D.ZackGarza · 2021. 2. 21. · Contents 7.2.13 NonstandardTopics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .68 8.1 RingTheory

Algebra Qualifying Exam Review

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Table of Contents 2

1 Topics and Remarks 131.1 General References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.2 Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.2.1 Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.3.1 Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.4.1 Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.5 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5.1 Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.6 Field Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.6.1 Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Group Theory 192.1 Big List of Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4 Conjugacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.4.1 Normal Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.5 Centralizing and Centers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.6 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.7 Special Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.7.1 Cyclic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.7.2 Symmetric Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.9 Counting Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.10 Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.11 Examples of Orbit-Stabilizer and the Class Equation . . . . . . . . . . . . . . . . . . 36

2.11.1 Left Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.11.2 Conjugation: The Class Equation and Burnside’s Lemma . . . . . . . . . . . 372.11.3 Conjugation on Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.11.4 Left Translation on Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3 Sylow Theorems 423.1 Statements of Sylow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.1.1 Sylow 1 (Cauchy for Prime Powers) . . . . . . . . . . . . . . . . . . . . . . . 433.1.2 Sylow 2 (Sylows are Conjugate) . . . . . . . . . . . . . . . . . . . . . . . . . . 443.1.3 Sylow 3 (Numerical Constraints) . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.2 Corollaries and Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

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3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.4 Automorphism Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.5 Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.6 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.7 Classification: Finitely Generated Abelian Groups . . . . . . . . . . . . . . . . . . . 523.8 Classification: Groups of Special Orders . . . . . . . . . . . . . . . . . . . . . . . . . 573.9 Series of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.10 Solvability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4 Ring Theory 634.1 Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.2 Important Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.3 Undergrad Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

4.3.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.3.2 Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.3.3 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

4.4 Types of Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.4.1 The Big Ones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.4.2 Others . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

4.5 Comparing and Transporting Ring Types . . . . . . . . . . . . . . . . . . . . . . . . 734.6 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.7 Structure Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.8 Zorn’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784.9 Unsorted . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

5 Number Theory 80

6 General Field Theory 806.1 Basics: Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.3 Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 846.4 Cyclotomic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 856.5 Misc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

7 Field Theory: Extensions and Towers 897.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 897.2 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 897.3 Separable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 917.4 Galois Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 987.5 Fundamental Theorem of Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . 1007.6 Quadratic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

8 Distinguished Classes 1018.0.1 Algebraic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1038.0.2 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

9 Galois Theory 1069.1 Showing Extensions are Galois . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

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9.2 Irreducibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1099.3 Computing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

9.3.1 Misc Useful Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1109.3.2 Transitive Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1119.3.3 Distinguishing Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1129.3.4 Density: Cycle Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1149.3.5 Discriminants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

9.4 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1179.4.1 Quadratics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1179.4.2 Cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1189.4.3 Quartics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1209.4.4 Cyclotomic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1229.4.5 Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

9.5 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

10 Modules 12710.1 Definitions and Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12710.2 Structure Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12810.3 Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12810.4 Free and Projective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12910.5 Classification of Modules over a PID . . . . . . . . . . . . . . . . . . . . . . . . . . . 13310.6 Algebraic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

11 Linear Algebra 13511.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

11.1.1 Matrix Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13911.2 Minimal / Characteristic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 14011.3 Finding Minimal Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14311.4 Other Canonical Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

11.4.1 Rational Canonical Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14311.4.2 Smith Normal Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14611.4.3 Using Canonical Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

11.5 Diagonalizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14711.6 Matrix Counterexamples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

11.6.1 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14911.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

12 Jordan Canonical Form 15112.1 Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15112.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

13 Representation Theory 155

14 Extra Problems 15614.1 Commutative Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15614.2 Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

14.2.1 Centralizing and Normalizing . . . . . . . . . . . . . . . . . . . . . . . . . . . 15614.2.2 Primes in Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

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14.2.3 p-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15814.2.4 Symmetric Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15914.2.5 Alternating Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16014.2.6 Dihedral Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16014.2.7 Other Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16014.2.8 Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16114.2.9 Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16114.2.10Series of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16114.2.11Misc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16214.2.12Nonstandard Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

14.3 Ring Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16314.3.1 Basic Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16314.3.2 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16314.3.3 Characterizing Certain Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . 16414.3.4 Misc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

14.4 Field Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16514.5 Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

14.5.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16614.5.2 Computations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

14.6 Modules and Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16714.7 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

15 Even More Algebra Questions 16715.1 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

15.1.1 Question 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16815.1.2 Question 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16815.1.3 Question 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16815.1.4 Question 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16815.1.5 Question 1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16815.1.6 Question 1.6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16815.1.7 Question 1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16915.1.8 Question 1.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16915.1.9 Question 1.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16915.1.10Question 1.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16915.1.11Question 1.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16915.1.12Question 1.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16915.1.13Question 1.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16915.1.14Question 1.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16915.1.15Question 1.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17015.1.16Question 1.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17015.1.17Question 1.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17015.1.18Question 1.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17015.1.19Question 1.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17015.1.20Question 1.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17015.1.21Question 1.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17015.1.22Question 1.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17015.1.23Question 1.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17115.1.24Question 1.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

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15.1.25Question 1.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17115.1.26Question 1.26. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17115.1.27Question 1.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17115.1.28Question 1.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17115.1.29Question 1.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17115.1.30Question 1.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17115.1.31Question 1.31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17215.1.32Question 1.32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17215.1.33Question 1.33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17215.1.34Question 1.34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17215.1.35Question 1.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17215.1.36Question 1.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17215.1.37Question 1.37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17215.1.38Question 1.38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17315.1.39Question 1.39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17315.1.40Question 1.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17315.1.41Question 1.41. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17315.1.42Question 1.42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17315.1.43Question 1.43 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17315.1.44Question 1.44 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

15.2 Classification of Finite groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17415.2.1 Question 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17415.2.2 Question 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17415.2.3 Question 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17415.2.4 Question 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17415.2.5 Question 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17415.2.6 Question 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17415.2.7 Question 2.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17515.2.8 Question 2.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17515.2.9 Question 2.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17515.2.10Question 2.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17515.2.11Question 2.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17515.2.12Question 2.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17515.2.13Question 2.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17515.2.14Question 2.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17515.2.15Question 2.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17615.2.16Question 2.16. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17615.2.17Question 2.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17615.2.18Question 2.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17615.2.19Question 2.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17615.2.20Question 2.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

15.3 Fields and Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17615.3.1 Question 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17615.3.2 Question 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17715.3.3 Question 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17715.3.4 Question 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17715.3.5 Question 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17715.3.6 Question 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

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Contents

15.3.7 Question 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17715.3.8 Question 3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17715.3.9 Question 3.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17715.3.10Question 3.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17815.3.11Question 3.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17815.3.12Question 3.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17815.3.13Question 3.13. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17815.3.14Question 3.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17815.3.15Question 3.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17815.3.16Question 3.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17815.3.17Question 3.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17915.3.18Question 3.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17915.3.19Question 3.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17915.3.20Question 3.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17915.3.21Question 3.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17915.3.22Question 3.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17915.3.23Question 3.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17915.3.24Question 3.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17915.3.25Question 3.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18015.3.26Question 3.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18015.3.27Question 3.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18015.3.28Question 3.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18015.3.29Question 3.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18015.3.30Question 3.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18015.3.31Question 3.31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18015.3.32Question 3.32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18015.3.33Question 3.33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18115.3.34Question 3.34. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18115.3.35Question 3.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18115.3.36Question 3.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18115.3.37Question 3.37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18115.3.38Question 3.38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18115.3.39Question 3.39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18115.3.40Question 3.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18115.3.41Question 3.41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18215.3.42Question 3.42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18215.3.43Question 3.43 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18215.3.44Question 3.44 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18215.3.45Question 3.45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18215.3.46Question 3.46 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18215.3.47Question 3.47 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18215.3.48Question 3.48 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18215.3.49Question 3.49 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18315.3.50Question 3.50 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18315.3.51Question 3.51 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18315.3.52Question 3.52 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18315.3.53Question 3.53 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18315.3.54Question 3.54 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

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Contents

15.3.55Question 3.55 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18315.3.56Question 3.56. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18315.3.57Question 3.57 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18415.3.58Question 3.58 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18415.3.59Question 3.59 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18415.3.60Question 3.60 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18415.3.61Question 3.61 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18415.3.62Question 3.62 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18415.3.63Question 3.63 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18415.3.64Question 3.64 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18515.3.65Question 3.65 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18515.3.66Question 3.66 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18515.3.67Question 3.67 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18515.3.68Question 3.68 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18515.3.69Question 3.69 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18515.3.70Question 3.70 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18515.3.71Question 3.71 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18515.3.72Question 3.72 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18615.3.73Question 3.73 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18615.3.74Question 3.74 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18615.3.75Question 3.75 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18615.3.76Question 3.76 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18615.3.77Question 3.77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18615.3.78Question 3.78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

15.4 Normal Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18715.4.1 Question 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18715.4.2 Question 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18715.4.3 Question 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18715.4.4 Question 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18715.4.5 Question 4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18715.4.6 Question 4.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18715.4.7 Question 4.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18715.4.8 Question 4.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18815.4.9 Question 4.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18815.4.10Question 4.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18815.4.11Question 4.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18815.4.12Question 4.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18815.4.13Question 4.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18815.4.14Question 4.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18815.4.15Question 4.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18815.4.16Question 4.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18915.4.17Question 4.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18915.4.18Question 4.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18915.4.19Question 4.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18915.4.20Question 4.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18915.4.21Question 4.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

15.5 Matrices and Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18915.5.1 Question 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

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Contents

15.5.2 Question 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19015.5.3 Question 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19015.5.4 Question 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19015.5.5 Question 5.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19015.5.6 Question 5.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19015.5.7 Question 5.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19015.5.8 Question 5.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19015.5.9 Question 5.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19015.5.10Question 5.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19115.5.11Question 5.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19115.5.12Question 5.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19115.5.13Question 5.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19115.5.14Question 5.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19115.5.15Question 5.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19115.5.16Question 5.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19115.5.17Question 5.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19115.5.18Question 5.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19215.5.19Question 5.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19215.5.20Question 5.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

15.6 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19215.6.1 Question 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19215.6.2 Question 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19215.6.3 Question 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19215.6.4 Question 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19215.6.5 Question 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19215.6.6 Question 6.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19315.6.7 Question 6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19315.6.8 Question 6.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19315.6.9 Question 6.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19315.6.10Question 6.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19315.6.11Question 6.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19315.6.12Question 6.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19315.6.13Question 6.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19315.6.14Question 6.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19415.6.15Question 6.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19415.6.16Question 6.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19415.6.17Question 6.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19415.6.18Question 6.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19415.6.19Question 6.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19415.6.20Question 6.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19415.6.21Question 6.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19415.6.22Question 6.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19515.6.23Question 6.23. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19515.6.24Question 6.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19515.6.25Question 6.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19515.6.26Question 6.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19515.6.27Question 6.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19515.6.28Question 6.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

Contents 9

Contents

15.6.29Question 6.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19615.6.30Question 6.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19615.6.31Question 6.31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19615.6.32Question 6.32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19615.6.33Question 6.33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19615.6.34Question 6.34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19615.6.35Question 6.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19615.6.36Question 6.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19715.6.37Question 6.37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19715.6.38Question 6.38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19715.6.39Question 6.39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19715.6.40Question 6.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19715.6.41Question 6.41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19715.6.42Question 6.42. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19715.6.43Question 6.43 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19715.6.44Question 6.44 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19815.6.45Question 6.45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19815.6.46Question 6.46 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19815.6.47Question 6.47 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19815.6.48Question 6.48 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19815.6.49Question 6.49 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19815.6.50Question 6.50 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19815.6.51Question 6.51 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19915.6.52Question 6.52 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

15.7 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19915.7.1 Question 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19915.7.2 Question 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19915.7.3 Question 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19915.7.4 Question 7.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19915.7.5 Question 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20015.7.6 Question 7.6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20015.7.7 Question 7.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20015.7.8 Question 7.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20015.7.9 Question 7.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20015.7.10Question 7.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

15.8 Representation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20015.8.1 Question 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20015.8.2 Question 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20115.8.3 Question 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20115.8.4 Question 8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20115.8.5 Question 8.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20115.8.6 Question 8.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20115.8.7 Question 8.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20115.8.8 Question 8.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20115.8.9 Question 8.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20115.8.10Question 8.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20215.8.11Question 8.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20215.8.12Question 8.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

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Contents

15.8.13Question 8.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20215.8.14Question 8.14. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20215.8.15Question 8.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20215.8.16Question 8.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20215.8.17Question 8.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20315.8.18Question 8.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20315.8.19Question 8.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20315.8.20Question 8.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20315.8.21Question 8.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20315.8.22Question 8.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20315.8.23Question 8.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20315.8.24Question 8.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20415.8.25Question 8.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20415.8.26Question 8.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20415.8.27Question 8.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20415.8.28Question 8.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20415.8.29Question 8.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20415.8.30Question 8.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20415.8.31Question 8.31. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20515.8.32Question 8.32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20515.8.33Question 8.33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20515.8.34Question 8.34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20515.8.35Question 8.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20515.8.36Question 8.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20515.8.37Question 8.37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

15.9 Categories and Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20615.9.1 Question 9.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20615.9.2 Question 9.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20615.9.3 Question 9.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

16 Appendix: Extra Topics 20616.1 Characteristic Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20616.2 Normal Closures and Cores . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

16.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20816.3 Nilpotent Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20816.4 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

17 UGA Fall 2019 Problem Sets 21017.1 Problem Set One . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

17.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21017.1.2 Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

17.2 Problem Set Two . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21617.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21617.2.2 Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

17.3 Problem Set Three . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21817.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21817.3.2 Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

Contents 11

Contents

17.4 Problem Set Four . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21917.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21917.4.2 Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

17.5 Problem Set Five . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22117.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22117.5.2 Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

17.6 Problem Set Six . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22317.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22317.6.2 Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

17.7 Problem Set Seven . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22417.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22417.7.2 Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

17.8 Problem Set Eight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22717.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22717.8.2 Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

17.9 Problem Set Nine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22917.9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22917.9.2 Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230

17.10Problem Set Ten . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23117.10.1Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23117.10.2Qual Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232

Bibliography 233

Contents 12

1 Topics and Remarks


Remark 1.0.1: (DZG) on the structure of these notes: these are extremely disorganized at themoment, and only reflect some amalgamation of all of the random notes I made to myself whilestudying for qualifying exams. As a result, things are bound to be out of order, and likely uselesspedagogically unless you’ve seen most of the material before. Moreover, this has been a long-running document (started in my undergrad years, so pre-2018), and since I’ve forgotten andrewritten certain things at various points, there may even be duplicated material (e.g. propositionsstated/proved in multiple places, repeated exercises or statements, etc).

In any case, I’d love to hear if you do find it useful! Readers are welcome to email me with anyquestions, comments, errors/typos, suggestions for improvement, or just to say hello!

Remark 1.0.2: Adapted from remark written by Roy Smith, August 2006:

“As a general rule, students are responsible for knowing both the theory (proofs) and practicalapplications (e.g. how to find the Jordan or rational canonical form of a given matrix, orthe Galois group of a given polynomial) of the topics mentioned.”

E 1.1 General References e

• David Dummit and Richard Foote, Abstract Algebra, Wiley, 2003. [1]

• Kenneth Hoffman and Ray Kunze, Linear Algebra, Prentice-Hall, 1971. [2]

• Thomas W. Hungerford, Algebra, Springer, 1974. [3]

• Roy Smith, Algebra Course Notes (843-1 through 845-3). [4]

– Note: scroll down the page to find links to his course notes.

E 1.2 Group Theory e

References: [1], [3], [4] “The first 6 chapters (220pages) of Dummit and Foote are excellent. All thedefinitions and proofs of these theorems on groups aregiven in Smith’s web based lecture notes for math 843part 1.”

Topics and Remarks 13


1.2.1 Topics

Chapters 1-9 of Dummit and Foote

• The first isomorphism theorem,

• Fundamental theorem of finite abelian groups

• Left and right cosets

• Normalizer

• Lagrange’s theorem

• Isomorphism theorems

• Lagrange’s Theorem

• Group generated by a subset

• Subgroups and quotient groups

• Fundamental homomorphism theorems

• Direct and semi-direct products

– Recognition of internal direct product– Recognition of semi-direct product

• Composite groups

• Structures of special types of groups such as:

– p-groups– Dihedral,♦ Cyclic groups♦ Free groups♦ Generators and relations

– Symmetric and Alternating groups♦ Cycle decompositions

• Group actions with applications to the structure of groups such as

– The Sylow Theorems♦ Proof of Sylow theorems

1.2 Group Theory 14


– Orbit stabilizer theorem– Orbits act on left cosets of subgroups– Action of G on itself by conjugation– Class equation– Cayley’s theorem

• The simple groups of order between 60 and 168 have prime order

• The simplicity of An, for n ≥ 5

• Solvable groups

• Subgroups of index p, the smallest prime dividing #G, are normal

• p-groups

• p2 groups are abelian

• Automorphisms

– Inner automorphisms

• An is simple for n ≥ 5

• Classification of groups of order pq

• Commutator subgroup

• Nilpotent groups

• Upper central series

• Lower central series

• Derived series

• Solvable groups

• Fratini’s argument

• The Jordan Holder theorem

The proof of Jordan-Holder is seldom tested on thequal**, but proofs are always of interest.

1.2 Group Theory 15


E 1.3 Linear Algebra e

References: [1],[2],[4]

1.3.1 Topics

• Determinants• Eigenvalues and eigenvectors• Cayley-Hamilton Theorem• Canonical forms for matrices• Linear groups (GLn,SLn,On,Un)• Duality

– Dual spaces,– Dual bases,– Induced dual map,– Double duals

• Finite-dimensional spectral theorem

E 1.4 Rings e


• DF chapters 13,14 (about 145 pages).

• Smith:

– 843-2, sections 11,12, and 16-21 (39 pages)– 844-1, sections 7-9 (20 pages)– 844-2, sections 10-16, (37 pages)

• DF Chapters 7, 8, 9.

1.4.1 Topics

• Properties of ideals and quotient rings

• The fundamental isomorphism theorems for rings

• I maximal iff R/I is a field

1.3 Linear Algebra 16


• Zorn’s lemma

– Every vector space has a basis– Maximal ideals exist– Construct algebraic field closures– Why it is unnecessary in countable or noetherian rings.

Smith discusses extensively in 844-1.

• Chinese Remainder Theorem

• Euclidean algorithm

• Primes and irreducibles

• Gaussian integers

• Localization of a domain

• Field of fractions

• Factorization in domains

• Factorization in Z[i]

• Characterizations and properties of special rings such as:

– Euclidean =⇒ PID =⇒ UFD– Domains♦ Primes are irreducible

– UFDs♦ Have GCDs♦ Sometimes PIDs

– PIDs♦ Noetherian♦ Irreducibles are prime♦ Are UFDs♦ Have GCDs♦ Results about PIDs (DF Section 8.2)♦ Example of a PID that is not a Euclidean domain (DF p.277)♦ Proof that a Euclidean domain is a PID and hence a UFD♦ Proof that Z and k[x] are UFDs (p.289 Smith, p.300 DF)♦ A polynomial ring in infinitely many variables over a UFD is still a UFD (Easy,

DF, p.305)– Euclidean domains♦ Are PIDs

1.4 Rings 17


• Gauss’s important theorem on unique factorization of polynomials:

– Z[x] is a UFD– R[x] is a UFD when R is a UFD

• Polynomial rings

• Polynomials

– Gauss’ lemma– Remainder and factor theorem– Eisenstein’s criterion (DF p.309) > Stated only for monic polynomials – proof of general

case identical. > See Smith’s notes for the full version.– Reducibility– Rational root test

• Cyclic product structure of (Z/nZ)×

Exercise in DF, Smith 844-2, section 18

• Gröbner bases and division algorithms for polynomials in several variables (DF 9.6.)

E 1.5 Modules e


1.5.1 Topics

• Fundamental homomorphism theorems for rings and modules

• Applications to the structure of:

– Finitely generated abelian groups– Canonical forms of matrices

• Classification of finitely generated modules over PIDs (with emphasis on Euclidean Domains)

• Modules over PIDs and canonical forms of matrices. DF sections 10.1, 10.2, 10.3, and 12.1,12.2, 12.3.

– Constructive proof of decomposition: DF Exercises 12.1.16-19

1.5 Modules 18

2 Group Theory

Smith 845-1 and 845-2: Detailed discussion of theconstructive proof.

E 1.6 Field Theory e

1.6.1 Topics


• Algebraic extensions of fields

• Properties of finite fields

• Separable extensions

• Fundamental theorem of Galois theory

• Computations of Galois groups

– of polynomials of small degree– of cyclotomic polynomials

• Solvability of polynomials by radicals

2 Group Theory

Remark 2.0.1: Summary of useful qual tips:

• Slightly obvious but good to remember:

– Subgroups of abelian groups are automatically normal.– If N is normal in G, then N is normal in any subgroup containing it.– If N ≤ G is the unique group of order #N , then N is normal (since any conjugate must

have the same size).– Using the subgroup correspondence: if L/H ≤ G/H then L ≤ G has size #(L/H)#H.

• Sizes and structure:

– Quotienting by bigger groups yields smaller indices:

1 ≤ H ≤ H ≤ K ≤ G apply[G : −] =⇒ #G = [G : 1] ≥ [G : H] ≥ [G : K] ≥ [G : G] = 1.

– x is central iff [x] = {e}.

1.6 Field Theory 19

2 Group Theory

– Unions aren’t (generally) subgroups, intersections always are.– Coprime order subgroups intersect trivially.– Distinct subgroups of order pn, pm can intersect trivially or in subgroups of order p`.

• Conjugacy:

– Sizes of conjugacy classes divide #G (by orbit-stabilizer).– Conjugate subgroups have equal cardinality.– Normal subgroups absorb conjugacy classes, and are thus unions of conjugacy classes.– Reasoning about conjugacy classes: in Sn they’re precisely determined by cycle type,

i.e. a partition of n.– Remembering the class equation: for literally any group action ϕ : G y X, one hasX = Fix(ϕ)

∐′Orb(xi) as a disjoint union of fixed points and nontrivial orbits, sinceorbits partition X. Then take your action to be G y G by ϕ : g.x := gxg−1 to getFix(ϕ) = Z(G) and Orb(xi) =

{gxig

−1}

= [xi] the conjugacy classes. Now apply orbitstabilizer to get Orb(x) ∼= G/Stab(x) where Stab(x) = Z(x) = CG(x) the centralizer.

• Cosets:

– Cosets partition a group.– Anything dealing with indices [G : H]: try just listing the cosets.– aH = bH ⇐⇒ ab−1 ∈ H.– Showing subgroup containment: K ⊆ H iff kH = H for all k ∈ K.

• Sylows:

– If Sp is normal, then Sp is characteristic. This is useful if H ≤ G and P ∈ Sylp(H) isnormal in H, then P is also normal in G.

E 2.1 Big List of Notation e

Remark 2.1.1(Notation): I use the following notation throughout:

Notation Definition

CG(x) Centralizer of an element:={g ∈ Γ

∣∣∣ [g, x] = 1}⊆ Γ

CG(H) Centralizer of an subgroup:={g ∈ Γ

∣∣∣ [g, x] = 1 ∀h ∈ H}

=⋂h∈H

CH(h) ⊆ G

C(H) Conjugacy Class:={ghg−1

∣∣∣ g ∈ G} ≤ G ⊆ GZ(G) Center

:={x ∈ G

∣∣∣ ∀g ∈ G, gxg−1 = x}⊆ G

NG(H) Normalizer:={g ∈ G

∣∣∣ gHg−1 = H}⊆ G

2.1 Big List of Notation 20

2 Group Theory

Notation Definition

Inn(G) Inner Automorphisms:={ϕg(x) := gxg−1

}⊆ Aut(G)

Out(G) Outer AutomorphismsAut(G)/ Inn(G) 7→Aut(G)

[gh] Commutator of Elements:= ghg−1 ∈ G

[GH] Commutator of Subgroups:=⟨{

[gh]∣∣∣ g ∈ G, h ∈ H}⟩ ≤ G

Ox, Gx Orbit of an Element:={gx∣∣∣ x ∈ X}

StabG(x), Gx Stabilizer of an Element:={g ∈ G

∣∣∣ gx = x}⊆ G

X/G Set of Orbits:={Gx

∣∣∣ x ∈ X} ⊆ 2X

Xg Fixed Points{x ∈ X

∣∣∣ ∀g ∈ G, gx = x}⊆ X

2X The powerset of X:= {U ⊆ X}

Remark 2.1.2: For any p dividing the order of G, Sylp(G) denotes the set of Sylow-p subgroupsof G.

E 2.2 Definitions e

Fact 2.2.1An set morphism that is either injective or surjective between sets of the same size is automatically abijection. It turns out that a group morphism between groups of the same size that is either injectiveor surjective is automatically a bijection, and the inverse is automatically a group morphism, sobijective group morphisms are isomorphisms.

Fact 2.2.2 (Bezout’s Identity)If a, b ∈ Z with gcd(a, b) = d, then there exist s, t ∈ Z such that

as+ bt = d.

This d can be computed using the extended Euclidean algorithm.

2.2 Definitions 21

2 Group Theory

Remark 2.2.3: Useful context clue! In particular, this works when a, b are coprime and d = 1,since you can write x1 = xas+bt = xasxbt to get interesting information about orders of elements. Ifyou see “coprime” in a finite group question, try the division algorithm.

Definition 2.2.4 (Order)The order of an element g ∈ G, denoted n := o(g), is the smallest n ∈ Z≥0 such that gn = e.

Exercise 2.2.5 (?)Show that the order of any element in a group divides the order of the group.

Definition 2.2.6 (Group Presentation)An expression of the form G =

⟨S∣∣∣ R⟩ where S is a set of elements and R a set of words

defining relations means that G := F [S]/cln(R) where F [S] is the free group on the set S andcln(R) is the normal closure, the smallest normal subgroup of F [S] containing R.

Remark 2.2.7: Finding morphisms between presentations: if G is presented with generators giwith relations ri and H is any group containing elements hi also satisfying ri, there is a groupmorphism

ϕ : G→ H

gi 7→ hi ∀i.

Why this exists: the presentation yields a surjective morphism π : F (gi)→ G with G ∼= F (gi)/ kerπ.Define a map ψ : F (gi)→ H where gi 7→ hi, then since the hi satisfy the relations ri, kerπ ⊆ kerψ.So ψ factors through kerπ yielding a morphism F/ kerπ → H.

E 2.3 Subgroups e

Definition 2.3.1 (Subgroup)A subset H ⊆ G is a subgroup iff

1. Closure: HH ⊂ H2. Identity: e ∈ H3. Inverses: g ∈ H ⇐⇒ g−1 ∈ H.

Definition 2.3.2 (Subgroup Generated by a Subset)If H ⊂ G, then 〈H〉 is the smallest subgroup containing H:

〈H〉 = ∩{H∣∣∣ H ⊆M ≤ G}M =

{h±1

1 · · ·h±1n

∣∣∣ n ≥ 0, hi ∈ H}

where adjacent hi are distinct.

Definition 2.3.3 (Commutator)The commutator subgroup of G is denoted [G,G] ≤ G. It is the subgroup generated by all

2.3 Subgroups 22

2 Group Theory

elementary commutators:

[G,G] :=⟨aba−1b−1

∣∣∣ a, b ∈ G⟩ .It is the smallest normal subgroup N E G such that G/N is abelian, so if H ≤ G and G/H isabelian, H ⊆ [G,G].

Note that elements in [G,G] are generally productsof elementary commutators, and not elementarythemselves, since we’re taking the group generatedby them.

Proposition 2.3.4(One-step subgroup test).If H ⊆ G and a, b ∈ H =⇒ ab−1 ∈ H, then H ≤ G.

Proof (of the one-step subgroup test).

• Identity: a = b = x =⇒ xx−1 = e ∈ H• Inverses: a = e, b = x =⇒ x−1 ∈ H.• Closure: let x, y ∈ H, then y−1 ∈ H by above, so xy = x(y−1)−1 ∈ H.

�

Exercise 2.3.5 (On subgroups)

• Show that the intersection of two subgroups is again a subgroup.

– Hint: one-step subgroup test.

• Show that if H := Cm,K := Cn ≤ G are cyclic, then H ∩K = Cd where d := gcd(m,n).• Show that the intersection of two subgroups with coprime orders is trivial.• Show that the union of two subgroups H,K is a subgroup iff H ⊂ K, and so is generally

not a subgroup.• Show that subgroups with the same prime order are either equal or intersect trivially.• Important for Sylow theory: show (perhaps by example) that if S1, S2 are distinct

subgroups of order pk, then it’s possible for their intersection to be trivial or for themto intersect in a subgroup of order p` for 1 ≤ ` ≤ k − 1.

• Give a counterexample where H,K ≤ G but HK is not a subgroup of G.

E 2.4 Conjugacy e

Definition 2.4.1 (Conjugacy class)The conjugacy class of h is defined as

C(h) :={ghg−1

∣∣∣ g ∈ G} .

2.4 Conjugacy 23

2 Group Theory

Remark 2.4.2: [e] = {e} is always in a conjugacy class of size one – this is useful for counting anddivisibility arguments. Conjugacy classes are not subgroups in general, since they don’t generallycontain e. However, by orbit-stabilizer and the conjugation action, their sizes always divide theorder of G.

Useful qual fact: [x] = {x} ⇐⇒ x ∈ Z(G), i.e. having a trivial conjugacy class is the same asbeing central.

Definition 2.4.3 (Conjugate subgroups)Two subgroups H,K ≤ G are conjugate iff there exists some g ∈ G such that gHg−1 = K.Note that all conjugate subgroups have the same cardinality.

Exercise 2.4.4 (?)Show that the size of a conjugacy class divides the order of a group.

Exercise 2.4.5 (?)Show that if H < G is a proper subgroup, then

⋃g∈G

gHg−1 ⊂ G is a proper subset.

Hint: consider the intersection and count. TryOrbit-stabilizer?

Solution:Strategy: bound the cardinality. All conjugates of H have the same cardinality, say #H = m.Suppose there are n distinct conjugates of H. Then they intersect only at the identity, socount their elements:

#⋃g∈G

gHg−1 = 1 + n(m− 1).

Use that n = [G : NG(H)] by Orbit-Stabilizer, and NG(H) ≤ G =⇒ n ≤ n′ := [G : H]. Nownote n′m = #H[G : H] = #G by Lagrange:

#⋃g∈G

gHg−1 = 1 + n(m− 1)

≤ 1 + n′(m− 1)= 1 + n′m− n′

= 1 + #G− n′

= #G− (n′ − 1)< #G ⇐⇒ n′ := [G : H] > 1.

Exercise 2.4.6 (?)Show that normal groups absorb conjugacy classes: if N E G and [gi] is a conjugacy class ing, either [gi] ⊆ N or [gi] ∩N = ∅.

Exercise 2.4.7 (?)Prove that the size of a conjugacy class of gi is the index of its centralizer, [G : Z(gi)] := [G :

2.4 Conjugacy 24

2 Group Theory

CG(gi)].

2.4.1 Normal Subgroups

Definition 2.4.8 (Normal subgroup)A subgroup N ≤ G is normal iff gH = Hg for every g ∈ G, or equivalently gHg−1 = H forall g, so H has only itself as a conjugate. We denote this by N E G. Equivalently, for everyinner automorphism ψ ∈ Inn(G), ψ(N) = N .

Proposition 2.4.9(Normal iff disjoint union of conjugacy classes).N E G ⇐⇒ N =

∐′[hi] is a disjoint union of conjugacy classes, where the index set for thisunion is one hi from each conjugacy class.

Proof (?).Note that C(hi) =

{ghig

−1∣∣∣ g ∈ G}, and ghig−1 ∈ H since H is normal, so C(hi) ⊆ G for all

i. Conversely, if C(hi) ⊆ H for all hi ∈ H, then ghig−1 ∈ H for all i and H is normal.�

Exercise 2.4.10 (?)

• Show that if H,K E G and H ∩K = ∅, then hk = kh for all h ∈ H, k ∈ K.• Show that if H,K E G are normal subgroups that intersect trivially, then [H,K] = 1 (sohk = kh for all k and h).

Exercise 2.4.11 (?)Prove that if G is a p-group, every subgroup N E G intersects the center Z(G).

Hint: use the class equation.

Solution:Easy solution:

• Use that #H mod p = 1 since H ≤ G and G is a p-group.• Then use that H is a union of conjugacy classes, and since e ∈ H there is at least one

class of size 1, so

#H = #∐′[hi] = #[e] +

′∑#[hi]

=⇒ 0 ≡ #H ≡ 1 +′∑

#[hi] mod p,

and since each #[hi] divides #H, not all can be of size p` since then the sum would be0 mod p. So at least one other #[hi] = 1, making that hi central.

Another solution:

2.4 Conjugacy 25

2 Group Theory

• Idea: use the class equation to force p to divide #(H ∩ Z(G)). Applying it to H yields

H = Z(H)∐mi=1[hi],

where the [hi] are conjugacy classes of size greater than 1.

• Now use that Z(H) = Z(G) ∩H, and since p divides the LHS the result will follow if pdivides the size of the disjoint union on the RHS.

• This is true because each #[hi] 6= 1 and [hi] divides #H which divides #G which is apower of p. So p

∣∣ #[hi] for each i.

E 2.5 Centralizing and Centers e

Definition 2.5.1 (Centralizer)The centralizer of an element is defined as

Z(h) := CG(h) :={g ∈ G

∣∣∣ ghg−1 = h},

the elements of G the stabilize h under conjugation.The centralizer of a subset H is defined as

Z(H) := CG(H) :=⋂h∈H

CG(h) :={g ∈ G

∣∣∣ ghg−1 = h ∀h ∈ H},

the elements of G that simultaneously stabilize all of H pointwise under conjugation.

Definition 2.5.2 (Normalizer)

NG(H) ={g ∈ G

∣∣∣ gHg−1 = H}

=⋃M∈S

M, S :={H∣∣∣ H E M ≤ G

}Contrast to the centralizer: these don’t have to fix H pointwise, but instead can permuteelements of H.

Remark 2.5.3: CG(S) E NG(H) for any H.

Definition 2.5.4 (Center)The center of G is defined as

Z(G) ={g ∈ G

∣∣∣ [g, h] = e∀h ∈ H}

={g ∈ G

∣∣∣ Z(g) = G},

the subgroup of central elements: those g ∈ G that commute with every element of G.

2.5 Centralizing and Centers 26

2 Group Theory

Exercise 2.5.5 (?)

• Show that if G/Z(G) is cyclic then G is abelian.• Show that G/N is abelian iff [G,G] ≤ N .• Show that normal subgroups are not necessarily contained in Z(G).

– Hint: consider the order 3 subgroup of S3.

Solution:The G/Z(G) theorem:

• Write H := Z(G) and G/H = 〈xH〉 as a cyclic quotient.• Fix a, b ∈ G, then aH = xnH and bH = xmH.• So ax−n = h1, bx

−m = h2 where the hi are now central.• Now write ab = (xnh1)(xmh2) = ba by commuting everything.

E 2.6 Cosets e

Proposition 2.6.1(Tower law for subgroups).

K ≤ H ≤ G =⇒ [G : K] = [G : H][H : K].

Proposition 2.6.2(Quotients by bigger subgroups yield smaller quotients).If H ≤ K ≤ G, then

#G = [G : 1] ≥ [G : H] ≥ [G : K] ≥ [G : G] = 1.

In particular, If H,K ≤ G are just arbitrary, since H ∩K ≤ H,K we have [H : H ∩K] ≥ [G :H] and [G : K].

Proof (?).Write G/H ∩K := G/J = {h1J, · · · , hmJ} as distinct cosets where m := [G : H] and the hiare all in H. Then i 6= j =⇒ hih

−1j 6∈ H ∩ K, but hih−1

j ∈ H since subgroups are closedunder products and inverses, which forces hih−1

j 6∈ K. So hiK 6= hjK, meaning there are atleast m cosets in G/K, so [G : K] ≥ m.

�

Proposition 2.6.3(Cosets are equal or disjoint).Any two cosets xH, yH are either equal or disjoint.

Proof (?).

• x ∈ xH, since e ∈ H because H is a subgroup and we can take h = e to get x = xe :=

2.6 Cosets 27

2 Group Theory

xh ∈ xH.

• The reverse containment is clear, so G =⋃x∈G

xH is a union of its cosets.

• Suppose toward a contradiction that ` ∈ xH ∩ yH we’ll show xH = yH.

• Write ` = xh1 = yh2 for some hi, then

xh1 = yh2 =⇒ x = yh2h−11

xh3 ∈ xH =⇒ xh3 = (yh2h−11 )h3 ∈ yH,

so xH ⊆ yH.

• A symmetric argument shows yH ⊆ xH.a

�aSee full argument: D&F p.80.

Theorem 2.6.4(The Fundamental Theorem of Cosets).

aH = bH ⇐⇒ a−1b ∈ H ⇐⇒ b−1a ∈ H.

Proof (?).a

aH = bH ⇐⇒ a ∈ bH ⇐⇒ a = bh for some h ⇐⇒ b−1a = h ⇐⇒ ba−1 ∈ H.

�aSee full argument: D&F p.80.

Definition 2.6.5 (Index of a subgroup)The index [G : H] of a subgroup H ≤ G is the number of left (or right) cosets gH.

Remark 2.6.6(Common coset trick): If you can reduce a problem to showing X ⊆ H, it sufficesto show xH = H for all x ∈ X.

Remark 2.6.7: Cosets form an equivalence relation and thus partition a group. Nice trick: writeG/H = {g1H, g2H, · · · , gnH}, then G =

∐i≤ngiH.

Theorem 2.6.8(Counting Cosets).If H E G and G is finite then

[G : H] = |G/H| = |G||H|

.

2.6 Cosets 28

2 Group Theory

Exercise 2.6.9 (?)Show that if G is finite then |G|/|H| = [G : H].

E 2.7 Special Groups e

Definition 2.7.1 (The Dihedral Group)A dihedral group of order 2n is given by

Dn =⟨r, s

∣∣∣ rn, s2, rsr−1 = s−1⟩

=⟨r, s

∣∣∣ rn, s2, (rs)2⟩

The r is a cycle of length n, and s is a reflection.Examples of explicit cycle presentations:

• D4 = 〈(1, 2, 3, 4), (1, 3)〉 which is a 2π/4 rotation and a reflection through the diagonalline y = −x in a square.

• D5 = 〈(1, 2, 3, 4, 5), (1, 5)(2, 4)〉 which is a 2π/5 rotation and a reflection about the linethrough vertex 3 in a pentagon.

Definition 2.7.2 (The Quaternion Group)The Quaternion group of order 8 is given by

Q =⟨x, y, z

∣∣∣ x2 = y2 = z2 = xyz = −1⟩

=⟨x, y

∣∣∣ x4 = y4, x2 = y2, yxy−1 = x−1⟩

Mnemonic: multiply clockwise to preserve sign, counter-clockwise to negate sign. Everythingsquares to −1, and the triple product is −1:

−1

i

k j

ki=j

ij=k

jk=i

ik=−j

kj=−iji=−k

ijk=−1

Link to Diagram

2.7 Special Groups 29

https://q.uiver.app/?q=WzAsNSxbMiwyLCJpIl0sWzMsNV0sWzAsNSwiayJdLFs0LDUsImoiXSxbMiwwLCItMSJdLFswLDMsImtpPWoiLDJdLFszLDIsImlqPWsiLDJdLFsyLDAsImprPWkiLDJdLFsyLDMsImlrPS1qIiwyLHsiY3VydmUiOjUsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDAsImtqPS1pIiwyLHsiY3VydmUiOjUsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDIsImppPS1rIiwyLHsiY3VydmUiOjUsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDQsImlqaz0tMSIsMl1d

2 Group Theory

Definition 2.7.3 (Transitive Subgroup)A subgroup H ≤ Sn is transitive iff its action on {1, 2, · · · , n} is transitive, i.e. for each pair(i, j) there is some element σ ∈ H such that σ(i) = j. Note that σ may not fix other elements,and can have other effects!

Definition 2.7.4 (p-groups)If |G| = pk, then G is a p-group.

2.7.1 Cyclic Groups

Theorem 2.7.5(Subgroups of Cyclic Groups).G is cyclic of order n := #G iff G has a unique subgroup of order d for each d dividing n.

Proof (?).⇐= : Use that

∑d∣∣n ϕ(d) = n, and that there are at most ϕ(d) elements of order d, forcing

equality.=⇒ : If G = 〈a〉 with an = e, then for each d

∣∣ n take Hd :=⟨and

⟩for existence.

�

Exercise 2.7.6 (?)

• Show that any cyclic group is abelian.• Show that every subgroup of a cyclic group is cyclic.• Show that

ϕ(n) = n∏p

∣∣∣n(

1− 1p

).

• Compute Aut(Z/nZ) for n composite.• Compute Aut((Z/pZ)n).

2.7.2 Symmetric Groups

Definition 2.7.7 (The symmetric group)The transposition presentation:

Sn :=⟨σ1, · · · , σn−1

∣∣∣ σ2i , [σi, σj ] (j 6= i+ 1), σiσi+1σi = σi+1σiσi+1

⟩.

Definition 2.7.8 (The sign homomorphism)


2 Group Theory

Writing a cycle as a product of transpositions, the map defined by

sgn : Sn → (Z/2,+)∏i≤2k

(aibi) 7→ 0

∏i≤2k+1

(aibi) 7→ 1.

Remark 2.7.9:• The kernel is the alternating group:

– Even cycles– For a single cycle: has odd length– Have an even number of even length cycles.– Can be written as an even number of transpositions.– Examples: (1, 2, 3) or (1, 2)(3, 4) in S4.– Non-examples: (1, 2) or (1, 2, 3, 4) in S4, since they have an odd number of even length

cycles.

