in this chapter. One reason is that inorder to construct a point on the lemniscate, we only need r. This might seem obviousin that we get the desired point (and its mirror images about the x- and y-axes) byintersecting the lemniscate with the circle of radius r. But the lemniscate is actuallyunnecessary. In other words, if 0 < r < 1 is constructible in the sense of Section 10.1,then so are the x- and y-coordinates of the four points on the lemniscate of distance rfrom the origin, To see this, we use (x2 +y2)2 = x2 — y2 and r2 = x2 +y2. This givesthe equations

r4=x2—y2 and r2=x2+y2.Solving for x and y in terms of r, we obtain

and

Since the constructible numbers form a subfield of C closed under square roots, wesee that x and y are constructible when r is. Thus, to prove that a given point onthe lemniscate is constructible by straightedge and compass, it suffices to show thatthe corresponding polar coordinate r is constructible. Also note that the converseholds: if x andy are constructible, then so is r = +y2. We have thus proved thefollowing result.

Proposition 15.1.1 Let P be a point on the lemniscate, and let r be the distance fromP to the origin. Then P can be constructed by straightedge and compass if and onlyif r is a constructible number •

466 THE LEMNISCATE

B. Arc Length of the Lemniscate. The n-division points of the lemniscateare defined in terms of arc length. Hence we need to study the arc length of thelemniscate. By (15.1), the polar equation of the lemniscate is

r2 = cos(20).

If we focus on the first quadrant, then we get the picture:

Solving the above equation for 0 gives 9 = cos1 (r2). This makes 9 into a functionof r. Note that 9 decreases from to 0 as r increases from 0 to I.

Recall that arc length in Cartesian and polar coordinates is given by

ds = + dy2 = + r2d02.

It follows that the arc length of the lemniscate from the origin to the point in the firstquadrant with polar coordinates (ro, Oo) is given by

dO2arclength=j dr.

Differentiating r2 = cos(20) with respect to r gives 2r = —sin(20) . so that

1 +r2(ç) = I +r2(_ sin(20)) = 1 +

sin2 (28) 1 — cos2(29) = 1 — r4, we obtain

2 dO 2 r4 1l+r =l+l_r4Hence our arc length formula becomes

(15.2) arc length= f v'i1 r4

dr.

The integral (15.2) is improper when r0 = 1. Since it converges (see Exercise 3),(1 r4)"2dr is the arc length of the first-quadrant portion of the lemniscate. In

the eighteenth century, this number was denoted where w is a variant of the Greekletter ir. Thus

tz7=2 IJo

1


It follows that the arc length of the lemniscate is and the arc length betweensuccessive n-division points is

We will write (15.2) as

(15.3)

s represents the arc length along the lemniscate from the origin to the point inthe first quadrant with polar coordinates (r, 0). Then (15.3) expresses s as a functionof r. Following Abel the inverse function will be written r = so that

(15.4)

SinceO<r< 1 correspondsto

we see that is defined on the interval [0, In Section 15.2 we will extend toa periodic function on R, and in Section 15.3 we will further extend to a doublyperiodic meromorphic function on C.

In particular, when n > 4, the first n-division point of the lemniscate lies in the firstquadrant. Since its arc length from the origin is Proposition 15.1.1 implies thatthe first n-division point is constructible by straightedge and compass if and only if

is a constructible number. In Section 15.2 we will develop multiplication formulasfor n Z and use them to show that:

•( is the root of a polynomial with coefficients in Z.

• ( is constructible if and only if all n-division points are constructible bystraightedge and compass.

In Section 15.5 we will consider the Galois group of the extension

Q(i) C

The appearance of i = is unexpected but will make perfect sense once we studythe complex multiplication formulas for ((n + im)s), n + im E Z[i), in Section 15.4.Using this and some clever ideas of Eisenstein, we will then be able to prove Abel'stheorem on the lemniscate.

Mathematical Notes

Here are comments about two topics from this section.

• Integrals and Inverse Functions. The definition of Abel's function r =involves an integral defining s in terms of r and then an inverse function to get r interms of s.

468 THE LEMNISCATE

The idea of an inverse function of an integral is more common than you mightexpect. For example, one standard definition of is to first define the naturallogarithm via the integral

x>0,

and then define? to be the inverse function of ln(x). So? is the inverse function ofan integral. Another example from calculus is the indefinite integral

I = sin1(x) + C.

In terms of definite integrals, this can be written

The inverse function of sin (x). So sin (x) is the inverse functionof an integral.

Now comes an intriguing idea. Suppose that we knew neither sin nor Howcan we understand the integral 1(1 — x2)_h/2 dx? One way would be to define sin(x)to be the inverse function of

Furthermore, if we define cos(x) = and = then allstandard properties of sin(x) and cos(x) can be derived from these definitions.

One way to regard (15.4) is that Abel's function p(s) is obtained by applying thesame idea to the integral (1 x4) — "2dx. There are many analogies between sin(x)and and it is possible to develop the properties of these functions in parallel.This is done nicely in [12, pp. 240—243]. See also [16, pp. 1—9].

• Elliptic Integrals and Elliptic Functions. The integral f(l —x4)1/2dx is anexample of an elliptic integral, and elliptic functions are inverse functions of ellipticintegrals. In general, an elliptic integral is an indefinite integral of the form

•

/ J

where A(x),B(x),C(x),D(x) are polynomials in x, and P(x) is a polynomial in x ofdegree 3 or 4. If P(x) has degree 1 or 2, then the integral (15.5) can be evaluated usingstandard techniques of integration. So elliptic integrals, where P(x) has degree 3 or4, are the next integrals to consider. It follows that 1(1 — x4) —

1/2 dx is an especiallysimple example of an elliptic integral.

We will say more about elliptic integrals and elliptic functions in the next section.For now, we conclude with another example of an elliptic integral. A standard ellipse


with center at the origin is given by an equation of the form + = 1. Assume

that a = 1 and 0< b <1. If we set k = '/1—b2, then in Exercise 4 you will showthat the ellipse

+ = 1

has arc length given by

(15.6)

This special case of (15.5) is where the "elliptic" in "elliptic integral" comes from.

Historical Notes

The lemniscate first appeared in the mathematical literature as part of the ovals ofCassini, described by the French astronomer Cassini in 1680. In Cartesian coordi-nates, the ovals are the family of curves defined by the equation

(15.7) ((x—a)2+y2)((x+a)2+y2) =b4.

The lemniscate we've been studying corresponds to a = b = (Exercise 5). Ingeneral, a < b gives a dumbbell-shaped curve and a > b gives two ovals, as in thefollowing picture:

Unaware of Cassini's work, in 1694 Jacob (or James) Bernoulli gave the equationof the lemniscate as

xx+yy =a'/xx—yy.

He described the curve as having "the form of a figure 8 on its side, as of a band foldedinto a knot, or of a lemniscus, or of a knot of a French ribbon." Here, "lemniscus" is aLatin word (taken from the Greek) meaning a hanging ribbon attached to the garlandworn by the winner of an athletic contest.

Bernoulli was led to this curve by an indirect route. In 1691 he encountered theintegral 2 (1 — t4) — "2th' (this should look familiar) in his study of the elastic curve.To represent this geometrically, he looked for a curve defined by an algebraic equation

whose arc length equals In 1694 he showed that the lemniscate

470 THE LEMNISCATE

has the desired arc length, using the polar description of the lemniscate, as we didearlier in the section. The reader should consult [2] for a discussion of the elasticcurve and Bernoulli's priority dispute with his brother Johann, who independentlydiscovered the lemniscate in 1694 in a different context.

Bernoulli's use of polar coordinates to compute the arc length of the lemniscaterepresents the first use of arc length in polar coordinates. It is ironic that calculusstudents study the lemniscate in one part of the course and arc length in polarcoordinates in another, but they never put the two together, since the resulting integralcannot be evaluated by the usual methods of calculus.

Our discussion shows that the lemniscate and its arc length were well known by thebeginning of the eighteenth century. Thus Abel's theorem on dividing the lemniscateinto arcs of equal length deals with a topic familiar to the mathematical communityof the time.

We will say more about elliptic integrals and elliptic functions in the HistoricalNotes to the next section.


Exercise!. Prove that the numbers described in Abel's theorem at the beginning of the chapterare precisely those in Theorem 10.2.1, provided we replace "product of several numbers" with"product of distinct numbers" in Abel's statement of the theorem.

Exercise 2. Show that in polar coordinates, the equation of the lemniscate is r2 = cos(20).

Exercise 3. Prove that the two improper integrals J'0' (1 — t4) — "2dt and f°1 (1 — t4) — 1/2 dt

converge.

Exercise 4. Prove the arc length formula stated in (15.6).

ExerciseS. Show that (15.7) reduces to (x2 +y2)2 = x2 —y2 when a = b =

Exercise 6. Let n > 0 be an odd integer, and assume that the n-division points of the lemniscatecan be constructed with straightedge and compass. Prove that the same is true for the 2n-division points. Your proof should include a picture.

Exercise 7. Recall that in Greek geometry, the ellipse is defined to be the locus of all pointswhose sum of distances to two given points is constant. Suppose instead we consider the locusof all points whose product of distances to two given points is constant. Show that this leadsto (15.7) when the given points are (a,0), (—a,0) and the constant is b4.

15.2 THE LEMNISCATIC FUNCTION

In (15.4) we defined Abel's function by

(15.8) r = p(s) s = I dt.j0

Since s represents arc length from the origin along the first quadrant portion of thelemniscate, we see that is defined on [0, where

THE LEMNISCATIC FUNCTION 471

w=2j 1dt

In this section, we will extend ço(s) to a function on R of period 2w and show that itsatisfies some remarkable addition and multiplication formulas. We will also applythese formulas to straightedge-and-compass constructions on the lemniscate.

A. A Periodic Function. Our first task is to define p(s) as a function of period2w on lIi. We will do this by extending the arc length interpretation of p(s) given inSection 15.1.

The arc length parametrization of the lemniscate is defined by sending a realnumber s to the point P on the lemniscate such that:

• Ifs 0, then P is the origin.

• If s > 0, then move from the origin into the first quadrant portion of the lemniscateand continue along the curve until we reach the point P whose cumulative arclength from the origin is s.

• If s < 0, then move from the origin into the third quadrant portion and continueuntil we reach the point P whose cumulative arc length from the origin is —s.

We call s the signed arc length variable of the lemniscate. When is large, wemay need to ioop around several times before reaching the point P. Since the totalarc length of the lemniscate is 2w, we see that s and s ±2w give the same point onthe lemniscate for any s R. This is similar to measuring angles on the unit circle,where s and s ± 2ir give the same point on the circle.

The lemniscate is r2 = cos(20) in polar coordinates. Recall that r is allowed tobe negative as well as positive or zero. We will restrict 0 to lie in so thato < r < 1 gives the right half of the lemniscate and —1 r 0 gives the left half.We call r the polar distance of the corresponding point on the lemniscate. Strictlyspeaking, r is really the signed polar distance, since r is negative on the left half ofthe lemniscate. We will use the shorter term "polar distance" for simplicity.

Now consider (15.8). It is easy to see that the signed arc lengths satisfies

= jrdt

for s and—i r 1. This implies that(15.8) can be used to define p(s)for — <s < In other words, for s in this range, Abel's function p(s) is simplythe polar distance (with the above convention on r) of the point on the lemniscatewith signed arc length s.

It is now easy to extend to all of R: given s e R, ço(s) is the polar distance ofthe point on the lemniscate whose signed arc length from the origin is s. Thus

p(s) = r,

Abel's Function

p(—s) = —p(s),

=

= 1

472 THE LEMNISCATE

where s and r are related according to the diagram:

S

Note that p(s) has period since s and s + give the same point on thelemniscate. Furthermore, p(s) also satisfies the following identities:

(15.9)

The first follows because s and —s correspond to points on the lemniscate symmetricabout the origin, and the second follows because s and — s correspond to pointssymmetric about the x-axis (recall that each half of the lemniscate has length zv). SeeExercise 1 for the details. Using the arc length interpretation of p(s), one can showthat p(s) is infinitely differentiable for all s R, though we omit the proof.

The function sin ( has the same period and amplitude as p(s) (check this). Ifwe plot p(s) and s < then we get the following graphs:

The Sine Function

The function sin(x) satisfies identities similar to (15.9), but the full theory of sin (x)requires sin'(x) = cos(x) as well. The same is true for p(s), where we will useWe will also need the following crucial identity.

Proposition 15.2.1 Let cp(s) be defined as above. Then

Proof: Using (15.9) and the periodicity of p(s), you will show in Exercise 2 that itsuffices to prove that


To derive this equation, first observe that (15.8) gives the identity

1s=fDifferentiating each side with respect to s and using the Chain Rule and the Funda-mental Theorem of Calculus, we obtain

1=(s)

(be sure you understand why s = is excluded). It follows that

= 0 s <

For s = note that — p4(s) vanishes at since = 1. Since 1 is themaximum value of p(s) (can you explain why?), we see that also vanishesat This completes the proof. .

Other properties of (s) will be developed in Exercise 3, and in Exercise 4 youwill adapt the method used in Proposition 15.2.1 to derive the standard trigonometricidentity cos2(x) = 1 sin2(x).

B. Addition Laws. The addition law for sin(x) states that

sin(x+y) = sin(x)cos(y) + cos(x)sin(y).

For the addition law goes back to Euler, who in 1753 proved the identity

(15.10)

To state this in terms of let x, y, and z represent the three integrals in (15.10), sothat p(y) $, and = 'y. Then x + y = z implies that

which when combined with = and = t3 gives

(1511) (

+—

Furthermore, — p4(x) = for 0 <x < by Proposition 15.2.1. Thus

ço(x),o'(y) + ço(y)p'(x)(15.12)

2 2 (y)

474 THE LEMNISCATE

Rather than use Euler's result, Abel gave a different proof that (15.12) holds for allx,y E R. You will explore Abel's argument in Exercise 5.

In Exercise 6 you will use = to show that (15.12) implies thesubtraction law

—

_______________

y

When this is combined with (15.12), we easily obtain the identity

(y)(15.13) ço(x+y)+ço(x—y)=

This will be useful later in the section.The addition laws give some nice straightedge-and-compass constructions.

Example 15.2.2 Let us divide the lemniscate into eight pieces of length =Here is a picture of ro = ( and the 8-division points:

This picture and Proposition 15.1.1 show that to construct the 8-division points, weneed only construct r0. Since = 1, the addition law (15.11) implies that

— —

—

Solving this equation in Maple or Mathematica shows that the unique real positivesolution is given by

= I .643594.

This is obviously constructible and hence gives the desired construction.

The reasoning behind Example 15.2.2 can be generalized as follows.

Proposition 15.2.3 If is constructible, then so is

Proof: Setting x = y in (15.12) gives the duplication formula

(15.14)=

1


Let ro = and a = Then (15.14) and = 1— imply that

2_ — —re)a

I —

To solve this equation for r0, let t E C satisfy

(15.15)

2 2 4

(15 16)it = = r0(l —r0)

=a21—t4

Solving (15.16) fort2 by the quadratic formula shows that t2 is constructible becausea is, and then solving (15.15) for r0 completes the proof. .

The formulas (15.15) and (15.16) in the above proof seem to come out of nowhere.In Section 15.4 we will use complex multiplication and 2 = (1 + i) (1 i) to "factor"the duplication formula for into (15.15) and (15.16). So these formulas willeventually make perfect sense.

Here is a slightly more complicated example.

Example 15.2.4 Dividing the lemniscate into six pieces of equal length gives thefollowing picture:

To computer0 = = first observe that(15.13) with 2x andxin placeof x andy gives

== 1

'

121+ )

476 THE LEMMSCATE

where the last line uses (15.14). Using = 1 — p4(x) and a bit of algebra, weobtain the tripling formula

(15.17) =

Since = 0, substituting x = into (15.17) shows that r0 = satisfies

+ — 3 = 0,

which is easily seen to have the unique real positive solution

ro = 3 .825379.

This is clearly constructible and hence gives the desired straightedge-and-compassconstruction. <II>

C. Multiplication by Integers. The doubling and tripling formulas

—

(3 \( )

from (15.14) and (15.17) can be generalized to formulas that express ço(nx) in termsof and p'(x) for any positive integer n.

Theorem 15.2.5 Given an integer n > 0, there are relatively prime polynomialsZ[u] such that

f

is then

ço(nx) = (x).

Furthermore, = 1.

Proof: We will prove the theorem by induction on n. Setting P1 (u) = Qi (u) = 1

gives the desired formula for n = 1. For n = 2, note that (15.14) can be written

=1


Thus the theorem holds for n = 2 with P2 (u) = 1, Q2(u) = 1 + u. Now assume that itholds for n — 1 and n. Using (15.13) with nx and x in place of x andy, we obtain

co((n+ l)x) = —co((n l)x) +1

If n is even, then n 1 is odd, so that our inductive hypothesis implies that

P, (x)/

( \=+

(x))

(x) (x) and clearing denominators, this simplifies to

Pn+i4 (x))

where

(15.18) = —u))

and 1(u) is given by a similar recursive formula (see Exercise 7). It follows thatPn+i (u), (u) E Z[uI by our inductive hypothesis. Note also that (0) = 1

follows from Q,, (0) = Q,, = 1. Finally, dividing (u) and 1(u) by theirgreatest common divisor shows that we may assume that they are relatively prime inZ[u]. In Exercise 7 you will show i(O) = 1 continues to hold, after multiplying

by —1 if necessary.The case when n is odd is similar and will be covered in Exercise 7. •

Theorem 15.2.5 has some nice consequences concerning the division points onthe lemniscate. The polar distances of the n-division points are

m=0,l,...,n—1.

When n is odd, the periodicity of and Theorem 15.2.5 imply that

p1 4[ 2w2w 2w

so that the polar distance is a root of when n is odd. In Exercise 8

you will show that when n is even, the polar distances are roots of — u2).We call these polynomials the n-division polynomials. We have thus proved the

following corollary of Theorem 15.2.5.

Corollary 15.2.6 Let n Z be positive. Then the polar distances of the n-divisionpoints of the lemniscate are roots of the n-division polynomials defined above. •

478 THE LEMNISCATE

We also have the following result about straightedge-and-compass constructions.

Corollary 15.2.7 Let n be a positive integer such that is constructible.

(a) is constructiblefor every mE Z.

(b) The n-division points of the lemniscate are constructible by straightedge andcompass.

(c) If in addition ( is constructible for a positive integer m, then so is (where N = lcm(n,m).

Proof: If is constructible, then so is since =Part (a) is obvious for n = 1 and 2, so we may assume that n > 2.

Whm >0, Theorem 15.2.5 implies that is

and with coefficients in Z. In Exercise 9 you will show that the denominatoris nonvanishing, since n> 2 and the polynomials Pm(u), Qm(u) in Theorem 15.2.5 arerelatively prime. Hence (me) is constructible, since the constructible numbersform a subfield of C.

The case m = 0 is obvious, and m < 0 follows from m > 0 because is an oddfunction. This completes the proof of part (a).

Part (b) follows immediately from part (a) and Proposition 15.1.1.For part (c), let d = gcd(n,m). Then N = lcm(n,m) = It follows that if

integers v satisfy + un = d, then

fL—+V—— =(itm+vn)— =d— =n m nm nm N

By part (a), and are constructible, and—as above—the same istrue for and Then the addition law (15.12) expresses asa rational expression with coefficients in Z in the constructible numbers given by thevalues of and at /i and v Since the denominator of this rational expressionis the nonzero number

it follows that is constructible. U

Parts (b) and (c) of Corollary 15.2.7 imply that if the n-division points and m-division points of the lemniscate are constructible by straightedge and compass, thenthe same is true of the N-division points for N = lcm(n,m). This fact will be usefulin Section 15.5.

Here are some applications of Corollary 15.2.7.

Example 15.2.8 Since = 0, Proposition 15.2.3 implies that is con-structible for n 0. Then part (b) of Corollary 15.2.7 shows that thepoints of the lemniscate can be constructed by straightedge and compass.


Example 15.2.9 When n = 5, one can show that

P5(p4(x))= where

(15.19)P5(u) = u6+50u5 — 125u4+300u3— 105u2—62u+5,

Q5(u) = 1 +SOu— 125u2+300u3— 105u4—62u5+5u6

(see [14, p. 821). Note the "reverse symmetry" of the coefficients of P5(u) and Q5(u).For the 5-division points of the lemniscate, the discussion preceding Corollary 15.2.6implies that r0 ( is a root of the 5-division polynomial uP5 (u4). Thus

0 = roP5(rg) = ro(rg4 + — + — + 5).

You will show in Exercise 10 that the real positive solutions are constructible, thoughthis is not obvious from the above equation. By Corollary 15.2.7, it follows that the5-division points are constructible by straightedge and compass.

This discussion makes it clear that understanding the n-division points of thelemniscate is intimately related to the multiplication formula for But tounleash the full power of these formulas, we will need to extend to a function of acomplex variable. This is will be done in the next section.

Historical Notes

Although the link between arc length and the lemniscate goes back to Bernoulli,the first person to make substantial progress in this area was Fagnano. In 1718 heproved the case a = of Euler's addition law (15.10), namely

1 P 1

210 dt= j dt when

= 1 + a4

Using this and other results, Fagnano was able to divide one arch of the lemniscateinto two, three, and five segments of equal length by straightedge and compass.Fagnano's results and methods are discussed in [1].

Things got really interesting when Fagnano's papers were submitted to the BerlinAcademy as part of his application for membership. Euler was asked to read thesepapers in December 1751, and by 1753 he was able to show that Fagnano's duplicationformula was a special case of (15.10). More importantly, he also realized thatcould be replaced by where P(t) is any separable polynomial of degree 4 withreal coefficients. This led to the theory of elliptic integrals, which was developedat great length by Lagrange and Legendre. Eventually the integrals were put in thestandard form

(15.20) f 1

dO,J \/l_k2sin2O

480 THE LEMNISCATE

which after the substitution t = sinO gives

(15.21) 1 dt.J —t2)(1 —k2t2)

We call k the modulus, so that f(1 — t4)—'/2 dt corresponds to the modulus k = i.

The first person to consider the inverse function of f (1 — t4) — "2dt was Gaussin 1797, though this work was not published until after his death in 1855. Abeland Jacobi introduced the inverse functions of elliptic integrals in 1827. In Abel'sgreat paper Recherches sur les fonctions elliptiques [Abel, Vol. I, pp. 263—3881, heconsidered the inverse function of an elliptic integral of the form

I dt,J —c2t2)(l +e2t2)

so that c = e = 1 gives the lemniscatic function we've been studying. Jacobi, on theother hand, used the integral (15.20) and wrote its inverse function as 0 = amu. ThussinO = sinamu is the inverse function of (15.21). These days, we write sinamu assn(u, k), or simply sn(u) if the modulus is understood, though Mathematica writessn(u,k) as JacobiSN[u,k2J. In the text, we used JacobiSN[u, —11 to draw the graphof the lemniscatic function = sn(u,i).

One of the critical discoveries of Gauss, Abel, and Jacobi is that inverse functionsof elliptic integrals are doubly periodic functions of a complex variable. We willconsider a special case of this in the next section. More on the history of ellipticintegralscanbefoundin[1], [7, pp. 3—16], and [12, pp.267—268]. Aniceintroductionto the duplication formula (15.14) appears in [16].


Exercise 1. Give a careful proof of (15.9) using the hints given in the text.

Exercise 2. Supply the details needed to complete the proof of Proposition 15.2.1.

Exercise 3. Here are some useful properties of (s).

(a) p(s) has period 2w. Explain why this implies that the same is true for(b) p(s) is an odd function by (15.9). Explain why this implies that is even.(c) Use (15.9) to prove that —s) =(d) Use Proposition 15.2.1 to prove that p"(s) =

Exercise 4. Suppose that we define sin(x) by y = sin(x) x = —t2)'/2dt. Thendefine cos(x) to be sin' (x). Use the method of Proposition 15.2.1 to prove the standardtrigonometric identity cos2(x) = 1 — sin2(x).

Exercise 5. Here is Abel's proof of the addition law for(a) Let g(x,y) be differentiable on R2, and set h(u, v) = (u + v), (u — v)). Use the Chain

Rule to prove that

= —


(b) Use part (a) to show that g(x,y) = g(x+y,O) if and only if = on

(c) Prove the addition law for by applying part (b) to

() —

_______________

gx,y—

Part (d) of Exercise 3 will be useful.

Exercise 6. Show that the subtraction law

(x— )——

follows from the addition law together with (15.9) and Exercise 3.

Exercise 7. The proof of Theorem 15.2.5 uses induction on n.(a) Assume that n is even. In (15.18), we gave a formula for (u) in terms of and

Derive the corresponding formula for

(u) satisfy all of the conditions of the theorem ex-cept for the requirement that they be relatively prime. Since Z[u] isa UFD, we canwrite = = where E Z[u] andPn(u),Qn(u) are relatively prime. Prove that we can assume that = 1 and thatPn(u),Qn(u) satisfy all conditions of Theorem 15.2.5.

(c) Complete the inductive step of the proof when n is odd.

Exercise 8. Let n be even, and let (u) be the polynomial from Theorem 15.2.5. Completethe proof of Corollary 15.2.6 by showing that the polar distances of the n-division points ofthe lemniscate are roots of (I — u2).

Exercise 9. This exercise is concerned with the proof of Corollary 15.2.7.(a) Suppose that P(u), Q(u) E 7Z[u] are relatively prime and Q(0) = 1. Prove that uP(u4) and

Q(u") have no common roots in any extension of Q.

(b) Fix x in Rand m >0 in Z, and let Pm(u),Qm(u) E Z[u] be as in Theorem 15.2.5. Thus= Prove that 0 when 0.

(c) Show that 0 when n >2 is in Z and conclude that 0.

Exercise 10. The polar distances of the 5-division points of the lemniscate satisfy the equation

0 = + — + — — + 5).

This equation was first derived by Fagnano in 1718.(a) Show that the corresponding to the 10-division points also satisfy this equation.

(b) Use Maple or Mathematica to show that this equation factors as

and that the only positive real solutions are

Explain (with a picture) how these solutions relate to the 5- and 10-division points.

482 THE LEMNISCATE

Exercise 11. Use sin(x + y) = sinxcosy + siny cosx to show that if E [0, 1], then

1 i 1

f dt+f dt

= fwhere -y is the real number defined by

=

Note the similarity to (15.10).

Exercise 12. Show that the substitution t = sin8 transforms (15.20) into (15.21), and use thisto prove carefully that p(u) = sin am(u) when the modulus is k = i.

15.3 THE COMPLEX LEMNISCATIC FUNCTION

Corollary 15.2.6 implies that the polar distances r = of the n-division pointsof the lemniscate are roots of the n-division polynomials. To prove Abel's theoremon the lemniscate, we need to represent all roots of these polynomials using Sincemany of the roots are complex, this requires that we follow Gauss and Abel andextend to a function defined on C.

