Computer Algebra - Fachbereich Mathematikboehm/lehre/1213_CA/ca.pdf · 2017. 4. 30. · Computer Algebra Lecture Notes, Summer Term 2017 Janko B ohm April 30, 2017

Computer Algebra

Lecture Notes, Summer Term 2017

Janko Bohm

April 30, 2017

Contents

1 Introduction 11.1 What is computer algebra and why should we do it? 11.2 Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4 Symbolic Integration . . . . . . . . . . . . . . . . . . . . 151.5 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . 181.6 Algebraic Geometry . . . . . . . . . . . . . . . . . . . . 191.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2 Ideals, varieties and Grobner bases 302.1 Ideals and varieties . . . . . . . . . . . . . . . . . . . . . 302.2 Introduction to the ideal membership problem and

Grobner bases . . . . . . . . . . . . . . . . . . . . . . . . 382.3 Monomial orderings . . . . . . . . . . . . . . . . . . . . 432.4 Division with remainder and Grobner bases . . . . . . 492.5 Computing Grobner bases . . . . . . . . . . . . . . . . 562.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3 Computing with ideals and algebraic sets 633.1 Sums and intersections of ideals . . . . . . . . . . . . . 633.2 Elimination of variables . . . . . . . . . . . . . . . . . . 653.3 Projection and elimination . . . . . . . . . . . . . . . . 673.4 Polynomial maps between algebraic sets . . . . . . . . 693.5 Rational maps between varieties . . . . . . . . . . . . . 743.6 Ideal quotients . . . . . . . . . . . . . . . . . . . . . . . 813.7 Solving algebraic systems with a finite set of solutions 843.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

4 Modules over principal ideal domains 904.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . 904.2 The elementary divisor algorithm . . . . . . . . . . . . 914.3 Modules and presentations . . . . . . . . . . . . . . . . 98

1

CONTENTS 2

4.4 Finitely generated modules over principal ideal do-mains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

4.5 Fundamental theorem of finitely generated abeliangroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

4.6 The Jordan normal form . . . . . . . . . . . . . . . . . 1094.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

5 Free resolutions and invariants 1155.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . 1155.2 Free resolutions . . . . . . . . . . . . . . . . . . . . . . . 1165.3 Projective algebraic sets and graded ideals . . . . . . 1185.4 Graded modules and the Hilbert function . . . . . . . 1275.5 Graded free resolutions . . . . . . . . . . . . . . . . . . 1305.6 Construction of the Hilbert polynomial . . . . . . . . 1325.7 Monomial orderings for modules . . . . . . . . . . . . . 1365.8 Division with remainder and Grobner bases for mod-

ules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1395.9 Computing Grobner bases of submodules . . . . . . . 1445.10 Buchberger’s criterion and computation of syzygy

modules . . . . . . . . . . . . . . . . . . . . . . . . . . . 1465.11 Computing free resolutions . . . . . . . . . . . . . . . . 1535.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

6 Computing with modules 1656.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . 1656.2 Computing kernels . . . . . . . . . . . . . . . . . . . . . 1656.3 Computing subquotients . . . . . . . . . . . . . . . . . 1676.4 Hom and the Hilbert scheme . . . . . . . . . . . . . . . 1706.5 Computing Hom . . . . . . . . . . . . . . . . . . . . . . 1726.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

7 Primary decomposition 1827.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . 1827.2 Irreducible decomposition of monomial ideals . . . . . 1827.3 Existence of primary decomposition . . . . . . . . . . . 1867.4 Algorithm for primary decomposition of monomial

ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1887.5 Uniqueness of primary decomposition . . . . . . . . . 1907.6 Computing primary decomosition in general . . . . . 1937.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

CONTENTS 3

8 Normalization 1958.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . 1958.2 Basics on normalization . . . . . . . . . . . . . . . . . . 1968.3 Finiteness of normalization . . . . . . . . . . . . . . . . 2008.4 Grauert-Remmert criterion . . . . . . . . . . . . . . . . 2028.5 Normalization algorithm . . . . . . . . . . . . . . . . . 2058.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

List of Figures

1.1 Gaussian elimination for the intersection of two lines 21.2 Buchberger’s algorithm for the intersection of two

ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Hyperboloids and double cone . . . . . . . . . . . . . . 51.4 Kummer quartic . . . . . . . . . . . . . . . . . . . . . . 51.5 Tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . 61.6 Togliatti quintic . . . . . . . . . . . . . . . . . . . . . . 61.7 Barth sextic . . . . . . . . . . . . . . . . . . . . . . . . . 71.8 Icosahedron . . . . . . . . . . . . . . . . . . . . . . . . . 71.9 Octahedron . . . . . . . . . . . . . . . . . . . . . . . . . 121.10 Generic point on octahedron. . . . . . . . . . . . . . . 131.11 Harmonic oscillator . . . . . . . . . . . . . . . . . . . . 171.12 Solution to the harmonic oscillator . . . . . . . . . . . 171.13 Graph of a rational function. . . . . . . . . . . . . . . . 201.14 Elliptic curve . . . . . . . . . . . . . . . . . . . . . . . . 211.15 Group structure on an elliptic curve. . . . . . . . . . . 221.16 Elliptic curve over F7 . . . . . . . . . . . . . . . . . . . 231.17 Icosahedron with numbering of the vertices. . . . . . . 28

2.1 Elliptical arc . . . . . . . . . . . . . . . . . . . . . . . . 312.2 Reducible affine algebraic set . . . . . . . . . . . . . . . 382.3 Twisted cubic . . . . . . . . . . . . . . . . . . . . . . . . 392.4 Surface containing the twisted cubic . . . . . . . . . . 392.5 A convex hull . . . . . . . . . . . . . . . . . . . . . . . . 482.6 D> for the ordering lp . . . . . . . . . . . . . . . . . . . 482.7 Grobner fan of the line. . . . . . . . . . . . . . . . . . . 49

3.1 Four points as the intersection of two pairs of lines . 653.2 Projection of the twisted cubic . . . . . . . . . . . . . . 673.3 Projection of a hyperbola . . . . . . . . . . . . . . . . . 683.4 Whitney umbrella . . . . . . . . . . . . . . . . . . . . . 703.5 Steiner surface over the reals . . . . . . . . . . . . . . . 723.6 Rational parametrization of the circle. . . . . . . . . . 77

4

LIST OF FIGURES i

3.7 Node . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.8 Ideal quotient . . . . . . . . . . . . . . . . . . . . . . . . 843.9 Curve with triple point . . . . . . . . . . . . . . . . . . 88

5.1 Projective space P2(R). . . . . . . . . . . . . . . . . . . 1195.2 Mapping A2(R) into P2(R) by stereographic projec-

tion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1215.3 Parabola x − y2 = 0 in A2(R). . . . . . . . . . . . . . . 1225.4 Projective parabola . . . . . . . . . . . . . . . . . . . . 1235.5 Projective parabola as a subset of the unit disc. . . . 1245.6 Hilbert function of K[x0, x1, x2]. . . . . . . . . . . . . . 1335.7 Hilbert function of R(−2)⊕R(−3) for R =K[x0, x1, x2].134

6.1 Deformation of three lines into an elliptic curve (realpicture). . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

6.2 A tangent to a conic . . . . . . . . . . . . . . . . . . . . 171

7.1 Simplicial complex with three maximal faces. . . . . . 194

List of Symbols

Sn Symmetric group on n symbols . . . . . . 5Z Integers . . . . . . . . . . . . . . . . . . . . 8gcd Greatest common divisor . . . . . . . . . . 9π(x) Density of prime numbers . . . . . . . . . 10Stab(j) Stabilizer of j . . . . . . . . . . . . . . . . . 12Gx Orbit of x under the action of G . . . . . 13LC(f) Lead coefficient of f , linear case . . . . . . 18L(f) Lead monomial of f , linear case . . . . . . 18spoly(f, g) S-pair of f and g, linear case . . . . . . . . 18An(K) Affine space of dimension n over the field

K . . . . . . . . . . . . . . . . . . . . . . . . 19V (f1, ..., fn) Affine algebraic set define by f1, ..., fn . . 20Γ(g) Graph of g . . . . . . . . . . . . . . . . . . . 20E(Q) Q-rational points of E . . . . . . . . . . . . 22Fp Finite field with p elements . . . . . . . . . 22F×p Prime residue class group of Fp . . . . . . 22

(cp) Legendre symbol . . . . . . . . . . . . . . . 24

(cn) Jacobi symbol . . . . . . . . . . . . . . . . . 25

Fn n-th Fermat prime . . . . . . . . . . . . . . 27⟨S⟩ Ideal generated by the set S . . . . . . . . 30I(S) Ideal of the set S . . . . . . . . . . . . . . . 31S Zariski closure of S . . . . . . . . . . . . . 31deg(f) Degree of the polynomial f , univariate case 34LC(f) Lead coefficient of f , univariate case . . . 34LT(f) Lead term of f , univariate case . . . . . . 34L(f) Lead monomial of f , univariate case . . . 34Spec(R/I) Spectrum of R/I . . . . . . . . . . . . . . . 35√

(I) Radical of I . . . . . . . . . . . . . . . . . . 36d(r) Euclidean norm of r . . . . . . . . . . . . . 40lp Lexicographical ordering . . . . . . . . . . 45dp Degree reverse lexicographical ordering . . 45ls Negative lexicographical ordering . . . . . 45L(f) Leading monomial of f . . . . . . . . . . . 46

ii

LIST OF SYMBOLS iii

LC(f) Leading coefficient of f . . . . . . . . . . . 46LT(f) Leading term of f . . . . . . . . . . . . . . 46>w Weighted degree ordering with weight vec-

tor w . . . . . . . . . . . . . . . . . . . . . . 47D> Characteristic set of > . . . . . . . . . . . . 47convHull(D) Convex hull of D . . . . . . . . . . . . . . . 47inw(f) Initial term of f with respect to the weight

vector w . . . . . . . . . . . . . . . . . . . . 49L(G) Leading ideal of G with respect to a fixed

monomial ordering . . . . . . . . . . . . . . 51NF(−,G) Normal form modulo G . . . . . . . . . . . 51NF Normal form . . . . . . . . . . . . . . . . . 51lcm Least common multiple . . . . . . . . . . . 56spoly(f, g) S-polynomial of f and g . . . . . . . . . . . 56Im Elimination ideal of I eliminating the first

m variables . . . . . . . . . . . . . . . . . . 67πm Projection forgetting the first m coordi-

nates . . . . . . . . . . . . . . . . . . . . . . 67im(ϕ) Image of ϕ . . . . . . . . . . . . . . . . . . . 69ker(f) Kernel of the ring homomorphism f . . . 73quot(R) Quotient field of R . . . . . . . . . . . . . . 78D(ϕ) Domain of definition of ϕ . . . . . . . . . . 78I ∶ J Ideal quotient of I by J . . . . . . . . . . . 81U1 ⊕ ...⊕Un Direct sum of the submodules Ui . . . . . 105pA Minimal polynomial of A . . . . . . . . . . 110χA Characteristic polynomial of A . . . . . . 110J (λ, e) Jordan block e × e with eigenvalue λ . . . 112L(v) Leading monomial of a vector v . . . . . . 138LC(v) Leading coefficient of a vector v . . . . . . 138LT(v) Leading term of a vector v . . . . . . . . . 138NF(−,G) Normal form modulo G . . . . . . . . . . . 140NF Normal form . . . . . . . . . . . . . . . . . 140L(G) Leading module of G with respect to a

fixed monomial ordering . . . . . . . . . . 142lcm Least common multiple of monomials in

a free module . . . . . . . . . . . . . . . . . 144spoly(f, g) S-polynomial of vectors f and g . . . . . . 144Syz(G) Syzygy module of G . . . . . . . . . . . . . 147SG Syzygy matrix of G . . . . . . . . . . . . . 147s(gi, gj) Syzygy induced by a successful Buchberger

test NF(spoly(gi, gj),G) = 0 . . . . . . . . 148>G Schreyer ordering induced by > and G . . 148

LIST OF SYMBOLS iv

subquot(A,B) Subquotient with generators A and rela-tions B . . . . . . . . . . . . . . . . . . . . . 167

HomR(M,N) Module of R-homomorphisms M → N . . 170Hn Hilbert scheme of subschemes of Pn . . . . 171Hn,P Hilbert scheme of subschemes of Pn with

Hilbert polynomial P . . . . . . . . . . . . 171Ass(I) Associated primes of I . . . . . . . . . . . 192Min(I) Minimal associated primes of I . . . . . . 192I∆ Stanley-Reisner ideal of ∆ . . . . . . . . . 193A Normalization of Noetherian domain A . 195X Normalization of variety X . . . . . . . . . 195S−1A Localization of A with denominators in S 197AP Localization of A at a prime P . . . . . . 197N(A) Non-normal locus of A . . . . . . . . . . . 198V(J) Scheme theoretic vanishing locus of J . . 198I(Y ) Scheme theoretic zero ideal of Y . . . . . . 198CA Conductor of A . . . . . . . . . . . . . . . . 198Jac(I) Jacobian ideal of I . . . . . . . . . . . . . . 199Sing(A) Singular locus of A . . . . . . . . . . . . . . 199

LIST OF SYMBOLS v

1

Introduction

1.1 What is computer algebra and why

should we do it?

Computer algebra has two fundamental goals: Provide algorithmsfor computations with algebraic structures, like fields, vector spaces,rings, ideals, and modules to the computer. And use the algorithmsand their implementations to solve mathematical problems in the-ory and applications. Here, computations usually refer to exact,that is, symbolic ones. However, in some cases, numerical compu-tations can be helpful in obtaining exact results.

So why is it useful to implement algebra in the computer? Ofcourse, there are practical problems, that can be solved by com-puter algebra, for example, in cryptography, robotics, algebraicstatistics, computational biology, and physics. On the other hand,experiments with the computer allow you to get an insight into the-oretical problems and test conjectures. In many settings, you caneven obtain theoretical results by handling just a single special caseby computer. Let us say, we want to prove that the determinant ofthe matrix

At =⎛⎜⎝

t − 1 1 −1t t2 + 1 t + 1t t2 t + 2

⎞⎟⎠∈ C[t]3×3

is non-zero as a polynomial without computing it. For example,the determinant may be too complicated (which is of course notthe case in the example). However, it may be possible to computeAt0 for a fixed t0 ∈ C. Since substitution is a ring homomorphism,

1

1. INTRODUCTION 2

it is sufficient to find one t0 such that detAt0 ≠ 0, for example,

detA0 = det⎛⎜⎝

−1 1 −10 1 10 0 2

⎞⎟⎠= −2

It follows that the determinant, as a continuous function, will benon-zero in an open neighbourhood of t0. In fact, we then knowthat it is non-zero for all but finitely many values of t, since 0 ≠detAt ∈ C[t] has only finitely many zeros. This means that thedeterminant is non-zero on an open set in the so called Zariskitopology, that is, on the complement of the zero set of a system ofpolynomial equations.

In the line of this example, our main focus will be on computa-tions in commutative algebra, specifically on all sorts of algorithmsconcerned with polynomial rings. Here, the fundamental buildingblock is Buchberger’s algorithm for computing Grobner bases,which generalizes Gaussian elimination. Recall that Gaussianelimination transforms multivariate linear systems of equations intorow echelon form

2x + y = 1x + 2y = −1

↦2x + y = 1

−3x = −3

(see Figure 1.1), from which we can read off the solution (x, y) =(1,−1).

Figure 1.1: Gaussian elimination for the intersection of two lines

Buchberger’s algorithm generalizes this idea to higher degreepolynomial equations, for example, it transforms

2x2 − xy + 2y2 − 2 = 02x2 − 3xy + 3y2 − 2 = 0

↦3y + 8x3 − 8x = 04x4 − 5x2 + 1 = 0

For most computations we will use the open source computer alge-bra system Singular [4], which is being developed at TU Kaisers-

1. INTRODUCTION 3

lautern. Singular can be either downloaded or conveniently ac-cessed in an online interface.1

Example 1.1.1 We can do the above Grobner basis calculation inSingular by the following code:ring R=0,(y,x),lp;

ideal I = 2x2-xy+2y2-2, 2x2-3xy+3y2-2;

std(I);

[1]=4x4-5x2+1

[2]=3y+8x3-8x

The ring definition specifies the characteristic of the prime field (so0 corresponds to Q), the variables, and an ordering of the variables(lp). To make an analogy to Gaussian elimination, by the orderingyou can tell the system, in which order you want to eliminate thevariables (for example, if you want a right or left upper triangu-lar matrix as row echelon form). An ideal represents a system ofpolynomial equations, and std refers to the term standard basis,which, in the setup considered here, is synonymous to Grobner ba-sis. From the resulting system we can read off the four solutions(x, y) = (±1,0) , (±1

2 ,±1), see Figure 1.2.

Figure 1.2: Buchberger’s algorithm for the intersection of two el-lipses

Example 1.1.2 These plots can be generated in the general pur-pose (commercial) computer algebra system Maple [12] by the codewith(plots):

p1:=implicitplot(2*x^2-x*y+2*y^2-2,x=-2..2,y=-2..2):

p2:=implicitplot(2*x^2-3*x*y+3*y^2-2,x=-2..2,y=-2..2):

1See https://www.singular.uni-kl.de:8003/

https://www.singular.uni-kl.de:8003/

1. INTRODUCTION 4

display(p1,p2,view=[-2..2,-2..2],thickness=2);

p3:=implicitplot(4*x^4-5*x^2+1,x=-2..2,y=-2..2):

p4:=implicitplot(2*x^2-3*x*y+3*y^2-2,x=-2..2,y=-2..2):

display(p3,p4,view=[-2..2,-2..2],thickness=2);

As already done in these two plots, we will motivate the alge-braic concepts by connecting multivariate systems of polynomialequations to the geometry of their set of solutions, the associatedalgebraic variety. This connection is called algebraic geome-try, an important branch of mathematics and one of the key appli-cations of commutative algebra. Of course algebraic varieties canget much more interesting than 4 points. First of all, they canhave higher dimension, for example, the ellipses considered abovehave dimension 1 and are called curves. One algebraic equationin three variables will give you an example of what is called analgebraic surface, that is, a variety of dimension 2. We will learnabout an algebraic characterization of the dimension of an algebraicvariety using Grobner bases. Another interesting feature algebraicvarieties can have are singularities, which are points that do nothave a well-defined tangent plane. Considering as an example thehyperboloids Ht ⊂ R3 given

x2 + y2 − z2 + t = 0

varying with the parameter t ∈ R, the tangent plane at the point(x, y, z) is given by the affine space

(x, y, z) + ker(x, y,−z),

since (x, y,−z) is the normal vector of the surface at that point.So every point of Ht has a well-defined tangent plane, providedt ≠ 0. However, if t = 0, then (0,0,0) lies on the hyperboloid (whichnow becomes a double cone) and at this point the dimension of thekernel is not 2 but 3. So Ht is non-singular for t ≠ 0, but has asingularity at (0,0,0) for t = 0, see Figure 1.3. These plots weregenerated by the visualization tool Surfer [8], which provides aneasy-to-use graphical interface to the raytracing program Surf [6]for plotting algebraic surfaces in 3-space. It can also be called fromSingular.