• The fiber over 1 is everything else:

– Odd cycles– For a single cycle: has even length– Have an odd number of even length cycles.– Can be written as an odd number of transpositions

Mnemonic: the cycle parity of a k-cycle is the usualinteger parity of k − 1.

Definition 2.7.10 (Alternating Group)The alternating group is the subgroup of even permutations, i.e.

An :={σ ∈ Sn

∣∣∣ sgn(σ) = 0}.

These are the permutations with an even number of even length cycles.

Proposition 2.7.11(An is generated by 3-cycles).For n ≥ 3, An is generated by 3-cycles.

Proof (?).Every 3-cycle (abc) is even, and thus in An. Given an arbitrary even permutation (t1 . . . t2k),it decomposes into a product of an odd number of transpositions (t2j−1t2j). So it suffices towrite every such transposition as a 3-cycle. There are only 3 cases the occur:

• (ab)(ab) = ()• (ab)(ac) = (abc)• (ab)(cd) = (abc)(adc).

�


2 Group Theory

Example 2.7.12(Explicit alternating group):

A3 = {id, (1, 2, 3), (1, 3, 2)} ,

which has cycle types (1, 1, 1) and (3).

A4 ={id,(1, 3)(2, 4), (1, 2)(3, 4), (1, 4)(2, 3),(1, 2, 3), (1, 3, 2),(1, 2, 4), (1, 4, 2),(1, 3, 4), (1, 4, 3),(2, 3, 4), (2, 4, 3)},

which has cycle types (1, 1, 1, 1), (2, 2), (3, 1).

A5 is too big to write down, but has cycle types

• (1, 1, 1, 1, 1)• (2, 2, 1)• (3, 1, 1)• (5)

Fact 2.7.13 (Some useful facts)

• σ ◦ (a1 · · · ak) ◦ σ−1 = (σ(a1), · · ·σ(ak))• Conjugacy classes are determined by cycle type• The order of a cycle is its length.• The order of an element is the least common multiple of the sizes of its disjoint cycles.• Disjoint cycles commute.• An≥5 is simple.

E 2.8 Exercises e

Exercise 2.8.1 (?)

• Show that if G is a finite group acting transitively on a set X with at least two elements,then there exists g ∈ G which fixes no point of X.

• Let p be prime. For each abelian group K of order p2, how many subgroups H ≤ Z×3

are there with Z3/H ∼= K?

• Let #G = pq, with p, q distinct primes. Show that G has a nontrivial proper normalsubgroup, and if p 6≡ 1 mod q and q 6≡ 1 mod p then G is abelian.

2.8 Exercises 32

2 Group Theory

• Let G be a finite group and let p be the smallest prime dividing #G, and assume G hasa normal subgroup of order p. Show that H ⊂ Z(G).

• Let G be finite and P a Sylow 2-subgroup. Assume P is cyclic and generated by x. Showthat the sign of the permutation of G corresponding to x 7→ gx is 1, and deduce that Ghas a nontrivial quotient of order 2.

E 2.9 Counting Theorems e

Theorem 2.9.1(Lagrange’s Theorem).

H ≤ G =⇒ #H∣∣ #G.

Moreover, there is an equality [G : H] = #G/#H when G is finite.

Proof (of Lagrange’s theorem).Write G/H = {g0H, g1H, · · · , gNH} for some N := [G : H]. Since cosets are equal or disjointand have equal cardinality,

G =∐k≤NgkH =⇒ #G =

∑k≤N

# (gkH) =∑k≤N

#H = N#H,

so #G = N#H, #H divides #G and N = [G : H] divides #G.�

Corollary 2.9.2(?).

#G = #(G/H)#H := [G : H] #H,

or written another way,

#(G/H) = #G/#H.

Corollary 2.9.3.The order of every element divides the size of G, i.e.

g ∈ G =⇒ o(g)∣∣ o(G) =⇒ g|G| = e.

4! Warning 2.9.4There do not necessarily exist H ≤ G with |H| = n for every n

∣∣ |G|. Counterexample: takeG = A5, then #G = 5!/2 = 60 but G has no subgroup of order 30. If it did, this would be index 2and thus normal, but An≥5 is simple.

2.9 Counting Theorems 33

2 Group Theory

Another direct counterexample: |A4| = 12 but has no subgroup of order 6. If such an H existed, itcan’t contain every 3-cycle, since A4 is generated by 3-cycles. For x any 3-cycle not in H, use that#A4/H = 2 and consider H,xH, x2H. x 6∈ H, so H 6= xH, but two must be equal:

• x2H = H: use x2 = x−1 since x3 = e, but x ∈ H =⇒ x−1 ∈ H, E• xH = x2H: the fundamental theorem of cosets forces x−1x2 ∈ H, so x ∈ H. E

Theorem 2.9.5(Cauchy’s Theorem).For every prime p dividing |G|. there is an element (and thus a subgroup) of order p.

This is a partial converse to Lagrange’s theorem,and strengthened by Sylow’s theorem.

Proof (?).See https://kconrad.math.uconn.edu/blurbs/grouptheory/cauchypf.pdf.

�

E 2.10 Group Actions e

Definition 2.10.1 (Group Action)An action of G on X is a group morphism

ϕ : G×X → X

(g, x) 7→ gx

or equivalently

ϕ : G→ Aut(X)g 7→ (x 7→ ϕg(x) := g · x)

satisfying

1. e · x = x2. g · (h · x) = (gh) · x

Remark 2.10.2(Reminder of notation): For a group G acting on a set X,

Notation Definition

O(x) = Gx ={g · x

∣∣∣ g ∈ G} ⊆ X Orbit

Stab(x) = Gx ={g ∈ G

∣∣∣ g · x = x}≤ G Stabilizer

X/G ⊆ 2X Set of OrbitsFix = XG =

{x ∈ X

∣∣∣ g · x = x ∀g ∈ G}⊆ X Set of Fixed

Points

2.10 Group Actions 34

https://kconrad.math.uconn.edu/blurbs/grouptheory/cauchypf.pdf

2 Group Theory

Note that being in the same orbit is an equivalence relation which partitions X, and G actstransitively if restricted to any single orbit. Also, x ∈ Fix iff Orb(x) = {x} and StabG(x) = G.

Fact 2.10.3For any group action, the kernel is the intersection of all stabilizers:

kerψ =⋂x∈X

Gx.

Definition 2.10.4 (Transitive Group Action)A group action Gy X is transitive iff for all x, y ∈ X there exists a g ∈ G such that g ·x = x.Equivalently, the action has a single orbit.

Proposition 2.10.5(Orbit Stabilizer Isomorphism).If Gy X transitively, then for any choice of x ∈ X there is an isomorphism of sets given by

Φ : G/Gx∼−→ X

gGx 7→ g y x.

Proof (of orbit stabilizer).

• Injectivity: Φ(gGx) = Φ(hGx) ⇐⇒ g y x = hy x ⇐⇒ gh−1 y x = x ⇐⇒ gh−1 ∈Gx ⇐⇒ gGx = hGx.

• Well-definedness: use gGx = hGx ⇐⇒ gh−1 ∈ Gx ⇐⇒ g−1h y x = x. Theng(g−1h) y x = g y x on one hand, and on the other (gg−1)hy x = hy x, so

Φ(hGx) := hy x = (gg−1)hy x = g(g−1h) y x = g y x = Φ(gGx).

• Surjectivity: equivalent to the action being transitive.

�

Proposition 2.10.6(Stabilizers of all orbit reps are conjugate).If X ∈ G-Set, then for any points xi ∈ X in the same orbit, the stabilizers Gx0 and Gx1 areconjugate.Note that if G acts transitively, this says all stabilizers are conjugate.

Proof (that stabilizers are conjugate).

• Fix x ∈ X and y ∈ Orb(x), so g.x = y for some g.• Let Hx := Stab(x) and Hy := Stab(y), the claim is that Hx = g−1Hyg.

2.10 Group Actions 35

2 Group Theory

• Now just check:

h ∈ Hx ⇐⇒ hx = x

⇐⇒ hg−1y = g−1y

⇐⇒ ghg−1y = y

⇐⇒ ghg−1 ∈ Hy

⇐⇒ h ∈ g−1Hyg,

so Hx = g−1Hyg.

�

Theorem 2.10.7(Orbit-Stabilizer).

#Gx = [G : Gx] = #G/#Gx if G is finite.

Mnemonic: G/Gx∼= Gx.

E2.11 Examples of Orbit-Stabilizer and the

Class Equation e

Remark 2.11.1(The fixed-point count trick): A useful mnemonic: for any group action ϕ :Gy X, using that orbits partition X we always have

X = Fix(ϕ)′∐x

Orb(x),

where Fix(ϕ) is the union of all orbits of size 1, and the remaining union is over distinct nontrivialorbits, taking one representative x from each.

Proposition 2.11.2(Simple groups with a nontrivial subgroup embed into symmetricgroups).An application of group actions: if G is simple, H < G proper, and [G : H] = n, then thereexists an injective map ϕ : G ↪→ Sn.

Proof .

• Define a group action ϕ : Gy G/H := {eH, g1H, · · · , gn−1H} acting on the n cosets ofH by left-translation g.(gkH) = (ggk)H.

• Then use that Sym(G/H) ≤ Sn, so imϕ ≤ Sn is a subgroup.

• Since G is simple and kerϕ ≤ G, we have kerϕ = 1, G. If kerϕ = 1, ϕ is injective andwe’re done.

2.11 Examples of Orbit-Stabilizer and the Class Equation 36

2 Group Theory

• Otherwise kerϕ = G, and acting on eH yields gH = H for all g, forcing H = G andn = 1, contradicting that H < G is proper. E

�

2.11.1 Left Translation

Example 2.11.3(The left translation action: trivial): Let G act on itself by left translation,where ϕ : g 7→ (h 7→ gh).

• The orbit Orb(x) = G is the entire group.

– This action is transitive.

• The set of fixed points Fix(ϕ) ={g ∈ G

∣∣∣ gx = x ∀x ∈ G}

= {e} is just the identity.

• The stabilizer Stab(x) ={g ∈ G

∣∣∣ gx = x}

= {e} is just the identity.• The kernel is the identity.• Orbit stabilizer just says G ∼= G/ {e}.

2.11.2 Conjugation: The Class Equation and Burnside’s Lemma

Example 2.11.4(Conjugation yields centers/centralizers): Let G act on itself by conjugation,so ϕ : g.x = gxg−1.

• The orbit Orb(g) = [g] is the conjugacy class of g.

– Thus the action is transitive iff G has only one single conjugacy class, which can onlyhappen if #G = 1, 2. On the other extreme, the orbits are all size 1 iff G is abelian.

• The set of fixed points Fix(ϕ) = Z(G) is the center.• The stabilizer is Stab(g) = Z(g), the centralizer of g in G.• The kernel is the intersection of all centralizers, i.e. again the center Z(G).• Orbit-stabilizer says [g] = G/Z(g), so the size of a conjugacy class is the index of the

centralizer.

Remark 2.11.5: Worth reiterating: [G : Z(g)] is the number of elements in the conjugacy class[g], and each g ∈ Z(G) has a singleton conjugacy class [g] = {g}. Applying the fixed-point count


2 Group Theory

trick and substituting in orbit-stabilizer yields

G = Fix(ϕ)′∐x

Orb(x)

= Z(G)′∐g

[g]

= Z(G)′∐g

G

Z(g) .

Now taking cardinalities yields the class equation:

Corollary 2.11.6(The Class Equation).

#G = #Z(G) +∑

One g fromeach nontrivialconj. class

[G : Z(g)]

As a reminder,

Z(g) ={h ∈ G

∣∣∣ hgh−1 = g}

is the centralizer of g

Z(G) ={h ∈ G

∣∣∣ hgh−1 = g ∀g ∈ G}

=⋂g∈G

Z(g) is the center of G.

Exercise 2.11.7 (Applications of the class equation)

• Show that p groups have nontrivial centers.• Show that groups of order p2 are abelian.

Solution:p-groups have nontrivial centers:

• Abusing notation by identifying sets with their cardinalities, the class equation says

G = Z(G) +′∑g

[G : Z(g)] where the terms in the sum are all bigger than 1.

• Reducing mod p yields 0 = Z(g) + 0, since p must divide [G : Z(g)] when [G : Z(g)] > 1because G = [G : Z(g)]Z(g) and p divides the LHS.

• So p divides Z(g), making Z(g) nontrivial.

p2 groups are abelian:

• Z(G) = 1, p, p2, and by above we know Z(G) 6= 1. If Z(G) = p2 we’re done, so assumeZ(G) = p.

• Then G/Z(G) = p and groups of order p are cyclic, so the G/Z(G) theorem applies andG is abelian.


2 Group Theory

Corollary 2.11.8(Burnside’s Lemma).For G a finite group acting on X,

#X/G = 1#G

∑g∈G

#Fix(g),

where X/G = {Orb(x1), · · · ,Orb(xn)} is the set or orbits and Fix(g) ={x ∈ X

∣∣∣ gx = x}are

the fixed points under g.Slogan: the number of orbits is equal to the averagenumber of fixed points.

Proof (of Burnside’s Lemma).Strategy: form the set A :=

{(g, x) ∈ G×X

∣∣∣ g y x = x}and write/count it in two different

ways. Write Stab(x) ={g ∈ G

∣∣∣ gx = x}and Fix(g) =

{x ∈ X

∣∣∣ gx = x}.

First union over G, where the inner set lets x vary:

A =∐g0∈G

{(g0, x)

∣∣∣ g0x = x}∼=∐g0∈G

{g0} × Fix(g0) ⊆ G×X.

Then union over X, where the inner set lets g vary:

A =∐x0∈X

{(g, x0)

∣∣∣ gx0 = x0}∼=

∐x0∈X

Stab(x0)× {x0} ⊆ G×X.

Taking cardinalities, and using the fact that {p} × A ∼= A as sets for any set A, we get thefollowing equality ∑

g0∈G#Fix(g0) = #A =

∑x0∈X

#Stab(x0).

Now rearrange orbit-stabilizer:

Orb(x0) = G/Stab(x0) =⇒ #Stab(x0) = #G/#Orb(x0),

and use this to rewrite the RHS: ∑g0∈G

#Fix(g0) =∑x0∈X

#Stab(x0)

=∑x0∈X

#G#Orb(x0)

= #G∑x0∈X

1#Orb(x0)

=⇒ 1#G

∑g0∈G

#Fix(g0) =∑x0∈X

1#Orb(x0) ,

so it suffices to show the right-hand side sum is the number of orbits, #(X/G).


2 Group Theory

Proceed by partitioning the sum up according to which orbit each term comes from:

∑x0∈X

( 1#Orb(x0)

)=

∑Orb(x0)∈X/G

∑y∈Orb(x0)

( 1#Orb(x0)

)=

∑Orb(x0)∈X/G

( 1#Orb(x0)

) ∑y∈Orb(x0)

1

=∑

Orb(x0)∈X/G

( 1#Orb(x0)

)#Orb(x0)

=∑

Orb(x0)∈X/G1

= #(X/G).

�

2.11.3 Conjugation on Subgroups

Example 2.11.9(?): Let G act on X :={H∣∣∣ H ≤ G} (its set of subgroups) by conjugation.

• The orbit O(H) ={gHg−1

∣∣∣ g ∈ G} is the set of conjugate subgroups of H.

– This action is transitive iff all subgroups are conjugate.

• The fixed points Fix(G) form the set of normal subgroups of G.

• The stabilizer Stab(H) = NG(H) is the normalizer of H in G.

• The kernel is the intersection of all normalizers: kerϕ =⋂H≤G

NG(H).

• Applying Orbit-stabilizer yields that the number of conjugates is the index of the normalizer:

#{gHg−1

∣∣∣ g ∈ G} = [G : NG(H)].

2.11.4 Left Translation on Cosets

Example 2.11.10(?): For a fixed proper subgroup H < G, let G act on its cosets X := G/H :={gH

∣∣∣ g ∈ G} by left translation.

• The orbit O(xH) = G/H, the entire set of cosets.


2 Group Theory

– Note that this is a transitive action, since the trivial coset eH ∈ G/H and its orbit isgH as g ranges over G, hitting every coset representative.

• The stabilizer Stab(xH) = xHx−1, a conjugate subgroup of H.

– This is becauseStab(xH) =

{g ∈ G

∣∣∣ gxH = xH}

={g ∈ G

∣∣∣ x−1gx ∈ H}

={g ∈ G

∣∣∣ gx ∈ xH}={g ∈ G

∣∣∣ g ∈ xHx−1}

= xHx−1.

• There are no fixed points, i.e. Fix(G) = ∅, since the action is transitive.

• The kernel of this action is kerϕ =⋂g∈G

gHg−1, the intersection of all conjugates of H,

sometimes called the normal core of H.

– Note that if kerϕ = G then H is normal, and if kerϕ = 1 then at least one conjugatedoesn’t intersect H nontrivially.

Proposition 2.11.11(Application of translation action on cosets).If G is a finite group and p := [G : H] is the smallest prime dividing #G, then H E G.

Proof (?).

• Let ϕ : Gy X := {xH}, noting that #X = p and Sym(X) ∼= Sp.

• Then K := kerϕ, and importantly K ⊇ H since K is the intersection of stabilizers, andcontains Stab(eH) ⊇ H since gH = H =⇒ g ∈ H.

• Since G is finite and K ≤ G, we have #(G/K) dividing #G, since

[G : K] = #(G/K) = #G/#K =⇒ #G = #(G/K)#K.

• Now

G/K ∼= K ′ ≤ Sp =⇒ #(G/K)∣∣ p!.

• So #(G/K) divides gcd(#G, p!) = p, using that p was the minimal prime dividing #G.This forces #(G/K) to be 1 or p.

• If it’s p:

– Then p = [G : K] = [G : H] and since K ⊇ H this forces K = H. Kernels areautomatically normal, so we’re done.


3 Sylow Theorems

• If it’s 1:

– Then [G : K] = 1 and K = kerϕ = G.– Identifying kerϕ =

⋂xH∈G/H

Stab(xH), we have Stab(xH) = xHx−1 = G for all x,

which says H is normal.

�

Exercise 2.11.12 (?)Prove the Poincaré theorem for groups: if H ≤ G are possibly infinite groups with finite indexn := [G : H], then there exists an N E H where [N : H] < n!.

3 Sylow Theorems

Remark 3.0.1: Useful facts:

• Counting contributions to #G from Sylp(G): writing #G = pkm so that #Sp = pk, usingthat every order p element is in some Sp one gets at least np(`− 1) for some constant ` > 1.

– Warning: every Sp is the same size, so it’s tempting to take ` := #Sp = pk. But thisonly works if one knows the Sp intersect trivially, e.g. if k = 1. Otherwise, the best onecan do without more information ` = p, i.e. the Sp all intersect trivially or in subgroupsof order p.

– Warning: This isn’t quite a count of elements of order p, since elements in Sp can haveorders pk′ for other k′ ≤ k.

• When counting: just leave the identity out of every calculation, and add it back in as a +1for the final count.

Definition 3.0.2A p-group is a group G such that every element is order pk for some k. If G is a finite p-group,then |G| = pj for some j.

Lemma 3.0.3(Congruences for fixed points).If Gy X for G a p-group, then letting Fix(G) :=

{x ∈ X

∣∣∣ gx = x}, one has

#X ≡ #Fix(G) mod p.

Proof (?).

• Use the fixed-point count trick:

#X = #Fix(G) +′∑x

#Orb(x).

Sylow Theorems 42

3 Sylow Theorems

Note that the result follows immediately by reducing mod p if the sum is zero mod p.• Letting x be an element with a nontrivial orbit, we have #Orb(x) > 1, so Stab(x) 6= G

since orbit-stabilizer would yield #Orb(x) = [G : Stab(x)] = 1.• Now use that #Orb(x) = #G/#Stab(x) = pk/p` where 0 < ` < k with strict inequalities.

So #Orb(x) = pk−` 6= 1, and p divides its size.

�

E 3.1 Statements of Sylow e

For full proofs (some of which I’ve borrowed), seeKeith Conrad’s notes: https: // kconrad. math.uconn. edu/ blurbs/ grouptheory/ sylowpf. pdf

Remark 3.1.1: Some setup and notation: assume

• |G| = pkm where (p,m) = 1,• Sp a Sylow-p subgroup, and• np the number of Sylow-p subgroups.

3.1.1 Sylow 1 (Cauchy for Prime Powers)

Theorem 3.1.2(Sylow 1).

∀pn dividing |G|, there exists a subgroup of size pn.

Slogan 3.1.3Sylow p-subgroups exist for any p dividing |G|, and are maximal in the sense that every p-subgroupof G is contained in a Sylow p-subgroup. If |G| =

∏pαii , then there exist subgroups of order pβii

for every i and every 0 ≤ βi ≤ αi. In particular, Sylow p-subgroups always exist.

Proof (of Sylow 1: left translation).

• Let #G = pkm. Idea: Induct up by showing that if #H = pi for i ≤ k, one can producta bigger subgroup H ⊇ H with [H : H] = p. This makes #H = pi+1.

• Let H ≤ G so that H is a p-group.• Let H y G/H by left-translation.• Use the lemma that #(G/H) ≡ FixH(G/H) mod p• Identify FixH(G/H) = NG(H), since fixing xH means gxH = xH =⇒ gHg−1 ⊆

3.1 Statements of Sylow 43

https://kconrad.math.uconn.edu/blurbs/grouptheory/sylowpf.pdf

https://kconrad.math.uconn.edu/blurbs/grouptheory/sylowpf.pdf

3 Sylow Theorems

H =⇒ gHg−1 = H for all g ∈ G.

xH ∈ FixH(G/H) ⇐⇒ gxH = xH∀g ∈ H⇐⇒ x−1gxH ∈ H∀g ∈ H⇐⇒ x−1Hx−1 = H

⇐⇒ x ∈ NG(H),

so FixH(G/H) ={gH

∣∣∣ g ∈ N(H)}

= NG(H)/H are cosets whose representatives arenormalizers of H.

• Since H E NG(H), these cosets form a group.• We have [G : H] = #(NG(H)/H), and if i < k then p divides [G : H].• So NG(H)/H is a p-group and has a subgroup L of order p by Cauchy.• Use the subgroup correspondence: L ≤ NG(H)/H corresponds to some L′ ≤ G withH ⊆ L′ ⊆ NG(H) and L = L′/H. Now use that #L = p implies #(L′/H) = [L′ : H] = p,so #L′ = [L′ : H]#H = p#H = pi+1 as desired.

�

3.1.2 Sylow 2 (Sylows are Conjugate)

Theorem 3.1.4(Sylow 2).All Sylow-p subgroups Sp are conjugate, i.e.

Sip, Sjp ∈ Sylp(G) =⇒ ∃g such that gSipg−1 = Sjp

Corollary 3.1.5.

np = 1 ⇐⇒ Sp E G.

Proof (of Sylow 2).

• Let S1, S2 ∈ Sylp(G), and let S1 y G/S2 by left-translation.

• Use the lemma:

#(G/S2) ≡ FixS1(G/S2) mod p.

• [G : S2] = m is coprime to p, so there is a fixed point, say xS2 where gxS2 = xS2 for allg ∈ S1.

gxS2 = xS2∀g ∈ S1

=⇒ gx ∈ xS2∀g ∈ S1

=⇒ S1x ⊆ xS2

=⇒ S1 ⊆ xS2x−1,


3 Sylow Theorems

where we now get equality since these sets have the same cardinality.

�

3.1.3 Sylow 3 (Numerical Constraints)

Theorem 3.1.6(Sylow 3).

1. np∣∣ m , and in particular, np ≤ m.

2. np ≡ 1 mod p.

3. np = [G : NG(Sp)] where NG is the normalizer.

Proof (of Sylow 3).

• np ≡ 1 mod p:

– Fix a P ∈ Sylp(G), and let P y S := Sylp(G) by conjugation.– Apply the lemma to get np ≡ FixS(P ) mod p. The claim is that there is just one

fixed point.– If Q ∈ FixS(P ), then pQp−1 = Q for all p ∈ P , so P normalizes Q and P ⊆NG(Q) ≤ G.

– Then P,Q ∈ Sylp(NG(Q)), which by Sylow II are conjugate.– Since Q E NG(Q), there is only one conjugate of Q, and P = Q.– So P is the only fixed point.

• np∣∣ m:

– Let G y X := Sylp(G) by conjugation; this is transitive by Sylow II and there isone orbit.

– Then #X must divide #G, so np divides #G = pkm.– Using np ≡ 1 mod p, we can’t have np

∣∣ pk, and so np must divide m.

• np = [G : NG(P )] for any P ∈ Sylp(G):

– LetGy Sylp(G) by conjugation and apply orbit-stabilizer to get np = [G : Stab(P )]– Identify Stab(P ) = NG(P ).

�

E 3.2 Corollaries and Applications e


3 Sylow Theorems

Corollary 3.2.1.By Sylow 3, p does not divide np.

Proposition 3.2.2.Every p-subgroup of G is contained in a Sylow p-subgroup.

Proof .Let H ≤ G be a p-subgroup. If H is not properly contained in any other p-subgroup, it is aSylow p-subgroup by definition. Otherwise, it is contained in some p-subgroup H1. Inductivelythis yields a chain H ( H1 ( · · ·, and by Zorn’s lemma H := ∪iH i is maximal and thus aSylow p-subgroup.

�

E 3.3 Exercises e

Exercise 3.3.1 (?)

• Let G be a group of order p with v and e positive integers, p prime, p > v, and v is nota multiple of p. Show that G has a normal Sylow p-subgroup.

E 3.4 Automorphism Groups e

Fact 3.4.1Some facts about common automorphism groups, and how to count and reason about them.There’s a good reference here: https://www.whitman.edu/documents/Academics/Mathematics/SeniorProject_BrianSloan.pdf

Let ϕ be the totient function, and note that a cyclic group Cn has precisely ϕ(n) choices of generators.One can compute

ϕ(p) = p− 1ϕ(pk) = pk−1(p− 1)

ϕ(pkq`) = ϕ(pk)ϕ(q`) when gcd(q, p) = 1.

• Automorphisms of cyclic groups are completely known:

Aut(Cn) ∼= C×n ,

which has size ϕ(n) but is not generally isomorphic to Cϕ(n)

3.2 Corollaries and Applications 46

https://www.whitman.edu/documents/Academics/Mathematics/SeniorProject_BrianSloan.pdf

https://www.whitman.edu/documents/Academics/Mathematics/SeniorProject_BrianSloan.pdf

3 Sylow Theorems

4! Warning 3.4.2Warning: C×n is not always cyclic!! For example, C×8 ∼= C2

2 6= C4. In fact, C×n cyclic iff n =2, 4, pk, 2pk for p an odd prime.

• For p an odd prime, Aut(Cp) ∼= C×p∼= Cp−1 is cyclic.

• For pk an odd prime power, Aut(Cpk) ∼= Cϕ(pk) is cyclic.

• For 2k with k ≥ 1, C×2k ∼= C2 × C2k−2 .

• If G,H have coprime order then Aut(G×H) ∼= Aut(G)×Aut(H). One can then compute a

general order by factoring n =∏k=1

pnkk to get a decomposition

Cn = C∏`

k=1 pnkk

=∏k=1

Cpnkk,

and thus

Aut(Cn) ∼= Aut(∏k=1

Cpnkk

)

∼=∏k=1

Aut(Cpnk

k

)∼=∏k=1

C×pnkk

∼= C×2` ×∏k=1pk 6=2

C×pnkk

∼= (C2 × C2`−2)×∏k=1pk 6=2

Cmk mk := ϕ(pnkk )

∼= (C2 × C2`−2)×∏k=1pk 6=2

Cmk mk := pnk−1k (pk − 1).

• Aut(Cnp ) ∼= GLn(Fp) which has size

# GLn(Fp) =n−1∏k=0

(pn − pk) = (pn − 1)(pn − p)(pn − p2) · · · (pn − pn−1).

• Counting homs: # HomGrp

(Cn, Cm) = gcd(n,m).

• If σ ∈ Aut(H) and τ ∈ Aut(N), then N oψ H ∼= N oτ◦ψ◦σ H.

3.4 Automorphism Groups 47

3 Sylow Theorems

• So if GLn shows up in a semidirect product, it suffices to consider similarity classes of matrices(i.e. just use canonical forms).

• Inn(G) ∼= G/Z(G).

Example 3.4.3(?): Some examples of writing automorphism groups as products of cyclic groups:

Aut(C22·3) ∼= Aut(C22)×Aut(C3) ∼= (C2 × C22−2)× Cϕ(3) = (C2)× C2

Aut(C23·3) ∼= Aut(C23)×Aut(C3) ∼= (C2 × C23−2)× Cϕ(3) = (C2 × C2)× C2

Aut(C24·3) ∼= Aut(C24)×Aut(C3) ∼= (C2 × C24−2)× Cϕ(3) = (C2 × C22)× C2

Aut(C22·7) ∼= Aut(C22)×Aut(C7) ∼= (C2 × C22−2)× Cϕ(7) ∼= C2 × C6

Aut(C2·3·5) ∼= Aut(C2)×Aut(C3)×Aut(C5) ∼= 1× Cϕ(3) × Cϕ(5) ∼= C2 × C4.

E 3.5 Isomorphism Theorems e

Theorem 3.5.1(1st Isomorphism Theorem).If ϕ : G→ H is a group morphism then

G/ kerϕ ∼= imϕ.

Note: for this to make sense, we also have

• kerϕ E G• imϕ ≤ G

Corollary 3.5.2.If ϕ : G→ H is surjective then H ∼= G/ kerϕ.

Theorem 3.5.3(Diamond Theorem / 2nd Isomorphism Theorem).If S ≤ G and N E G, then

SN

N∼=

S

S ∩Nand |SN | = |S||N |

|S ∩N |.

3.5 Isomorphism Theorems 48

3 Sylow Theorems

Figure 1: The 2nd “Diamond” Isomorphism Theorem

Remark 3.5.4: For this to make sense, we also have

• SN ≤ G,• S ∩N E S,

If we relax the conditions to S,N ≤ G with S ∈ NG(N), then S ∩N E S (but is not normal in G)and the 2nd Isomorphism Theorem still holds.

Theorem 3.5.5(Cancellation / 3rd Isomorphism Theorem).Suppose N,K ≤ G with N E G and N ⊆ K ⊆ G.

1. If K ≤ G then K/N ≤ G/N is a subgroup2. If K E G then K/N E G/N .3. Every subgroup of G/N is of the form K/N for some such K ≤ G.


3 Sylow Theorems

4. Every normal subgroup of G/N is of the form K/N for some such K E G.5. If K E G, then we can cancel normal subgroups:

G/N

K/N∼=G

K.

Theorem 3.5.6(The Correspondence Theorem / 4th Isomorphism Theorem).Suppose N E G, then there exists a correspondence:

{H < G

∣∣∣ N ⊆ H} {H∣∣∣ H <

G

N

}{Subgroups of Gcontaining N

}{Subgroups of thequotient G/N

}.

In words, subgroups of G containing N correspond to subgroups of the quotient group G/N .This is given by the map H 7→ H/N .

Fact 3.5.7N E G and N ⊆ H < G =⇒ N E H.

E 3.6 Products e

Proposition 3.6.1(HK Subgroup Theorem).If H,K ≤ G and H ≤ NG(K) (or K E G) then HK ≤ G is a subgroup.

Theorem 3.6.2(Chinese Remainder Theorem).

gcd(p, q) = 1 =⇒ Z/pZ× Z/qZ ∼= Z/pqZ.

Theorem 3.6.3(Recognizing Direct Products).We have G ∼= H ×K when

1. H,K E G

2. G = HK.

3. H ∩K = {e} ⊂ G

Note: can relax to [h, k] = 1 for all h, k.

Exercise 3.6.4 (?)Prove the “recognizing direct products” theorem. Can the conditions be relaxed?

3.6 Products 50

3 Sylow Theorems

Remark 3.6.5: Things are particularly nice when the orders of H and k are coprime. For 3,x ∈ H ∩K implies that the order of x divides gcd(#H,#K) = 1, so H ∩K = {e}. Thus for 2, oneonly needs that #(HK) = #G.

Proof (?).With these conditions, the following map is an isomorphism:

Γ : H ×K → G

(h, k) 7→ hk.

• This is a group morphism by condition (1):

Γ(h1, k1)Γ(h2, k2) := (h1k1)(h2k2) = h1(k1h2)k2

= h1(h2k1)k2

= (h1h2)(k1k2):= Γ((h1, k1)(h2, k2)).

• This is surjective by condition (2)• This is injective by condition(3) and checking the kernel:

ker Γ ={

(h, k)∣∣∣ hk = 1G, hk = 1G

}=⇒ h = k−1 =⇒ hk ∈ K ∩H = {1G} .

�

Theorem 3.6.6(Recognizing Generalized Direct Products).

We have G ∼=n∏i=1

Hi when

• Hi E G for all i.

• G = H1 · · ·Hn

• Hk ∩H1 · · · Hk · · ·Hn = ∅

Note on notation: intersect Hk with the amalgamleaving out Hk.

Theorem 3.6.7(Recognizing Semidirect Products).We have G ∼= N oψ H when

• N E G

• G = NH

• H y N by conjugation via a map

ψ : H → Aut(N)h 7→ h(−)h−1.

3.6 Products 51

3 Sylow Theorems

Relaxed condition: H,N E G for direct product,or just H ≤ G for a semidirect product.

E3.7 Classification: Finitely Generated

Abelian Groups e

Definition 3.7.1 (Invariant Factor Decomposition)If G is a finitely generated abelian group, then there is a decomposition

G ∼= Zr ×m∏k=1

Cnk where n1∣∣ · · · ∣∣ nm,

into a free group and a finite number of cyclic groups, where r ∈ Z≥0 is unique and the ni areuniquely determined.

Definition 3.7.2 (Elementary Divisor Decomposition)If G is a finitely generated abelian group, then there is a unique list of not necessarilydistinct prime powers pekk such that

G ∼= Zr ×m∏k=1

Cekpk,

where r ∈ Z≥0 is uniquely determined.

Proposition 3.7.3(Converting between elementary divisors and invariant factors).Given any presentation of a group as a product of cyclic groups G =

∏Zi/mi, with the mi

not necessarily distinct,

• Factor all of the mi into prime powers, keeping the exponents intact.• Organize into a table whose columns correspond to individual primes pi.

– Within an individual column for the prime pk, write all terms of the form pekk (withexponents intact)

– Arrange the terms from lowest at the top to highest at the bottom. Push everythingdown so that the bottom-most rows are all filled out.

• For elementary divisors, just list out all of elements of the table individually, runningacross rows.

• For invariant factors, iterate a process of taking the largest of each prime power (i.e. thebottom row) at each step, deleting that row, and continuing in the same fashion.

Note: this sounds much more complicated than itactually is. Try it!

Example 3.7.4(Abstract Example):

3.7 Classification: Finitely Generated Abelian Groups 52

3 Sylow Theorems

Suppose G is given to you as a product of cyclic groups whose sizes factor in the following way

pe11 p

e21 p

e31 · p

f12 p

f22 · p

g13 p

g23 p

g33 · p

h14 .

Assemble these into a table, grouped by prime factor pi, being careful not to separate primes fromtheir exponents:

p1 p2 p3 p4

pe11 pg1

3pe2

1 pf12 pg2

3pe3

1 pf22 pg3

3 ph14

For elementary divisors: take columns, which just amounts to listing them again:

Z/pe11 × Z/pe2

1 × Z/pe31

× Z/pf12 × Z/pf2

2× Z/pg1

3 × Z/pg23 × Z/pg3

3

× Z/ph14 .

For invariant factors: take rows (grouped by CRT)

Z/(pe3

1 pf22 p

g33 p

h14

)× Z/

(pe2

1 pf12 p

g23

)× Z/ (pe1

1 pg13 ) .

Example 3.7.5(of putting a group in invariant factor form):

G = Z2 × Z2 × Z2 × Z3 × Z3 × Z52

I’ll use a shortcut for the table: instead of listing columns, I just list the prime powers for a singlep in increasing order in the same cell. Then just always take the largest prime power in each cell ateach stage:

p = 2 p = 3 p = 5

2, 2, 2 3, 3 52

=⇒ nm = 52 · 3 · 2


3 Sylow Theorems

p = 2 p = 3 p = 5

2, 2 3 ∅

=⇒ nm−1 = 3 · 2

p = 2 p = 3 p = 5

2 ∅ ∅

=⇒ nm−2 = 2

and thus the invariant factor form is

G ∼= Z2 × Z3·2 × Z52·3·2

Example 3.7.6:

G := Z2 × Z2 × Z23 × Z52·7

Make the table by factoring the order of each cyclic piece, being careful not to combine terms thatcome from distinct summands (e.g. not combining the two copies of 21), and to keep exponentsfrom factorizations intact as a single term (e.g. the 23):

2 5 7

2223 52 7

Reading across rows from bottom to top (and using CRT to merge everything within a row) yieldsinvariant factors on the LHS below. Reading down columns, left to right (merging nothing) yieldselementary divisors on the right-hand side below

Z2 × Z2 × Z23·52·7∼= Z2 × Z2 × Z23 × Z52 × Z7.

Proposition 3.7.7(Number of abelian groups is given by products of partitionnumbers).

If #G := n =m∏k=1

pekk , then there are exactlym∏k=1

P (ek) abelian groups of order n, where P is

the integer partition function.


3 Sylow Theorems

Example 3.7.8(of an integer partition): One can compute P (6) = 11, where all of the partitionsare given by

[6],[5, 1],[4, 2],[4, 1, 1],[3, 3],[3, 2, 1],[3, 1, 1, 1],[2, 2, 2],[2, 2, 1, 1],[2, 1, 1, 1, 1],[1, 1, 1, 1, 1, 1].

Remark 3.7.9: In practice, it is easy to list all of the partitions out for a given n, but it’s alsouseful to have a systematic way to generate them and actually check that you have them all.

Proposition 3.7.10(Formula for partitions).There is a recurrence relation

Pk(n) = Pk(n− k) + Pk−1(n− 1),

which follows from the fact that one can obtain a partition of n with k parts by either

• Taking a partition of n − k into k parts and adding 1 to each part, e.g. [1, 1, 1, 3] 7→[2, 2, 2, 4]

• Taking a partition of n − 1 into k − 1 parts and adding a new standalone part 1,e.g. [1, 1, 2, 5] 7→ [1, 1, 2, 5, 1].

Summing over k yields the following, which can be recursed:

P (n) =n∑k=1

Pk(n− k) + P (n− 1)

=n∑k=1

Pk(n− k) +n−1∑k=1

Pk(n− 1− k) + P (n− 2)

= · · · ,

where Pk(m) = 0 for k > m and Pm(m) = 1.


3 Sylow Theorems

Example 3.7.11(?): One can compute that P (5) = 7, and the formula recovers this:

P (5) =5∑

k=1Pk(5− k) + P (4)

= (P1(4) + P2(3)) + P (4)= (P1(4) + P2(3)) + (P1(3) + P2(2)) + P (3)= (P1(4) + P2(3)) + (P1(3) + P2(2)) + (P1(2)) + P (2)= (P1(4) + P2(3)) + (P1(3) + P2(2)) + (P1(2)) + (P1(1) + P (1))= (1 + 1) + (1 + 1) + (1) + (1 + 1)= 7.

Note that you could just stop at the third line, since P (3) = 3 is easy to enumerate: [1, 1, 1], [1, 2], [3].

Example 3.7.12(Applying this to classifying groups): Suppose #G = n = p3q4. Computethat p(3) = 3 and p(4) = 5, so there should be 15 abelian groups of this order. Enumerate thepartitions:

• For 3: [1, 1, 1], [1, 2], [3]• For 4: [1, 1, 1, 1], [1, 2, 1], [1, 3], [2, 2], [4]

Now for every distinct pair taking one from the first line and one from the second, we get agroup of that order. A partition of m of the form [a, b, c, · · · ] contributes a group of the formZma × Zmb × Zmc · · ·.

Crossing [1, 1, 1] with everything:

• (Zp × Zp × Zp)× (Zq × Zq × Zq × Zq) 7→[1, 1, 1]× [1, 1, 1, 1]• (Zp × Zp × Zp)×

(Zq × Zq2 × Zq

)7→[1, 1, 1]× [1, 2, 1]

• (Zp × Zp × Zp)×(Zq × Zq3

)7→[1, 1, 1]× [1, 3]

• (Zp × Zp × Zp)×(Zq2 × Zq2

)7→[1, 1, 1]× [2, 2]

• (Zp × Zp × Zp)× Zq4 7→[1, 1, 1]× [4]

Crossing [1, 2] with everything:

•(Zp × Zp2

)× (Zq × Zq × Zq × Zq) 7→[1, 2]× [1, 1, 1, 1]

•(Zp × Zp2

)×(Zq × Zq2 × Zq

)7→[1, 2]× [1, 2, 1]

•(Zp × Zp2

)×(Zq × Zq3

)7→[1, 2]× [1, 3]

•(Zp × Zp2

)×(Zq2 × Zq2

)7→[1, 2]× [2, 2]

•(Zp × Zp2

)× Zq4 7→[1, 2]× [4]

And so on!


3 Sylow Theorems

E 3.8 Classification: Groups of Special Orders e

General strategy: find a normal subgroup (usually a Sylow) and use recognition of semidirectproducts.

• Keith Conrad: Classifying Groups of Order 12• Order pqr: ?• Order p2q: ?

Proposition 3.8.1(Classification of groups of order p).Every group G of prime order p ≥ 2 is cyclic and thus isomorphic to Z/p.

Proof (?).Supposing that g 6= e, it generates a cyclic subgroup H := 〈g〉 ≤ G of order dividing p byLagrange. Since g 6= e, #H = p = #G.

�

Proposition 3.8.2(Classification of groups of order p2).Every group G of order p2 is abelian, and thus isomorphic to either Cp2 or C2

p .