Abel began by considering for y e R. We know that r = is the inversefunction of y = 1 — t4) — "2dt. The change of variables t = iu shows that

çir1 1

I dt=i I du=zy..Jo Jo

This suggests that can be defined to be = ir = Then Abel used theaddition law to define + iy) as

( + )—

coX ZY—

Since = easily implies that = (see Exercise 1), the formulafor + iy) simplifies to

(15.22) = iy)

To make Abel's approach rigorous, we will define using (15.22). OverL is periodic and defined everywhere; over C, we will see that is doublyperiodic and has poles. The properties of will play a crucial role in Sections 15.4and 15.5.

This section will assume familiarity with standard topics from complex analysis,including the Cauchy—Riemann equations and Laurent series. We will refer to [131,though the results we need are in most introductory texts on the subject.

A. A Doubly Periodic Function. As above, we define iy) usingequation (15.22). Here are some basic properties of this function.

THE COMPLEX LEMNISCATIC FUNCTION 483

Proposition 15.3.1 The function satisfies the following:(a) is analytic for all z (m + in) m, n odd.(b) The addition law

— cp(z)ço'(w) +'p(z+w) —

1

holds for all z, w E C such that both sides are defined.(c) ForzECandm,nEZ, we have

=

Proof: First observe that p(z) is defined whenever the denominator 1 (y)in (15.22) is nonzero. The polar distance interpretation shows that p2(x) 1

for all x IR, with equality if and only if x is an odd multiple of Hence isdefined on the open set = {z E C z (m + in) m, n Z odd}.

Write iy) = u(x,y)+ iv(x,y), where u(x,y) and v(x,y) are the real andimaginary parts of the right-hand side of( 15.22). It is easy to see that u(x,y) and v(x, y)are differentiable on as functions of x,y, since is infinitely differentiable on

Furthermore, using the identity (x) = 1 for x E R, it is straightforwardto verify that u(x,y) and v(x,y) satisfy the Cauchy—Riemann equations

auav Ou avaxay' 9y 9x

(see Exercise 2). By [13, 1.5.8], it follows that is analytic on ftFor part (b), let z and w be complex variables, and define

+(w)

When x0 E R is fixed, + w) and g(xo,w) are analytic in w and coincide whenwE IR by the addition law (15.12). By the Identity Theorem [13, 6.1.1], +w) =g(xo, w) for all w such that both are defined. It follows that when w0 e C is fixed,

+ wo) and g(z, wo) are analytic in z and coincide when z E IR and both are defined.Using the Identity Theorem again, we see that + wo) = g(z, WO) for all z such thatboth are defined. Since wo E C is arbitrary, this proves the addition law.

The proof of part (c) requires a series of facts about and that you willverify in Exercise 2. We begin with the following table of values:

x

0 0 1

(15.23) 1 00 —1

1 02

We also need the identities for z E C:

(15.24)=

(zz) = p (z).

484 THE LEMNISCATE

Earlier, these identities were "proved" in order to motivate (15.22). Here, they areinstead rigorous consequences of (15.22).

Since and have period 2w on IR, (15.23) and (15.24) easily imply that

1525'p(mw) = cc(mwi) = 0,

Using the addition law, it is now straightforward to show that

(15.26) = and ço(z+nwi) =

for m, n e Z. The desired identity for + mw + nwi) follows immediately. .Part (c) of Proposition 15.3.1 implies that is doubly periodic:

(15.27) (1 + i)w) = (1 — i)w).

Note that the periods (1 + i)w and (1 — i)w are linearly independent over R. Thepicture is the following:

(15.28)

The dots in this picture are the complex numbers in the set

—i)wlm,nEZ}.

This is the period lattice of Double periodicity means that once we know thevalues of for all z in one of the tilted squares, we know its values for all z E C.

B. Zeros and Poles. Our next task is to study the zeros and poles of Recallthat Zo E C is a simple zero of an analytic function g(z) if g(zo) = 0 and g'(zo) 0.

This is equivalent to saying that the power series expansion of g(z) at Zo is

g(z) at

2wi


As defined in [13, 3.3.2], Zo is a simple pole of a meromorphic function g(z) if theLaurent expansion of g(z) at Zo is

00a_i

g(z)=Z Zo

n=O

Theorem 15.3.2 is meromorphic on C with the following zeros and poles:(a) The zeros are all simple and occur at z 00 (m + in)rzr form, n E Z.(b) The poles are all simple and occur at z = (m + in) form, n odd.

Proof: Since =0 and = 1, part (c) of Proposition 15.3.1 easily impliesthat has a simple zero at (m + for all m, n E 7L.

Using the addition law together with (15.23) and (15.24), we see that

=2

Similarly,

= ±i1—p2(z)

(see Exercise 3). Multiplying these two equations gives the remarkable identity

=

z with z + and using + to') (prove this), we obtain

(15.29) =

Now assume p(zo) = 0. Then + (1 + is undefined by (15.29). Hence

m,nodd

by Proposition 15.3.1. It follows easily that Zo is one of our known simple zeros.To analyze the poles we write (15.29) as

Since has a simple zero at z 0, we see that has simple poles at z = (1 ± i)Using the double periodicity of we conclude that has simple poles at (m + in)for m, n odd. Then we are done, since these are the only possible singularities ofby Proposition 15.3.1.

Our next result will play an important role in the next section.

486 THE LEMNISCATE

Theorem 15.3.3 Fix a complex number Wo. Then the equation = wo has asolution Zo E C. Furthermore, is one solution, then all solutions are given by

z + m,n E Z.

Proof: Let g(z) be analytic in a region Il C C, and let C C Il be a simple closedcurve, oriented counterclockwise. The Zero—Pole Theorem [13, 6.2.11 says that ifg(z) has no zeros or poles on C, then

27r, c g(z)

where Z is the numbers of zeros of g(z) inside C, counted with multiplicity, and P isthe numbers of poles of g(z) inside C, also counted with multiplicity.

The function g(z) = — has the same poles as which are (m +m,n odd, by Theorem 15.3.2. This means that we cannot use the tilted squares from(15.28). However, since the zeros of g(z) are isolated (see [13, 6.1.21), we can shiftone of the squares to the left as pictured below to obtain a curve C such that g(z) hasneither zeros nor poles on C:

(15.30)

The open circles are poles of g(z) and are simple by Theorem 15.3.2. Exactly twolie in the interior of C, so that P = 2.

Since g(z) = — wo has periods (1 ± the same is true for g'(z) andg'(z)/g(z). Opposite edges of C differ by (1 ± so that g'(z)/g(z) takes the samevalues on opposite edges. Hence the integrals along opposite edges cancel, sincethey have opposite orientations. This gives

Z—2=Z—P=--1-----27r1 Jc g(z)

We conclude that inside C, g(z) = — wo has either two simple zeros or one doublezero. In particular, = w0 must have a solution zo inside C.


From Zo and m, n E Z, Proposition 15.3.1 gives the additional solution

+ (m+ = = wo,

where the second equality follows since is odd. We must show that there areno other solutions. Let D be the region enclosed by C (including the boundary).Translating D by elements of the period lattice 2 = {(m + n is even}covers the entire complex plane. In particular, —Zo + has a translate by 2 that liesin the interior of D, i.e., there are m, n E Z with m + n even such that

(15.31) (_lyi+m+lzo+ ((m+

lies inside the curve C. If (15.31) differs from Zo, then we have found all zeros ofg(z) = — w0 inside C. Since every other zero has a translate by .2' that lies insideC, it follows that all solutions = w0 have the desired form. Finally, if (15.31)coincides with Zo, then it is easy to see that

a,bEZ,a+bodd.

In Exercise 4 you will show that this implies that g'(zo) = = 0. By what weproved above, it follows that Zo is the only zero of g(z) inside C. As before, weconclude that the solutions have the desired form. •

Mathematical Notes

Two ideas implicit in this section require further comment.

Elliptic Functions. By Proposition 15.3.1, is a meromorphic function on Cwith periods (1 + i)w, (1 — i)w that are linearly independent over R. In general, anelliptic function is a meromorphic function on C with periods Wi, w2 that are linearlyindependent over R. While the basic ideas of elliptic functions go back to Abel andJacobi, these days most texts follow the approach of Weierstrass, who defined theWeierstrass p-function to be

1 / 1 1

2 2Z

m,nEZ (z — (nwi + mw2)) (nwi + mw2)

For example, if we let pi(z) denote the p-function with periods (1 + (1 —

(this is the notation of [15]), then one can show that

p1(z) ,(15.32) and

Furthermore, the relation1 —

translates into the relation

(15.33) pç2(z)

488 THE LEMNISCATE

In general, the p-function = satisfies

(15.34) —g3,

where and g3 are constants determined by the periods Wi, W2. There is also anaddition law for p(z + w). Introductions to elliptic functions can be found in [3, §101,

[9, Ch. 3], [12, Sec. 8.3], [14, Ch. 2], [16, Ch. 1], and [20, Ch. 9].

• Elliptic Curves. The primary geometric object of this chapter is the lemniscate,which is the curve defined by the equation (x2 +y2)2 = x2 —y2. However, the ellipticfunctions we've been studying lead to other curves of interest. For example, therelation

= 1 —p4(z)

shows that the map z '—* parametrizes the curvey2 = 1 —x4. Similarly,the relation (15.33) for the Weierstrass p-function (z) shows that z '—÷ (z), (z))parametrizes the curve

y2 4x3 +x,

and for a general gc-function, (15.34) shows that z '—* (z)) parametrizes

y2 = 4x3 g2x g3.

These are elliptic curves. They have an intrinsic addition law compatible with theaddition law of the p-function. Some of the most important theorems and conjecturesof modern number theory involve elliptic curves. Introductions to this wonderful areaof mathematics can be found in [8], [10], [14], [181, and [20].

Historical Notes

In the Historical Notes to Section 15.2, we saw that Abel's theory of ellipticfunctions began with the integral

I —c2t2)(l +e2t2)dt.

He denoted the inverse function by and then defined f(x) = and

F(x) = These functions are related via = f(x)F(x). Abel gaveaddition laws for f(x), F(x) and multiplication formulas for f(nx),F(nx) similar in spirit to Theorem 15.2.5. He also extended these functions tofunctions of z E C and determined their periods, zeros, and poles. Abel's paper[Abel, Vol. I, pp. 263—388] contains many wonderful formulas and is fun to read.

Jacobi developed a similar theory based on the integral (15.21). He definedfunctions sin amx, cos amx, and amx, later simplified to sn(x), cn(x), and dn(x).His version of the theory became very influential, though it was eventually supersededby the p-function introduced by Weierstrass in 1882. One nice result of Weierstrassis that every elliptic function with the same periods as is a rational function in

COMPLEX MULTIPLICATION 489

and So once the period lattice is fixed, only two elliptic functions areneeded in order to get all others.

Gauss anticipated most of the work of Abel and Jacobi on elliptic functions butnever published his results. As he wrote in 1828 [Gauss, Vol. X.l, p. 248],

I shall most likely not soon prepare my investigations on transcendental functionsthat I have had for many years—since 1798—because I have many other mattersthat must be cleared up. Herr Abel has now, I see, anticipated me and relievedme of the burden in regard to one third of these matters, particularly since hecarried out all these developments with great concision and elegance.

What led Gauss and Abel to work over the complex numbers? It appears that theywere inspired to define for z C in order to represent all roots of the n-divisionpolynomials of the lemniscate. The high degree of these polynomials suggests thatthe roots cannot be all real. In the next section, we will use the theory of complexmultiplication to describe the roots of the n-division polynomials.

More on the history of elliptic functions can be found in [3] and [7]. A classictreatment of the Jacobian elliptic functions appears in [21].


Exercise 1. Suppose that g(z) is an analytic function satisfying g(iz) = ig(z). Prove thatg'(iz) = g'(z).

Exercise 2. This exercise is concerned with the proof of Proposition 15.3.1.(a) Prove that + iy), as defined by (15.22), satisfies the Cauchy—Riemann equations.(b) Prove (15.23), (15.24), (15.25), and (15.26).

Exercise 3. Prove the formula for (z ± i) stated in the proof of Theorem 15.3.2.

Exercise 4. Prove that (z) vanishes at all points of form (m + in) m + n odd.

Exercise 5. A useful observation is that an identity for p proved over automatically becomesan identity over C.(a) Prove this carefully, using results from complex analysis such as [13, 6.1.1].(b) Explain why = 1 — holds for all z E C.

Exercise 6. By Theorem 15.3.3, = if and only if z = (_1)m+nZO + (m + in)tti.Following Abel, prove this using (15.13).

15.4 COMPLEX MULTIPLICATION

By Exercise 5 of Section 15.3, the multiplication formulas for x E extendto give formulas for z C. Over C we also have the formula (15.24) given by

(15.35) = i = VCT.

So besides multiplying by n e we can also multiply by i. Combining these withthe addition law gives formulas for ((n + im)z), where n + im E Z[i] is any Gaussianinteger. In other words, has complex multiplication by Z[i].

490 THE LEMNISCATE

Before developing the general theory, let us give an example to illustrate the powerof complex multiplication.

Example 15.4.1 In Exercise 1 you will use the addition law together with (15.35)and = to prove that

=

(z)LIZ)

—

These are simple examples of complex multiplication.To see the relevance of (15.36), square each side and apply = 1 —

This gives

+i)z)2iço2(z)

(15.37) (p

2i21 . \

l—(p4(z)

The surprise is that we've seen disguised versions of these formulas in the proof ofProposition 15.2.3. To explain why, let r0 = and a = (p(xo) as in the proof.Then set t = + and apply the first formula of (15.37) to obtain

2 = (1 i) (1 + i), the second formula of (15.37) implies that

—2i —2t2a2 = 2(xo) = (p2((l —i)(1

= 1 —(p4((l =

The above two equations are (15.15) and (15.16) from the proof of Proposition 15.2.3.Earlier, they seemed to appear out of nowhere, but now that we know complexmultiplication, they are no longer so mysterious. 4>

The proof of Proposition 15.2.3 used the duplication formula for (p(2x). Exam-ple 15.37 shows that factoring 2 in Z[iI enables us to factor the duplication formulainto equations that are simpler to understand. We will use similar factorizations inSection 15.5 when we prove Abel's theorem on the lemniscate.

The theory of complex multiplication gives formulas for (p(/3z), where z e C and= n + im is a Gaussian integer. In this section we will first review some

basic facts about Z[i] and then derive formulas for (p(13z), paying special attention tothe case when 1.3 is prime in Z{i].

A. The Gaussian Integers. The ring of Gaussian integers is defined by

Z[iI E Z}.


The units ofZ[i] formthegroupZ[i]* = {±l,±i} =I

e=O,l,2,3},andnonzeroGaussian integers a,fi are associate if a =for some Furthermore,Z[iI is a UFD with the following primes (up to associates):

• 2 = (1 +i)(l i), where 1 + i and 1 — iare associate primes in Z[i].• When p 3 mod 4 is prime in Z, p is also prime in Z[i].• When pm 1 mod4 is prime in Z, there are a,b E Z such that p =a2+b2 =

(a + bi) (a — bi), where a + bi and a — bi are nonassociate primes in Z[i].

Also, Z[iI is a PID, so that every ideal is of the form /3Z[i] for some /3 E Z[i]. All ofthese facts are proved in most books on abstract algebra. See, for example, [Herstein,Sec. 3.81.

Given a E Z [i], we say that /3 -y mod a if a divides /3 — -y in Z[i]. To understandthe quotient ring Z[i]/aZ[iI, recall that a = a + ib E Z[i] has norm

N(a) = al2 =a2+b2 E Z

such that N(a/3) = N(a)N(/3). Then we have the following result.

Lemma 15.4.2 Let a be a nonzero element of Then:(a) Z[iI/aZ[i] is a finite ring with N(a) elements.(b) If a is prime, then is the finite field

Z[iI/aZ[iI

Proof: You will prove this in Exercises 2 and 3. •

We say that a Gaussian integer a + bi E Z[iI is odd if a + b is odd and even if a + bis even. If a,/3 E Z[iJ, then

a/3 is odd a and /3 are odd,

(15.38) a+ /3 is even a,/3 are both even or both odd,

a is even 1 + i divides a

(see Exercise 4). Since 1 + i is prime in Z[iJ, the last line of (15.38) can be stated as

a is odd 1 + i and a are relatively prime.

B. Multiplication by Gaussian Integers. When n E 7Z, Theorem 15.2.5 ex-presses in terms of when n is odd and in terms of and when nis even. Here, we will generalize on the former case by giving formulas for interms of when /3 E is odd.

In one sense, the formulas are easy—the proof of Theorem 15.4.4 given belowshows that they are simple consequences of the addition law, the multiplicationformulas for from Theorem 15.2.5, and the identity = However,in order to prove Abel's theorem on the lemniscate, we need to understand the finestructure of these formulas.

Here is an example to illustrate the issues involved.

492 THE LEMNISCATE

Example 15.4.3 In Exercise 5 you will use the addition formula to derive the formula

—(—2 + i)ço8(z) — + 2 + i

+ 11Z1 —— + 1

The numerator and denominator have a common factor that can be canceled. InExercise 5 you will show that this leads to the simpler formula

(15.39) (2+z)z = —iço(z)(—1 +2i)ço4(z) +

We pulled out a factor of — ito ensure that the numerator is monic and the denominatorhas constant term 1. Note also the "reverse symmetry" of the coefficients of numeratorand denominator. This will be important below.

The following theorem generalizes the formula (15.39) for ((2 + i)z).

Theorem 15.4.4 Let /3 e Z[i] be odd. Then there exist relatively prime polynomialsin the polynomial ring Z{i][uI ande E {O, l,2,3} such that:

(a) For all z C, we have

p(z)

(b) /3 mod 2(1 +i).(c) and Q13(u) have degree d = (N(/3) — l)/4, where N(/3) is the norm of/3.(d) The roots of the 13-division polynomial uP13(u4 ) are the complex numbers (az)

for E Z[i] odd.(e) Pfi(u) is monic, = 1, and Qfi(u) where d isfmm part (c).

Before beginning the proof, let us explain what the theorem says aboutwhen /3 E Z[i] is odd. Let Z[i}[u] be the polynomials given in thetheorem. Parts (c) and (e) imply that Pfi(u),Q$(u) can be written in the form

==ud((l/u)d+aI(l/u)d_t+...+ad)

=l+alu+...+adud,

where d = (N(/3) l)/4 and a1,... ,ad E Z[i]. This is the "reverse symmetry"mentioned above. Then the complex multiplication formula for p(/3z) can be written

+ +ad1 + . . .

where /3 mod 2(1 + i) by part (b). Here is an example.


Example 15.4.5 Suppose that = 2 + i. Since d = (N(f3) — 1 )/4 = (5 — 1 )/4 = 1

and /3 —i mod 2(1 + i), the above formula reduces to

4

=

where aI E Z{iJ. Comparing this to (15.39), we see that a1 = —1 + 2i.

The following lemma will be useful in the proof of Theorem 15.4.4.

Lenuna 15.4.6 Let E Z[i] be odd. Then the set

has precisely N(/3) elements and consists of all complex numbers of the form

aEZ[i] odd.

Proof: First observe that if a E Z[i] is odd, then E since == 0, where the last equality is by Theorem 15.3.2. Going the other way,

suppose that = 0. Then Theorem 15.3.2 implies that

i3z=(a+ib)tv, a,bEZ.

Leta=a+ibE 7Z{il. so that = Ifais odd, then weare done. On the other hand, if a is even, then — a is odd. Using the identity

— z) = from (15.9), we obtain

This shows that the elements of have the desired form.To determine the size of Rfl, fix E where a E Z{i] is odd. We claim

that a is unique modulo /3Z[iI. To see why, suppose that

= E Z[iI odd.

By Theorem 15.3.3, there is a + ib E Z[i] such that

= +

This implies that

Since a, and /3 are odd, a + ib is even by (15.38). Thus = 1 and hence

a= a+(a+ib)/3,

so that a and a give the same element of Z [iJ //3Z[iI. Since every coset of Z[iI //3Z[iJcan be represented by an odd Gaussian integer (given any a, either a or a + /3 isodd), it follows that

= Z[ij//3Z[i]I =N(/3),where the last equality follows from Lemma 15.4.2. •

494 THE LEMNISCATE

We now turn to the proof of the theorem.

Proof of Theorem 15.4.4: We will prove the theorem in five steps.

Step 1: Existence of (u) and (u) for all /3. Given /3 e we claim that thereare polynomials E Z[i}[u] such that = 1 and

(15.40) = when /3 is odd,(z))

and

),(15.41) = (z), when /3 is even.(z))

We will prove the formulas (15.40) and (15.41) using the multiplication formulasfrom Theorem 15.2.5 together with the identities

ço(iz) =

— (1t)z1

— 1 —p4(z)(15.42) 2 (/3z) '(z)

l)z) = 1)z)+ 1

+ i)z) = i)z)+ 1

We already know the first and second lines, and the third and fourth lines follow fromthe first and (15.13) (see Exercise 6).

The formulas for ço(iz) and + i)z) from (15.42) satisfy (15.40) and (15.41).From here, repeated use of the third line of (15.42) shows that for all integers n > 0,

there are polynomials (u), (u) E Z[i] [u] that give the desired formula for((n + i)z). The argument is similar to what we did in the proof of Theorem 15.2.5.

In particular, when n is even, we get the recursion

= + (1 — u))

similar to (15.18). This makes it easy to show that = 1 for all n 0 even,and the argument that = 1 for n � 0 odd is similar.

Now fix an integer n � 0. We have formulas for + i)z) (just proved) and(by Theorem 15.2.5). Then repeated use of the fourth line of (15.42) shows

that for all m > 0, there are polynomials Pn+im(U),Qn+im(u) E Z[i][uI that give thedesired formula for ((n + im)z) and satisfy (0) = 1. See Exercise 7.

Hence we have formulas for p((n + im)z) for all integers n, m > 0. Then

=

(n + im)z) = —p((n + im)z),

ço(m — in)z) = i(n + im)z) = —iço((n + im)z)

make it easy to construct the desired (u), (u) E Z[i] [u] for all /3 E Z[i].


Step 2: Remove Common Factors. For the rest of the proof, we will assume that /3is odd. The polynomials constructed in Step 1 might have a commonfactor. Since Z[i] is a UFD, the same is true for Z[i] [uJ by Theorem A.5.6. Thus

Pfi(u) = and =

where E Z[i][u] and are relatively prime. Since(0) = 1, we can multiply (u) , (u), (u) by suitable units in 7L[i]t = {± 1, ±i}

so that = 1. Since /3 is odd, we have

C13=

Hence we may assume P13(u),Q13(u) are relatively prime in Z[iI[u] with Q13(0) = 1.

Step 3: The Constant In Exercise 8 you will show that (Z[iI/2(l + i)Z{i])* ={+[l],±[i]}, so that /3 E {O, l,2,3}. MultiplyingP13(u)by a suitable unit of Z[iJ*, we obtain the equation

(15.44) = i(z))

In Exercise 8 you will also show that

(15.45)

This will be useful later in the proof. It follows that the relatively prime polynomialsP13(u),Q13(u) E Z{i]{u] satisfy parts (a) and (b) of the theorem together with thecondition (0) = 1 from part (e). Steps 4 and 5 will show that (u), (u) satisfythe remaining conditions of the theorem.

Step 4: The Roots of uP13(u4). We will use Lemma 15.4.6 to determine the roots ofthe /3-division polynomial A13(u) = uP13(u4). Also let B13(u) = Q13(u4). Since /3 isodd, (15.44) implies that

(15.46)=

B13

In Exercise 9 you will show that A13 (u) and B13(u) have no common roots in C, sinceQ13(0) = 1 and P13(u),Q13(u) are relatively prime in 7L{i][u]. Using this and (15.46),it follows that

A13(ço(z)) = 0 = 0.

Since any root of A13(u) is of the form for some z E C by Theorem 15.3.3, weconclude that the roots of A13 (u) form the set

496 THE LEMNISCATE

from Lemma 15.4.6. Then the lemma implies that the roots can be written in theform described in part (d) of the theorem.

We next show that all roots of A13 (u) have multiplicity 1. Assume that =is a multiple root. Then A13 (uo) = A'13 (uo) 0, and hence B13 (UO) 0 by the previousparagraph. Differentiating (15.46) with respect to z and substituting z = Zo gives

= —B'13(uo)A13(uo)çc'(zo)=0

B13(uo)

(note that is defined because is). Since 0, has a multiplezero at /3Z0. This is impossible by Theorem 15.3.2. Hence A13(u) has simple roots.

We conclude that the degree of A13 (u) is the number of elements in R13. ByLemma 15.4.6, it follows that A13(u) = uP13(u4) has degreeN(/3), so that P13(u) hasdegree d = (N(/3) — l)/4. This proves part (c) for P13(u).

Step 5: Relate P13 and Once we show that

(15.47) Q13(u) = u°'P13(l/u), d = (N(j3) — 1)/4,

it will follow immediately that Q13(u) has degree d and P13(u) is monic (since Q13(u)has constant term 1). Thus we need only prove (15.47) to complete the proof.

The identity (15.29) implies that

=—i=i3.

Setting w = z + (1 + i) we obtain

(15.48) = i3.

In Exercise 10 you will use (15.48) and mod 2(1 + i) to show that

(15.49) =

Then

(15 50)= = =

where the first equality uses (15.48) and (15.49), the second uses (15.44) with win place of z, and the third follows by raising (15.48) to the fourth power to obtain

= Comparing (15.50) with (15.44), we conclude that

Q13(l/u4) — P13(u4)

P13(l/u4) — Q13(u4)

as rational functions in u with coefficients in Q(i). Thus

Q13(1/u) — P13(u)

P13(l/u) —


Recall from Step 4 that d = (N(8) — l)/4. In Exercise 11you will show that the above equation implies that

(15.51) =

A E Q (i). However, if we evaluate (15.44) at z = and

use (15.45) and = 1, then we obtain

—

( l)

Thus = 0. Then substituting u = 1 into (15.51) implies that A = 1,sothat = This completes the proof. •

Here are two examples of Theorem 15.4.4 from earlier in the chapter.

Example 15.4.7 When = 3, equation (15.17) gives

8 —3

In the notation of Theorem 15.4.4, this means

P3(u) = u2 + 6u — 3 and (u) = u2P3 (1/u) = 1 + 6u — 3u2.

These polynomials have degree (N(3) — l)/4 = 2. Note also that = —1, since3m—i mod 2(l+i).

When /3 = 5, equation (15.19) gives

= where

P5(u)=u6+50u5— 125u4+300u3— 105u2—62u+5,

Qs(u) = 1 +50u— 125u2-I-300u3— 105u4—62u5-I-5u6.

These polynomials have degree (N(S) — 1)/4 = 6 and satisfy Q5(u) = u6P6(1/u).Furthermore, we have = 1, since 5 ml mod 2(1+i).

In general, one can show that when n > 0 is in Z, the polynomials and (u)from Theorem 15.4.4 lie in

C. Multiplication by Gaussian Primes. When /3 is an odd prime in Z[i],Theorem 15.4.4 has the following important refinement due to Eisenstein. Thisresult will play a crucial role in the proof of Abel's theorem.