Using this tool, also Figure 1.4 was generated, which shows theKummer quartic surface in 3-space, which has the maximumnumber 16 of isolated singularities, a surface given by polynomialof degree 4 in 3 variables can have.

If you look closely, you see that the configuration of singularitiesof the Kummer quartic is invariant under the symmetry group of the

1. INTRODUCTION 5

Figure 1.3: Hyperboloids and double cone

Figure 1.4: Kummer quartic

tetrahedron (which is S4), see Figure 1.5. The Togliatti quinticshown in Figure 1.6 has the maximum number of 31 singularities,which a degree 5 surface can have and comes with the symmetryof a 5-gon. Again, the Barth sextic has the maximum number of65 singularities on a degree 6 surface, see Figure 1.7. It comes withthe symmetry of an icosahedron, see Figure 1.8.

Before turning to a short summary of the algebraic and geomet-ric foundations necessary for the things to come, we take a quicklook of what else is out there in computer algebra. This appe-tizer to computer algebra in number theory, group theory, calculusand algebraic geometry will hopefully stimulate the appetite at thebeginning of the meal.

1. INTRODUCTION 6

Figure 1.5: Tetrahedron

Figure 1.6: Togliatti quintic

1. INTRODUCTION 7

Figure 1.7: Barth sextic

Figure 1.8: Icosahedron

1. INTRODUCTION 8

1.2 Numbers

One of the most important algorithms in mathematics is Euclideanalgorithm for finding the greatest common divisor. In a generalizedform, it will be present explicitly or implicitly in many algorithmswe will discuss later on. So let us invest a little bit of time in areminder on the remainder.

Lemma 1.2.1 (Division with remainder) For a, b ∈ Z, b ≠ 0,there are q, r ∈ Z with

a = b ⋅ q + r

and 0 ≤ r < ∣b∣.

Proof. Without loss of generality b > 0. The set

{w ∈ Z ∣ b ⋅w > a} ≠ ∅

has a smallest element w. Then set

q ∶= w − 1 r ∶= a − qb

1. INTRODUCTION 9

Theorem 1.2.2 (Euclidean Algorithm) Suppose a1, a2 ∈ Z/ {0}.Successive division with remainder terminates

a1 = q1a2 + a3

⋮

aj = qjaj+1 + aj+2

⋮

an−2 = qn−2an−1 + an

an−1 = qn−1an + 0

andgcd (a1, a2) = an.

Reading the equations backwards

an = an−2 − qn−2an−1

⋮

a3 = a1 − q1a2

gives a representation

gcd (a1, a2) = x ⋅ a1 + y ⋅ a2

with x, y ∈ Z.

Proof. We have ∣ai+1∣ < ∣ai∣ for i ≥ 2 so after finitely many stepsai = 0. Then an divides an−1, hence also an−2 = qn−2an−1 + an andinductively an−2, ..., a1. If t is a divisor of a1 and a2, then also ofa3, ..., an.

Example 1.2.3 We compute the gcd of 36 and 15:

36 = 2 ⋅ 15 + 6

15 = 2 ⋅ 6 + 3

6 = 2 ⋅ 3 + 0

hence gcd (36,15) = 3. Furthermore, we can express gcd (36,15) asa Z linear combination of 36 and 15:

3 = 15 − 2 ⋅ 6 = 15 − 2 ⋅ (36 − 2 ⋅ 15) = 5 ⋅ 15 + (−2) ⋅ 36

Many other number theoretic concepts will show up later in amore general context, for example, factorization in the context ofprimary decomposition.

1. INTRODUCTION 10

Definition 1.2.4 An element p ∈ Z>1 is called prime number, ifp = a ⋅ b, a, b ∈ Z≥1 implies a = 1 or b = 1.

Theorem 1.2.5 (Fundamental theorem of arithmetic) Everynumber n ∈ Z/ {0,1,−1} has a unique representation

n = ±1 ⋅ pr11 ⋅ ... ⋅ prrr

with prime factors p1 < ... < pr and ri ∈ N.

Algorithm 1.2.6 (Trial division) Let n ∈ Z be composite. Thesmallest prime factor p of n satisfies

p ≤m ∶= ⌊√n⌋ .

If we know all primes p ≤m, then we can test p ∣ n by division withremainder and, hence, factor n.

Algorithm 1.2.7 (Sieve of Eratosthenes) We can find all primenumbers smaller than n in the following way: Note all numbers from2 to n. Starting with p = 2, delete all a ⋅ p for a > 1, and continuewith the next largest number p which not has been deleted. Notethat p is prime, since it is not a multiple of smaller prime. Stop ifp >

√n.

We will discuss in Exercise 1.3 an analogue of this in the caseof univariate polynomials.

Example 1.2.8 We compute all primes ≤ 15

2 3 4 5 6 7 8 9 10 11 12 13 14 152 3 5 7 9 11 13 152 3 5 7 11 13

In the first step we delete all multiples of 2, in the second step allmultiples of 3. All remaining numbers are prime, since 5 >

√15.

One can even describe the distribution of the primes over allintegers:

Theorem 1.2.9 (Prime number theorem) For x ∈ R>0 let

π(x) = ∣{p ≤ x ∣ p ∈ N prime}∣

Then

limx→∞

π (x)x

ln(x)= 1

1. INTRODUCTION 11

In Exercise 1.4 we will test the prime number theorem.Computer algebra in this spirit has many theoretical applica-

tions in number theory and algebraic geometry and practical appli-cations, for example, in coding theory or RSA public key crypto-graphy.

In general, number theory explores the properties of numbers,most importantly the interaction of addition and multiplication.This leads to many problems which are easy to formulate, buthighly non-trivial to solve. The most famous one is Fermat’s lasttheorem of 1637: There is no (non-trivial) integer solution of theequation

xn + yn = zn

for n ≥ 3. With the help of a computer one can test Fermat’s lasttheorem for very large n (using the theoretical result that you onlyhave to test it for so called irregular primes). Fermat’s last theoremwas finally proven in 1995 (by A. Wiles) after 350 years of work ofmany people, which led to many new concepts in mathematics. Oneof the leading systems in number theory is Pari/gp, see [15]. Itcan calculate, for example, with p-adic numbers, which have playedan important role in the proof of Fermat’s last theorem. Like we dowith power series, we consider two numbers as close to each other,if they differ by a high power of a prime.

Example 1.2.10 Using the prime p = 7 up to powers less than 5we calculate in Pari/gp the square root of 569:gp > n=569 + O(7^5)

%1 = 2 + 4*7 + 4*7^2 + 7^3 + O(7^5)

gp > sqrt(n)

%2 = 3 + 4*7 + 7^2 + 3*7^3 + 7^4 + O(7^5)

gp > factor((3 + 4*7 + 7^2 + 3*7^3 + 7^4)^2-569)

%2 = [7 5] [733 1]

so we observe, that

(3 + 4 ⋅ 7 + 72 + 3 ⋅ 73 + 74)2− 569 = 12 319 531 = 733 ⋅ 75.

Also Maple has very efficient implementations of algorithms innumber theory:

Example 1.2.11 Using Maple you can easily show, that the n-thFermat number Fn = 22n + 1 is prime for 0 ≤ n ≤ 4 and compositefor 5 ≤ n ≤ 8, for example:> ifactor((2^(2^6)+1))

274177 * 67280421310721

1. INTRODUCTION 12

Using distributed computing one can take this a little bit further.However, it is unknown, whether there are more Fermat primesthan the 5 known ones. See also Exercise 1.4.

1.3 Groups

The concept of groups has important applications in almost everyfield of mathematics, including algebraic geometry and commuta-tive algebra. It allows us to describe symmetries in mathematicalobjects and problems, reducing a more complicated problem to asimpler one. From the practical point of view this can speed upcomputations.

Consider, for example, the symmetry group G of the octahedron(Figure 1.9), which contains all rotations, reflections, and rotoreflec-tions, which map the octahedron to itself. Numbering the vertices,

Figure 1.9: Octahedron

we can identify any symmetry with an element of the symmetricgroup S6. For example, the rotation by 90 degrees along the axisthrough 1 and 6 is given by (2,3,4,5) ∈ S6 using cycle notation.

Let us say, we want to compute the stabilizers of the vertices ofthe octahedron, that is, the subgroups

Stab(j) = {σ ∈ G ∣ σ(j) = j} .

By the action of G, it is sufficient to determine Stab(1), since allstabilizers can be identified by conjugation of groups

Stab(j) = σ−1 Stab(1)σ,

1. INTRODUCTION 13

where σ ∈ G with 1 = σ(j). So instead of 6 computations, by takingthe symmetry of the problem into account, we only have to do one.

To compute the group order of Stab(1), we use:

Theorem 1.3.1 (Orbit-counting theorem) Let G be a groupacting on a set X by

G ×X →X

(that is, ex = x and (g ○ h)x = g(hx) for all x ∈ X and g, h ∈ G).Fix x ∈X and write

Gx = {gx ∣ g ∈ G}

for the orbit of x and

Stab(x) = {g ∈ G ∣ gx = x} .

for the stabilizer of x. Then

∣G∣ = ∣Gx∣ ⋅ ∣Stab(x)∣ .

Hence

∣Stab(1)∣ =∣G∣

6,

and, if p is a point which does not lie on any symmetry plane oraxis,

∣G∣ = ∣Stab(p)∣ ⋅ ∣Gp∣ = 1 ⋅ (6 ⋅ 8) = 48,

see Figure 1.10, so∣Stab(1)∣ = 8.

It is easy to guess elements of G and Stab(1), but how do we know,

Figure 1.10: Generic point on octahedron.

1. INTRODUCTION 14

whether they really generate the respective groups, equivalently,that they generate groups of the correct order? We can leave thistedious calculation to the open source computer algebra systemGap [14], the leading software for group theory:

Example 1.3.2 We use Gap to prove that

Stab(1) = ⟨(2,3,4,5), (2,4)⟩

gap> Stab1:=Group((2,3,4,5),(2,4));;

gap> Size(Stab1);

8gap> Elements(Stab1);

[(),(3,5),(2,3)(4,5),(2,3,4,5),(2,4),(2,4)(3,5),

(2,5,4,3),(2,5)(3,4)]

In the same way, we can find generators of the whole symmetrygroup:gap> G:=Group((2,3,4,5),(1,3)(5,6));;

gap> Size(G);

48

HenceG = ⟨(2,3,4,5), (1,3)(5,6)⟩ .

See Exercise 1.5 for the symmetry group of the icosahedron.Gap implements algorithms for computing with subgroups of

symmetric groups. As we have seen, given a set of generators, it candetermine the group order, and the set of all elements. It can alsodetermine, whether two such groups are isomorphic, and specifyan isomorphism. Computing inside symmetric groups seems like arestriction, however, this is sufficient for many purposes, becauseof the following:

Theorem 1.3.3 (Cayley) Every group G is isomorphic to a sub-group of the group S (G) of bijections G→ G.

Proof. The action of G on itself induced by the group operation

G ×GÐ→ G

(g, h)↦ g ○ h

yields a group homomorphism

ϕ ∶ G → S (G)

g ↦ (G → Gh ↦ g ○ h

)

1. INTRODUCTION 15

and sinceKer(ϕ) = {g ∈ G ∣ gh = h ∀h ∈ G} = {e}

ϕ is injective. So by the homomorphism theorem

G = G/Ker(ϕ) ≅ Im(ϕ) ⊂ S (G) .

For finite groups this just amounts to writing down the grouptable. For g ∈ G the bijection ϕ(g) maps the e-th row of the grouptable to the g-th row.

Example 1.3.4 For G = Z/3 = {0,1,2} with group table

+ 0 1 2

0 0 1 21 1 2 02 2 0 1

we haveϕ(1) = (0,1,2) ∈ S (G)

using cycle notation.

1.4 Symbolic Integration

In addition to the specialized systems like Singular, Gap andPari/gp, there are general purpose computer algebra systems likethe commercial systems Maple [12], and Mathematica [16], andthe open source systems Maxima [13], Reduce [10], and Axiom[2]. They are usually less powerful in the specific areas, but pro-vide a larger set of algorithms to manipulate symbolic expressions.Most noteably there is Risch’s algorithm for symbolic integra-tion of functions which are compositions of rational functions, ex-ponentials, logarithm, radicals, and trigonometric functions. It isbased on the theorem of Liouville, which uses the following defi-nitions: A differential field is field K with a differentiation mapK → K, f ↦ f ′, which satisfies the usual rules (f + g)′ = f ′ + g′

and (fg)′ = f ′g + fg′. Denote by Const(K) = {f ∈K ∣ f ′ = 0} itssubfield of constants. An elementary extension K ⊂ E is afield extension, which can be obtained by a sequence of extensions,which are either algebraic, exponential, that is, transcendental ofthe form F ⊂ F (g) with

g′ = g ⋅ f ′

1. INTRODUCTION 16

for some f ∈ K , or logarithmic, that is, transcendental of theform F ⊂ F (g) with

g′ =f ′

f

for some f ∈K.

Theorem 1.4.1 (Liouville) Let K be a differential field with fieldof constants C = Const(K), and F ∈ K. If there is an elementaryextension K ⊂ E and G ∈ E with

G′ = F

then there are constants c1, ..., cn ∈ C in an algebraic field extensionof C, and functions v ∈K, u1, ..., un ∈K(c1, ..., cn) such that

G = v +∑iai lnui.

This tells us, that if an elementary function has an elementaryintegral, then this integral can be written in the functions usedto write down the integrand and their logarithms. Based on thisobservation, Risch’s algorithm decides the existence of G, and com-putes G, if it exists. However note, that the only almost completeimplementation of Risch’s algorithm is that of Axiom.

Example 1.4.2 We have ∫ xndx = 1

n+1xn+1 provided n ≠ −1 and

∫1xdx = lnx. On the other hand, the function e−x

2does not have

an elementary integral.

Example 1.4.3 We compute the area A inside the ellipse

(x

a)

2

+ (y

b)

2

= 1

using Maple:F:=int(2*b/a*sqrt(a^2-x^2),x);

F:=b/a*sqrt(a^2-x^2)*x+a*b*arctan(x/sqrt(a^2-x^2))

and hence A = πab.

In this spirit, general purpose computer algebra systems areheavily used for solving differential equations and, hence are verypopular in physics (most noteably Reduce).

Example 1.4.4 Consider the damped harmonic oscillator describedby the ordinary differential equation (ODE)

x′′(t) = −a ⋅ x(t) − b ⋅ x′(t).

1. INTRODUCTION 17

Figure 1.11: Harmonic oscillator

Here t is the time, x(t) the position, x′(t) the speed, and x′′(t) theacceleration of the object, see Figure 1.11.

Choosing the parameters a = 5 and b = 1, and the boundaryconditions x(0) = 1 (starting position) and x′(0) = 1 (starting speed)we solve this differential equation using Maple:ode := diff(x(t),t$2)+diff(x(t),t)+5*x(t);d2

dt2x(t) +ddtx(t) + 5 ⋅ x(t)

dsolve(ode);

x(t) = C1e− 1

2t sin(

√192 t) +C2e

− 12t cos(

√192 t)

f:=subs({ C1=1, C2=1},subs(%,x(t)));f ∶= e−

12t sin(

√192 t) + e−

12t cos(

√192 t)

plot(f,t=0..5);

The plot command will generate Figure 1.12 showing the solutionx(t).

–1

0

1

1 2 3 4 5

Figure 1.12: Solution to the harmonic oscillator

1. INTRODUCTION 18

1.5 Linear Algebra

In addition to the Euclidean algorithm, the second key algorithmgeneralized by Buchberger’s algorithm is Gaussian reduction. Wecan reformulate Gaussian reduction on a homogeneous linear sys-tem of equations over a field K in the following way:

Algorithm 1.5.1 (Gauss) Consider non-zero homogeneous lin-ear polynomials f1, ..., fn ∈ K[x1, ..., xm]. Choose an ordering ofthe variables (without loss of generality x1 > x2 > ... > xm). DefineL(fi) as the largest monomial of fi and by LC(fi) its coefficient.

As long as there are fi and fj with L(fi) = L(fj) replace fj bythe S-pair

spoly(fi, fj) = LC(fi) ⋅ fj − LC(fj) ⋅ fi.

If fj = 0 then delete fj.Sort the set of fj by the size of L(fj).

This algorithm terminates with a row echelon form. If we re-duce the resulting polynomials by those with smaller lead monomi-als, and divide all polynomials by their lead coefficients, we obtainthe (unique) reduced row echelon form. When discussing Grobnerbases, we will see how the special case of linear equations generalizesto the higher degree setup.