Proof (?).Quotient by the center to get m := #G/Z(G) ∈

{1, p, p2

}. By cases:

• Since G is a p-group, G has nontrivial center, so m 6= 1• If m = p, then G/Z(G) is cyclic and thus G is abelian by the G/Z(G) theorem.• If m = p2, Z(G) = G and G is abelian, done.

�

Proposition 3.8.3(Classification of groups of order pq).If G is a group of order pq where without loss of generality q < p, then

1. If q∣∣- p− 1 then G is cyclic and G ∼= Sp × Sq ∼= Cpq.

2. If q∣∣ p − 1 then G ∼= Sq oψ Sp where Sp E G and ψ : Sq → Aut(Sp), and G has a

presentation

G ∼=⟨a, b

∣∣∣ ap, bq, bab−1 = a`⟩

` 6≡ 1 mod p `q ≡ 1 mod p.

Proof (of pq theorem, case 1).

• Suppose q < p.• Apply the Sylow theorems to p:

3.8 Classification: Groups of Special Orders 57

https://kconrad.math.uconn.edu/blurbs/grouptheory/group12.pdf

3 Sylow Theorems

– np ∼= 1 mod p =⇒ np ∈ {1, p+ 1, 2p+ 1, · · ·}.– np

∣∣ q =⇒ np ∈ {1, q}.– Since 1 < q < p < p+ 1, this forces np = 1

• Suppose q∣∣- p− 1 and apply the Sylow theorems to q:

– nq ≡ 1 mod q =⇒ nq ∈ {1, q + 1, 2q + 1, · · ·}– nq

∣∣ p =⇒ nq ∈ {1, p}– Now note that if nq 6= 1, then nq = p and p is of the form kq + 1 for some k.– Use of assumption: then p = kq + 1 ⇐⇒ p − 1 = kq ⇐⇒ q

∣∣ p − 1, which isprecisely what we assumed is not the case.

• So np = nq = 1 and Sp, Sq E G.• Apply recognition of direct products:

– Sp, Sq ≤ G: check.– Sp, Sq E G: check.– Sp ∩ Sq = {e}: check, because they are coprime order.– SpSq = G: follows from a counting argument:

#SpSq = #Sp#Sq# (Sp ∩ Sq)

= pq

1 = #G.

If G is finite, then AB ≤ G with #AB = #G implies AB = G.

�

Proof (of pq theorem, case 2). • Suppose q∣∣ p−1, the previous argument for Sp works, but

the argument for Sq doesn’t, so we get a semidirect product.• Work up to isomorphism:

Sp ∼= Z/p =⟨a∣∣∣ ap⟩ E G

Sq ∼= Z/q =⟨b∣∣∣ bq⟩ ≤ G.

• We have

G ∼= Z/q oψ Z/p ψ : Z/q → Aut (Z/p)

=⇒ G ∼=⟨a, b

∣∣∣ ap, bq, aba−1 = ψ(b) = b`⟩

for some `.

– Since Z/q is cyclic, such a morphism is determined by the image of the generator[1]q ∈ Z/q.

– Note that [1]q 7→ idZ/p is such a morphism, and yields the direct product again.

• Identify Aut(Z/p) ∼=((Z/p)× ,×

)∼= (Z/(p− 1),+).

• So we need to classify morphisms

ψ : Z/q → Z/(p− 1).


3 Sylow Theorems

– Consider imψ ≤ Z/(p− 1).– Sending [1]q to the identity in Aut(Z/p) yields the direct product again, so pick

nontrivial morphisms.– Since # imψ

∣∣ q which is prime, its order is equal to q.– Since q

∣∣ p− 1 and Z/(p− 1) is cyclic of order p− 1, by Cauchy’s theorem there isa unique subgroup of order q, say Cq ≤ Z(p− 1)

– We can send [1]q to [α]p−1 ∈ Z/(p − 1) where α is any generator of Cq, of whichthere are ϕ(q) = q − 1 nontrivial choices.

• Thus there are q − 1 distinct nontrivial choices for the action ψ : Z/q → Z/(p− 1).

Claim: All choices yield isomorphic semidirect products.

• Use that G := Aoψ N with ψ : A→ Aut(N) is an Aut(N) and Aut(A) module, wheref ∈ Aut(N) and π ∈ Aut(A) act in the following ways:

π y Aoψ N = Aoψ◦π N

f y Aoψ N = Aoγf◦ψ N.

where

γf : Aut(N)→ Aut(N)ψ 7→ f ◦ ψ ◦ f−1.

– These actions preserve the group isomorphism type of G

• However, since Cq ≤ Z/(p − 1) and Aut(Cq) ∼= Z/(q − 1), there are exactly q − 1automorphisms of the image Cq, say {πk}q−1

k=1.• So ψ ◦ πk : Z/q → Z/(p − 1) for 1 ≤ k ≤ q − 1 yields q − 1 distinct actions, and we’re

done.

�

Lemma 3.8.4(Frattini’s Argument).If N E G and P ∈ Sylp(H) then G = NG(P )H.

Proof (?).

• Let g ∈ G, then since P ≤ H E G we have gPg−1 ⊆ gHg−1 = H.• So P ′ := gPg−1 ∈ Sylp(H) for all g, and since Sylows in H are all conjugate, we can

write P ′ = h−1Ph−1 for some h ∈ H.• This says hPh−1 = gPg−1 and thus P = (g−1h)P (h−1g) = (h−1g)−1P (h−1g).• But then g−1h ∈ NG(P ) so g ∈ NG(P )H.

�

Lemma 3.8.5(p groups are solvable).Every finite p group is solvable.


3 Sylow Theorems

Proof (?).

• By induction on k in #G = pk: if #G = p then G is abelian and automatically solvable.• Inductively, for #G = pk, now consider Z(G) 6= 1 since we’re in a p-group.• If G/Z(G) is abelian, use the general fact: H solvable and G/H solvable implies G

solvable.

– Here Z(G) and G/Z(G) are both abelian and thus solvable.

• Otherwise G/Z(G) is a p-group of size pk−1 and thus solvable by hypothesis.

�

Lemma 3.8.6(pq groups have normals the size of the biggest prime).If #G = pq with p < q distinct primes, then G has a normal subgroup of size q.This is immediate from Sylow theory: [nq]q = 1, nq

∣∣ p, p < q forces nq = 1.

Proposition 3.8.7(PQR Theorem).If |G| = pqr where p < q < r are distinct primes then G is solvable.

Proof (?).Idea:

• Get a normal subgroup R of order r, so #(G/R) = pq.• Get a normal subgroup Q1 of order q in G/R, which corresponds to Q E G of order qr

containing R. Note that R E Q since normality descends to subgroups.• Now G → Q → R → 1 is a subnormal series whose quotients are all cyclic and thus

abelian:

– #(G/Q) = pqr/qr = p,– #(Q/R) = qr/r = q,– #(R/1) = r,

�

Remark 3.8.8: Proof of first claim: let m := #G = pqr, then G has a normal subgroup of order r.

• Claim: at least one of the Sylows for p, q, or r is normal.

– If none of the Sylow p, q, r groups are normal, then nr ≥ r and np ≥ q. Counting thecontributions from just Sylq(G) and Sylp(G) yields

nq(q − 1) + nr(r − 1) ≥ pr(q − 1) + pq(r − 1) = pqr + p(qr − q − r).

– If this is to be at most m, it must be that qr−q−r is negative (since p > 1 and otherwisethis would yield more than pqr elements).


3 Sylow Theorems

– But if this holds,

qr − q − r ≤ 0 ⇐⇒ q(r − 1) ≤ r ⇐⇒ q ≤ r

r − 1 .

But q > 2 be assumption, and 1 ≤ r

r − 1 ≤ 2 for any number r. E.

– So there is one of Sp, Sq, Sr that is normal in G.

• Now if Sr is normal we’re done, so suppose not and nr > 1. Claim: we can get anothersubgroup of order r

– Let N be the normal Sylow, so either N ∈ Sylp(G) or N ∈ Sylq(G).

– Then G/N has order r` for either ` = q or ` = p respectively.

– In either case, ` < r. Using the lemma, G/N has a normal subgroup of size r, sayR/N ≤ G/N .

– Then by the subgroup correspondence theorem, R corresponds to a normal subgroupR′ E G of size r` with r < `.

– Applying the same lemma to R′ immediately yields a normal subgroup R′′ of order r inR′

– Now use that R′′ charR′ since Sylows are characteristic, and R′ E G, so R′′ E G too.

E 3.9 Series of Groups e

Definition 3.9.1 (Normal Series)A normal series of a group G is a sequence G → G1 → G2 → · · · such that Gi+1 E Gi forevery i.

Definition 3.9.2 (Central Series)A central series for a group G is a terminating normal series G→ G1 → · · · → {e} such thateach quotient is central, i.e. [G,Gi] ≤ Gi−1 for all i.

Definition 3.9.3 (Composition Series)A composition series of a group G is a finite normal series such that Gi+1 is a maximalproper normal subgroup of Gi.

Theorem 3.9.4(Jordan-Holder).Any two composition series of a group have the same length and isomorphic composition factors(up to permutation).

3.9 Series of Groups 61

3 Sylow Theorems

Definition 3.9.5 (Simple Groups)A group G is simple iff H E G =⇒ H = {e} , G, i.e. it has no non-trivial proper subgroups.

Proposition 3.9.6.If G is not simple, then G is an extension of any of its normal subgroups. I.e. for any N E G,G ∼= E for some extension of the form N → E → G/N .

Definition 3.9.7 (Lower Central Series)Set G0 = G and Gi+1 = [G,Gi], then G0 ≥ G1 ≥ · · · is the lower central series of G.

Mnemonic: “lower” because the chain is descending.Iterate the adjoint map [−, G], if this terminates thenthe map is nilpotent, so call G nilpotent!

Definition 3.9.8 (Upper Central Series)Set Z0 = 1, Z1 = Z(G), and Zi+1 ≤ G to be the subgroup satisfying Zi+1/Zi = Z(G/Zi).Then Z0 ≤ Z1 ≤ · · · is the upper central series of G.Equivalently, since Zi E G, there is a quotient map π : G → G/Zi, so define Zi+1 :=π−1(Z(G/Zi)) (?).

Mnemonic: “upper” because the chain is ascending.“Take higher centers”.

Definition 3.9.9 (Derived Series)Set G(0) = G and G(i+1) = [G(i), G(i)], then G(0) ≥ G(1) ≥ · · · is the derived series of G.

E 3.10 Solvability e

Remark 3.10.1: A useful way to extract normal subgroups: let G act on literally anything byϕ : G→ Aut(X). Then kerϕ E G is always a normal subgroup. Some examples:

• Gy G by x 7→ gx.• Gy {H ≤ G} by H 7→ gH or H 7→ gHg−1.• Gy

{Sylp(G)

}for a fixed p by Sp 7→ gSpg

−1.• Gy H for H E G by inner automorphisms h 7→ ghg−1.

Definition 3.10.2 (Solvable)A group G is solvable iff G has a terminating normal series with abelian composition factors,i.e.

G := Gn > Gn−1 > · · · > G2 > G1 := {e} with Gi/Gi+1 abelian for all i.

Remark 3.10.3: If G = Gal(L/K) is a Galois group corresponding to a polynomial f , then G issolvable as a group iff f is solvable in radicals: there is a tower of extensions K = F0 ⊂ F1 ⊂ F2 ⊂

3.10 Solvability 62

4 Ring Theory

· · · ⊂ Fm = L where

1. Fi = Fi−1(αi) where αmii ∈ Fi−1 for some power mi ∈ Z≥0, and2. Fm ⊇ SF(f) contains a splitting field for f .

Theorem 3.10.4(Characterization of Solvable).A group G is solvable iff its derived series terminates.

Theorem 3.10.5(Sn is Almost Always Solvable).If n ≥ 4 then Sn is solvable.

Fact 3.10.6Some useful facts about solvable groups:

• G is solvable iff G has a terminating derived series.• Solvable groups satisfy the 2 out of 3 property• Abelian =⇒ solvable• Every group of order less than 60 is solvable.

4 Ring Theory

Proposition 4.0.1(Subring criteria).A subset S ⊆ R is a subring iff

• (S,+) forms an abelian subgroup (so closed under addition and contains inverses)• (S, ·) forms a submonoid (so closed under multiplication)

Proposition 4.0.2(Ideal Operations).

• I + J ={i+ j

∣∣∣ i ∈ I, j ∈ J} = 〈I, J〉 is the smallest ideal containing I and J .

• IJ =

∑k≤N

xkyk∣∣∣ xk ∈ I, yk ∈ J,N ∈ Z≥0

is the ideal generated by all finite sums of

products.• I ∩ J is an ideal, I ∪ J is generally not an ideal• Ideals are comaximal if I + J = 〈1〉.• If I + J = 〈1〉 then I ∩ J = IJ .

Definition 4.0.3 (Ideal generated by a set)The ideal generated by {a, b} is defined as

〈a, b〉 := Ra+Rb :={r1a+ r2b

∣∣∣ ri ∈ R} .

Ring Theory 63

4 Ring Theory

More generally for a set S = {sk},

〈S〉 :=#S∑k=1

Rsk :={∑

rksk∣∣∣ rk ∈ R, sk ∈ S} .

Example 4.0.4(?): • 〈p, q〉 = 〈gcd(p, q)〉 E Z.

E 4.1 Isomorphism Theorems e

Remark 4.1.1: These are all basically the same for modules.

Proposition 4.1.2(First Isomorphism Theorem).For any ring morphism f : A→ B there is SES of rings

0→ ker f → A→ im(f)→ 0,

and thus A/ ker f ∼= im f . If f is surjective, then A/ ker f ∼= B.More traditionally stated:

• kerϕ ∈ Id(A)• imϕ ≤ B is a subring (not necessarily an ideal)• R/ kerϕ ∼= imϕ.

Proposition 4.1.3(Second Isomorphism Theorem).Let R ∈ Ring, S ≤ R, I ∈ Id(R), then there is an isomorphism:

S + I

I∼−→ S

S ∩ I.

Where it’s also true that this statement makes sense:

• S + I ≤ R is a subring.• S ∩ I E S

Proposition 4.1.4(Third Isomorphism Theorem).For I ∈ Id(R), the canonical quotient map ϕ : R→ R/I induces a bijective correspondence:{

J∈Id(R)∣∣∣ J⊇I} Id(R/I)

J := ϕ−1(J) 7→JJ 7→ J := ϕ(J),

where ϕ : R→ R/I is the canonical quotient morphism.More traditionally:


4 Ring Theory

• If S, I ∈ Id(R) with S containing I then

S/I ≤ R/I.

• Every ideal in Id(R/I) is of the form S := S/I for some S ∈ Id(R) containing I.

• If I, J ∈ Id(R) with I ⊆ J ⊆ R then there is an isomorphism

R/I

J/I∼−→R

J.

Moreover, A ≤ R is a subring containing I iff A/I ∈ Id(R/I).

Exercise 4.1.5 (?)Show that if J ∈ Id(R) (with J ⊇ I) is radical/prime/maximal iff J ∈ Id(R/I) is radi-cal/prime/maximal.

E 4.2 Important Techniques e

Proposition 4.2.1(Fields are simple).R ∈ Field ⇐⇒ Id(R) = {0, R}.

Proof (?).=⇒ : If 0 6= x ∈ I E R, using that R• = R×, x is a unit. So x−1 ∈ R, and xx−1 := 1 ∈ I soI = R.⇐= : Let x ∈ R•, then Rx = R so 1 ∈ Rx and 1 = rx for some r ∈ R. This forces x = r−1.

�

Proposition 4.2.2(Showing ideals are maximal/prime with quotients).

• R/m is a field ⇐⇒ m ∈ mSpec(R) is maximal.• R/p is an integral domain ⇐⇒ p ∈ Spec(R) is prime.• R/J is reduced ⇐⇒ J is radical.

Proof (of 1).Use the ideal correspondence theorem: Id(R/m) are ideals of R containing m:

R/m ∈ Field⇐⇒ 6 ∃J/m ∈ Id(R/m)• such that J ∈ Id(R)⇐⇒ 6 ∃m ( J ( R

⇐⇒ J ∈ mSpec(R).

�

4.2 Important Techniques 65

4 Ring Theory

Proof (of 2).⇐= : Show xy = 0 with x 6= 0 forces y = 0. Let x, y ∈ p ∈ SpecR, so x = a+ I, y = b+ I forsome a, b ∈ R. If xy = 0 mod p with y 6= 0 mod p, we can check

xy = (a+ p)(b+ p) := (ab) + p = 0 + p =⇒ ab ∈ p.

Since p is prime and x 6= 0 =⇒ a 6∈ p, so b ∈ p. But then

y := b+ p = 0 + p = 0 mod p.

=⇒ : Let a, b ∈ R with xy ∈ p, we want to show that if x 6∈ p then y ∈ p. Note x 6∈ p ⇐⇒x ∼= 0 mod p. Setting x := a+ p, y := b+ p yields

xy := (a+ p)(b+ p) := ab+ p = 0 mod p.

Since R/p is a domain, assuming x 6= 0 mod p we have y = 0 mod p, so y ∈ p.�

Remark 4.2.3: Note that this yields a quick proof that mSpecR ⊆ SpecR, using that Field ≤IntDomain:

I maximal ⇐⇒ R/I ∈ Field =⇒ R/I ∈ IntDomain ⇐⇒ I prime.

Fact 4.2.4If m is maximal and x ∈ R \m then m +Rx = R = 〈1〉.

E 4.3 Undergrad Review e

Remark 4.3.1: Notation:

• 〈a〉 := Ra :={ra∣∣∣ r ∈ R} is the ideal generated by a single element.

• R = 〈1〉 is equivalently the ideal generated by 1.

4.3.1 Basics

Definition 4.3.2 (Ring)A ring is a triple (R,+, ·) ∈ CRing such that

• (R,+) ∈ AbGrp,• (R, ·) ∈ Mon• Distributivity: a(b+ c) = ab+ ac.

4.3 Undergrad Review 66

4 Ring Theory

Example 4.3.3(of rings): Some of the most important examples of rings:

• The usual suspects: Z,Q

– Their analogs: number fields K := Q(ζ), their rings of integers ZK or OK ,

• Gaussian integers Z(i)• Fields k = Fpn ,R• Fraction fields of rings ff(R), e.g. ff(Z) = Q.• Polynomial rings R[x1, x2, · · · , xn], particularly for R = k a field• Power series rings R[x1, x2, · · · , xn].

– Formal power series rings R[[x1, x2, · · · , xn]].

• Zp :={a/b

∣∣∣ p ∣∣- b} the ring of p-adic integers• Rings of germs, e.g. C∞(X,Y ) where f ∼ g iff there exists some U ⊆ X with f |U = g|U .

Definition 4.3.4 (Ring Morphism)A morphism f ∈ CRing(X,Y ) satisfies:

• f(1X) = 1Y• f(a(b+ c)) = f(a)f(b) + f(a)f(c)

Remark 4.3.5: Important notes:

• ker f := f−1({0}).• A bijective ring morphisms is automatically an isomorphism in CRing.• ker f E X is an ideal, but im f ≤ Y is only a subring in general.• For any ideal I E R there is a quotient map R→ R/I, it’s useful to write cosets as a+ I.• For quotients, x ≡ ymod I ⇐⇒ x− y ∈ I.

Definition 4.3.6 (Ideal)An ideal I E R is a subset where (I,+) ≤ (R,+) ∈ Grp is a subgroup and for x ∈ R, i ∈ I,xi ∈ I. Equivalently,

• RI ⊆ I• I + I ⊆ I

Note that 0 is in every ideal.

Definition 4.3.7 (Characteristic)Using that every ring has a Z-Mod structure, the characteristic of a ring R is the smallest n

such that ny 1R = 0R, i.e.n∑i=1

1R = 0R.


4 Ring Theory

4.3.2 Elements

Definition 4.3.8 (Divisibility of Elements)An element r ∈ R is divisible by q ∈ R if and only if there exists some c ∈ R such that r = qc.In this case, we sometimes write q

∣∣ r.Definition 4.3.9 (Units)An element r ∈ R is a unit if r

∣∣ 1: there exists an s ∈ R such that rs = sr = 1. Then r−1 := sis uniquely determined, and the set of units (R×, ·) ∈ AbGrp forms a group.

Definition 4.3.10 (Irreducible Element)An element r ∈ R is irreducible iff

r = ab =⇒ a ∈ R× or b ∈ R×

Definition 4.3.11 (Prime Element)An element p ∈ R is prime iff

a, b ∈ R× \ {0} , ab∣∣ p =⇒ a

∣∣ p or b∣∣ p.

Fact 4.3.12If R is an integral domain, prime =⇒ irreducible. If R is a UFD, then irreducible =⇒ prime, sothis is an iff.

Definition 4.3.13 (Associate Elements)a, b ∈ R are associates iff there exists a u ∈ R× such that a = ub. Equivalently, a

∣∣ b andb∣∣ a.

4.3.3 Ideals

Example 4.3.14(of specs):

• Id(Z) ={〈m〉

∣∣∣ m ∈ Z≥0}

• mSpecZ ={〈p〉

∣∣∣ p 6= 0 is prime}

• SpecZ = mSpecZ ∪ {〈0〉}.• For k a field and f ∈ k[x1, · · · , xn] irreducible, 〈f〉 ∈ Spec k[x1, · · · , xn].

– m :={f =

∑I

aIxI ∈ k[x1, · · · , xn]

∣∣∣ a0 = 0}∈ mSpec k[x1, · · · , xn] (i.e. this is the ideal

of polynomials with no constant term).


4 Ring Theory

Proposition 4.3.15(Proper ideals contain no units).If I E R is a proper ideal ⇐⇒ I contains no units.

Proof .r ∈ R× ∩ I =⇒ r−1r ∈ I =⇒ 1 ∈ I =⇒ x · 1 ∈ I ∀x ∈ R.

�

Proposition 4.3.16.If I1 ⊆ I2 ⊆ · · · are ideals then ∪jIj is an ideal.

Definition 4.3.17 (Irreducible Ideal)An ideal I E R is irreducible if it can not be written as the intersection of two larger ideals,i.e. there are not J1, J2 ⊇ I such that J1 ∩ J2 = I.

Definition 4.3.18 (Prime Ideal)p is a prime ideal ⇐⇒

ab ∈ p =⇒ a ∈ p or b ∈ p.

Proposition 4.3.19(Prime implies irreducible for UFDs).In R a UFD, an element r ∈ R is prime ⇐⇒ r is irreducible.

Example 4.3.20(of why the converse doesn’t hold): For R an integral domain, prime =⇒irreducible, but generally not the converse. Take R := k[x, y]/

⟨x2 − y3

⟩∼= k[x2, y3], which is a

domain, But here [x2] = [y3] as equivalence classes where [y3] is irreducible since every element inr ∈ R has degy(r) = 0, 3, 6, · · ·. But [y3] is not prime since it divides [x2] but doesn’t divide [x].

Definition 4.3.21 (Prime Spectrum)The prime spectrum (or just the spectrum) of R is defined as

Spec(R) ={p E R

∣∣∣ p is prime}.

Definition 4.3.22 (Maximal Ideal)An ideal m is maximal iff whenever I E R with m ( I a proper containment then I = R.

Example 4.3.23(Some counterexamples): Some examples. Reminder: maximal always impliesprime, and for PIDs, prime and nonzero implies maximal. Maximals quotient to fields, primes todomains.

• Prime and maximal:

– pZ ∈ Id(Z). Maximal (and thus prime) since Z/p is a field and a domain.– 〈2, x〉 ∈ Id(Z[x]). Maximal (and thus prime) Z[x]/ 〈2, x〉 ∼= Z/2 is a field and a domain.

• Prime but not maximal:


4 Ring Theory

– 〈0〉 ∈ Id(Z), since mZ ⊇ 〈0〉 for any m.– 〈x〉 ∈ R[x] over any integral domain since R[x]/ 〈x〉 ∼= R is a domain (making it maximal),

but R can be chosen not to be a field (making it non-prime).

• Not prime, not maximal:

– mZ ∈ Id(Z), since m composite implies Z/m is not a domain since it has nonzero zerodivisors. For example, in Z/6, [3] is a zero divisors since [2][3] = 0.

• Useful examples:

– mSpecZ = {pZ} and SpecZ = {pZ} ∪ 〈0〉.– mSpecC[x] =

{x− a

∣∣∣ a ∈ C}, since over a PID 〈α〉 is maximal iff α is irreducible, and

over C irreducibles are degree 1.– mSpec k[x1, · · · , xn] =

{〈x− a1, x− a2, · · · , x− an〉

∣∣∣ ak ∈ k}.• A ring with no maximal ideals: the Prüfer p-group Z(p∞) =

{ζpk}∞k=1

with the trivial ring

structure xy = 0. The subgroups are Hk :={ζpk}, which form an increasing chain that

doesn’t stabilize.

Definition 4.3.24 (Max Spectrum)The max spectrum of R is defined as

mSpec(R) ={m E R

∣∣∣ m is maximal}.

Example 4.3.25(An irreducible element that is not prime.): 3 ∈ Z[√−5]. Check norm to

see irreducibility, but 3∣∣ 9 = (2 +

√−5)(2−

√−5) and doesn’t divide either factor.

Example 4.3.26: Maximal ideals of R[x] are of the form I = (x− ai) for some ai ∈ R.

E 4.4 Types of Rings e

Definition 4.4.1 (Division ring or skew field)A division ring is any (potentially noncommutative) ring R for which R\{0} ⊂ R×, i.e. everynonzero element is a unit and thus has a multiplicative inverse.

Definition 4.4.2 (Zero Divisor)An element r ∈ R is a zero-divisor iff there exists an a ∈ R \ {0} such that ar = ra = 0,i.e. r

∣∣ 0. Equivalently, the map

r· : R→ R

x 7→ rx

fails to be injective.

4.4 Types of Rings 70

4 Ring Theory

Definition 4.4.3 (Integral Domain)A ring is an integral domain if and only if it has no nonzero zero divisors:

a, b ∈ R \ {0} , ab = 0 =⇒ a = 0orb = 0.

Example 4.4.4(of integral domains): Examples of integral domains: Z, k[x1, x2, · · · , xn]. Non-examples: Z/6,Mat(2× 2; k)

Definition 4.4.5 (Field)A field is a commutative division ring, i.e. every nonzero element is a uni, i.e. every nonzeroelement is a unit

Exercise 4.4.6 (?)Show that TFAE:

• A ∈ Field• A is a simple ring, so Id(A) = {0, A}.• If B ∈ Field is nonzero then every ring morphism A→ B is injective.

Remark 4.4.7: Every field is an integral domain, but e.g. Z is an integral domain that is not afield.

4.4.1 The Big Ones

Definition 4.4.8 (Principal Ideal)An ideal I E R if principal if there exists an a ∈ R such that I = 〈a〉, i.e. I = Ra.

Definition 4.4.9 (Principal Ideal Domain)A ring R is a principal ideal domain iff every ideal is principal.

Exercise 4.4.10 (?)Let R be a PID.

• Show primes are maximal, so SpecR ⊆ mSpecR and nonzero ideals are prime iff maximal.• Show that R is Noetherian.• Show that every element is a finite product of irreducibles.• Show that in a PID, every maximal ideal is generated by an irreducible element.• Show that not Z is Noetherian but not Artinian.

– Hint: take a chain nZ ⊇ n2Z ⊇ · · ·.

Definition 4.4.11 (Unique Factorization Domain)A ring R is a unique factorization domain iff R is an integral domain and every r ∈ R \{0}


4 Ring Theory

admits a decomposition

r = un∏i=1

pi

where u ∈ R× and the pi irreducible, which is unique up to associates.

Definition 4.4.12 (Euclidean Domain)An integral domain R is Euclidean if R admits a degree function d : R→ Z≥0 such that forall x, y ∈ R there exist q, r ∈ R with x = qy + r and either f(r) < f(y) or r = 0.

4.4.2 Others

Definition 4.4.13 (Noetherian)A ring R is Noetherian if the ACC holds: every ascending chain of ideals I1 ≤ I2 · · · stabilizesin the sense that there exists some N such that IN = IN+1 = · · ·.

Definition 4.4.14 (Reduced Ring)A ring R is reduced if R contains no nonzero nilpotent elements.

Definition 4.4.15 (Local Ring)A ring R is local iff it contains a unique maximal ideal m, so mSpecR = {0,m}. As aconsequence, there is a uniquely associated residue field κ := R/m.

Exercise 4.4.16 (?)Show that if R is a nonzero ring where every element is either a unit or nilpotent, then R islocal.

Exercise 4.4.17 (?)Show that if p ∈ SpecR then R [p−1] is local.

Exercise 4.4.18 (?)Suppose m ∈ mSpecR is a proper maximal ideal. Show that under either of the following twoconditions, R is local:

• R \m ⊆ R×, so every element of R \m is a unit, or• 1 + m ⊆ R×

Solution: • Sketch: m must contain every non-unit.

– If I 6= R then I contains no units, so I ⊆ N := R \ R×, i.e. I is contained in thenon-units. But N ⊆ m since no element of m is a unit and no element of R \m is anon-unit.

• Sketch: show that every r ∈ R \m is a unit and apply the first part.


4 Ring Theory

– If r ∈ R \ m then 〈r,m〉 = R = 〈1〉 so rt + m = 1 for some t ∈ R,m ∈ m, sort = 1−m ∈ 1 + m ⊆ R× by assumption. Now apply (1).

Definition 4.4.19 (Dedekind Domains)A Dedekind domain is an integral domain for which the monoid Id(R) of nonzero idealsof R satisfies unique factorization: every ideal can be decomposed uniquely into a product ofprime ideals.

Exercise 4.4.20 (?)Show that a Dedekind domain R is a PID iff R is a UFD.

Definition 4.4.21 (Valuation Ring)A valuation ring is an integral domain R such that for every x ∈ ff(R), x ∈ R or x−1 ∈ R.

Definition 4.4.22 (Discrete Valuation Rings)A discrete valuation ring or DVR is a local PID with a unique maximal ideal.

Definition 4.4.23 (Regular ring)A commutative ring R is regular if R is Noetherian and for every p ∈ SpecR the localizationR [p−1] is a regular local ring: it has a maximal ideal m which admits a minimal generating setof n elements where n is the Krull dimension of R [p−1].

Remark 4.4.24: Motivation: if R = OX,x is the ring of germs at x of an algebraic variety X, thenR is regular iff X is nonsingular at x.

E4.5 Comparing and Transporting Ring

Types e

Proposition 4.5.1(Characterizations of Rings).

• R a commutative division ring =⇒ R is a field• R a finite integral domain =⇒ R is a field.• F a field ⇐⇒ F[x] is a PID.• F is a field ⇐⇒ F is a commutative simple ring.• R is a UFD ⇐⇒ R[x] is a UFD.• R a PID =⇒ R[x] is a UFD• R a PID =⇒ R Noetherian

Exercise 4.5.2 (?)Show that R[x] a PID ⇐⇒ R is a field.

Solution:Hint: take r ∈ R, then 〈r, x〉 = 〈f〉 for some f . Write r = fp and x = fq for p, q ∈ R[x], show

4.5 Comparing and Transporting Ring Types 73

4 Ring Theory

deg f = 0 and deg q = 1. Write f = c a constant, q(x) = ax + b to get c(ax + b) = x =⇒ca = 1 =⇒ c ∈ R× =⇒ 〈f〉 = R[x]. Conclude by writing 1 = ar1(x) + xr2(x), evaluate atx = 0 to get a−1 = r1(0).

Example 4.5.3(?): A polynomial ring over a PID is not necessarily a PID: take 〈2, x〉 E Z[x].

Proposition 4.5.4(Big chain of inclusions).Fields ⊂ Euclidean domains ⊂ PIDs ⊂ UFDs ⊂ Integral Domains ⊂ Rings

Remark 4.5.5: Sketch proofs of the inclusions:

• Field =⇒ Euclidean: given x, y we need to write x = qy + r, so just take q = y−1 and r = 0.

• Euclidean =⇒ PID: to divide is to contain, and the Euclidean algorithm terminates to yielda gcd. Alternatively, pick an element a ∈ I of minimal degree. If I 6= Ra pick b ∈ Ra that adoesn’t divide and write b = aq + r with d(r) < d(a). Then r = b− aq ∈ I. E

• PID =⇒ UFD:

– To get existence, use that PIDS are Noetherian, maximals are generated by irreducibles,and irreducibles are prime. Write a = a1b1 a proper factorization to get a propercontainment 〈a〉 ⊂ 〈a1〉 that eventually stabilizes to yield an irreducible factor ar. Usethe same idea to write a as finitely many irreducible factors.

– To get uniqueness, write a =∏

pi =∏

qi as primes and divide everything over.

Example 4.5.6(showing these inclusions are strict):• A Euclidean Domain that is not a field: k[x] for k a field.

– Proof : Use that k a field implies k[x] is a PID, and PID implies UFD. But this is not afield since the element x is not invertible.

• A PID that is not a Euclidean Domain: Z[

1 +√−19

2

].

– Proof : complicated.

• A UFD that is not a PID: Z[x].

– Proof : Z a UFD implies Z[x] is a UFD, but 〈2, x〉 = 2Z[x]+xZ[x] ={∑

rixi∣∣∣ r0 ∈ 2Z

}is not principal. Why: if 〈2, x〉 = 〈f〉 and f is constant, then every polynomial in thisideal has even coefficients and thus misses g(x) := x. Otherwise, deg f ≥ 1 and we miss2, which has degree zero.

• An integral domain that is not a UFD: Z[√−5]

– Proof : (2 +√−5)(2−

√−5) = 9 = 3 · 3, where all factors are irreducible (check norm).

• A ring that is not an integral domain: Z/4

– Proof : [2]4 is a zero divisor since [2]4[2]4 = [0]4.

4.5 Comparing and Transporting Ring Types 74

4 Ring Theory

E 4.6 Radicals e

Definition 4.6.1 (Radical of an Ideal)For an ideal I E R, the radical

√I :=

{r ∈ R

∣∣∣ rn ∈ I for some n ≥ 0},

so xn ∈ I =⇒ x ∈√I.

An ideal is radical iff√I = I.

Remark 4.6.2: In general, “radical” refers to “bad elements” of some type to be quotiented out,not necessarily

√−.

Definition 4.6.3 (Nilpotent)An element r ∈ R is nilpotent if rn = 0 for some n ∈ Z≥0.

Fact 4.6.4The binomial expansion works in any ring:

(a+ b)n =∑k≤n

(n

k

)akbn−k.

This is useful when considering nilpotents or radicals.

Exercise 4.6.5 (?)Notation: let N or N(R) be the set of nilpotents in R. Let ZD or ZD(R) be the set of zerodivisors. Let U,U(R), R× be the units of R.

• Show that every nilpotent is either zero or a zero divisor.

– Solution: am = 0 with a 6= 0 and m > 1, then xxm−1 = 0, so xm−1 is a nontrivialelement annihilating x.

• Show that R commutative and unital and x nilpotent implies 1+x is a unit, and moreoverN +R× = R× (the sum of a nilpotent and unit is a unit).

– Solution: expand 1/(1 + x) =∞∑k=0

(−x)k =n∑k=0

(−x)k := f(x), so (1 + x)f(x) = 1.

Now use that RN = N since xn = 0 implies (rx)n = rxrx · · · rx = rnxn = 0.Taking n+ u ∈ N +R×, then u+ n = u−1(1 + u−1n) ∈ R×R× since u−1n ∈ N and1 + u−1 ∈ R× by the first part.

• Show that f(x) =∑

akxk ∈ R[x] iff f ∈ R[x]× ⇐⇒ a0 ∈ R×, ak>1 ∈ N .

– Solution: use that if ak is nilpotent, akxk is nilpotent. Then a0 a unit at a1xnilpotent implies a0 + a1x is a unit, and inductively f is a unit. If f is a unit,

4.6 Radicals 75

4 Ring Theory

take fg = 1 with f =n∑k=0

akxk and g =

m∑k=0

akxk. Write fg(x) =

n+m∑k=0

ckxk where

ck =k∑j=0

ajbk−j . Using fg = 1, c0 = a0b0 = 1 so a0, b0 are units, and proceed

inductively by descending coefficients, checking that anbm is the r = 0 case.

• Show that f(x) ∈ N(R[x]) ⇐⇒ ak ∈ N(R) for all k.

– Solution: f nilpotent with f(x) =∑

akxk implies fm = 0, and check the leading

term amn xnm. Induct down: f, anxn nilpotent implies f−anxn nilpotent. Conversely,

if anii = 0, use that N(R) E R form an ideal.

• Show that f ∈ ZD(R[x]) ⇐⇒ f 6= 0 and rf(x) = 0 for some r ∈ R.

Definition 4.6.6 (Nilradical)The nilradical of R ∈ CRing is

√0R :=

{x ∈ R

∣∣∣ x is nilpotent}.

Exercise 4.6.7 (Quotient by nilradical is reduced)Show

√0R E R is an ideal and A/

√0R is reduced.

Solution:

• R√

0R ⊆ R: For r nilpotent of order n and x ∈ R, xr is nilpotent since

(xr)n = (xr)(xr) · · · (xr) = xnrn = xn0 = 0.

• R2 ⊆ R, for r, s ∈√

0R write rn = sm = 0, then

(r + s)N =∑k≥0

(N

k

)rksN−k,

so just choose N large enough so that either k > n or N − k > m always holds,e.g. N := n+m− 1.

• R/√

0R has no nonzero nilpotents: Take r ∈ R/√

0R for some r ∈ R, then ϕ(rn) =ϕ(r)n = rn. So

rn = 0 mod√

0R ⇐⇒ rn ≡ 0 mod√

0R ⇐⇒ rn ∈√

0R ⇐⇒ r ∈ 0R.

Exercise 4.6.8 (?)Show that the nilradical is the intersection of all prime ideals.

Solution:See A&M 1.8

Write P as the intersection of all prime ideals of R.

4.6 Radicals 76

4 Ring Theory

√0R ⊆ P : Suppose r ∈

√0R so rn = 0 and let p ∈ SpecR. Then use that 0 ∈ I for any ideal:

rn = 0 ∈ p =⇒ r ∈ p since p is prime.√

0Rc ⊆ P c: Fix f non-nilpotent, we want to show f is not in any prime ideal. set S ⊆ R to

be all ideals I such that f>0 6∈ I. Apply Zorn’s lemma: S 6= ∅ since 0 ∈ S, so after ordering Iby inclusions S contains a maximal p which we claim is prime. If a, b ∈ pc then p + 〈a〉 andp + 〈b〉 supsetp strictly, and by maximality they aren’t in S. So there exist m,n such thatfm ∈ p + 〈a〉 and fn ∈ p + 〈b〉. Then fm+n ∈ p + 〈ab〉, so p + 〈ab〉 is not in S. Thus ab 6∈ p sof 6∈ p. Letting p be arbitrary yields f 6∈ P .

Definition 4.6.9 (Jacobson Radical)The Jacobson radical J(R) is the intersection of all maximal ideals, i.e.

J(R) =⋂

m∈mSpecRm.

Exercise 4.6.10 (?)Show x ∈ J(R) ⇐⇒ 1− xR ⊆ R×.

E 4.7 Structure Theorems e

Definition 4.7.1 (Simple Modules)A module M is simple iff every submodule M ′ ≤ M is either 0 or M . A ring R is simple ifand only if it is simple as an R-module, i.e. there are no nontrivial proper ideals.

Definition 4.7.2 (Semisimple Modules)A module M is semisimple if and only if it admits a decomposition

M =⊕j∈J

Mj

with each Mj simple.

Theorem 4.7.3(Artin-Wedderburn?).If R is a nonzero, unital, semisimple ring then

R ∼=m⊕i=1

Mat(ni, Di),

a finite sum of ni × ni matrix rings over division rings Di.

Corollary 4.7.4.If M is a simple ring over R a division ring, the M is isomorphic to a matrix ring.

4.7 Structure Theorems 77

4 Ring Theory

Theorem 4.7.5(Wedderburn).Every finite division ring is a field, i.e. finite division rings must be commutative.

E 4.8 Zorn’s Lemma e

Definition 4.8.1 (Chain in a poset)In a poset, a chain is a totally ordered subset. An upper bound on a subset S of a poset Xis any x ∈ X such that s ≤ x for all s ∈ S.

Theorem 4.8.2(Zorn’s Lemma).If P is a poset in which every chain has an upper bound, then P has a maximal element.

Remark 4.8.3: You can always form a subset poset, and restrict with any sub-collection thereofwith a set predicate. To use Zorn’s lemma, you need to take an arbitrary chain in your poset X,produce an upper bound U (e.g. by taking a union), and showing that U is still in X (i.e. it stillsatisfies the right predicate).

Theorem 4.8.4(Krull).Every ring has a proper maximal ideal, and any proper ideal is contained in a maximal ideal.

Proposition 4.8.5(Existence of maximal ideals).Every proper ideal is contained in a maximal ideal.

Proof .Let 0 < I < R be a proper ideal, and consider the set

S ={J∣∣∣ I ⊆ J < R

}.

Note I ∈ S, so S is nonempty. The claim is that S contains a maximal element M .S is a poset, ordered by set inclusion, so if we can show that every chain has an upper bound,we can apply Zorn’s lemma to produce M .Let C ⊆ S be a chain in S, so C = {C1 ⊆ C2 ⊆ · · ·} and define C = ∪iCi.C is an upper bound for C: This follows because every Ci ⊆ C.C is in S: Use the fact that I ⊆ Ci < R for every Ci and since no Ci contains a unit, Cdoesn’t contain a unit, and is thus proper.

�

Exercise 4.8.6 (?)Show that every non-unit of R is contained in a maximal ideal.

Solution:This follows because if x ∈ R \ R×, then Rx E R and Rx 6= R implies R/Rx 6= 0. Thenthere exists some m ∈ mSpecR/Rx, and by the correspondence theorem this lifts to some

4.8 Zorn’s Lemma 78

4 Ring Theory

m ∈ mSpecR containing Rx.