498 THE LEMNISCATE

Theorem 15.4.8 Let E Z{i] be an odd prime, and let

= ud+aiud_I + +ad E Z[i][uJ, d = (N(fi)— l)/4,

be the corresponding polynomialfrom Theorem 15.4.4. Then:(a) a1,... ,ad are divisible by 1.3 and ad = where 13 mod 2(1 + i).(b) is irreducible over Q(i).

Proof: Our proof will follow [14] and is based on Eisenstein's original proof from1850 [Eisenstein, pp. 556—619]. We first observe that the Schönemann—Eisensteincriterion, stated in Theorem 4.2.3 for polynomials in Z[u] and primes in Z, also appliesto polynomials in Z [i] [ul and primes in Z[iI. You will prove this in Exercise 12. Thenpart (a) implies that

Pfl(u4) = + +ad e Z[i][uI

satisfies the criterion for the Gaussian prime 13 and hence is irreducible over Q(i).Thus part (b) of the theorem follows from part (a).

Proving part (a) will be harder. Since 13 is odd, Theorem 15.4.4 implies that

1552 —

_____________________

( . ) p(z)(z)

where the coefficients a1,... ,ad E Z[i] depend on 13. To prove part (a), we willanalyze the relation between ai,.. . ,ad and 13 by expanding each side of (15.52) as apower series in z.

Several power series will appear in the proof. The first comes from

which we write as

to emphasize the dependence on 13. This rational function is analytic at u = 0 (thedenominator doesn't vanish at 0) and hence has a power series expansion

d (Q\ d—1eu k

(15.53) 1-f-al(/3)u+•••+ad(/3)u"k=O

= bo(/3) + b1 (/3)u + b2(/3)u2 +....

In Exercise 13 you will prove that bk(/3) E Z[i] for all k. This follows because theconstant term in the denominator is 1 and the other coefficients lie in Z[il. Using thepower series (15.53), the multiplication formula (15.52) can be written

(15= + ...)= + b1 + +....


The second power series comes from Since is analytic at z 0, itcan be expanded in a power series in z. In Exercise 14 you will use ço(iz) =p'(O) = 1, and = 1 — to prove that the power series has the form

(15.55) c1eQ.1=1

You will also show that = and c2 = th• Then replacing z with /3z in (15.55)gives the third power series

(15.56) = = /3z + + c2/39Z9 +

From here, the proof proceeds in three steps. Here is an overview of what we willdo in each step:

• Step 1. Derive a formula for bk([3) in terms of /3 that holds for all odd /3 E Z[i].This will follow by substituting the series for and into (15.54).

• Step 2. Prove that /3 divides bo(/3),... ,bd_I(/3) when /3 is an odd prime. Thiswill be done by analyzing the formula of Step I using a clever idea of Eisenstein.

• Step 3. Relate aI , ad(fi) to bo(/3),. . . , bd_J (/3) and conclude that /3 dividesal ,ad(fi). This will follow easily from (15.53).

We now turn to the first step.

Step 1: Express bk(/3) in terms of /3. If we substitute (15.55) and (15.56) into theidentity (15.54), then we obtain

/3z+ci/35z5 +C2/39Z9+" = bo(/3)(z+ciz5 +

(15.57)

b2(/3)(z+ciz5 +c2Z9 + .. .)9

When we expand the right-hand side of (15.57), a given power of z appears onlyfinitely often, since all terms of

(z+cIz5+c2Z9+...)41+I =z4J+l(l+cIz4+c2zS+...)4i+l

have degree � 4] + 1 in z. In Exercise 15 you will show that up to degree 9 in z, theright-hand side of (15.57) begins with

(15.58) bo(/3)z+ (bo(/3)cj +bi(/3))z5 + (boCB)c2 +5b1(/3)ci +b2(/3))z9 +

Since this equals /3z + c1 /35z5 + c2/39z9 + ..., comparing coefficients gives

ci/35 = bo(/3)ci +bi(/3),

C2/39 bo(/3)c2+5b1(/3)ci +b2(fi),

500 THE LEMNISCATE

and then solving for bo(8),b1 (/3),b2(/3) yields

bo($) =8,

b2(/3) = /3(c2fi8

These equations hold for all odd /3 Z[i]. We will see below that bo(/3) = /3 is veryimportant.

In general, one can prove (see Exericse 16) that for any k, there is a polynomialSk(U) E Q[u] of degree 4k such that

(15.59) bk(/3)=/3Sk(/3), /3EZ[i]odd.

This follows because the c1 all lie in Q. The crucial thing here is that the samepolynomial Sk(U) works for all odd /3. For example, since C! = the aboveequations imply that

b1(/3) =/3S1(/3), Si(u) =

Step 2: Prove that /3 divides bo(/3),...,bd_I(/3) when /3 is an odd prime. Theequation (15.59) seems to imply that bk(/3) is a multiple of /3 for all k � 0. Theproblem is that Sk (u) E Q [u] need not have integer coefficients, as shown by S1 (u).Hence we need to study the denominators of the coefficients of Sk(u).

Let Sk be the least common multiple of these denominators. Then

Sk(u) =

where sk E Z\{0}, Tk(u) e Ziul, and ±1 are the only integers dividing sk and allcoefficients of Tk(U). Eisenstein observed that if a E Z[i] is an odd prime, then

(15.60) aIsk N(a)<4k+1.

To prove this, first observe that (15.59) implies that

(15.61) Skbk(/3)/3Tk(/'3), /3EZ[i]odd.

We noted above that bk(/3) always lies in Z[i]. This means that if an odd Gaussianprime a divides sk, then a also divides /3Tk(/3). It follows that

(15.62) /3Tk(/3)mOmoda, /3EZ[i]odd.

Then consider the following:

• Since a is odd, the proof of Lemma 15.4.6 shows that elements of Z[i]/aZ[iI areof the form [/3], /3 odd. Thus (15.62) implies that the reduction of uTk (u) modulo ais a polynomial with at least IZ[i]/aZ[i] roots.

• Since a divides sk, the definition of Sk shows that the reduction of uTk(u) moduloa is a nonzero polynomial of degree at most 4k + 1. Hence the reduction has atmost 4k+ 1 roots since Z[i]/aZ[i] is field by Lemma 15.4.2.


These bullets imply that <4k+ 1. However, IZ[i]/aZ[i]I byLemma 15.4.2. Thus <4k+ 1, and(15.60) follows.

Now fix an odd Gaussian prime Then (15.60), applied to /3, tells us that

Note thatN(/3) > 4k+ 1 if and only if k <d = (N(/3)— 1)/4. It follows that /3tskfork = 0,...,d— 1. Since /3 is prime, (15.61) implies that divides bk(/3) fork =0,...,d— 1. This is what we needed to prove.

Step 3: Relate a1 ,ad(/3) to bo(/3),... ,bd_I This is easy, for if we write(15.54) in the form

+ . . . +ad(13))

=

and multiply out the right-hand side, then comparing coefficients of the powers of ugives the equations

= bo(/3),

zad_1(/3) = ai(13)bo(/3) +b1(/3),

iead2(/3) = a2(/3)bo(/3) +ai(/3)bi (/3)

=ad_l(/3)bo(/3)+ad_2(/3)bl(/3)+..•+bd_I(/3).

The a1(/3) lie in Z[i], and bo(/3),... ,bd_I(/3) are divisible by /3 by Step 2. It follows

that in the above equations, the right-hand side is always divisible by /3. This showsthat /3 divides a! (/3),... ,ad(/3), since is a unit. Furthermore, we proved earlier thatbo(j3) /3, so that the first equation implies that ad(/3) = This completes theproof of part (a).

Mathematical Notes

Here are some further comments about complex multiplication.

• Complex Multiplication. In our discussion of elliptic functions in Section 15.3,we mentioned that the Weierstrass p-function p(z; WI,W2) for periods Wi, w2 hasan addition law. It follows easily that it also satisfies multiplication formulas forn E Z that generalize Theorem 15.2.5. However, the ga-function rarely has complexmultiplication. More precisely, WI, W2) has complex multiplication by some

/3 E C \ Z if and only if W2/WI is a root of a quadratic polynomial with integer

coefficients. This means that W2/WI lies in an imaginary quadratic field, whichis a field of the form for some m > 0 in 7L. For example, the periods

= (1 — i)w,w2 = (1 + of Abel's function have ratio

— (1 — i)w—

502 THE LEMNISCATE

which is a root of x2 + 1 = 0. So the associated imaginary quadratic field is Q(i).In general, elliptic functions with complex multiplication have a deep relation toimaginary quadratic fields. This is discussed in books such as [3], [11], [17], and[201. This is also related to class field theory, which will be discussed in theMathematical Notes to Section 15.5.

Historical Notes

In addition to the general theory of elliptic functions, Abel also considered thelemniscatic function we've been studying. Let m + pi E Z[i] be odd, and setx = Then Abel states complex multiplication by m + as "one has

ço(m+pi)8=x.T,

where T is a rational function of x4" [Abel, Vol. I, p. 354]. As an example, he writesthe formula for complex multiplication by 2+ i as

2—2x8+i(l—6x4+x8) 1—2i—x"

l—2x4+5x8

This is remarkably close to what we did in Example 15.4.5.Eisenstein also has an important role to play in this story since he was the first

to prove Theorems 15.4.4 and 15.4.8. Here is an extract of a letter that he wrote toGauss in 1847 [Eisenstein, p. 845]:

If m = a + bi is an odd complex integer, p is its norm and

Uy= =

is the algebraic integral of the equation

then I have further shown that for a two-term complex prime number m thecoefficients of the numerator, except for the last which is a complex unit, and thecoefficients of the denominator, except for the first which = 1, are all divisibleby m. I conjecture that the theorem is also correct, when m is a one-term primenumber

Here, a "two-term" odd complex prime is m = a + bi such that p = a2 + b2 is primein Z with p 1 mod 4, and a "one-term" complex prime is a prime in Z such thatp 3 mod 4. In this letter, Eisenstein could prove part (a) of Theorem 15.4.8 onlyin the "two-term" case, though later he obtained a general proof. Also, if we thinkof as the inverse function of the elliptic integral 1(1 — t4) —1/2 dt, then it shouldbe clear that the displayed equation in Eisenstein's letter refers to the multiplicationformula for ço(mz) in terms of

The clearest statement of Eisenstein's irreducibility criterion appears in a paperhe wrote in 1850 [Eisenstein, p. 542], where we find the following theorem:


If in a polynomial F(x) of x of arbitrary degree whose coefficient of thehighest term is = 1, and all following coefficients are (real, complex) integers,in which a certain (real resp. complex) prime number m appears, if in additionthe last coefficient is = where e represents a number not divisible by m; thenit is impossible to bring F(x) into the form

+

where and v 1, jt+ = the degree of F(x), and all a and bare (real resp.complex) integers; and the equation F(x) = 0 is accordingly irreducible.

The reason Eisenstein states the theorem for both Z and Z [i] is that he probablydiscovered it first over Z[i] in his study of complex multiplication on the lemniscateand then realized that it also applies over 7Z.

When we discussed the Schönemann—Eisenstein criterion in the Historical Notesto Section 4.2, it was easy to explain what led Schönemann to the criterion, namely,does reducibility modulo p imply reducibility modulo p2. But as we've seen inthis chapter, it was a much richer mathematical context that led Eisenstein to hisdiscovery. See [4] for more on Eisenstein and his criterion.



Exercise 2. Let a E Z[iI be nonzero. The goal of this exercise is to prove part (a) ofLemma 15.4.2, which asserts that Z[i]/a7Z[i] I = N(a). The idea is to forget multiplicationand think of Z[iJ and 7L[i}/aZ[iI as groups under addition. Let m be the greatest commondivisor of the real and imaginary parts of a, so that a = m(a + bi), where gcd(a, b) = 1. Thenpick c,dEZsuch thatad—bc= 1.(a) Show that the map Z[i] —+ Z Z defined by

p/d, —b)+v(—c,a) = ('id — va)

is a group isomorphism under addition.(b) Show that the map of part (a) takes a and ia to (in, 0) and (—m(ac + bd) ,m(a2 + b2)),

respectively. Then use this to show that the map takes aZ[i] C Z[i] to the subgroup

C

(c) Use part (b) to conclude that IZ[i]/aZ[ijI = N(a).

Exercise 3. Prove part (b) of Lemma 15.4.2.


Exercise 5. Derive the two formulas for + i)z) stated in Example 15.4.3.

Exercise 6. Prove the third and fourth lines of (15.42).

Exercise 7. Supply the details omitted in the proof of Step 1 of Theorem 15.4.4.

Exercise 8. Consider the finite ring ZL[i] /2(1 + i)Z[i], and let E Z[i] be odd.(a) Prove that (7L[i] /2(1 + i)Z[iI)

* = {± [1], + [ii }. and then explain why this implies that{0,1,2,3}.

(b) Prove that

504 THE LEMNISCATE

Exercise 9. Suppose that we have relatively prime polynomials (u), (u) E Z[i] [u] such

that = 1. Prove that and have no common roots inC.

Exercise 10. Let w = z+(l +i) to show that

=

Exercise!!. Let F be a field, and letA(u) , B(u) E F [u] be nonzero relatively prime polynomialssuch that

B(l/u) A(u)A(1/u) B(u)

in F(u). Let d = deg(A). Prove that d = deg(B) and that there is a constant A E F* such thatu"A(l/u) = AB(u).

Exercise 12. Let E Z[i] be prime, and let f = aou" + +ad E Z[i][u]. Provethe Schönemann—Eisenstein criterion over which states that if fi tao, i3lai,. . . , lad, and

tad, then f is irreducible over Q(i).

Exercise 13. Prove that the coefficients bk(13) defined in (15.54) lie in Z[iJ.

Exercise 14. The function is analytic at z = 0 and hence has a power series expansion.(a) In Exercise 3 of Section 15.2, you used (z) = 1

— (z) to show that (z) = (z).Use these two identities to prove by induction that for every n 1, there is a polynomial

E Z[u] such that equals if n is even and jf n isodd.

(b) Use part (a) to prove that the coefficients of the power series expansion of at z = 0

lie in Q.(c) Use part (b) and p(iz) = iço(z) to show that = EQ.(d) Show that = 1, Cl = andc2 = th•Exercise 15. Show carefully that (15.58) follows from (15.57).

Exercise 16. Prove that for each integer k � 0 there exists a polynomial Sk(u) E Q[u] of degree4k such that (15.59) holds for all odd E Z[i].

Exercise 17. Let n E 7Z be an odd integer. Prove that n mod 2(1 + i). Thisshows that when n is an odd integer, we have = in the formula for givenin Theorem 15.4.4.

15.5 ABEL'S THEOREM

In this final section of the book, we will prove Abel's theorem about straightedge-and-compass constructions on the lemniscate. The tools used will include Galoistheory and the theory of complex multiplication developed in Section 15.4.

A. The Lemniscatic Galois Group. Let n be an odd positive integer and consider

L=

We will see that the Galois group of Q(i) C L involves the group

ABELS THEOREM 505

of units in Z[i] /nZ[i]. Since Z[i] is a PID, a coset [a] lies in (Z[i] /nZ[i]) *if and onlyif a is relatively prime to n in Z[i] (see Exercise 1).

Theorem 15.5.1 Q(i) C L is a Galois extension and there is a one-to-one grouphomomorphism

Gal(L/Q(i)) —÷ (Z[i]/nZ[i])t.

In particular; Gal (L/Q(i)) is Abelian.

Proof: Let = be the n-division polynomial defined in part (d) ofTheorem 15.4.4. The theorem tells us that the roots of (u) are given by

(15.63) aEZ[i] odd

and the proof of Lemma 15.4.6 shows that for each root, the associated a E 7Z[i] isunique modulo nZ[iJ.

Since each a in (15.63) is odd, the complex multiplication formula forgiven by Theorem 15.4.4 shows that is a rational function in withcoefficients in Q(i). It follows that splits completely in L =Since one of the roots is ( it follows immediately that L is the splitting field of

over Q(i). Thus Q(i) C L is a Galois extension.Now take a E Gal(L/Q(i)). Then a(p(?)) is a root of and hence is one

of the numbers (15.63). Thus there is a e Z[i] odd such that

(15.64)

a is unique modulo n7L[iJ.In Exercise 2 you will use Theorem 15.4.4 to show that if/3 E Z[i] is odd, then

(15.65)

a is relatively prime ton. Let m be the order of a in Gal (L/Q(i)),so that atm is the identity. Then repeatedly applying (15.65) yields

By uniqueness, we conclude that

1 am mod n.

Hence a is relatively prime to n in Z[i], so that a '—+ [a] gives a well-defined map

(15.66) Gal (L/Q(i)) —+ (Z[i]/n7L[i])

If a andr map to a and /3, respectively, then (15.65) easily implies thatar maps to aj3, which shows that the map is a group homomorphism.

Furthermore, if [a] = [/3] in (Z[i]/nZ[i])*, then

a—/3+(a-Fib)n

506 THE LEMNISCATE

where a + ibis even because and n are odd. Then Proposition 15.3.1 implies that

from which we conclude that the map is one-to-one since ( generates L overQ(i). This completes the proof. U

Since Abelian groups are solvable, one corollary of Theorem 15.5.1 and Chapter 8is that the coordinates of the n-division points of the lemniscate are expressible byradicals over Q. (You will prove this assertion carefully in Exercise 3.)

The homomorphism (15.66) constructed in Theorem 15.5.1 is the lemniscaticanalog of the homomorphism

(Z/nZ)"

studied in Chapter 9. We will say more about this analogy in the Mathematical Notesat the end of the section.

B. Straightedge-and-Compass Constructions. We now have the tools neededto prove Abel's theorem on the lemniscate.

Theorem 15.5.2 Let n be a positive Then the following are equivalent:(a) The n-division points of the lemniscate can be constructed using straightedge

and compass.(b) ( is constructible.(c) n is an integer of the form

where s 0 is an integer and p1, . . . ,Pr are r 0 distinct Fermat primes.

Proof: The implication (a) (b) is easy, since is the polar distance of ann-division point. The converse (b) (a) follows from part (b) of Corollary 15.2.7.

The proof of (c) (b) will be a nice application of Theorem 15.5.1 together withsome results of Section 15.2. We first observe that by part (c) of Corollary 15.2.7,

is constructible provided that

are constructible.

Since ( is constructible by Proposition 15.2.3, we need only show that p (is constructible when p is a Fermat prime.

By part (a) of Corollary 15.2.7, is constructible whenever is. Forthe latter, Theorem 15.5.1 gives the Galois extension Q(i) CL = with

Gal (L/Q(i)) a subgroup of (Z[i]/pZ[i])*.

In Exercise 4 you will use the methods of Chapter 10 to prove that if

(15.67) (z[i]/pz[ifl* = a power of 2,

ABEUS THEOREM 507

then is constructible.

We will show that (15.67) holds whenever p = 22 + 1 is a Fermat prime. Thecase p = 3 is easy (see Exercise 5). If p > 3, then m> 1, so that

p = +1 i)

where fi,fi are nonassociate primes in 7Z[ij of norm p. In this case, Exercise 6 andLemma 15.4.2 give isomorphisms

Z[i]/pZ[i] = x IF,, x IF,,.

Thus= 1F x IF,fl = (p— 1)2

This proves (15.67) for all Fermat primes and completes the proof of (c) (b).It remains to prove (b) (c). This is where we will use the irreducibility result

proved in Theorem 15.4.8. Let n be an integer such that ( is constructible. Wemay assume that n> 1 since the theorem is trivially true when n = 1. Furthermore,the doubling formula (15.14) implies that we may assume that n is odd (be sure youcan explain why), and Proposition 15.2.3 shows that is constructible.

Let p be a prime dividing n. Then p is odd because n is. Let be a complex primesuchthatp=,8 3 mod4andp 1 mod4. E Z[i] isodd(since n and are), so that is an odd multiple of This makes it easy to showthat

E

(see Exercise 7). It follows that is constructible, since i and are. ByCorollary 10.1.8 from Chapter 10, the minimal polynomial of over Q hasdegree equal to a power of 2. Then the Tower Theorem shows that the minimalpolynomial of over Q(i) also has degree equal to a power of 2.

Theorem 15.4.4 implies that is a root of It is easy to see that0 (see Exercise 8), so that is a root of Since is an odd

prime, has degree — 1 by Theorem 15.4.4 and is irreducible over Q(i)by Theorem 15.4.8. This proves that the minimal polynomial of over Q(i) hasdegree — 1.

When p = for p 3 mod 4, we have N($) — 1 = p2 — 1 = (p + 1) (p — 1). Oneeasily sees that this is a power of 2 if and only if p = 3 (see Exercise 9). On the otherhand, when p = fi/3 for p 1 mod 4, we have — 1 = p — 1, which is a powerof 2 if and only if p is a Fermat prime.

Thus the only primes dividing n are Fermat primes. To complete the proof of thetheorem, we need to show that p2 In cannot occur. So assume that p2 In, where p isprime. Then there is an odd complex prime /3 such that /32 In. By Exercise 7,

which implies as above that u0 is constructible. Hence the degree of its minimalpolynomial over Q(i) is a power of 2. We will prove that the minimal polynomialhas degree N(/3) (N(/3) — 1). This is not a power of 2 since N(/3) = p or p2.

508 THE LEMNISCATE

Since /3 is odd, Theorem 15.4.4 implies that

n I 4—

——1 —l UOQ(4).

Since is a root of this formula for gives the equation

If we write = U" + +ad, d = (N(/3) — 1)/4, then clearing denom-inators in the above equation shows that is a root of

P(u) + (u4)4 + . . . +

This has coefficients in Z[i] and degree 4d(4d+ 1) = N(/3)(N(13) — 1), sinceE Z[iI[u] have degree d. Furthermore, Theorem 15.4.8 implies that

(15.68)

Thus P13(u) ud mod /3. Using this and (15.68), we see that

P(u) mod /3,

since 4d(4d+ 1) = N(/3)(N(/3) — 1). Furthermore, = 1 by Theorem 15.4.4,so that the constant term of P(u) is

Theorem 15.4.8 shows that ad is not divisible by 132, so that by the Schönemann—Eisenstein criterion over Q(i) (proved in Exercise 12 of Section 15.4), P(u) is ir-reducible over Q(i). Thus the minimal polynomial of uo over Q(i) has degreeN(/3)(N(/3) 1). The proof is now complete. •

Mathematical Notes

Here are comments about some ideas related to this section.

• The Lemniscatic Galois Extension. Let n e Z be odd and positive. The field

L=

played an important role in our treatment of the lemniscate. This field has a nicerelation to the elliptic curve y2 = 4x3 + x discussed in the Mathematical Notes toSection 15.3. To explain this, first note the surprising fact that e L. You willprove this in Exercise 10. This means that

(15.69)

ABELS THEOREM 509

Then, using the formulas (15.32) and (15.33), one can show that L is the extensionof Q(i) generated by the x- and y-coordinates of the (1 + i)n-torsion points on theelliptic curve y2 = 4x3 + x.

In general, extensions generated by torsion points of elliptic curves are an impor-tant topic in number theory. See [18] or [20] for a nice introduction.

• Abelian Extensions of Q(i). In Theorem 15.5.1, we constructed a one-to-onegroup homomorphism

Gal (L/Q(i)) —+ (Z[iI/nZ[i])*

when n is odd and positive. Thus Q(i) C L is an Abelian extension. A remarkablefact is that as n ranges over all positive integers, the fields L defined in (15.69) containall Abelian extensions of Q(i), in the sense that if Q(i) c K is a Galois extensionwith Abelian Galois group, then there is an integer n> 0 such that

Q(i) c K c L=

The proof of this result uses class field theory and complex multiplication. See, forexample, [17, Ch. II, Example 5.8].

• Class Field Theory. A number field K is a finite extension of Q. The main goalof class field theory is to describe all Abelian extensions of K. For example, whenK = Q, the Kronecker—Weber Theorem from the Historical Notes to Section 6.5states that every Abelian extension of Q is a subfield of the cyclotomic extension

for some n. Similarly, we noted above that every Abelian extension of Q(i) isa subfield of the lemniscatic extension (15.69) for some n.

The general version of class field theory describes Abelian extensions of a numberfield K, though the description uses the language of algebraic number theory and isnot as explicit as for K = Q or Q(i). See [3, §8], [11, Sec. 8.4], or [17, §11.3] for abrief review of class field theory. In the special case of an imaginary quadratic fieldK, the theory of complex multiplication uses certain elliptic curves to give an explicitdescription of the Abelian extensions of K and their Galois groups. This is described

10] and [17, Ch. II].For example, the theory of ray class fields implies that if n is odd, then

Q(i) CL'

is a Galois extension with Galois group

(15.70) Gal (L'/Q(i)) (Z{iJ/(i +i)nZ[i])*/{±[1],+[i]}.

Using (15.70), one can get a shorter proof of Theorem. 15.5.2 that doesn't requirethe hard work of Theorems 15.4.4 and 15.4.8. This is closely related to the eleganttreatment of Abel's theorem given in [15].

The theory of elliptic curves is an important and beautiful area of number the-ory. There are also many unsolved problems of great interest. But only certain

510 THE LEMNISCATE

elliptic curves—those with complex multiplication—have the special link to Abelianextensions of quadratic number fields.

• Origami and the Lemniscate. Besides straightedge-and-compass constructionson the lemniscate, one can also use origami from Section 10.3 to divide the lemiscateinto arcs of equal length. As explained in [5], the answer is almost the same as forthe circle. The proof uses the class field theory developed in [15].

Historical Notes

The story of this section begins with Article 335 of Disquisitiones [6], whereGauss introduces his theory of geometric constructions and cyclotomic fields. Hethen goes on to say

The principles of the theory that we are going to explain actually extend muchfarther than we will indicate. For they can be applied not only to circular functionsbut just as well to other transcendental functions, e.g. to those that depend onthe integral f [1 / 1 — x4)1 dx and also to various types of congruences. Since,however, we are preparing a large work on those transcendental functionswe have decided to consider only circular functions here.

Gauss's "large work" never appeared, though the reference to the lemniscate wouldhave been unmistakable to any nineteenth-century reader.

Abel was clearly intrigued by Gauss's remark. He read Disquisitiones carefullyand understood Gauss's method for solving cyclotomic equations by radicals. Healso defined a version of the function for the integral

I —c2t2)(1 +e2t2)dt

and gave formulas for multiplication by n. The resulting n-division polynomials leadto certain algebraic equations, and one of Abel's goals in Recherches sur lesfonctionselliptiques [Abel, Vol. I, pp. 263—388] is to determine whether these equations aresolvable by radicals. Abel notes on page 352 that the n-division polynomial,

taken in its full generality, is probably not solvable algebraically for arbitraryvalues of e and c, but nevertheless, there are particular cases when one can solveit completely

For Abel, the case of greatest interest was the lemniscate given by e = c = 1, thoughhe also knew that e = and e = c(2 ± give polynomials that are solvableby radicals. From the modem point of view, these "particular cases" correspond toelliptic curves with complex multiplication.