Example 1.5.2 We solve the system

f1 = x1 + x2 + x5 = 0f2 = x1 + x2 + 2x3 + 2x4 + x5 = 0f3 = x1 + x2 + x3 + x4 + x5 = 0

Gaussian elimination yields

f1 = x1 + x2 + + x5 = 0spoly(f1, f2) = 2x3 + 2x4 = 0spoly(f1, f3) = x3 + x4 = 0

and, since the S-pair of the last two vanishes

x1 + x2 + + x5 = 02x3 + 2x4 = 0

Using matrix notation, we can do the Gaussian elimination inMaple by the following code:with(LinearAlgebra):

1. INTRODUCTION 19

A := <<1,1,1>∣<1,1,1>∣<0,2,1>∣<0,2,1>∣<1,1,1>>:GaussianElimination(A);⎡⎢⎢⎢⎢⎣

1 1 0 0 10 0 2 2 00 0 0 0 0

⎤⎥⎥⎥⎥⎦

Utilizing the Grobner basis engine of Singular, we can do thesame computation as follows:ring R = 0,(x(1..5)),lp;

ideal I = x(1) + x(2) + x(5),

x(1) + x(2) + 2*x(3) + 2*x(4) + x(5),

x(1) + x(2) + x(3) + x(4) + x(5);

option(redSB);

std(I);

[1] = x(3) + x(4)

[2] = x(1) + x(2) + x(5)

See also Exercise 1.6.

Remark 1.5.3 Solving an inhomogeneous linear system of equa-tions

fi =∑jaijxj − ci = 0

can be reduced to the homogeneous case by homogenizing the systemto

fi =∑jaijxj − ciy = 0

with a homogenization variable y. See also Exercise 1.7. We willcome back to this idea later in the context of higher degree polyno-mial equations.

1.6 Algebraic Geometry

If we pass from linear systems to polynomial equations of higherdegree things become much more interesting. Algebraic geometrystudies the set of solutions of such systems. It will be the mainmotivation of the algorithms we encounter in commutative algebra.

Definition 1.6.1 The affine space of dimension n over the fieldK is defined as

An(K) =Kn

By this notation we express that we (usually) do not care aboutthe structure of Kn as a K-vector space.

1. INTRODUCTION 20

Definition 1.6.2 An affine algebraic set is the common zeroset

V (f1, ..., fr) = {p ∈ An(K) ∣ f1(p) = 0, ..., fr(p) = 0}

of polynomials f1, ..., fr ∈K[x1, ..., xn].An affine algebraic set X is called irreducible, if it cannot be

written as X =X1 ∪X2 with affine algebraic sets Xi ⫋X. Then wealso call X an affine algebraic variety.

Example 1.6.3 Affine algebraic varieties commonly known alsooutside algebraic geometry are V (1) = ∅, V (0) = Kn, the set ofsolutions of a linear system of equations

A ⋅ x − b = 0

or the graph

Γ (g) = V (x2 ⋅ b (x1) − a (x1)) ⊂ A2(K)

of a rational function

g =a

b∈K(x1).

For example, the graph of g(x1) =x31−1

x1is

V (x2x1 − x31 + 1) ⊂ A2(K),

as depicted in Figure 1.13 for K = R.

Figure 1.13: Graph of a rational function.

1. INTRODUCTION 21

Example 1.6.4 If f ∈ K[x1, ..., xn] is non-zero and non-constant,then V (f) ⊂ An(K) is called a hypersurface. For example, thesurfaces shown in the Figures 1.4, 1.6, and 1.7 are hypersurfacesin A3(R).

Note, that V (f) is irreducible, if and only if f is irreducible. InExercise 1.10 we will discuss, in the case that K is a finite field, aprobabilistic test for checking whether f is irreducible.

For the remainder of this section we will take a closer look at el-liptic curves. After that, we will discuss the algorithmic foundationfor computating with algebraic sets in the next chapter.

Example 1.6.5 Not every curve in A2(K) is a graph, for example,the circle

V (x21 + x

22 − 1) ⊂ A2(R)

or the elliptic curve

V (x22 − x

31 − x

21 + 2x1 − 1) ⊂ A2(R)

are not, see Figure 1.14.

Figure 1.14: Elliptic curve

Example 1.6.6 In general, an elliptic curve is of the form E =V (f) where f ∈ K[x1, x2] is an irreducible polynomial of degree

3 and V (f, ∂f∂x ,∂f∂y) = ∅, which guarantees that the curve does not

have a singularity. The set of points of an elliptic curve has the

1. INTRODUCTION 22

Figure 1.15: Group structure on an elliptic curve.

following remarkable property: It comes with the structure of anabelian group with the operation as specified in Figure 1.15. Canyou tell what is the neutral element?

Example 1.6.7 If we consider elliptic curves over the field Q

E(Q) = {p ∈ A2(Q) ∣ f (p) = 0}

(also called the Q-rational points of the elliptic curve over C),we close the circle back to number theory: The Theorem of Mordellshows, that E(Q) as a group is finitely generated. The structure ofa finitely generated group is known: It is the product of factors Zand Z/n (that is, of cyclic groups). The question about the numberof Z-factors of E(Q) is the topic of ongoing research (for example,on the conjecture of Birch and Swinnerton-Dyer, which in fact isbased on computer experiments, and is one of the one million dollarproblems).

An important application of elliptic curves is public-key crypto-graphy. Given a crypto system based on the discrete logarithm inthe prime residue class group F×p (like Diffie-Hellman key exchange),one can replace F×p by the group E (Fp) ⊂ A2(Fp) of points of anelliptic curve over a finite field Fp. This allows one to achieve thesame strength with a much smaller key size.

One can show that (for p ≥ 5) in a suitable coordinate systemevery elliptic curve can be written as

E (Fp) = {(x, y) ∈ A2(Fp) ∣ f (x, y) = 0} ∪ {O}

1. INTRODUCTION 23

Q

P

P+Q

-P-Q

–3

–2

–1

0

1

2

3

–3 –2 –1 0 1 2 3

Figure 1.16: Elliptic curve over F7

with a polynomial

f = (x3 + a ⋅ x + b) − y2 ∈ Fp [x, y]

with 4a3+27b2 ≠ 0 modp. HereO denotes the point at infinity, whichlies on any line parallel to the y-axis and is the neutral element ofthe group.

Example 1.6.8 Figure 1.16 depicts in black the points of the curve

x3 − 2x + 1 − y2 = 0

over F7 (where y is the vertical axis). The grey dots are the pointsof the line

x − 2y − 1 = 0

through the points P,Q ∈ E (F7). The sum P + Q is obtained asthe reflection of the third point of intersection − (P +Q) of the linewith the curve. Note: If we substitute x = 2y + 1 into f , we obtaina polynomial of degree 3 in one variable

y3 + 4y2 + 2y = 0

which has 3 zeros y = 0,1,2 corresponding to the 3 points of inter-section (1,0), (3,1) and (−2,2). This is a special case of Bezout’stheorem.

1. INTRODUCTION 24

In the context of cryptography it is of course important to knowthe group order (that is, the number of points) of E (Fp). Weanswer this question using the computer algebra system Magma,which is the only closed source system devoted to algebraic geome-try, number theory and combinatorics. The curve is specified by thevector [0,0,0, a, b]F := FiniteField(7,1);

E := EllipticCurve([Zero(F),Zero(F),Zero(F),-2,1]);

E;

Elliptic Curve defined by y^2 = x^3 + 5*x +1 over GF(7)

#E;

12

Note that this includes O, not visible in Figure 1.16. Of course wecan also do larger examples:F := FiniteField(NextPrime(10^30),1);

E := EllipticCurve([Zero(F),Zero(F),Zero(F),-2,1]);

#E;

999999999999999242601621761120

How does this work? Let p be an odd prime. The Legendresymbol is defined for c ∈ Z by

(c

p) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 if y2 ≡ cmodp is solveable and p ∤ c−1 if y2 ≡ cmodp is not solveable0 if p ∣ c

With this notation, given x ∈ Fp we have

(x3 + a ⋅ x + b

p) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 ⇔ if there are 2 points (x,−) ∈ E (Fp)−1 ⇔ if there is no point (x,−) ∈ E (Fp)0 ⇔ if there is 1 point (x,−) ∈ E (Fp)

that is, there are exactly

1 + (x3 + a ⋅ x + b

p)

points of the form (x,−) ∈ E (Fp). Hence we obtain (not forgettingthe point at infinity O):

Theorem 1.6.9 On the elliptic curve

E (Fp) = {(x, y) ∈ F2p ∣ x3 + a ⋅ x + b = y2} ∪ {O}

there are exactly

∣E (Fp)∣ = p + 1 + ∑x∈Fp

(x3 + a ⋅ x + b

p)

points.

1. INTRODUCTION 25

See also Exercise 1.8. Theorem 1.6.9 immediately implies:

Corollar 1.6.10 For the number of points of an elliptic curve E (Fp)it holds

1 ≤ ∣E (Fp)∣ ≤ 2p + 1

More precisely there is the following theorem (which we cannotprove here):

Theorem 1.6.11 (Hasse) The number of points of an elliptic curveE (Fp) satisfies

p + 1 − 2√p ≤ ∣E (Fp)∣ ≤ p + 1 + 2

√p

One can show, that all possible values are indeed attained. InExercise 1.9 we verify this theorem for a finite set of primes.

To compute ∣E (Fp)∣ it remains to evaluate the Legendre symbol,which one can do as follows. To obtain an easier algorithm, we firstgeneralize the Legendre symbol:

Definition 1.6.12 If n = p1 ⋅ ... ⋅ pr with odd primes pi and c ∈ Z,then define the Jacobi symbol as

(c

n) ∶=

r

∏i=1

(c

pi).

Note, that Singular (like most other computer algebra sys-tems) has a command to compute the Jacobi symbol.

Remark 1.6.13 If b ≡ cmodn, then

(b

n) = (

c

n).

For any b and c

(b ⋅ c

n) = (

b

n)(c

n).

Theorem 1.6.14 (Law of quadratic reciprocity) If b, c > 0 areodd, then

(b

c) = (−1)

(b−1)(c−1)4 (

c

b)

and

(2

c) = (−1)

c2−18 .

1. INTRODUCTION 26

A proof can be found in most books on algebra or number the-ory. The formulas of Remark 1.6.13 and Theorem 1.6.14 yield arecursive algorithm to compute the Jacobi symbol.

Example 1.6.15 We apply this algorithm:

(55

103) = −(

103

55) = −(

48

55) = −(

2

55)

4

(3

55)

= (55

3) = (

1

3) = 1

Hypersurfaces X = V (f) ⊂ An(K), as considered above, aregiven by a single non-constant equation f ∈ K[x1, ..., xn]. We alsospeak of an algebraic set of codimension 1. The set of 4 points inA2(R) we considered in Figure 1.2 was given by two equations

2x2 − xy + 2y2 − 2 = 02x2 − 3xy + 3y2 − 2 = 0

and has codimension 2. Once we give a precise definition of thedimension dimX of a variety X, we can define the codimension asthe difference

codimX = n − dimX.

We will see that, in general, the codimension can differ from thenumber of equations. As in the case of linear equations and vectorspaces, the first step is to write down an independent set of equa-tions, that is, a Grobner basis. Even then, the codimension can besmaller than the number of equations, however, we will describe analgorithm to compute the dimension (and hence the codimension)from a Grobner basis.

1.7 Exercises

Exercise 1.1 Install Singular, Gap, Pari/Gp, and Surferand try the examples featured in Chapter 1. Check out the onlinehelp pages.

Exercise 1.2 1) Implement the Euclidian Algorithm for com-puting the greatest common divisor in Z. Test your imple-mentation at examples.

2) Use your implementation to cancel

90189116021

18189250063.

1. INTRODUCTION 27

Exercise 1.3 Let p be a prime and Fp = Z/p the field with p ele-ments.

1) Use an analogue of the sieve of Eratosthenes to find all irre-ducible polynomials in F2[x] of degree ≤ 3.

2) Factor x5 + x2 + x + 1 ∈ F2[x] into a product of irreduciblepolynomials.

3) Determine all elements of

K = F2 [x] / (x2 + x + 1) ,

the addition table of K, and the multiplication table of K.Prove that K is a field. What is the characteristic of K?

Exercise 1.4 Test the prime number theorem:

1) Write a procedure to compute

π (x) = ∣{p ≤ x ∣ p ∈ N prime}∣

for x > 0.

2) Plot π(x)x and 1

ln(x) and compare for large x.

3) Show that, according to the prime number theorem, the totalexpected number of Fermat primes is finite.

Recall, that a Fermat prime is a prime number of the form

Fn = 22n + 1

with n ∈ N0.

Exercise 1.5 Let G be the symmetry group of the icosahedron (Fig-ure 1.17).

1) Compute the group order of G.

2) Find generators of G as a subgroup of S12. Prove your claimusing Gap.

Exercise 1.6 Let A ∈ Qn×m be a matrix. Using the Singularcommand Grobner basis command std for ideals, implement a pro-cedure which computes the reduced row echelon form of A.

You may also want to use the following commands, see the on-line help: ring, matrix, proc, int, nrows, ncols, def, basering,imap, poly, ideal, for, option(redSB), diff, maxideal, setring,return.

1. INTRODUCTION 28

Figure 1.17: Icosahedron with numbering of the vertices.

Exercise 1.7 Using the idea of homogenization, describe an algo-rithm for describing the set of solutions of an inhomogeneous sys-tem of linear equations, which is based on the algorithm for solvinga homogeneous system.

Exercise 1.8 Let p ≥ 5 be a prime, and

Ea,b (Fp) = {(x, y) ∈ A2(Fp) ∣ fa,b (x, y) = 0} ∪ {O}

an elliptic curve given by

fa,b = x3 + a ⋅ x + b − y2 ∈ Fp [x, y] .

with 4a3 + 27b2 ≠ 0 modp. Write a procedure, which computes thenumber of points ∣Ea,b (Fp)∣ of the elliptic curve.

Note: You can simply test all points of A2(Fp), or make use ofthe Singular command Jacobi.

Exercise 1.9 Using your procedure from Exercise 1.8, prove Hasse’stheorem

p + 1 − 2√p ≤ ∣Ea,b (Fp)∣ ≤ p + 1 + 2

√p

for all primes p with 5 ≤ p ≤ 53. Verify that ∣Ea,b (Fp)∣ takes allpossible values.

Exercise 1.10 Let p be a prime and f ∈ Fp[x1, ..., xn] a non-constantpolynomial. For every point x ∈ An(Fp) the polynomial f can takep possible values f(x). Assume that an irreducible polynomial ftakes these values with equal probability 1

p for randomly chosen x.

1. INTRODUCTION 29

1) What is the probability that f(x) = 0 for random x under theassumption that f is the product of two irreducible polynomi-als.

2) Derive from this a probabilistic test for irreducibility.

3) Try out examples and compare with the Singular commandfactorize or the Maple command Factor() mod p.

2

Ideals, varieties andGrobner bases

2.1 Ideals and varieties

We begin with an easy, but important, observation about the alge-braic set V (f1, ..., fs) with fi ∈ R =K [x1, ..., xn]:

If f1 (p) = 0, ..., fs (p) = 0 for p ∈ An(K), then also any R-linearcombination of the fi vanishes on p, that is,

(s

∑i=1

ri ⋅ fi)(p) =s

∑i=1

ri (p) fi (p) = 0

for all ri ∈ R. Hence, V (f1, ..., fs) depends only on the ideal

⟨f1, ..., fs⟩ = {∑si=1rifi ∣ ri ∈ R} ⊂ R,

generated by f1, ..., fs. Recall:

Definition 2.1.1 Let R be a commutative ring with 1. An idealis a non-empty subset I ⊂ R with

a + b ∈ I

ra ∈ I

for all a, b ∈ I and r ∈ R.If S ⊂ R then

⟨S⟩ = {∑finiterifi ∣ ri ∈ R, fi ∈ S}

is the ideal generated by S.

30

2. IDEALS, VARIETIES AND GROBNER BASES 31

Figure 2.1: Elliptical arc

Recall, that the definition of an ideal is motivated in algebraby the following: For a subgroup I ⊂ R the additive group R/Ibecomes a ring with multiplication induced by that of R if andonly if I is an ideal (prove this as an easy exercise).

By the above observation it is natural to consider, instead ofthe vanishing locus of a set of equations, the vanishing locus of anideal:

Definition 2.1.2 If I ⊂K [x1, ..., xn] then

V (I) = {p ∈Kn ∣ f (p) = 0 ∀f ∈ I}

is called the vanishing locus of I.

This is indeed an affine variety, because any ideal I ⊂ k [x1, ..., xn]is finitely generated, as we will prove in Theorem 2.1.7.

Definition 2.1.3 Let S ⊂ An(K) be a subset. Then

I(S) = {f ∈K [x1, ..., xn] ∣ f (p) = 0 ∀p ∈ S}

is (as we have seen above) an ideal, the vanishing ideal of S.

Example 2.1.4 Consider the elliptical arc

S = {(x1, x2) ∈ An(R) ∣ x21 + 2x2

2 = 1 and x1, x2 ≥ 0}

shown in black in Figure 2.1. We have

I(S) = (x21 + 2x2

2 − 1)

hence V (I(S)) is the complete ellipse, the smallest algebraic setcontaining S. This is the closure

S = V (I(S))


of S in the so called Zariski topology:The Zariski topology on An(K) has as closed sets the V (I)

for ideals I ⊂ K [x1, ..., xn]. See also Exercise 2.2, which you needto show that this indeed gives a topology.

By I and V inclusion reversing maps

{affine algebraic sets X ⊂ An(K)}I⇄V

{ideals in K[x1, ..., xn]}

between the set of algebraic subsets of An(K) and the set of idealsof K [x1, ..., xn] are given. It remains to show that any ideal I ⊂K[x1, ..., xn] is finitely generated, that is, there are finitely manyf1, ..., fs ∈ R with I = ⟨f1, ..., fs⟩. We begin with a characterizationof these ideals:

Theorem 2.1.5 Let R be a commutative ring with 1. The follow-ing conditions are equivalent:

1) Every ideal I ⊂ R is finitely generated .

2) Every ascending chain

I1 ⊂ I2 ⊂ I3 ⊂ ... ⊂ In ⊂ ...

of ideals terminates, that is, there is an m, such that

Im = Im+1 = Im+2 = ...

3) Every non-empty set of ideals has a maximal element withrespect to inclusion.

If R satisfies these conditions, then R is called Noetherian.