E 4.9 Unsorted e

Fact 4.9.1Division algorithm for Euclidean domains.

todo

Definition 4.9.2 (Field of fractions)For R ∈ CRing an integral domain, the field of fractions of R can be constructed as

ff(R) := (R×R•) / ∼ (a, s) ∼ (b, t) ⇐⇒ at− bs = 0R.

Checking transitivity requires having no nonzerozero divisors.

Definition 4.9.3 (Localization)For R ∈ CRing and S ⊆ R a multiplicatively closed subset, so RS ⊆ S and 1R ∈ S, thelocalization of R at S can be constructed as

R [S−1] := (R× S) / ∼ (a, s) ∼ (b, t) ⇐⇒ ∃u ∈ S (at− bs)u = 0R.

Why the u: use in proof of transitivity.

Universal property.

4! Warning 4.9.4There is a canonical ring morphism

R→ R [S−1]

x 7→ x

1 ,

but this may not be injective.

Remark 4.9.5: For integral domains R,

ff(R) ∼= R [(R•)−1] .

Theorem 4.9.6(Hilbert Basis Theorem).

todo

4.9 Unsorted 79

6 General Field Theory

Definition 4.9.7 (Primary Ideal)An ideal I E R is primary iff whenever pq ∈ I, p ∈ I and qn ∈ I for some n.

5 Number Theory

Proposition 5.0.1(Properties of the norm).Let K be a number field and N : K → Z be its norm function.

• N(ab) = N(a)N(b)• a

∣∣ b ∈ K =⇒ N(a)∣∣ N(b) ∈ Z.

• a ∈ K× ⇐⇒ N(a) = ±1.


Remark 6.0.1: The most useful tricks of the trade:

• #Gm(GF(pk)) = pk − 1, since every element is invertible except 0. You can use this to cookup strong numerical constraints on orders of elements. E.g. if a17 = 1 in some finite field ofsize pk, o(a) divides 17 and o(a) divides pk − 1, so o(a) divides gcd(17, pk − 1).

• Multiplicativity in towers can force numerical divisibility constraints. E.g. if α is a root ofany irreducible f , take the tower SF(α, k)/k(α)/k: then the degree of min

α,k(x) ∈ k[x] divides

the degree of the extension [SF(α, k) : k].•

E 6.1 Basics: Polynomials e

Definition 6.1.1 (Reducible and Irreducible Polynomials)For F a field, a polynomial f ∈ F[x] is reducible if and only if f can be factored as f(x) =g(x)h(x) for some g, h ∈ F[x] with deg g,deg h ≥ 1 (so g, h are nonconstant). f is irreducibleif f is not reducible.

Definition 6.1.2 (Primitive Polynomials)For R a UFD, a polynomial p ∈ R[x] is primitive iff the greatest common divisors of itscoefficients is a unit.

Number Theory 80


Theorem 6.1.3(Gauss’ Lemma).Let R be a UFD and F its field of fractions. Then a primitive p ∈ R[x] (so e.g. p monic) isirreducible in R[x] ⇐⇒ p is irreducible in F [x].More precisely, if p = AB is reducible in F [x], then there exist r, s ∈ F such that rA, sB ∈ R[x]and p = (rA)(sB) is a factorization in R[x].

Corollary 6.1.4.A primitive polynomial p ∈ Q[x] is irreducible ⇐⇒ p is irreducible in Z[x].

E 6.2 Definitions e

Definition 6.2.1 (Characteristic)

The characteristic of a ring R is the smallest integer p such thatp∑

k=11 = 0.

Proposition 6.2.2(Freshman’s Dream).If ch k = p then (a+ b)p = ap + bp and (ab)p = apbp.

Definition 6.2.3 (Fixed Field)For H ≤ Aut

Fieldsk(L),

LH :={` ∈ L

∣∣∣ σ(l) = `}.

Definition 6.2.4 (Prime Subfield)The prime subfield of a field F is the subfield generated by 1.

Theorem 6.2.5(Characterization of Prime Subfields).The prime subfield of any field is isomorphic to either Q or Fp for some p.

Definition 6.2.6 (Field Automorphisms)

Aut(L/k) ={σ : L→ L

∣∣∣ σ|k = idk}.

Definition 6.2.7 (Embeddings and Lifts)Let k denote a field, and L/k extension. Every field morphism is an embedding (injection). Anembedding of k-algebras L ↪→ L′ will refer to any ring morphism over k, i.e. a field morphismthat restricts to the identity on k:

6.2 Definitions 81


L L′

k

Link to DiagramMore generally, we can ask for lifts of any map σ : k → k′:

L L′

k k′σ

Link to DiagramMost often, we’ll take σ : k → k to be the identity.

Definition 6.2.8 (Perfect Fields)The following are equivalent:

• k is a perfect field.

• If ch k > 0, the Frobenius is an automorphism of k, so kp = k.

• Every finite extension F/k is separable.

• Every irreducible polynomial p ∈ k[x] is separable.

Example 6.2.9(of a non-perfect field): Example of a non-perfect field: Fp(t). Use that f(x) :=xp − t is irreducible in Fp(t)[x] but not separable.

Proposition 6.2.10(Characterization of perfect fields).k is perfect (using the irreducible implies separable condition) if either

• ch k = 0 or• ch k = p > 0 and kp = k.

Proof (?).For ch k = 0, use that irreducible implies separable.For ch k = p, show that kp 6= k ⇐⇒ irreducible does not imply separable, so there exists aninseparable irreducible.

• Supposing kp 6= k, choose a ∈ k not a pth power.

• Note that f(x) := xp − a has only one root in k: in a splitting field, any root r satisfies

6.2 Definitions 82

https://q.uiver.app/?q=WzAsMyxbMCwwLCJMIl0sWzIsMCwiTCciXSxbMSwyLCJrIl0sWzIsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJib3R0b20ifX19XSxbMiwxLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d

https://q.uiver.app/?q=WzAsNCxbMCwwLCJMIl0sWzIsMCwiTCciXSxbMCwyLCJrIl0sWzIsMiwia19cXHNpZ21hIl0sWzAsMSwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCJcXHNpZ21hIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwwLCIiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDEsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzUsNCwiIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==


rp = a, so

xp − a = xp − rp = (x− r)p.

• Note f is irreducible: its only possible divisors are (x− r)m for m ≤ p. Expanding yields

(x− r)m =m∑k=0

(m

k

)xm−k(−r)k = xm +

(m

1

)xm−1(−r)m + · · · ,

so the coefficient of xm−1 is −mr ∈ k.

• Thus if (x − r)m has a nontrivial divisor in k[x] then m must be in k×, forcing r ∈ k.But then rp = a ∈ k, E.

�

Remark 6.2.11(Numerical Invariants): Let K/k be an extension.

[K : k] = dimVectk K

is the dimension of K as a k-vector space. Automorphisms of fields over K are defined as

AutFieldsk

(K) := Aut(K/k) :={σ : K → K ′

∣∣∣ σ|k = idk},

so lifts of the identity on k, and

{K : k} := # Aut(K/k).

If K/k is finite, normal, and separable,

Gal(K/k) := Aut(K/k),

where in general

{K : k} ≤ [K : k]

with equality when L/k is Galois.

Fact 6.2.12

• All fields are simple rings (no proper nontrivial ideals).

– Thus every field morphism is either zero or injective.

• The characteristic of any field k is either 0 or p a prime.• If L/k is algebraic, then min(α,L) divides min(α, k).

6.2 Definitions 83


Proposition 6.2.13(Towers are multiplicative in degree).Let L/F/k be a finite tower of field extensions.

[L : k] = [L : F ][F : k].

E 6.3 Finite Fields e

Theorem 6.3.1(Construction of Finite Fields).GF(pn) ∼=

Fp(f) where f ∈ Fp[x] is any irreducible of degree n, and GF(pn) ∼= F[α] ∼=

spanF

{1, α, · · · , αn−1

}for any root α of f .

Proposition 6.3.2(Prime Subfields of Finite Fields).Every finite field F is isomorphic to a unique field of the form GF(pn) and if chF = p, it hasprime subfield Fp.

Proposition 6.3.3(Containment of Finite Fields).GF(p`) ≤ GF(pk) ⇐⇒ ` divides k.

Proposition 6.3.4(Identification of Finite Fields as Splitting Fields).GF(pn) is the splitting field of ρ(x) = xp

n − x, and the elements are exactly the roots of ρ.

Proof .Every element is a root by Cauchy’s theorem, and the pn roots are distinct since its derivativeis identically −1.todo

�

Proposition 6.3.5(Splits Product of Irreducibles).Let ρn := xp

n − x. Then f(x)∣∣ ρn(x) ⇐⇒ deg f

∣∣ n and f is irreducible.

Corollary 6.3.6.xp

n − x =∏

fi(x) over all irreducible monic fi ∈ Fp[x] of degree d dividing n.

Proof .⇐= :

• Suppose f is irreducible of degree d.• Then f

∣∣ xpd − x, by considering F [x]/ 〈f〉.• Thus xpd − x

∣∣ xpn − x ⇐⇒ d∣∣ n.

=⇒ :

6.3 Finite Fields 84


• α ∈ GF(pn) ⇐⇒ αpn − α = 0, so every element is a root of ϕn and deg min(α,Fp)

∣∣ nsince Fp(α) is an intermediate extension.

• So if f is an irreducible factor of ϕn, f is the minimal polynomial of some root α of ϕn,so deg f

∣∣ n.• ϕ′n(x) = pnxp

n−1 6= 0, so ϕn is squarefree and thus has no repeated factors. So ϕn is theproduct of all such irreducible f .

�

Proposition 6.3.7(Finite fields are not algebraically closed).If F is a finite field then F 6= F .

Proof .If k = {a1, a2, · · · an} then define the polynomial

f(x) := 1 +n∏j=1

(x− aj) ∈ k[x].

This has no roots in k.�

E 6.4 Cyclotomic Polynomials e

Definition 6.4.1 (Euler’s Totient Function)

ϕ(n) := #{k ≤ n

∣∣∣ gcd(k, n) = 1}.

Remark 6.4.2:• ϕ(p) = p− 1, because every number k ≤ p− 1 is coprime to p.• ϕ(pk) = pk−pk−1, since there are pk total numbers less than pk, most of which are coprime top. The ones to remove are those dividing pk: the only divisors of pk are p` for 0 ≤ ` ≤ k, andgcd(pk,m) = p` whenever m = tp for t = 1, 2, 3, · · · , pk−1 (i.e. m is divisible by some powerof p, so the pk−1 multiples of p are possible).

• ϕ is multiplicative (arithmetically, so only on prime powers!)

6.4 Cyclotomic Polynomials 85


Example 6.4.3(Some totient values):

ϕ(1) = 1ϕ(2) = 1ϕ(3) = 2ϕ(4) = 2ϕ(6) = 2ϕ(8) = 4

.

Definition 6.4.4 (Cyclotomic Polynomials)Let ζn = e2πi/n, then the nth cyclotomic polynomial is given by

Φn(x) =n∏k=1

(j,n)=1

(x− ζkn

)∈ Z[x],

which is a product over primitive roots of unity. It is the unique irreducible polynomial whichis a divisor of xn − 1 but not a divisor of xk − 1 for any k < n.Note that deg Φn(x) = ϕ(n) for ϕ the totient function.

Definition 6.4.5 (Cyclotomic Field)Any subfield of SF(xn − 1) is a cyclotomic field.

Proposition 6.4.6(Computing Cyclotomic Polynomials).Computing Φn:

1.

Φn(z) =∏d∣∣n

d>0

(zd − 1

)µ(nd )

where

µ(n) ≡

0 if n has one or more repeated prime factors1 if n = 1(−1)k if n is a product of k distinct primes,

2.

xn − 1 =∏d|n

Φd(x) =⇒ Φn(x) = (xn − 1)

∏d|nd<n

Φd(x)

−1

,

so just use polynomial long division.

6.4 Cyclotomic Polynomials 86


Fact 6.4.7 (computing cyclotomic polynomials, special cases and examples)

Φp(x) = xp−1 + xp−2 + · · ·+ x+ 1Φ2p(x) = xp−1 − xp−2 + · · · − x+ 1

k∣∣ n =⇒ Φn(x) = Φn

k

(xk)

Φ1(z) = z − 1Φ2(z) = z + 1Φ4(z) = z2 + 1Φ6(z) = z2 − z + 1Φ8(z) = z4 + 1.

Proposition 6.4.8(Splitting Fields of Cyclotomic Polynomials).The splitting field of xm − 1 is Q(ζm) for ζm any primitive root of unity, and

Gal(Q(ζm)/Q) ∼= (Z/mZ)×.

Theorem 6.4.9(Kronecker-Weber).If K/Q is an abelian extension, then K ⊆ Q(ζm) for some m.

E 6.5 Misc e

Definition 6.5.1 (Elementary Symmetric Functions)

todo

E 6.6 Exercises e

Exercise 6.6.1 (?)If f ∈ k[x]irr with ch k = p, then there is a unique separable g ∈ k[x]irr such that f(x) = g(xpk)for some unique k.

Exercise 6.6.2 (?)

6.5 Misc 87


Show

x` − 1∣∣ xm − 1 ⇐⇒ `

∣∣ m.Solution:=⇒

• Write m = `q + r with 0 ≤ r < `.• Write

p(x) = xm − 1x` − 1 = xlq+r − 1

x` − 1 = xrxlq − 1x` − 1 + xr − 1

x` − 1 = q(x) + xr − 1x` − 1 ,

where p, q are polynomial by divisibility.• So the remaining ratio must be polynomial, but since r < ` is strict this forces r = 0.

Thus `∣∣ m.

I don’t like this proof!

⇐= :

• Write m = `q + r, then r = 0 by divisibility.• Then xm − 1 = x`q − 1 := zq − 1 where z := x`.• Use that z − 1

∣∣ zq − 1, so x` − 1∣∣ x‘− 1 = xm − 1.

Exercise 6.6.3 (?)Show that if f ∈ Fp[x]irr is degree d,

f∣∣ xpn − x ⇐⇒ d

∣∣ n.Solution:⇐= :

• If d∣∣ n, xd − 1

∣∣ xn − 1 by a previous exercise, and so pd − 1∣∣ pn − 1.

• So xpd−1 ∣∣ xpn−1, now multiply through by x.• Claim: f

∣∣ xpd−1, from which the result immediately follows.• For α any root of f , Fp(α) is a finite field of size pd since [Fp(α) : Fp] = d.• So Fp(α) ∼= GF(pd), which is the splitting field of xpd − x.• Thus α is a root of xpd − x. Iterating over all roots yields the divisibility statement.

=⇒ :

• If f∣∣ gn(x) := xp

n − x, then every root α of f is a root of gn.• So Fp(α) ⊆ GF(pn).• The result follows from the computation

n = [GF(pn) : Fp]= [GF(pn) : Fp(α)] · [Fp(α) : Fp]= kd.

6.6 Exercises 88

7 Field Theory: Extensions and Towers


E 7.1 Basics e

Remark 7.1.1: Galois is defined as normal and separable.

Definition 7.1.2 (Simple Extensions)An extension L/k is simple iff L = K(α) for some α ∈ L.

Theorem 7.1.3(Primitive Element Theorem).Every finite separable extension is simple.

Corollary 7.1.4.GF(pn) is a simple extension over Fp.

Definition 7.1.5 (Algebraic Extension)A field extension L/k is algebraic iff every α ∈ L is the root of some polynomial f ∈ k[x].

Theorem 7.1.6(Finite Extensions are Algebraic).Every finite extension is algebraic.

Proof (that finite extensions are algebraic).If K/F and [K : F ] = n, then pick any α ∈ K and consider 1, α, α2, .... This yields n + 1elements in an n-dimensional vector space, and thus there is a linear dependence

f(α) :=n∑j=1

cjαj = 0.

But then α is the root of the polynomial f .�

E 7.2 Normal Extensions e

Definition 7.2.1 (Normal Field Extension)Let L/k be an extension. Then TFAE:

• L/k is normal.

• Every irreducible polynomial f ∈ k[x] that has one root in L has all of its roots in L,and thus splits in L[x].

Field Theory: Extensions and Towers 89


– So if α ∈ L then every Galois conjugate αk ∈ L as well.. Thus either f splits in Lor f has no roots in L.

Example 7.2.2(of normal extensions):• Useful trick: if [L : k] = 2 then L/k is automatically normal.

• Useful trick: if L/K/k, then K/k is normal iff Gal(L/K) E Gal(L/k).

• K := Q(213 ) is not normal, since K ⊂ R but (x3 − 2) =

∏k

x− ζk3 213 with ζ3, ζ

23 ∈ C.

– Another reason: an embedding σ : K → k can send 213 to any other root of x3 − 2.

• Q(√

2,√

3) is normal over Q, since it it is finite and splits f(x) := (x2 − 2)(x2 − 3), which is aseparable polynomial.

• L := Q(214 ) is not normal, since it is finite but not the splitting field of any polynomial.

• Q(ζk) is normal for ζk any primitive kth root of unity.

• A normal non-separable extension: Fp(x, y)/Fp(xp,yp). This has finite degree p2 but infinitelymany subfields?

Proposition 7.2.3(Characterization of normal algebraic extensions).For L/k algebraic: let k be an algebraic closure containing L, then L/k is normal iff everyk-embedding σ : L→ k satisfies im σ = L, so σ is a k-automorphism of L:

k

L σ(L) = L

k k

σ

Link to Diagram

Definition 7.2.4 (Normal Closure)If K/k is algebraic, then there is an extension Nk/K such that Nk/k is normal and Nk/K/kis a tower. Nk is referred to as the normal closure of K/k.

Proposition 7.2.5(Characterization of finite normal extensions as splitting fields).An extension L/k is finite and normal ⇐⇒ L is the splitting field of some polynomial f ∈ k[x].

7.2 Normal Extensions 90

https://q.uiver.app/?q=WzAsNSxbMCw0LCJrIl0sWzAsMiwiTCJdLFsyLDQsImsiXSxbMiwwLCJcXGJhcntrfSJdLFsyLDIsIlxcc2lnbWEoTCkgPSBMIixbMTIwLDYwLDYwLDFdXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDIsIiIsMSx7ImxldmVsIjoyLCJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzIsNCwiIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwzLCIiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsxLDMsIlxcc2lnbWEiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsxLDQsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn0sImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dXQ==


Proof (?).=⇒ :

• Write L = k(a1, · · · , an) by finiteness.• Let mi be the minimal polynomials of the ai.• By normality, the mi split in L[x].• Then L is the splitting field of f(x) :=

∏i

mi(x).

⇐= :

• Suppose L/k = SF(f), and pick any monic m ∈ L[x]irr with a root a ∈ L, so that m isthe minimal polynomial of a.

• Toward showing m splits in L: let M = SF(m), we’ll show M = L.

• To show that for any root b ∈M we have b ∈ L, it suffices to show [L(b) : L] = 1.The strategy: use that [L(a) : L] = 1 since a ∈ L by assumption, and try to relate thetwo degrees.

• We have L/k, and a number of towers to work with:

[L(a) : k] = [L(a) : k(a)][k(a) : k] = [L(a) : L][L : k]

[L(b) : k] = [L(b) : k(b)][k(b) : k] = [L(b) : L][L : k].

• In the first set of equalities, note that k(a)/k ∼= k(b)/k since a, b are conjugate roots overk. Moreover L(a)/k(a) ∼= L(b)/k(b) since both are splitting fields for f .

• Thus [L(a) : k] = [L(b) : k], which forces [L(a) : L] = [L(b) : L] after dividing by [L : k].But [L(a) : L] = 1.

�

Proposition 7.2.6.|Aut(L/k)| ≤ [L : k] with equality precisely when L/k is normal.

E 7.3 Separable Extensions e

Definition 7.3.1 (Separable polynomials)A polynomial f ∈ k[x] is separable iff f has no repeated roots.

Example 7.3.2(of separable and inseparable polynomials):• x2 − 1 is separable over Q, but inseparable over F2 since it factors as (x− 1)2.• (x2 − 2)2 is inseparable over Q

7.3 Separable Extensions 91


• x2 − t is inseparable over F2(t).• f(x) := xn − 1 is inseparable over Fp when p

∣∣ n.– Otherwise, f ′ = nxn−1 has only x = 0 as roots, whereas 0 is not a root of f , so f is

separable.

• f(x) := xp − t is not separable over Fp(t): it is irreducible by Eisenstein, but has only thesingle root t

1p .

• f(x) := xpn − x is separable over Fp, since f ′(x) = −1 has no roots at all.

Definition 7.3.3 (Separable Field Extension)Let L/k be a field extension, α ∈ L be algebraic over k, and f(x) := min(α, k). The followingare equivalent

• L/k is a separable extension.• Every element α ∈ L is separable over k, so α has separable minimal polynomial m(x)

in some splitting field of m.• Every finite subextension L′/k is separable.

Fact 7.3.4If α ∈ K/k is separable, then α is separable in any larger field L/K/k since the minimal polynomialover the larger field will divide the minimal polynomial over the smaller field.

Proposition 7.3.5(Separability test: gcd with derivative).f is separable iff gcd(f, f ′) = 1, so f, f ′ share no common roots. Moreover, the multiple rootsof f are precisely the roots of gcd(f, f ′).

Proof (of separability test).6 =⇒ : Suppose f has a repeated root ri, so its multiplicity is at least 2. Then

f(x) = (x− r)2g(x) =⇒ f ′(x) = 2(x− r)g(x) + (x− r)2g′(x),

so r is a root of f ′.6 ⇐= : Suppose r is a root of f, f ′. Write f(x) = (x− r)p(x) and f ′(x) = (x− r)p′(x) + p(x).Rearranging, f ′(x) − (x − r)p′(x) = p(x), and since r is a root of the LHS it’s a root of theRHS. So p(x) = (x− r)q(x) and f(x) = (x− r)2q(x), making r a repeated root.

�

Proposition 7.3.6(Separability test: identically zero derivative).f ∈ k[x]irr is inseparable (so f has a repeated root) iff f ′(x) ≡ 0.

Proof (of separability test).Assume f is monic, then f is inseparable iff f, f ′ have a common root a. So (x − a)

∣∣ q :=gcd(f, f ′), and since f is irreducible, it must be the minimal polynomial of a. Since f ′(a) = 0,this forces f ′

∣∣ f , and since deg f ′ = deg f − 1 < deg f this forces f ′ ≡ 0.�



Proposition 7.3.7(Derivative completely detects separability).

• For any field k, f ∈ k[x] is separable ⇐⇒ f ′ 6≡ 0 ∈ k[x].• For ch k = 0, irreducible implies separable.• For ch k = p, irreducibles f(x) are inseparable iff f(x) = g(xp) for some g ∈ k[x].

Thus for an irreducible polynomial f ,

f separable ⇐⇒ gcd(f, f ′) = 1 ⇐⇒ f ′ 6≡ 0 ⇐⇒ ch k=pf(x) = g(xp).

Proof (?). • First part:

– 6 A =⇒ 6 B:♦ Let f be irreducible, and suppose f is separable. If d(x) := gcd(f, f ′) 6= 1, thenf ′ can not divide f since f is irreducible, so f divides f ′. But deg f ′ < f andf∣∣ f ′ forces f ′ ≡ 0.

– 6 B =⇒ 6 A:♦ If f ′ ≡ 0, then d(x) := gcd(f, f ′) = gcd(f, 0) = f 6= 1 and f is not separable.

• Second part:

– If ch k = 0 and f is irreducible, then deg f ≥ 2 and deg f ′ ≥ 1 so f ′ 6= 0 and f isthus separable.

• Third part:

– ⇐= : If f(x) = g(xp) then f ′(x) = g′(xp) · pxp−1 ≡ 0.

– =⇒ : Let f be irreducible and inseparable, so f ′ ≡ 0 ∈ k[x]. Then f(x) :=n∑k=0

akxk

implies f ′(x) :=n∑k=1

kakxk−1, which is zero iff kak ≡ 0 so p divides kak. So ak 6≡ 0

forces p∣∣ k, so f = a0 + apx

p + a2px2p + · · ·.

�

Corollary 7.3.8(Inseparable iff polynomial in characteristic powers).If f ∈ k[x]irr and p := ch k > 0, then f inseparable ⇐⇒ f(x) = q(xpn) for some unique n.

Proof (of inseparable characterization).=⇒ :Use that f is inseparable iff f ′ ≡ 0. The claim is that f ′ ≡ 0 in characteristic p iff all exponentspresent in f are divisible by p. If f ′ ≡ 0, write

f(x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0

=⇒ f ′(x) = nanxn−1 + (n− 1)an−1x

n−2 + · · ·+ a1

≡ 0,

which forces iai = 0 for all i. For any ai 6= 0, this forces i ≡ 0 mod p, so ai can only be nonzero



when p∣∣ i, so i = kp for some k. So reindex to write

f(x) = a0 + a1xp + a2x

2p + · · ·+ anxnp =

(b0 + b1x+ b2x

2 + · · ·+ bnxn)p∈ k[x],

using (c+d)p = cp+dp in characteristic p, and taking bi := a1p

i ∈ k So f ′ ≡ 0 =⇒ f(x) = q(xp)where q(t) :=

∑bit

i.⇐= : If f(x) = q(xp) for some q, the previous calculation shows q has multiple roots, thus sodoes f , so f is inseparable.

�

Fact 7.3.9 (Irreducible implies separable in characteristic zero)If ch k = 0 and f ∈ k[x]irr, then f is automatically separable.

Why this is true: assuming f is irreducible, gcd(f, f ′) = 1 or f . It can’t be f , since f∣∣ f ′ would

force deg f = deg f ′ = 0 and make f a constant. So this gcd is 1.

Fact 7.3.10 (Irreducible implies separable for perfect fields)

• Use that irreducible polynomial f must have distinct roots, by the argument above. (In fact,it is the minimal polynomial of its roots.)

• Toward a contradiction, suppose f is irreducible but inseparable.

• Then f(x) = g(xp) for some g(x) :=∑

akxk.

• Since Frobenius is bijective, write ak = bpk for some bk, then

f(x) =∑

akxpk =

∑bpkx

pk =(∑

bkxk)p,

making f reducible. E

Fact 7.3.11 (finite extensions of perfect fields are separable)A finite extension of a perfect field is automatically separable, and one only needs to show normalityto show it’s Galois.

Proposition 7.3.12(Simplifications of separability for finite extensions).If L/k is a finite extension, then, TFAE:

• L/k is separable.• L = k(α) for α a separable element.• L = k(S) for S some set of separable elements



• [L : K] = [L : K]s, i.e. the separable degree equals the actual degree.

[L : k] = {L : k} := # AutFieldsk

(L).

Proposition 7.3.13(Separable splitting fields are Galois).If L/k is separable, then

[L : k] = {L : k} .

If L/k is a splitting field, then

[L : K] = # AutFieldsk

(L) := #Gal(L/k).

Proposition 7.3.14(Irreducible polynomials have separable splitting fields).Irreducible polynomials have distinct roots after passing to a splitting field.

Proposition 7.3.15(Algebraic extensions of perfect fields are separable).If ch k = 0 or k is finite, then every algebraic extension L/k is separable.

Proposition 7.3.16(Irreducible implies separable for perfect fields).If k is a perfect field, then every irreducible f ∈ k[x]irr is automatically separable.

Proof (?).If ch k = 0 and f is irreducible, then since deg f ′ < deg f and f is irreducible we must havegcd(f, f ′) = 1 and f is separable.If ch k = p > 0, then if f is irreducible and inseparable then f(x) = g(xp) for some g. Write

g(x) =∑

akxk, and since k is perfect, write bk := a

1p

k , then

f(x) =∑

akxpk =

∑bpkx

pk =(∑

bkxk)p,

so f is reducible. E.�

Definition 7.3.17 (Separable degree)The separable degree of an extension L/k is defined by fixing an embedding σ : k ↪→ k (thealgebraic or separable closure) and letting [L : k]s be the number of embeddings σ′ : L→ k:

L k

k k

σ′

σ

Link to Diagram


https://q.uiver.app/?q=WzAsNCxbMCwyLCJrIl0sWzIsMCwiXFxiYXJ7a30iXSxbMCwwLCJMIl0sWzIsMiwiayJdLFsyLDEsIlxcc2lnbWEnIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifSwiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzAsMiwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMywxLCJcXHNpZ21hIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMCwzLCIiLDIseyJsZXZlbCI6Miwic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==


Proposition 7.3.18(Separability is transitive.).If L/K/k, then L/K is separable and K/k is separable ⇐⇒ L/k is separable:

L L

K K

k k

Link to Diagram

Proof (?).Use that L/k is separable ⇐⇒ [L : k] = [L : k]s.⇐= :

• By definition, every α ∈ L is separable over k.• K/k is separable:

– Since K ⊆ L, any α ∈ K is also separable over k.

• L/K is separable:

– If α ∈ L, then minα,k

(x) is a separable polynomial over some splitting field.

– Use that L/k implies minα,L

(x) divides minα,k

(x), so the former is separable, done.

=⇒ :

• Now use that the separable degree is multiplicative in towers.

• If all extensions in sight are finite, this direction is immediate:

[L : k]s = [L : K]s[K : k]s = [L : K][K : k] = [L : K].

• For the infinite case, want to show every α ∈ L is separable over k. It suffices to show αis contained in some finite separable subextension. The strategy:


https://q.uiver.app/?q=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


α ∈ L

k(α, S) 3 α

f ∈ K[x] K

F := k(α, S) ∩K f ∈ F [x]

ks

s

s

s

Link to Diagram

• Let f(x) := minα,K

(x) be the minimal polynomial of α over the intermediate extension K,

which by assumption is separable since L/K is separable.

– So f ∈ K[x], and letting S be the finite set of coefficients of f , S ⊆ K.– Note that each coefficient s ∈ S is separable over k since K/k is separable by

assumption.

• Set F := k(α, S) ∩K. Note K/k is separable and F ⊆ K, so F/k is separable.• Moreover k(α, S)/F is separable, since the minimal polynomial of α over F is still f .• Now k(α, S)/F/K is a tower of finite extensions where k(α, S)/F and F/k are separable,

so this reduces to the finite case.

�

Proposition 7.3.19(Separability has the compositing property).E/k and F/k are separable ⇐⇒ EF/k is separable.

Proof (?).⇐= : Separability always descends to subfields, and E ≤ EF,F ≤ EF .=⇒ :

• Write E = k(S) for some finite set S. Then EF = F (S).• Use that k(S)/k is separable iff s ∈ S is a separable element for all s.

– Since E/k is separable, each s ∈ S is separable over k.

• Since F/k is separable, each s ∈ S is separable over F .• So F (S)/F is separable.• Now use the tower F (S)/F/k to obtain F (S)/k separable, which is EF/k.

�


https://q.uiver.app/?q=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


E 7.4 Galois Extensions e

Definition 7.4.1 (Galois Extension and Galois Group)Let L/k be a finite field extension. The following are equivalent:

• L/k is a Galois extension.

• L is normal, and separable.

• The fixed field LH of H := Aut(L/k) is exactly k.

• L is the splitting field of a separable polynomial p ∈ K[x].

• L is a finite separable splitting field of an irreducible polynomial.

• There is a numerical equality:

# AutFieldsk

(L) = [L : k] = {L : k} ,

where {E : F} is the number of isomorphisms to any field lifting idF :

In this case, we define the Galois group as

Gal(L/k) := AutFieldsk

(L/k).

Fact 7.4.2For L/k algebraic and ch k = 0, L/k is Galois ⇐⇒ L/k is normal.

Proposition 7.4.3(Galois is upper transitive, characterization of when lower tran-sitivity holds).If L/k is Galois, then L/F is always Galois. Moreover, F/k is Galois if and only ifGal(L/F ) E Gal(L/k)

7.4 Galois Extensions 98


L

F

k

Galois

Galois

Galois

Link to diagramIn this case,

Gal(F/k) ∼=Gal(L/k)Gal(L/F ) .

Proposition 7.4.4(?).Let L/K/k with L/k Galois. Then

K/k is Galois ⇐⇒ Gal(L/K) E Gal(L/k),

and moreover Gal(K/k) = G.

Proof (?).

• Note separability is distinguished, so K/k is separable.• K/k is Galois ⇐⇒ F/k is normal (since we already have separability).• ⇐⇒ σ(K) = K for all σ ∈ G• ⇐⇒ σHσ−1 = H for all σ ∈ G.

So H is normal and G/H is a group. For the isomorphism, take

ρ : Gal(L/k)→ Gal(K/k)ρ 7→ ρ|K .

This is well-defined since by normality σ(K) = K. Any f ∈ ker ρ is the identity on K, sof ∈ Gal(L/K) and kerϕ = H. Since L/K is Galois, every f ∈ Gal(K/k) lifts to Gal(L/k),making ρ surjective.

�

Example 7.4.5(?):• Q(ζ3, 21/3) is normal but Q(21/3) is not since the irreducible polynomial x3 − 2 has only one

root in it.• Q(21/3) is not Galois since its automorphism group is too small (only of size 1 instead of 3?).• Q(21/4) is not Galois since its automorphism group is too small (only of size 2 instead of 4).

However, the intermediate extensions Q(21/4)/Q(21/2) and Q(√

2)/Q are Galois since theyare quadratic. Slogan: “Being Galois is not transitive in towers.”

7.4 Galois Extensions 99

https://q.uiver.app/?q=WzAsMyxbMCwwLCJMIl0sWzAsMiwiRiJdLFswLDQsImsiXSxbMCwyLCJcXHRleHR7R2Fsb2lzfSIsMCx7ImN1cnZlIjotMywic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDEsIlxcdGV4dHtHYWxvaXN9IiwwLHsiY3VydmUiOi0yLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJzcXVpZ2dseSJ9LCJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzAsMSwiXFx0ZXh0e0dhbG9pc30iLDIseyJjdXJ2ZSI6Miwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn0sImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=


• A quadratic extension that is not Galois: SF(x2 + y) ∈ F2(y)[x], which factors as (x−√y)2,making the extension not separable.

E 7.5 Fundamental Theorem of Galois Theory e

Theorem 7.5.1(Fundamental Theorem of Galois Theory).Let L/k be a Galois extension, then there is a correspondence:

{Subgroups H≤Gal(L/k)}{Fields F suchthat L/F/k

}H → {EH := The fixed field of H}{

Gal(L/F ):={σ∈Gal(L/k)

∣∣∣ σ(F )=F}}← F

• This is contravariant with respect to subgroups/subfields.

• [F : k] = [G : H], so degrees of extensions over the base field correspond to indices ofsubgroups.

• [K : F ] = |H|

• L/F is Galois and Gal(K/F ) = H

• F/k is Galois ⇐⇒ H is normal, and Gal(F/k) = Gal(L/k)/H.

• The compositum F1F2 corresponds to H1 ∩H2.

• The subfield F1 ∩ F2 corresponds to H1H2.

Remark 7.5.2: A trick for remembering the degree/index correspondence:

K 1

E H := Gal(K/E)

F G := Gal(K/F )

[E:F ]

[K:E]

[K:F ]

[H:1]

[G:H]

[G:1]

Gal(K/−)

Link to Diagram

Theorem 7.5.3(Splitting + Perfect implies Galois).

7.5 Fundamental Theorem of Galois Theory 100

https://q.uiver.app/?q=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

8 Distinguished Classes

• If ch k = 0 or k is finite, then k is perfect.

• k = C,R,Q,Fp are perfect, so any finite normal extension is Galois.

• Every splitting field of a polynomial over a perfect field is Galois.

Proposition 7.5.4(Composite Extensions).If F/k is finite and Galois and L/k is arbitrary, then FL/L is Galois and

Gal(FL/L) = Gal(F/F ∩ L) ⊂ Gal(F/k).

E 7.6 Quadratic Extensions e

Proposition 7.6.1(Classification of quadratic extensions).If F is a field with ch(F) 6= 2 and E/F is a degree 2 extension, then E is Galois and E = F (

√a)

for some squarefree a ∈ F.

Corollary 7.6.2(Quadratic extensions of rationals).If E/Q is a quadratic extension, E = Q(√q) for some q ∈ Q squarefree. Explicitly, use theprimitive element theorem to write E = Q(α), let f be the minimal polynomial, then takeq = b2− 4ac. One can do slightly better by writing b2− 4ac = a/b so that

√b2 − 4ac =

√ab/b

and taking q = ab.

Proposition 7.6.3(?).For Fp a finite field of prime order, all quadratic extensions E/Fp are isomorphic.


See http: // math. wsu. edu/ students/jstreipel/ notes/ galoistheory. pdf

Definition 8.0.1 (Distinguished Classes)A collection of field extensions S is distinguished iff

1. (Transitive property) For any tower L/K/k, the extension L/k ∈ S ⇐⇒ L/K ∈ S(upper transitivity) and K/k ∈ S (lower transitivity):

7.6 Quadratic Extensions 101

http://math.wsu.edu/students/jstreipel/notes/galoistheory.pdf

http://math.wsu.edu/students/jstreipel/notes/galoistheory.pdf


L L

K ⇐⇒ K

k k

Link to Diagram

2. (Lifting property) Lifts of distinguished extensions are distinguished: K/k ∈ S and L/kany extension =⇒ LK/L ∈ S:

LK

L K

k

∴

Link to Diagram

3. (Compositing property) Whenever L/k,K/k ∈ S, the amalgam LK/k ∈ S as well:

LK

L K

k

∴

Link to DiagramOne is supposed to think of LK/L as a “lift” ofK/k.

Example 8.0.2(of distinguished classes): The following classes of extensions are distinguished:

• Algebraic.• Finite.• Separable.• Purely inseparable.• Finitely generated.• Solvable.

Distinguished Classes 102

https://q.uiver.app/?q=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

https://q.uiver.app/?q=WzAsNCxbMSwyLCJrIl0sWzIsMSwiSyJdLFswLDEsIkwiXSxbMSwwLCJMSyJdLFswLDEsIiIsMCx7ImNvbG91ciI6WzAsNjAsNjBdLCJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9LCJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMCwyLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6ImJvdHRvbSJ9fX1dLFsyLDMsIlxcdGhlcmVmb3JlIiwwLHsiY29sb3VyIjpbMCw2MCw2MF0sInN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsxLDMsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d

https://q.uiver.app/?q=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


4! Warning 8.0.3Normal extensions are not distinguished, since they fail the forward implication for (lower) transi-tivity. However, they do have the (forward implication) upper transitive, lifting, and compositingproperties.

As a consequence, Galois extensions are also not distinguished.

Fact 8.0.4 (Normal/Algebraic/Galois extensions are upper transitive)For L/F/k: L/k normal/algebraic/Galois =⇒ L/F normal/algebraic/Galois.

8.0.1 Algebraic Extensions

Proposition 8.0.5(Transitivity of algebraic extensions, forward implication).If L/K/k (not necessarily finite) with L/K and K/k both algebraic, then L/k is algebraic.

Proof (?).

• We want to show every α ∈ L is algebraic over k, and it suffices to show α is algebraicover some finite subextension k(S).

• Pick α ∈ L, then α is algebraic over K by assumption, so it is a root of some f ∈ K[x].• Let S be the finitely many coefficients of f , then α is algebraic over k(S).• Note that k(S)/k is finite and thus algebraic, and k(S, α)/k(s) is finite and also algebraic,

so we’re reduced to the finite case.• It suffices to show k(S, α)/k(s)/k is finite, which follows from multiplicativity of degrees.

�

Remark 8.0.6: If L/K/k with α algebraic over L, then α is algebraic over K and minα,L

divides minα,K

(so minimal polynomials only get smaller in extensions).

8.0.2 Normal Extensions

Corollary 8.0.7(Normality satisfies the lifting property).E1/k normal and E2/k normal =⇒ E1E2/k normal and E1 ∩ E2/k normal.



E1E2

E1 E2

E1 ∩ E2

k

normalnormalnormal

normal

Link to diagram

Issues with Normal Towers

Example 8.0.8(Normal extensions are not transitive: failure of lower transitivity,forward implication): One can similarly produce towers where the total extension is normal butthe lower iterate is not normal: take

L/K/k := Q(213 , ζ3)/Q(2

13 )/Q.

Now K/k isn’t normal, since Gal(L/k) = S3 but Gal(L/K) = Z/2 6 E S3.

Another example: let L/k be any algebraic extension that isn’t normal, and take Nk to be thenormal closure to get Nk/L. Concretely, NQ/Q(2

13 )/Q works.

Example 8.0.9(Normal extensions are not transitive: failure of reverse implication):One can produce towers of successively normal extensions whose total extension is not normal in acheap way: take

L/K/k := Q(214 )/Q(2

12 )/Q.

Each iterate is normal since it’s quadratic, but the overall extension misses complex roots and isthus not normal.

Proposition 8.0.10(Normal extensions are upper transitive, forward implication).For L/k finite,


https://q.uiver.app/?q=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


L L

K =⇒ K

k k

Normal

Normal

Link to Diagram

Proof (Finite case).

• Use the fact that for finite extensions, L/k is normal and separable ⇐⇒ L is thesplitting field of a separable polynomial f ∈ k[x].

• Now regard f as a polynomial in K[x]; then L is still the splitting field of f over K,done.

Alternatively,

• Let α ∈ L be a root of f ∈ K[x] with f irreducible, it suffices to show all roots of f arein L.

• Let m ∈ K[x] be the minimal polynomial of α over K, and let m′ ∈ k[x] be the minimalpolynomial of α over k.

• Since L/k is normal and α ∈ L, m′ splits in L.• Minimal polynomials are divisible in towers, so m divides m′. Since m′ splits in L, so

must m.

�

Proof (General case).