As an example of Abel's methods, let be the lemniscatic function and fix aprime p 1 mod 4. Write p = 4ii + 1 = a2 + a, /3 E 7L. Then, on pages 357 and358, Abel asserts that

one has an equationR=0

of degree = 2u, whose roots are

ABELS THEOREM 511

where for brevity one supposes ö =Given this, one can easily solve the equation R = 0, by aid of the method of

M. Gauss.Letting E be a primitive root of a2 + j32, I say that one can express the roots

as follows:

Here, w is what we call and is an integer whose congruence class modulop = a2 + generates the multiplicative group F; (z[iI/(a + /3i)Z[i]) ".

This quotation shows Abel using a + /3i to study the p-division points on thelemniscate. Furthermore, the roots listed above have an important structure. We mayassume that r is odd, so that the multiplication formula for easily implies that

=

for some rational function 0(u) with coefficients in Z. It follows that if we let= then the roots of R = 0 can be written

ill \1121 \n31 \ n2ui \

where the exponents refer to composition, i.e., 02(xo) = 0(0(xo)), etc. Compare this

with Abel's 1829 paper Mémoire sur une classe particulière d'équations résolublesalgébriquement, where he says that radical solutions exist

if all of the roots of an equation can be expressed by

x,Ox,02x,03x,...O"1x, where

Ox being a rational function of x, and Ox, 02x,... the functions of the same formas Ox, taken two times, three times, etc.

(See [Abel, Vol. I, pp. 478—4791.) Abel proves that any equation whose roots satisfythis condition is solvable by radicals. These quotations show that Abel's conditionarises naturally from his work on the lemniscate.

We saw in Section 6.5 that Abel considered a more general class of equationsin his "classe particulière" paper. Rather than assume as above that all of the rootsare generated by iterating a single rational function, suppose instead that f(x) = 0,

f F [x], has a root x0 with that property that any other root is of the form 0, (xo) forsome rational function 0, E F(u). If we further assume that

for all i and j, then Theorem 6.5.3 implies that the Galois group of the splitting fieldoff over F is Abelian and hence solvable by radicals over F by Chapter 8.

The Historical Notes to Section 6.5 describe how Abel's equations led to themodern Abelian group via the nineteenth century Abelian equation. But in Chapter 6,we didn't know what led Abel to these particular equations. Now we do—it was hiswork on the lemniscate! Thus the term "Abelian group," known to every beginningalgebra student, has an unexpectedly rich history.

512 THE LEMNISCATE

Kronecker was the first to realize the full power of the equations described byAbel. In the 1853 paper where he introduced the term "Abelian equation," Kroneckerconjectured that all Abelian extensions of Q are contained in cyclotomic extensions(this is the Kronecker—Weber Theorem from Section 6.5), and he also asserts thefollowing [Kronecker, Vol. IV, p. 11]:

There also exists a close relation between the roots of Abelian equations whosecoefficients are complex integers of the form a + and the roots of equationsarising from the division of the lemniscate

Kronecker speculated that similar results might hold over imaginary quadratic fields.He called this "mein liebsten Jugendtraum" ("the dearest dream of my youth") in aletter written to Dedekind in 1880 [Kronecker, Vol. V. p. 455]. The first completeproofs of the theorems of class field theory and complex multiplication were givenby Tagaki and Fueter in the 1 920s. A nice discussion of Gauss, Abel, Eisenstein, andKronecker appears in [19, Ch. 3 and 4]. See also [12, Sec. 8.6].


Exercise 1. Let /3 E Z[iI be nonzero. Then a E Z[iI gives {aJ E Z[iI//3Z[iI. Prove that[a] E (Z[i]//3Z[i])* if and only if a is relatively prime to

Exercise 2. As in the proof of Theorem 15.5.1, let Uo = and assume that a EGal (L/Q(i)) satisfies a(uo) = where a E Z[i] is odd. Use the multiplication formulafor /3 E Z[i] odd to prove (15.65).

Exercise 3. Use Theorem 15.5.1 and Chapter 8 to prove that the x- and y-coordinates of then-division points of the lemniscate are expressible by radicals over Q.

Exercise 4. Give a careful proof that (15.67) implies that is constructible.

Exercise 5. Prove that (Z[j]/3Z[j])*I = 8.

Exercise 6. Let a, j3 E 7Z[ij be nonzero and relatively prime. Prove the Chinese RemainderTheorem for Z[i], which asserts that there is a ring isomorphism

ZZ[iJ/a137Z[i] ?Z[i]/a7L[i] x 7L[i]/$Z[iI.

Exercise 7. When evaluating the multiplication formula for a complex number zo,one needs to worry about poles and vanishing denominators.(a) Let a E ZL{i] be odd, and assume that zo is a pole of neither nor Prove

carefully that 0 and that

= lQa('p4(zo)Y

Exercise 9 of Section 15.4 will be useful.(b) Let n be odd, and let p be a prime dividing n. Then let /3 be a Gaussian prime such that

1 mod4. Usepart(a)toprovecarefullythat

E

Theorem 15.3.2 will be helpful.

REFERENCES 513

(c) Let n be odd, and let p be a prime such that p2 divides n. Also define as in part (b).Prove that

Exercise 8. Let be an odd prime. Prove that 0.

Exercise 9. Let p E Z be prime. Prove that p2 — 1 is a power of 2 if and only if p = 3.

Exercise 10. Let n E Z be odd and positive, and let L = Use (15.9) and themultiplication law for cp((n — l)z) to prove that EL.

REFERENCES

1. R. Ayoub, The lemniscate and Fagnano 's contributions to elliptic integrals, Arch. Hist.Exact Sci. 29(1984), 131—149.

2. D. Cox, The arithmetic—geometric mean of Gauss, L'Ens. Math. 30 (1984), 275—330.

3. D. Cox, Primes of the Formx2 + ny2, Wiley, New York, 1989.

4. D. Cox, Why Eisenstein proved the Eisenstein criterion and why Schönemann discoveredit first, Amer. Math. Monthly 118 (2011), 3—2 1.

5. D. Cox and J. Shurman,, Geometry and number theory on clovers, Amer. Math. Monthly112 (2005), 682—704.

6. C. F. Gauss, Disquisitiones Arithmeticae, Leipzig, 1801. Republished in 1863 as VolumeI of [Gaussj. French translation, Recherches Arithmétiques, Paris, 1807. Reprint byHermann, Paris, 1910. German translation, Untersuchungen über Höhere Arithmetik,Berlin, 1889. Reprint by Chelsea, New York, 1965. English translation, Yale U. P., NewHaven, 1966. Reprint by Springer, New York, Berlin, Heidelberg, 1986.

7. C. Houzel, Fonctions elliptiques et intégrales abéliennes, in Abrégé d'Histoire desMathématiques, edited by J. Dieudonné, Hermann, Paris, 1978, 1—113.


9. G. A. Jones and D. Singerman, Complex Functions: An Algebraic and Geometric View-point, Cambridge U. P., Cambridge, 1987.

10. N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer, New York,Berlin, Heidelberg, 1984.

11. S. Lang, Elliptic Functions, Addison-Wesley, Reading, MA, 1973.

12. E Lemmermeyer, Reciprocity Laws, Springer, New York, Berlin, Heidelberg, 2000.

13. J. E. Marsden and M. J. Hoffman, Basic Complex Analysis, Third Edition, W. H. Freeman,New York, 1999.

14. V. Prasolov and Y. Solovyev, Elliptic Functions and Elliptic Integrals, AMS, Providence,RI, 1997.

15. M. Rosen, Abel's theorem on the lemniscate, Amer. Math. Monthly 88 (1981), 387—395.

16. C. L. Siegel, Topics in Complex Function Theory, Vol. 1, Wiley, New York, 1969.

514 THE LEMNISCATE

17. J. H. Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Springer, NewYork, Berlin, Heidelberg, 1994.

18. J. H. Silverman and J. Tate, Rational Points on Elliptic Curves, Springer, New York,Berlin, Heidelberg, 1992.

19. S. 0. Kronecker's Jugentraum and Modular Functions, Gordon and Breach, NewYork, 1991.

20. L. C. Washington, Elliptic Curves: Number Theory and Cryptography, Chapman andHall/CRC, Boca Raton, FL, 2003.

21. E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, Fourth Edition, Cam-bridge U. P., Cambridge, 1963.

APPENDIX A

ABSTRACT ALGEBRA

This appendix summarizes most of the abstract algebra needed for Galois theory.Section A. 1 reviews basic material on groups, rings, fields, and polynomials. Mostof this material should be familiar, though it might be a good idea to review thenotation. Before beginning Chapter 1, readers should also review complex numbersand the nth roots of unity from Section A.2.

The other sections cover a variety of topics. Section A.3 discusses polynomialswith coefficients in Q. Section A.4 deals with group actions, which are used in severalplaces in the text. Section A.5 includes the Sylow theorems, the Chinese RemainderTheorem, the multiplicative group of a field, and unique factorization domains.

A.1 BASIC ALGEBRA

We recall some basic material from abstract algebra.

A. Groups. We assume that the reader is familiar with groups and subgroups. Weusually write the group operation in a group G as gh for g, h E G, and the identityelement is denoted e. If G is finite, then IGI is called the order of G.


516 ABSTRACT ALGEBRA

If g E G, then the order of g, denoted o(g), is the smallest positive integer n suchthat gfl = e, if it exists. If gfl e for all positive integers n, then o(g) = 00.

Given a subgroup H of a group G, the left coset determined by g E G is

gH = {gh I h E H}

and the right coset determined by g is

Hg = {hg I h H}.

Two left cosets g1H and g2H are equal if and only if H. Similarly,Hg1 = Hg2 if and only if g2gj' H.

The left cosets gH of H C G partition G into disjoint subsets. Furthermore, if His finite, then each left coset has the same number of elements as H, i.e., IgHI = HIfor all g e H. If G is also finite, then the number of cosets is finite. This leads toLagrange 's Theorem, which is stated as follows.

Theorem A.1.1 If H is a subgroup of a finite group G, then HI divides IGI. •

The quotient IGI/IHI equals the number of left cosets. This number is the indexof H in G and is denoted [G HI. We discuss Lagrange's version of Theorem A.!.in Chapter 12.

The above statements also apply to right cosets. In general, the partition of Ginto right cosets can differ from its partition into left cosets. Galois was the firstto recognize the importance of when these partitions agree. This happens when thesubgroup H is normal. As is well known,

HisnormalinG gH=Hgforallg€G gHg'=Hforallg€G.

When H C G is normal, the left (= right) cosets form a group under the operationg1H . g2H = gIg2H. This is called the quotient group and is denoted G/H. Theidentity element of G/H is the coset eH = H.

Example A.1.2 The integers modulo n under addition form the quotient group Z/n7L.Elements of Z/nZ are sometimes called congruence classes. The congruence classof i Z is denoted [i] 7L/nZ.

We also assume that the reader knows the definition of group homomorphismG1 —* G2. Given such a its kernel is

= {g G1 I = e2},

where e2 is the identity of G2, and its image is

g€Gi}.

Then is a normal subgroup of G1 and is a subgroup of G2.

BASIC ALGEBRA 517

If a group homomorphism —+ is one-to-one and onto, then the inversefunction G2 —* is also a group homomorphism. Thus is a gmup isomor-phism. In this situation, we often write G1 G2.

Given a group homomorphism G1 —+ G2, the Fundamental Theorem of GroupHomomorphisms relates and Im(ço) as follows.

Theorem A.1.3 Let : G1 —* be a group homomorphism. Then there is a uniquegroup isomorphism such that = cp(g) for allgEGi. U

A group G is cyclic if there is g E G such that G = {g1 1 E Z}. When G is cyclic,recall that

ifGis infinite,

ifIGI=n<oo.

We have the following result about the subgroups of a cyclic group.

Theorem A.1.4 Let G be a cyclic group. Then:(a) Every subgroup of G is cyclic.

(b) If = n < 00, then for every positive divisor d of n, G has a unique subgroupoforderd. •

One way to create cyclic groups is to pick g E G and consider the subgroupgenerated by g, namely

(g) = {g' I E 7L}.

If g has finite order o(g) <00, then (g) is a cyclic group of order o(g). More generally,(S) C G denotes the subgroup generated by a subset S C G.

If G is a finite group, then applying Lagrange's Theorem to (g) C G shows thato(g) divides IGI. A partial converse is the following classic theorem of Cauchy.

Theorem A.1.5 If a prime p divides the order IGI ofafinite group G, then G has anelement of order p. •

For us, one of the most important groups is the symmetric group This is

the group of permutations of n objects, usually thought of as elements of the set{ 1,. . . n}. Thus is the set of functions

= {o-: {1,... ,n} —* {1, . . . ,n} o is one-to-one and onto},

where the group operation is given by composition of functions, and the identityelement is the identity function e(i) = i for 1 i n.

If a E is given by o-(j) = i3 for j = I, . . . , n, then following Cauchy, we writea in the form

(1 2 no.= I . .\ti 12


Also recall cycle notation. Given distinct numbers ii,... ,i1 E {l,. .. ,n} with I � 2,the i-cycle a = (i1 j2 . . . ii) E is the permutation defined by

a(ii)a(i2) =

(A.l)a(i1_ i) =

a(ii) = ii,o-(i) =i, {i1,...,i1}.

Note also that

= = (i3...i1i1i2) = = (i1i1i2...i,1).

As usual, a 2-cycle is called a transposition. Every element of can be writtenuniquely as a product of disjoint cycles.

When multiplying cycles, it is important to remember that the operation is com-position of functions. For example, consider

(345)(l23)(l2) = (1453).

When we apply the left-hand side, we first operate using (12), then using (123), andfinally using (345). So we move right to left through the cycles, while inside anindividual cycle, we move in the opposite direction (e.g., (345) takes 4 to 5). Notethat some books use different conventions for multiplying cycles.

Also recall the identity

(A.2) (i1 i2 . = (ii i1)(i1 i1_1) . (ii i3)(i1 i2),

which expresses an 1-cycle as a product of I — 1 transpositions.A permutation a E 5n is even if it is a product of an even number of transpositions,

and odd otherwise. It follows from (A.2) that an 1-cycle is even when i is odd, and isodd when 1 is even. The sign of a is defined by

1+1, ifaiseven,(A.3) sgn(a)

if a is odd.

Note that sgn : Sn —* {+l} is a group homomorphism.The most important subgroup of is the alternating group An, which is the

subgroup consisting of all even permutations. It is a normal subgroup of Sn ofindex 2. This follows from An = Ker(sgn).

Example A.1.6 Note that

S3 = {e,(12),(13),(23),(123),(132)},A3 = {e,(123),(132)} = ((123)).

BASIC ALGEBRA 519

Furthermore, one can show that

S3, A3, ((12)), ((13)), ((23)), {e}

are all subgroups of S3.

A group G is Abelian if the group operation is commutative, i.e., if gh = hg forall g, h E G. Recall that every subgroup of an Abelian group is normal. The reasonfor the name "Abelian" involves the Galois theory of an interesting class of equationsstudied by Abel. This is explained in Sections 6.5, 8.5, and 15.5.

Given groups G and H, their direct product, or more briefly product, is theset G x H = {(g,h)

Ig E G,h E H} with group operation (g,h)(g',h') = (gg',hW).

Products enable us to create new groups from old ones and are used in structuretheorems, such as Theorem A. 1.7 below. The Mathematical Notes to Section 6.4introduce a generalization of the direct product called the semidirect product.

Most courses in abstract algebra prove the following structure theorem for finiteAbelian groups.

Theorem A.1.7 Every finite Abelian group is isomorphic to a product of cyclicgroups of prime power order.

Another important group is the dihedral group of order 2n. This group isgenerated by elements g of order n and h of order 2 such that hgh' = g'. Somebooks write as For us, the subscript is the order of the group.

B. Rings. The reader should also be familiar with rings and ideals from abstractalgebra. For us, all rings are commutative and have a multiplicative identity. Wewrite the additive identity of a ring R as 0 and the multiplicative identity as 1.

Since R is commutative, a subset I c R is an ideal if and only if I is a subgroupunder addition and ra E I whenever r E R and a E I.

An ideal I is principal if there is r E R such that I = {rsI

s E R}. We say thatr generates I. Principal ideals of R are denoted either rR or (r). More generally,

The cosets of an ideal! in Rare sets of the form r+l = {r+sI

s E I} for r ER.Two cosets r +1 and s + I are equal if and only if r — s El. The set of all cosets isdenoted R/I and is a ring under the operations

(r+I)+(s+l) = (r+s)+I,(r+I) . (s+I) = rs+I.

We call R/I a quotient ring. Since R is commutative with a multiplicative identity,the same is true for R/I. The additive and multiplicative identities of RI! are 0 + I = Iand 1 + I, respectively.

Example A.1.8 Every ideal of 7L is principal, so that the ideals of Z are nZ for integersn � 0. Integers modulo n under addition and multiplication form the quotient ringZ/nZ, where the congruence class [ii is the coset i + n7L.


For us, a ring homomorphism R —* S is a function satisfying the usual conditionscp(r+s) = -i-p(s) and = for all r,s ER. In this book, allring homomorphisms preserve the multiplicative identity, unless explicitly statedotherwise. This means that = where 'R and 1s are the multiplicativeidentities of R and S, respectively.

Given such a its kernel is

= =O},

and its image is

IrER}.

Then Ker(ço) is an ideal of R and is a subring of S.If a ring homomorphism R —* S is one-to-one and onto, then the inverse function

S —÷ R is also a ring homomorphism. Thus is a ring isomorphism. In thissituation, we often write : R S.

Given a ring homomorphism : R —+ 5, the Fundamental Theorem of Ring Ho-momorphisms is as follows.

Theorem A.1.9 Let : R —* S be a ring homomorphism. Then there is a uniquering homomorphism such that + = for allrER. U

An integral domain is a ring R such that rs = 0, r, s E R, implies that r = 0 ors = 0. Section A.5 will discuss a special class of integral domains called uniquefactorization domains.

Example A.1.1O The ring of integers Z is an integral domain, but Z/6Z is not, since[21 [3] = [6] = [0], yet [2] and [3] are nonzero in Z/6Z.

C. Fields. Afield F is a ring such that every nonzero element has a multiplicativeinverse. To avoid trivial examples, we assume that 0 1 in F. Commonly used fieldsinclude:

Q = the field of rational numbers,

R = the field of real numbers,

C = the field of complex numbers.

Note that a field is always an integral domain. Also recall that the only ideals of afield F are {0} and F itself.

One way to create fields is via the field offractions of an integral domain R. Thisis defined to be the set

BASIC ALGEBRA 521

where we regard r/s and t/u as equal if and only if ru = St. This becomes a ringunder the operations

If r/s 0, then the multiplicative inverse of r/s is sir. Thus K is a field. We call Kthe "field of fractions" of R, though the term "quotient field" is also used.

Note that the function

defined by = r/ 1 is a one-to-one ring homomorphism, so that R p(R). Inthis situation, we usually identify R with This allows us to regard an integraldomain R as a subset of its field of fractions K.

Example A.1.11 The field of fractions of Z is the field of rational numbers Q. <I>

A second important method for creating fields is by means of maximal ideals.An ideal M c R is maximal if M R and for all ideals J of R, M C J C R impliesJ = M or J = R. Most abstract algebra courses prove the following theorem thatcharacterizes maximal ideals in terms of their quotient rings.

Theorem A.1.12 Let M be an ideal of a ring R. Then R/M is afield and only (1Mis a maximal ideal. •

For ZL, we can determine the maximal ideals as follows.

Example A.1.13 One easily checks that nZ C mZ if and only if m divides n. Itfollows that pZ is a maximal ideal of Z if and only if p is prime (be sure you seewhy). By Theorem A.1.12, Z/p7L is a field. It is customary to denote this field by

This field has p elements. In Chapter 11, we describe all finite fields. <II>

Theorem A. 1.12 is used in Chapter 3 to prove that any polynomial with coefficientsin a field has roots in some possibly larger field.

We next discuss the characteristic of a field F. Given a positive integer n, define

n times

where 1 is the multiplicative identity of F.The distributive law implies that (n. 1)(m. 1) = (nm). 1. If 1 = 0 for some

positive n, then let p be the least such number. We claim that p is prime. This is easyto see, for if we had p = ab with 0 < a, b < p, then

0= p. 1= (ab) .1= (a 1)(b . 1).


Since F is an integral domain, we would have a• 1 0 or 1 = 0, which wouldcontradict the minimality of p. Thus p is prime.

Because of this, we say that F has characteristic 0 if n• 1 0 for all positiveintegers n and has characteristic p if p. 1 0 and p is prime.

Thus Q, R, and C all have characteristic 0, while has characteristic p. Ingeneral, Galois theory is easier in characteristic 0 than in characteristic p.

D. Polynomials. A polynomial in x with coefficients in a field F is an expression

where ,a1,a0 e F. If 0, then we say that f has degree n, writtendeg(f) = n. If = 1, then we say that f is monic.

1ff and g are nonzero polynomials, then fg is also nonzero, since F is an integraldomain. It follows easily that

(A.4) deg(fg) =deg(f)+deg(g).

Notice also that we have not defined the degree of the zero polynomial. One mightbe tempted to set deg(0) = 0, but this would not be consistent with (A.4) (do you seewhy?). For this reason, we prefer to leave deg(0) undefined.

The set of all polynomials in x with coefficients in F forms a ring F[x] underaddition and multiplication of polynomials. Note that F[x] is an integral domain.

The following division algorithm is proved in most abstract algebra texts.

Theorem A.1.14 Let f,g e F[xl, and assume that g is nonzero. Then there arepolynomials q, r E F[x] such that

f=qg+r, where r=Oordeg(r) <deg(g).

Furthennore, q and r are unique. •

As an application of this theorem, consider the case when g = x — a for somea E F. The division algorithm implies that f = q• (x — a) + r where r E F (be sureyou see why). Evaluating this equation at x = a yields

f(a)=q(a).0+r,

so that r f(a). Thus f = q• (x — a) +f(a). This leads to the following result.

Corollary A.1.15 Given f e F[x] anda E F, the linearpolynomialx—a is afactorof f if and only if f(a) 0, i.e., if a is a root of f. •

Using this corollary and induction, one easily obtains the following bound on the

number of roots of a polynomial.

Corollary A.1.16 Let f E F[x] be nonconstant. Then f has at most deg(f) roots inthefieldF.

BASIC ALGEBRA 523

In Chapter 3, we show that by going to a larger field, a polynomial f E hasexactly deg(f) roots, provided that we take the multiplicities of the roots into account.

Another application of Theorem A. 1.14 is the Euclidean algorithm for computingthe greatest common divisor (or gcd) of two polynomials f, g E F [x], at least one ofwhich is nonzero. Recall that gcd(f, g) is the monic polynomial of maximum degreein F{x] which divides both f and g. If g 0, we compute gcd(f, g) by repeatedlyapplying the division algorithm until we get a zero remainder:

f=qog+ro, deg(ro) <deg(g),

g==qlro+rI, deg(r1) <deg(ro),

r0 = q2rI + r2, deg(r2) <deg(ri),

= + rfl+2,

Then one can prove that gcd(f, g) is the monic polynomial obtained by multiplyingrfl+2 by a suitable constant. On the other hand, if g = 0, then one easily sees thatgcd(f, 0) f. In general, the greatest common divisor has the following threeproperties:

• For any h e F[x], h divides gcd(f,g) h divides both f and g.• gcd(f,g) = 1 f and g are relatively prime in F[x].• gcd(f,g)=Af+BgforsomeA,BeF[x].

One can also use Theorem A. 1.14 to determine the ideals of F [xJ.

Theorem A.1.17 Every ideal of F[x] is of the form (f) = {fg g e F[x]}for somefEF[x].

This is proved in most abstract algebra courses. Recall that the basic idea of theproof is that if! C F[xI is a nonzero ideal, then pick f I \ {0} of minimal degree.Then one proves I = (f) using the division algorithm.

In general, an integral domain in which every ideal is principal is called a principalideal domain, or PID. It follows that Z and F [x] are both PIDs.

One can also find unique generators for ideals in F[x]. For the zero ideal, theunique generator is of course 0. For nonzero ideals, we can use monic polynomialsto give unique generators as follows.

Proposition A.1.18 Every nonzero ideal of F[xI can be written uniquely as (f) wheref is monic. •

Be sure you can prove this proposition.In the ring of integers Z, prime numbers play a central role. For F [x], the corre-

sponding objects are irreducible polynomials. Recall that a nonconstant polynomialin F [x] is irreducible over F if it is not a product of polynomials in F [xl of strictlysmaller degree.


An important result proved in most abstract algebra texts is that every nonconstantpolynomial in F [xJ can be factored into a product of irreducibles, where the factor-ization is unique up to order and multiplication by constants. In the terminology ofSection A.5, F[x] is a unique factorization domain, or UFD.

Another important result is that the ideal (f) C F{xJ is maximal if and only if thepolynomial f F[x] is irreducible over F. This is proved in Chapter 3 when westudy the existence of roots.

In general, it is not easy to test whether a given polynomial f e is irreducibleover F. When deg(f) = 2 or 3, any nontrivial factorization off must have a factorof degree 1. By Corollary A. 1.15, having a factor of degree 1 in F {xI is equivalent tohaving a root in F. Thus we have proved the following.

Lemma A.L19 1ff E F[x] has deg(f) = 2 or 3, then f is irreducible over F andonly has no roots in F. •

See Sections A.3, A.5, and 4.2 for more about factorization.

A.2 COMPLEX NUMBERS

In this appendix, we take a naive point of view and regard C as the set of numbersa + bi, where i = and a,b E JR. A rigorous algebraic construction of C ispresented in Chapter 3. Given z = a + bi, we define

Re(z) = a the real part of z,

Im(z) = b the imaginary part of z,

= a — bi the complex conjugate of z.

Furthermore, the absolute value of z = a + bi is

A. Addition, Multiplication, and Division. Addition and multiplication ofcomplex numbers are defined by

(a+bi)+(c+di)= (a+c)+(b+d)i,(a+bi)(c+di) = (ac—bd)+(bc+ad)i

and satisfy

(A.5)

Under these operations, C is a ring with additive identity 0 = 0 + Oi and multi-plicative identity 1 = 1 + Oi. To see that C is afield, note that if z = a + bi 0 (whichmeans that a and b are not both 0), then

1 1 1 a—bi a—bi a b— = = = — z C.z a+bi a+bi a—bi a2+b2 a2+b2 a2+b2

COMPLEX NUMBERS 525

If we think of z = a + bi as the point (a, b) in the plane, we can also represent zusing polar coordinates (r, 0). Since r = v'a2 + b2 = Izi, we get the picture

(a,b)

Polar coordinates in the complex plane C

In this situation, we follow Euler and define

(A.6) e'9 =cosO+isinO.