These rings are called Noetherian after Emmy Noether (1882-1935), who has formulated the general structure theory for this classof rings and used this to give a simpler and more general proof ofthe theorems of Kronecker and Lasker.Proof. (1)Ô⇒ (2): Let I1 ⊂ I2 ⊂ ... be a chain of ideals. Then

I = ⋃∞j=1Ij

is also an ideal: If a, b ∈ I, then there are j1, j2 ∈ N with a ∈ Ij1 ,b ∈ Ij2 , and

a + b ∈ Imax(j1,j2) ⊂ I


By (1) the ideal I is finitely generated, hence there are a1, ..., as ∈ Iwith I = ⟨f1, ..., fs⟩. For every fk there is a jk with fk ∈ Ijk . For

m ∶= max{jk ∣ k = 1, ..., s}

we have f1, ..., fs ∈ Im, so

I = ⟨f1, ..., fs⟩ ⊂ Im ⊂ Im+1 ⊂ ... ⊂ I

and henceIm = Im+1 = ...

(2) Ô⇒ (3): Assume that (3) does not hold. Then there is a setM of ideals, such that for every I ∈M there is an I ′ ∈M with I ⫋ I ′

strictly contained. Hence, by induction, we obtain a sequence

I1 ⫋ I2 ⫋ I3 ⫋ ...

of ideals in M , which does not terminate, that is, (2) is not satisfied.(3)Ô⇒ (1): Let I be an arbitrary ideal. The set

M = {I ′ ⊂ I ∣ I ′ finitely generated}

is non-empty, for example, ⟨0⟩ ∈ M . Let J be a maximal elementof M . So there are f1, ..., fs ∈ J with J = ⟨f1, ..., fs⟩. We show thatI = J : If this is not true, then there is an f ∈ I/J with

J ⫋ ⟨f1, ..., fs, f⟩ ⊂ I.

This contradicts the maximality of J .

Example 2.1.6 1) The ring of integers Z is Noetherian, sinceall ideals of Z are of the form

⟨n⟩ = nZ = {nk ∣ k ∈ Z}

and, hence, are finitely generated (by a single element). SeeExercise 2.1.

2) A field K only has the ideals (0) and K = (1), in particular,K is Noetherian.

3) If R is Noetherian and I ⊂ R an ideal, then the quotient ringR/I is Noetherian:

Let π ∶ R → R/I be the canonical epimorphism. If J ⊂ R/Ian ideal then by assumption π−1(J) = ⟨f1, ..., fs⟩, and J =⟨π(f1), ..., f(fs)⟩.


4) The polynomial ring K[x1, x2, ...] in infinitely many variablesis not Noetherian. Also the ring

R = {f ∈ Q[X] ∣ f(0) ∈ Z}

of polynomials in Q[X] with integer values at 0 is not Noethe-rian. See Exercise 2.4.

Hilbert has shown in 1890, that the polynomial ring K [x1, ...xn]over a field K is Noetherian:

Theorem 2.1.7 (Hilbert’s basis theorem) If R is a Noetherianring, then also R [x] is Noetherian.

Using that a field K and the ring of integers Z are Noetherian,by induction on the number n of variables

R [x1, ..., xn] = R [x1, ..., xn−1] [xn]

we obtain:

Corollar 2.1.8 Let K be a field. Then the polynomial rings K [x1, ...xn]and Z [x1, ...xn] in n variables are Noetherian.

The fact that K [x1, ...xn]is Noetherian, is the basis of all algo-rithms, we will discuss.

For the proof of Theorem 2.1.7 we consider the lead coefficientsin R of polynomials in R [x]. If

f = akxk + ... + a1x + a0 ∈ R [x]

with ak ≠ 0 then the degree of f is deg (f) = k, its lead coeffi-cient is LC (f) = ak, its lead term LT (f) = akxk, and its leadmonomial L (f) = xk.Proof. Assume R [x] is not Noetherian. Then there is an idealI ⊂ R [x] which is not finitely generated. Let f1 ∈ I with deg (f1)minimal, f2 ∈ I/ ⟨f1⟩ mit deg (f2) minimal, and inductively

fk ∈ I/ ⟨f1, ..., fk−1⟩

with deg (fk) minimal. Then

deg (f1) ≤ deg (f2) ≤ ... ≤ deg (fk) ≤ ...

and we obtain an ascending chain of ideals in R

⟨LC (f1)⟩ ⊂ ⟨LC (f1) ,LC (f2)⟩ ⊂ ... ⊂ ⟨LC (f1) , ...,LC (fk)⟩ ⊂ ...


We show that the inclusions are strict (and hence R is not Noethe-rian): Assume

⟨LC (f1) , ...,LC (fk)⟩ = ⟨LC (f1) , ...,LC (fk+1)⟩

Then we can write

LC (fk+1) =k

∑j=1

bj LC (fj)

with bj ∈ R. Hence

g ∶=k

∑j=1

bj ⋅ xdeg(fk+1)−deg(fj) ⋅ fj

∈ ⟨f1, ..., fk⟩

has the same lead term as fk+1, so

deg (g − fk+1) < deg (fk+1) ,

a contradiction, since fk+1 was chosen to have minimal degree.So any algebraic set can be represented by an ideal, and any

ideal gives an algebraic set. However, the V -map is not injective,for example,

V (x) = V (x2) ⊂ A1(K).

There are two ways to remedy this situation. One possibility isto generalize our notation of an algebraic set: Given I ⊂ R =K[x1, ..., xn] we replace V (I) by the spectrum

Spec(R/I) = {P ⊂ R/I ∣ P prime ideal}

and consider R/I as the ring of function on Spec(R/I). Togetherwith the Zariski topology we obtain a generalization of an alge-braic set, called a scheme. An easier approach is to restrict theclass of ideals in consideration. To determine that class of ideals, itis, astonishingly, enough to find out, under which conditions an al-gebraic set is empty. This is characterized by the following theoremof Hilbert (which we cannot prove here):

Theorem 2.1.9 (Weak Nullstellensatz) Let K be an algebraicallyclosed field and I ⊂K[x1, ..., xn] an ideal. Then

V (I) = ∅⇐⇒ I =K[x1, ..., xn]

Remark 2.1.10 The condition, that K is algebraically closed, isnecessary. For example, V (x2 + y2 + 1) ⊂ A2(R) is empty (it is notempty over C).


From Theorem 2.1.9 we obtain:

Theorem 2.1.11 (Strong Nullstellensatz) Let K be an alge-braically closed field and I ⊂K[x1, ..., xn] an ideal. Then

I(V (I)) =√I.

where √I = {f ∈K[x1, ..., xn] ∣ ∃a ∈ N with fa ∈ I}

denotes the radical of I.

Proof. According to the basis theorem, write I = ⟨f1, ..., fs⟩. Forf ∈ I(V (I)) consider

J = ⟨I, y ⋅ f − 1⟩ ⊂K[x1, ..., xn, y].

Since f vanishes at any common zero of f1, ..., fs, and, hence, y ⋅f − 1 does not, we have V (J) = ∅. So by Theorem 2.1.9 J =K[x1, ..., xn, y], that is, there are ci, d ∈K[x1, ..., xn, y] with

1 = c1 ⋅ f1 + ... + cs ⋅ fs + d ⋅ (y ⋅ f − 1).

Substituting y = 1f makes the coefficients to ci(x1, ..., xn,

1f ). So

multiplying with a sufficiently high power a of f cancels the de-nominators and yields fa ∈ I.

The other inclusion is easy.

Definition 2.1.12 An ideal I ⊂ K[x1, ..., xn] is called a radicalideal, if I =

√I.

Theorem 2.1.11 shows that, if K is algebraically closed,

{affine algebraic sets X ⊂ An(K)}I⇄V

{radical ideals in K[x1, ..., xn]}

is a one-to-one correspondence. In Exercise 2.3 we will prove, thatan algebraic set X = V (I) is irreducible, if and only if I(V (I)) isprime. This is true over any field K. If K is algebraically closed,then, by the strong Nullstellensatz, V (I) is irreducible iff I(V (I)) =√I is prime. In particular, if I is prime then V (I) is irreducible.

Note that this is not true in general if K is not algebraically closed.So for K algebraically closed we obtain a one-to-one correspondenceof varieties (irreducible algebraic sets) and prime ideals:

{affine varieties X ⊂ An(K)}I⇄V

{prime ideals in K[x1, ..., xn]}


If K is algebraically closed, the points correspond to the maximalideals, that is, we have a one–to-one correspondence

{maximal ideals of K[x1, ..., xn]}V⇄I

Kn

⟨x − a1, ..., x − an⟩ (a1, ..., an)

Recall, that an ideal P ⫋ R of a commutative ring R with 1 iscalled prime ideal, if ∀a, b ∈ R it holds

a ⋅ b ∈ P Ô⇒ a ∈ P or b ∈ P .

The ideal P is called maximal ideal, if for all ideals I ⊂ R it holds

P ⊂ I ⫋ RÔ⇒ P = I.

Recall also the following, standard and easy to prove, characteriza-tion of prime and maximal ideals:

Theorem 2.1.13 Let R be a commutative ring with 1 and I ⫋ Ran ideal. Then it holds:

1) I prime ⇐⇒ R/I is an integral domain.

2) I maximal ⇐⇒ R/I is a field.

Example 2.1.14 1) The ideal ⟨x2⟩ ⊂K[x1, x2] is a prime ideal,because

K[x1, x2]/ (x2) ≅K[x1]

is an integral domain. On the other hand, ⟨x1 ⋅ x2⟩ is not aprime ideal, since

x1 ⋅ x2 = 0 ∈K[x1, x2]/I

and x1, x2 ≠ 0. Geometrically, the prime ideals ⟨x1⟩ and ⟨x2⟩correspond to the coordinate axes and ⟨x1 ⋅ x2⟩ to their union

V (x1 ⋅ x2) = V (x1) ∪ V (x2)

2) The ideal ⟨x2 − x21⟩ ⊂K [x1, x2] is a prime ideal, since

K [x1, x2] / ⟨x2 − x21⟩ → K[t]x1 ↦ tx2 ↦ t2

is an isomorphism and K[t] is an integral domain.


Figure 2.2: Reducible affine algebraic set

The idealI = ⟨(x2 − x

21) ⋅ (x1 − x

22)⟩

is not prime, and

V (I) = V (x2 − x21) ∪ V (x1 − x

22),

see Figure 2.2.

In fact, any radical ideal can be written as an intersection ofprime ideals, more generally, any ideal as an intersection of, socalled, primary ideals. We will discuss in detail an algorithm whichcomputes this primary decomposition.

Example 2.1.15 The ideal ⟨x1, x2⟩ ⊂K[x1, x2] is a maximal ideal,since K[x1, x2]/ ⟨x1, x2⟩ ≅K is a field.

See also the Exercises 2.5 and 2.6.So the bottom line is: Any geometric problem concerning affine

algebraic sets, can be translated into a problem concerning idealsin polynomial rings.

2.2 Introduction to the ideal member-

ship problem and Grobner bases

Suppose we want to obtain information about a variety V (I) ⊂An(K) specified by an ideal I = ⟨f1, ..., fs⟩ ⊂ K[x1, ..., xn] whichagain is given by generators f1, ..., fs ∈ K[x1, ..., xn]. For exam-ple, we may want to determine, whether V (I) is contained in the


hypersurface V (f). Equivalently, we have to determine whetherf ∈ ⟨f1, ..., fs⟩. This question is called the ideal membershipproblem and appears as a fundamental buildung block in manymore advanced algorithms.

Example 2.2.1 Consider the twisted cubic curve C = V (I) definedby I = ⟨y − x2, z − x3⟩, see Figure 2.3. By definition, C is contained

Figure 2.3: Twisted cubic

in the hypersurfaces V (y − x2) and V (z − x3). However, is it alsocontained in the hypersurface V (z − xy)? Figure 2.4suggests yes,

Figure 2.4: Surface containing the twisted cubic

and we easily find a representation

z − xy = (−x) ⋅ (y − x2) + 1 ⋅ (z − x3) .

How to find such a representation in a systematic way?


Before solving the ideal membership problem in general, let usfirst discuss two special settings, the linear and the univariate case.

Example 2.2.2 Let f1, ..., fs, f ∈K[x1, ..., xn] be linear. Then test-ing f ∈ I = ⟨f1, ..., fs⟩ is easy and can be done in the following twosteps:

1) Apply Algorithm 1.5.1 to obtain linear equations g1, ..., gr inrow echelon form, so L(g1) > ... > L(gr).

2) For i = 1, ..., r do

If L(f) = L(gi) then

f = f − LC(f)LC(gi)gi

If f = 0 then return true else return false.

As a second special case, consider higher degree equations in asingle variable. The polynomial ring K[x] in one variable over afield K is an example of a Euclidean domain:

Definition 2.2.3 A Euclidean domain is an integral domain Rtogether with a map (called Euclidean norm)

d ∶ R/ {0}Ð→ N0

such that for any a, b ∈ R/ {0} there exist g, r ∈ R with

1) a = g ⋅ b + r and

2) r = 0 or d (r) < d (b).

Example 2.2.4 The ring of integers Z with d (n) = ∣n∣ and thepolynomial ring K[X] in one variable X over a field K with d (f) =deg (f) is Euclidean.

There are many more Euclidean domains, for example, Z [i]with

d (a1 + i ⋅ a2) = ∣a1 + i ⋅ a2∣2= a2

1 + a22.

The Euclidean algorithm given in Theorem 1.2.2 and its proofcarry over directly to any Euclidean domain by replacing the abso-lute value by d.

Theorem 2.2.5 Euclidean domains are principal ideal domains(any ideal is principal, that is, generated by a single element).


Proof. Let I ⊂ R be an ideal in a Euclidean domain. The idealI = ⟨0⟩ is principal. Otherwise, there is a non-zero b ∈ I with d (b)minimal. Let a ∈ I be arbitrary and a = g ⋅ b + r with r = 0 ord (r) < d (b). By a, b ∈ I also r ∈ I. As d(b) was chosen minimal, weget r = 0 and, hence, a ∈ ⟨b⟩. This proves I ⊂ ⟨b⟩ ⊂ I.

Corollar 2.2.6 If R is a principal ideal domain, and f1, ..., fs ∈ R,then

⟨f1, ..., fs⟩ = ⟨gcd (f1, ..., fs)⟩

Proof. As R is a principal ideal domain,

⟨f1, ..., fs⟩ = ⟨d⟩

with d ∈ R, hence d ∣ fi for all i. On the other hand, there are xi ∈ Rwith

d = x1f1 + ... + xsfs.

So every divisor of all fi divides d. Hence

d = gcd (f1, ..., fs) .

Recall that the gcd is only unique up to units in R.Hence, the ideal membership problem translates into the follow-

ing characterization

f ∈ ⟨f1, ..., fs⟩ ⇐⇒ gcd (f1, ..., fs) divides f .

Example 2.2.7 We test whether

x3 + x ∈ I = ⟨x4 − 1, x4 − 3x2 − 4⟩ ⊂ Q[x]

The Euclidean algorithm yields

x4 − 3x2 − 4 = 1 ⋅ (x4 − 1) + (−3x2 − 3)x4 − 1 = x2 ⋅ (x2 + 1) + (−x2 − 1)

= (x2 − 1) ⋅ (x2 + 1) + 0

henceI = ⟨x2 + 1⟩

and division with remainder

x3 + x = x ⋅ (x2 + 1) + 0,

shows that, x3 + x ∈ I.

So what about the general case?


Example 2.2.8 Suppose we want to check whether

x2 − y2 ∈ ⟨x2 + y, xy + x⟩ .

In order to do division with remainder, we have to decide whichterm of a polynomial is the lead term. Ordering by degree is notsufficient, consider xy2 + x2y. For example, we could order themonomials in a lexicographic way, that is, like the words in a tele-phone book. Then

L(x2 + y) = x2 L(xy + x) = xy

and the usual strategy for division with remainder would give

x2 − y2 = 1 ⋅ (x2 + y) + (−y2 − y)x2 + y−y2 − y

The lead terms we write in bold face red. So the remainder is −y2−y ≠ 0, however

x2 − y2 = −y (x2 + y) + x (xy + x) ∈ ⟨x2 + y, xy + x⟩ .

The problem is caused by the cancelling of the lead terms in thisexpression. How to resolve the problem?

Simply add to the set of generators all polynomials, which canbe obtained by canceling lead terms. The result is what is called aGrobner basis. In the example we would add x2 − y2 and then

−y2 − y = (x2 − y2) + (−1) ⋅ (x2 + y).

Finally, we could get rid of x2 − y2 or x2 + y as it is sufficient tokeep one generator for each possible lead monomial. This results ina minimal Grobner basis

x2 − y2, xy + x, y2 + y

orx2 + y, xy + x, y2 + y.

The second one is the unique reduced Grobner basis, which canbe obtained by removing terms in tail(f) = f − LT(f) which aredivisable by some lead monomial. For any of these Grobner bases,the division of x2 − y2 will give remainder zero: For the first onethis is trivial, since x2 − y2 is already an element of the Grobnerbasis. For the second one, we can continue the above calculation,resulting in the expression

x2 − y2 = 1 ⋅ (x2 + y) + (−1) ⋅ (y2 + y) + 0

with remainder 0.


Indeed, we will show in general, that when dividing f by aGrobner basis g1, ..., gr, division will give remainder zero if and onlyif f ∈ ⟨g1, ..., gr⟩. We begin by formalizing this concept:

2.3 Monomial orderings

For monomials we use multi-index notation xα = xα11 ⋅ ... ⋅ xαn

n withthe exponent vector α = (α1, ..., αn) ∈ Nn

0 .

Definition 2.3.1 A monomial ordering (or semigroup or-dering) on the semigroup of monomials in the variables x1, ..., xnis an ordering > with

1) > is a total ordering

2) > respects multiplication, that is,

xα > xβ ⇒ xαxγ > xβxγ

for all α,β, γ.

Definition and Theorem 2.3.2 A global ordering is a mono-mial ordering > with the following equivalent properties

1) > is a well ordering

(that is, any non-empty set of monomials has a smallest ele-ment)

2) xi > 1 ∀i.

3) xα > 1 for all 0 ≠ α ∈ Nn0 .

4) If xβ ∣ xα and xα ≠ xβ then xα > xβ

(that is, > refines divisibility).