• Suppose L/K/k with L/k normal, we want to show L/K is normal.• Use the embedding characterization, it suffices to show that every embedding σ : L ↪→ K

satisfies im σ = L:


https://q.uiver.app/?q=WzAsNyxbMCwwLCJMIl0sWzAsMiwiSyJdLFswLDQsImsiXSxbMiwyLCJcXGltcGxpZXMiXSxbNCwwLCJMIl0sWzQsMiwiSyJdLFs0LDQsImsiXSxbMSwwXSxbMiwxXSxbMiwwLCJcXHRleHR7Tm9ybWFsfSIsMix7ImN1cnZlIjo0LCJjb2xvdXIiOlswLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fSxbMCw2MCw2MCwxXV0sWzYsNV0sWzUsNF0sWzUsNCwiXFx0ZXh0e05vcm1hbH0iLDIseyJjdXJ2ZSI6MywiY29sb3VyIjpbMCw2MCw2MF0sInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX0sWzAsNjAsNjAsMV1dXQ==

9 Galois Theory

K = k

L σ(L)

K K

k k

σ

σ

Link to Diagram

• Now just use the fact that k = K, and since k ⊆ K, any K-morphism is also ak-morphism.

• Since L/k is normal, σ(L) = L and L/K is thus normal.

�

9 Galois Theory

Some useful exercises and solutions: https: // feog.github. io/ chap4. pdf

Remark 9.0.1:• Given x :=

√a+√b, to find a minimal polynomial consider x2, x3, · · · and try to get a linear

combination. Then check if its irreducible.

– General strategy here: try to isolate radicals on one side, then raise both sides to thatpower.

• To find a minimal polynomial for an element α, figure out the dimension of Q(α)/Q – say it’sn, then 1, α, · · · , αn must be a Q-linearly dependent set, so you compute these powers andfiddle with Q coefficients (or invert a matrix).

• Useful trick: for x :=√a +√b, compute x, x2, x3, x4 and write them in terms of the basis{

1,√a,√b,√ab}. Then put this linear system into a matrix and invert:

Av = c := A[1,√a,√b,√ab]

=[x, x2, x3, x4

].

Galois Theory 106

https://q.uiver.app/?q=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

https://feog.github.io/chap4.pdf

https://feog.github.io/chap4.pdf

9 Galois Theory

Once you get A−1x = b, read off the first row dotted against b to get a polynomial in x.

• In general: take α, sort out the degree n of the extension Q(α)/Q, and use the basis1, α, α2, · · · , αn−1.

• A trick to remember how degrees, indices and sizes match up: L/K/F corresponds to 1/H/G,and [L : K] = [H : 1] = #H, [F : K] = [G : H], [L : F ] = [G : 1] = #G, etc.

• Trick: once you find SF(f)/Q, if any subextension is not normal over Q, then G can not beabelian.

– Example: f(x) = x3 − 2 splits in Q(ζ3, 213 ) which has a non-normal subextension Q(2

13 ),

forcing G = S3.

• If αβ ∈ Q, then α ∈ Q(β) and vice-versa (I think).

• Checking subgroup lattices: https://hobbes.la.asu.edu/groups/groups.html

• De-nesting radicals:

Remark 9.0.2: Assume all extensions here are algebraic and finite. Let f ∈ Q[x] with n := deg f .

Galois Theory 107

https://hobbes.la.asu.edu/groups/groups.html

9 Galois Theory

Theorem 9.0.3(The Algorithm).

• Show your extension is Galois (normal and separable)

– Show f is irreducible and separable.

• Find the degree of the extension d, since then #G = d.

– Note that in general, G ≤ Sn and n 6= d, #G 6= n.

• Obtain n∣∣ d := #G

∣∣ n! and G ≤ Sn is a transitive subgroup, list possibilities.• Rule out cases or determine the group completely by finding cycle types.

Example 9.0.4(Of using the algorithm): Consider f(x) := x5 − 9x+ 3, let L := SF(f)/Q.

• f is irreducible: Apply Eisenstein with p = 3.• f is separable:

– Q is perfect, so irreducible implies separable.

• L is Galois:

– L/Q is a finite extension over a perfect field and thus automatically separable.– L/Q is the splitting field of a separable polynomial, and thus normal.

• Since L is Galois, #G = d := [L : Q], so try to compute the degree by computing the splittingfield (and its degree) explicitly.

– Here: difficult! The roots are complicated.

• Since L is Galois, G ≤ S5 is a transitive subgroup. Possibilities:

S5, A5, F5 ∼= C5 o C4, D5, C5.

• Claim: G = S5.

– Reduce mod 2: (x2 + x+ 1)(x3 + x2 + 1), yielding a cycle type (2, 3). This rules out♦ C5, D5 since 3

∣∣- 5, 10.♦ A5, since this is an odd number of even length cycles.♦ F5 since 3

∣∣- 20.– So this only leaves S5.

E 9.1 Showing Extensions are Galois e

Fact 9.1.1Showing your polynomial is irreducible:

• Eisenstein (including shifting/inverting tricks, see section below)• To show f is irreducible, it suffices to show it is irreducible over any Fp[x].

9.1 Showing Extensions are Galois 108

9 Galois Theory

• A quadratic with no real roots is irreducible.

Showing your polynomial is separable:

• Show directly that f has distinct roots in k by factoring it.• For perfect fields, irreducibles are automatically separable.• For f irreducible, f is separable iff f ′(x) 6≡ 0.

Showing your extension is separable:

• Splitting fields of separable polynomials are automatically separable and normal (and thusGalois).

• Algebraic extensions of a perfect field are automatically separable.

– In particular, extensions over Q or any ch k = 0 are separable, and one only needs toshow normality.

• (Hard) Show [L : k]s = [L : k].• (Hard) Use that separability is a distinguished class.

Showing your extension is normal:

• Show that L/k is finite and the splitting field of some separable polynomial.

Showing your extension K/k is Galois:

• Show normality and separability.• Show K is the splitting field of a separable polynomial (“separable splitting field”)• Automatic when K/k is algebraic and a finite field, since it’s the splitting field of xpn − x.

E 9.2 Irreducibility e

Proposition 9.2.1(Consequence of Chebotarev density: checking irreducibility modp).If f ∈ Z[x] is monic and there exists any prime p such that f mod p is irreducible in Fp[x],then f irreducible in Q[x].

Remark 9.2.2: Finding a good prime for this is hard, but irreducibility can be checked exhaustivelyin small fields: just enumerate all polynomials and try polynomial long division.

Example 9.2.3(using irreducibility mod p): f(x) := x4 + x + 1 is irreducible in Z[x], sincechecking manually in F2[x] shows that 0, 1 are not roots mod 2 so there is not linear factor. Manuallydividing a1x

2 + a2x+ a3 for ai ∈ 0, 1 leaves remainders, so there are no quadratic factors.

9.2 Irreducibility 109

9 Galois Theory

Theorem 9.2.4(Eisenstein’s Criterion).If

f(x) =n∑i=0

αixi = anx

n + an−1xn−1 + · · ·+ a1x+ a0 ∈ Q[x].

then f will be irreducible over Q[x] (and thus over Z[x] by Gauss’ lemma) if ∃p such that

• p divides every coefficient except an and• p2 does not divide a0.

Note that if f is monic, it suffices to find any prime dividing all of the non-leading terms.

Remark 9.2.5(Shifting): If f(x+ a) satisfies Eisenstein for any p, then f is irreducible. This isgenerally because ∆f(x) = ∆f(x+a), and if p works for Eisenstein on any f then p

∣∣ ∆f .

Example 9.2.6(of shifting): Set f(x) := x2 + x+ 2, then f(x+ 3) = x2 + 7x+ 14 and Eisensteinapplies at p = 7.

Remark 9.2.7(Inverting): If n := deg(f) and xnf(1/x) is irreducible, then f is irreducible. Notethat this is just reversing the coefficients.

Example 9.2.8(Of inverting): Take f(x) := 2x5 − 4x2 − 3, then for g(x) := 3x5 + 4x2 − 2Eisenstein applies with p = 2.

Remark 9.2.9(mod p reduction checks to find a good p for Eisenstein): If f(x) ≡ b(x +a)n mod p for some p where n := deg f , then Eisenstein may work on f(x − a) using the prime p.Note the change in signs/reverse translation.

In other words, reduce mod p for various p, and if any p collapses f to a power of a linear factor,use that p for Eisenstein.

Example 9.2.10(of mod p reduction checks): Check

f(x) := x3 + x2 − 48x+ 128 f(x) ≡ (x− 3)3 mod 5,

and Eisenstein on f(x+ 3) with p = 5 works.

E 9.3 Computing e

9.3.1 Misc Useful Facts

Once you’ve confirmed that you have a Galois extension, some useful tricks are available:

9.3 Computing 110

9 Galois Theory

Fact 9.3.1 (Degrees of extensions)

• The size #G(f) is the degree [SF(f) : Q].• The degree of [Q(α) : Q] is the degree of min

α(x), or any irreducible polynomial with α as a

root.

– Note that Q(α) 6= SF(f) in general.

• If f =∏

(x− ri), then SF(f) contains every Q(ri). Thus

[Q(ri) : Q] = d =⇒ d∣∣ [SF(f) : Q].

– Note that d 6= deg f in general!

Fact 9.3.2 (Random tricks)

• Reminder of rational roots test: for f(x) = anxn + · · · + a0, rational roots are of the form

p0/pn where pi∣∣ ai.

• Gal(L/k) permutes the roots of any irreducible polynomial in k[x]. In particular, if L = SF(f)with f reducible, then G must send roots of irreducible factors to conjugates of the samefactor.

• Q(ζa) = Q(ζb) ⇐⇒ a = 2b and b is odd.• If there are k complex conjugate pairs (accounting for 2k roots) then G contains a cycle

(1, 2)(3, 4) · · · (2k − 1, 2k).• If all exponents are even, f(r) = 0 ⇐⇒ f(−r) = 0, so roots occur in pairs (r,−r).

– Pairs are preserved by G in the sense that every σ ∈ G satisfies either {r,−r} 7→ {r,−r}or {r,−r} 7→ {s,−s} for another pair.

– Example: x4 − 5x2 + 5 has two pairs.

9.3.2 Transitive Subgroups

Proposition 9.3.3(Galois groups are transitive subgroups).If f ∈ k[x] is irreducible, then Gal(SF(f)/k) ≤ Sn is always a transitive subgroup, i.e. it actstransitively on the set of roots.

Corollary 9.3.4.

n∣∣ #Gal(K/Q)

∣∣ n!.

Why: Gal(K/Q) ∼= G ≤ Sn, and Lagrange yields #H∣∣ n!. Note that G acts on R the set of n

roots, and since it acts transitively, R is a single orbit. By orbit stabilizer, Or ∼= G/StabG(r)and thus

#G = #Or ·#StabG(r),

9.3 Computing 111

9 Galois Theory

so both terms on the right-hand side patently divide #G

Fact 9.3.5 (Table of transitive subgroups of Sn for qual-sized n)Write Cn for the cyclic group of order n. The following are transitive subgroups of Sn for small n,where blue groups are nonabelian:

n in Sn Transitive Subgroups Sizes

1 1 12 S2 ∼= C2 23 S3 ∼= D3, A3 ∼= C3 6,34 S4, A4, D4, C4, C

22 24,12,8,4,4

5 S5, A5, F5 ∼= C5 o C4, D5, C5 120,60,20,10,5

Other useful facts:

• #Dn = 2n, #Sn = n!,#An = n!/2, and #F5 = 20.

• For degree 8 extensions (which sometimes arise as quadratic extensions of degree 4 extensions):Q8 ≤ S8 is transitive and nonabelian of order 8, and has presentation

Q8 =⟨α, β

∣∣∣ α4 = β4 = 1, αβα = β, β2 = α2⟩.

Note that Q8 ≤ S8 but Q8 6≤ S<7.

• F5 has presentation ⟨a, b

∣∣∣ a5, b4, bab−1 = a2⟩.

9.3.3 Distinguishing Groups

Material borrowed from https: // kconrad. math.uconn. edu/ blurbs/ galoistheory/ galoisSnAn.pdf

Remark 9.3.6(Distinguishing groups):By n in G ≤ Sn:

n = 4:

• C22 vs C4:

– C22 has two elements of order 2, the latter does not. So a cycle of type (2, 2) forces C2

2 .

9.3 Computing 112

https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisSnAn.pdf



9 Galois Theory

• S4 vs A4:

– S4 contains a Sylow-2 subgroup of order 8 (which divides 4! = 24) but A4 does not sinceit’s of order 4!/2 = 12 and 8 6

∣∣ 12.– If G contains a transposition, then G = S4 orD4, since A4 doesn’t contain a transposition.

• D4 vs Q8:

• 5 roots:

– S5 is generated by any transposition and any 5-cycle.– Sn is generated by (a, b) and (1, 2, · · · , n) ⇐⇒ gcd(b − a, n) = 1. In particular, the

(1, 2) and any length n cycle works.

Fact 9.3.7 (Recognizing cycle types)The following are the cycle types that can occur:

•

Proposition 9.3.8(Recognizing An or Sn).Useful fact: if G ≤ Sn for n prime contains a 2-cycle and a p-cycle, then G ∼= Sn. Note that forn not prime, a transposition and an n-cycle isn’t enough, since one needs the specific n-cycle(1, 2, · · · , n) in general.If n > 2 and G contains a 3-cycle and an n-cycle, then G = An or Sn. Note that by Orbit-Stabilizer n

∣∣ #G, and if n is prime then by Cauchy there is an n-cycle (but this is not alwaysthe case). In fact, it suffices to find a k-cycle for any k ≥ n/2, which can be found by reducingmod p and examining cycle types.Moreover, if G contains a 2-cycle (transposition), then G = Sn.

Remark 9.3.9(Other useful facts to reason about An):

9.3 Computing 113

9 Galois Theory

• Alternating groups have even numbers of cycles of even length.• Elements in An either have cycle type with an even number of even lengths (including 0).• A4 does not contain a subgroup isomorphic to C2

2 .

Fact 9.3.10Some useful generating sets: see https://kconrad.math.uconn.edu/blurbs/grouptheory/genset.pdf

9.3.4 Density: Cycle Types

Proposition 9.3.11(A consequence of Chebotarev Density: reading cycles fromreduction mod p).

For any p 6∣∣ ∆, writing f(x) =

m∏i=1

fi(x) mod p, G contains a cycle of type

(deg f1, deg f2, · · · , deg fm). Equivalently, if f := f mod p, then G(f) ≤ G(f) is a subgroup.

4! Warning 9.3.12Warning: this only works if the fi are distinct, i.e. there are no repeated factors in the factorizationmod p.

Example 9.3.13(Ruling out choices by existence of cycle types): You can use this to ruleout types of groups using Lagrange’s theorem: if you find a cycle of length m which doesn’t divide#H, then H isn’t a possibility! Example: deg f = 5 with exactly one conjugate pair of roots, thenthere is a 5-cycle σ := (1, 2, 3, 4, 5) because 5

∣∣ #G and a 2-cycle τ := (1, 2) coming from complex

9.3 Computing 114

https://kconrad.math.uconn.edu/blurbs/grouptheory/genset.pdf

https://kconrad.math.uconn.edu/blurbs/grouptheory/genset.pdf

9 Galois Theory

conjugation. There check that a1 := στσ−1 = (1, 5) and a1τ = (1, 5, 2) is a 3-cycle, so 3∣∣ #G. This

rules out F5 which is of order 20, since 3 6∣∣ 20.

Example 9.3.14(Finding cycle types by reducing mod p): Consider f(x) = x5 + 2x + 1.Reducing to F3 yields no roots, so f is irreducible, and moreover G(f) contains a 5-cycle. Reducingto F7 splits f as (x+ 2)(x+ 3)(x3 + 2x2 + 5x+ 5), so G(f) contains a 3-cycle.

Example 9.3.15(of using density): Take f(x) := x6 + x4 + x+ 3, then

f(x) ≡ (x+ 1)(x2 + · · · )(x3 + · · · ) mod 2 =⇒ type (1, 2, 3) ∈ Gf(x) ≡ x(x+ 2)(x4 + · · · ) mod 3 =⇒ type (1, 1, 4) ∈ G

.

Example 9.3.16(of using density): Take f(x) := x4 + x+ 1, then

f(x) ≡ x4 + x+ 1 mod 2 =⇒ type (4)f(x) ≡ (x− 1)(x3 + x2 + x− 1) mod 3 =⇒ type (1, 3)

.

So G contains a 4-cycle and a 3-cycle. This is enough to show G = A4.

Example 9.3.17(?): Let f(x) = x6 + x4 + x + 3, reduce mod 11 to get a cycle type (1, 5). SoG ≤ S6 contains a 5-cycle, where 5 > n/2 := 6/2 = 3, meaning G = An, Sn. Now reduce mod p forvarious p to look for a cycle type of the form (2, 1, 1, · · · ) or (3, 1, 1, · · · ). This is hard, but f mod 2has type (1, 2, 3) and ((a, b)(c, d, e))3 = (a, b), so G contains a transposition and thus G = Sn = S6.

Example 9.3.18(?): Let f(x) = x7− x− 1, reduce mod 2 to get a 7-cycle, and mod 3 to get (2, 5).Then use (2, 5)5 = (2, 1, 1, · · · ) to get a transposition, So G = S7.

Example 9.3.19(?): Let f(x) := x7 − 7x+ 10. Reducing mod 3 yields (2, 5) and (2, 5)5 = (2, · · · )and have a transposition. Since 5 > n/2 = 7/2, G = S7.

9.3.5 Discriminants

Definition 9.3.20 (Discriminant)For f =

∑akx

k monic,

∆f =∏i<j

(ri − rj)2.

Note that ∆ = 0 when f has a repeated root.For cubics,

9.3 Computing 115

9 Galois Theory

• ∆ > 0 =⇒ 3 distinct real roots• ∆ < 0 =⇒ 1 real root and 1 conjugate pair.

Example 9.3.21(How to actually write this product): For f a cubic:

∆f = (r1 − r2)2(r1 − r3)2

(r2 − r3)2.

For f a quartic:

∆f = (r1 − r2)2(r1 − r3)2(r1 − r4)2

(r2 − r3)2(r2 − r4)2

(r3 − r4)2.

In general, for a degree n polynomial this will have n(n− 1)/2 terms.

Remark 9.3.22: In general,

G ⊆ An ⇐⇒√

∆ ∈ k,

i.e. ∆ is a perfect square in the ground field k.

Some special cases of discriminant values:

• Quadratics:

f(x) = ax2 + bx+ c =⇒ ∆ = b2 − 4ac.

• Cubics:

– General:

f(x) = ax3 + bx2 + cx+ d =⇒ ∆ = bšcš− 4acş− 4bşd− 27ašdš + 18abc.

♦ Note that you can depress a general cubic by substituting t = x− b

3a , yielding

f(t) = tş + pt+ q =⇒ ∆ = −4pş− 27qš.

Remark 9.3.23: Some useful facts:

• ∆ = 0 ⇐⇒ f has a repeated root.• G ↪→ An ⇐⇒ ∆ is a perfect square in k.

9.3 Computing 116

9 Galois Theory

E 9.4 Worked Examples e

Fact 9.4.1

Example 9.4.2(Indirect: exactly one conjugate pair of roots): If deg f = 5 with exactly3 real roots and one non-real complex conjugate pair, then G(f) = S5. G contains a transposi-tion, namely complex conjugation on the conjugate pair. This already implies G 6= A5, since atransposition is an odd number of even cycles.

The claim is that G contains an element of order 5, i.e. a 5-cycle, which is enough to generate S5.This follows because

• Galois acts transitively, so there is a length 5 orbit.• By Orbit-Stabilizer, 5 divides #G.• By Sylow, there is an element of order 5.

So G = S5.

9.4.1 Quadratics

Example 9.4.3(Classifying quadratics): Every degree 2 extension L/k is Galois, except possiblyin characteristic 2:

• If α ∈ L \ k then minα

(x) ∈ L[x] must split in L[x], so L is automatically a splitting field.

– Why? α ∈ L =⇒ minα

(x) = (x− α)g(x) which forces deg(g) = 1.

• If ch(k) 6= 2, then L is separable since

minα

(x)′ = 2x+ · · · 6≡ 0,

Remark 9.4.4: One can complete the square for quadratics:

f(x) = x2 + αx+ β =(x− α

2

)2+ β − α2

4 . .

Thus it suffices to consider quadratics of the form x2 + a.

Example 9.4.5(Quadratics):• G(x2 −m) = C2 for m not a perfect square.

– x2 −m = (x+√m)(x−

√m), so the splitting field is Q(

√m) of degree 2.

9.4 Worked Examples 117

9 Galois Theory

– Since G ≤ S2 and has order 2, G = S2 ∼= C2.– Concretely, take m = 2, then G = {id, τ} where τ :

√2 7→ −

√2, and correspondingly

a+ b√

2 7→ a− b√

2.

• G((x2 − 2)(x2 − 3)) = C2 × C2.

– Since G must permute irreducible factors, labeling the roots r1, r2 = ±√

2 and r3, r4 =±√

3, we have

G ⊆ 〈(1, 2), (3, 4)〉 = {id, (1, 2), (3, 4), (1, 2)(3, 4)} ∼= C2 × C2.

– #G = 4, taking the tower Q(√

2,√

3)/Q(√

2)/Q and noting√

3 6∈ Q(√

2) which makeseach step degree 2. So this forces G ∼= C2 × C2.

9.4.2 Cubics

Remark 9.4.6: Tricks/reminders:

• Try the rational roots test to check irreducibility, since reducible implies there’s a linear factor.• Nice situation: one real and two complex roots. Try Calculus and MVT to reason about real

roots. This immediately yields S3.• Otherwise, 3 real roots (or no easy way to check root types). Try discriminant classification:

put in the form t3 + pt+ q, potentially using t = x− b/3a, then ∆ = −4p3 − 27q2.

Remark 9.4.7(Easy cycles in odd/prime degrees with basic Calculus): If n := deg f is odd,then f has at least one real root. If f has two non-real roots, then G contains a transposition. If nis prime, then G contains an n-cycle (by transitivity and Cauchy), forcing G = Sn.

Example 9.4.8(Of using easy cycles to get S3): Let f(x) = x3 − 2.

Then f ′(x) = 3x2, so f is monotone increasing. By the MVT, checking f(−2) < 0 and f(2) > 0, fhas a single real root in [−2, 2]. The other two must be a complex conjugate pair.

Alternatively, just factor the darn thing: f(x) = (x − ω)(x − ζ3ω)(x − ζ23ω) where ω := 2

13 and

ζ33 = 1.

So G(f) contains a transposition, and since deg f = 3 is prime, G contains a 3-cycle and G(f) = S3.

Example 9.4.9(Of using easy cycles to get S3): Let f(x) = x3 − 4x+ 5.

Then f ′(x) = 3x2 − 4 = 0 when x = ±√

4/3. Checking f ′′(x) = 6x yields a max at −√

4/3 and a

min at√

4/3. Checking

f(4/3) = (4/3)3 − 4(4/3) + 5 = (64/27)− (16/3) + 5 = 55/27 > 0,

so knowing the general shape of a cubic, there is exactly one real root, somewhere in (−∞,−√

4/3).So G(f) = S3.


9 Galois Theory

Proposition 9.4.10(Classification for cubics).Away from ch k = 2, Galois groups of cubics are entirely determined by discriminants:There are only two possibilities: S3 or A3 ∼= C3.

• If√

∆ ∈ k, then G ∼= A3.• Otherwise, G ∼= S3.

Example 9.4.11(of discriminants of cubics):

Example 9.4.12(Cubics (manually)):• G(x3 + x+ 1) = S3:

– Irreducible because it has no rational roots (by the rational roots test)– f ′(x) = 3x2 + 1 > 0 so f increases everywhere and can only have one real root r, so

Q(r)/Q = deg f = 3.


9 Galois Theory

– The other roots are a non-real conjugate pair w,w, so Q(w, r)/Q(r) = deg f(x)/(x−r) =2.

– So [SF(f) : Q] = 6, and the only transitive subgroup of order 6 in S3 is S3 itself.

Example 9.4.13(Cubics (using ∆)):• G(x3 − x+ 1) = S3:

– This is already a depressed cubic, so use ∆ = −4(−1)3 − 27(1) = −23.– ∆ = −23 6∈ Q2, so G 6≤ A4 which forces G = S4.

• G(x3 − 3x+ 1) = A3:

– This is already a depressed cubic, so ∆ = −4(−3)3 + 27(1) = −3(27) = 81.– ∆ = 81 ∈ Q2, so G ≤ A3.

9.4.3 Quartics

Definition 9.4.14 (Resolvent of a quartic)If

f(x) = x4 + a3x3 + a2x

2 + a1x+ a0

then define the resolvent of f by

R4(t) = t3 − a2t2 + (a1a3 − 4a0) t+ 4a0a2 − a2

1 − a0a32.

Alternatively, it can be defined in terms of the roots ri:

(x− (r1r2 + r3r4)) (x− (r1r3 + r2r4)) (x− (r1r4 + r2r3)) .

For depressed quartics,

f(X) = X4 + cX + d =⇒ R3(X) = X3 − 4dX − c2.

Proposition 9.4.15(Classification for quartics).The Galois groups of irreducible quartics can be determined using discriminants, resolvents,and checking irreducibility:

• If√

∆ ∈ Q, then G = A4, C22 .

– If resolvent is irreducible: A4. Otherwise C22 .

• If√

∆ 6∈ Q then G = S4, D4, C4.

– Is resolvent is irreducible: S4. Otherwise, D4 or C4, argue by cycle types (or if f isirreducible in Q(

√∆), D4).


9 Galois Theory

• Summary:

A flow chart summarizing the full process:

See Hungerford 273 for classification.

Example 9.4.16(Quartics using resolvent cubics):• G(x4 − x− 1) = S4:

– Check f is irreducible in F2[x].


9 Galois Theory

– R3(t) = t4 + 4t− 1

• G(x4 + 8x+ 12) = A4:

– The resolvent cubic is x3 − 48x+ 64, which has no rational roots.– Now check

∆ = (−27)(84) + (256)(123) = (81)(212) ∈ Q2,

so G = A4.

• G(x4 + 3x+ 3) = D4:

– The resolvent cubic is g(x) = x3 − 12x + 9 = (x − 3)(x2 + 3x − 3) and ∆ = 33527, soG = C4, D4.

– Check D := ∆g = 21.– Check if g is irreducible in Q(

√21): suppose x4 + 3x+ 3 =

(x2 + ax+ b

) (x2 − ax+ c

),

then −a2 + b+ c = 0, a(c− b) = 0, bc = 3♦ From a(c − b) = 0, if a = 0 then b = −c and c2 = 3, but

√−3 6∈ Q(D). Otherwise

c = b and c2 = 3, but√

3 6∈ Q(D).– So G = D4.

9.4.4 Cyclotomic Fields

Example 9.4.17(Cyclotomic Fields): Gal(Q(ζn)/Q) ∼= (Z/n)× and is generated by maps of theform ζn 7→ ζjn where (j, n) = 1. I.e., the following map is an isomorphism:

(Z/n)× → Gal(Q(ζn),Q)[r] 7→ (ϕr : ζn 7→ ζrn)

Fact 9.4.18The splitting field of xp − 1 is Q(ζp), and the splitting field of xp + 1 is Q(ζ2p).

• xp − a factors asp−1∏k=0

(x− ζkpω) where ω := a1p , so this splits in Q(ζp, ω) which has degree

ϕ(p) · deg minω

(x) = (p− 1)p.

– This yields two cyclic subgroups Cp−1, Cp where Cp E G, and thus some semidirectproduct Cp−1 y Cp.

• xp + a factors asp−1∏k=0

(x− ζkpω) for ω := (−a)1p .

Also use that splitting fields over Q are always normal, so it suffices to check that f is separableand irreducible to show extensions are Galois.


9 Galois Theory

Example 9.4.19(xn − a): Degree 3:

• G(x3 − 2) : S3

– The roots are ζk3ω for 0 ≤ k ≤ 3, ω := 213 .

– The splitting field is Q(ω, ζ3) which has degree 3ϕ(3) = 6.– The possibilities are G = A3 ∼= C3, S3, and order 6 forces G = S3.– Useful alternative:♦ Note that there is exactly one real root and one conjugate pair, so G contains a

transposition (23).♦ There is a 3-cycle (123) given by fixing ω and sending ζ3 7→ ζ3ω, and this is enough

to generate D3 ∼= S3.

Degree 4:

• G(x4 − 1) = C2:

– The roots are ζk4 for 0 ≤ k ≤ 3.– The splitting field is Q(ζ4) = Q(i) which has degree ϕ(4) = 2.– But this is not a reducible polynomial! Use that that Galois is defined as Aut(SF(f)/Q)

and quadratic extensions are Galois.

• G(x4 − 2) = D4:

– The roots are ζk4ω for 0 ≤ k ≤ 3, where ζ4 = i, ω = 214 .

♦ Explicitly, ri ∈ {ω, iω,−ω,−iω}– The splitting field is Q(ω, ζ4), which has degree 4ϕ(4) = 8 since min

ζ4= x2 + 1, which is

still irreducible over Q(ω) ⊆ R.– D4 ≤ S4 is the only transitive subgroup of order 8.– Useful note on bounding the size:♦ Any σ ∈ G must preserves roots of x4 − 2 but also x2 + 1. So there are at most 4

possibilities for σ(ω), and at most 2 for σ(ζ4), so #G ≤ 8 and G 6= S4.– Explicitly, there is a 4-cycle σ = (1, 2, 3, 4) generated by ω 7→ ζ4ω and a 2-cycle (2, 4)

given by complex conjugation, and this generates D4 since gcd(4− 2, 4) 6= 1.♦ Why this is a 4-cycle: check σ(i) = i, and:

σ(r1) = r2 = ir1

σ(r2) = σ(ir1) = ir2 = r3

σ(r3) = σ(ir2) = ir3 = i(ir2) = −r2 = r4.

General cases:

• G(xp − 1) = C×p :

– ?

Example 9.4.20(xn + a):


9 Galois Theory

• G(x4 + 1) = C22 :

– This is irreducible because it is irreducible (by having no roots) mod 3.– The roots are ζk8 for k = 1, 3, 5, 7 coprime to 8, since this is Φ8(x).– The splitting field is Q(ζ8) = Q(i,

√2), noting that ζ8 = e

2πi8 = e

πi4 = cos(π/4) +

i sin(π/4) = (1/2)(√

2 + i√

2) so we have containment and both are degree ϕ(8) = 4extensions.

– This restricts to C4, C22 .

– Reduce mod 5 to get (x2 + 2)(x2 + 2) of cycle type (2, 2), forcing C22 .

• G(x4 + 2) = D4:

– The roots are ζk8ω for ω = 214 , k = 1, 3, 5, 7 coprime to 8.

– The splitting field is Q(ζ8, ω) = Q(ζ4, ω).

• G(x4 + 3) = D4:

– The roots are ζk8 , ω for ω = 314 , k = 1, 3, 5, 7 coprime to 8.

– The splitting field is Q(ω, ζ8) of degree ϕ(8) = 8

9.4.5 Finite Fields

Example 9.4.21(Finite Fields): Gal(Fpn/Fp) ∼= Z/ 〈n〉, a cyclic group generated by powers ofthe Frobenius automorphism:

ϕp : Fpn → Fpnx 7→ xp

See D&F p.566 example 7.

E 9.5 Lattices e

Fact 9.5.1 (Common lattices of subgroups)n = 2:

• D2 ∼= 1• A2 ∼= 1• S2 ∼= C2

n = 3:

• D3 ∼= S3• A3 ∼= C3.

9.5 Lattices 124

9 Galois Theory

•

•

n = 4:

•

•

9.5 Lattices 125

9 Galois Theory

•

n = 5:

•

•

•

Misc:

9.5 Lattices 126

10 Modules

•

10 Modules

E 10.1 Definitions and Basics e

Definition 10.1.1 (R-modules)Four properties:

• r(x+ y) = rx+ ry• (r + s)x = rs+ sx• (rs)x = r(s(x))• 1Rx = x

Note that M is additionally an R-algebra if the multiplication map is R-bilinear and so givenby m : M⊗R2 →M satisfying

r.m(a⊗ b) = m(r.a⊗ b) = m(a⊗ r.b) ∀r ∈ R, a, b ∈M.

Proposition 10.1.2(The one-step submodule test).N ⊆ M is an R-submodule iff N is nonempty and for every r ∈ R and x, y ∈ N , we haverx+ y ∈ N .

Definition 10.1.3 (Module Morphisms)A map f : M → N is a morphism of modules iff f(rm+ n) = rf(m) + f(n).

Proposition 10.1.4(One-step module morphism test).A map ϕ : M → N is a morphism in R-Mod iff

ϕ(r.x+ y) = r.ϕ(x) + ϕ(y) ∈ N ∀r ∈ R, x, y ∈M.

Remark 10.1.5: Quotients of modules are easier to reason about additively, writing M/N ={x+N} as cosets. Then (x+N) + (y +N) = (x+ y) +N and (x+N)(y +N) = (xy) +N .

Modules 127

10 Modules

Definition 10.1.6 (Simple modules)A module is simple iff it has no nontrivial proper submodules.

Definition 10.1.7 (Indecomposable modules)A module M is decomposable iff it admits a direct sum decomposition M ∼= M1 ⊕M2 withM1,M2 6= 0. An indecomposable module is defined in the obvious way.

Definition 10.1.8 (Cyclic modules)A module M is cyclic if there exists a single generator m ∈M such that M = mR := 〈m〉.

E 10.2 Structure Theorems e

Proposition 10.2.1(Isomorphism theorems).

M/ kerϕ ∼= imϕ

A+B

B∼=

A

A ∩BM/A

B/A∼=M

B{Submodules of M

containing N

} {Submodules of M/N}

A A/N.

Note that the lattice correspondence commutes with sums and intersections of submodules.

Proposition 10.2.2(Recognizing direct sums).If M1,M2 ≤M are submodules, then M = M1 ⊕M2 if the following conditions hold:

• M1 +M2 = M• M1 ∩M2 = 0

E 10.3 Exact Sequences e

Definition 10.3.1 (Exact Sequences)A sequence of R-module morphisms

0 d1−→ Ad2−→ B

d3−→ C → 0

is exact iff im di = ker di+1.

10.2 Structure Theorems 128

10 Modules

Remark 10.3.2: Note that C ∼= B/d1(A) always, but B is not a direct sum of the outer termsunless the sequence splits.

Definition 10.3.3 (Split Exact Sequences)A short exact sequence

ξ : 0→ Ad1−→ B

d2−→ C → 0

has a right-splitting iff there exists a map s : C → B such that d2 ◦ s = idC . ξ has aleft-splitting iff there exists a map t : B → A such that t ◦ d1 = idA.

Proposition 10.3.4(Equivalent conditions for splitting SESs).Let ξ : 0→ A

d1−→ Bd2−→ C → 0 be a SES, then TFAE

• ξ admits a right-splitting s : C → B.• C is projective.• ξ admits a left-splitting t : B → A.• A is injective.• ξ is isomorphic to a SES of the form 0→ A→ A⊕ C → C → 0.

Proof (?).Right-splitting implies direct sum:

• Use that B ⊂ ker d2 + im s, writing b = (b− sd2(b)) + sd2(b) and noting

d2(b− sd2(b)) = d2(b)− d2sd2(b) = d2(b)− d2(b) = 0.

• Show ker d2 ∩ im s = 0, writing b with d2(b) = 0 and b = s(c) for some c yields

0 = d2(b) = d2s(c) = idC(c) = c.

�

E 10.4 Free and Projective Modules e

Definition 10.4.1 (Free Module)A free module M is a module satisfying any of the following conditions:

• A universal property: There is a set B and a set map M ι−→ B such that every set mapB N−→ lifts:

10.4 Free and Projective Modules 129

10 Modules

M

B Nf

fι

Link to Diagram

• Existence of a basis:There is linearly independent (so

∑riβi = 0 =⇒ ri = 0) spanning set (so m ∈M =⇒

m =∑

riβi ) of the form B := {βi}i∈I ,

• Direct sum decomposition:M decomposes as M ∼=

⊕i∈I

βiR, a sum of cyclic submodules.

Example 10.4.2(A non-free module): Z/6 is a Z-module that is not free, since the element [3]is a torsion element, where 2[3] = [6] = [0]. This uses the fact that free modules over a PID aretorsionfree.

Definition 10.4.3 (Free rank)If a module M is free, the free rank of M is the cardinality of any basis.

Proposition 10.4.4(?).Every free R-module admits a basis (spanning R-linearly independent set).

Definition 10.4.5 (Torsion and torsionfree)An element m ∈M is a torsion element if there exists a nonzero r ∈ R such that rm = 0M .A module M is torsion-free if and only if for every x ∈M , mx = 0M =⇒ m = 0M , i.e. Mhas no nonzero torsion elements. Equivalently, defining Mt :=

{m ∈M

∣∣∣ ∃r ∈ R, rm = 0M}

as the set of all torsion elements, M is torsion free iff Mt = 0. If Mt = M , we say M is atorsion module.

Proposition 10.4.6(Free implies torsionfree).For R an integral domain, any finitely generated free R-module M is torsionfree.

Proof (that free implies torsionfree).

• If M is finitely generated, write M = 〈X〉 with X := {x1, · · · , xm} and #X <∞ a finitegenerating set.

• Since M is free, there is some maximal subset of generators B := {x1, · · · , xn} ⊆ Xwhere n ≤ m that is linearly independent.

• Consider N ≤ M defined by 〈B〉; this is a basis for N and makes N free. The claim isnow that M ∼= N , so that any maximal linearly independent subset of generators is allof X.


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10 Modules

• If N 6∼= M , set Bc := X \ B = {xn+1, · · · , xm} to be all generators for M that the basisB misses.

• Then Bc ∪ {xj} for any n+ 1 ≤ j ≤ m has a linear dependence, and rjxj +n∑k=1

rnxn = 0

for some rj 6= 0 implies rjxj = −n∑k=1

rnxn.

• Let r be the product of all of the scalars obtained this way, so r =m∏

k=n+1rj , and consider

the submodule rX ≤ N ≤M . We get rM ≤ N ≤M since X is a generating set for M ,so it now suffices to show rM ∼= M .

• Just define a map ϕr : M � rM where m 7→ rm, and note kerϕr ={m ∈M

∣∣∣ rm = 0}

= 0 since M is torsionfree. So M = M/ kerϕr ∼= rM .

�

Example 10.4.7(A torsionfree module that is not free): Q ∈ Z-Mod is torsionfree, but notfree as a Z-module. This follows because any two elements a/b, p/q are in a single ideal, since takingd := gcd(b, q) we have 1/a = 1/d+ · · · 1/d and similarly p/q = 1/a+ · · ·+ 1/a, so these are in 〈1/d〉.So any basis has size one, which would mean Q = {±1/d,±2/d, · · ·} which in particular doesn’tinclude the average of the first two terms.

Definition 10.4.8 (Projective Modules)A module P is projective iff it satisfies any of the following conditions:

• A universal property: for every surjective N g−→M and P f−→M , the following lift exists:

P

N M 0g

f∃f

Link to Diagram

• Direct summand:P is a direct summand of a free module F , so F = P ⊕ T for some module T ≤ F .

• Splitting:For every SES 0 → A → B → P → 0, there is a right section P → B such thatP → B → P = idP .

Note that this implies B ∼= im(P → B)⊕ ker(B →P ).

• Exactness:The (always left-exact) covariant hom functor Hom(P,−) is right-exact.


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10 Modules

Remark 10.4.9: There is a nice way to remember the right diagrams for injective and projectivemodules. The slogan is that morphisms out of a projective module can be pulled back throughepimorphisms/surjections, and morphisms into an injective module can be pushed forward throughmonomorphisms/injections.

P

0 A B C 0

I

Pull back through surjection

Push forward through injections

Link to Diagram

Proposition 10.4.10(Free implies projective).Any free M ∈ R-Mod is projective.

Proof (?).

• Let M be free, so that the universal property gives us this diagram:

M

B Nf

fι

Link to Diagram

• To show M is projective, we need to produce a lift in the following diagram, where B,Care arbitrary:

M

B C 0

f∃f

g

Link to Diagram

• It suffices to produce a map B → B, since the universal property then provides M → B.Here’s the schematic:


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10 Modules

B {ei}

M

B C 0

{g−1f(ei)

}{f(ei)}

f∃f

g

Link to Diagram

• Here we write B := {ei}, included into M , and mapped by f to C. Then use surjectivityto choose preimages in B under g arbitrarily, and this defines a morphism B → B.

�

Example 10.4.11(Projective 6 =⇒ free): Let R1, R2 be two nontrivial rings and set R := R1⊕R2.Then R1, R2 are projective R-modules by construction, but each factor contains R-torsion: settinge := (0, 1) ∈ R we have ey R1 = 0R1 . Since free implies torsionfree, R1 can not be a free R-module.

E 10.5 Classification of Modules over a PID e

Proposition 10.5.1(STFGMPID).Let M be a finitely generated modules over a PID R. Then there is an invariant factordecomposition

M ∼= Fm⊕i=1

R/(ri) where r1∣∣ r2

∣∣ · · ·and similarly an elementary divisor decomposition:

M ∼= Fn⊕i=1

R/ 〈peii 〉

where F is free of finite rank and the pi are not necessarily distinct primes in R.

Proposition 10.5.2(Principal Ideals are Free).If I E R is an ideal of R, then I is a free R-module iff I is a principal ideal.

10.5 Classification of Modules over a PID 133

https://q.uiver.app/?q=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

10 Modules

Proof (?).=⇒ :Suppose I is free as an R-module, and let B = {mj}j∈J ⊆ I be a basis so we can writeM = 〈B〉. Suppose that |B| ≥ 2, so we can pick at least 2 basis elements m1 6= m2, andconsider

c = m1m2 −m2m1,

which is also an element of M . Since R is an integral domain, R is commutative, and so

c = m1m2 −m2m1 = m1m2 −m1m2 = 0M

However, this exhibits a linear dependence between m1 and m2, namely that there existα1, α2 6= 0R such that α1m1 + α2m2 = 0M ; this follows because M ⊂ R means that we cantake α1 = −m2, α2 = m1. This contradicts the assumption that B was a basis, so we musthave |B| = 1 and so B = {m} for some m ∈ I. But then M = 〈B〉 = 〈m〉 is generated by asingle element, so M is principal.⇐= : Suppose M E R is principal, so M = 〈m〉 for some m 6= 0M ∈M ⊂ R.Then x ∈ M =⇒ x = αm for some element α ∈ R and we just need to show thatαm = 0M =⇒ α = 0R in order for {m} to be a basis for M , making M a free R-module.But since M ⊂ R, we have α,m ∈ R and 0M = 0R, and since R is an integral domain, wehave αm = 0R =⇒ α = 0R or m = 0R. Since m 6= 0R, this forces α = 0R, which allows {m}to be a linearly independent set and thus a basis for M as an R-module.