The relation between polar and Cartesian coordinates implies that a = jzlcos0 andb = zlsino. Hence

z=a+bi= zlcosO+IzlsinOi= zle'°.

This is the polar representation of z. In Exercise 1, you will prove that

= IzI wi,(A.7)

=

It follows that if z = Izi e8 and w = Iwi e4', then the polar representation of zw is

(A.8) zw = izi iwi

Thus we multiply lengths and add angles when we multiply two complex numbers.

B. Roots of Complex Numbers. We next consider the roots of the polynomial— a, where a e C and n e 7L is positive. The solutions of x'2 — a = 0 are called

the nth roots ofa. To describe the nth roots, write our given complex number a asa = iai We will assume that a 0, so that iai is positive. We seek a complexnumber w such that w" = a. If we write w = wi then Exercise 2 implies that

(A.9) w'1 =

so that the equation w" = a becomes

I = ale'°.

This equation is clearly satisfied if = iai and = 0, i.e., if iwI = and= 0/n. In other words, the complex number

(A.l0) w=


is an nth root of a.In polar coordinates, we can change the angle by an integer multiple of 2ir without

changing the point. For the poiar representation a = this means that we canwrite a = ale O+2irm) for any m E Z. Then, if we apply (A.lO) to this representationof a, we get the nth root

(A.ll) w=

As we vary m e Z, we claim that this gives precisely n distinct nth roots of a.To prove this, note that (0 + 2irmi)/n and (9 + 27rm2)/n differ by an integer

multiple of 2ir if and only if m1 m2 mod n. Hence, in (A. 11), we can assume thatm=O,1,...,n— l,whichgivesthenthroots

(A.12) .,

Note that 9/n, (0 + 2ir)/n, . . . , (9 + 2ir(n — 1))/n are n distinct angles in the planesince no two differ by an integer multiple of 27r. Thus we have proved the following.

Proposition A.2.1 Every a 0 in C has n distinct nth roots (A. 12). These are theroots of the polynomial — a E C{x]. •

By Corollary A. 1.15, each root gives a linear factor of x" a. This implies that

(A.13) x" —a (x— ja[eiOIPz) ... (x—

We can simplify the above formulas using the nth roots of unity. If we set

ç =

then (A.9) implies that = It follows that when a = 1, (A.l2) shows thatthe roots off — 1 are given by

itt2 tn—I

These are the nth roots of unity. In this case, the factorization (A.13) becomes

(A.l4)

Returning to the nth roots of a E C, we can now simplify (A.12). By (A.7),

= =

Then the nth roots of a given by (A. 12) can be written as

w1, wherew, =

and the factorization (A. 13) simplifies to

a = —(A.15)

= (x -Wi) (X wi) (x (x wi).

COMPLEX NUMBERS 527

For small n, the root of unity = = + is easy to work out.For example, standard facts from trigonometry imply that

(2 = cOSlr+iSifl7r = —1,

= ==i.

Thus the square roots of unity are 1,—i; the cube roots are and the fourthroots are ± 1, ±i. We often denote the cube root of unity (3 by w. Exercise 6 belowwill show how the formula for w = follows from the quadratic formula.

Roots of unity appear in several places in the text: in Chapter 8, where we studysolvability by radicals; in Chapter 9, where we compute the minimal polynomialof and in Chapter 10, where we explore Gauss's work on the constructibility ofregular polygons.

Exercises for Section A.2

Exercise 1. Prove (A.7).

Exercise 2. Let z = zI e'9 be the polar representation of z E C. Prove that

= n >0 in Z,

using induction on n and (A.8).

Exercise 3. This exercise will discuss De Moivre 'sformula, which states that

= cosnO+isinn9, n >0 in Z.

(a) Show that De Moivre's formula follows from (A.6) and Exercise 2.(b) Use De Moivre's formula and the binomial theorem for n = 4 and 5 to express cos46,

sin4O, cos5O, and sin5O in terms of cosO and sin9.(c) Use De Moivre's formula and the binomial theorem to prove that cosnO can be written as

a polynomial in cos 0 with integer coefficients.

Exercise 4. Use a calculator to find a seventh root of 3 + 2i. Note that 0 = tan1

Exercise 5. For n = 4, 5, and 6, draw a picture to show how the nth roots of unity form thevertices of a regular n-gon inscribed in the unit circle in the complex plane.

Exercise 6. The cube root of unity w = is a root ofx3 — 1 = (x — l)(x2 +x+ 1). Use thequadratic formula to show that w and w2 are given by (—1 ±

Exercise 7. Use = 1 and (,, = 1 to show that = =

Exercise 8. This exercise will derive an explicit formula for the fifth root of unity using thefactorizationx5 —1 = (x— l)(x4+x3+x2+x+ 1).(a) Use Exercise 7 to show that ifx = then x+ l/x =(b) Explain whyx = satisfiesx2 +x+ 1 + 1/x+ l/x2 = 0. Then show thaty =x+ 1/xis

arootofy2+y— 1.


(c) Use part (b) to conclude that = + Then show that

i

4 2

Chapter 10 explains how this relates to the straightedge-and-compass construction of theregular pentagon.

Exercise 9. In this exercise, you will give two proofs of the identity

n>OinZ.

(a) Show that this identity follows from (A. 14) by comparing the coefficients of x"1.(b) Give a second proof using the factorization x" — 1 = (x — 1) (x"t + + 1).

(c) More generally, use part (b) to show that m 0 mod n implies that

Also determine the sum on the left-hand side when m 0 mod n.

Exercise 10. The eighth root of unity (8 is given by = +i).(a) Show that the eighth roots of unity are given by + 1, +i, (± 1 ± i).

(b) Use the factorization of x4 + 1 given at the end of Section A.3 to show that

and explain how this factorization relates to part (a).

A.3 POLYNOMIALS WITH RATIONAL COEFFICIENTS

We next discuss the polynomial ring Q [x]. In this case, we often take a polynomialwith rational coefficients and multiply it by a constant to clear denominators, givinga polynomial with integer coefficients. In general, we let Z[x] denote the ring ofpolynomials in x with coefficients in Z.

As is well known, we can describe the rational roots off E Z[x] as follows.

Proposition A.3.1 Letf = + +ao E 7L[x] be nonconstant. If p/q E Q is aroot off, where p, q E 7Z are relatively prime, then plao and qlan.

Note that combining Lemma A. 1.19 and Proposition A.3. 1 gives an algorithmfor deciding whether a polynomial in Q[xI of degree 2 or 3 is irreducible over Q.In Section 4.2, we show that a similar algorithm exists when the degree is greaterthan 3. The crucial result, due to Gauss, is that we can reduce factorization in Q[x]to factorization in Z[x]. This is Gauss's Lemma, which is stated as follows.

Theorem A.3.2 Suppose that f E Z{x] is nonconstant and that f = gh where g, h EQ[x]. Then there is a nonzero 8 e Q such that g = ög and h = 8'h have integercoefficients. Thus f = in Z[x].

POLYNOMIALS WITH RATIONAL COEFFICIENTS 529

Proof: Let s e Z be a common denominator for the coefficients of g, so that gfor E 7L[x]. Then let r e Z be the greatest common divisor of the coefficients of

Factoring out r enables us to write

I rg = —gi =

S S

where E Z{x] has relatively prime coefficients. Similarly, we can write

h=

where h1, h2 e and h2 has relatively prime coefficients.Let 5 = and observe that

Sg= EZ[x],

= =

If we can show that sulrt, then h2 E Z[x]. This willprove the theorem.

Hence it remains to show that sulrt. For this purpose, we will prove that if p isprime, then pa su implies that ii (do you see why this implies sujrt?). So pick aprime p and suppose that su. Then write

(A.16)

Since b1, . . . , have no nontrivial common factors, we can find an index i � 0such that pbo,... and Similarly, there is an index j � 0 such thatpjco,...,pcj_i and p{cj.

Multiplying the expressions for and h2 given in (A. 16), we see that the coefficientof in g2h2 is = boc,+3 + We can write this in the followingform:

(A.17) + . . . + . . . +bi+,co.

pdividesbo ,...,b1_1 pdividescj1

Since p b1c1, this shows that is relatively prime to p. Thus gcd(pa, = 1.

Next observe that f = gh = implies that

(A.18) suf = rtg2h2.

Since p° Isu and f Z[x], we see that pa divides the coefficient ofxHj on the left-handside of (A. 18). However, the coefficient of on the right-hand side of (A. 18) is

and it follows that divides Since pa is relatively prime to weconclude that pa must divide ii, which is what we needed to show. This completesthe proof. •

We will generalize Gauss's Lemma in Section A.5.


A.4 GROUP ACTIONS

Some of the most interesting groups arise as the symmetries of particular mathe-matical objects. This leads to the notion of a gmup action. Here is the precisedefinition.

Definition A.4.1 Let G be a gmup and X be a set. Then an action of G on X is afunction G xX —+ X, written (g,x) g such that(a)

(b) g.(h.x)= (gh).xforallg,h E GandxEX.

Here are some simple examples of group actions.

Example A.4.2 The symmetric group acts on the set { 1,2, ... , n}. If a E and

iE{l,2,...,n},thena.iisjusta(i). <ii>

Example A.4.3 Let GL(n, R) be the set of invertible n x n matrices with real entries.This is a group under matrix multiplication and acts on in the following way: ifA E GL(n,R) and v e then A . v is the matrix productAv, where we think of v asa column vector.

Example A.4.4 Let S' = {e'°I

0 R} be the set of complex numbers of absolutevalue 1. This is a group under multiplication of complex numbers, and S' acts on Cby multiplication.

We next define some important concepts related to a group action.

Definition A.4.5 Let a gmup G act on a set X, and let x E X.(a)

The is the subgmup = {g E G I g = x}.

Here are some examples of orbits and isotropy subgroups.

Example A.4.6 In the action of on X = {1, 2,..., n}, the orbit of any i e X is allof X, and the isotropy subgroup of i consists of all permutations which fix i. Do yousee why the isotropy subgroup is isomorphic to Sn_i?

Example A.4.7 Let H (a) be the cyclic subgroup generated by a E Then Hacts on X = { 1,2,. . . , n}. In Exercise 1, you will show that the orbits of this actioncorrespond to the decomposition of a into a product of disjoint cycles.

Example A.4.8 In the action of S1 on C, consider a point z 0. Then the orbit S' . z

of z is the circle of radius Izi centered at the origin, and the isotropy subgroup of z istrivial.

If G acts on X, then one can easily show that

GROUP ACTIONS 531

is an equivalence relation on X whose equivalence classes are the orbits of G. Itfollows that X is a disjoint union of orbits. Furthermore, the isotropy subgroups ofpoints on the same orbit are related as follows

(A.19) Gg.x

g E G and x E X. Proofs of these assertions can be found in Section 1.12 ofVolume I of [Jacobson].

The Fundamental Theorem of Group Actions relates orbits to cosets of the isotropysubgroup as follows.

Theorem A.4.9 Let G act on X, and let G be the isotropy subgroup of x E X.Then:

(a) There is a one-to-one correspondence

{left cosets of G

G is finite, then= G•xI.

Thus IGI= sothatlGl is divisible by both IGxI and IG.xI.

Proof: Let G/GX =Ig E G} be the set of left cosets of the isotropy subgroup.

Then define : G/GX —* G x by

We first need to show that this map is well defined. If = then = g2h

for some h E (be sure you know why). Then

gi.x(g2h).xg2.(h.x)zzg2.x,

where the second equality follows from Definition A.4. 1, and the third follows fromh E This proves that is well defined.

Since every y e G . x is of the form y = g x for some g E G, we see that

Thus p is onto. To show that is also one-to-one, suppose that =By the definition of this implies that

gi

Using the properties of group actions, we obtain

Thus E so that = Hence is one-to-one. From here, the restof the theorem follows easily.


Chapter 12 discusses the special case of this theorem discovered by Lagrange inhis study of the roots of polynomials.

The following definition is used in Section 6.3 when we study how the Galoisgroup acts on the roots of a polynomial.

Definition A.4.1O The action of G on X is transitive for every x,y E X, there isg E Gsuch

More on group actions may be found in Section 1.12 of Volume I of [Jacobson].

Exercises for Section A.4

Exercise 1. As in Example A.4.7, let H = (a) be the cyclic subgroup generated by a EAssume that T e and that

is the decomposition of a into a product of disjoint cycles. Suppose that ri = (ii . . . i1)

(a) Use (A. 1) to prove that {ii,. . . ,i1 } is the orbit of i1 under the action of H.(b) Explain why Theorem A.4.9 implies that I divides the order of a.

Exercise 2. In the group action considered in Example A.4.8, find the orbit and isotropysubgroup of 0 E C.

Exercise 3. The symmetric group 53 is sometimes introduced as the symmetry group of anequilateral triangle A.(a) In the language of this section, explain how S3 acts on A. You may assume that the

vertices of A are labeled 1,2,3.(b) For each subgroup of S3 given in Example A. 1.6, determine all points p E A whose

isotropy subgroup is the given subgroup of S3. Also describe the orbit of p.

Exercise 4. Given a group G, define g . h = for g, h E G. Prove that this is a groupaction of G on itself. We say that G acts on itself by conjugation. Then:(a) Prove that the orbit G g is the conjugacy class Cg of g and that the isotropy subgroup is

the subgroup C(g) consisting of all elements of G that commute with g.(b) Let G be finite. Prove that [G : C(g)] = CgI.

Exercise 5. Prove that a group G acts transitively on a set X if and only if G x = X for allxEX if and only ifGx=X for somexEX.

A.5 MORE ALGEBRA

Here are some further results about groups, rings, fields, and polynomials that willbe used in the text.

A. The Sylow Theorems. Let G be a finite group, and let p be a prime dividingthe order of G. Then a subgroup H C G is called p-Sylow subgroup if IH =where p dividing IGI. Here is the basic result concerningSylow subgroups.

MORE ALGEBRA 533

Theorem A.5.1 Letp be a prime dividing the order of afinite group G. Then:(a) (First Sylow Theorem) G has a p-Sylow subgroup.(b) (Second Sylow Theorem) Any two p-Sylow subgroups of G are conjugate in G.(c) (Third Sylow Theorem) Let N be the number of p-Sylow subgroups of G. Then

I modp, andNdivides IGI.

A proof of the First Sylow Theorem can be found in [Herstein, Thm. 2.12.11,and the Second and Third Sylow Theorems are proved in [Herstein, Thms. 2.12.2and 2.12.3 and Lem. 2.12.61.

The Sylow Theorems have some nice applications in Chapters 8 and 14.

B. The Chinese Remainder Theorem. The following result will be useful inseveral places in the text. Given a positive integer n, let E 7L/nZ denote thecongruence class of a modulo n.

Lemma A.5.2 Let n and m be relatively prime positive integers. Then the map[aJnm F—* [a]m) gives a well-defined ring isomorphism

Z/nmZ Z/nZ x 7L/mZ.

Proof: If [ainm = [binm, then nmja — b, from which we conclude that [aim) =([bin, [bim). Hence the map is well defined, and it is easy to see that it is a ringhomomorphism. Furthermore, if [a]nm is in the kernel, then nla and mja, whichimplies that since n and m are relatively prime. Thus = [0mm, so that themap is one-to-one. It is then onto, since both rings have order nm. •

C. The Multiplicative Group of a Field. Given a field F, its multiplicative groupis F* = F \ {0}, which is a group under multiplication by the definition of field. Thefact that a polynomial of degree m has at most m roots in a field implies the followinginteresting property of F*.

Proposition A.5.3 Let G C F* be a finite subgroup of the multiplicative group of afield F. Then G is cyclic.

Proof: First observe that G is Abelian because F is a field. Then Theorem A. 1.7implies that G is isomorphic to a product of cyclic groups, say

G Z/miZ x x Z/mrZ,

where m1,. . . ,mr are integers> 1. Thus GI = m1 If r = 1, then we are done.So assume that r 2.

Let m = lcm(m,, . . . , be the least common multiple of the m,. It is then easy toverify that gm = 1 for every g E G. Since G is a subgroup of F*, it follows that everyg E G is a root of xm — 1 E F[xI. Hence this polynomial has at least GI = mi mrroots in F. But, as noted above, xm — 1 has at most m roots in F, since F is a field.Thus

m=lcm(mi,...,mr)�mi...mr,


which clearly implies that lcm(mi,... , = mi This in turn implies thatmi,... , m,. are pairwise relatively prime (be sure you understand why). However, ifn and m are relatively prime, then by Lemma A.5.2, there is a ring isomorphism

7L/nZ x 7L/mZ 7L/nm7L,

which is also a group isomorphism if we forget multiplication. Using this repeatedly,we obtain a group isomorphism

This completes the proof of the proposition. .D. Unique Factorization Domains. Given a ring R, a unit of R is an element ofR that has a multiplicative inverse in R. The set of all units of R is denoted R*. Notethat R* is a group under multiplication.

Now let R be an integral domain. We say that r E R is irreducible if it is not a unitand r = ab, a,b ER, implies that a orb is in R*.

Example A.5.4 Given a field F, the units of the polynomial ring F [x} are the nonzeroelements of F, i.e., F [xl * = F*. Furthermore, the irreducible elements of F [x], asdefined above, are precisely the irreducible polynomials of F [x].

Here is the precise definition of unique factorization domain.

Definition A.5.5 An integral domain R is a unique factorization domain, or UFD,if the following two conditions hold:(a) Every nonzero element of R is either a unit or a product of irreducibles.(b) If r1 = s1 where r1,... , si,. . . ,sl E R are irreducible, then k = 1,

and there is a permutation o E Sk such that for each 1 i k there is a unita, E R* such that r, =

a of integers 7L. Another important class ofexamples come from polynomial rings. Here is the basic result.

Theorem A.5.6 Let R be a UFD, and let R[x] be the ring of polynomials in a variablex with coefficients in R. Then R[x] is a UFD. U

A proof can be found in [Herstein, Thm. 3.11.1] or [Jacobson, Vol. I, Thm. 2.25].This result implies, for example, that Z[x] is a UFD. In Chapter 2 we will discussthe ring F[xi,... of polynomials in x1,... ,x,, with coefficients in F. UsingTheorem A.5.6 and induction on the number of variables, it is straightforward toprove the following.

Corollary A.5.7 1fF is afield, then F[xi,.. . is a UFD. •

In the course of proving Theorem A.5.6, one needs the following generalizationof Gauss's Lemma (Theorem A.3.2).

MORE ALGEBRA 535

Theorem A.5.8 Let R be a UFD with field offractions K. Suppose that f E R[xJ isnonconstant and that f = gh where g,h E K[x]. There is a nonzero 8 E K such thatg = äg and h = 8'h have coefficients in R. Thus f = in R[x]. •

The proof is identical to the proof of Theorem A.3.2 given in Section A.3. An im-mediate corollary of Theorem A.5 .8 is that if f E R[xJ is irreducible and nonconstant,then it is also irreducible in K[x]. Furthermore, if f, g E R[xJ are relatively prime andnonconstant, then Theorem A.5.8 implies that they are also relatively prime in K[x].

APPENDIX B

HINTS TO SELECTED EXERCISES

This appendix contains hints to selected exercises in the text.

Section 1.1 (pages 9—10)

Exercise 2. Hint: Explain why = w2.

Exercise 3. Hint: By choosing the correct square root of q2, show that Cardan's formulasreduce = = and y3 = when p = 0.

Exercise 7. Hint: First show that all three polynomials give the same zi but a different Z2.

Exercise 8. Hint: Use Example 1.1.1 and Exercise 7.


Exercise!. (c) Hint: f'(yi) = 3(yi — a)(yt — andf(a) = (a —yi)(a —y2)(a—y3).

Exercise 2. (a) Hint: Use (1.22).

Exercise 3. (a) Hint: Remember that yi ,y2,y3 are distinct.

Exercise 4. (b) Hint: Part (c) of Exercise 1 and $ 0 imply that f(a) and are nonzero.

Exercise 6. Hint: By part (a) of Exercise 3, > 0 implies that p 0.


538 HINTS TO SELECTED EXERCISES

Exercise 7. Hint: Dividing y3 — l5y —4 by y —4 leads to a quadratic equation.

Exercise 11. (b) Hint: Use Exercise 3 of Section A.2. (c) Hint: w =

Section 2.1 (page 30)

Exercise 1. Hint: Give a proof by contradiction. Unique factorization will be useful. You mayassume that x andy are irreducible in F [x, y].

Exercise 3. (c) Hint: Let al = = = —a in the corollary.


Exercise 1. Hint: Let ii < <ir. If 1, then the exponent of xi in isO.

Exercise 2. (b) Hint: Use part (a).

Exercise 4. Hint: Express the coefficients of (2.17) in terms of ai , a2, a3 evaluated at the a1.

Exercise 7. Hint: If E is fixed, you will need to explain why ra ranges over all elementsof S, as a does.

Exercise 9. (c) Hint: Use the well-ordering property of the nonnegative integers, which statesthat any strictly decreasing sequence of nonnegative integers is finite.

Exercise 11. Hint: Use the method of Example 2.2.6 and Exercise 4.

Exercise 16. (c) Hint: You can't use the formulas of Chapter 1. So you need to compute(1 w)2(l —w2)2(w—w2)2.

Exercise 18. Hint: Use the Newton identities and explain why every sk is a polynomial in thea1 with coefficients in Z.

Exercise 20. Hint: Suppose that a2 = P(si,.. . ,s4. Then evaluate this at xi = = = 0


Exercise 1. Hint: To find the roots of y3 + 2y2 — 3y + 5, use the Mathematica command

+ — 3y + 5 == O,y]]

or the Maple command

O,y,complex);

Note that f solve normally only finds real roots, but by specifying the complex option, it willfind all roots, real and complex.

Exercise 4. Hint: If the roots are xi ,x2,x3, then one way for this to happen is Xi = (x2 +x3)/2,which gives the equation 2xi — x2 — X3 = 0. There are two other ways this can happen. Thentake the product of all three ways.


Exercise 2. Hint: Use Theorem 2.4.4.

HINTS TO SELECTED EXERCISES 539

Exercise 5. Hint: Use Proposition 2.4.1.

Exercise 8. (e) Hint: F has characteristic 2. (1) Hint: Use Exercise 5.

Exercise 10. (a) Hint: F[ai, . . . , F[ui, . . . , is a UFD.


Exercise 2. Hint: The definition of ring homomorphism given in Section A. 1 requires that ppreserve the multiplicative identity. Also remember that a homomorphism is one-to-one if andonly if its kernel is {0}.


Exercise 2. (c) Hint: Solve the second equation of part (b) for y, and substitute the result intothe first. Also remember that b 0.

Exercise 3. Hint: Apply the IVT to x2 — a on a suitably chosen interval.

Exercise 5. (b) Hint: This follows from Lemma 3.2.3 and Exercise 4 with IR replaced by F.

Exercise 6. Hint: Use part (b) of Exercise 1.


Exercise 1. Hint: When f(a) = 0 for f F[x] nonzero, what equation is satisfied by 1/cs?

Exercise 4. Hint: First use Lemma 4.1.9 to show that F(ai,... , C F(ai, . . . , cs,). Thenuse the lemma a second time.

Exercise 6. Hint: Show that the ring homomorphism F[xi,. . . —* F[ai, . .. , a,,] given byx, '—+ a, is an isomorphism. Then explain why this extends to an isomorphism of the fields offractions.

Exercise 7. (a) Hint: Remember that g(a) 0.

Exercise 8. (a) Hint: First explain why it suffices to show that You may assumethat and are irrational. (b) Hint: What is a —


Exercise 1. (c) Hint: How many roots does h — g have? What is its degree? Corollary A. 1.16will be useful.

Exercise 5. (b) Hint: Note that (2' and (12 are also roots of x24 — 1. What are theminimal polynomials of these numbers?

Exercise 9. Hint: Suppose that (g/h)" = t, where g, h E k[tj are relatively prime. Show thatgP = th" would imply that first g and then h are divisible by t.


Exercise 6. Hint: Compute [F(a, /3) : F] in two ways.



Exercise 2. (a) Hint: Consider the field extensions

Q C C C C ... C L.

Exercise 3. (a) Hint: Use Gauss's Lemma. (b) Hint: The minimal polynomial of w over Q is

x2 + x + 1.

Exercise 6. Hint: If an element of F(x) is algebraic over F, write it as p/q where p, q E FLx}are relatively prime. If a E F, then clear denominators anduse unique factorization to conclude that p, q E F.

Exercise 7. Hint: Take a E L and consider the minimal polynomial of a over F.

Exercise 9. Hint: Use Exercise 3.


Exercise 3. Hint: Use Exercise 4 of Section 4.3 to show that L =F(a) for some a E L. Thenlet f be the minimal polynomial of a over F.

Exercise 4. Hint: Consider —w, where w =

Exercise 5. Hint: In Section 4.2, we used Maple and Mathematica to factor fin L[x].

Exercise 6. Hint: Compute +

Exercise 7. (b) Hint: If a E L is a root off, then compute f(a + 1).

Exercise 8. (b) Hint: Use Proposition 4.2.5 and the method of Exercise 5 of Section 4.3.

Exercise 9. (a) Hint: Combine [L: F] = n! with the proof of Theorem 5.1.5.

Exercise 11. (a) Hint: Consider F C F(a) CL, where a E Lisa root ofF.

Exercise 13. Hint: Apply the proposition to Q CL. Part (a) of Exercise 7 from Section 4.1will be useful.


Exercise 3. (c) Hint: Compute (x — a)3.

Exercise 4. Hint: Use Theorem 4.4.10 and Exercise 1 from Section 4.4.


Exercise 2. (a) Hint: Treat the cases p = 2 and p> 2 separately.

Exercise 3. (b) Hint: Use Lemma 5.3.10.

Exercise 4. Hint: Recall how and are obtained from You may assume thatis a polynomial in Z[ai,. ..

Exercise 6. Hint: Remember that the given polynomial need not be irreducible.

Exercise 7. (b) Hint: Look at the exponents of the nonzero terms of 1.


Exercise 8. (a) Hint: Write F = k(u) (t) and use Exercise 9 of Section 4.2. (b) Hint: Rememberthat k has characteristic 3.

Exercise 9. (b) Hint: Express j3 as a polynomial in a.

Exercise 13. Hint: Use Lemma 5.3.5.


Exercise 4. (c) Hint: Use part (b) and Example 5.4.4.

Exercise 5. (a) Hint: First explain why a + .\I3 and a + p/3 lie in F(a + )tf3).

Exercise 7. Hint: Use Theorem 5.4.1 and the previous exercise.

Section 6.1 (pages 129—1 30)

Exercise 4. (b) Hint: Think about kernels of ring homomorphisms and ideals of fields.

Exercise 6. Hint: What is

Exercise 7. (a) Hint: See part (b) of Exercise 4. (b) Hint: Regard L as a vector space over F,and show that a is a linear map. Now use standard results from linear algebra.


Exercise 3. (a) Hint: Use the method of Exercise 5 of Section 4.3.

Exercise 6. Hint: Use Exercise 11 of Section 5.1.