If xi < 1 ∀i, then > is called a local ordering.

Proof. The implications (1) ⇒ (2) ⇒ (3) ⇒ (4) are easy, seeExercise 2.7. With respect to (4) ⇒ (1), we have to prove thatany non-empty set of monomials has only finitely many minimalelements with respect to divisibility. Then, by assumption (4) weonly have to consider those minimal elements, and, since > is a totalordering, among them there is a smallest.


Lemma 2.3.3 (Dickson, Gordan) Any non-empty set of mono-mials in the variables x1, ..., xn has only finitely many minimal ele-ments with respect to divisibility.

Proof. Let M ≠ ∅ be a set of monomials in the variables x1, ..., xn,and ⟨M⟩ ⊂ K[x1, ..., xn] the ideal generated by the elements ofM . By the Hilbert basis theorem 2.1.7 we have ⟨M⟩ = ⟨f1, ..., fs⟩with polynomials fi = ∑

uj=1 ri,jmj where ri,j ∈ K[x1, ..., xn] and

m1, ...,mu ∈M . Hence

⟨M⟩ ⊂ ⟨m1, ...,mu⟩ ⊂ ⟨M⟩ .

Among the m1, ...,mu consider the minimal elements with respectto divisibility.

The ideal we have encountered in the proof is an example of amonomial ideal:

Definition 2.3.4 An ideal I ⊂K[x1, ..., xn] is called a monomialideal, if it is generated by monomials.

Corollar 2.3.5 Every monomial ideal has a unique set of mini-mal generators consisting of monomials.

Proof. See the proof of Lemma 2.3.3 (or apply the lemma to theset of monomials in the ideal).

In the proof we have also encountered the following trivial, butimportant, observation:

Lemma 2.3.6 Let I = ⟨M⟩ be a monomial ideal generated by themonomials in M . If f ∈ I, then every term of f is in I.

In particular, if f ∈ I is a monomial, then there is an m ∈ Mwith m ∣ f .

Proof. If f = ∑uj=1 rjmj ∈ I with rj ∈ K[x1, ..., xn] and mj ∈ M ,

any term of f is a term of some (perhaps several) rjmj and hencea multiple of some mj.

We discuss some specific monomial orderings, there are manymore. First note:

Example 2.3.7 In one variable all global orderings are equivalentto the ordering defined by x > 1, all local orderings to that definedby x < 1.

Definition 2.3.8 The following definitions yield global monomialorderings:


1) Lexicographical ordering:

xα > xβ ⇐⇒ the leftmost nonzero entry of α − β is positive.

In Singular this ordering is abbreviated as lp.

2) Degree reverse lexicographical ordering:

xα > xβ ⇔ degxα > degxβ or degxα = degxβ and ∃1 ≤ i ≤ n ∶αn = βn, ..., αi+1 = βi+1, αi < βi.

In Singular this ordering is abbreviated as dp.

An example of a local ordering is the negative lexicographicalordering:

xα > xβ ⇐⇒ the leftmost nonzero entry of α − β is negative.In Singular this ordering is abbreviated as ls.

Example 2.3.9 For lp on the monomials in x, y, z we have (iden-tifying monomials and exponent vectors)

x = (1,0,0) > y = (0,1,0) > z = (0,0,1)

xy2 = (1,2,0) > (0,3,4) = y3z4

x3y2z4 = (3,2,4) > (3,1,5) = x3y1z5

on the other hand, for dp we get

x = (1,0,0) > y = (0,1,0) > z = (0,0,1)

xy2 = (1,2,0) < (0,3,4) = y3z4

x3y2z4 = (3,2,4) > (3,1,5) = x3yz5

and for ls

x = (1,0,0) < y = (0,1,0) < z = (0,0,1)

xy2 = (1,2,0) < (0,3,4) = y3z4

x3y2z4 = (3,2,4) < (3,1,5) = x3yz5

In Singular we can compare monomials as follows:ring R=0,(x,y,z),lp;

x>y;1

y>z;1

xy2>y3z4;1

x3y2z4>x3yz51


ring R=0,(x,y,z),dp;

x>y;1

y>z;1

xy2>y3z4;0

x3y2z4>x3yz51

ring R=0,(x,y,z),ls;

1>z;1

z>y;1

y>x;1

xy2>y3z4;0

x3y2z4>x3yz50

Our definitions in the linear and univariate case generalize di-rectly:

Definition 2.3.10 With respect to a given monomial ordering >,for any polynomial f = ∑α cαx

α the leading monomial is thelargest monomial xα with cα ≠ 0 and is denoted by L (f). Further-more, we denote by LC(f) = cα the leading coefficient, and byLT (f) = cαxα the leading term.

Example 2.3.11 Using lp we have

L(5x2y + yx2) = x2y.

In Singular we compute the lead term, monomial and coeffi-cient as follows:ring R=0,(x,y,z),lp;

poly f = 5x2y+xy2;

lead(f);

5x2y

leadcoef(f);

5

leadmonom(f);

x2y


Definition 2.3.12 A monomial ordering > is called a weighteddegree ordering if there is some w ∈ Rn with non-zero entriessuch that

wα > wβ ⇒ xα > xβ.

Example 2.3.13 If > is any monomial ordering and w ∈ Rn, then>w given by

xα >w xβ⇔ wα > wβ or (wα = wβ and xα > xβ)

is a monomial ordering.The ordering >, which takes over if the w-weights of xα and xβ

are equal, is called tie-break ordering.By construction, >w is a weighted degree ordering, it is global if

wi > 0 ∀i and it is local if wi < 0 ∀i.

For the purpose of explicit computations, which will only involvea finite number of monomials, we can represent any monomial or-dering by a weight vector:

Proposition 2.3.14 Given a monomial ordering > and a finite setM of monomials in the variables x1, ..., xn, there is a weight vectorw ∈ Zn with

xα > xβ ⇐⇒ wα > wβ

for all xα, xβ ∈M .We can choose w such that wi > 0 if xi > 1 and wi < 0 if xi < 1.

Proof. Consider

D> = {α − β ∈ Zn ∣ xα > xβ} .

If α1 − β1, α2 − β2 ∈D> then

xα1+α2 = xα1xα2 > xβ1xα2 > xβ1xβ2 = xβ1+β2 ,

henceδ1, δ2 ∈D>⇒ δ1 + δ2 ∈D>.

So for all λi ∈ N and δi ∈D>

∑iλiδi ∈D>.

As 0 ∉D>, we get ∑iλiδi ≠ 0 for all λi ∈ Q>0, hence

0 ∉ convHull(D> ∩M) = {∑ri=1λiδi ∣ δi ∈D, λi ∈ Q≥0, ∑

ri=1λi = 1} .


Figure 2.5: A convex hull

For an example in Figure 2.5 the grey area is the convex hull of theblack points. By 0 ∉ convHull(D> ∩M)), there is a w ∈ Zn suchthat convHull(D>) is contained in the half space where the linearform δ ↦ wδ takes positive values. Hence,

wδ > 0 for all δ ∈D>.

For the second statement observe: This w satisfies wi > 0 ifxi > 1 and wi < 0 if xi < 1, provided that 1, x1, ..., xn ∈M .

Example 2.3.15 For the lexicographical ordering > in the vari-ables x1, x2, the set D> is plotted in Figure 2.6.The figure also shows

–4

–2

0

2

4

–4 –2 0 2 4

Figure 2.6: D> for the ordering lp

the line wδ = 0, where w is a weight vector representing lp on allmonomials of degree ≤ 4. For more examples see Exercise 2.9.

You can imagine that a Grobner basis computation for a fixedideal will only involve finitely many monomials due to the Noethe-rian property of the polynomial ring. The equivalent weight vectorsw form a cone and these cones fit nicely together in what is called


the Grobner fan (that is, the intersection of two such cones isagain one of the cones). Figure 2.7 shows the Grobner fan clas-sifying the non-equivalent weight vectors for f = x + y + 1 and theinitial term of each cone. Given a weight-vector w, the initial terminw(f) is the sum of all terms of f with maximal w-weight. Thenthe Grobner fan of ⟨f⟩ consists of the (closures of) cones

{w ∣ inw(f) = g}

for any possible g. This definition can be generalized to ideals withmore than one generator.

If you have heard about tropical geometry, the tropical varietyof an ideal can be considered as a subfan of the Grobner fan. In theexample, it consists of those faces such that inw(f) is not a mono-mial. So the three black half-lines in the figure form the tropicalvariety of a line.

Figure 2.7: Grobner fan of the line.

2.4 Division with remainder and Grobner

bases

Write R = K[x1, ..., xn]. Given a global monomial ordering, it ispretty clear, how to do division with remainder. Algorithm 2.4.1divides f ∈ R by g1, ..., gs ∈ R.


Algorithm 2.4.1 Division with remainder

Input: f ∈ R, g1, ..., gs ∈ R, > be a global ordering on the monomialsof R.

Output: An expression

f = q + r = ∑si=1aigi + r

such that L(r) is not divisible by any L(gi).1: q = 02: r = f3: while r ≠ 0 and L(gi) ∣ L(r) for some i do4: Cancel the lead term of r:5: a = LT (r)

LT (gi)6: q = q + a ⋅ gi7: r = r − a ⋅ gi

Proof. In every step the lead term of r becomes smaller withrespect to >, so the algorithm terminates, since > is a well-ordering.

Remark 2.4.1 The assumption that > is global is necessary for thetermination: If we divide x by x−x2 using the local ordering x < 1,then Algorithm 2.4.1 will compute

x = 1 ⋅ (x − x2) +x2

= (1 + x) ⋅ (x − x2) +x3

⋮

= (∑∞i=0x

i) ⋅ (x − x2) + 0

that is, the geometric series expansion of

x

x − x2=

1

1 − x.

So the algorithm works as expected, but does not give an answerafter finitely many steps. We will come back to this (however, thesolution is pretty obvious, clear the denominator 1 − x).

Example 2.4.2 Using lp divide x2y + x by y − 1 and x2 − 1.

x2y + x = x2 (y − 1) + 1 ⋅ (x2 − 1) + x + 1x2y − x2

x2 + xx2 − 1x + 1


We will now show that Algorithm 2.4.1 solves the ideal mem-bership problem, provided we divide by a Grobner basis.

Definition 2.4.3 Given a monomial ordering > and a subset G ⊂R, we define the leading ideal of G as

L>(G) = ⟨L(f) ∣ f ∈ G/{0}⟩ ⊂ R,

the monomial ideal generated by the lead monomials. If the choiceof > is clear, we also write L(G).

Given an ideal I the ideal L(I) will contain all possible leadmonomials obtainable by cancelling lead term, hence we define:

Definition 2.4.4 (Grobner basis) Let I be an ideal and > a globalmonomial ordering. A finite set 0 ∉ G ⊂ I is called a Grobner ba-sis of I with respect to >, if

L(G) = L(I).

Note that the inclusion ⊂ is true for any subset. The existenceof a Grobner basis is easy:

Theorem 2.4.5 Every ideal I ⊂ R has a Grobner basis.

Proof. Since L(I) is finitely generated, L(I) = ⟨m1, ...,ms⟩ withmonomials mi. Furthermore, mi is divisible by some L(gi) for somegi ∈ I, see Lemma 2.3.6. Hence

L(I) = ⟨L(g1), ..., L(gs)⟩ ,

so g1, ..., gs form a Grobner basis of I.From the definition it is not clear whether a Grobner basis of

I is indeed a set of generators of I. Solving the ideal membershipproblem will answer also this question. To clarify the behaviour ofAlgorithm 2.4.1, we first formalize its properties:

Definition 2.4.6 Given a list G = (g1, ..., gs) ⊂ R, a normalform is a map NF(−,G) ∶ R → R with

1) NF(0,G) = 0.

2) If NF(f,G) ≠ 0 then L(NF(f,G)) ∉ L(G).

3) For all 0 ≠ f ∈ R there are ai ∈ R with

f −NF(f,G) = ∑si=1aigi

and L(f) ≥ L(aigi) for all i with aigi ≠ 0.


We also say that NF is a normal form, if NF(−,G) is a normalform for all G.

Lemma 2.4.7 Algorithm 2.4.1 is a normal form. It is called theBuchberger normal form.

Proof. We map f to NF(f,G) ∶= r. If the algorithm returns re-mainder r ≠ 0 then L(r) is not divisible by any L(gi), so L(r) ∉L(G) by Lemma 2.3.6. Condition (3) is clear, since in every itera-tion of the algorithm L(a ⋅ gi) ≤ L(f).

Using Grobner bases, we can now decide the ideal membershipproblem:

Theorem 2.4.8 (Ideal membership) Let I ⊂ R be an ideal andf ∈ R. If G = (g1, ..., gs) is a Grobner basis of I and NF is a normalform, then

f ∈ I ⇐⇒ NF(f,G) = 0.

Proof. Write f = ∑i aigi + r with r = NF(f,G). If r = 0 thenf = ∑i aigi ∈ ⟨G⟩ ⊂ I. On the other hand, if r ≠ 0 then by Definition2.4.6 (2.)

L(r) ∉ L(G) = L(I).

So, by definition of the lead ideal,

r ∉ I,

hence f = ∑i aigi + r ∉ I.

Lemma 2.4.9 If J ⊂ I ⊂ R are ideals with L(J) = L(I) then I = J .

Proof. Let G = (g1, ..., gs) be a Grobner basis of J , NF a normalform, f ∈ I and f = ∑i aigi + r with r = NF(f,G). So r ∈ I. If r ≠ 0,then by Definition 2.4.6 (2.)

L(r) ∉ L(G) = L(J) = L(I).

By the definition of the lead ideal, we have r ∉ I, a contradiction.

Corollar 2.4.10 If G is a Grobner basis of I, then

I = ⟨G⟩ .

Proof. We have L(I) = L(G) ⊂ L(⟨G⟩) ⊂ L(I), so G is a Grobnerbasis of ⟨G⟩ ⊂ I and L(⟨G⟩) = L(I). Equality follows from Lemma2.4.9.


Example 2.4.11 The generators of the ideal I = ⟨x2 − 1, y − 1⟩ al-ready form a Grobner basis with respect to lp: Since x ∉ L(I) (Ex-ercise) we have

L(I) = ⟨x2, y⟩ .

Of course, the result of division with remainder depends onthe monomial ordering. But, even if we fix a monomial orderingand divide by a Grobner basis, the result may not be uniquelydetermined:

Example 2.4.12 At every step of Algorithm 2.4.1, there can beseveral choices of gi such that L(gi) ∣ L(f). We divide x2y + x bythe Grobner basis G = (y − 1, x2 − 1), using lp as in Example 2.4.2.However we now prefer x2 − 1 over y − 1 when possible:

x2y + x = y ⋅ (x2 − 1) + x + yx2y − yx + y

So depending on the choice made, the remainder will be x + 1 orx + y.

To have a uniquely determined remainder, we proceed as follows:

Definition 2.4.13 A polynomial f ∈ R is called reduced with re-spect to a set G ⊂ R, if no term of f is contained in L(G).

A normal form is called reduced normal form, if NF(f,G)is reduced with respect to G.

Algorithm 2.4.2 yields a reduced normal form, the reducedBuchberger normal form.


Algorithm 2.4.2 Reduced division with remainder

Input: f ∈ R, g1, ..., gs ∈ R, > be a global ordering on the monomialsof R.


f = q + r = ∑si=1aigi + r

such that not term of r is divisible by any L(gi).1: q = 02: r = 03: h = f4: while h ≠ 0 do5: if L(gi) ∣ L(h) for some i then6: Cancel the lead term of h:7: a = LT (h)

LT (gi)8: q = q + a ⋅ gi9: h = h − a ⋅ gi10: else11: Put the lead term into the remainder:12: r = r +LT (h)13: h = h −LT (h)

Example 2.4.14 So, also putting terms into the remainder in theintermediate steps, we can continue in Example 2.4.12:

x2y + x = y ⋅ (x2 − 1) +x + 1 ⋅ (y − 1) + 1x2y − yx + yyy − 11

which leads to the same remainder x+1 as in Example 2.4.2. Indeed,the remainder is now unique:

Theorem 2.4.15 Let > be a global ordering, I ⊂ R an ideal, f ∈ Rand G a Grobner basis of I. If NF is a reduced normal form, thenNF(f,G) is uniquely determined by >, f and I. We then also writeNF(f, I).

Proof. Write G = (g1, ..., gs) and suppose that

f = ∑si=1aigi + r

= ∑si=1a

′igi + r

′


Thenr − r′ = ∑

si=1(ai − a

′i)gi ∈ ⟨G⟩ = I

(using Corollary 2.4.10). So, if r − r′ ≠ 0, then L(r − r′) ∈ L(I) =L(G). Since L(r − r′) is a monomial of r or r′, this would meanthat r or r′ is not reduced with respect to G.

Example 2.4.16 In Singular we can compute the reduced Buch-berger normal form in Example 2.4.14 by:ring R=0,(x,y),lp;

ideal I = x2-1,y-1;

We first check that the generators of I form a Grobner basis:I=std(I);

I;

I[1]=x-1

I[2]=x2-1

reduce(x2y+x,I);

x+1

The non-reduced version is called by reduce(-,-,1).Singular makes the choices in the algorithms for you in a cleverway, however, you cannot influence this.

Remark 2.4.17 Even when using a reduced normal form and aGrobner basis, although the remainder is unique, the generated ex-pression in the generators may not be. In the Examples 2.4.2 and2.4.14 we obtain

x2y + x = y ⋅ (x2 − 1) + 1 ⋅ (y − 1) + x + 1

andx2y + x = x2 (y − 1) + 1 ⋅ (x2 − 1) + x + 1

respectively.When dividing f by G = (g1, ..., gs), we can obtain a unique

expressionf = ∑

si=1aigi + r

by requiring that no term of aiL(gi) is divisible by any L(gj) forj < i.

In the example, the first expression would be returned for G =(x2 − 1, y − 1), and the second for G = (y − 1, x2 − 1).


2.5 Computing Grobner bases

In Section 2.2 we have already seen the basic idea for computinga Grobner basis of an ideal. We will now turn this idea into analgorithm. Again write R =K[x1, ..., xn] and fix a global monomialordering >.