�

Remark 10.5.3: This says every module M decomposes as M ∼= FM ⊕Mt where FM is free (andthus torsionfree) and Mt is torsion, and moreover FM ∼= M/Mt.

That M/Mt is torsionfree: suppose r(m+Mt) = Mt, so rm ∈Mt is torsion. Then r′(rm) = 0 forsome r′, making m torsion, and m+Mt = Mt is the zero coset.

That FM ∼= M/Mt: take the SES 0→Mt →M → F → 0 to get F ∼= M/Mt. This splits since F isfree and thus projective, so F ∼= M ⊕Mt.

E 10.6 Algebraic Properties e

Definition 10.6.1 (Module structure on tensor products)

r y (m⊗ n) := (r y m)⊗ n.

Proposition 10.6.2(?).

10.6 Algebraic Properties 134

11 Linear Algebra

If dimk V,dimkW <∞ then there is an isomorphism

V ∨ ⊗k W∼−→ Hom

k-Mod(V,W )

v ⊗ w 7→ v(−)w.

Proposition 10.6.3(?).If either of dimk V,dimkW is finite, then

V ∨ ⊗k W∨∼−→ (V ⊗W )∨

v ⊗ w 7→ (x⊗ y 7→ v(x)w(y)).

Proposition 10.6.4(?).

Homk-Mod

(V,W ) ∼−→ Homk-Mod

(W,V )∨

T 7→ Tr(T ◦ −).

Proposition 10.6.5(?).If T : V ↪→W is injective, then T ⊗ 1X : V ⊗X ↪→W ⊗X is also injective for any X. ThusF (−) = (−⊗X) is right-exact for any X.

Example 10.6.6(Computing tensor products): Z/2⊗Z Z/3 = 0:

0 Z⊗Z Z/3 Z⊗Z Z/3 Z/3⊗Z Z/2

0 Z/3 Z/3 0(−×2)

(−×2)×1

proj2 proj2 ∼=

Link to Diagram

11 Linear Algebra

Remark 11.0.1: Algorithm for SNF: D&F page 479.

Remark 11.0.2: Some definitions:

• At is the usual transpose.• A† is the conjugate transpose.

Linear Algebra 135

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11 Linear Algebra

• A matrix is A† is adjoint to A iff 〈Ax, y〉 =⟨x, A†y

⟩.

– A is self-adjoint iff A is an adjoint for itself, so 〈Ax, y〉 = 〈x, Ay〉.

• A is symmetric iff A = At.

– A is orthogonal iff AtA = AAt = I

• A is Hermitian iff A† = A.

– A is normal iff AA† = A†A.– A is unitary iff A†A = AA† = I.

Fact 11.0.3 (Undergrad reminders)

detM =∏σ∈Sn

ε(σ)n∏i=1

ai,σ(i).

For example,

det

a11 a12 a13a21 a22 a23a31 a32 a33

=a11a22a33 + a12a23a31 + a13a21a32

−a13a22a31 − a12a21a33 − a11a23a32.

Let minorA(i, j) denote A with the ith row, jth column deleted.

One can expand determinants along rows:

det(A) =n∑j=1

(−1)i+jaij det minorA(i, j).

Also useful, a matrix can be inverted by computing the adjugate:

A−1 = 1detA adj(A) adj(A)ij := (−1)i+j det minorA(j, i).

The eigenvalues of an upper-triangular matrix are exactly the diagonal entries, and the determinantis their product. More generally, the determinant is always the product of the eigenvalues, and thetrace is the sum of the eigenvalues, so tr(A) =

∑λi and det(A) =

∏λi.

Matrices can be block-multiplied when all dimensions are compatible:[A BC D

] [E FG H

]=[AE +BG AF +BHCE +DG CF +DH

].

Linear Algebra 136

11 Linear Algebra

Note that if any of these matrix multiplications don’tmake sense, the results won’t be valid!

If A is upper triangular, some entries of Ak can be computed easily:

A :=

a1 ∗

. . .

0 an

=⇒ Ak =

ak1 ∗

. . .

0 akn

.

Traces of products can be commuted: tr(AB) = tr(BA), so similar matrices have identical tracessince tr(PJP−1) = trPP−1J = tr J .

The coefficients of the characteristic polynomial are elementary symmetric functions in the eigen-values:

χA(t) = tn −(∑

i

λi

)tn−1 +

∑i<j

λiλj

tn−2 + · · · ±(∏

i

λi

).

Example 11.0.4(of polynomial long division): Consider f(x) := x3 − 6x2 + 12x− 8, then anyrational root is in {±8,±4,±2,±1}. Testing f(2) = 0 works, and dividing by x− 2 yields

Linear Algebra 137

11 Linear Algebra

The rest can be factored by inspection:

f(x) = (x− 2)(x2 − 4x+ 4) = (x− 2)3.

E 11.1 Definitions e

Remark 11.1.1: The main powerhouse: for T : V → V a linear transformation for V ∈ Vectk,map to V ∈ k[x]-Mod by letting polynomials act via p(x) · v := p(T )(v). Using that k[x] is a PIDiff k is a field, and we can apply the FTFGMPID to get two decompositions:

V ∼=n⊕i=1

k[x]/ 〈qi(x)〉 qi(x)∣∣ qi+1(x)

∣∣ · · ·V ∼=

m⊕j=1

k[x]/ 〈pi(x)ei〉 with pi not necessarily distinct.

• The qi are the invariant factors of T

– qi is the minimal polynomial of T restricted to Vi := k[x]/ 〈qi(x)〉.– The largest invariant factor qn is the minimal polynomial of T .

– The productn∏i=1

qi(x) is the characteristic polynomial of T .

• The pi are the elementary divisors of T .

– Todo: what can you read off of this. . . ?

Definition 11.1.2 (Nondegenerate Bilinear Form)

todo

Definition 11.1.3 (Quadratic Form)

todo

Definition 11.1.4 (Gram Matrix)

todo

Definition 11.1.5 (Normal Matrix)A matrix A ∈ Mat(n× n;C) is normal iff A†A = AA† where A† is the conjugate transpose.

Definition 11.1.6 (Semisimple)A matrix A over k is semisimple iff A is diagonalizable over kAlg, the algebraic closure.

11.1 Definitions 138

11 Linear Algebra

Definition 11.1.7 (Nilpotent)A matrix A over k is nilpotent iff Ak = 0 for some k ≥ 1.

Idea: upper triangular matrices.

Definition 11.1.8 (Unipotent)A element A in a ring R is unipotent iff A− 1 is nilpotent.

Idea: an upper-triangular matrix with ones on thediagonal.

Proposition 11.1.9(Triangular Decomposition).Any linear map T : V → V over a perfect field decomposes as T = S +N with S semisimple(diagonal), N nilpotent, and [DN ] = 0. If T is invertible, then T decomposes as T = SUwhere S is semisimple, U is unipotent, and [UN ] = 0.

Proposition 11.1.10(Perp of sum is intersection of perps).

(∑Wi

)⊥=⋂(

W⊥i

).

Definition 11.1.11 (Similar Matrices)Two matrices A,B are similar (i.e. A = PBP−1) ⇐⇒ A,B have the same Jordan CanonicalForm (JCF).

Definition 11.1.12 (Equivalent Matrices)Two matrices A,B are equivalent (i.e. A = PBQ) ⇐⇒

• They have the same rank,

• They have the same invariant factors, and

• They have the same (JCF)

11.1.1 Matrix Groups

Definition 11.1.13 (General Linear Group)

GLn(R) ={A∣∣∣ A = A

}.

Proposition 11.1.14(Order of GLn).

todo

11.1 Definitions 139

11 Linear Algebra

Definition 11.1.15 (Special Linear Group)

SLn(C) :={A∣∣∣ detA = 1

}.

Definition 11.1.16 (Orthogonal Group)

On(C) :={A∣∣∣ AtA = AAt = I

}.

Dimension: n(n− 1)/2.

Definition 11.1.17 (Special Orthogonal Group)

SOn(R) ={A∣∣∣ AAt = I

}= ker(GLn(R)→ k×).

Definition 11.1.18 (Unitary Group)

Un(C) :={A∣∣∣ A†A = AA† = 1

}.

Definition 11.1.19 (Special Unitary Group)

SUn(C) :={A ∈ Un(C)

∣∣∣ detA = 1}.

Definition 11.1.20 (Symplectic Group)

Sp2n(C) :={A ∈ GL2n(C)

∣∣∣ AtJA = J}

J :=[

0 1n1n 0

].

Matrix group definitions.

E 11.2 Minimal / Characteristic Polynomials e

Proposition 11.2.1(Useful computational trick).

11.2 Minimal / Characteristic Polynomials 140

11 Linear Algebra

A trick for finding characteristic polynomials:

χA(t) =n∑k=0

(−1)k tr(∧k

A

)tn−k

= tn − tr (A) tn−1 + tr(∧2

A)tn−2 − · · · ± tr

(∧n−1A)t∓ det(A),

using that

∧0A := 1∧1A := A

tr(∧n

A)

= det(A).

Moreover, the intermediate traces are easy to compute by hand:

tr(∧`

A

)=∑

det(M `),

where the sum is taken over all ` × ` principal minors: determinants of the(n`

)principal

matrices which are obtained by choosing ` diagonal entries to keep and and deleting the rowsand columns for every entry not chosen. Equivalently, one can select n − ` diagonal entriesand delete the corresponding row/column for each.

Example 11.2.2(?):


11 Linear Algebra

To factor this polynomial, the rational roots test can be useful: for f(t) = antn+· · ·+a1t+a0,

rational roots are of the form p/q where p∣∣ an and q

∣∣ a0. Note that this simplifies greatly forf monic! Once you have a root, apply polynomial long division to get a smaller problem,and hopefully this continues to work until it’s factored.

Remark 11.2.3: Fix some notation:

minA

(x) : The minimal polynomial of A

χA(x) : The characteristic polynomial of A.

Definition 11.2.4 (Minimal polynomial)The minimal polynomial of a linear map T is the unique monic polynomial min

T(x) of

minimal degree such that minT

(T ) = 0.

Definition 11.2.5 (Characteristic polynomial)The characteristic polynomial of A is given by

χA(x) = det(A− xI)) = det(SNF (A− xI)).

Fact 11.2.6If A is upper triangular, then det(A) =

∏i

aii

Theorem 11.2.7(Cayley-Hamilton).The minimal polynomial divides the characteristic polynomial, and in particular χA(A) = 0.

Proof (?).By minimality, min

Adivides χA. Every λi is a root of min

A(x): Let (vi, λi) be a nontrivial

eigenpair. Then by linearity,

minA

(λi)vi = minA

(A)vi = 0,

which forces minA

(λi) = 0.�


11 Linear Algebra

E 11.3 Finding Minimal Polynomials e

Proposition 11.3.1(How to find the minimal polynomial).Let m(x) denote the minimal polynomial A.

1. Find the characteristic polynomial χ(x); this annihilates A by Cayley-Hamilton. Thenm(x)

∣∣ χ(x), so just test the finitely many products of irreducible factors.

2. Pick any v and compute Tv, T 2v, · · ·T kv until a linear dependence is introduced. Writethis as p(T ) = 0; then min

A(x)

∣∣ p(x).

E 11.4 Other Canonical Forms e

Proposition 11.4.1(?).Let T : V → V be a linear map where n := dimk V . TFAE:

• There exists a basis {ei} of V such that

T (ei) ={ei−1 i ≥ 20 i = 1.

• There exists a cyclic vector v such that{T kv

∣∣∣ k = 1, 2, · · · , n}form a basis for V .

• Tn−1 6= 0

• dimk kerT ` = ` for each 1 ≤ ` ≤ n.

• dimk kerT = 1.

11.4.1 Rational Canonical Form

Corresponds to the Invariant Factor Decomposition of T .

Definition 11.4.2 (Companion Matrix)Given a monic p(x) = a0 + a1x+ a2x

2 + · · ·+ an−1xn−1 + xn, the companion matrix of p is

11.3 Finding Minimal Polynomials 143

11 Linear Algebra

given by

Cp :=

0 0 . . . 0 −a01 0 . . . 0 −a10 1 . . . 0 −a2...

. . ....

0 0 . . . 1 −an−1

.

Proposition 11.4.3(Equivalent conditions for cyclic vectors).Let V be finite dimensional and T ∈ GL(V ). TFAE:

• V ∼= k[x]/ 〈p〉 as a k[x]-module.• V admits a cyclic vector v where p(x) is minimal degree monic polynomial such thatp(x) y v = 0

• V is a cyclic k[x]-module with annihilator ideal generated by p(x).• T is similar to the companion matrix of p(x).• min

T(x) = χT (x).

• T has exactly one invariant factor.• RCF(T ) has a single block.

Proposition 11.4.4(Minimal equals characteristic iff cyclic).χA(x) = min

A(x) iff A admits a cyclic vector.

Proof (?).6 =⇒ : In general, min

A

∣∣ χA, so suppose they’re not equal. Set n := degχA, then if n′ :=

deg minA

< n, using that minA

(A) = 0 this exhibits a linear dependence in{v,Av, · · · , An′v

}for any v. In particular, since n > n′, any set {v,Av, · · · , Anv} has a linear dependence.

=⇒ : Apply the structure theorem to write V ∼=m⊕i=1

k[x]/ 〈pi〉. Since χA(x) =∏

pi(x) and

minA

(x) = pm(x), this forces m = 1 – one way to see this is that dimk V =m∑i=1

dimk k[x]/ 〈pi〉,

where degχA = dimk V and deg minA

= dim k[x]/ 〈pm〉. For these to be equal, this forcesdimk k[x]/ 〈pi〉 = 0 for 1 ≤ i ≤ m − 1, making V a cyclic k[x]-module. So V = k[x] y v forsome v ∈ V , which is the desired cyclic vector, and

V ={f(x).v

∣∣∣ f ∈ k[x]}

= spank{Akv

∣∣∣ k ≥ 0}.

By Cayley-Hamilton, χA(A) = 0 and so An is a linear combinations of Ak for 0 ≤ k ≤ n− 1,so V = spank

{Akv

∣∣∣ 0 ≤ k ≤ n− 1}.

�

Proposition 11.4.5(Rational Canonical Form).RCF(A) is a block matrix where each block is the companion matrix of an invariant factor of

11.4 Other Canonical Forms 144

11 Linear Algebra

A.

Remark 11.4.6: Thus the blocks of RCF(A) biject with invariant factors of A. Note that anycompanion matrix is already in RCF.

Proof (Derivation of RCF).

• Let k[x] y V by p(x) y v := p(T )(v), making V into a finitely generated torsionk[x]-module.

– Note that k[x]-submodules are exactly T -invariant subspaces.

• k a field implies k[x] a PID, so apply structure theorem to obtain an invariant factordecomposition

V ∼= k[x]/ 〈χT (x)〉 ∼=m⊕i=1

k[x]/ 〈pi(x)〉 p1(x)∣∣ p2(x)

∣∣ · · · pm(x).

• Since each factor is submodule, each corresponds to a T -invariant subspace Vi where piis the minimal polynomial of T restricted to Vi.

– The largest invariant factor pm is the minimal polynomial of T , their product is thecharacteristic polynomial. This follows because pm(x) y V = 0, since pi

∣∣ pm forall i, forcing min

A

∣∣ pm by minimality.

• Write V ∼=m⊕i=1

Vi as a k[x]-module, where Vi := k[x]/ 〈pi(x)〉, then T is a block matrixm⊕i=1

Ti where Ti is the restriction of T to Vi:

T1 0 0 · · · 00 T2 0 · · · 0...

. . ....

0 · · · Tn

.

• It suffices to determine the form of a single Mi, so without loss of generality supposem = 1 so V = V1 = k[x]/ 〈p(x)〉 is a cyclic k[x]-module with deg p(x) = n.

• χM (x) = minM

(x) ⇐⇒ there exists a cyclic vector v, so the set {vi}n−1i=0 :={

v, Tv, T 2v, · · · , Tn−1v}is a basis for V1.

– If there is any linear independence, this gives a polynomial relationn′∑i=1

aiTiv = 0 for

some n′ < n, but then q(x) :=n′∑i=1

aixi is a polynomial annihilating T , contradicting

the minimality of p(x).


11 Linear Algebra

– So this yields n linearly independent vectors in kn, so it’s a basis.

• What is Mi in this basis? Check where basis elements are mapped to by T , noting that

p(T ) =n∑i=1

aiTiv = Tn + an−1T

n−1v + an−2Tn−2 + · · ·+ a1Tv + a0v = 0,

using the minimal polynomial we can write

– Tv0 = v1– Tv2 = T 2v0– Tv3 = T 3v0– · · ·– Tvn−2 = Tn−1v– Tvn−1 = Tnv = −an−1T

n−1v− · · · − a1Tv− a0v

• So we have

M1 =

0 −a01 0 −a1

1 0 −a2. . . 0

...1 −an−1

.

�

11.4.2 Smith Normal Form

Fact 11.4.7For A ∈ Mat(m×n;R) over R any PID, SNF(A) is a matrix whose diagonal entries are the invariantfactors. How to compute SNF(A): take A = diag(ai) where ai = di/di−1 and di is the gcd of thedeterminants of all i× i minors of A. A ∼ B are similar ⇐⇒ SNF(A) = SNF(B).

11.4.3 Using Canonical Forms

Lemma 11.4.8(?).The characteristic polynomial is the product of the invariant factors, i.e.

χA(x) =n∏j=1

fj(x).

Lemma 11.4.9(?).


11 Linear Algebra

The minimal polynomial of A is the invariant factor of highest degree, i.e.

minA

(x) = fn(x).

Proposition 11.4.10(?).For a linear operator on a vector space of nonzero finite dimension, TFAE:

• The minimal polynomial is equal to the characteristic polynomial.

• The list of invariant factors has length one.

• The Rational Canonical Form has a single block.

• The operator has a matrix similar to a companion matrix.

• There exists a cyclic vector v such that spank{T jv

∣∣∣ j = 1, 2, · · ·}

= V.

• T has dimV distinct eigenvalues

E 11.5 Diagonalizability e

Remark 11.5.1: Notation: A† denotes the conjugate transpose of A.

Lemma 11.5.2(?).Let V be a vector space over k an algebraically closed and A ∈ End(V ). Then if W ⊆ V is aninvariant subspace, so A(W ) ⊆W , the A has an eigenvector in W .

Theorem 11.5.3(The Spectral Theorem).

1. Hermitian (self-adjoint) matrices (i.e. A† = A) are diagonalizable over C.2. Symmetric matrices (i.e. At = A) are diagonalizable over R.

Remark 11.5.4: In fact, A is symmetric ⇐⇒ SpecA forms an orthonormal basis.

Proof (of spectral theorem).

• Suppose A is Hermitian.

• Since V itself is an invariant subspace, A has an eigenvector v1 ∈ V .

• Let W1 = spank {v1} ⊥.

• Then for any w1 ∈W1,

〈v1, Aw1〉 = 〈Av1, w1〉 = λ〈v1, w1〉 = 0,

so A(W1) ⊆W1 is an invariant subspace, etc.

11.5 Diagonalizability 147

11 Linear Algebra

• Suppose now that A is symmetric.

• Then there is an eigenvector of norm 1, v ∈ V .

λ = λ〈v, v〉 = 〈Av, v〉 = 〈v, Av〉 = λ =⇒ λ ∈ R.

�

Proposition 11.5.5(Simultaneous Diagonalizability).A set of operators {Ai} pairwise commute ⇐⇒ they are all simultaneously diagonalizable.

Proof (?).By induction on number of operators

• An is diagonalizable, so V =⊕

Ei a sum of eigenspaces• Restrict all n− 1 operators A to En.• The commute in V so they commute in En• (Lemma) They were diagonalizable in V , so they’re diagonalizable in En• So they’re simultaneously diagonalizable by I.H.• But these eigenvectors for the Ai are all in En, so they’re eigenvectors for An too.• Can do this for each eigenspace.

Full details here�

Theorem 11.5.6(Characterizations of Diagonalizability).M is diagonalizable over F ⇐⇒ min

M(x,F) splits into distinct linear factors over F, or

equivalently iff all of the roots of minM

lie in F.

Proof (?).=⇒ : If min

Afactors into linear factors, so does each invariant factor, so every elementary

divisor is linear and JCF (A) is diagonal.⇐= : If A is diagonalizable, every elementary divisor is linear, so every invariant factor factorsinto linear pieces. But the minimal polynomial is just the largest invariant factor.

�

E 11.6 Matrix Counterexamples e

Example 11.6.1(?): A matrix that:

• Is not diagonalizable over R but diagonalizable over C

• Has no eigenvalues over R but has distinct eigenvalues over C

11.6 Matrix Counterexamples 148

https://kconrad.math.uconn.edu/blurbs/linmultialg/minpolyandappns.pdf#page=9

11 Linear Algebra

• minM

(x) = χM (x) = x2 + 1

M =(

0 1−1 0

)∼(−1√−1 0

0 1√−1

).

Example 11.6.2(?): A matrix that:

• Is not diagonalizable over C,

• Has eigenvalues [1, 1] (repeated, multiplicity 2)

• minM

(x) = χM (x) = x2 − 2x+ 1

M =(

1 10 1

)∼(

1 10 1

).

Example 11.6.3(?): Non-similar matrices with the same characteristic polynomial(0 00 0

)and

(0 00 0

)

Here χA(x) = χB(x) = x2, but they are not conjugate since their JCFs differ (note that they’realready in JCF!)

Example 11.6.4(?): A full-rank matrix that is not diagonalizable: 1 1 00 1 10 0 1

.

Example 11.6.5(?): Matrix roots of unity, i.e. representations of i:

M1 :=[0 −11 0

]M2 :=

[0 1−1 0

].

11.6.1 Counting

11.6 Matrix Counterexamples 149

11 Linear Algebra

Proposition 11.6.6(Size of GLn(Fp)).

|GLn(Fp)| = (pn − 1)(pn − p)(pn − p2) · · · (pn − pn−1).

It suffices to count ordered bases of Fnp :

• Choose v1: there are p choices for each coefficient, but leave out the vector 0, so pn − 1choices.

• Choose any v2 6= λv1, so pn − p choices.• Choose any nonzero v3 6= λv1 + ηv2, so pn − p2 choices.• Etc.

E 11.7 Exercises e

Exercise 11.7.1 (?)Show that normal matrices are diagonalizable.

Exercise 11.7.2 (?)Consider the Vandermonde matrix:

A :=

1 · · · 1λ1 · · · λk...

...

λk−11 · · · λk−1

k

.

Show that

detA =∏i<j

(λi − λj).

Exercise 11.7.3 (?)Show that a nonzero nilpotent matrix A is not diagonalizable over any field. Some useful facts:

• SpecA = {0}, since Ax = λx =⇒ An = λnx, so An = 0 forces λ = 0. This forcesJCF(A) to be strictly upper-triangular.

• minA

(x) = xn.• If A were diagonalizable, JCF(A) = 0.

Exercise 11.7.4 (?)Prove Cayley-Hamilton in the following way. Let V = span {v1, · · · ,vn} and define the ithflag as FiliV := span {v1, · · · ,vi} for all 1 ≤ i ≤ n, and set Fil0V := {0}. Show that if if A is

11.7 Exercises 150

12 Jordan Canonical Form

upper triangular, then A(FiliV ) ⊆ FiliV . Now supposing vi are eigenvectors for λi, show that

(A− λnI)FilnV ⊆ Filn−1V

(A− λn−1I)(A− λnI)FilnV ⊆ Filn−2V

...∏i

(A− λn−iI)FilnV ⊆ Fil0V = {0} .

Conclude that χA(A) = 0.


Useful reference: https: // mattbaker. blog/ 2015/07/ 31/ the-jordan-canonical-form/

E 12.1 Facts e

Fact 12.1.1The JCF corresponds to elementary divisors.

Make more precise..

Proposition 12.1.2(JCF Algorithm for generalized eigenvectors).The following algorithm always works for computing JCF(A):

• Compute and factor the characteristic polynomial as χA(x) =∏i

(x− λi)mi .

• For each λi, find the constant ì such that

· · · rank(A− λiI)ì−1 > rank(A− λiI)ì= rank(A− λiI)ì+1= rank(A− λiI)ì+1= · · · .

• Find as many usual eigenvectors vi as you can. The number of eigenvectors you findwill be dimEλi . Suppose you just get one, v.

• Solve the systems:

(A− λiI)v1 = v =⇒ v1 =?(A− λiI)2v2 = v1 =⇒ v2 =?

...

,

Jordan Canonical Form 151

https://mattbaker.blog/2015/07/31/the-jordan-canonical-form/

https://mattbaker.blog/2015/07/31/the-jordan-canonical-form/


which can be solved by putting the vi in an augmented matrix and computing the RREF.• This terminates in at most ì steps, and these vectors correspond to a single Jordan

block.• If there are other eigenvectors w, · · · for λi, repeating this process yields a Jordan block

for each of them. Assemble P by placing these vi in the appropriate columns.

Lemma 12.1.3(JCF from Minimal and Characteristic Polynomials).Writing Spec(A) = {(λi,mi)},

minA

(x) =∏i

(x− λi)ì

χA(x) =∏

(x− λi)mi

Eλi = dim(A− λiI)

• The roots both polynomials are precisely the eigenvalues λi of A.

– mi are the algebraic multiplicities.– dimEλi are the geometric multiplicities.

• ì ≤ mi by Cayley-Hamilton.

• ì is

– The size of the largest Jordan block associated to λia, and– The “stabilizing constant”.

• mi, associated to the characteristic polynomial, is

– The sum of sizes of all Jordan blocks associated to λi,– The number of times λi appears on the diagonal of JCF (A),– The dimension of the generalized eigenspace V λi .

• dimEλi is

– The number of Jordan blocks associated to λi– The number of (usual) eigenvector associated to λi, i.e. the dimension of their span.

• A is diagonalizable iff dimEλi = mi for all i.aThis is because (x− λi)ì annihilates a Jordan block of size ì, along with any blocks of size k ≤ ì.

Example 12.1.4(?): Suppose A is 5× 5 with

minA

(t) = (t− 4)2(t+ 6)

χA(t) = (t− 4)3(t+ 6)2.

Some deductions:

• For λ = 4:

12.1 Facts 152


– The total size of all blocks is 3– The largest block is size 2– So this yields J1 ⊕ J2.

• For λ = −6:

– The total size of all blocks is 2– The largest block is size 1– So this must be J1 ⊕ J1

4! Warning 12.1.5The data of min

A(t), χA(t) is not enough to uniquely determine JCF(A). Counterexample: there

are two distinct possibilities for 4 × 4 matrices with minA

(t) = t2 and χA(t) = t4, namely J2 ⊕ J2

and J2 ⊕ J1 ⊕ J1.

Lemma 12.1.6(?).The elementary divisors of A are the minimal polynomials of the Jordan blocks.

Remark 12.1.7: Writing Ann(v) as the annihilator of v, a generalized eigenvector for the pair(λi,v) for a matrix A is any operator in the space

√Ann(v), where we view V as a k[x]-module

using p(x) y v := p(A)(v). So

Ann(v) :={q(x) ∈ k[x]

∣∣∣ q(x) y v = 0}

={q(x) ∈ k[x]

∣∣∣ q(A)(v) = 0}.

Now use that w is an eigenvector for A with eigenvalue λi ⇐⇒ A − λiI ∈ Ann(w), and is ageneralized eigenvector iff

(A− λiI)k ∈ Ann(w) for some k ⇐⇒ A− λiI ∈√

Ann(w).

We can then write

V λi :={

v ∈ V∣∣∣ (A− λiI)nv = 0 for some n

}={

v ∈ V∣∣∣ (A− λiI)n ∈ Ann(v)

}={

v ∈ V∣∣∣ A− λiI ∈ √Ann(v)

},

and the theorem is that V ∼=⊕i

V λi . It also turns out that V λi = ker(A− λiI)n for n := dimV .

Proof (of generalized eigenspace decomposition). • Suppose χA(x) =∏

(x− λi)ni .• Define V j := ker(A− λiI)n as the generalized eigenspace for each i.• Fix j and define hj(x) =

∏i 6=j

(x − λi)ni , the characteristic polynomial with the λj term

deleted.• Define W j := im(hj(A)), then the claim is W j ⊆ V j

12.1 Facts 153


– This follows because 0 = χA(A) = (A−λjI)njhj(A), so in fact W j ⊆ ker(A−λj)nj .

• Claim:∑

V j = V :

– Let v ∈ V be arbitrary, then by Euclid’s algorithm write∑i

fihi = 1 since the hi

are coprime.– Thus

∑fi(A)hi(A) = I =⇒

(∑fi(A)hi(A)

)(v) = v =⇒ v ∈

∑W j

• Claim: the sum is direct.

– It suffices to show 0 =∑

wi with wi ∈W i implies wi = 0 for all i.– Use that hj(wi) = 0 for i 6= j since wi ∈W i := ker(A− λII)ni .– Write wi =

∑fj(A)hj(A)wi, which collapses to fi(A)hi(A)wi.

– So fi(A)hi(A)(∑

wi)

= 0 =⇒ wi = 0.

�

Messy indexing.

E 12.2 Exercises e

Exercise 12.2.1 (?)Prove Cayley-Hamilton using the JCF.

Exercise 12.2.2 (?)Prove the rank-nullity theorem using JCF.

Exercise 12.2.3 (?)Compute JCF(A) for

A :=

1 −1 0−1 4 −1−4 13 −3

.Solution: • det(A) = 0

• tr(A) = 2• tr(

∧2A) = 1

• χA(t) = t3 − 2t2 + t• e1 = [1, 1, 3]• e2 = [1, 0,−1]

– e2,1 = [−3,−1, 0].

12.2 Exercises 154

13 Representation Theory

Exercise 12.2.4 (?)Determine JCF(B) for

B :=

5 −1 0 09 −1 0 00 0 7 −20 0 12 −3

.

13 Representation Theory

Theorem 13.0.1(Schur’s Lemma).If M ∈ G-Mod is an irreducible representation of G with dimkM <∞ and k = k, then thereis an isomorphism

M∼−→ Aut

G(M,M).

Representation Theory 155

14 Extra Problems

Theorem 13.0.2(Maschke’s Theorem).Let k be a field with ch(k) not dividing #G. Then any finite-dimensional representation of Gdecomposes into a direct sum of irreducible representations.

Definition 13.0.3 (Characters)The character of a representation M is the trace of the map

Tg : M →M

m 7→ g y m.

14 Extra Problems

E 14.1 Commutative Algebra e

• Show that a finitely generated module over a Noetherian local ring is flat iff it is free usingNakayama and Tor.

• Show that 〈2, x〉 E Z[x] is not a principal ideal.

• Let R be a Noetherian ring and A,B algebras over R. Suppose A is finite type over R andfinite over B. Then B is finite type over R.

E 14.2 Group Theory e

14.2.1 Centralizing and Normalizing

• Show that CG(H) ⊆ NG(H) ≤ G.

• Show that Z(G) ⊆ CG(H) ⊆ NG(H).

• Given H ⊆ G, let S(H) =⋃g∈G

gHg−1, so |S(H)| is the number of conjugates to H. Show

that |S(H)| = [G : NG(H)].

– That is, the number of subgroups conjugate to H equals the index of the normalizer ofH.

• Show that Z(G) =⋂a∈G

CG(a).

Extra Problems 156

14 Extra Problems

• Show that the centralizer GG(H) of a subgroup is again a subgroup.

• Show that CG(H) E NG(H) is a normal subgroup.

• Show that CG(G) = Z(G).

• Show that for H ≤ G, CH(x) = H ∩ CG(x).

• Let H,K ≤ G a finite group, and without using the normalizers of H or K, show that|HK| = |H||K|/|H ∩K|.

• Show that if H ≤ NG(K) then HK ≤ H, and give a counterexample showing that thiscondition is necessary.

• Show that HK is a subgroup of G iff HK = KH.

• Prove that the kernel of a homomorphism is a normal subgroup.

14.2.2 Primes in Group Theory

• Show that any group of prime order is cyclic and simple.

• Analyze groups of order pq with q < p.

Hint: consider the cases when p does or does notdivide q − 1.

– Show that if q does not divide p− 1, then G is cyclic.– Show that G is never simple.

• Analyze groups of order p2q.

Hint: Consider the cases when q does or does notdivide p2 − 1.

• Show that no group of order p2q2 is simple for p < q primes.

• Show that a group of order p2q2 has a normal Sylow subgroup.

• Show that a group of order p2q2 where q does not divide p2 − 1 and p does not divide q2 − 1is abelian.

• Show that every group of order pqr with p < q < r primes contains a normal Sylow subgroup.

– Show that G is never simple.

14.2 Group Theory 157

14 Extra Problems

• Let p be a prime and |G| = p3. Prove that G has a normal subgroup N of order p2.

– Suppose N = 〈h〉 is cyclic and classify all possibilities for G if:♦ |h| = p3

♦ |h| = p.Hint: Sylow and semidirect products.

• Show that any normal p- subgroup is contained in every Sylow p-subgroup of G.

• Show that the order of 1 + p in(Z/p2Z

)×is equal to p. Use this to construct a non-abelian

group of order p3.

14.2.3 p-Groups

• Show that every p-group has a nontrivial center.

• Show that every p-group is nilpotent.

• Show that every p-group is solvable.

• Show that every maximal subgroup of a p-group has index p.

• Show that every maximal subgroup of a p-group is normal.

• Show that every group of order p is cyclic.

• Show that every group of order p2 is abelian and classify them.

• Show that every normal subgroup of a p-group is contained in the center.

Hint: Consider G/Z(G).

• Let OP (G) be the intersection of all Sylow p-subgroups of G. Show that Op(G) E G, ismaximal among all normal p-subgroups of G

• Let P ∈ Sylp(H) where H E G and show that P ∩H ∈ Sylp(H).

• Show that Sylow pi-subgroups Sp1 , Sp2 for distinct primes p1 6= p2 intersect trivially.

• Show that in a p group, every normal subgroup intersects the center nontrivially.


14 Extra Problems

14.2.4 Symmetric Groups

Specific Groups

• Show that the center of S3 is trivial.• Show that Z(Sn) = 1 for n ≥ 3• Show that Aut(S3) = Inn(S3) ∼= S3.• Show that the transitive subgroups of S3 are S3, A3• Show that the transitive subgroups of S4 are S4, A4, D4,Z2

2,Z4.• Show that S4 has two normal subgroups: A4,Z2

2.• Show that Sn≥5 has one normal subgroup: An.• Z(An) = 1 for n ≥ 4• Show that [Sn, Sn] = An• Show that [A4, A4] ∼= Z2

2• Show that [An, An] = An for n ≥ 5, so An≥5 is nonabelian.

General Structure

• Show that an m-cycle is an odd permutation iff m is an even number.• Show that a permutation is odd iff it has an odd number of even cycles.• Show that the center of Sn for n ≥ 4 is nontrivial.• Show that disjoint cycles commute.• Show directly that any k-cycle is a product of transpositions, and determine how many

transpositions are needed.

Generating Sets

• Show that Sn is generated by any of the following types of cycles:


14 Extra Problems

– Show that Sn is generated by transpositions.– Show that Sn is generated by adjacent transpositions.– Show that Sn is generated by {(12), (12 · · ·n)} for n ≥ 2– Show that Sn is generated by {(12), (23 · · ·n)} for n ≥ 3– Show that Sn is generated by {(ab), (12 · · ·n)} where 1 ≤ a < b ≤ n iff gcd(b− a, n) = 1.– Show that Sp is generated by any arbitrary transposition and any arbitrary p-cycle.

14.2.5 Alternating Groups

• Show that An is generated 3-cycles.• Prove that An is normal in Sn.• Argue that An is simple for n ≥ 5.• Show that Out(A4) is nontrivial.

14.2.6 Dihedral Groups

• Show that if N E Dn is a normal subgroup of a dihedral group, then Dn/N is again a dihedralgroup.

14.2.7 Other Groups

• Show that Q is not finitely generated as a group.• Show that the Quaternion group has only one element of order 2, namely −1.


14 Extra Problems

14.2.8 Classification

• Show that no group of order 36 is simple.• Show that no group of order 90 is simple.• Classifying all groups of order 99.• Show that all groups of order 45 are abelian.• Classify all groups of order 10.• Classify the five groups of order 12.• Classify the four groups of order 28.• Show that if |G| = 12 and has a normal subgroup of order 4, then G ∼= A4.• Suppose |G| = 240 = s4 · 3 · 5.

– How many Sylow-p subgroups does G have for p ∈ {2, 3, 5}?– Show that if G has a subgroup of order 15, it has an element of order 15.– Show that if G does not have such a subgroup, the number of Sylow-3 subgroups is either

10 or 40.Hint: Sylow on the subgroup of order 15 and semidi-rect products.

14.2.9 Group Actions

• Show that the stabilizer of an element Gx is a subgroup of G.• Show that if x, y are in the same orbit, then their stabilizers are conjugate.• Show that the stabilizer of an element need not be a normal subgroup?• Show that if Gy X is a group action, then the stabilizer Gx of a point is a subgroup.

14.2.10 Series of Groups

• Show that An is simple for n ≥ 5

• Give a necessary and sufficient condition for a cyclic group to be solvable.

• Prove that every simple abelian group is cyclic.

• Show that Sn is generated by disjoint cycles.

• Show that Sn is generated by transpositions.

• Show if G is finite, then G is solvable ⇐⇒ all of its composition factors are of prime order.

• Show that if N and G/N are solvable, then G is solvable.

• Show that if G is finite and solvable then every composition factor has prime order.


14 Extra Problems

• Show that G is solvable iff its derived series terminates.

• Show that S3 is not nilpotent.

• Show that G nilpotent =⇒ G solvable

• Show that nilpotent groups have nontrivial centers.

• Show that Abelian =⇒ nilpotent

• Show that p-groups =⇒ nilpotent

14.2.11 Misc

• Prove Burnside’s theorem.

• Show that Inn(G) E Aut(G)

• Show that Inn(G) ∼= G/Z(G)

• Show that the kernel of the map G→ Aut(G) given by g 7→ (h 7→ ghg−1) is Z(G).

• Show that NG(H)/CG(H) ∼= A ≤ Aut(H)

• Give an example showing that normality is not transitive: i.e. H E K E G with H not normalin G.

14.2.12 Nonstandard Topics

• Show that H char G⇒ H EG

Thus “characteristic” is a strictly stronger conditionthan normality

• Show that H char K char G⇒ H char G

So “characteristic” is a transitive relation for sub-groups.

• Show that if H ≤ G, K E G is a normal subgroup, and H char K then H is normal in G.

So normality is not transitive, but strengthening oneto “characteristic” gives a weak form of transitivity.


14 Extra Problems

E 14.3 Ring Theory e

14.3.1 Basic Structure

• Show that if an ideal I E R contains a unit then I = R.• Show that R× need not be closed under addition.

14.3.2 Ideals

• ? Show that if x is not a unit, then x is contained in some maximal ideal.

Problem 14.3.1 (Units or Zero Divisors)Every a ∈ R for a finite ring is either a unit or a zero divisor.

Solution:

• Let a ∈ R and define ϕ(x) = ax.• If ϕ is injective, then it is surjective, so 1 = ax for some x =⇒ x−1 = a.• Otherwise, ax1 = ax2 with x1 6= x2 =⇒ a(x1 − x2) = 0 and x1 − x2 6= 0• So a is a zero divisor.

Problem 14.3.2 (Maximal implies prime)Maximal =⇒ prime, but generally not the converse.

Solution: • Suppose m is maximal, ab ∈ m, and b 6∈ m.

• Then there is a containment of ideals m ( m + (b) =⇒ m + (b) = R.

• So

1 = m+ rb =⇒ a = am+ r(ab),

but am ∈ m and ab ∈ m =⇒ a ∈ m.Counterexample: (0) ∈ Z is prime since Z is a domain, but not maximal since it is properlycontained in any other ideal.

• Show that every proper ideal is contained in a maximal ideal• Show that if x ∈ R a PID, then x is irreducible ⇐⇒ 〈x〉 E R is maximal.• Show that intersections, products, and sums of ideals are ideals.• Show that the union of two ideals need not be an ideal.• Show that every ring has a proper maximal ideal.• Show that I E R is maximal iff R/I is a field.• Show that I E R is prime iff R/I is an integral domain.

14.3 Ring Theory 163

14 Extra Problems

• Show that ∪m∈maxSpec(R) = R \R×.• Show that maxSpec(R) ( Spec(R) but the containment is strict.• Show that every prime ideal is radical.• Show that the nilradical is given by

√0R =

√(0).

• Show that rad(IJ) = rad(I) ∩ rad(J)• Show that if Spec(R) ⊆ maxSpec(R) then R is a UFD.• Show that if R is Noetherian then every ideal is finitely generated.

14.3.3 Characterizing Certain Ideals

• Show that for an ideal I E R, its radical is the intersection of all prime ideals containing I.• Show that

√I is the intersection of all prime ideals containing I.

Problem 14.3.3 (Jacobson radical is bigger than the nilradical)The nilradical is contained in the Jacobson radical, i.e.

√0R ⊆ J(R).

Solution:Maximal =⇒ prime, and so if x is in every prime ideal, it is necessarily in every maximalideal as well.

Problem 14.3.4 (Mod by nilradical to kill nilpotents)R/√

0R has no nonzero nilpotent elements.