Exercise 4. Hint: Write down the roots off explicitly, and determine how the elements ofGal(L/Q), as described in the proof of Theorem 6.2.1, act on the roots. Then look at thecorresponding permutations in S4.

Exercise 6. Hint: Use Theorem A.4.9 from Section A.4.


Exercise 4. Hint: Can you find an inverse function?


Exercise 3. Hint: Lemma 6.1.3 will be useful.

Exercise 4. Hint: Use the nth roots of unity, and show that (x) can be chosen to be x' providedthe roots of x" — 1 = 0 are labeled appropropriately.




Exercise 5. Hint: Use part (b) of Proposition 7.1.7 twice.


Exercise 7. Hint: This is similar to Exercise 5.

Exercise 8. Hint: Let ai = a, a2,. , be as in the definition of h, and consider the subgroupH = {a E GaI(L/F) I a(a) = a}. Then study how the left cosets of H act on a.

Exercise 12. (a) Hint: Do not use the Theorem of the Primitive Element—give a direct proof.(b) Hint: Use Lemma 5.3.5.


Exercise 2. Hint: Show that a'Gal(L/aK)cr C Gal(L/K) follows from a'(aK) = K.

Exercise 7. Hint: This follows from the argument used to prove (a) (b) in Theorem 7.2.5.

Exercise 9. (b) Hint: Use Exercise 7.


Exercise 5. (b) Hint: Use the automorphism L L which sends t to it.

Exercise 7. (a) Hint: Proposition 4.2.5 will be useful.

Exercise 9. Hint for (a) (b): If Gal(L/F) = {e, a, r, crr}, then consider the fixed fields of(a) and (T). See also Exercise 12 of Section 7.1.

Exercise 10. Hint for (c) (a): If Q(a) then let L = Q(a, /3), and show that thereare a,r E Gal(L/Q) such that Q(a) is the fixed field of (a), and Q(/3) is the fixed field of(i-). Then see where Q(a + /3) fits in the Galois correspondence, and explore how a, r, ar actona+/3.

Exercise 11. Hint: Use the Galois correspondence and Exercise 8 of Section 7.2.

Exercise 13. Hint: Show that the Galois closure constructed in Proposition 7.1.7 can berealized as a subfield of L. Then use the Galois correspondence and Exercise 12.

Exercise 14. Hint: First use part (d) of Exercise 4 of Section 6.2 to show thatis Abelian.


Exercise 7. Hint: Use the Galois correspondence and Proposition 6.3.7.


Exercise 2. Hint: R = F[y] is a UFD with field of fractions K = F(y).

Exercise 3. (a) Hint: In A and B, the coefficient of each power of xis a rational function in y.

Exercise 12. (c) Hint: In part (b), we "broke" one of the symmetries of the polyhedron bymoving some of the vertices. To obtain the groups in part (c), you need to "break" some of thesymmetries in a similar way.


Exercise 4. Hint: The proof is similar to part (a) of Exercise 2.



Exercise 1. (b) Hint: See Exercise 8 of Section 7.3.

Exercise 2. Hint: Use Definition 8.2.1, the Tower Theorem, and : = 3.

Exercise 3. (a) Hint: Consider the intersection of all subfields of L containing K1 and K2.

Exercise 4. (a) Hint: Adapt the proof of Theorem 7.1.7 to show that F(ai,. . . , is a Galoisclosure. (b) Hint: Use Proposition 5.1.8.


Exercise 2. Hint: Splitting fields.

Exercise 3. Hint: See Exercise 9 in Section A.2.

Exercise 4. (b) Hint: Remember that E F,_

Exercise 7. (a) Hint: Use Exercise 6.

Exercise 8. (a) Hint: As in Exercise 3 of Section 7.4, c4 + = 2A. Also, when usingthe computer algebra system, you should write w as (—1 + (b) Hint: Recall that

= xi +x2 + x3. Also, what is l+w + w2?


Exercise 1. Hint: Use Cauchy's Theorem (Theorem A. 1.5).

Exercise 2. Hint: When i,j,k,1 are distinct, verify that (ij)(kl) = (ijk)(jkl). You will alsoneed to consider the case when i, j, k, I are not distinct.

Exercise 5. Hint: If a, are elements of H different from e, then what can you say abouta2 ar T2?

Exercise 6. Hint: If H1 C G/H is normal, then what can you say about 1(Hi) whereG —* G/H takes g to the coset gH? See Exercises 3 and 4 from Section 8.1.


Exercise 1. Hint: Suppose that F C L1 C Mi, where F C M1 is radical. Explain why wecan assume that is the splitting field of some polynomial g E Fix]. Then let L2 C M2be a splitting field of g regarded as a polynomial in L2 {x]. Prove that F C M2 is radical.Theorem 5.1.6 will be useful.



Exercise 3. Hint: If you use a computer to draw the graph of f, it seems clear that there arefour real roots. To make this rigorous, you should use the Intermediate Value Theorem.

Exercise 5. (a) Hint: What is (a + i)" in characteristic p?

Exercise 6. (a) Hint: Unique factorization.

Exercise 7. Hint: Follow the proof of (b) (a) of Theorem 8.3.3. You will also need toexplain why primitive mth roots of unity exist for all m not divisible by p.



Exercise 1. When n = 1, note that [0] = [1] in 7L/17L, so that [0] is in (Z/1Z).

Exercise 5. Hint: Maple and Mathematica can factor polynomials over Q. Also, what is thedegree of (x)? (See Exercise 13 for another approach to computing (x).)

Exercise 6. (a) Hint: Analyze the proof of Theorem 2.2.2. (b) Hint: Use Lemma 5.3.10 andLemma 9.1.2.

Exercise S. (f) Hint:

Exercise 11. Hint: Use (9.4).

Exercise 15. Hint: Use Exercise 14 and (9.4).

Exercise 16. (b) Hint: Use part (a) and Lemma 9.1.1.


Exercise 6. Hint: Use Lemma 9.2.4 and Exercise 9 of Section A.2.

Exercise 9. (b) Hint: To five decimal places, (—1 + = 1.56155 and (—1 — =—2.56155, yet the quadratic fonnula says that (8, 1) = (—1 + for some choice of sign.(c) Hint: When computing —4(4,3), do not compute (4,1)2 using part (b). Rather,use Proposition 9.2.9 to express (4,1)2 in terms of(4,2), (4,3), and (4,17) = 4.

Exercise 10. Hint: See Exercise 3.

Exercise 11. (a) Hint: [gC] generates H1, and [ge/q] generates Hjq. (b) Hint: First prove that[Lf(w) : = [L1 : using the method of Exercise 5 of Section 4.3. Then look atthe argument used in the proof of the General Case of Theorem 8.3.3. (c) Hint: First usePropostion 9.2.6 to study how cr' acts on f-periods.

Exercise 12. (a) Hint: Use Exercise 16 of Section 9.1.

Exercise 13. Hint: First show that H8 c (Z/17Z) is the subgroup of squares. Then, for p{a,explain why x2 a mod 17 has a solution if and only if [a] E H8.

Exercise 14. (a) Hint: Label the roots as = for i = 1,... ,n — 1.


Exercise 4. (a) Hint: You can think of as the line through the points (ui , Vi) and (U2, v2) inthe plane R2 Consider the cases UI = U2 and UI separately.

Exercise 5. (a) Hint: Do you remember how to construct an equilateral triangle?

Exercise 6. Hint: Argue as in Example 10.1.9 that such a trisection implies that cos 20°is constructible. Then use = and the identity cos(30) = 4cos3O — 3cosO fromSection 1.3.

Exercise 7. Hint: Combine the construction used in Proposition 7.1.7 with the fact that C isalgebraically closed.

Exercise 8. (a) Hint: Mimic the proof of Theorem 10.1.6.

Exercise 10. (d) Hint: Use part (c) and x = a/3.

Exercise 12. (b) Hint: r/3.



Exercise 1. Hint: Note that if m is odd, then xm + 1 = xm — (—

Exercise 5. (a) Hint: Use and Theorem 10.1.6. (b) Hint: Use Exercise 16 ofSection 9.1.


Exercise 1. (a) Hint: Qi and Q2 are the reflections of P1 and P2 about 1.

Exercise 3. (c) Hint: How does part (b) relate to (10.8)?

Exercise 4. Hint: Use implicit differentiation.

ExerciseS. (b) Hint: What are the roots of x3 + 2x+ 1? (c) Hint: Use (10.11).

Exercise 6. Hint: The distance between ai and a2 equals the distance between the reflectionsof these points about £.

Exercise 13. Hint: Let M be the midpoint of QR. Use the circle with center M and radius 1/2to show that LQMP = 2LQRP.

Exercise 14. (a) Hint: A perpendicular from R to the x-axis will meet the x-axis at a point S.This gives f\ROS. Then let T bisect the segment and prove that is congruent to

Exercise 15. (b) Hint: This is very challenging. A solution can be found on page 128 of [15]in the references to Chapter 10.


Exercise 2. Hint: See Section A. I.

Exercise 3. Hint: Show that x x whenever m divides n.

Exercise 9. (a) Hint: Use Exercise 4 of Section 9.1. (b) Hint: (p, f) = pZ[a] + fZ[aI.

Exercise 11. Hint: If a is a root of fin some splitting field, then (a) : = n.


Exercise 4. (b) Hint: Divide by the smallest power of p appearing in the formula and thenwork modulo p.

ExerciseS. Hint: If atm = 1, then write m = p5d where s 0 and gcd(d,p) = 1.

Exercise 10. (b) Hint: Explain why f and are relatively prime for i j.

Exercise 11. (b) Hint: Show that each is a field.

Exercise 12. (c) Hint: What is the factorization of x" — x?

Exercise 17. Use Theorem 11.2.7.


Exercise 1. Hint: Look at the proof of (7.1).


Exercise 2. (b) Hint: Another root of the Ferrari resolvent is Y2 = (23) = Xi X3 + x2x4. HowareH(yi) and H(y2) related?

Exercise 5. (a) Hint: For the choice of sign with as roots, the constant term is Butwhat is the constant term of the corresponding equation in (12.11)? Then do the same for theother equation.

Exercise 9. (b) Hint: First show that a E H, using the fact that multiplication bya permutes the elements of H. Then show that o•. p (p for all a H, using the fact that theexponents are distinct. Remember that two polynomials are different if one has a term whichdoesn't appear in the other.

Exercise 11. Hint: If you have done Exercise 19, then use part (f) of that exercise. Otherwise,follow the proof of Theorem 12.1.10, and pick (p E L with H as isotropy subgroup. Then let

be the distinct rational functions a (p for a E An. Show that s = [An : H] and that

G= {a E An Ia. (pi= (pi for alli= 1,... ,s}

is a normal subgroup of An. Then use Theorem 8.4.3 and the argument of Theorem 12.1.10.

Exercise 12. (b) Hint: Use the Galois correspondence.

Exercise 13. Hint: First explain why a a = for some j.

Exercise 14. (b) Hint: Look at the proof of Theorem 12.1.4.

Exercise 20. Hint: Use part (f) of the previous exercise.

Exercise 21. (a) Hint: What obvious subgroup of Sn has (n — 1)! elements? (b) Hint: Exercise 3.


Exercise 1. Hint: If W1 C W2 U U Wm, then intersect both sides with W1 and use the inductiveassumption.

Exercise 3. Hint: In the second proof of Theorem 12.1.6, replace (pi and i/,,, i = 1,. . . , s,

with Vt,. and a E S,,. Thus the polynomial '11(x) from (12.5) will be a sum of n! terms.Show that 111(x) E F [x] by arguments similar to those used in the proof of Proposition 5.2.1 inSection 5.2. Be sure that your argument explains where the separability of s(y) is used.

Exercise 4. Hint: Let g be the minimal polynomial of /3, and let M be a splitting field of fgover F.

Exercise 6. (d) Hint: Use (8.3).

Exercise 9. Hint: Use Exercise 14 and F C KflL C K.

Exercise 12. (b) Hint: Use what you did in Exercise 10.

Exercise 15. (a) Hint: See the proof of Theorem 7.2.7.



Exercise 1. Hint: We are in characteristic 0.

Exercise 5. (a) Hint: A nonzero polynomial in one variable of degree N has at most Nroots in afield. Also, when n> 1, note that g = . . . where at least onegj E F[xi,.. . is nonzero.

Exercise 6. (b) Hint: Use (12.7). (c) Hint: Use Exercise 4 of Section 9.1.


Exercise!. (a) Hint: Take a E Gal(L/F) and let T = cbi(a). Also, explain why = /37—1(1).

Exercise 5. Hint: If g doesn't split completely over F, then show that the roots of g in somesplitting field are b,u± b,u,v E F and v F2.

Exercise 6. (a) Hint: Consider (x ) (x — c52) and (x — (x —

Exercise 15. Hint: First show that GI = [L: Fl is a power of 2.


Exercise 3. (b) Hint: Let i- = (i1 . . .i5). Show that a(ii) = 1k implies that a(i2) = 1k+1 andso on. Then show that a = (d) Hint: How many 5-cycles are there in S5? (e) Hint:Remember that 5 divides GI.

Exercise 8. (a) Hint: Explain why T u1 = ±Uj for some j, and prove that some evenpermutation takes u to u3. Then use Exercise 7 to determine the sign.

Exercise 9. (a) Hint: See the proof of part (a) of Theorem 13.1.1. (e) Hint: Comparing thecoefficients of y5, y4, andy3 gives equations which can be solved using Maple or Mathematicato express b2,b4,b6 in terms of a,b. Then compare the coefficients of y2 and substitute theformulas for b2, b4, b6 to get an equation involving only a and b. Now factor. A differentargument is needed in characteristic 3.

Exercise 10. (c) Hint: Use the method described in the hint to part (e) of Exercise 9. Here,you will need a different argument in characteristic 5.

Exercise 16. (e) Hint: Use Section 7.4.


Exercise 1. (a) Hint: First multiply f by a suitable integer so that ao,.. . , E Z. Then multiplyby


Exercise 5. (b) Hint: Use Galois theory.

Exercise 8. (b) Hint: Let 'p2 = (234) . 'p and = (34) . Show that G1 . 'p = andG1•'p2 = {±'p2,±1p3}. (c)Hint: Study the action of G1 on'p, 'p2, and 'p3, where 'p2 and 'p3are defined in the hint to part (b). (d) Hint: Use the hints to parts (b) and (c).

Exercise 9. Hint: By linear algebra, the map sending a matrix to the corresponding linear mapis one-to-one.


Exercise 10. (a) Hint: If F is a field, then every subspace of F'1 of dimension n — I is definedby an equation aix! + + = 0 where ai, . . . , a, E F are not all zero. Furthermore, thisequation is unique up to multiplication by a nonzero element of F. (c) Hint: If V C F3 hasdimension 2, then it has a basis which can be completed to a basis of F3. Now use part (b).

Exercise 11. Hint: For a field F, let be the subgroup of GL(n, F) consisting of allnonzero multiples of the identity matrix. Then PGL(n, F) = GL(n, F) =SL(n, F)/(F'1l, fl SL(n, F)).

Exercise 13. (b) Hint: Compute isotropy subgroups.


Exercise 4. (a) Hint: Use Section 2.3 to express the universal version of s,, in terms of0i, a2, a3,y. Then specialize to the coefficients off. You will be surprised at the size of thepolynomials involved.

Exercise 6. (a) Hint: If f = gh in Q[xi, . . . then pick positive integers r, s as small aspossible such that rg,sh E Z[xi,. .. ,x,]. Now apply unique factorization to rsf = (rg)(sh),and remember that a prime p E Z is irreducible (in the UFD sense) in Z[xi, . . . ,x,]. (b) Hint:Take an irreducible factorization of in 7L[ui,. . . , u,,,y], and apply part (a) together with thefact that isaUFD.

Exercise 10. (b) Hint: Look at subgroups of S4 with four elements.


Exercise 2. (b) Hint: m o(g) + n IG : H] = I.

Exercise 4. Hint: Remember that g• hH = (gh)H gives an action of G on the set of left cosets

of H in G.

Exercise 8. (a) Hint: By adjoining roots one at time, prove that the splitting field satisfies[L: F] <40. (b) Hint: Regard the roots as the nonzero vectors in and pick roots y suchthat-y=o+/3.


Exercise 2. (a) Hint: Show that S3 S2 contains a transposition.

Exercise 3. (d) Hint: Use part (b) of Exercise 7 of Section 14.3.

Exercise 8. (b) Hint: For a E Gal(L/F), we have r E Sm such that a(R,) = What isthe kernel of the map sending a to r? The Galois correspondence from Chapter 7 will also beuseful.

Exercise 9. (a) Hint: Use Proposition 9.2.8 with f = 1 and f' = f.Exercise 10. Hint: What power of p divides (p2)!?


Exercise 15. Hint: Show that G11n = GI, where G, is the isotropy subgroup of i E {l,. . . ,n}.Given a subgroup C H C G, consider the subsets (i-H) . i, 'r E G. Show that these partition{ 1,... , n} into blocks that are stable under the action of G.

Exercise 16. (b) Hint: Use the congruence classes modulo p in Z/p2Z.



Exercise 2. (c) Hint: Consider the map 'YAv A. (d) Hint: Semidirect products are discussedin the Mathematical Notes to Section 6.4, and (6.10) does the special case when n = 1 andq=p.

Exercise 3. (a) Hint: Consider the map which sends 7Ac,y to O••

Exercise 4. (c) Hint: Use linear algebra.

ExerciseS. (b) Hint: You need to study what happens when N CA x {eB} orN C {eA} x B.

(c)Hint: Conjugate ENby EA x B.

Exercise 6. (a) Hint: Explain why conjugation by g gives an automorphism of N.

Exercise 7. (b) Hint: Consider

Exercise 12. (b) Hint: Use part (c) of Exercise 2.

Exercise 13. (b) Hint: Given two (n — 2)-tuples of elements of { 1,. . . , n} consisting of distinctpoints, show that there are exactly two elements T,r' E S,, which map one (n — 2)-tuple to theother. Then show that T and differ by a transposition.

Exercise 14. (a) Hint: A 2 x 2 matrix has determinant 0 if and only if either its first column iszero or its first column is nonzero and the second column is a multiple of the first. (e) Hint:Apply the Fundamental Theorem of Group Actions to GL(3, F2) acting on \ {0}.

Exercise 15. Hint: First show that has p + 1 lines through the origin, and explain whyPSL(2, permutes these lines. Then study the cases p = 2 and p = 3 in detail.

Exercise 16. Hint: In the proof of Proposition 14.3.10, replace A and its conjugates with theminimal normal subgroups of G.

Exercise 17. Hint: Use induction and the action of GL(n, F \ {0}.




Exercise 9. (a) Hint: Use Lemma 14.4.3. (b) Hint: To analyze the kernel, write m E Ker(çb)as a linear combination of '2 and g. For the image, use the Fundamental Theorem of GroupHomomorphisms. (d)Hint: Forg = showthatC(g) = r,s E r2 +s2 0}.Then use the Fundamental Theorem of Group Homomorphisms to determine the size of thekernel.

Exercise 12. (c) Hint: The multiplicative group of a finite field is cyclic. (d) Hint: The onlyhard case is when p = 7. Show that M2 C M3 implies that M2 maps to A4 C S4 when you mapto PGL(2,F7). Then use part (a). (e) Show that the image of (M2)o in PGL(2,F5) has order 8and contains the subgroup H of Lemma 14.4.4. Recall that subgroups of index 2 are normal.

Exercise 13. Hint: Consider the map F; x G0 —+ defined by (A,g) '—* Ag.



Exercise 9. (a) Hint: Gauss's Lemma implies that P(u),Q(u) are relatively prime inThus R(u)P(u)+S(u)Q(u) = 1 for some R(u),S(u) e Q[u]. Now replace u with u4.

Exercise 11. Hint: Let a = sinx and = siny.


Exercise 2. (b) Hint: Use Proposition 15.2.1 and part (d) of Exercise 3 of Section 15.2.

Exercise 6. Hint: Use the identity (15.13) with x = (z + zo) and y = (z zo). See also[Abel, Vol. I, pp. 277—278].

Section 15.4 (page 503—504)

Exercise 3. Hint: See the proof of (a) (b) of Proposition 3.1.1. Also recall if a E Z[i] isprime, then a = /9'y, f3,-y E Z[i], implies thatfl or-y is a unit, i.e., lies in

Exercise 9. Hint: See the hint for Exercise 9 of Section 15.2.

Exercise 12. Hint: Follow the proof of Theorem 4.2.3. You will also need Gauss's Lemmaover Z[i], which holds by Theorem A.5.8 because Z[i] is a UFD.

Exercise 13. Hint: Use the geometric series 1/(1 + x) = 1 to write

l+alu+...+adud

The series on the right-hand side makes sense because

(alu+ ... + aaud)Ic= uk (ai+ +

a given power of u appears in only finitely many terms.

Exercise 14. (d) Hint: For Co = 1, use = 1, and for Ci = use = I —

with = ... and = 1 + Sciz4 + .. What is the coefficient of z4?


Exercise 1. Hint: Use the fact that is a PID to prove that if a, fi are relatively prime, thenya+6/3= lforsomey,öEZ[i].

Exercise 5. Hint: Use part (b) of Lemma 15.4.2.

Exercise 6. Hint: Show that the obvious map Z[i]/af3Z[i] —+ x 7L[il//37L[i] isone-to-one. Then use part (a) of Lemma 15.4.2.

APPENDIX C

STUDENT PROJECTS

The material in the latter part of the book lends itself well to independent projects. Inthis appendix, we suggest some topics that students might find interesting. Most ofthe projects listed here are reasonably short, though a few are more ambitious. Manyare based on optional sections of the text. Here is the list:

• Abelian Equations. These equations and their relation to Abelian groups arediscussed in Sections 6.5 and 8.5. The goal would be to explain why commutativegroups are called "Abelian." A more ambitious version of this project wouldinvolve looking at the Historical Notes to Section 15.5.

• Automorphisms and Geometry. There are several projects involving Section 7.5:

Theorem 7.5.3 gives a classic argument from Galois theory and leads tosome nice geometric examples of Galois groups.Another project would be to study linear fractional transformations andstereographic projection.More ambitious projects would be to classify finite subgroups of PGL(2, C),explore some invariant theory, or give a proof of Lüroth's theorem. See theMathematical Notes to Section 7.5.


552 STUDENT PROJECTS

• The Casus Irreducibilus. As explained in Section 1.3, Cardan's formulas for acubic with real roots involve complex numbers. The goal of this project wouldprove that complex numbers are unavoidable, as explained in Section 8.5.

• Gauss and Roots of Unity. Here are two projects based on Section 9.2:o A project could use Gauss's theory of periods to work out the Galois corre-

spondence for a cyclotomic extension Q c p prime.o Another project would be to focus on the properties of periods and derive

Gauss's amazing formula for cos(2ir/17) given in (9.19).

• Squarable Lunes. Read references [18] and [20] from Section 10. ito learn moreabout the squarable lunes mentioned in the Historical Notes to the section.

• Regular n-gons. Theorem 10.2.1 characterized the n's for which a regular n-goncan be constructed by straightedge and compass. The proof used the irreducibilityof the cyclotomic polynomial 1,, (x), which is not easy to prove. A nice project,based on Exercises 2—6 of Section 10.2, is to give a more elementary proof ofTheorem 10.2.1 that uses the Schönemann—Eisenstein criterion.

• Origami. Section 10.3 leads to several possible projects:o Use origami constructions to trisect angles and duplicate the cube.o There is also the Galois theory of origami, presented in Theorem 10.3.6.o A student could also focus on Exercise 18 of Section 10.3, which character-

izes the n's for which a regular n-gon can be constructed by origami.The references for Section 10.3 can be used for other projects involving origami.

• Polynomials over Finite Fields. There are two projects based on Section 11.2:o The first would be to prove the formula for the number of monic irreducible

polynomials in given by Theorem 11.2.4. See (11.10) and Exercise 8.o The second would be to study cyclotomic polynomials modulo a prime p

and, following Exercise 17, give a proof of the irreducibility of thatdoes not use the Schönemann—Eisenstein criterion.

• Lagrange. Section 12.1 lends itself to several projects:o It is fun to work out the solution of quartic given in (12.11) (Ferrari) and

(12.17) (Euler). Exercise 18 is relevant.o One project would be to explain how Lagrange's formula for the degree of a

resolvent (Theorem 12.1.4) relates to Lagrange's Theorem in group theory.o A student could explore Lagrange's version of Galois correspondence for

the universal extension. See Theorems 12.1.6 and 12.1.9 and Exercise 9.o A nice project would be to explain how the affine linear group AGL( 1, F,,) is

implicit in Lagrange's work. This involves a study of the Lagrange resolvents(12.19). See also Exercise 15.

o Finally, a student could explain how Theorem 12.1.10 messes up Lagrange'sinductive strategy for finding the roots of a polynomial.

• Galois. Here are some projects based on Section 12.2:o A student could summarize Lagrange's approach from Section 12.1 and

explain how Galois went beyond Lagrange.

STUDENT PROJECTS 553

o A good project is to explain how Galois thought about his group and how itrelates to the modem notion of Galois group. See Theorem 12.2.3.

o Another project is to explain Galois's strategy for finding radical expressionsfor roots. See Example 12.2.6 and Exercises 7 and 8.

• Kronecker. We proved the existence of splitting fields in Theorem 3.1.4, and inthe Historical Notes to Section 3.1, we gave credit to Kronecker. A project basedon Section 12.3 would give Kronecker's construction of splitting fields. Thisrequires Exercises 4—8 plus the Galois resolvents defined in Section 12.2.

• Quartic Polynomials. A student could report on the Galois group of quartics,following Section 13.1. A more ambitious version of the project would includematerial from Section 13.3 on quartics in all characteristics.

• Quintic Polynomials. Similarly, a student could report on the Galois group ofquintics, following Section 13.2. A more substantial project would be to studythe roots of quintics that are solvable by radicals, using the references mentionedin the Mathematical Notes to Section 13.2.

• Computing Galois Groups. Here are more projects based on Chapter 13:o A student could study how resolvents relate to Cardan's or Ferrari's formulas

for the roots of a cubic or quartic and also to the Galois group of a quartic orquintic. The centerpiece of this project would be Section 13.3.

o One project would be work out the details of why GL(3,lF2) is the Galoisgroup of x7 — 154x +99 over Q. This involves Proposition 13.3.9 andExample 13.3.10.

o A computer project would be to determine various Galois groups over Q byfactoring modulo p for various primes p, as explained in Section 13.4.

• Polynomials of Prime Degree. Section 14.1 is a lovely continuation of the ideasof Chapter 8. Theorem 14.1.1 on solvable polynomials of prime degree is one ofthe great theorems of Galois. This is a very accessible project.