Definition 2.5.1 The syzygy polynomial or S-polynomial off, g ∈ R is defined as

spoly(f, g) =lcm(L(f), L(g))

LT (f)f −

lcm(L(f), L(g))

LT (g)g ∈ R.

Doing all possible cancelations of lead terms, Algorithm 2.5 com-putes a Grobner basis.

Algorithm 2.5.1 Buchberger

Input: I = ⟨g1, ..., gs⟩ ⊂ R an ideal, > a global monomial ordering,and NF a normal form.

Output: A Grobner basis of I with respect to >.1: G = {g1, ..., gs}2: repeat3: H = G4: for all f, g ∈H do5: r = NF(spoly(f, g),H)6: if r ≠ 0 then7: G = G ∪ {r}8: until G =H

Proof. If r ≠ 0 then L(r) ∉ L(H) by Definition 2.4.6(2.), hence

L(H) ⫋ L(H ∪ {r}).

So, by the Noetherian property of R, the algorithm terminates.To show that the final result is a Grobner basis, we prove:

Theorem 2.5.2 (Buchberger’s criterion) If I ⊂ R is an ideal,NF a normal form and G = (g1, ..., gs) a list of elements of I, thenthe following conditions are equivalent:

1) G is a Grobner basis of I.

2) NF(f,G) = 0 for all f ∈ I.

3) I = ⟨G⟩ and NF(spoly(gi, gj),G) = 0 for all i ≠ j.


Proof. (1)⇒ (2)⇒ (3): If G is a Grobner basis we get NF(f,G) =0 for all f ∈ I by Theorem 2.4.8. Furthermore, spoly(gi, gj) ∈ I.

(2)⇒ (1): If f ∈ I then by NF(f,G) = 0 and Definition 2.4.6(3.)we have an expression

f = ∑si=1aigi

with L(f) ≥ L(aigi) for all i. So there has to be an i with L(f) =L(aigi), which implies that L(gi) ∣ L(f). Hence L(f) ∈ L(G).

(3)⇒ (1): We have to show that if f ∈ I then L(f) ∈ L(G). Byassumption there are ai ∈ R with

f = ∑si=1aigi.

Clearly L(f) ≤ maxiL(aigi). If L(f) < maxiL(aigi), then somelead terms of summands in∑

si=1aigi cancel, say L(ai1gi1) and L(ai2gi2).

By assumption we have a division expression

0 = spoly(gi1 , gi2) −∑si=1cigi.

Such a relation is called a syzygy. Subtracting a multiple ofthis equality from f = ∑

si=1aigi we obtain a new expression for

f with smaller maxiL(aigi). After finitely many steps L(f) =maxiL(aigi), hence L(gi) ∣ L(f) for some i, that is, L(f) ∈ L(G).

For the step (3) ⇒ (1) we have given more like a sketch of aproof. The theory of Grobner bases of modules will enable us togive a very elegant proof later.

Example 2.5.3 Using Buchberger’s criterion we can easily checkthat G = (x2−1,y−1) is a Grobner basis of I = ⟨G⟩: For the S-pair

s = y(x2 − 1) − x2(y − 1) = x2 − y

division with remainder gives NF(s,G) = 0:

x2 − y = 1 ⋅ (x2 − 1) − 1 ⋅ (y − 1) + 0x2 − 1−y + 1−y + 10

Example 2.5.4 We apply Buchberger’s algorithm to compute aGrobner basis of

I = ⟨t2 − x, t3 − y, t4 − z⟩ ⊂ k [t, z, y, x]


for the lexicographical ordering t > z > y > x. In each step the firstcolumn denotes the coefficients in the syzygy polynomial and thesecond the division with remainder.

↱ −tx + y ↱ −t2x + z ↱ −ty + z ↱ t3y − zxt2 − x t t2 x −tyt3 − y −1 tt4 − z −1 −1 xtx − y 1 −t3 −yz − x2 −1 −1 xty − x2 1y2 − x3 −1

Writing this in terms of syzygies

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

t2 − xt3 − yt4 − ztx − yz − x2

ty − x2

y2 − x3

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

t

⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

t t2 + x 0 −ty−1 0 t 00 −1 −1 x1 0 0 −t3 − y0 −1 −1 x0 0 1 00 0 0 y2 − x3

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= 0.

As an exercise, do the divisions with remainder and show that allremaining syzygy polynomials reduce to 0. So a Grobner basis of Iis given by

G = (t2 − x, t3 − y, t4 − z, tx − y,z − x2, ty − x2,y2 − x3).

However, do we need all these polynomials?

Definition 2.5.5 A Grobner basis G = (g1, ..., gs) is called mini-mal, if L(gi) ∤ L(gj) for all i ≠ j.

If, in addition, LC(gi) = 1 and tail(gi) is reduced with respectto G for all i, then, G is called reduced.

Remark 2.5.6 From any Grobner basis we can obtain a minimalone by deleting elements.

Proof. Given a Grobner basis G = (g1, ..., gs), by Lemma 2.3.3, theset {L(gi) ∣ i} has a unique subset of minimal elements with respectto divisibility. This subset clearly generates the same ideal, and,hence, the corresponding gi form a Grobner basis.

Theorem 2.5.7 Let > be a global ordering. Every ideal has aunique reduced Grobner basis (up to permutation of the elements).


Proof. Suppose G and H are reduced Grobner bases of the idealI. By

L(G) = L(I) = L(H)

and since G and H are minimal, for any g ∈ G there is an h ∈ Hwith L(g) = L(h). Then

s = g − h = tail(g) − tail(h)

and, as no term of the tails is divisible by any lead term, we have

s = NF(s,G).

Finally NF(s,G) = 0 by Theorem 2.4.8 and s ∈ I.

Remark 2.5.8 If G = (g1, ..., gs) is minimal and NF is a reducednormal form, then H = (h1, ..., hs) with

hi = NF(gi, (g1, ..., gi−1, gi+1, ..., gs))

is the reduced Grobner basis of I.

Proof. If G is minimal, then L(gi) is not divisible by any L(gj) fori ≠ j, so L(gi) = L(hi). By construction, tail(hi) is reduced withrespect to hj for j ≠ i. Moreover, no term of tail(hi) is divisible byL(hi), since by Definition and Theorem 2.3.2 the global orderingrefines divisibility.

Example 2.5.9 In Example 2.5.4 a minimal Grobner basis is

G = (t2 − x, tx − y,z − x2, ty − x2,y2 − x3).

Example 2.5.10 For Example 2.5.4 we can compute a minimalGrobner basis using Singular as follows:ring R=0,(t,z,y,x),lp;

ideal I = t2-x,t3-y,t4-z;

std(I);

[1]=y2-x3

[2]=z-x2

[3]=tx-y

[4]=ty-x2

[5]=t2-x

In this example, the result is already reduced.


Example 2.5.11 In general, the Grobner basis returned std maynot be reduced. To force Singular to compute the reduced Grobnerbasis, we set the option redSB:ring R=0,(x,y),lp;

ideal I = x+y,y;

std(I);

[1]=y

[2]=x+y

option(redSB);

std(I);

[1]=y

[2]=x

2.6 Exercises

Exercise 2.1 Show the following:

1) Any ideal of Z is of the form

⟨n⟩ = nZ = {nk ∣ k ∈ Z}

with n ∈ Z.

2) Let ⟨n⟩ ⊂ Z be an ideal and n > 0. Then

(n) is a prime ideal⇐⇒ n is a prime number.

Exercise 2.2 Let I, J ⊂K[x1, ..., xn] be ideals. Show that

V (I + J) = V (I) ∩ V (J)

andV (I ⋅ J) = V (I ∩ J) = V (I) ∪ V (J).

Recall that I ∩ J ,

I + J = {f + g ∣ f ∈ I and g ∈ J}

I ⋅ J = ⟨f ⋅ g ∣ f ∈ I and g ∈ J⟩

are ideals.

Exercise 2.3 1) Let K be a field and X ⊂ An(K) an affinealgebraic set. Prove that X is irreducible if and only if I(X) ⊂K[x1, ..., xn] is prime.


2) Show that, if K is algebraically closed and I is prime, thenV (I) is irreducible.

3) Give an example of a prime ideal I ⊂K[x1, ..., xn] over a fieldK such that V (I) is reducible.

4) For K algebraically closed, prove that

{affine varieties X ⊂ An(K)}I⇄V

{prime ideals in K[x1, ..., xn]}

is a one-to-one correspondence.

Exercise 2.4 Prove that the following rings are not Noetherian:

1) The polynomial ring K[x1, x2, ...] in infinitely many variables.

2) The ringR = {f ∈ Q[X] ∣ f(0) ∈ Z}

of polynomials with integer values at 0.

Exercise 2.5 Let K be a field and (a1, ..., an) ∈Kn. Prove that

⟨x1 − a1, ..., xn − an⟩ ⊂K [x1, ..., xn]

is a maximal ideal.Hint: Develop a multivariate division with remainder.

Exercise 2.6 Determine all maximal ideals of R [x].

Exercise 2.7 Let > be a semigroup ordering on the monomialsin the variables x1, ..., xn. Prove that the following conditions areequivalent:

1) > is a well ordering

(that is, any non-empty set of monomials has a smallest ele-ment).

2) xi > 1 ∀i.

3) xα > 1 for all 0 ≠ α ∈ Nn0 .

4) If xβ ∣ xα and xα ≠ xβ then xα > xβ

(that is, > refines divisibility).

Exercise 2.8 Sort all monomials of degree ≤ 2 in the variablesx, y, z with respect to the monomial orderings lp, dp, and ls.


Exercise 2.9 If > is a monomial ordering on x1, x2 define

D> = {α − β ∣ xα > xβ} ⊂ Z2

1) Plot the set D> for the monomial orderings lp and dp.

2) Assuming that we consider only monomials of degree ≤ d, findweight vectors u = (u1, u2) which represent lp and dp, respec-tively.

What about the weight ordering >w with weight vector w ∈ Z2

and tie break ordering lp?

Exercise 2.10 Let I = ⟨f1, f2, f3⟩ ⊂ R[x, y, z] be the ideal generatedby

f1 = x2 + y2 − 1

f2 = x2 + z2 − 1

f3 = x + y + z

1) Compute the reduced Grobner basis of I with respect to lp.

2) Deduce from this, that V (I) consists of 4 points.

Exercise 2.11 Use Buchberger’s algorithm to compute the reducedGrobner basis of

I = ⟨t2 − x, t3 − y, t4 − z⟩ ⊂ k [t, z, y, x]

for the lexicographical ordering t > z > y > x.

Exercise 2.12 Let I ⊂ K[x1, ..., xn] be an ideal and NF a reducednormal form. Show that

K[x1, ..., xn]/I → K ⟨xα ∣ xα ∉ L(I)⟩

f ↦ NF(f, I)

is an isomorphism of K-vectorspaces.

Exercise 2.13 Let

I = ⟨y + x2 + 1, xy + y2 + y⟩ ⊂ R[x, y]

and fix the monomial ordering lp.

1) Find the minimal generators of L(I).

2) Show that 1, x, y, y2 form a vectorspace basis of R[x, y]/I.

3) Compute the multiplication table of R[x, y]/I with respect tothis basis.

3

Computing with ideals andalgebraic sets

We now discuss how to use Buchberger’s algorithm to computewith algebraic sets, or equivalently with ideals. We have alreadyseen, how to decide, whether an algebraic set is contained in ahypersurface by solving the ideal membership problem.

3.1 Sums and intersections of ideals

Given two ideals I, J ⊂ R =K[x1, ..., xn], by

V (I) ∩ V (J) = V (I + J)

the intersection of algebraic sets corresponds to the sum of ideals,see Exercise 2.2. If

I = ⟨f1, ..., fs⟩ J = ⟨g1, ..., gr⟩

thenI + J = ⟨f1, ..., fs, g1, ..., gr⟩ ,

so computing the sum is trivial. Note, however, that even if (f1, ..., fs)and (g1, ..., gr) are Grobner bases, (f1, ..., fs, g1, ..., gr) may not be.

On the other hand, by

V (I) ∪ V (J) = V (I ∩ J)

the union of algebraic sets corresponds to the intersection of ideals(see again Exercise 2.2), which can be obtained as follows:

Algorithm 3.1.1 Let I, J ⊂ R be ideals and

I = ⟨f1, ..., fs⟩ J = ⟨g1, ..., gr⟩

63

3. COMPUTING WITH IDEALS AND ALGEBRAIC SETS 64

With

L = ⟨t ⋅ f1, ..., t ⋅ fs, (1 − t) ⋅ g1, ..., (1 − t) ⋅ gr⟩ ⊂ R[t]

it holdsI ∩ J = L ∩R

Proof. If f ∈ I ∩ J , then

R ∋ f = t ⋅ f − (1 − t) ⋅ f ∈ L.

On the other hand if f ∈ L ∩R, then

f(x, t) = t∑si=1ai(x, t)fi(x) + (1 − t)∑

rj=1bj(x, t)gj(x)

is independent of t, hence

I ∋ ∑si=1ai(x,1)fi(x) = f(x,1) = f(x,0) = ∑

rj=1bj(x,0)gj(x) ∈ J .

To turn this result into an algorithm, we need a way to computethe intersection of L with the subring R ⊂ R[t]. Before discussingthis fundamental problem in the next section, we try out ideal in-tersection in Singular:

Example 3.1.2 Consider the points

P1 = V (⟨x, y⟩) P2 = V (⟨x − 1, y⟩)P3 = V (⟨x, y − 1⟩) P4 = V (⟨x − 1, y − 1⟩)

in A2K. Then the union P1 ∪ P2 ∪ P3 ∪ P4 is defined by the ideal

⟨x, y⟩ ∩ ⟨x − 1, y⟩ ∩ ⟨x, y − 1⟩ ∩ ⟨x − 1, y − 1⟩ .

We compute a Grobner basis of this intersection using Singular:ring R=0,(x,y),lp;

ideal I1 = x,y;

ideal I2 = x-1,y;

ideal I3 = x,y-1;

ideal I4 = x-1,y-1;

std(intersect(I1,I2,I3,I4));

[1]=y2-y

[2]=x2-x

Hence

⟨x, y⟩ ∩ ⟨x − 1, y⟩ ∩ ⟨x, y − 1⟩ ∩ ⟨x − 1, y − 1⟩ = ⟨x2 − x, y2 − y⟩ ,

so the four points are represented as the intersection of two pairsof parallel lines, see Figure 3.1.


Figure 3.1: Four points as the intersection of two pairs of lines

In the context of primary decomposition, we will discuss, howto do the converse, that is, how to represent a given ideal as anintersection.

In Exercise 6.1 we will develop an algorithm for computing in-tersections of ideals with the help of syzygyies.

3.2 Elimination of variables

We now discuss how to intersect an ideal with a polynomial subring.Geometrically this corresponds to the projection of algebraic sets.

Definition 3.2.1 A monomial ordering > on

R =K[x1, ..., xm, xm+1, ..., xn]

is called elimination ordering for x1, ..., xm if for all f ∈ R itholds

L(f) ∈K[xm+1, ..., xn]⇒ f ∈K[xm+1, ..., xn].

Example 3.2.2 The lexicographic ordering with x1 > ... > xn isclearly an elimination ordering (for any 1 <m < n).

Algorithm 3.2.3 Let I ⊂ R be an ideal, > an elimination orderingfor x1, ..., xm, and G a Grobner basis of I with respect to >. Then

H = {g ∈ G ∣ L(g) ∈K[xm+1, ..., xn]}

is a Grobner basis of J = I ∩K[xm+1, ..., xn].


Proof. If g ∈ H ⊂ G ⊂ I, so L(g) ∈ K[xm+1, ..., xn], then by Defini-tion 3.2.1 we have g ∈K[xm+1, ..., xn], so g ∈ J . This shows H ⊂ J .

From this we get L(H) ⊂ L(J). For the other inclusion: If

f ∈ J ⊂ I

then L(f) ∈ L(I) = L(G), hence there is a g ∈ G with

L(g) ∣ L(f).

Since L(f) ∈ K[xm+1, ..., xn], also L(g) ∈ K[xm+1, ..., xn], that is,g ∈ H. This proves that L(J) ⊂ L(H), and, hence, H is a Grobnerbasis of J .

Example 3.2.4 Consider the twisted cubic

C = V (I) ⊂ A3K

defined byI = ⟨y − x2, z − x3⟩ ⊂K[x, y, z].

We compute the intersection

P = I ∩K[y, z]

by computing a Grobner basis of I with respect to lp with x > y > z:ring R=0,(x,y,z),lp;

ideal I = y-x2,z-x3;

std(I);

[1]=y3-z2

[2]=xz-y2

[3]=xy-z

[4]=x2-y

HenceP = ⟨y3 − z2⟩

The curve V (P ) corresponds to the image of C under the projectionmap

A3K → A2

K, (x, y, z)↦ (y, z)

see Figure 3.2. The elimination ideal can also be computed with thefollowing command (no matter what the monomial order is):eliminate(I,x);

[1]=y3-z2

To eliminate several variables, put their product into the secondargument.

In the next section we discuss the relation between the geometricconcept of projection and the algebraic concept of elimination.


Figure 3.2: Projection of the twisted cubic

3.3 Projection and elimination

For any ideal I ⊂ K [x1, ..., xn], for 0 ≤ m < n consider the elimi-nation ideal

Im = I ∩K[xm+1, ..., xn]

and the projection

πm ∶ AnK → An−m

K

πm (a1, ..., an) = (am+1, ..., an)

forgetting the first m coordinates.

Example 3.3.1 For

I = ⟨x2 − 1, y2 − 4⟩

V (I) = {(1,2) , (−1,2) , (1,−2) , (−1,−2)}

then π1 (V (I)) = {−2,2} and I1 = ⟨y2 − 4⟩. We will see how to usethe idea of projection to solve algebraic systems with a finite set ofsolutions.

Example 3.3.2 The projection of an algebraic set may not be analgebraic set: For the hyperbola S = V (xy − 1) we have π1 (S) =A1K/ {0}, see Figure 3.3.