Solution:

a+√

0R nilpotent =⇒ (a+√

0R)n := an +√

0R =√

0R=⇒ an ∈

√0R

=⇒ ∃` such that (an)` = 0=⇒ a ∈

√0R.

Problem 14.3.5 (Nilradical is intersection of primes)The nilradical is the intersection of all prime ideals, i.e.

√0R =

⋂p∈Spec(R)

p

Solution:

•√

0R ⊆ ∩p:

• x ∈√

0R =⇒ xn = 0 ∈ p =⇒ x ∈ p or xn−1 ∈ p.

14.3 Ring Theory 164

14 Extra Problems

• R \√

0R ⊆ ∪p(R \ p):

• Define S ={I E R

∣∣∣ an 6∈ I for any n}.

• Then apply Zorn’s lemma to get a maximal ideal m, and maximal =⇒ prime.

14.3.4 Misc

• Show that localizing a ring at a prime ideal produces a local ring.• Show that R is a local ring iff for every x ∈ R, either x or 1− x is a unit.• Show that if R is a local ring then R \R× is a proper ideal that is contained in the Jacobson

radical J(R).• Show that if R 6= 0 is a ring in which every non-unit is nilpotent then R is local.• Show that every prime ideal is primary.• Show that every prime ideal is irreducible.

E 14.4 Field Theory e

General Algebra

• Show that any finite integral domain is a field.• Show that every field is simple.• Show that any field morphism is either 0 or injective.• Show that if L/F and α is algebraic over both F and L, then the minimal polynomial of α

over L divides the minimal polynomial over F .• Prove that if R is an integral domain, then R[t] is again an integral domain.• Show that ff(R[t]) = ff(R)(t).• Show that [Q(

√2 +√

3) : Q] = 4.

– Show that Q(√

2 +√

3) = Q(√

2−√

3) = Q(√

2,√

3).

• Show that the splitting field of f(x) = x3 − 2 is Q( 3√2, ζ2).

Extensions?

• What is [Q(√

2 +√

3) : Q]?• What is [Q(2

32 ) : Q]?

• Show that if p ∈ Q[x] and r ∈ Q is a rational root, then in fact r ∈ Z.• If {αi}ni=1 ⊂ F are algebraic over K, show that K[α1, · · · , αn] = K(α1, · · · , αn).• Show that α/F is algebraic ⇐⇒ F (α)/F is a finite extension.• Show that every finite field extension is algebraic.• Show that if α, β are algebraic over F , then α± β, αβ±1 are all algebraic over F .• Show that if L/K/F with K/F algebraic and L/K algebraic then L is algebraic.

14.4 Field Theory 165

14 Extra Problems

Special Polynomials

• Show that a field with pn elements has exactly one subfield of size pd for every d dividing n.• Show that xpn − x =

∏fi(x) over all irreducible monic fi of degree d dividing n.

• Show that xpd − x∣∣ xpn − x ⇐⇒ d

∣∣ n• Prove that xpn − x is the product of all monic irreducible polynomials in Fp[x] with degree

dividing n.• Prove that an irreducible π(x) ∈ Fp[x] divides xpn − x ⇐⇒ deg π(x) divides n.

E 14.5 Galois Theory e

14.5.1 Theory

• Show that if K/F is the splitting field of a separable polynomial then it is Galois.• Show that any quadratic extension of a field F with ch(F ) 6= 2 is Galois.• Show that if K/E/F with K/F Galois then K/E is always Galois with g(K/E) ≤ g(K/F ).

– Show additionally E/F is Galois ⇐⇒ g(K/E) E g(K/F ).– Show that in this case, g(E/F ) = g(K/F )/g(K/E).

• Show that if E/k, F/k are Galois with E ∩ F = k, then EF/k is Galois and G(EF/k) ∼=G(E/k)×G(F/k).

14.5.2 Computations

• Show that the Galois group of xn − 2 is Dn, the dihedral group on n vertices.• Compute all intermediate field extensions of Q(

√2,√

3), show it is equal to Q(√

2 +√

3), andfind a corresponding minimal polynomial.

• Compute all intermediate field extensions of Q(214 , ζ8).

14.5 Galois Theory 166

15 Even More Algebra Questions

• Show that Q(213 ) and Q(ζ32

13 )

• Show that if L/K is separable, then L is normal ⇐⇒ there exists a polynomial p(x) =n∏i=1

x− αi ∈ K[x] such that L = K(α1, · · · , αn) (so L is the splitting field of p).

• Is Q(213 )/Q normal?

• Show that GF(pn) is the splitting field of xpn − x ∈ Fp[x].• Show that GF(pd) ≤ GF(pn) ⇐⇒ d

∣∣ n• Compute the Galois group of xn − 1 ∈ Q[x] as a function of n.• Identify all of the elements of the Galois group of xp − 2 for p an odd prime (note: this has a

complicated presentation).• Show that Gal(x15 + 2)/Q ∼= S2 o Z/15Z for S2 a Sylow 2-subgroup.• Show that Gal(x3 + 4x+ 2)/Q ∼= S3, a symmetric group.

E 14.6 Modules and Linear Algebra e

• Prove the Cayley-Hamilton theorem.• Prove that the minimal polynomial divides the characteristic polynomial.• Prove that the cokernel of A ∈ Mat(n× n,Z) is finite ⇐⇒ detA 6= 0, and show that in this

case |coker(A)| = |det(A)|.• Show that a nilpotent operator is diagonalizable.• Show that if A,B are diagonalizable and [A,B] = 0 then A,B are simultaneously diagonaliz-

able.• Does diagonalizable imply invertible? The converse?• Does diagonalizable imply distinct eigenvalues?• Show that if a matrix is diagonalizable, its minimal polynomial is squarefree.• Show that a matrix representing a linear map T : V → V is diagonalizable iff V is a direct

sum of eigenspaces V =⊕i

ker(T − λiI).

• Show that if {vi} is a basis for V where dim(V ) = n and T (vi) = vi+1 modn then T isdiagonalizable with minimal polynomial xn − 1.

• Show that if the minimal polynomial of a linear map T is irreducible, then every T -invariantsubspace has a T -invariant complement.

E 14.7 Linear Algebra e

Sort out from module section.


14.6 Modules and Linear Algebra 167


Remark 15.0.1: (DZG): These all come from a random PDF I found, but I couldn’t find theoriginal author/source!

E 15.1 Groups e

15.1.1 Question 1.1

What is a normal subgroup? Can you get some natural map from a normal subgroup? Whattopological objects can the original group, normal subgroup, and quotient group relate to?

15.1.2 Question 1.2

Prove that a subgroup of index two is normal.

15.1.3 Question 1.3

Find all normal subgroups of A4.

15.1.4 Question 1.4

Give an interesting example of a non-normal subgroup. Is SO(2) normal inside SL2(R)?

15.1.5 Question 1.5

Is normality transitive? That is, is a normal subgroup of a normal subgroup normal in the biggestgroup?

15.1.6 Question 1.6.

Define a solvable group. Give an example of a solvable nonabelian group.

Show A4 is solvable. Do the Sylow theorems tell you anything about whether this index 3 subgroupof A4 is normal?

15.1 Groups 168


15.1.7 Question 1.7

Define lower central series, upper central series, nilpotent and solvable groups.

15.1.8 Question 1.8

Define the derived series. Define the commutator. State and prove two nontrivial theorems aboutderived series.

15.1.9 Question 1.9

Prove that SL2(Z) is not solvable.

15.1.10 Question 1.10

What are all possible orders of elements of SL2(Z)?

15.1.11 Question 1.11

Can you show that all groups of order pn for p prime are solvable? Do you know how to do this forgroups of order prqs?

15.1.12 Question 1.12

Suppose a p-group acts on a set whose cardinality is not divisible by p (p prime). Prove that thereis a fixed point for the action.

15.1.13 Question 1.13

Prove that the centre of a group of order pr (p prime) is not trivial.

15.1.14 Question 1.14

Give examples of simple groups. Are there infinitely many?

15.1 Groups 169


15.1.15 Question 1.15

State and prove the Jordan-Holder theorem for finite groups.

15.1.16 Question 1.16

What’s Cayley’s theorem? Give an example of a group of order n that embeds in Sm for some msmaller than n.

Give an example of a group where you have to use Sn.

15.1.17 Question 1.17

Is A4 a simple group? What are the conjugacy classes in S4? What about in A4?

15.1.18 Question 1.18

Talk about conjugacy classes in the symmetric group Sn.

15.1.19 Question 1.19

When do conjugacy classes in Sn split in An?

15.1.20 Question 1.20

What is the centre of Sn? Prove it.

15.1.21 Question 1.21

Prove that the alternating group An is simple for n ≥ 5.

15.1.22 Question 1.22

Prove the alternating group on n letters is generated by the 3-cycles for n ≥ 3.

15.1 Groups 170


15.1.23 Question 1.23

Prove that for p prime, Sp is generated by a p-cycle and a transposition.

15.1.24 Question 1.24

What is the symmetry group of a tetrahedron? Cube? Icosahedron?

15.1.25 Question 1.25

How many ways can you color the tetrahedron with C colors if we identify symmetric colorings?

15.1.26 Question 1.26.

What is the symmetry group of an icosahedron? What’s the stabiliser of an edge?

How many edges are there? How do you know the symmetry group of the icosahedron is the sameas the symmetry group of the dodecahedron?

Do you know the classification of higher-dimensional polyhedra?

15.1.27 Question 1.27

Do you know what the quaternion group is? How many elements are there of each order?

15.1.28 Question 1.28

What is the group of unit quaternions topologically? What does it have to do with SO(3)?

15.1.29 Question 1.29

What’s the stabiliser of a point in the unit disk under the group of conformal automorphisms?

15.1.30 Question 1.30

What group-theoretic construct relates the stabiliser of two points?

15.1 Groups 171


15.1.31 Question 1.31

Consider SL2(R) acting on R2 by matrix multiplication. What is the stabiliser of a point? Does itdepend which point? Do you know what sort of subgroup this is? What if SL2(R) acts by Möbiustransformations instead?

15.1.32 Question 1.32

What are the polynomials in two real variables that are invariant under the action of D4, thesymmetry group of a square, by rotations and reflections on the plane that the two variablesform?

15.1.33 Question 1.33

Give an interesting example of a subgroup of the additive group of the rationals.

15.1.34 Question 1.34

Talk about the isomorphism classes of subgroups of Q. How many are there? Are the ones you’vegiven involving denominators divisible only by certain primes distinct? So that gives you thecardinality. Are these all of them?

15.1.35 Question 1.35

Is the additive group of the reals isomorphic to the multiplicative group of the positive reals? Isthe same result true with reals replaced by rationals?

15.1.36 Question 1.36

What groups have nontrivial automorphisms?

15.1.37 Question 1.37

A subgroup H of a group G that meets every conjugacy class is in fact G. Why is that true?

15.1 Groups 172


15.1.38 Question 1.38

Let G be the group of invertible 3× 3 matrices over Fp, for p prime. What does basic group theorytell us about G?

How many conjugates does a Sylow p-subgroup have? Give a matrix form for the elements in thissubgroup.

Explain the conjugacy in terms of eigenvalues and eigenvectors. give a matrix form for the normaliserof the Sylow p-subgroup.

15.1.39 Question 1.39

Let’s look at SL2(F3). How many elements are in that group? What is its centre? Identify PSL2(F3)as a permutation group.

15.1.40 Question 1.40

How many elements does gl2(Fq) have? How would you construct representations?

What can you say about the 1-dimensional representations? What can you say about simplicity ofsome related groups?

15.1.41 Question 1.41.

A subgroup of a finitely-generated free abelian group is?

A subgroup of a finitely-generated free group is..? Prove your answers.

15.1.42 Question 1.42

What are the subgroups of Z2?

15.1.43 Question 1.43

What are the subgroups of the free group F2? How many generators can you have?

Can you find one with 3 generators? 4 generators? Countably many generators?

15.1 Groups 173


Is the subgroup with 4 generators you found normal? Why? Can you find a normal one?

15.1.44 Question 1.44

Talk about the possible subgroups of Z3. Now suppose that you have a subgroup of Z3. Whattheorem tells you something about the structure of the quotient group?

E 15.2 Classification of Finite groups e

15.2.1 Question 2.1

Given a finite abelian group with at most n elements of order divisible by n, prove it’s cyclic.

15.2.2 Question 2.2

Suppose I asked you to classify groups of order 4. Why isn’t there anything else? Which of thosecould be realised as a Galois group over Q?

15.2.3 Question 2.3

State/prove the Sylow theorems.

15.2.4 Question 2.4

Classify groups of order 35.

15.2.5 Question 2.5

Classify groups of order 21.

15.2.6 Question 2.6

Discuss groups of order 55.

15.2 Classification of Finite groups 174


15.2.7 Question 2.7

Classify groups of order 14. Why is there a group of order 7? Are all index-2 subgroups normal?

15.2.8 Question 2.8

How many groups are there of order 15? Prove it.

15.2.9 Question 2.9

Classify all groups of order 8.

15.2.10 Question 2.10

Classify all groups of order p3 for p prime.

15.2.11 Question 2.11

What are the groups of order p2? What about pq? What if q is congruent to 1 mod p?

15.2.12 Question 2.12

What are the groups of order 12? Can there be a group of order 12 with 2 nonisomorphic subgroupsof the same order?

15.2.13 Question 2.13

How would you start finding the groups of order 56? Is there in fact a way for Z/7Z to act on agroup of order 8 nontrivially?

15.2.14 Question 2.14

How many abelian groups are there of order 36?

15.2 Classification of Finite groups 175


15.2.15 Question 2.15

What are the abelian groups of order 16?

15.2.16 Question 2.16.

What are the abelian groups of order 9? Prove that they are not isomorphic. groups of order 27?

15.2.17 Question 2.17

How many abelian groups of order 200 are there?

15.2.18 Question 2.18

Prove there is no simple group of order 132.

15.2.19 Question 2.19

Prove that there is no simple group of order 160. What can you say about the structure of groupsof that order?

15.2.20 Question 2.20

Prove that there is no simple group of order 40.

E 15.3 Fields and Galois Theory e

15.3.1 Question 3.1

What is the Galois group of a finite field? What is a generator? How many elements does a finitefield have? What can you say about the multiplicative group? Prove it.

15.3 Fields and Galois Theory 176


15.3.2 Question 3.2

Classify finite fields, their subfields, and their field extensions. What are the automorphisms of afinite field?

15.3.3 Question 3.3

Take a finite field extension Fnp over Fp. What is Frobenius? What is its characteristic polynomial?

15.3.4 Question 3.4

What are the characteristic and minimal polynomial of the Frobenius automorphism?

15.3.5 Question 3.5

What’s the field with 25 elements?

15.3.6 Question 3.6

What is the multiplicative group of F9?

15.3.7 Question 3.7

What is a separable extension? Can Q have a non-separable extension? How about Z/pZ? Whynot? Are all extensions of characteristic 0 fields separable? Of finite fields? Prove it.

Give an example of a field extension that’s not separable.

15.3.8 Question 3.8

Are there separable polynomials of any degree over any field?

15.3.9 Question 3.9

What is a perfect field and why is this important? Give an example of a non-perfect field.



15.3.10 Question 3.10

What is Galois theory? State the main theorem. What is the splitting field of x5− 2 over Q? Whatare the intermediate extensions? Which extensions are normal, which are not, and why? What arethe Galois groups (over Q) of all intermediate extensions?

15.3.11 Question 3.11

What is a Galois extension?

15.3.12 Question 3.12

Take a quadratic extension of a field of characteristic 0. Is it Galois? Take a degree 2 extension ontop of that. Does it have to be Galois over the base field? What statement in group theory can youthink of that reflects this?

15.3.13 Question 3.13.

Is Abelian Galois extension transitive? That is, if K has abelian Galois group over E, E has abelianGalois group over F , and K is a Galois extension of F , is it necessarily true that Gal(K/F ) is alsoabelian? Give a counterexample involving number fields as well as one involving function fields.

15.3.14 Question 3.14

What is a Kummer extension?

15.3.15 Question 3.15

Say you have a field extension with only finitely many intermediate fields. Show that it is a simpleextension.

15.3.16 Question 3.16

Tell me a condition on the Galois group which is implied by irreducibility of the polynomial. Whathappens when the polynomial has a root in the base field?



15.3.17 Question 3.17

What is the discriminant of a polynomial?

15.3.18 Question 3.18

If we think of the Galois group of a polynomial as contained in Sn, when is it contained in An?

15.3.19 Question 3.19

Is Q( 3√21) normal? What is its splitting field? What is its Galois group? Draw the lattice ofsubfields.

15.3.20 Question 3.20

What’s the Galois group of x2 + 1 over Q? What’s the integral closure of Z in Q(i)?

15.3.21 Question 3.21

What’s the Galois group of x2 + 9?

15.3.22 Question 3.22

What is the Galois group of x2 − 2? Why is x2 − 2 irreducible?

15.3.23 Question 3.23

What is the Galois group of

Q(√

2,√

3) /Q?

15.3.24 Question 3.24

What is the Galois group of

Q (√n1, · · · ,

√nm) /Q(

√n1 + · · ·+

√nm)?



15.3.25 Question 3.25

What are the Galois groups of irreducible cubics?

15.3.26 Question 3.26

If an irreducible cubic polynomial has Galois group NOT contained in A3, does it necessarily haveto be all of S3?

15.3.27 Question 3.27

Compute the Galois group of x3 − 2 over the rationals.

15.3.28 Question 3.28

How would you find the Galois group of x3 + 2x+ 1? Adjoin a root to Q. Can you say somethingabout the roots of x3 + 3x+ 1 in this extension?

15.3.29 Question 3.29

Compute the Galois group of x3 + 6x+ 3.

15.3.30 Question 3.30

Find the Galois group of x4 − 2 over Q.

15.3.31 Question 3.31

What’s the Galois group of x4 − 3?

15.3.32 Question 3.32

What is the Galois group of x4 − 2x2 + 9?



15.3.33 Question 3.33

Calculate the Galois group of x5 − 2.

15.3.34 Question 3.34.

Discuss sufficient conditions on a polynomial of degree 5 to have Galois group S5 over Q and proveyour statements.

15.3.35 Question 3.35

Show that if f is an irreducible quintic with precisely two non-real roots, then its Galois group isS5.

15.3.36 Question 3.36

Suppose you have a degree 5 polynomial over a field. What are necessary and sufficient conditionsfor its Galois group to be of order divisible by 3? Can you give an example of an irreduciblepolynomial in which this is not the case?

15.3.37 Question 3.37

What is the Galois group of x7 − 1 over the rationals?

15.3.38 Question 3.38

What is the Galois group of the polynomial xn − 1 over Q?

15.3.39 Question 3.39

Describe the Galois theory of cyclotomic extensions.

15.3.40 Question 3.40

What is the maximal real field in a cyclotomic extension Q(ζn)/Q?



15.3.41 Question 3.41

Compute the Galois group of p(x) = x7 − 3.

15.3.42 Question 3.42

What Galois stuff can you say about x2n − 2?

15.3.43 Question 3.43

What are the cyclic extensions of (prime) order p?

15.3.44 Question 3.44

Can you give me a polynomial whose Galois group is Z/3Z?

15.3.45 Question 3.45

Which groups of order 4 can be realised as a Galois group over Q?

15.3.46 Question 3.46

Give a polynomial with S3 as its Galois group.

15.3.47 Question 3.47

Give an example of a cubic with Galois group S3.

15.3.48 Question 3.48

How do you construct a polynomial over Q whose Galois group is Sn? Do it for n = 7 in particular.



15.3.49 Question 3.49

What’s a Galois group that’s not Sn or An?

15.3.50 Question 3.50

Which finite groups are Galois groups for some field extension?

15.3.51 Question 3.51

What Galois group would you expect a cubic to have?

15.3.52 Question 3.52

Draw the subgroup lattice for S3.

15.3.53 Question 3.53

Do you know what the quaternion group is? How many elements are there of each order? Suppose Ihave a field extension of the rationals with Galois group the quaternion group. How many quadraticextensions does it contain? Can any of them be imaginary?

15.3.54 Question 3.54

Suppose you are given a finite Galois extension K/Q by f(x) ∈ Z[x] such that deg(f) = n andGal(K/Q) = Sn. What can you say about the roots?

15.3.55 Question 3.55

How many automorphisms does the complex field have? How can you extend a simple automorphism√2 7→ −

√2 of an algebraic field into C? How can you extend a subfield automorphism? What

feature of C allows you to?

15.3.56 Question 3.56.

Can it happen that a proper subfield of C is isomorphic to C? How?



15.3.57 Question 3.57

Consider the minimal polynomial f(x) for a primitive mth root of unity. Prove that if p dividesf(a) for some integer a and gcd(p,m) = 1 then m divides p− 1. Use this fact to show that thereare infinitely many primes congruent to 1 modm.

15.3.58 Question 3.58

What is Dirichlet’s theorem about primes in arithmetic progression? What can you say about thedensity of such primes?

15.3.59 Question 3.59

How many irreducible polynomials of degree six are there over F2?

15.3.60 Question 3.60

Can you have a degree 7 irreducible polynomial over Fp? How about a degree 14 irreduciblepolynomial?

15.3.61 Question 3.61

How many irreducible polynomials are there of degree 4 over F2?

15.3.62 Question 3.62

For each prime p, give a polynomial of degree p that is irreducible over Fp. You can do it in a“uniform” way.

15.3.63 Question 3.63

Can we solve general quadratic equations by radicals? And what about cubics and so on? Whycan’t you solve 5th degree equations by radicals?



15.3.64 Question 3.64

Talk about solvability by radicals. Why is S5 not solvable? Why is A5 simple?

15.3.65 Question 3.65

For which n can a regular n-gon be constructed by ruler and compass?

15.3.66 Question 3.66

How do you use Galois theory (or just field theory) to prove the impossibility of trisecting an angle?Doubling a cube? Squaring a circle?

15.3.67 Question 3.67

Which numbers are constructible? Give an example of a non-constructible number whose degree isnevertheless a power of 2.

15.3.68 Question 3.68

State and prove Eisenstein’s Criterion.

15.3.69 Question 3.69

Why is (xp − 1)/(x− 1) irreducible over Q?

15.3.70 Question 3.70

Can you prove the fundamental theorem of algebra using Galois theory? What do you need fromanalysis to do so?

15.3.71 Question 3.71

What are the symmetric polynomials?



15.3.72 Question 3.72

State the fundamental theorem of symmetric polynomials.

15.3.73 Question 3.73

Is the discriminant of a polynomial always a polynomial in the coefficients? What does this haveto do with symmetric polynomials?

15.3.74 Question 3.74

Find a non-symmetric polynomial whose square is symmetric.

15.3.75 Question 3.75

Let f be a degree 4 polynomial with integer coefficients. What’s the smallest finite field in which fnecessarily has four roots?

15.3.76 Question 3.76

Define p-adic numbers. What is a valuation?

15.3.77 Question 3.77

What’s Hilbert’s theorem 90?

15.3.78 Question 3.78

Consider a nonconstant function between two compact Riemann Surfaces. How is it related toGalois theory?



E 15.4 Normal Forms e

15.4.1 Question 4.1

What is the connection between the structure theorem for modules over a PID and conjugacy classesin the general linear group over a field?

15.4.2 Question 4.2

Explain how the structure theorem for finitely-generated modules over a PID applies to a linearoperator on a finite dimensional vector space.

15.4.3 Question 4.3

I give you two matrices over a field. How would you tell if they are conjugate or not? What theoremare you using? State it. How does it apply to this situation? Why is k[x] a PID? If two matricesare conjugate over the algebraic closure of a field, does that mean that they are conjugate over thebase field too?

15.4.4 Question 4.4

If two real matrices are conjugate in Mat(n×n,C), are they necessarily conjugate in Mat(n×N,R)as well?

15.4.5 Question 4.5

Give the 4× 4 Jordan forms with minimal polynomial (x− 1)(x− 2)2.

15.4.6 Question 4.6

Talk about Jordan canonical form. What happens when the field is not algebraically closed?

15.4.7 Question 4.7

What are all the matrices that commute with a given Jordan block?

15.4 Normal Forms 187


15.4.8 Question 4.8

How do you determine the number and sizes of the blocks for Jordan canonical form?

15.4.9 Question 4.9

For any matrix A over the complex numbers, can you solve B2 = A?

15.4.10 Question 4.10

What is rational canonical form?

15.4.11 Question 4.11

Describe all the conjugacy classes of 3× 3 matrices with rational entries which satisfy the equationA4 −A3 −A+ 1 = 0. Give a representative in each class.

15.4.12 Question 4.12

What 3 × 3 matrices over the rationals (up to similarity) satisfy f(A) = 0, where f(x) = (x2 +2)(x− 1)3? List all possible rational forms.

15.4.13 Question 4.13

What can you say about matrices that satisfy a given polynomial (over an algebraically closedfield)? How many of them are there? What about over a finite field? How many such matrices arethere then?

15.4.14 Question 4.14

What is a nilpotent matrix?

15.4.15 Question 4.15

When do the powers of a matrix tend to zero?

15.4 Normal Forms 188


15.4.16 Question 4.16

If the traces of all powers of a matrix A are 0, what can you say about A?

15.4.17 Question 4.17

When and how can we solve the matrix equation exp(A) = B? Do it over the complex numbersand over the real numbers. give a counterexample with real entries.

15.4.18 Question 4.18

Say we can find a matrix A such that exp(A) = B for B in SLn(R). Does A also have to be inSLn(R)? Does A need to be in SLn(R)?

15.4.19 Question 4.19

Is a square matrix always similar to its transpose?

15.4.20 Question 4.20

What are the conjugacy classes of SL2(R)?

15.4.21 Question 4.21

What are the conjugacy classes in GL2(C)?

E 15.5 Matrices and Linear Algebra e

15.5.1 Question 5.1

What is a bilinear form on a vector space? When are two forms equivalent? What is an orthogonalmatrix? What’s special about them?

15.5 Matrices and Linear Algebra 189


15.5.2 Question 5.2

What are the possible images of the unit circle under a linear transformation of R2?

15.5.3 Question 5.3

Explain geometrically how you diagonalise a quadratic form.

15.5.4 Question 5.4

Do you know Witt’s theorem on real quadratic forms?

15.5.5 Question 5.5

Classify real division algebras.

15.5.6 Question 5.6

Consider the simple operator on C given by multiplication by a complex number. It decomposesinto a stretch and a rotation. What is the generalisation of this to operators on a Hilbert space?

15.5.7 Question 5.7

Do you know about singular value decomposition?

15.5.8 Question 5.8

What are the eigenvalues of a symmetric matrix?

15.5.9 Question 5.9

What can you say about the eigenvalues of a skew-symmetric matrix?



15.5.10 Question 5.10

Prove that the eigenvalues of a Hermitian matrix are real and those of a unitary matrix are unitary.

15.5.11 Question 5.11

Prove that symmetric matrices have real eigenvalues and can be diagonalised by orthogonal matri-ces.

15.5.12 Question 5.12

To which operators does the spectral theorem for symmetric matrices generalise?

15.5.13 Question 5.13

Given a skew-symmetric/skew-Hermitian matrix S, show that U = (S + I)(S − I)− 1 is orthogo-nal/unitary. Then find an expression for S in terms of U .

15.5.14 Question 5.14

If a linear transformation preserves a nondegenerate alternating form and has k as an eigenvalue,prove that 1/k is also an eigenvalue.

15.5.15 Question 5.15

State/prove the Cayley–Hamilton theorem.

15.5.16 Question 5.16

Are diagonalisable N × N matrices over the complex numbers dense in the space of all N × Nmatrices over the complex numbers? How about over another algebraically closed field if we usethe Zariski topology?

15.5.17 Question 5.17

For a linear ODE with constant coefficients, how would you solve it using linear algebra?



15.5.18 Question 5.18

What can you say about the eigenspaces of two matrices that commute with each other?

15.5.19 Question 5.19

What is a Toeplitz operator?

15.5.20 Question 5.20

What is the number of invertible matrices over Z/pZ?

E 15.6 Rings e

15.6.1 Question 6.1

State the Chinese remainder theorem in any form you like. Prove it.

15.6.2 Question 6.2

What is a PID? What’s an example of a UFD that is not a PID? Why? Is k[x] a PID? Why?

15.6.3 Question 6.3

Is C[x, y] a PID? Is 〈x, y〉 a prime ideals in it?

15.6.4 Question 6.4

Do polynomials in several variables form a PID?

15.6.5 Question 6.5

Prove that the integers form a PID.

15.6 Rings 192


15.6.6 Question 6.6

Give an example of a PID with a unique prime ideal.

15.6.7 Question 6.7

What is the relation between Euclidean domains and PIDs?

15.6.8 Question 6.8

Do you know a PID that’s not Euclidean?

15.6.9 Question 6.9

Give an example of a UFD which is not a Euclidean domain.

15.6.10 Question 6.10

Is a ring of formal power series a UFD?

15.6.11 Question 6.11

Is a polynomial ring over a UFD again a UFD?

15.6.12 Question 6.12

What does factorisation over Q[x] say about factorisation over Z[x]?

15.6.13 Question 6.13

Give an example of a ring where unique factorisation fails.

15.6 Rings 193


15.6.14 Question 6.14

Factor 6 in two different ways in Z[√−5] Is there any way to explain the two factorisations? Factor

the ideal generated by 6 into prime ideals.

15.6.15 Question 6.15

What’s the integral closure of Z in Q(i)?

15.6.16 Question 6.16

Find all primes in the ring of Gaussian integers.

15.6.17 Question 6.17

What is a ring of integers? What does “integral over Z” mean?

15.6.18 Question 6.18

Let O be the ring of integers of Q(d), where d > 0. What can you say about the quotient of O byone of its prime ideals?

15.6.19 Question 6.19

Do you know about Dedekind domains and class numbers?

15.6.20 Question 6.20

Talk about factorisation and primes in a polynomial ring. What is irreducibility? For what ringsR is it true that R[x1, · · · , xn] is a unique factorisation domain? What is wrong with uniquefactorisation if we don’t have a domain? Now, PIDs are Noetherian, but are there UFDs which arenot?

15.6.21 Question 6.21

What is the radical of an ideal? What is special about elements in the nilradical?

15.6 Rings 194


15.6.22 Question 6.22

Define the “radical” of an ideal. Prove it is an ideal. Prove that the ideal of all polynomialsvanishing on the zero set of I is

√I.

15.6.23 Question 6.23.

Do you know what the radical is? Use the fact that the intersection of all prime ideals is the setof all nilpotent elements to prove that F [x] has an infinite number of prime ideals, where F is afield.

15.6.24 Question 6.24

What are the radical ideals in Z?

15.6.25 Question 6.25

Give a prime ideal in k[x, y]. Why is it prime? What is the variety it defines? What is theNullstellensatz? Can you make some maximal ideals?

15.6.26 Question 6.26

State/describe Hilbert’s Nullstellensatz. Sketch a proof.

15.6.27 Question 6.27

What is an irreducible variety? Give an example of a non-irreducible one.

15.6.28 Question 6.28

What are the prime ideals and maximal ideals of Z[x]?

15.6 Rings 195


15.6.29 Question 6.29

Is the following map an isomorphism?

Z[t]/ 〈tp − 1〉 → Z[w]t 7→ w where wp = 1.

15.6.30 Question 6.30

Describe the left, right, and two-sided ideals in the ring of square matrices of a fixed size. Nowidentify the matrix algebra Mat(n × n,K) with End

K(V ) where V is an n-dimensional K-vector

space. Try to geometrically describe the simple left ideals and also the simple right ideals via thatidentification.

15.6.31 Question 6.31

Give examples of maximal ideals in K = R×R×R× · · ·, the product of countably many copies ofR. What about for a product of countably many copies of an arbitrary commutative ring R?

15.6.32 Question 6.32

Consider a commutative ring, R, and a maximal ideal I, what can you say about the structure ofR/I? What if I were prime?

15.6.33 Question 6.33

Define “Noetherian ring”. give an example.

15.6.34 Question 6.34

Prove the Hilbert basis theorem.

15.6.35 Question 6.35

What is a Noetherian ring? If I is an ideal in a Noetherian ring with a unit, what is the intersectionof In over all positive integers n?

15.6 Rings 196


15.6.36 Question 6.36

What is the Jacobson radical? If R is a finitely-generated algebra over a field what can you sayabout it?

15.6.37 Question 6.37

Give an example of an Artinian ring.

15.6.38 Question 6.38

State the structure theorem for semisimple Artinian rings.

15.6.39 Question 6.39

What is a semisimple algebra? State the structure theorem for semisimple algebras.

15.6.40 Question 6.40

What is a matrix algebra?

15.6.41 Question 6.41

Does L1 have a natural multiplication with which it becomes an algebra?

15.6.42 Question 6.42.

Consider a translation-invariant subspace of L1. What can you say about its relation to L2 as aconvolution algebra?

15.6.43 Question 6.43

State the structure theorem for simple rings.

15.6 Rings 197


15.6.44 Question 6.44

Do you know an example of a local ring? Another one? What about completions?

15.6.45 Question 6.45

Consider the space of functions from the natural numbers to C endowed with the usual law ofaddition and the following analogue of the convolution product:

(f ∗ g)(n) =∑d∣∣n f(d)g

(n

d

).

Show that this is a ring. What does this ring remind you of and what can you say about it?

15.6.46 Question 6.46

Prove that any finite division ring is a field (that is, prove commutativity). Give an example of a(necessarily infinite) division ring which is NOT a field.

15.6.47 Question 6.47

Prove that all finite integral domains are fields.

15.6.48 Question 6.48

Can a polynomial over a division ring have more roots than its degree?

15.6.49 Question 6.49

Classify (finite-dimensional) division algebras over R.

15.6.50 Question 6.50

Give an example of a C-algebra which is not semisimple.

15.6 Rings 198


15.6.51 Question 6.51

What is Wedderburn’s theorem? What does the group ring generated by Z/5Z over Q look like?

What if we take the noncyclic group of order 4 instead of Z/5Z? The quaternion group instead ofZ/5Z?

15.6.52 Question 6.52

Tell me about group rings. What do you know about them?

E 15.7 Modules e

15.7.1 Question 7.1

How does one prove the structure theorem for modules over PID? What is the module and what isthe PID in the case of abelian groups?

15.7.2 Question 7.2

If M is free abelian, how can I put quotients of M in some standard form? What was crucial aboutthe integers here (abelian groups being modules over Z)? How does the procedure simplify if thering is a Euclidean domain, not just a PID?

15.7.3 Question 7.3

Suppose D is an integral domain and the fundamental theorem holds for finitely-generated modulesover D (i.e. they are all direct sums of finitely many cyclic modules).

Does D have to be a PID?

15.7.4 Question 7.4

Classify finitely-generated modules over Z, over PIDs, and over Dedekind rings.

15.7 Modules 199


15.7.5 Question 7.5

Prove a finitely-generated torsion-free abelian group is free abelian.

15.7.6 Question 7.6.

What is a tensor product? What is the universal property? What do the tensors look like in thecase of vector spaces?

15.7.7 Question 7.7

Now we’ll take the tensor product of two abelian groups, that is, Z-modules. Take Z/pZ and Z/qZ,where p and q are distinct primes. What is their tensor product?

15.7.8 Question 7.8

What is a projective module?

15.7.9 Question 7.9

What is an injective module?

15.7.10 Question 7.10

Do you know an example of a flat module?

E 15.8 Representation Theory e

15.8.1 Question 8.1

Define “representation” of a group. Define “irreducible representation”. Why can you decomposerepresentations of finite groups into irreducible ones? Construct an invariant inner product.

15.8 Representation Theory 200


15.8.2 Question 8.2

State and prove Maschke’s theorem. What can go wrong if you work over the real field? What cango wrong in characteristic p?

15.8.3 Question 8.3

Do you know what a group representation is? Do you know what the trace of a group representationis?

15.8.4 Question 8.4

State/prove/explain Schur’s lemma.

15.8.5 Question 8.5

What can you say about characters? What are the orthogonality relations? How do you usecharacters to determine if a given irreducible representation is a subspace of another given repre-sentation?

15.8.6 Question 8.6

What’s the relation between the number of conjugacy classes in a finite group and the number ofirreducible representations?

15.8.7 Question 8.7

What is the character table? What field do its entries lie in?

15.8.8 Question 8.8

Why is the character table a square?

15.8.9 Question 8.9

If χ(g) is real for every character χ, what can you say about g?



15.8.10 Question 8.10

What’s the regular representation?

15.8.11 Question 8.11

Give two definitions of “induced representation”. Why are they equivalent?

15.8.12 Question 8.12

If you have a representation of H, a subgroup of a group G, how can you induce a representationof G?

15.8.13 Question 8.13

If you have an irreducible representation of a subgroup, is the induced representation of the wholegroup still irreducible?

15.8.14 Question 8.14.

What can you say about the kernel of an irreducible representation? How about kernels of directsums of irreducibles? What kind of functor is induction? Left or right exact?

15.8.15 Question 8.15

What is Frobenius reciprocity?

15.8.16 Question 8.16

Given a normal subgroup H of a finite group G, we lift all the representations of G/H to represen-tations of G.

Show that the intersection of the kernels of all these representations is precisely H. What can yousay when H is the commutator subgroup of G?



15.8.17 Question 8.17

If you have two linear representations π1 and π2 of a finite group G such that π1(g) is conjugate toπ2(g) for every g in G, is it true that the two representations are isomorphic?

15.8.18 Question 8.18

Group representations: What’s special about using C in the definition of group algebra?

Is it possible to work over other fields?

What goes wrong if the characteristic of the field divides the order of the group?

15.8.19 Question 8.19

Suppose you have a finite p-group, and you have a representation of this group on a finite-dimensionalvector space over a finite field of characteristic p. What can you say about it?

15.8.20 Question 8.20

Let (π, V ) be a faithful finite-dimensional representation of G. Show that, given any irreduciblerepresentation of G, the nth tensor power of GL(V ) will contain it for some large enough n.

15.8.21 Question 8.21

What are the irreducible representations of finite abelian groups?

15.8.22 Question 8.22

What are the group characters of the multiplicative group of a finite field?

15.8.23 Question 8.23

Are there two nonisomorphic groups with the same representations?



15.8.24 Question 8.24

If you have a Z/5Z action on a complex vector space, what does this action look like? What aboutan S3 action? A dihedral group of any order?

15.8.25 Question 8.25

What are the representations of S3? How do they restrict to S2?

15.8.26 Question 8.26

Tell me about the representations of D4. Write down the character table. What is the 2-dimensionalrepresentation? How can it be interpreted geometrically?

15.8.27 Question 8.27

How would you work out the orders of the irreducible representations of the dihedral group Dn?

Why is the sum of squares of dimensions equal to the order of the group?

15.8.28 Question 8.28

Do you know any representation theory? What about representations of A4?

Give a nontrivial one. What else is there? How many irreducible representations do we have? Whatare their degrees? Write the character table of A4.

15.8.29 Question 8.29

Write the character table for S4.

15.8.30 Question 8.30

Start constructing the character table for S5.



15.8.31 Question 8.31.

How many irreducible representations does Sn have?

What classical function in mathematics does this number relate to?

15.8.32 Question 8.32

Discuss representations of Z, the infinite cyclic group. What is the group algebra of Z?

15.8.33 Question 8.33

What is a Lie group? Define a unitary representation. What is the Peter–Weyl theorem? What isthe Lie algebra? The Jacobi identity? What is the adjoint representation of a Lie algebra? Whatis the commutator of two vector fields on a manifold?

When is a representation of Z completely reducible? Why?

Which are the indecomposable modules?

15.8.34 Question 8.34

Talk about the representation theory of compact Lie groups. How do you know you have a finite-dimensional representation?

15.8.35 Question 8.35

How do you prove that any finite-dimensional representation of a compact Lie group is equivalentto a unitary one?

15.8.36 Question 8.36

Do you know a Lie group that has no faithful finite-dimensional representations?

15.8.37 Question 8.37

What do you know about representations of SO(2)? SO(3)?


16 Appendix: Extra Topics

E 15.9 Categories and Functors e

15.9.1 Question 9.1

Which is the connection between Hom and tensor product? What is this called in representationtheory?

15.9.2 Question 9.2

Can you get a long exact sequence from a short exact sequence of abelian groups together withanother abelian group?

15.9.3 Question 9.3

Do you know what the Ext functor of an abelian group is? Do you know where it appears? Whatis Ext(Z/mZ,Z/nZ)? What is Ext(Z/mZ,Z)?


Proposition 16.0.1(NC Theorem).NG(H)/CG(H) is isomorphic to a subgroup of Aut(H).

Definition 16.0.2 (Normalizers Grow)If for every proper H < G, H E NG(H) is again proper, then “normalizers grow” in G.

E 16.1 Characteristic Subgroups e

Slogan 16.1.1Normality is not transitive!

I.e. if H E G and N E H, it’s not necessarily the case that N E G.

15.9 Categories and Functors 206


Definition 16.1.2 (Characteristic Subgroups)A subgroupH ≤ G is characteristic in G, writtenH chG, iff for every ϕ ∈ Aut(G), ϕ(H) ≤ H.Equivalently, ϕ(H) = H. I.e. H is fixed (not necessarily pointwise) under every automorphismof the ambient group G.

Remark 16.1.3(Characteristic isn’t equivalent to normalcy): Characteristic subgroups arenormal, because ψg(−) := g(−)g−1 is an (inner) automorphic of G. Not every normal subgroup ischaracteristic: take G := H1 ×H2 and ψ(x, y) = (y, x).

Proposition 16.1.4(Fixing transitivity of normality).Characteristic subgroups of normal subgroups are normal, i.e. if H E G and N chH, thenN E G.

Proof (?).A chB E C =⇒ A E C:

• A chB iff A is fixed by every ψ ∈ Aut(B)., WTS cAc−1 = A for all c ∈ C.• Since B E C, the automorphism ψ(−) := c(−)c−1 descends to an element of Aut(B).• Then ψ(A) = A since A chB, so cAc−1 = A and A E C.