• Solvable Permutation Groups. Chapter 14 has lots of material for projects:o A student could explore how wreath products first arose in the context of

Galois theory, following the Historical Notes to Section 14.2.o A student could discover why Galois invented finite fields, explained in the

Historical Notes to Section 14.4.o A student could explore the structure of primitive solvable permutation

groups, as described in Section 14.3. The math is surprisingly deep.o A nice project would be to classify imprimitive solvable permutation groups

of degree p2, p prime, based on Section 14.2. This is surprisingly easy.o A harder project would be to classify primitive solvable permutation groups

of degree p2, p prime, based on Section 14.4.

• The Lemniscate. Abel's theorem about geometric constructions on the lemniscateinvolves a rich combination of Galois theory, complex analysis, and number theory.A report on Abel's theorem in Section 15.5 could be the basis for a nice project.A more ambitious project would be to digest all of Chapter 15.

REFERENCES

The references are divided into three categories and are cited in the text using the last nameof the author. Each chapter has its own references cited numerically.

A. Books and Monographs on Galois Theory

[Artini E. Artin, Galois Theory, Univ. of Notre Dame Press, Notre Dame, 1942. Reprint byDover, New York, 1998.

[Chebotarev] N. 0. Chebotarev, Grundzüge der Galois'schen Theorie von N. Tschebotaröw,translated and edited by H. Schwerdtfeger, Noordhoff, Groningen, 1950. Note: Thecorrect English transliteration is "Chebotaryov." We use "Chebotarev" because it is theversion most often used in the literature.

[Dehn] E. Dehn, Algebraic Equations, Columbia U. P., New York, 1930.

[Dickson] L. E. Dickson, Introduction to the Theory of Algebraic Equations, Wiley, NewYork, 1903. Reprinted in Congruence of Sets and Other Monographs, Chelsea, NewYork, 1967.

[Edwards] H. M. Edwards, Galois Theory, Springer, New York, Berlin, Heidelberg, 1984.

[Escofier] J.-P. Escofier, Galois Theory, Springer, New York, Berlin, Heidelberg, 2001.

[Garling] D. J. H. Garling, A Course in Galois Theory, Cambridge U. P., Cambridge, 1986.

[Hadlock] C. R. Hadlock, Field Theory and Its Classical Problems, Carus Monographs,Volume 19, MAA, Washington, DC, 1978.

Galois Theory. Second Edition. By David A. Cox 555Copyright © 2012 John Wiley & Sons, Inc.

556 REFERENCES

[Jordan I] C. Jordan, Traité des substitutions et des equations algébriques, Gauthier-Villars,Paris, 1870.

[Postnikovl M. M. Postnikov, Fa.nya [Foundations of Galois Theory],Gosudarstv. Izdat. Fiz.-Mat. Lit., Moscow, 1960. Enlarged edition, 1963. English trans-lations of the 1960 edition by Hindustan Pub. Corp., New Delhi, and Gordon & Breach,New York, 1961; Pergamon Press, Oxford, 1962; and Noordhoff, Groningen, 1962.Unpublished English translation of the additions to the 1963 edition by A. Shenitzer.

[Stewart] I. Stewart, Galois Theory, Third Edition, Chapman & Hall/CRC, Boca Raton, FL,2003.

[Swallow] J. Swallow, Exploratory Galois Theory, Cambridge U. P., Cambridge, 2004.

[Tignol] J.-P. Tignol, Galois' Theory ofAlgebraic Equations, Longman Scientific and Techni-cal, Harlow, England, and Wiley, New York, 1988. Corrected reprint by World Scientific,Singapore, 2001.

B. Books on Abstract Algebra

[Grillet] P. Grillet, Algebra, Wiley, New York, 1999.

[Herstein] I. N. Herstein, Topics in Algebra, Second Edition, Wiley, New York, 1975.

[Jacobson] N. Jacobson, Basic Algebra, Volumes I and II, W. H. Freeman, San Francisco,1980.

[van der Waerden] B. L. van der Waerden, Algebra, Volumes I and II, Springer, New York,Berlin, Heidelberg, 1991.

C. Collected Works

[Abel] N. H. Abel, (Euvres completes de Niels Henrik Abel, edited by L. Sylow and S. Lie,Grøndahl & SØn, Christiana, 1881.

[Cauchy] A. L. Cauchy, (Euvres completes, Series 1, Volume X, Gauthier-Villars, Paris, 1897.

[Eisenstein] F. G. Eisenstein, Mathematische Werke, Volume II, Chelsea, New York, 1975.Reprint by Amer. Math. Soc., Providence, RI, 1989.

[Galois] E. Galois, Ecrits et mémoires mathématiques d 'Evariste Galois, edited by R. Bourgneand J.-P. Azra, Gauthier-Villars, Paris, 1962.

[Gauss] C. F. Gauss, Werke, K. Ges. Wiss., Göttingen, 1863—1927.

[Jordan2] C. Jordan, cEuvres de Camille Jordan, Volume I, edited by J. Dieudonné, Gauthier-Villars, Paris, 1961,

[Kronecker] L. Kronecker, Werke, Leipzig, 1895—1931. Reprint by Chelsea, New York, 1968.

[Lagrange] J. L. Lagrange, (Euvres de Lagrange, Volume 3, Gauthier-Villars, Paris, 1869.

INDEX

This index uses the following conventions:• Page numbers of definitions are underlined.• Page numbers of theorems are in bold.

A

see group, alternatingAGL(n, F), see affine linear groupA['L(n, F), see affine semilinear groupAbe, H., 284Abel, N. H., 80, 87, 143—144,217, 273, 418, 440,

463—464,467, 474, 480,482, 487—489,502, 510—511, 519, 556

unsolvability of quintic, 200, 214, 220, 330

Abelian equation, l44—145, 217—218, 551relation to Abelian group, 144-145, 218, 511relation to the lemniscate, 511

Abelian group, see group, AbelianAbel's function, xxvii, 467—468, 470—471, 472,

480, 501, 506, 510addition law for, 473, 480-483, 484—485,489,

492analogy with sinx, 468, 472—473,480, 482complex multiplication formulas for, 489,

491-492,494—498,504—505,508,512according to Abel, 502, 510

Galois Theory, Second Edition. By David A. CoxCopyright © 2012 John Wiley & Sons, Inc.

according to Eisenstein, 502for odd Gaussian primes, 498, 502

relation to Abelian equations, 510-511doubly periodic, 482, 484—486

period lattice, 484, 487multiplication formulas for, 476, 479, 489, 491,

494, 511

doubling, 474,476, 490, 507quintupling, 479, 497tripling, 476,497

over C, xxviii, 482—486, 489

power series of, 499

real period of, 472subtraction law for, 474,481zeros and poles of, 485—486

Abel's Theorem on the Lemniscate, 463—464, 467,470,482, 490—491,497, 504, 506, 509

Abu Ali Hasan ibn al-Haytham, 283accessory irrationality, 339, 341action ofagroup, 48, 318, 530

transitive, 135, 532See also Fundamental Theorem of, Group

Actions

557

558 INDEX

addition lawfor Abel's function, see Abel's function,

addition law forfor elliptic integrals, see elliptic, integral,

addition law forfor sinx, 473, 482

adjunction, 75—76, 80, 348See also extension, field

affine linear group, xxvii, 430—431,435,437,441,458

according to Galois, 417, 440one-dimensional, xxvi, 137, 140—142, 167, 186,

194, 329, 369, 37 1—373, 376—377,381—386, 399, 414—417,424,429,444—445, 457, 459, 552

order of, 440, 443translation subgroup of, 430,432, 434two-dimensional, '141 145, 450—451, 454,

459—460affine linear transformation, 137, 141,430affine semilinear group, xxvii, 431, 441

according to Galois, 440one-dimensional, 444, 453, 457—459

affine semilinear transformation, 431algebraically

closed, see field, algebraically closedindependent, 37, 40, 79—80

algebraicclosure, 97coding theory, 297geometry, 29integer, 97number, 96number theory, 79, 85, 97, 366, 509over a field, 73, 89quantity, xxvii, 348—350, 353

algebraische Grösse, 348algebra over afield, 93algorithm

Berlekamp's, 305, 308—309division, 89, 305—306, 522—523

refined, 233—234, 236Euclidean, 81, 523for computing Galois groups, 386—389, 402

according to Maple, 389, 396according to Jordan, 386according to Kronecker, 401—402

for factoring polynomials, 83, 349Alhazen, see Abu All Hasan ibn al-HaythamAlperin, R. C., 288analytic function, 483Apollonius, 283Archbold, J. W., 288Archimedes, 282—283

See also spiral of Archimedesarc length, 272, 466, 470—471,479

signed, 471arrangement, 251, 343, 346

See also permutation, according to GaloisArs Magna, 9, 19Artin, E., xix, 67,94, 119, 129,218—219,225,316,

555Artin symbol, 407Aschbacher, M., 461ASCII, 297automorphism, see field, automorphismAuckly, D., 288Ayoub, R. G., 227, 356, 513

B

Babylonians, 8Bachman, G., 254Baragar, A., 282, 288Bashmakova, I. G., 23, 70, 227, 356Berlekamp, E., 308, 310Berndt, B. C., 410Bernoulli, Jacob, 469,479Bernoulli, Johann, 470Bertrand, J., 331Betti,E.,l60Bézout, E., 329binomial theorem, 4, 30, 85, 306, 527blocks, 419Bombelli, R., 9, 17, 19, 22Brent, R. p., 310Bring, E., 381Bring—Jerrard quintic, see quintic polynomial,

Bring—Jerrard formBnoschi quintic, see quintic polynomial, Brioschi

formBrown, E., 462Bruin,N., 396,410Buchmann, J. A., 254Burgess, D. A., 248, 254Burnside, W., 194, 227, 410Burnside's Theorem, 194—196, 279Butler, M. C. R., 308, 310

C

C, see field, of complex numbersC, xxvi, 180

see field, of constructible numbersCannon, J. J., 410Cantor, G., 354Capelli, A., 87Cardan, G., 9, 19, 23, 61, 329Cardan's formulas, 7, 197, 217, 239, 277,

325—326, 537via Galois theory, 207—208, 210

Carter Edwards, B., 288Cartesian product, xxiii

INDEX 559

Cassini, G. D., 469casus irreducibilis, 17, 19, 87. 220, 222, 226, 265,

552Cauchy, A. L., 50—51,56—57,60,62,93,330—331,

353, 556Cauchy—Riemann equations, 482—483,489Cauchy's Theorem, 139, 223, 415, 517, 543Cayley,A., 171,382Cayley—Hamilton Theorem, 447, 454, 459Cayley table, 170—172, 434centralizer of an element, 427, 446,451—455,

460-461centralizer of a subgroup, 447, 459Chain Rule, 473, 480Chebotarev, N. G., 267, 555Chebotarev Density Theorem, 407—408Chevalier, A., 440, 459Chinese Remainder Theorem, 533

over the Gaussian integers, 512Chrystal, G., 53class field theory, 407, 502, 509, 512Cleveland, J., 288code word, 297coding theory, 297

algebraic, 297Cohen, H., 98, 410compass, 264

Euclidean, 264See also straightedge, and compass

complex conjugation, 63, 127, 139See also complex number, conjugate of

complex multiplication, 145, 407, 467, 475,489—490, 501, 503—504,509—510,512

See also Abel's function, complexmultiplication formulas for

complex numberabsolute value of, xxiv, 524addition of, 524conjugate of, xxiv, 524imaginary part of, xxiv, 524multiplication of, 524—525nth root of, 525—526polar representation of, xxiv, 525real part of, xxiv, 524

See also field, of complex numberscomposition

factors, 214series, 194,214—215

conchoid of Nicomedes, 282—283, 287congruence class, 516, 533conic, 280

dual, 282nondegenerate, 281

conic section, 283See also conic

conjugacy class, 407—409, 532

conjugatecomplex, see complex, conjugatefield, see field, conjugatesubgroup, see subgroup, conjugate

Conrad, K., xxi, 390, 410constructible number, 256

See also field, of constructible numbers andstraightedge, and compass

Conway, J., 410Coolidge, J. L., 288coset, 57—58, 60

left, xxiv, 516, 531of an ideal, 519representative, 60, 333right, xxiv, 516

Cox, D., 53, 70,98, 123, 310,410,513cubic polynomial, 4, 329—330

Cardan's solution of, 7See also Cardan's formulas

Galois group of, 169, 357origami solution of, 275, 279overR, 15

See also casus irreducibilisreduced, 4trigonometric solution of, 18, 20

cubic resolvent, see resolvent polynomial, cubiccycle, 518

multiplication of, 518cycle type, 407, 409cyclotomic

extension, see extension, cyclotomicpolynomial, see polynomial, cyclotomic

D

see group, dihedralD'Alembert, J.-R., 61Dedekind, R., 80, 93, 99, 128—129, 146, 236, 299,

310, 349, 353, 400,404, 512Dedekind's Theorem, 404, 407degree

of an extension, xxv, 89of a polynomial,separable, 116total, 26,41transcendence, 40

Dehn, E., 555del Ferro, S., 9, 329Delian problem, see duplicating the cubeDe Moivre, A., 377, 381De Moivre's formula, 527derivative, see formal derivativeDescartes, R., 329determinant, 459Dickson, L. E., 299, 310, 356,441,462, 555Diffie—Hellman key exchange protocol, 248Digital Signature Algorithm, 248

560 INDEX

Diophantine equation, 365, 396direct product, 519directrix, 280, 283

See also parabola, directrix ofDinchiet, P. G. L., 310discrete log, 248discriminant, 23, xxv, 46, 48, 352—355

as a resultant, 115cubic, 8,12, 15, 41,46, 207of a monic polynomial, xxv, 47, 110, 167—168quartic, 52, 358—359quintic, 371, 373

Disquisitiones Arithmeticae, see Gauss,Disquisitiones

Divaani-Aazar, K., xixdividers, 264division algorithm, see algorithm, divisionDixon, J. D., 462Dobbs, D. E., 53Dobson, E., 462Dorodnov, A. W., 267Dorwart, H. L., 99doubly transitive, see permutation group, doubly

transitiveDoyle, P., 368, 410Dresden, G. P., 254Dummit, D., 410Dunham, W., 356duplicating the cube, 262, 266—267, 269, 283Dürer, A., 287

E

eccentricity, 280. 283Edwards, H. M., xix, 349, 354, 356, 462, 555eigenvalues, 441Eisenstein, F. G., 87, 464, 467, 498—499, 502, 512,

556Eisenstein criterion, see Schönemann—Eisenstein

criterionelastic curve, 469Elkies, N. D., 396, 410ellipse, 280, 283, 470

relation to elliptic integrals, 468elliptic

curve, 366, 380, 396,488, 508—510addition law for, 488torsion points of, 509

function, 145, 368, 468, 487—488, 501according to Abel, 480,488, 510according to Jacobi, 480, 488

integral, 468,479,488, 502addition law for, 473, 479modulus of, 480, 482

Emert,J.W., 288entire function, 350

See also polynomial

Erbach,D.W.,410error-correcting codes, 297Escofier, i-P., 555Euclidean algorithm, see algorithm, EuclideanEudoxus, 283Euler, L., 61, 64—65, 67—68, 79, 97, 237, 271, 297,

329—330, 356, 377, 381,473-474,479Euler 4k-function, xxiv, 75, 230, 237, 270—271,

304, 310evaluation map, 26

See also homomorphism, ringEves, H., 288exponent vector, 39expressible

by radicals, 215—217by real radicals, 221, 223—224, 265, 362

extensionAbelian, 509algebraic, 95cyclotomic, 132, 229, 270, 509, 552

Galois group of, 235, 506degree of, see degree, of an extensionfield, xxv, 58, 73, 344finite, 89, 95lemniscatic, 504—505, 508

Galois group of, 505, 509normal, 108, 130, 148, 155of groups, 141,214problem in group theory, 141, 214purely inseparable, 117, 123, 165purely transcendental, 79radical, 197—200, 204, 206, 339, 543real radical, 221, 223separable, 111, 116, 149—150, 165, 199solvable, 197, 199, 204, 206, 226solvable by radicals, 197, 325, 339totally ramified, 85universal, see universal, extension

F

]Fq, see finite field

F, xxvi, 180

factoring polynomialsover a number field, 8 1—84over Q, 81, 83—84

See also algorithm, for factoring polynomialsusing Maple, see Maple, factorusing Mathematica, see Mathematica, Factor

Fagnano, C. 0., 273, 479, 481Faltings, G, 396Felt, W., 195, 227Felt—Thompson Theorem, 195, 215Fermat, P., 79, 271, 297Fermat

number, xxvi, 271, 274prime, 270—271, 274, 464, 506—507

INDEX 561

Fermat's Last Theorem, 97Fermat's Little Theorem, 230, 271

Euler's generalization, 237Ferran, L., 9,61Ferrari resolvent, see resolvent polynomial, FerrariFerrari's formulas, see quartic polynomial,

Ferrari's solution ofFibonacci, 9, 23field, 348, 520

algebraically closed, 67, 69, 97, 219, 225automorphism, 119, 125characteristic of, 521compositum of, xxvi, 198, 200—201conjugate, xxvi, 154, 159, 161, 198, 325—326extension, see extension, fieldfinite, see finite fieldfixed, xxvi, 147formally real, 67history of, 79imaginary quadratic, 501, 509intermediate, 149, 162—163, 165isomorphism of, see isomorphism, fieldmultiplicative group of, 533number, 79, 81, 509of algebraic numbers, xxv, 96—97of characteristic 0,42,111—112,117,121, 144,

199, 201, 215, 316, 354, 377, 413,425,436, 457, 522, 547

of characteristic p.111, 113—1 14, 117, 121,165, 224—225, 291, 355, 522

of complex numbers, xxiii, 55—56, 60, 219,520,524

of constructible numbers, xxvi, 257, 259, 261,263, 268, 362, 465, 467, 474,476,478—479,506—507, 512

of fractions, 26, 62, 520of ongami numbers, xxvi, 277—278, 280-281of Pythagorean numbers, xxvi, 265, 268of rational functions, xxv, 26, 80

in one variable, 175of rational numbers, xxiii, 520of real numbers, xxiii, 520ordered, 69Pythagorean, 265, 268ray class, 509real closed, 67, 69splitting, 60, 101, 130

according to Kronecker, 350, 352, 400existence of, 102uniqueness of, 105

Fine, B., 70Fineman, J., xixfinite field, xxv, xxvii, 121, 291, 344,404,458,

491, 521, 552equations over, 296existence of, 292, 298

Galois correspondence of, 296Galois group of, 294—295, 298,404, 431, 440,

453,455history of, 297—298, 300, 307, 439, 553irreducible polynomial over, 301, 306

number of, 301—302, 308multiplicative group of, 123, 298uniqueness of, 292

Fischer, J., 410Florido, 9focus, 280, 283

See also parabola, focus ofFontana, N., see Tartagliaformal derivative, 110Frobenius complement, 417Frobenius group, 417—418Frobenius homomorphism, xxvii, 114, 295, 404,

407Fueter, R., 512Fundamental Theorem of

Algebra, 31, 51, 61—62,64,66,68—69,97, 139,330, 351, 355

via Galois theory, 218, 263Calculus, 473Galois Theory, 161, 163, 169, 171Group Actions, 161, 186, 319, 368, 415, 430,

531, 549Group Homomorphisms, 140, 179, 517, 549Ring Homomorphisms, 77, 292,520Symmetric Polynomials, 13, 31,36, 38, 316,

352, 400over the integers, 234, 236, 405

G

GL(n, F), see general linear groupGal(L/F), see Galois, groupGalois, E., 14, 62, 79—80, 108—109, 121—122, 128,

135, 141, 145, 152, 160, 208, 214,220,238, 251, 253, 297, 299—300, 331,334—337, 343—344, 347, 350, 353—354,357, 370, 382, 413, 417, 426, 439—440,458, 552, 556

strategy, 209, 341—343Galois

closure, 151—152, 198—199,542—543correspondence, 159, 163, 172, 205, 209, 219,

238, 249, 279, 296, 321, 323, 344,542,546, 548

See also Fundamental Theorem of, GaloisTheory

extension, 131,149—150, 152field, 293group, xxvi, 404, 407

according to Galois, 337, 339, 346, 553algorithm for computing, see algorithm, for

computing Galois groups

562 INDEX

of an extension. 125—126, 130See also extension, cyclotomic and extension,

lemniscatic and finite field and universal,extension

of a polynomial, 128, 135, 216See also cubic polynomial and quartic

polynomial and quintic polynomialspecialization of, 139

resolvent, see resolvent polynomial, GaloisGAP, 396, 410Garling, D. J. H., 555Gauss, C. E, 31, 39, 50—51, 61, 68, 87, 99, 145,

238, 242, 245, 254, 270, 288, 298, 307,311, 330, 347, 350—351, 353, 355, 426,464,480,482,489, 502, 512—513,527—528, 556

mathematical diary, 236, 272, 296Disquisitiones, 87, 235, 239, 246, 248—251,

272, 297,510Gaussian integers, xxviii, 489—490, 502—503, 550

associate, 491even, 491,493—494norm of, 491—493,496—498, 500—501, 503, 507odd, 491—495, 498, 500, 502—503, 505,

507—508, 512primes of, 491,498, 500—504, 507, 512units of, 491,495

Gauss's Lemma, 83, 234—237, 398, 528—529, 535,540, 550

Gauss sum, 249gcd, see greatest common divisorgeneral linear group, xxvii, 296, 396, 530, 548

as vector space automorphisms, 453—454, 461irreducible subgroup of, 436, 450—452over a finite field, 430,438, 441,444,456, 549

solvable subgroups of, 457subgroups of, 461

three-dimensional, 393—394, 396, 438, 549two-dimensional, xxvi, 178, 185—186, 443—447,

450—453, 457,460, 549genus, 396geometric construction, see compass and marked

ruler and ongami and straightedgeSee also duplicating the cube and squaring

the circle and trisecting the angleSee also lemniscate, geometric constructions

on and regular, n-gon, constructiblegeometric series, 550Girard, A., 38, 61Girstmair, K., 410Glashan, J. C., 382Goldbach,C.,271Goldschmidt, D. M., 311Goppa code, 297Gorenstein, D., 462Gray, J. J., 254, 288, 311

greatest common divisorof integers, xxiv, 478, 503of polynomials, 110, 112, 292—293, 301,

306—307, 309, 477, 523Grillet, p., 556Gröbner basis, 45group, 139, 515

Abelian, 428, 435, 506, 511, 519, 542relation to Abelian equation, 144—145, 218,

511structure of, 519

action, see action of a groupaffine linear, see affine linear groupaffine semilinear, see affine semilinear groupalmost simple, 439alternating, xxiv, 49, 135, 168, 170, 183, 211,

331, 358, 369, 373, 376, 380, 384, 391,399,406,432, 518

center of, 196, 456, 461cohomology, 141cyclic, 135, 183, 187, 428, 517, 533decomposition, 408dihedral, xxiv, 135, 164, 177, 183, 187, 320,

359, 369, 391, 423, 519extraspecial, 461Frobenius, 417-418Galois, see Galois, groupgeneral linear, see general linear grouphomomorphism of, see homomorphism, groupinertia, 408isomorphism of, see isomorphism, groupKlein four-, 131, 187, 359, 391Mathieu, 437metacyclic, 141—142of automorphisms of afield, 173of prime power order, 194, 196, 218, 263of rotations of the sphere, xxvi, 181order of, 515permutation, see permutation groupproduct of, 135, 141—142,519projective linear, see projective linear groupprojective special linear, see projective special

linear groupquaternion, 135, 461quotient of, see quotient, groupsimple, 194, 210—212, 331, 394, 397,432,

438—439,441—443socle of, see socle of a groupsolvable, 191—192, 194—196, 204—205, 209,

212,216,218—219,279,329,344,371,425, 437, I'll '115, 450, 457, 459, 506

See also permutation group, solvablespecial linear, see special linear groupsporadic, 437subgroup of, see subgroupsymmetric, see symmetric group

INDEX 563

symmetryof a geometric figure, 184, 186, 368, 427,

449, 532, 542of a polynomial, xxvii, 371, 387, 397of a set, xxvii, 433

Guy, R. K., 254

H

Hack, F., 366Hadlock, C. R., 555Hall, M., 462Hamilton, W. R., 55, 61Hanks, R., 53heptagon, see regular, heptagonHermes, J., 272Hermite, C., 98Herstein, I. N., 556Hubert, D., 67, 171Hilbert Irreducibility Theorem, 140, 171Hippias of Elis, 267Hippocrates of Chios, 267Hoffman,M.J.,5l3Holder, 0. L., 214, 222, 226-227Holt, D. F., 410homomorphism

group, 516image of, xxiv—xxv, 516, 520kemel of, xxiv—xxv, 516,520ring, 26, 58, 520, 539

Houzel, C., 513Hudde, J., 329Hull, T., 284, 289Hulpke, A., xix, xxi, 410Huppert, B., 458—459, 462hyperbola, 280, 283

icosahedron, 184, 187, 368, 380ideal, 29, 300, 349, 353—354, 519

maximal, 57, 521principal, 519, 523

generator of, 519ideal number, 353identification, 58—59identity map, xxividentity matrix, 179, 396, 438Identity Theorem, 483imprimitive

permutation group, see permutation group,imprimitive

polynomial, see polynomial, imprimitiveindex of a subgroup, see subgroup, index oflino, A., xxiimage, xxivIntegritäts-Bereich, 349

See also ring, integral domain

Intermediate Value Theorem (IVT), 63, 66, 69,539, 543

intersection ofconics, 279, 281—282, 286

relation to origami, 281parabolas, 267, 269, 283

invariant theory, 184, 551inverse function of an integral, 468, 480,488inverse Galois problem, 170, 267

overQ, 171—172inverse image, xxivIreland, K., 254, 311, 513lsaacs, I. M., 222, 227Ishibashi, Y., 410Islamic mathematicians, 8, 283isomorphism

field, 103, 105, 125, 130group, 517ring, 77, 520

J

Jacobi, C. G. J., 47, 51, 382, 480,487-488Jacobson, N., 556Jensen, C. U., 188Jerrard,G.,381Jones, G. A., 188, 513Jordan canonical form, 441Jordan, C., 134—135, 14.4—145, 214, 218, 236, 331,

343, 418,441,459,556strategy, 386

Jordan—HOlder Theorem, 194, 214

K

k-transitive, see permutation group, k-transitive

Kajiwara, T., xxiKappe,L.-C.,411

Kearnes, K., 248, 254Khayyam, 0., 9Kiernan, B. M., 99, 146, 188, 227, 356King, R. B., 411Klein, F., 185, 188,411Klein four-group, see group, Klein four-Kleiner, I., 99Kobayashi, S., 411Koblitz, N., 311, 513Kreimer, H. F., 411

Kreuzer, M., 123

Krick, T., xix

Kronecker, L., 62, 80, 87, 93, 144—145, 218, 236,

331, 347, 349, 400—401,407,418, 512,

553, 556

Fundamentalsatz, 62, 350—352Grundzuge, 62, 347, 349—350, 353Jugendtraum, 512

Kronecker—Weber Theorem, 145, 509, 512

564 INDEX

Kummer, E., 353

L

Lagrange, J. L., 14, 37, 50, 62, 68, 79, 122, 141,148, 153, 172,202,209,219,235,297,332, 335—337, 347, 350—351, 353—354,364, 382, 479, 516, 532, 552, 556