Hence, the right question is, how to describe the closure of theprojection in the Zariski topology. Recall that the closed sets of


the Zariski topology are the algebraic sets. For any subset S ⊂ AnK

the Zariski closure isS = V (I(S)).

In particular, for an algebraic set S, we have S = V (I(S)).

Figure 3.3: Projection of a hyperbola

Theorem 3.3.3 If K =K and I ⊂K [x1, ..., xn] is an ideal, then

πm(V (I)) = V (Im)

for all 0 ≤m < n.

Proof. If (am+1, ..., an) = πm(a1, ..., an) ∈ πm(V (I)) and f ∈ Im,then, considering f as an element of I ⊂ R, we have

f(am+1, ..., an) = f(a1, ..., an) = 0,

henceπm(V (I)) ⊂ V (Im).

Since V (Im) is closed, this also holds true for the closure.Any g ∈ I (πm(V (I))) ⊂ K[xm+1, ..., xn] can be considered as

an element of K[x1, ..., xn] and vanishes on V (I). So by The-orem 2.1.11 (Nullstellensatz) there is an a with ga ∈ I, so ga ∈I ∩K[xm+1, ..., xn] = Im. Hence, again by Theorem 2.1.11,

I (πm(V (I))) ⊂√Im = I(V (Im)),


so by applying V we get

V (Im) = V (Im) = V (I(V (Im))) ⊂ V (I (πm(V (I)))) = πm(V (I)).

Example 3.3.4 Considering the ideal

I = ⟨x − st, y − t, z − s2⟩ ⊂K[s, t, x, y, z]

we compute the projection on the variables x, y, z using Singular:ring R=0,(s,t,x,y,z),lp;

ideal I = x-st, y-t, z-s2;

std(I);

[1]=x2-y2z

[2]=t-y

[3]=sy-x

[4]=sx-yz

[5]=s2-z

Henceπ2 (V (I)) = V (x2 − y2z).

We will now discuss how to interpret this calculation.

3.4 Polynomial maps between algebraic

sets

Proposition 3.4.1 Suppose K =K and we are given a polynomialmap

ϕ ∶ AmK → An

K

t = (t1, .., tm)↦ (f1(t), ..., fn(t))

with fi ∈ k [t1, .., tm]. Then for the closure of the image of ϕ it holds

im(ϕ) = V (Jm)

where

J = ⟨x1 − f1, ..., xn − fn⟩ ⊂ k [t1, .., tm, x1, ..., xn]

Proof. With the graph

Γ(ϕ) = {(t, x) ∈ An+mK ∣ x = ϕ(t)} = V (J).


we haveπm(V (J)) = πm(Γ(ϕ)) = im (ϕ) .

Using Theorem 3.3.3 it follows

im (ϕ) = V (Jm) .

Example 3.4.2 The Whitney umbrella is the image of the map

ϕ ∶ A2K → A3

K

(s, t) ↦ (st, t, s2)

Applying Proposition 3.4.1, we have already determined in Example3.3.4 that the

im (ϕ) = V (x2 − y2z).

Figure 3.4 shows the Whitney umbrella over the reals K = R.

Figure 3.4: Whitney umbrella

In general we define:

Definition 3.4.3 A polynomial parametrization of the alge-braic variety X ⊂ An

K is a morphism ϕ ∶ AmK → An

K with im(ϕ) =X.

Computing the ideal of the image of a polynomial parametriza-tion is also referred to as implicitation, since we determine implicitequations for the image of the parametrization.


Example 3.4.4 Given a homogeneous linear system of equations,using Gaussian elimination we can compute a parametrization ofthe solution set.

For example, consider the linear system of equations from Ex-ample 1.5.2

f1 = x1 + x2 + x5 = 0f2 = x1 + x2 + x3 + x4 + x5 = 0

Gaussian elimination yields

x1 + x2 + + x5 = 0x3 + x4 = 0

So we obtain a basis of the vector space of solutions

V (⟨f1, f2⟩) = ⟨

⎛⎜⎜⎜⎜⎜⎜⎝

−10001

⎞⎟⎟⎟⎟⎟⎟⎠

,

⎛⎜⎜⎜⎜⎜⎜⎝

00−110

⎞⎟⎟⎟⎟⎟⎟⎠

,

⎛⎜⎜⎜⎜⎜⎜⎝

−11000

⎞⎟⎟⎟⎟⎟⎟⎠

⟩

and, hence, a parametrization

ϕ ∶ A3K → A5

K

(t1, t2, t3) ↦ (−t1 − t3, t3,−t2, t2, t1)

withV (⟨f1, f2⟩) = im(ϕ).

Even more general we define:

Definition 3.4.5 A map ϕ ∶X → Y between algebraic sets X ⊂ AmK

and Y ⊂ AnK, which is induced by a polynomial map ϕ′ ∶ Am

K → AnK

with ϕ′(X) ⊂ Y is called a morphism.

Proposition 3.4.6 If X = V (I) ⊂ AmK with an ideal I ⊂K[t1, ..., tm]

andϕ ∶ X → An

K

t = (t1, .., tm)↦ (f1(t), ..., fn(t))

is a morphism, thenim(ϕ) = V (Jm)

where

J = ⟨I⟩ + ⟨x1 − f1(t), ..., xn − fn(t)⟩ ⊂ k [t1, .., tm, x1, ..., xn] .


Proof. Again

Γ(ϕ) = {(t, x) ∈ An+mK ∣ x = ϕ(t), t ∈X} = V (J)

is the graph, and we can argue as in the proof of Proposition 3.4.1.

Example 3.4.7 The image of the sphere

S = V (t21 + t22 + t

23 − 1) ⊂ A3

K

under the morphism

ϕ ∶ S → A3K

(t1, t2, t3) ↦ (t1t2, t1t3, t2t3)

is the Steiner surface. Figure 3.5 shows an image for K = R.By computing a Grobner basis of

J = ⟨t21 + t22 + t

23 − 1, x − t1t2, y − t1t3, z − t2t3⟩

with respect to an elimination ordering for t1, t2, t3, by Proposition3.4.6 we get

im(ϕ) = V (x2y2 + x2z2 + y2z2 − xyz).


Figure 3.5: Steiner surface over the reals


The elimination ideal can be interpreted as the kernel of a ringhomomorphism. Recall that the kernel of a ring homomorphismis an ideal (in fact, the ideal definition is motivated by the keyproperties of kernels).

Proposition 3.4.8 Let I ⊂ K[t1, ..., tm] be an ideal. The kernelker(f) of a ring homomorphism

f ∶ K[x1, ..., xn] → K[t1, ..., tm]/Ixi ↦ fi

is an ideal, andker(f) = Jm

with

J = ⟨I⟩ + ⟨x1 − f1(t), ..., xn − fn(t)⟩ ⊂K [t1, .., tm, x1, ..., xn] .

Note that J is independent of the representatives fi of the classesfi, since we add I.Proof. Write I = ⟨g1, ..., gs⟩ ⊂K[t1, ..., tm]. We have

h(x) ∈ ker(f)

⇔ h(f1(t), ..., fn(t)) ∈ I

⇔ ∃ai(t) ∈K[t1, ..., tm] ∶ h(f1(t), ..., fn(t)) = ∑si=1ai(t)gi(t)

⇔ h(x) ∈ J ∩K [x1, ..., xn]

For the last equivalence observe that, by Taylor’s formula,

h(x) = h(f1(t), ..., fn(t)) +∑ni=1ci(x, t) ⋅ (xi − fi(t))

with some ci(t, x) ∈K [t1, .., tm, x1, ..., xn].So Proposition 3.4.6 can be written in a more intrinsic way as

follows: If X = V (J) and

ϕ ∶ AmK → An

K

t = (t1, .., tm)↦ (f1(t), ..., fn(t))

thenϕ(X) = V (ker(f)).

On the level of schemes, we would simply define the image as thescheme defined by ker(f).


Example 3.4.9 By computing a kernel, the ideal of the Steinersurface in Example 3.4.7 can be determined as follows:ring S = 0,(x,y,z),dp;

ring R = 0,(t1,t2,t3),dp;

ideal I = t1^2+t2^2+t3^2-1;

qring Q = std(I);

map f = S, ideal(t1*t2, t1*t3, t2*t3);

setring S;

kernel(Q,f);

[1]=x2y2+x2z2+y2z2-xyz

Note, that the ring homomorphism f is defined in the target ringQ (and gives a reference to the source ring S). On the other handthe kernel is computed in the source ring S (and you have to givea reference to the target ring Q).

3.5 Rational maps between varieties

More generally, one can consider maps given by rational functions.This is indeed necessary, for example to parametrize a circle, seeExercise 3.2. We discuss how to determine the image of a rationalmap. This is the algorithmic basis of birational geometry, whichstudies algebraic varieties up to birational maps, that is, rationalmaps which admit a rational inverse.

Definition 3.5.1 A rational map

ϕ = (f1

g1

, ...,fngn

) ∶ AmK An

K

is a tuple of rational functions figi∈K (t1, .., tm). Hence it defines a

mapϕ ∶ Am

K/Z → AnK

t = (t1, .., tm)↦ (f1(t)g1(t) , ...,fn(t)gn(t))

where AmK/Z is the locus on which all component functions are de-

fined, that is, Z = V (g1 ⋅ ... ⋅ gn). The image of ϕ is defined as

im (ϕ) = ϕ(AmK/Z).

The dashed arrow indicates that the map is not defined on thewhole of Am

K . Of course any polynomial map is also a rational map.

Proposition 3.5.2 Suppose K =K,

ϕ = (f1

g1

, ...,fngn

) ∶ AmK An

K


is a rational map, and g = g1 ⋅ ... ⋅ gn. Then for the closure of theimage of ϕ it holds

im (ϕ) = V (Jm+1)

where

J = ⟨g1x1 − f1, ..., gnxn − fn, 1 − gs⟩ ⊂K[s, t1, .., tm, x1, ..., xn].

The idea of the proof is simple: The equation 1−gs removes thesolutions (t, x) of the equations gixi − fi with some gi (x) = 0.Proof. With the map

ι ∶ AmK/Z → An+m+1

K

ι(t)↦ (g(t)−1, t, ϕ(t))

we obtain a commutative diagram

An+m+1K

AmK/Z

ϕ-

ι

-

AnK

πm+1

-

It holdsι(Am

K/Z) = V (J)

since (s, t, x) ∈ V (J) if and only if

gi(t) ≠ 0∀i and xi =fi(t)

gi(t)∀i and s =

1

g(t)

that is (s, t, x) ∈ ι(AmK/Z). Applying πm+1 we get

ϕ(AmK/Z) = πm+1(V (J))

and hence, by Theorem 3.3.3,

ϕ(AmK/Z) = πm+1(V (J)) = V (Jm+1) .

Remark 3.5.3 With the notation as in Proposition 3.5.2, the graphof ϕ is

Γ(ϕ) = V (J ∩K[t1, .., tm, x1, ..., xn]).


Example 3.5.4 Consider the rational map

ϕ ∶ A1K A2

K

t ↦ (1−t2t2+1 ,

2tt2+1)

defined on the complement of Z = V (t2 + 1). To determine theimage, we calculate a Grobner basis of

I = ⟨(t2 + 1)x − (1 − t2) , (t2 + 1) y − 2t, 1 − (t2 + 1)2s⟩

for lp with s > t > x > y using Singular:ring R=0,(s,t,x,y),lp;

ideal I = (t2+1)*x-(1-t2), (t2+1)*y-2t, 1-(t2+1)^2*s;

option(redSB);

std(I);

[1]=x2+y2-1

[2]=ty+x-1

[3]=tx+t-y

[4]=4s-2x+y2-2

Hence, the closure of the image of ϕ

C = im (ϕ) = V (I2) = V (x2 + y2 − 1)

is a circle.

Definition 3.5.5 A rational parametrization of the algebraicvariety X ⊂ An

K is a rational map ϕ ∶ AmK An

K with im(ϕ) =X.

In general the parametrization problem is much more compli-cated. Indeed, most varieties do not admit a rational parametriza-tion. For the circle, however, the problem is easy:

Remark 3.5.6 To find ϕ, we do stereographic projection from apoint P on the circle C to a line, for example, from P = (−1,0) to{x = 0}. So consider the linear system of all lines

Lt = {y − t (x + 1) = 0} ⊂ A2K

through P , and t ∈ P1K = A1

K ∪ {∞}. Writing

Lt ∩C = {P,Pt}Lt ∩ {x = 0} = {Qt} = {(0, t)}

we defineϕ(Qt) = Pt,


see Figure 3.6. As an exercise do the calculation. Note that theaffine space A1

K is not sufficient to parametrize all points of C: Inorder to obtain P we have to consider the vertical line L∞ whichcorresponds to the point at infinity on A1

K. Hence, we have to passfrom affine to the projective space P1

K.This is a hint, that birational geometry should better be done

with projective varieties.

Figure 3.6: Rational parametrization of the circle.

Remark 3.5.7 The parametrization is indeed birational. The in-verse

ϕ−1 = (x

x + 1) ∶ A1

K C

is obtained by solving the equation of the line for t.

Like in the case of polynomial maps, we can generalize ratio-nal maps between affine spaces to rational maps between algebraicvarieties.

Definition 3.5.8 If I ⊂ K[x1, ..., xn] is a prime ideal and X =V (I) ⊂ An

K is the corresponding algebraic variety, then the coordi-nate ring

K[X] =K[x1, ..., xn]/I


of X is an integral domain. Hence, we can form the quotient field

K(X) = quot(K[X])

by calculating with fractions. It describes the rational functions onX and is called the rational function field of X.

The domain of definition D(q) of q ∈K(X) is

D(q) = {x ∈X ∣ ∃f , g ∈K[X] with g(x) ≠ 0 and q =f

g} .

So q is a well-defined map q ∶D(q)→ A1K.

Definition 3.5.9 A rational map

ϕ = (q1, ..., qn) ∶X AnK

is a tuple of rational functions qi ∈ K(X). The domain of defi-nition D(ϕ) of ϕ is

D(ϕ) =D(q1) ∩ ... ∩D(qn),

and the image of ϕ is

im(ϕ) = ϕ(D(ϕ)).

If Y ⊂ AnK is a variety, then a rational map

ϕ ∶X Y

is a rational map ϕ ∶X AnK with im(ϕ) ⊂ Y .

A straightforward combination of the ideas of Propositions 3.4.6and 3.5.2 generalizes these result to an algorithm for computing theimage of a rational map X An

K (we skip the proof):

Algorithm 3.5.10 Suppose K = K, I ⊂ K[t1, ..., tm] is a primeideal, X = V (I) ⊂ Am

K is the corresponding algebraic variety

ϕ = (f1

g1

, ...,fngn

) ∶X AnK

is a rational map, and g = g1 ⋅ ... ⋅ gn. Then for the closure of theimage of ϕ it holds

im (ϕ) = V (Jm+1)

where

J = ⟨I⟩+⟨g1x1 − f1, ..., gnxn − fn, 1 − gs⟩ ⊂ k [s, t1, .., tm, x1, ..., xn] .


Example 3.5.11 We compute the image of the node

X = V (t31 + t21 − t

22) ⊂ A2

K

(see Figure 3.7) under the rational map

t1t2∶X A1

K

using Singular:ring R=0,(s,t1,t2,x),lp;

ideal I = t1^3+t1^2-t2^2, t2*x-t1,1-t2*s;

std(I);

t2*x^3+x^2-1

t1-t2*x

s*x^2-s+x^3

s*t2-1

Hence the closure of the image

im (ϕ) = V (0) = A1K

is the whole of A1K.

The map ϕ is indeed birational: From the equations

t2x3 + x2 − 1 = 0

t1 − t2x = 0

we can read off the inverse

ϕ−1 = (−x2 − 1

x2,x2 − 1

x3) ∶ A1

K X

which yields a rational parametrization of X.See also Exercise 1.5.2.

Example 3.5.12 Let I = ⟨xy − zw⟩ ⊂ R = K[x, y, z,w] and X =V (I) ⊂ A3

K. Then K[X] = R/I is not a unique factorization do-main, for example,

xy = zw.

Hence we have two representations for the rational function

q =x

z=w

y∈K(X)

SoS/V (z) ⊂D(q) and S/V (y) ⊂D(q)

henceS/V (y, z) ⊂D(q).


–1

0

1

0 1

Figure 3.7: Node

Do we have equality? How to compute the domain of a rationalfunction in general? The following lemma describes the domain byan ideal:

Algorithm 3.5.13 Let X ⊂ AnK be a variety and q = f

g ∈ K(X).Then the domain of definition of q is the complement

D(q) =X/V (N)

of the vanishing locus of the ideal

N = {γ ∈K[x1, ..., xn] ∣ γ ⋅ f ∈ ⟨g⟩ + I(X)} .

Proof. The possible denominators of q are the elements of theideal

N = {γ ∈K[X] ∣ γ ⋅ q ∈K[X]}

It is the image of N under the canonical epimorphism

K[x1, ..., xn]→K[x1, ..., xn]/I(X) =K[X]

since

γ ∈K[X]⇔ γ ⋅f

g∈K[X]⇔ γ ⋅ f ∈ ⟨g⟩⇔ γ ⋅f ∈ ⟨g⟩+I(X)⇔ γ ∈ N .


How can we compute the ideal N using Grobner bases? It is aspecial case of an ideal quotient

I ∶ J = {w ∈ R ∣ w ⋅ J ⊂ I}

of two ideals I, J ⊂ R. Indeed,

N = (⟨g⟩ + I(X)) ∶ ⟨f⟩ .

3.6 Ideal quotients

We discuss how to compute ideal quotients using Grobner bases.Ideal quotients will also play a major role in primary decomposition.

Definition 3.6.1 Let R be a ring. For ideal I, J ⊂ R the idealquotient is

I ∶ J = {f ∈ R ∣ f ⋅ J ⊂ I} .

Remark 3.6.2 For all ideals I, J ⊂ R it holds

(1) I ∶ J is an ideal of R,

(2) I ⊂ (I ∶ J),

(3) J ⊂ I if and only if I ∶ J = ⟨1⟩.