�

Proposition 16.1.5(Centers are characteristic).For any group G,

Z(G) chG.

Proof (?).Let ψ ∈ Aut(H) and x = ψ(y) ∈ ψ(Z(H)) so y ∈ Z(H), then for arbitrary h ∈ H,

ψ(y)h = ψ(y)(ψ ◦ ψ−1)(h)= ψ(y · ψ−1(h))= ψ(ψ−1(h) · y) since ψ−1(h) ∈ H, y ∈ Z(H)= hψ(y).

�

E 16.2 Normal Closures and Cores e

Definition 16.2.1 (Normal Closure of a Subgroup)The smallest normal subgroup of G containing H:

HG := {gHg−1 : g ∈ G} =⋂{N : H ≤ N E G} .

16.2 Normal Closures and Cores 207


Definition 16.2.2 (Normal Core of a subgroup)The largest normal subgroup of G containing H:

HG =⋂g∈G

gHg−1 = 〈N : N E G & N ≤ H〉 = kerψ.

where

ψ : G→ Aut(G/H)g 7→ (xH 7→ gxH)

Theorem 16.2.3(Fratini’s Argument).If H E G and P ∈ Sylp(G), then HNG(P ) = G and [G : H] divides |NG(P )|.

16.2.1 Exercises

Exercise 16.2.4 (?)Show that Z(G) ≤ G is always characteristic.

Solution:Let ψ ∈ Aut(G). For one containment, we can show ψ(g) = h = hψ(g) for all ψ(g) ∈ ψ(G)and h ∈ G. This is a computation:

ψ(g)h = ψ(g)(ψψ−1)(h)= ψ(g)ψ(ψ−1(h))= ψ(ψ−1(h)g)= (ψψ−1)(h)ψ(g)= hψ(g).

This yields ψ(Z(G)) ⊆ Z(G). Applying the same argument to ψ−1 yields ψ−1(Z(G)) ⊆ Z(G).Since ψ is a bijection, ψψ−1(A) = A for all A ≤ G, so Z(G) ⊆ ψ(Z(G)).

E 16.3 Nilpotent Groups e

Definition 16.3.1 (Nilpotent)A group G is nilpotent iff G has a terminating upper central series.

Moral: the adjoint map is nilpotent.

Theorem 16.3.2(Characterization of Nilpotent Groups).G is nilpotent iff G has an upper central series terminating at G.

16.3 Nilpotent Groups 208


Theorem 16.3.3(Characterization of Nilpotent Groups).G is nilpotent iff G has a lower central series terminating at 1.

Theorem 16.3.4(Nilpotents Have All Sylows Normal).A group G is nilpotent iff all of its Sylow p-subgroups are normal for every p dividing |G|.

Theorem 16.3.5(Nilpotent Implies Maximal Normals).A group G is nilpotent iff every maximal subgroup is normal.

Proposition 16.3.6.For G a finite group, TFAE:

• G is nilpotent• Normalizers grow, i.e. if H < G is proper then H < NG(H).• Every Sylow-p subgroup is normal• G is the direct product of its Sylow p-subgroups• Every maximal subgroup is normal• G has a terminating Lower Central Series• G has a terminating Upper Central Series

Fact 16.3.7

• Nilpotent groups satisfy the 2 out of 3 property. Todo. Specify.

• G has normal subgroups of order d for every d dividing |G|

E 16.4 Rings e

Definition 16.4.1 (Gorenstein Rings)A commutative Noetherian ring R is Gorenstein iff R viewed as an R-module has finiteinjective dimension.

Example 16.4.2(Why care about Gorenstein rings?): If R ∈ gr Alg/k with dimk R <∞, thenR decomposes as R = R0 ⊕R1 ⊕ · · ·Rn with R0 := k, and R is Gorenstein iff R satisfies “Poincaréduality”: dimk R0 = dimk Rm = 1 and there is a perfect pairing Ri ⊗k Rn−j → Rn.

16.4 Rings 209

17 UGA Fall 2019 Problem Sets


E 17.1 Problem Set One e

17.1.1 Exercises

Problem 17.1.1 (Hungerford 1.6.3)If σ = (i1i2 · · · ir) ∈ Sn and τ ∈ Sn, then show that τστ−1 = (τ(i1)τ(i2) · · · τ(ir)).

Problem 17.1.2 (Hungerford 1.6.4)Show that Sn ∼= 〈(12), (123 · · ·n)〉 and also that Sn ∼= 〈(12), (23 · · ·n)〉

Problem 17.1.3 (Hungerford 2.2.1)Let G be a finite abelian group that is not cyclic. Show that G contains a subgroup isomorphicto Zp ⊕ Zp for some prime p.

Problem 17.1.4 (Hungerford 2.2.12.b.)Determine (up to isomorphism) all abelian groups of order 64; do the same for order 96.

Problem 17.1.5 (Hungerford 2.4.1)Let G be a group and A E G be a normal abelian subgroup. Show that G/A acts on A byconjugation and construct a homomorphism ϕ : G/A→ Aut(A).

Problem 17.1.6 (Hungerford 2.4.9).)Let Z(G) be the center of G. Show that if G/Z(G) is cyclic, then G is abelian.

Note that Hungerford uses the notation C(G) forthe center.

Problem 17.1.7 (Hungerford 2.5.6)Let G be a finite group andH E G a normal subgroup of order pk. Show that H is containedin every Sylow p-subgroup of G.

Problem 17.1.8 (Hungerford 2.5.9)Let |G| = pnq for some primes p > q. Show that G contains a unique normal subgroup ofindex q.

UGA Fall 2019 Problem Sets 210


17.1.2 Qual Problems

Problem 17.1.9Let G be a finite group and p a prime number. Let Xp be the set of Sylow-p subgroups of Gand np be the cardinality of Xp. Let Sym(X) be the permutation group on the set Xp.

1. Construct a homomorphism ρ : G→ Sym(Xp) with image a transitive subgroup (i.e. witha single orbit).

2. Deduce that if G is simple then the order of G divides np!.

3. Show that for any 1 ≤ a ≤ 4 and any prime power pk, no group of order apk is simple.

Solution:

1. Define the required group action by

ρ : G→ Sym(Xp)g 7→ (γg : P 7→ gPg−1).

The claim is that this action is transitive on Xp. This can be equivalently stated as

∀P ∈ Xp,∃g ∈ G,P ′ ∈ Xp

∣∣∣ gP ′g−1 = P.

However, by Sylow 2, all Sylow p−subgroups are conjugate to each other, and thus thiscondition is satisfied.

2. Suppose that G is simple, so that we have

H E G =⇒ H = {e} or H = G.

Note that Sym(Xp) = (np)!, and if we have an injective homomorphism Gϕ−→ Sym(Xp),

then |G| = |ϕ(G)|, since ϕ(G) ≤ Sym(Xp) will be a subgroup and thus have orderdividing (np)!, which proves the statement.Using the ϕ defined in (1), we can apply the first isomorphism theorem

G/ kerϕ ∼= imϕ ≤ Sym(Xp),

and so it suffices to show that kerϕ = {e}.Note that since kerϕ E G and G is simple, we can only have kerϕ = {e} or kerϕ = G.Towards a contradiction, suppose kerϕ = G.By definition, we have

kerϕ = {g ∈ G∣∣∣ γg = idXp}

= {g ∈ G∣∣∣ ∀P ∈ Xp, gPg

−1 = P}

=⋂

P∈XpNG(P ),

17.1 Problem Set One 211


and so the kernel of ϕ is the intersection of all of the normalizers of the Sylowp−subgroups.But this means that NG(P ) = G for every Sylow p−subgroup, which means that np = 1and there is a unique P which must be normal in G. Since G is simple, this forces P tobe trivial or the whole group.Towards a contradiction, suppose P = G. Then G is a p−group and thus has orderpn. But then G has normal subgroups of order pk for all 0 < k < n, contradicting thesimplicity of G.But the only other option is that P is trivial, whereas we know nontrivial Sylowp−subgroups exist by Sylow 1.Thus we can not have kerϕ = G, and so kerϕ is trivial as desired.

3. Suppose |G| = apk, where 1 ≤ a ≤ 4. Then by Sylow 3, we have np = 1 mod p and npdivides a. If a = 1, then np = 1, and so G can not be simple. Moreover, if p ≥ a, thennp ≤ a and np = 1 mod p forces np = 1 again.So we can restrict our attention to 2 ≤ a ≤ 4 and p = 2, 3, which reduces to checkingthe cases apk = 2(3k), 4(3k), or 3(2k) for k ≥ 1.If apk = 2(3k), we have n3 = 1 mod 3 and n3

∣∣ 2, which forces n3 = 1, so this can not bea simple group.Similarly, if apk = 4(3k), then n3 = 1 mod 3 and n3 divides 4, which forces n3 = 1 andthus G can’t be simple.If apk = 3(2k), then n2 = 1 mod 2 and n2 divides 3, so n2 = 1, 3. But then n3! = 6, andif k > 1, we have 3(2k) > 6 = n3!, so G can not be simple by the result in (2).If k = 1, then G is order 6, so G is isomorphic to either Z6 or S3. The group S3 is notsimple, since A3 E S3, and the only simple cyclic groups are of prime order, so Z6 is notsimple. This exhausts all of the possible cases.

Problem 17.1.10Let G be a finite group and let N E G, and let p be a prime number and Q a subgroup of Gsuch that N ⊂ Q and Q/N is a Sylow p−subgroup of G/N .

1. Prove that Q contains a Sylow p−subgroup of G.

2. Prove that every Sylow p−subgroup of G/N is the image of a Sylow p−subgroup of G.

Solution:Proof.

1. Since Q/N is a Sylow p−subgroup of G/N , we can write |G/N | = pkl where gcd(p, l) = 1,and |Q/N | = pk.

We can then write |G| = pnm where n ≥ l and l∣∣∣ m.



By the third isomorphism theorem, we have

G/N

Q/N∼= G/Q

and so ∣∣∣∣G/NQ/N

∣∣∣∣ = |G/N ||Q/N |

= pkl

pk= l

and so |G/Q| = l where (p, l) = 1, and thus

|G/Q| = |G|/|Q| = l =⇒ |G| = |Q| l.

We then have

pnm = |Q| l,

and since (p, l) = 1, it must be the case that pn divides |Q|. But since Q ≤ G, this meansthat Q itself must be a Sylow p− subgroup of G.

2. Let PN ∈ Syl(p,G/N). By the subgroup correspondence theorem, Pn = H/N for someH ≤ G such that N ⊆ H.

So choose PH ∈ Syl(p,H); the claim is that PH ∈ Syl(p,G) and that PHNN∼= PN , which

exhibits PN as the image of a Sylow p−subgroup of G.We first have PH ∈ Syl(p,G), which follows because we have [G/N,H/N ] = [G : PH ]from the fourth isomorphism theorem, and thus [G/N,PN ] = [G : PH ]. In particular,since PN is a Sylow p−subgroup, p does not divide [G/N,PN ] and thus p doesn’t divide[G : PH ], which makes PH a maximal p−subgroup in G and thus a Sylow p−subgroup.We then have PHN/N = PN , which follows because PH ≤ H =⇒ PHN/N ≤ H/N =PN ≤ G/N .However, it is also the case that PHN/N ∈ Syl(p,G/N). This follows because

1. PHN/N = PH/PH ∩N by the 2nd isomorphism theorem, so it is a p−group.2. PH ⊆ PHN ⊆ G =⇒ p doesn’t divide [G : PHN ], since PH is also a Sylowp−group of G and thus has maximal prime power dividing |G|.

3. N ⊆ PHN ⊆ G =⇒ [G/N : PHN/N ] = [G : PHN ]

Taken together, this says that PHN/N is a p-group and p doesn’t divide [G/N,PHN/N ],so it is a maximal p−subgroup and PHN/N ∈ Syl(p,G/N).But since PHN/N ≤ PN and |PHN/N | = |PN |, we must have PHN/N = PN as desired.

Problem 17.1.11Let G be a finite group and H < G a subgroup. Let nH be the number of subgroups of G thatare conjugate to H. Show that nH divides the order of G.



Solution:.* Let

CH = {gHg−1∣∣∣ g ∈ G}

be the conjugacy class of H, so |CH | = nH .We wish to show that nH divides |G|.Claim 1:

nH = [G : NG(H)],

where NG(H) ≤ G is the normalizer of H in G.Note that if this claim is true, then we can apply Lagrange’s theorem, which states

A ≤ G =⇒ |G| = [A : G] |A|,

which in this case translates to

|G| = [NG(H) : G] |NG(H)| = nH |NG(H)|.

Since nH divides the right-hand side, it must divide the left-hand side as well, which is preciselywhat we would like to show.Proof of Claim 1:The normalizer of H in G, written NG(H), is the largest subgroup of G containing H suchthat H E NG(H), i.e.

NG(H) = {g ∈ G∣∣∣ gHg−1 = H} ≤ G.

Now consider S, the set of left cosets of NG(H). Suppose there are k of them, so

[G : NG(H)] = |S| := k.

Then S can be written as

S = {g1NG(H), g2NG(H), · · · , gkNG(H)}.

where each gi is a distinct element of G yielding a distinct coset giNG(H). In particular, ifi 6= j, then gi 6= gj , and giNG(H) 6∈ gjNG(H).In particular, S acts on CH ,

S y CH

giNG(H) y H = giHg−1i ,

taking H to one of its conjugate subgroups.So define

K := {giHg−1i

∣∣∣ 1 ≤ i ≤ k}.

Note that K ⊆ CH , and has at most k elements.



We claim that K has k distinct elements, i.e. that each gi takes H to a distinct conjugatesubgroup. We have

giHg−1i = gjHg

−1j =⇒

g−1j giHg

−1i gj = H =⇒

(g−1j gi)H(g−1

j gi)−1 = H =⇒g−1j gi ∈ NG(H) =⇒gi ∈ gjNG(H) =⇒gi = gj ,

where the last line follows because we assumed that each coset contains at most one gi.Thus K has k distinct elements, and so

= k = |K| ≤ |CH | = nH .

We now claim that k ≥ nH as well.Let H ′ ∈ CH be any subgroup conjugate to H, so H ′ = gHg−1 for some g ∈ G. Then g = gifor some i, so g ∈ giNG(H).Thus g = gin for some n ∈ NG(H), but n ∈ NG(H) ⇐⇒ nHn−1 = H by definition, and sowe have

H ′ = gHg−1

= (gin)H(gin)−1

= gi(nHn−1)g−1i

= giHg−1i ∈ K.

Since H ′ ∈ CH was an arbitrary subgroup conjugate to H, this says that CH ⊆ K and thus

nH = |CH | ≤ |K| = k

Thus

[G : NG(H)] = k = |M | = |K| = nH ,

which is what we wanted to show.

Problem 17.1.12Let G = S5, the symmetric group on 5 elements. Identify all conjugacy classes of elements inG, provide a representative from each class, and prove that this list is complete.

Solution:

Claim 1: Conjugacy classes in Sn are completely determined by cycle type.This follows because of the result on homework 1, which says that for any two cycles τ, σ ∈ Sn,



we have

τ(s1 s2 · · · sk)τ−1 = (τ(s1) τ(s2) · · · τ(sk)).

In particular, this shows that the cycle type of a single cycle is invariant under conjugation.If an element σ ∈ Sn is comprised of multiple cycles, say σ = σ1 · · ·σ`, then

τ(σ)τ−1 = τ(σ1 · · ·σ`)τ−1 = (τσ1τ−1) · · · (τσ`τ−1),

which shows that the entire cycle type is preserved under conjugation. So each conjugacy classhas exactly one cycle type, and distinct classes have distinct cycle types, so this completelydetermines the conjugacy classes.Claim 2: Cycle types in Sn are in bijective correspondence with integer partitions of n.This follows because any integer partition of n can be used to obtain a canonical representativeof a conjugacy class of Sn: if n = a1 + a2 + · · · an, we simply take a cycle of length a1 the firsta1 integers in order, a cycle of length a2 containing the integers a1 + 1 to a2 in order, and soon.Conversely, any permutation can be written as a product of disjoint cycles, and when thecycles for fixed points are added in, every integer between 1 and n will appear, and the sumof the lengths of all cycles must sum to n. Thus taking the cycle lengths yields an integerpartition of n.All integer partitions of 5 are given below, along with a canonical representative of theassociated conjugacy class.

5 (1 2 3 4 5)4 + 1 (1 2 3 4)(5)3 + 2 (1 2 3)(4 5)

3 + 1 + 1 (1 2 3)(4)(5)2 + 2 + 1 (1 2)(3 4)(5)

2 + 1 + 1 + 1 (1 2)(3)(4)(5)1 + 1 + 1 + 1 + 1 (1)(2)(3)(4)(5)

E 17.2 Problem Set Two e

17.2.1 Exercises

Problem 17.2.1 (Hungerford 2.1.9)Let G be a finitely generated abelian group in which no element (except 0) has finite order.Show that G is a free abelian group.

17.2 Problem Set Two 216


Problem 17.2.2 (Hungerford 2.1.10)

1. Show that the additive group of rationals Q is not finitely generated.

2. Show that Q is not free.

3. Conclude that Exercise 9 is false if the hypothesis “finitely generated” is omitted.

Problem 17.2.3 (Hungerford 2.5.8)Show that if every Sylow p−subgroup of a finite group G is normal for every prime p, then Gis the direct product of its Sylow subgroups.

Problem 17.2.4 (Hungerford 2.6.4)What is the center of the quaternion group Q8? Show that Q8/Z(Q8) is abelian.

Problem 17.2.5 (Hungerford 2.6.9)Classify up to isomorphism all groups of order 18. Do the same for orders 20 and 30.

Problem 17.2.6 (Hungerford 1.9.1)Show that every non-identity element in a free group F has infinite order.

Problem 17.2.7 (Hungerford 1.9.3)Let F be a free group and for a fixed integer n, let Hn be the subgroup generated by the set{xn

∣∣∣ x ∈ F}. Show that Hn E F .


Problem 17.2.8List all groups of order 14 up to isomorphism.

Problem 17.2.9Let G be a group of order p3 for some prime p. Show that either G is abelian, or |Z(G)| = p.

Problem 17.2.10Let p, q be distinct primes, and let k denote the smallest positive integer such that p dividesqk − 1. Show that no group of order pqk is simple.

Problem 17.2.11Show that S4 is a solvable, nonabelian group.

17.2 Problem Set Two 217


E 17.3 Problem Set Three e

17.3.1 Exercises

Problem 17.3.1 (Hungerford 2.7.10)Show that Sn is solvable forn ≤ 4 but S3 and S4 are not nilpotent.

Problem 17.3.2 (Hungerford 2.8.3)Show that if N is a simple normal subgroup of a group G and G/N has a composition series,then G has a composition series.

Problem 17.3.3 (Hungerford 2.8.9)Show that any group of order p2q(for primes p, q) is solvable.

Problem 17.3.4 (Hungerford 5.1.1)Let F/K be a field extension. Show that

1. [F : K] = 1 iff F = K.

2. If [F : K] is prime, then there are no intermediate fields between F and K.

3. If u ∈ F has degree n over K, then n divides [F : K].

Problem 17.3.5 (Hungerford 5.1.8)Show that if u ∈ F is algebraic of odd degree overK, then so is u2, and moreoverK(u) = K(u2).


1. If F = Q(√

2,√

3), compute [F : Q] and find a basis of F/Q.

2. Do the same for Q(i,√

3, ζ3) where ζ3 is a complex third root of 1.

Problem 17.3.7 (Hungerford 5.1.16)Show that in C, the fields Q(i) ∼= Q(

√2) as vector spaces, but not as fields.


Problem 17.3.8Let R and S be commutative rings with multiplicative identity.

1. Prove that when R is a field, every non-zero ring homomorphism ϕ : R→ S is injective.

2. Does (a) still hold if we only assume that R is a domain? If so, prove it, and if not

17.3 Problem Set Three 218


provide a counterexample.

Problem 17.3.9Determine for which integers the ring Z/nZ is a direct sum of fields. Carefully prove youranswer.

Problem 17.3.10Suppose that R is a commutative ring. Show that an element r ∈ R is not invertible iff it iscontained in a maximal ideal.

Problem 17.3.11

1. Give the definition that a group G must satisfy the be solvable.

2. Show that every group G of order 36 is solvable.

Hint: You may assume that S4 is solvable.

E 17.4 Problem Set Four e

17.4.1 Exercises

Problem 17.4.1 (Hungerford 5.3.7)If F is algebraically closed andE is the set of all elements in F that are algebraic over a fieldK, then E is an algebraic closure of K.

Problem 17.4.2 (Hungerford 5.3.8)Show that no finite field is algebraically closed.Hint: if K = {ai}ni=0, consider

f(x) = a1 +n∏i=0

(x− ai) ∈ K[x]

where a1 6= 0.

Problem 17.4.3 (Hungerford 5.5.2)Show that if p ∈ Z is prime, then ap = a for all a ∈ Zp, or equivalently cp ≡ cmod p for allc ∈ Z.

Problem 17.4.4 (Hungerford 5.5.3)Show that if |K| = pn, then every element of K has a unique pth root in K.

17.4 Problem Set Four 219


Problem 17.4.5 (Hungerford 5.5.10)Show that every element in a finite field can be written as the sum of two squares.

Problem 17.4.6 (Hungerford 5.6.1)Let F/K be a field extension. Let charK = p 6= 0 and let n ≥ 1 be an integer such that(p, n) = 1. If v ∈ F and nv ∈ K, then v ∈ K.

Problem 17.4.7 (Hungerford 5.6.8)If charK = p 6= 0 and [F : K] is finite and not divisible by p, then F is separable over K.


Problem 17.4.8Suppose that α is a root in C of P (x) = x17 − 2. How many field homomorphisms are therefrom Q(α) to:

1. C,

2. R,

3. Q, the algebraic closure of Q?

Problem 17.4.9Let C/F be an algebraic field extension. Prove that the following are equivalent:

1. Every non-constant polynomial f ∈ F [x] factors into linear factors over C[x].

2. For every (not necessarily finite) algebraic extension E/F , there is a ring homomorphismα : E → C such that α

∣∣∣Fis the identity on F .

Hint: use Zorn’s Lemma.

Problem 17.4.10Let R be a commutative ring containing a field k, and suppose that dimk R <∞. Let α ∈ R.

1. Show that there exist n ∈ N and {c0, c1, · · · cn−1} ⊆ k such that

an + cn−1an−1 + · · ·+ c1a+ c0 = 0.

2. Suppose that (a) holds and show that if c0 6= 0 then a is a unit in R.

3. Suppose that (a) holds and show that if a is not a zero divisor in R, then a is invertible.

17.4 Problem Set Four 220


E 17.5 Problem Set Five e

17.5.1 Exercises

Problem 17.5.1 (Hungerford 5.3.5)Show that if f ∈ K[x] has degree n and F is a splitting field of f over K, the [F : K] dividesn!.

Problem 17.5.2 (Hungerford 5.3.12)Let E be an intermediate field extension in K ≤ E ≤ F .

1. Show that if u ∈ F is separable over over K, then u is separable over E.

2. Show that if F is separable over K, then F is separable over E and E is separable overK.

Problem 17.5.3 (Hungerford 5.3.13)Show that if [F : K] <∞, then the following conditions are equivalent:

1. F is Galois over K

2. F is separable over K and F is a splitting field of some polynomial f ∈ K[x].

3. F is a splitting field over K of some polynomial f ∈ K[x] whose irreducible factors areseparable.

Problem 17.5.4 (Hungerford 5.4.1)Suppose that f ∈ K[x] splits inF as

f =k∏i=1

(x− ui)ni

with the ui distinct and each ni ≥ 1. Let

g(x) =k∏i=1

(x− ui) =k∑i=1

vixi

and let E = K({vi}ki=1). Then show that the following hold:

1. F is a splitting field of g over E.

2. F is Galois over E.

3. AutE(F ) = AutK(F ).

17.5 Problem Set Five 221


Problem 17.5.5 (Hungerford 5.4.10 a/g/h)Determine the Galois groups of the following polynomials over the corresponding fields:

1. x4 − 5 over Q,Q(√

5),Q(i√

5).

2. x3 − 2 over Q.

3. (x3 − 2)(x2 − 5) over Q.

Problem 17.5.6 (Hungerford 5.6.11)If f ∈ K[x] is irreducible of degree m > 0 and char(K) does not divide m, then f is separable.


Problem 17.5.7Let E/F be a Galois field extension, and let K/F be an intermediate field of E/F . Show thatK is normal over F iff Gal(E/K) E Gal(E/F ).

Problem 17.5.8Let F ⊂ L be fields such that L/F is a Galois field extension with Galois group equal toD8 =

⟨σ, τ

∣∣∣ σ4 = τ2 = 1, στ = τσ3⟩. Show that there are fields F ⊂ E ⊂ K ⊂ L such that

E/F and K/E are Galois field extensions, but K/F is not Galois.

Problem 17.5.9Let f(x) = x3 − 7.

1. Let K be the splitting field for f over Q. Describe the Galois group of K/Q and theintermediate fields between Q and K. Which intermediate fields are not Galois over Q?

2. Let L be the splitting field for f over R. What is the Galois group L/R?

3. LetM be the splitting field for f over F13, the field with 13 elements. What is the Galoisgroup of M/F13?

17.5 Problem Set Five 222


E 17.6 Problem Set Six e

17.6.1 Exercises

Problem 17.6.1 (Hungerford 5.4.11)Determine all subgroups of the Galois group and all intermediate fields of the splitting (overQ) of the polynomial (x3 − 2)(x2 − 3) ∈ Q[x].

Problem 17.6.2 (Hungerford 5.4.12)Let K be a subfield ofR and let f ∈ K[x] be an irreducible quartic. If f has exactly 2 realroots, the Galois group of f is either S4 or D4.

Problem 17.6.3 (Hungerford 5.8.3)Let ϕ be the Euler function.

1. ϕ(n) is even for n > 2.

2. find all n > 0 such that ϕ(n) = 2.

Problem 17.6.4 (Hungerford 5.8.9)If n > 2 and ζ is a primitive nth root of unity over Q, then [Q(ζ + ζ−1) : Q] = ϕ(n)/2.

Problem 17.6.5 (Hungerford 5.9.1)If F is a radical extension field of K and E is an intermediate field, then F is a radicalextension of E.

Problem 17.6.6 (Hungerford 5.9.3)Let K be a field, f ∈ K[x] an irreducible polynomial of degree n ≥ 5 and F a splitting field off over K. Assume that Autk(F ) ' Sn. Let u be a root of f in F . Then,

1. K(u) is not Galois over K; [K(u) : K] = n and AutK(K(u)) = 1 (and hence solvable).

2. Every normal closure over K that contains u also contains an isomorphic copy of F .

3. There is no radical extension field E of K such that K ⊂ K(u) ⊂ E.


Problem 17.6.7

1. Let K be a field. State the main theorem of Galois theory for a finite field extensionL/K

17.6 Problem Set Six 223


2. Let ζ43 := e2πi/43. Describe the group of all field automorphisms σ : Q(ζ43)→ Q(ζ43).

3. How many proper subfields are there in the field Q(ζ43)?

Problem 17.6.8Let F be a field and let f(x) ∈ F [x].

1. Define what is a splitting field of f(x) over F .

2. Let F be a finite field with q elements. Let E/F be a finite extension of degree n > 0.Exhibit an explicit polynomial g(x) ∈ F [x] such that E/F is a splitting of g(x) over F .Fully justify your answer.

3. Show that the extension E/F in (2) is a Galois extension.

Problem 17.6.9Let K ⊂ L ⊂ M be a tower of finite degree field extensions. In each of the following parts,either prove the assertion or give a counterexample (with justification).

1. If M/K is Galois, then L/K is Galois

2. If M/K is Galois, then M/L is Galois.

E 17.7 Problem Set Seven e

17.7.1 Exercises

Problem 17.7.1 (Hungerford 4.1.3)Let I be a left ideal of a ringR, and let A be an R−module.

1. Show that if S is a nonempty subset of A, then

IS :={

n∑i=1

riai∣∣∣ n ∈ N∗; ri ∈ I; ai ∈ S

}

is a submodule of A.

Note that if S = {a}, then IS = Ia = {ra∣∣∣ r ∈ I}.

2. If I is a two-sided ideal, then A/IA is an R/I module with the action of R/I given by

(r + I)(a+ IA) = ra+ IA.

17.7 Problem Set Seven 224


Problem 17.7.2 (Hungerford 4.1.5)If R has an identity, then a nonzero unitary R-module is simple if its only submodules are 0and A.

1. Show that every simple R−module is cyclic.

2. If A is simple, every R−module endomorphism is either the zero map or an isomorphism.


1. Show that if A,B are R-modules, then the set HomR(A,B) is all R-module homomor-phisms A→ B is an abelian group with f + g given on a ∈ A by

(f + g)(a) := f(a) + g(a) ∈ B.

Also show that the identity element is the zero map.

2. Show that HomR(A,A) is a ring with identity, where multiplication is given by composi-tion of functions.Note that HomR(A,A) is called the endomorphism ring of A.

3. Show that A is a left HomR(A,A)-module with an action defined by

a ∈ A, f ∈ HomR(A,A) =⇒ f y a := f(a).

Problem 17.7.4 (Hungerford 4.1.12)Let the following be a commutative diagram of R-modules and R-module homomorphismswith exact rows:Prove the following:

1. If α1 is an epimorphisms and α2, α4 are monomorphisms then α3 is a monomorphism.

2. If α5 is a monomorphism and α2, α4 are epimorphisms then α3 is an epimorphism.

Problem 17.7.5 (Hungerford 4.2.4)Let R be a principal ideal domain,A a unitary left R-module, and p ∈ R a prime (and thusirreducible) element. Define

pA := {pa∣∣∣ a ∈ A}

A[p] := {a ∈ A∣∣∣ pa = 0}.

Show the following:

1. R/(p) is a field.

2. pA and A[p] are submodules of A.

3. A/pA is a vector space over R/(p), with

(r + (p))(a+ pA) = ra+ pA.



4. A[p] is a vector space over R/(p) with

(r + (p))a = ra.

Problem 17.7.6 (Hungerford 4.2.8)If V is a finite dimensional vector space and

V m := V ⊕ V ⊕ · · · ⊕ V (m summands),

then for each m ≥ 1, V m is finite dimensional and dimV m = m(dimV ).

Problem 17.7.7 (Hungerford 4.2.9)If F1, F2 are free modules of a ring with the invariant dimension property, then

rank(F1 ⊕ F2) = rankF1 + rankF2.


Problem 17.7.8Let F be a field and let f(x) ∈ F [x].

1. State the definition of a splitting field of f(x) over F .

2. Let F be a finite field with q elements. Let E/F be a finite extension of degree n > 0.Exhibit an explicit polynomial g(x) ∈ F [x] such that E/F is a splitting field of g overF . Fully justify your answer.

3. Show that the extension in (b) is a Galois extension.

Problem 17.7.9Let R be a commutative ring and letM be an R-module. Recall that for µ ∈M , the annihilatorof µ is the set

Ann(µ) = {r ∈ R∣∣∣ rµ = 0}.

Suppose that I is an ideal in R which is maximal with respect to the property there exists anonzero element µ ∈M such that I = Ann(µ).Prove that I is a prime ideal in R.

Problem 17.7.10Suppose that R is a principal ideal domain and I E R is an ideal. If a ∈ I is an irreducibleelement, show that I = Ra.



E 17.8 Problem Set Eight e

17.8.1 Exercises

Problem 17.8.1 (Hungerford 4.4.1)Show the following:

1. For any abelian group A and any positive integer m,

Hom(Zm, A) ∼= A[m] := {a ∈ A∣∣∣ ma = 0}.

2. Hom(Zm,Zn) ∼= Zgcd(m,n).

3. As a Z−module, Z∗m = 0.

4. For each k ≥ 1, Zm is a Zmk−module, and as a Zmk module, Z∗m ∼= Zm.

Problem 17.8.2 (Hungerford 4.4.3)Letπ : Z→ Z2 be the canonical epimorphism. Show that the induced map π : Hom(Z2,Z)→Hom(Z2,Z2) is the zero map. Conclude that π is not an epimorphism.

Problem 17.8.3 (Hungerford 4.4.5)Let R be a unital ring, show that there is a ring homomorphism HomR(R,R) → Rop whereHomR denotes left R−module homomorphisms. Conclude that if R is commutative, thenthere is a ring isomorphism HomR(R,R) ∼= R.

Problem 17.8.4 (Hungerford 4.4.9)Show that for any homomorphismf : A → B of left R−modules the following diagram iscommutative:where θA, θB are as in Theorem 4.12 and f∗ is the map induced on A∗∗ := HomR(Hom(A,R), R)by the map

f : Hom(B,R)→ HomR(A,R).

Problem 17.8.5 (Hungerford 4.6.2)Show that every free module over a unital integral domain is torsion-free. Show that theconverse is false.

Problem 17.8.6 (Hungerford 4.6.3)Let A be a cyclic R−module of order r ∈ R.

1. Show that if s is relatively prime to r, then sA = A and A[s] = 0.

2. If s divides r, so sk = r, then sA ∼= R/(k) and A[s] ∼= R/(s).

17.8 Problem Set Eight 227


Problem 17.8.7 (Hungerford 4.6.6)Let A,B be cyclic modules over Rof nonzero orders r, s respectively, where r is not relativelyprime to s. Show that the invariant factors of A⊕B are gcd(r, s) and lcm(r, s).


Problem 17.8.8Let R be a PID. Let n > 0 and A ∈Mn(R) be a square n× n matrix with coefficients in R.Consider the R-module M := Rn/im(A).

1. Give a necessary and sufficient condition forM to be a torsion module (i.e. every nonzeroelement is torsion). Justify your answer.

2. Let F be a field and now let R := F [x]. Give an example of an integer n > 0 and an n×nsquare matrix A ∈Mn(R) such that M := Rn/im(A) is isomorphic as an R−module toR× F .

Problem 17.8.9

1. State the structure theorem for finitely generated modules over a PID.

2. Find the decomposition of the Z−module M generated by w, x, y, z satisfying the rela-tions

3w + 12y + 3x+ 6z = 06y = 0

−3w − 3x+ 6y = 0.

Problem 17.8.10Let R be a commutative ring and M an R−module.

1. Define what a torsion element of M is .

2. Given an example of a ring R and a cyclic R−module M such that M is infinite and Mcontains a nontrivial torsion element m. Justify why m is torsion.

3. Show that if R is a domain, then the subset of elements of M that are torsion is anR−submodule of M . Clearly show where the hypothesis that R is a domain is used.

17.8 Problem Set Eight 228


E 17.9 Problem Set Nine e

17.9.1 Exercises


1. Show that the center of the ring Mn(R) consists of matrices of the form rIn where r isin the center of R.Hint: Every such matrix must commute with εij, the matrix with 1R in the i, j positionand zeros elsewhere.

2. Show that Z(Mn(R)) ∼= Z(R).


1. Show that if A,B are (skew)-symmetric then A+B is (skew)-symmetric.

2. Let R be commutative. Show that if A,B are symmetric, then AB is symmetric ⇐⇒AB = BA. Also show that for any matrix B ∈Mn(R), both BBt and B+Bt are alwayssymmetric, and B −Bt is always skew-symmetric.

Problem 17.9.3 (Hungerford 7.1.7)Show that similarity is an equivalence relation on Mn(R), and *equivalence* is an equivalencerelation on Mm×n(R).

Problem 17.9.4 (Hungerford 7.2.2)Show that an n×m matrix Aover a division ring D has an m×n left inverse B (so BA = Im)⇐⇒ rankA = m. Similarly, show A has a right m× n inverse ⇐⇒ rankA = n.


1. Show that a system of linear equations

a11x1 + a12x2+ · · ·+ a1mxm = b1

...

an1x1 + an2x2+ · · ·+ anmxm = bn

has a simultaneous solution ⇐⇒ the corresponding matrix equation AX = B has asolution, where A = (aij), X = [x1, · · · , xm]t, and B = [b1, · · · , bn]t.

2. If A1, B1 are matrices obtained from A,B respectively by performing the same sequenceof elementary row operations, then X is a solution of AX = B ⇐⇒ X is a solution ofA1X = B1.

17.9 Problem Set Nine 229


3. Let C be the n× (m+ 1) matrix given by

C =

a11 · · · a1m b1

·an1 · · · anm bn

.Then AX = B has a solution ⇐⇒ rankA = rankC and the solution is unique ⇐⇒rank(A) = m.

Hint: use part 2.

4. If B = 0, so the system AX = B is homogeneous, then it has a nontrivial solution⇐⇒ rankA < m and in particular n < m.

Problem 17.9.6 (Hungerford 7.2.5)Let R be a PID. For each positive integer r and sequence of nonzero ideals I1 ⊃ I2 ⊃ · · · ⊃ Ir,choose a sequence di ∈ R such that (di) = Ii and di

∣∣∣ di+1.For a given pair of positive integers n,m, let S be the set of all n ×m matrices of the form(Lr 00 0

)where r = 1, 2, · · · ,min(m,n) and Lr is a diagonal r×r matrix with main diagonal

di.Show that S is a set of canonical forms under equivalence for the set of all n ×m matricesover R.


Problem 17.9.7Let R be a commutative ring.

1. Say what it means for R to be a unique factorization domain (UFD).

2. Say what it means for R to be a principal ideal domain (PID)

3. Give an example of a UFD that is not a PID. Prove that it is not a PID.

Problem 17.9.8Let A be an n×n matrix over a field F such that A is diagonalizable. Prove that the followingare equivalent:

1. There is a vector v ∈ Fn such that v,Av, · · ·An−1v is a basis for Fn.

2. The eigenvalues of A are distinct.

17.9 Problem Set Nine 230


Problem 17.9.9Let x, y ∈ C and consider the matrix

M =

1 0 x0 1 0y 0 1

1. Show that [0, 1, 0]t is an eigenvector of M .

2. Compute the rank of M as a function of x and y.

3. Find all values of x and y for which M is diagonalizable.

E 17.10 Problem Set Ten e

17.10.1 Exercises

Problem 17.10.1 (Hungerford 7.3.1)Let B be an R-module. Show that if r + r 6= 0 for all r 6= 0 ∈ R, then an n-linear formBn → R is alternating ⇐⇒ it is skew-symmetric.

Problem 17.10.2 (Hungerford 7.3.5)If R is a field and A,B ∈Mn(R) are invertible then the matrix A+ rB is invertible for all buta finite number of r ∈ R.

Problem 17.10.3 (Hungerford 7.4.4)Show that if q is the minimal polynomial of a linear transformation ϕ : E → E with dimk E = nthen deg q ≤ n.

Problem 17.10.4 (Hungerford 7.4.8).)Show that A ∈Mn(K) is similar to a diagonal matrix ⇐⇒ the elementary divisors of A areall linear.

Problem 17.10.5 (Hungerford 7.4.10)Find all possible rational canonical forms for a matrix A ∈Mn(Q) such that

1. A is 6× 6 with minimal polynomial q(x) = (x− 2)2(x+ 3).

2. A is 7× 7 with q(x) = (x2 + 1)(x− 7).

Also find all such forms when A ∈ Mn(C) instead, and find all possible Jordan CanonicalForms over C.

17.10 Problem Set Ten 231


Problem 17.10.6 (Hungerford 7.5.2)Show that if ϕ is an endomorphism of a free k-module E of finite rank, then pϕ(ϕ) = 0.Hint: If A is the matrix of ϕ and B = xIn −A then

BaB = |B|In = pϕIn ∈Mn(k[x]).

If E is a k[x]-module with structure induced by ϕ, and ψ is the k[x]-module endomorphismE → E with matrix given by B, then

ψ(u) = xu− ϕ(u) = ϕ(u)− ϕ(u) = 0 ∀u ∈ E.


1. Let ϕ,ψ be endomorphisms of a finite-dimensional vector space E such that ϕψ = ψϕ.Show that if E has a basis of eigenvectors of ψ, then it has a basis of eigenvectors forboth ψ and ϕ simultaneously.

2. Interpret the previous part as a statement about matrices similar to a diagonal matrix.


Problem 17.10.8LetM ∈M5(R) be a 5×5 square matrix with real coefficients defining a linear map L : R5 → R5.Assume that when considered as an element of M5(C), then the scalars 0, 1 + i, 1 + 2i areeigenvalues of M .

1. Show that the associated linear map L is neither injective nor surjective.

2. Compute the characteristic polynomial and minimal polynomial of M .

3. How many fixed points can L have?(That is, how many solutions are there to the equation L(v) = v with v ∈ R5?)

Problem 17.10.9Let n be a positive integer and let B denote the n× n matrix over C such that every entry is1. Find the Jordan normal form of B.

Problem 17.10.10Suppose that V is a 6-dimensional vector space and that T is a linear transformation on Vsuch that T 6 = 0 and T 5 6= 0.

1. Find a matrix for T in Jordan Canonical form.

2. Show that if S, T are linear transformations on a 6-dimensional vector space V whichboth satisfy T 6 = S6 = 0 and T 5, S5 6= 0, then there exists a linear transformation Afrom V to itself such that ATA−1 = S.

17.10 Problem Set Ten 232

17 Bibliography

Bibliography

[1] David Steven. Dummit and Richard M. Foote. Abstract algebra. John Wiley and Sons, 2004.[2] Kenneth Hoffman and Ray Kunze. Linear Algebra. Prentice Hall, 1981.[3] Thomas W. Hungerford. Algebra. Springer, 2008.[4] Roy Smith. Algebra Notes by Roy Smith. url: https://www.math.uga.edu/directory/

people/roy-smith.

Bibliography 233

https://www.math.uga.edu/directory/people/roy-smith

https://www.math.uga.edu/directory/people/roy-smith

Documents

D.ZackGarza · 2021. 2. 21. · Contents 7.2.13 NonstandardTopics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .68 8.1 RingTheory