Réflexions, 14, 315—316, 329, 331strategy, 246, 251, 323, 325, 332, 341, 382

failure of, 326—327, 329—33 1, 552Lagrange

interpolation formula, 83, 88, 322resolvent, xxvi—xxvii, 203, 207, 209, 247, 251,

318, 328, 333, 552—553Lagrange's Theorem, 133, 206, 210, 3 19—320,

516—517, 552Lang, S., 513Laplace, P. S., 51, 64—65Laurent series, 482, 485Lazard,D.,4l11cm, see least common multipleleading term, see term, leadingleast common multiple, xxiv, 404, 415, 478, 500,

533Ledet, A., 188Legendre, A. M., 297, 479Legendre symbol, 249, 407Lemmermeyer, F.. 99, 513lemniscate, 145, 463, 469—470, 479, 488—489,

503, 508, 5 10-5 11, 553arc length, 466,479

parametrization of, 471signed, see signed arc lengthtotal, 467

Cartesian equation of, 463, 465, 469geometric constructions on, 272, 465, 467, 470,

474, 478—479, 504, 506n-division points of, 46.4—467, 470,

477,479,481-482,506, 511expressible by radicals, 506, 511—512Galois group of, 504, 506, 508

polar equation of, 465—466, 470—471See also Abel's function and Abel's Theorem

on the Lemniscate and extension,lemniscatic

lemniscatic function, see Abel's functionlemniscus, 469Lenstra, H. W., Jr., 289, 411Leonardo of Pisa, see Fibonaccilexicographic order, 39—40

graded, 31, 40, 44L'Hôspital, G. F. A., 283LidI, R., 296-297,311limaçon, 282Lindemann, F. von, 98, 263, 267linear fractional transformation, 178, 551

Little, J., 53, 70, 123Loehr, N., 462Loewy, A., 226lune, 266, 268

squarable, 267, 552Lüroth's Theorem, 184, 551Luther, E., 382

M

M1, xxvii, 444, 451,453—454,457, 459—460M2, xxvii, 445, 451-452, 457, 459—460, 549M3, xxvii, 450—451,455,457, 549MacDonald, I. G., 53

Magma, 396, 411Malfatti, F., 382Malle,G., 188Maple, 44,75,81, 113, 115, 336, 362, 375, 378,

385, 388, 395, 408,428,474,481,540,547alias, 82Basis, 45coeff, 45complex, 538factor, 75, 81f solve, 538galois, 279—280, 389, 396, 423gcd, 113Gcd, 115,293Groebner, 44lexdeg, 44mod, 115NormalForm, 45, 409resultant, 115RootOf, 82subs, 383

Marcus, D. A., 146marked ruler, 266, 279, 283, 286—287

and compass, 282, 385relation to origami, 280-281

See also straightedgeMarsden,J.E.,513Martin, G. E., 264, 277, 283, 289Mathematica, 42,75,82, 113, 115, 336, 362, 375,

378, 383, 385, 388, 395, 403, 408, 474,481, 540, 547

Coefficient, 44Eliminate, 43Extension, 82Factor, 75, 82

JacobiSN, 480

Modulus, 115

PolynomialGCD, 113, 115, 293

Resultant, 115

Root, 83

Solve, 538

SymmetricPolynomials, 43

SymmetricReduction, 43, 409

566 INDEX

cycle type of, 407, 409even, 518odd, 518of the roots, 11, 132—133, 337, 342, 404sign of, xxiv, 47,49, 168, 518

See also arrangement and substitutionpermutation group, 419

doubly transitive, 429—431, 439,441,443—444,446,459

solvable, 458imprimitive, 420—423,459

according to Galois, 426, 429solvable in 424

k-transitive, 437, 443multiply transitive, 437, 443primitive, 420,430—431, 435, 437-439,441,

'144 445, 450,458-459according to Galois, 426solvable in S,,2, 451

regular, 434, 439,442solvable, 427, 433, 435, 458, 553

imprimitive in 424in 5,,, 371, 415primitive in 451

transitive, see symmetric group, transitivesubgroup of

triply transitive, 444Pesic, P., 227PID, see ring, principal ideal domainPierpont, J., 288—289,411Pierpont prime, 288Plato, 283Pohig—Hellman symmetric key exponentiation

cipher, 248point at infinity, 180Poisson, S., 336polar coordinates, 465—466,470—471, 525—526polar distance, 471,477,481—483, 506polyhedra, 184, 187polynomial, 25, 522

/3-division, xxviii, 492,495, 498, 507characteristic, 441, 447, 452,459cyclotomic, xxv—xxvi, 75, 85, 231—232, 237,

262, 426,428,465, 510,544irreducibility of, 234, 236, 271modulo p, 303—305See also root of unity, primitive

elementary symmetric, xxv, 14, 28, 31, 36, 138,234, 316, 329, 382—383

general, 219Galois group of, see Galois, group of a

polynomialhomogeneous, 41imprimitive, 420,422, 457

according to Galois, 426, 428irreducible, 26,57, 81, 134,420, 523, 534

minimal, 74, 111monic, 522n-division, xxvii, 476—477,481—482,489,505,

510See also lemniscate, n-division points of

primitive, 310, 420, 436, 457according to Abel, 440according to Galois, 426,441

reducible, 83ring, see ring, polynomialroot of, 522resolvent, see resolvent polynomialseparable, 109, 130solvable by radicals, 144, 215—218, 224, 238,

246, 344, 371, 377, 385, 425,428, 506,510-512

imprimitive of degree p2, 425, 457, 553of degree p2, 345of degree p, 345, 370, 413, 416—418,553primitive, 436, 440, 458primitive of degree p2. 444, 457, 553

squarefree decomposition of, 117symmetric, 13, 30, 107, 316

See also Fundamental Theorem of, symmetricpolynomials

universal, see universal, polynomialPostnikov, M. M., 289, 556power series, 498, 501, 504power sums, 38,42

See also Newton's identitiesPrasolov, V., 513primitive

permutation group, see permutation group,primitive

polynomial, see polynomial, primitiveprimitive element, 119, 122

according to Galois, 121, 336—337according to Gauss, 239, 241according to Lagrange, 322Theorem of, 119, 121—122, 130, 162, 198,

219—220, 263, 332, 348, 354, 542primitive root

modulo p, 242, 248of a finite field, 298, 310of unity, see root of unity, primitive

principal ideal domain, see ring, principal idealdomain

product, see Cartesian product and group, productof and semidirect product

projective duality, 397projective linear group, xxvii, 396

over a finite field, 438over C, 181—183,449,461three-dimensional, 398two-dimensional, xxvi, 179, 185—186, 443—444,

446—447,450,455, 460-461,549, 551

INDEX 565

matrix algebra, 93Matzat, B. H., 188Maistrova, A. L., 411Mazur, B., xix, 23McKay, J., xix, 53, 410—411McMullen, C., 368, 410Meeks, K. I., 288Menaechmus, 267, 283meromorphic function, 485, 487minimal polynomial, see polynomial, minimalmira, 284mixed equation, 250Möbius

function, xxvii, 237, 302inversion formula, 303, 310

module system, 349, 351See also ideal

modulus, see elliptic, integral, modulus ofMohr—Mascheroni Theorem, 264Mollame, V., 226Mollin, R. A., 254monomial, 26Moore, E. H., 300Mora, T., 188Mordell Conjecture, 396Moreno, C. J., 297, 311Mortimer, B., 462multiply transitive, see permutation group,

multiply transitive

N

nth root, see root, nthnth root of unity, see root of unity, nthNakagawa, H., 411Nakamizo, T., 410natural irrationality, 339, 341

Theorem on, 340—341, 344n-division points

of the circle, 464of the lemniscate, see lemniscate, n-division

points ofNelson, R. B., 288Nemorarius, J., 287Newton, I., 38,46Newton identities, 38, 42, 538Newton's method, 368Nicomedes, 283, 286

See also conchoid of NicomedesNiederreiter, H., 296—297, 311Niven, I., 99, 311normal

closure, 152—153extension, see extension, normalform, 45subgroup, see subgroup, normal

normalizer, xxvi, 159, 161, 167, 370, 414,418,447,451—455,460—461

0

0, see field, of origami numberso(g), see order, of a group elementoctahedron, 181—182,184,186-187,427,449O'Nan—Scott Theorem, 438—439one-to-one correspondence, xxivorbit, 318, 530—531order

of a group, 515of a group element, xxiv, 516

origami, 274, 282, 284, 552number, 276, 278—281

See also field, of origami numbersO'Shea, D., 53, 70, 123Osofsky, B., 188ovals of Cassini, 469

p

.9, see field, of Pythagorean numbers(variant of ir), xxvii, 466

p-function, xxviii, 487-488, 501addition law for, 488, 501complex multiplication for, 501

See also elliptic, function4-function, see EulerPGL(n, F), see projective linear groupPSL(n, F), see projective special linear groupPambuccian, V., 289paperfolding, 277, 284

See also origamiPappus, 282—283, 286parabola, 267, 280, 283

directrix of, 275, 284—285focus of, 275, 284—285intersection of, see intersection of parabolassimultaneous tangents to, 274—275, 282, 285tangent line to, 275, 285

Parry, W., xix, 459Parshin, A. N., 70Pascal, B., 287Pascal, E., 287

See also Iimaconperiod, xxvi, 239—240, 249, 272, 552

expressible by radicals, 246—247, 250Galois action on, 240generalized, 242minimal polynomial of, 241product of, 243relation to Gauss sums, 249

period lattice, see Abel's function, doublyperiodic, period lattice

permutation, 517according to Galois, 338, 343

INDEX 567

projective plane, 283, 397projective special linear group, xxvii, 396, 548

over a finite field, 438simplicity of, 438, 441

three-dimensional, 438two-dimensional, 438, 443

finite subgroups of, 183, 458, 461,551pseudo-random number generator, 310pure equation, 250Pythagorean

field, see field, Pythagoreannumber, see field, of Pythagorean numbersTheorem, 265triple, 365, 367

Q

see field, of rational numbersquadratic

formula, 3,5, 13, 51, 64, 217, 252, 261, 277,475, 527,544

in characteristic 2, 52reciprocity, 87, 249residue, 407

quadratrix, 267—269quartic polynomial, 329—330, 334, 358, 361—362

Euler's solution of, 325—326, 330, 334, 360, 552Ferrari's solution of, 9, 217, 323—324, 325, 552Galois group of, 358, 363, 39 1—392, 553origami solution of, 279

quatemion group, see group, quaternionquaternions, 61quintic polynomial, 184, 330, 368, 553

Bnng—Jerrard form, 377, 379—380, 382, 385equivalent, 380

Brioschi form, 185,380Galois group of, 139, 371, 373, 376, 553icosahedral solution of, 185solvable by radicals, 371, 377, 382

formula for roots, 379unsolvability of, 185, 200, 214,217, 220, 330

quotientgroup, xxiv, 158, 160, 179, 516ring, xxv, 56—57, 60, 300, 349, 519

R

IR, see field, of real numbersradical, 197, 216

extension, see extension, radicalprime, 85

real, 221radix, 216Radloff, I., 188, 356. 462rational function, 26, 316

similar, 320, 323symmetric, 38, 40,52, 169, 316

rational integral algebraic function, 250

See also polynomialRationalitäts-Bereich, 348reduction modulo p. 404, 407

bad, 408regular

heptagon, 278, 283n-gon, 235, 256, 464

constructible, 270, 273, 552via origami, 288

resolvent polynomial, xxvii, 153, 317, 319, 323,330—331, 379, 387—388, 398, 418, 553

approximate, 387cubic, 5, 11,317factorization of, 393Ferrari, xxvii, 52, 324, 332, 334, 358—359,

36 1—362, 364, 386, 388—389, 546Galois, xxvii, 335, 337, 345,347,351,353,400importance of simple roots, 365, 386, 388, 418Kronecker, xxvii, 401Lagrange, see Lagrange, resolventquadratic, xxvii, 390—391relative, 365, 390, 398sextic, xxvii, 371—373, 378, 382, 386, 389

universal, 372restriction of a function to a subset, xxivresultant, xxvi, 115Richelot, F. J., 272Riemann sphere,Rigatelli, L., 356ring, 349, 519

division, 61homomorphism of, see homomorphism, ringintegral domain, 26,

irreducible element, 534isomorphism of, see isomorphism, ringof Gaussian integers, see Gaussian integerspolynomial, xxv, 26, 93, 522, 534principal ideal domain, 29,491, 523, 550quotient of, see quotient, ringunique factorization domain, 26, 59,401,409,

491,495, 520, 524,534,538,540,543,548, 550

unit of, 491, 534Robbiano, L., 123Roberval, G., 288Romanus, A., 20Roney-Dougal, C. M., 462root, 216

multiple, 15, 68, 109, 389—390,496multiplicity of, 109, 523nth, 525—526

See also root of unityof a polynomial, 522

existence of, 59, 61, 351—352, 521simple, 365

568 INDEX

root of unity, 262cube, xxv, 6minimal polynomial of, see polynomial,

cyclotomicin characteristic p, 224—226nth, xxiv, xxvi, 229, 257, 464, 526, 541primitive, 201, 204—206, 231, 236, 304, 543pth, 85, 136, 238

Rose, J. S., 462Rosenberger, G., 70Rosen, M., 254,311,462,513Row, T. S., 284, 289Rudakov, A. N., 70Ruffini, p., 214, 220, 330—331ruler, see marked ruler and straightedgeRunge,C., 382Ruppert, W. M., 385,411

S

SL(n, F), see special linear groupS', xxiv, 530

xxvi, 180S,,, see symmetric groupSamuel, P., 411Schönemann, 1., 87—88, 236, 298—299, 307, 353,

503Schönemann—Eisenstein criterion, 84—85, 88, 92,

136, 139, 271, 273, 398, 552according to Eisenstein, 502over the Gaussian integers, 498, 503—504, 508

Schreier, 0., 67, 119, 225Schur, I., 171semidirect product, 140, 142, 426,428,431,441,

519, 549semilinear group

one-dimensional, 453—455,461semilinear transformation, 453separable

degree, see degree, separableextension, see extension, separableover a field, 111polynomial, see polynomial, separable

Serre, i-P., 146Serret, J., 343sextic resolvent, see resolvent polynomial, sexticShenitzer, A., xixShort, M. W., 462Shurman, J., xix, 188, 288, 411,513Siegel, C. L., 5 13sign of a permutation, see permutation, sign ofsigned arc length, 471signed polar distance, 471Silverman, I. H., 411,514similar functions, see rational function, similarsimple group, see group, simplesimple pole, 485

simple zero, 484See also root, simple

Singerman, D., 188, 513Slavutin, E. I., 70Smirnova, G. S., 23, 70, 227, 356Smith, D. E., 70, 356socle of a group, 439,443

nonregular, 439regular, 439

Soicher, L., 411Solovyev, Y., 513solvable

by radicals, see polynomial, solvable by radicalsextension, see extension, solvablegroup, see group, solvable

Spearman, B. K., 410-411special linear group, xxvii, 396, 548

over a finite field, 438two-dimensional, 443, 461

spiral of Archimedes, 267, 269splits completely, 59, 107splitting field, see field, splittingsquaring the circle, 262, 266—269Starr, N., xixstarting configuration, 264Stauduhar, R. P., 387, 411Steinitz, E., 79, 122, 348, 354stereographic projection, 180,551Stevenhagen, P., 289,411Stewart, I., 556Stillwell, J., 411straightedge, 255

and compass, 235, 245, 255,464—465,467,470—471,474,476, 478—479, 504, 506

and dividers, 264See also marked ruler

Stubhaug, A., 227subgroup, 515

conjugate, 155, 358, 364, 366, 370, 373, 376,386, 388, 395, 398, 401,403,415,422,424, 435, 451,458, 531

index of, 159, 319, 330,415,418, 516

isotropy, 161,318, 320,437,443, 530—531,

546, 548

normal, 155, 158, 160, 331, 344, 516minimal, 432—433,435, 442,455, 461, 549

of a cyclic group, 517Sylow, 195, 219, 370, 383, 415—416,418,

427—428,448, 460, 532transitive, see symmetric group, transitive

subgroup ofsubstitution, 343

See also permutationSuprunenko, D. A., 462Swallow, J., 556

INDEX 569

Sylow Theorems, 195—196, 220, 370, 383,415—416,418,427, 460,533

symmetric group, xxiv, 11, 30, 135, 138—140, 183,186, 316,401,408, 517

index of subgroups of, 327, 330—331, 334nonnal subgroups of, 213solvable subgroups of, see permutation group,

solvablesubgroup of, see permutation grouptransitive subgroup of, 134, 136, 360, 363—364,

415, 419, 421,429—430,435, 441classification for 54, 363classification for S5, 368—370, 406classification for S7, 394—395classification for n < 32, 386classification for 458

symmetric polynomial, see polynomial, symmetricsymmetry group, see group, symmetry

T

2-3 tower, 277—278, 281Tagaki,T.,512Tartaglia, 9, 19, 329Tate, J., 411, 514term, 25

leading, 32, 39—40tetrahedron, 184, 187Thabit ibn Qurra, 283Theorem of the Primitive Element, see primitive

element, Theorem ofTheorem on Natural Irrationalities, see natural

irrationality, Theorem onThompson, J., 195, 227Tignol, J.-P., 556total degree, xxv, 26Tower Theorem, 91trace, 459transcendental

extension, see extension, purely transcendentalover afield, 73, 79

transitivegroup action, see action of a group, transitivesubgroup of see symmetric group, transitive

subgroup oftransposition, 518triangle inequality, 64trmnomial, 396

equivalent, 396of large degree, 310

trisecting the angle, 262, 264, 266—269, 283via intersection of conics, 286via marked ruler, 286via marked ruler and compass, 282, 285via origami, 274, 284

Tschirnhaus, E. W., 329, 380

Tschirnhaus transformation, 379—380, 381, 385,389

twice-notched straightedge, 279See also marked ruler and straightedge

U

UFD, see ring, unique factorization domainUnger, W. R., 462unique factorization domain, see ring, unique

factorization domainuniversal

extension, xxvi, 138, 169, 217, 316, 552Galois group of, 138

polynomial, xxv, 14, 37, 138, 169, 217, 219,316, 355

unsolvability for n � 5, 217unsolvability of quintic, see Abel, unsolvability of

quintic and quintic polynomial,unsolvability of

V

Vandermonde, C. A., 50-51,235,330-331,382Vandermonde determinant, 49van der Waerden, B. L., 356, 556van Roomen, A., see Romanus, A.Velleman, D., 70verging, 279, 282—283, 285, 287Videla, C. R., 289Viète, F., 9, 18, 20, 283, 329

See also cubic polynomial, trigonometricsolution of

Vlàdul, S. 0., 514

w

w, see root of unity, cube(variant of ir), xxvii, 466

Wantzel, P., 267, 272Waring, E., 34, 38, 53Warren, B., 411Washington, L. C., 514Watson, G. N., 514Weber, H., 80, 129, 144—145Weierstrass, K., 487—488Weierstrass p-function, see p-functionweight, 41Weisner, L., 53, 123Weisstein,E.,4l1well-ordering, 538Whittaker, E. T., 514Williams, K. S., 410—411Wine, D., 462Wolfram Research, 188, 411Wong, S., xixwreath product, xxvii, 421—423,425, 427—428,

432,439, 457, 548, 553

570 INDEX

Wussing, H., 227, 356

Y

Young, 0. P., 382

Yui,N., 188

z

see root of unity, nthsee root of unity, pth

Theorem, 486zeta function, 297Zimmerman, P., 310Zuckerman, H. S., 99, 311

PURE AND APPLIED MATHEMATICSA Wiley Series of Texts, Monographs, and Tracts

Founded by RICHARD COURANTEditors Emeriti: MYRON B. ALLEN III, DAVID A. COX, PETER HILTON,HARRY HOCHSTADT, PETER LAX, JOHN TOLAND

ADAMEK, HERRLICH, and STRECKER—Abstract and Concrete CatetoriesADAMOWICZ and ZBIERSKI—Logic of MathematicsAINSWORTH and ODEN—A Posteriori Error Estimation in Finite Element AnalysisAKIVIS and GOLDBERG—Conformal Differential Geometry and Its GeneralizationsALLEN and ISAACSON—Numerical Analysis for Applied Science

AlgebraATKINSON, HAN, and STEWART—Numerical Solution of Ordinary Differential

EquationsAUB1N—Applied Functional Analysis, Second EditionAZIZOV and IOKH VIDO V—Linear Operators in Spaces with an Indefinite MetricBASENER—Topology and Its ApplicationsBERG—The Fourier-Analytic Proof of Quadratic ReciprocityBERKOVITZ—Convexity and Optimization inBERMAN, NEUMANN, and STERN—Nonnegative Matrices in Dynamic SystemsBOYARINTSEV—Methods of Solving Singular Systems of Ordinary Differential

EquationsBRIDGER—Real Analysis: A Constructive ApproachBURK—Lebesgue Measure and Integration: An Introduction

* CARTER—Finite Groups of Lie TypeCASTILLO, COBO, JUBETE, and PRUNEDA—Orthogonal Sets and Polar Methods in

Linear Algebra: Applications to Matrix Calculations, Systems of Equations,Inequalities, and Linear Programming

CASTILLO, CONEJO, PEDREGAL, GARCIA, and ALGUACIL—Building and SolvingMathematical Programming Models in Engineering and Science

CHATELIN—Eigenvalues of MatricesCLARK—Mathematical Bioeconomics: The Mathematics of Conservation, Third EditionCOX—Galois Theory, Second Edition

tCOX—Primes of the Form x2 + ny2: Fermat, Class Field Theory, and ComplexMultiplication

*CURTIS and REINER—Representation Theory of Finite Groups and Associative Algebras*CURTIS and REINER—Methods of Representation Theory: With Applications to Finite

Groups and Orders, Volume ICURTIS and REINER—Methods of Representation Theory: With Applications to Finite

Groups and Orders, Volume IIDINCULEANU—Vector Integration and Stochastic Integration in Banach Spaces

*DUNFORD and SCHWARTZ—Linear OperatorsPart 1—General TheoryPart 2—Spectral Theory, Self Adjoint Operators in

Hilbert SpacePart 3—Spectral Operators

FARINA and RINALDI—Positive Linear Systems: Theory and Applications

*Now available in a lower priced paperback edition in the Wiley Classics Library.tNow available in paperback.

FATICONI—The Mathematics of Infinity: A Guide to Great IdeasFOLLAND—Real Analysis: Modern Techniques and Their ApplicationsFROLICHER and KRIEGL—Linear Spaces and Differentiation TheoryGARDINER—Teichmüller Theory and Quadratic DifferentialsGILBERT and NICHOLSON—Modem Algebra with Applications, Second Edition

*GRIFFITHS and HARRIS—Principles of Algebraic GeometryGRILLET—AlgebraGROVE—Groups and CharactersGUSTAFSSON, KREISS and OLIGER—Time Dependent Problems and Difference

MethodsHANNA and ROWLAND—Fourier Series, Transforms, and Boundary Value Problems,

Second Edition*HENpJCIApplied and Computational Complex Analysis

Volume 1, Power Series—Integration-—Conformal Mapping—Locationof Zeros

Volume 2, Special Functions—Integral Transforms—Asymptotics—Continued Fractions

Volume 3, Discrete Fourier Analysis, Cauchy Integrals, Constructionof Conformal Maps, Univalent Functions

* HILTON and WU—A Course in Modem Algebra*H0CHsTADT.....Integral EquationsJOST—Two-Dimensional Geometric Variational ProceduresKHAMSI and KIRI(—An Introduction to Metric Spaces and Fixed Point Theory

*KOBAYASHI and NOMIZU—Foundations of Differential Geometry, Volume I*KOBAYASHI and NOMIZU—Foundations of Differential Geometry, Volume IIKOSHY—Fibonacci and Lucas Numbers with ApplicationsLAX—Functional AnalysisLAX—Linear Algebra and Its Applications, Second EditionLOGAN—An Introduction to Nonlinear Partial Differential Equations, Second EditionLOGAN and WOLESENSKY—Mathematical Methods in BiologyLUI—Numerical Analysis of Partial Differential EquationsMARKLEY—Principles of Differential EquationsMORRISON—Functional Analysis: An Introduction to Banach Space TheoryNAYFEH—Perturbation MethodsNAYFEH and MOOK—Nonlinear OscillationsO'LEARY—Revolutions of GeometryO'NEIL—Beginning Partial Differential Equations, Second EditionPANDEY—The Hubert Transform of Schwartz Distributions and ApplicationsPETKO V—Geometry of Reflecting Rays and Inverse Spectral Problems

and Variational MethodsPROMISLOW—A First Course in Functional AnalysisRAO—Measure Theory and IntegrationRASSIAS and SIMSA—Finite Sums Decompositions in Mathematical AnalysisRENELT—Elliptic Systems and Quasiconformal MappingsRIVLIN—Chebyshev Polynomials: From Approximation Theory to Algebra and Number

Theory, Second EditionROCKAFELLAR—Network Flows and Monotropic OptimizationROITMAN—Introduction to Modem Set TheoryROSSI—Theorems, Corollaries, Lemmas, and Methods of Proof

Analysis on GroupsSENDO V—The Averaged Moduli of Smoothness: Applications in Numerical Methods

and Approximations


SENDOV and POPO V—The Averaged Moduli of SmoothnessSE WELL—The Numerical Solution of Ordinary and Partial Differential Equations,

Second EditionSEWELL—Computational Methods of Linear Algebra, Second EditionSHICK—Topology: Point-Set and GeometricSHISKOWSKI and FRINKLE—Principles of Linear Algebra With MapleTMSHISKOWSKI and FRJNKLE—Principles of Linear Algebra With Mathematica®

* SIEGEL—Topics in Complex Function TheoryVolume I—Elliptic Functions and Uniformization TheoryVolume 2—Automorphic Functions and Abelian IntegralsVolume 3—Abelian Functions and Modular Functions of Several Variables

SMITH and ROMANOWSKA—Post-Modern AlgebraDifferential Equations and the Finite Element Method

STADE—Fourier AnalysisSTAHL—Introduction to Topology and GeometrySTAHL—Real Analysis, Second EditionSTAKGOLD and HOLST—Green's Functions and Boundary Value Problems,

Third EditionSTANOYEVITCH—Introduction to Numerical Ordinary and Partial Differential

Equations Using MATLAB®*STOIçER_Differential Geometry

Vibrations in Mechanical and Electrical Systems*STO}(ERWater Waves: The Mathematical Theory with ApplicationsWATKINS—Fundamentals of Matrix Computations, Third EditionWESSELING—An Introduction to Multigrid Methods

tWHITHAM_Linear and Nonlinear WavesZAUDERER—Partial Differential Equations of Applied Mathematics, Third Edition


Documents

Galois Theory, Second Edition