For all ideals I1, ..., Is, J ⊂ R it holds

(4) (I1 ∩ ... ∩ Is) ∶ J = (I1 ∶ J) ∩ ... ∩ (Is ∶ J),

(5) J ∶ (I1 + ... + Is) = (J ∶ I1) ∩ ... ∩ (J ∶ Is),

For all ideals I, J,L ⊂ R it holds

(6) (I ∶ J) ∶ L = I ∶ (J ⋅L).

We do the straightforward proof in Exercise 3.5.For computational purposes we now focus on ideals in R =

K[x1, ..., xn]. By Remark 3.6.2.(5) and the ideal intersection algo-rithm 3.1.1, it is sufficient to compute ideal quotients by principalideals: If J = ⟨g1, ..., gr⟩ then

I ∶ J = I ∶ (⟨g1⟩ + ... + ⟨gr⟩) = (I ∶ ⟨g1⟩) ∩ ... ∩ (I ∶ ⟨gr⟩) .

Again using the ideal intersection algorithm, we can compute:


Algorithm 3.6.3 Let I ⊂ R be an ideal and g ∈ R. If we write

I ∩ ⟨g⟩ = ⟨g ⋅ f1, ..., g ⋅ fs⟩

with fi ∈ R, then it holds

I ∶ ⟨g⟩ = ⟨f1, ..., fs⟩ .

Proof. Any element of I ∩ ⟨g⟩ is a multiple of g, so

I ∩ ⟨g⟩ = ⟨g ⋅ f1, ..., g ⋅ fs⟩ .

By⟨f1, ..., fs⟩ ⋅ g ⊂ I

we have⟨f1, ..., fs⟩ ⊂ I ∶ ⟨g⟩ .

On the other hand, if f ∈ I ∶ ⟨g⟩, then f ⋅ g ∈ I ∩ ⟨g⟩, so

f ⋅ g = ∑iai ⋅ g ⋅ fi = g ⋅∑iai ⋅ fi

with ai ∈ R. Hence

f = ∑iaifi ∈ ⟨f1, ..., fs⟩ .

Example 3.6.4 We compute the domain of

q =x

z∈K(X)

where X = V (xy − zw) ⊂ A4K using Singular:

ring R=0,(x,y,z,w),dp;

ideal I = xy-zw;

quotient(I+ideal(z),ideal(x));

[1]=z

[2]=y

So the domain of q is the complement of the plane V (y, z) ⊂X.

We can interpret ideal quotients geometrically as follows:

Theorem 3.6.5 Let I, J ⊂ R be ideals.

1) If I ⊂ R is a prime ideal, and J /⊂ I is an ideal then I ∶ J = I.


2) If I = ⋂sl=1Il with Il prime and J is an ideal, then

I ∶ J = ⋂J /⊂IlIl

Proof. Let g ∈ J with g ∉ I. If f ∈ I ∶ J , that is, f ⋅ J ⊂ I, thenf ⋅ g ∈ I. Since I is a prime ideal, it follows that f ∈ I.

Using the first claim and Remark 3.6.2.(5) and (3), the secondclaim follows:

I ∶ J = (⋂J /⊂Il(Il ∶ J)´¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¶

Il

) ∩ (⋂J⊂Il(Il ∶ J)´¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¶

R

)

For prime ideals Il by the Nullstellensatz

J ⊂ Il ⇒ V (Il) ⊂ V (J)⇒√J ⊂ Il ⇒ J ⊂ Il

that is,J ⊂ Il⇔ V (Il) ⊂ V (J)

hence:

Corollar 3.6.6 If I = ⋂sl=1Il with Il ⊂ R prime and J ⊂ R is anideal, then

V (I ∶ J) = ⋃V (Il)/⊂V (J)V (Il)

This impliesV (I ∶ J) = V (I)/V (J)

Example 3.6.7 Consider the ideal

I = ⟨x2 − x, y2 − y⟩ = ⟨x, y⟩∩⟨x, y − 1⟩∩⟨x − 1, y⟩∩⟨x − 1, y − 1⟩ ⊂K[x, y]

withV (I) = {(0,0), (0,1), (1,0), (1,1)} .

For J = ⟨x − y⟩ we compute:ring R=0,(x,y),dp;

ideal I = x2-x, y2-y;

ideal J = x-y;

quotient(I, J);

[1]=x+y-1

[2]=y2-y

SoI ∶ J = ⟨x + y − 1, y2 − y⟩ ,

henceV (I ∶ J) = {(0,1), (1,0)} ,

see Figure 3.8.


Figure 3.8: Ideal quotient

3.7 Solving algebraic systems with a fi-

nite set of solutions

Suppose K = K. Given an ideal I ⊂ K[x1, ..., xn] with a finite setof solutions, how to compute the points of V (I) explicitly? Thenaive algorithm would be:

1) Project to each coordinate, that is, compute I ∩K[xi] = ⟨fi⟩by elimination (note that K[xi] is a principal ideal domain).

2) Compute the finite sets V (fi), for example by numerical tech-niques.

3) Check, which points of V (f1) × ... × V (fn) are contained inV (I).

Example 3.7.1 Consider the ideal

I = ⟨2x2 − xy + 2y2 − 2, 2x2 − 3xy + 3y2 − 2⟩

from Section 1.1.ring R=0,(x,y),lp;

ideal I = 2x2-xy+2y2-2, 2x2-3xy+3y2-2;

eliminate(I,x);

[1] = y3-y

eliminate(I,y);

[1] = 4x4 - 5x2 + 1

Hence, byI ∩K[y] = ⟨y3 − y⟩ = ⟨y(y + 1)(y − 1)⟩


and

I ∩K[x] = ⟨4x4 − 5x2 + 1⟩ = ⟨(x + 1)(x − 1)(2x + 1)(2x − 1)⟩

we get

V (I) ⊂ {0,−1,1} × {−1,1,−1

2,1

2}.

Testing these 9 points with respect to the generators of I yields

V (I) = {(1,0) , (−1,0) ,(1

2,1) ,(−

1

2,−1)} .

But we can do better. The cornerstone is the following theorem:

Theorem 3.7.2 Let K =K, I ⊂K [x1, ..., xn]. Then the followingare equivalent:

1) ∣V (I)∣ <∞

2) If G is a Grobner basis of I, then for every i there is a g ∈ Gwith

L(g) = xαii

and αi ≥ 0.

3) dimK(K [x1, ..., xn] /I) <∞.

Proof. For (1)⇒ (2) consider the projection

π ∶ V (I)→ A1K , (a1, ..., an)↦ ai

Thenf =∏t∈π(V (I))(xi − t) ∈ I(V (I)) =

√I,

hence fw ∈ I for some w ≥ 1, so L(fw) ∈ L(I).(2)⇔ (3) follows, since (by Exercise 2.12)

K [x1, ..., xn] /I ≅ K ⟨xα ∣ xα ∉ L(I)⟩

as K-vector spaces.(3)⇒ (1): Since dimK(K [x1, ..., xn] /I) <∞ the classes x0

i , x1i , x

2i , ...

are linearly dependent, so there is a 0 ≠ fi ∈ K[xi] with fi ∈ I. Asabove,

V (I) ⊂ V (f1) × ... × V (fn) .

In Exercise 3.8 we show that dimK(K [x1, ..., xn] /I) is a boundfor the number of solutions, and equality holds if I is radical.


Remark 3.7.3 By the previous theorem, if V (I) ≠ ∅, the minimalGrobner basis of I for lp with x1 > ... > xn contains equations

xα11 − g1 (x1, ..., xn)

xα22 − g2 (x2, ..., xn)

⋮

xαn−1n−1 − gn−1 (xn−1,xn)

gn (xn)

with αi > 0.To determine V (I) solve the equations for xn, ..., x1 and keep

those solutions, which satisfy the remaining Grobner basis elements.This can be done by numerical methods for solving univariate poly-nomial equations.

Example 3.7.4 Consider again, the intersection of two ellipsesfrom Section 1.1. As we have already seen there, for y > x we get:ring R=0,(y,x),lp;

ideal I = 2x2-xy+2y2-2, 2x2-3xy+3y2-2;

std(I);

[1]=4x4-5x2+1

[2]=3y+8x3-8x

Solving 4x4 − 5x2 + 1 = 0 yields

x = 1,−1,1

2,−

1

2and from the second equation we get

V (I) = {(1,0) , (−1,0) , (1

2,1), (−

1

2,−1)} .

On the other hand, for x > y,ring R=0,(x,y),lp;

ideal I = 2x2-xy+2y2-2, 2x2-3xy+3y2-2;

std(I);

[1]=y3-y

[2]=2xy-y2

[3]=2x2-3xy+3y2-2

so we gety = 0,1,−1.

Solving the third equality yields

V (I) ⊂ {(1,0) , (−1,0) , (1,1), (1

2,1), (−1,−1), (−

1

2,−1)} .

To obtain V (I), we discart the solutions (1,1), (−1,−1) which donot satisfy 2xy − y2 = 0.



3.8 Exercises

Exercise 3.1 Show that for

C = {(t2 − 1, t3 − t) ∣ t ∈ R} ⊂ A2R

we have I (C) = ⟨x3 + x2 − y2⟩. Make a plot of the curve.

Exercise 3.2 Show, that for char (K) ≠ 2 the circle C = V (x2 + y2 − 1) ⊂A2K does not admit a polynomial parametrization.

Exercise 3.3 Let

S = V (t21 + t22 + t

23 − 1) ⊂ A3

K

be the unit sphere, and consider the morphism

α ∶ S → A3K

(t1, t2, t3)↦ (t1t2, t1t3, t2t3)

Show, that

1) if K = R,C then

X = im (α) = V (x21x

22 + x

21x

23 + x

22x

23 − x1x2x3) ,

2) if K = C, then im(α) =X,

3) if K = R, then im(α) ⫋X.

Exercise 3.4 Compute the closure of the image of the curve (Fig-ure 3.9)

C = V (y3 − 3x2y − (x2 + y2)2) ⊂ A2R

under the rational map

ϕ = (x2

y2,x

y) ∶ C A2

R

Exercise 3.5 Let R = K[x1, ..., xn] be a polynomial ring over afield K and for ideals I, J ⊂ R

I ∶ J = {f ∈ R ∣ f ⋅ J ⊂ I}

the ideal quotient. Prove that:


–1

0

1

0 1

Figure 3.9: Curve with triple point

1) For all ideals I, J ⊂ R it holds

(a) I ∶ J is an ideal of R,

(b) I ⊂ (I ∶ J),

(c) J ⊂ I if and only if I ∶ J = ⟨1⟩.

2) For all ideal I1, ..., Is, J ⊂ R it holds

(a) (I1 ∩ ... ∩ Is) ∶ J = (I1 ∶ J) ∩ ... ∩ (Is ∶ J),

(b) J ∶ (I1 + ... + Is) = (J ∶ I1) ∩ ... ∩ (J ∶ Is).

3) For all ideals I, J,L ⊂ R it holds

(I ∶ J) ∶ L = I ∶ (J ⋅L).

Exercise 3.6 Let R =K[x1, ..., xn].

1) Using the Singular command eliminate, write a procedurethat computes I ∩ J for ideals I, J ⊂ R.

2) Suppose J = ⟨g⟩ ⊂ R is principal and I ⊂ R is an ideal. Apply-ing your ideal intersection command from (1), write a Sin-gular procedure which computes I ∶ J .

3) For any two ideals I, J ⊂ R, write a procedure that computesI ∶ J . Try out some examples.


Exercise 3.7 Let

I = ⟨xz − y2, x − yz⟩ ⊂ R[x, y, z]

1) Use your implementation from Exercise 3.6 to compute

J = I ∶ ⟨x⟩

andI ∶ J

Compare with the Singular command quotient.

2) Write I as an intersection of two prime ideals.

3) Make a plot of V (I) ⊂ A3R. Hint: Find a polynomial parametriza-

tion of V (J).

Exercise 3.8 Let K = K, and I ⊂ K [x1, ..., xn] with ∣V (I)∣ < ∞.Show that

∣V (I)∣ ≤ dimK (K [x1, ..., xn] /I)

and equality holds if I is radical.Hint: Use interpolation to find a basis.

Exercise 3.9 Let K = K, and I ⊂ K [x1, ..., xn] with ∣V (I)∣ < ∞.Then the following conditions are equivalent:

1) For all p, q ∈ V (I) with p ≠ q we have pn ≠ qn and I is radical.

2) The reduced Grobner basis of I with respect to the lexico-graphic ordering with x1 > ... > xn has the form

x1 − g1 (xn)

x2 − g2 (xn)

⋮

xn−1 − gn−1(xn)

gn (xn)

and gn (xn) is squarefree.

Hint: Use Exercise 3.8.

4

Modules over principalideal domains

4.1 Overview

In the same way as for ideals, we first discuss the special case ofmodules over principal ideal domains R like Z or the polynomialring K[x] in one variable over a field. Examples you know arevector spaces over a field K, or abelian groups G, which are exactlythe Z-modules with the scalar multiplication

Z ×G→ G, (n, g)↦ n ⋅ g = g + ... + g´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

n

Groups as Z-modules play an important role, for example, in thetheory of toric varieties (divisor class group).

For modules over principle ideal domains, we will describe an al-gorithm, which generalizes, like Buchberger’s algorithm, both Gaus-sian reduction and the Euclidean algorithm. An interesting classof R-modules (however not the most general, as we will see) areimages M = im(A) of matrices A ∈ Rn×m. To describe M and todecide, whether a given vector v ∈ Rn is contained in M , we requirea suitable normal form for A.

From linear algebra over a field K we know: If A ∈Kn×m, thenthere are changes of bases T ∈ GL (m,K) and S ∈ GL (n,K), whichtransform A to the normal form

S ⋅A ⋅ T =

⎛⎜⎜⎜⎝

1⋱ 0

10 0

⎞⎟⎟⎟⎠

=D

90

4. MODULES OVER PRINCIPAL IDEAL DOMAINS 91

(where the number of ones on the diagonal is the rank of A). Thenv ∈ im(A) if and only if S ⋅ v ∈ im(D), which in turn we can decideimmediately.

If we replace K by a principal ideal domain R, then we cannotexpect the normal form to be that simple any more: By the adjointmatrix formula for the inverse, a matrix T ∈ Rn×n is invertible ifand only if det (T ) is a unit. For example A = (2) ∈ Z1×1 already isin normal form, because Z× = {±1}. In general we can achieve that

S ⋅A ⋅ T =D =

⎛⎜⎜⎜⎝

d1

⋱ 0dr

0 0

⎞⎟⎟⎟⎠

∈ Rn×m

where di is a divisor of di+1 for all i and the di are uniquely deter-mined by A.

The chapter is organized as follows: We first prove this claim inthe elementary divisor theorem. The proof also gives an algorithmto obtain S, T and D. Using the normal form D we prove variousstructure results: Generalizing vector spaces over a field, we firstrecall some basic facts on modules over a ring, and then describefinitely generated modules over principle ideal domains. In partic-ular, we will discuss finitely generated abelian groups (Z-modules),and the Jordan normal form.

4.2 The elementary divisor algorithm

Remark 4.2.1 A matrix A ∈ Rn×n is invertible (that is, A ∈ GL (n,R))if and only if its determinant is a unit, that is,

det (A) ∈ R×

Proof. If A ⋅ A−1 = E, then det (A) ⋅ det (A−1) = 1. On the otherhand, if det (A) ∈ R×, then

A−1 =Aadj

det (A)∈ Rn×n

with the adjoint matrix Aadj = ((−1)i+j

det (Aji))i,j ∈ Rn×n, where

Aji is obtained by canceling the j-th row and i-th column of A.

Theorem 4.2.2 (Elementary divisor theorem) Let R be a prin-cipal ideal domain and A ∈ Rn×m. Then there are S ∈ GL (n,R) and


T ∈ GL (m,R) and r ≤ min (n,m) with

S ⋅A ⋅ T =D =

⎛⎜⎜⎜⎝

d1

⋱ 0dr

0 0

⎞⎟⎟⎟⎠

∈ Rn×m

andd1 ∣ d2, d2 ∣ d3 . . . dr−1 ∣ dr ≠ 0

The di are uniquely determined by A up to multiplication with units,and are called the elementary divisors of A, and D is called theSmith normal form of A.

Remark 4.2.3 In the special case where R = K is a field, we canachieve di = 1 and we obtain S as a product of row operations and Tas a product of column operations, see Section 1.5. Permuting rowsand columns, we can assume that a11 ≠ 0 (if A ≠ 0). Subtractingthe

a1,ja1,1

-multiple of the first column from the j-th column (and in

the same way for rows) we transform A to

⎛⎜⎜⎜⎝

a1,1 0 ⋯ 00⋮ ∗0

⎞⎟⎟⎟⎠

Finally, multiply the first column by 1a1,1

. The claim follows by

induction.

In a ring R it is in general not possible to forma1,ja1,1

as a1,1 may

not be a unit. We now prove Theorem 4.2.2 in the case that R isa Euclidean domain (see Definition 2.2.3). Here we can replace thedivision

a1,ja1,1

by the division with remainder. The proof yields an

algorithm to compute the Smith normal form.Proof. Let R be a Euclidean domain with norm d. We may assumethat A ≠ 0.

1) By permuting rows and columns we can assume that a1,1 ≠ 0and

d (a1,1) ≤ d (ai,j) or ai,j = 0

for all (i, j) ≠ (1,1).

2) If some entry a1,j in the first row (analogously for the firstcolumn) is not divisible by a1,1, then write by division withremainder

a1,j = q ⋅ a1,1 + r


with d (r) < d (a1,1). The case r = 0 does not happen, sinceby assumption a1,1 ∤ a1,j.

After subtracting q times the 1st column from the j-th col-umn, we obtain

d (a1,1) > d (a1,j)

Now go back to step (1). This process terminates, sinced (a1,1) becomes smaller in every run.

3) If, now, all entries of the first row and column are divisible bya1,1, by adding multiples of the first column, we can transformA into

⎛⎜⎜⎜⎝

a1,1 0 ⋯ 0

∗ ∗

⎞⎟⎟⎟⎠

and, by adding multiples of the first row, further to

⎛⎜⎜⎜⎝

a1,1 0 ⋯ 00⋮ A′

0

⎞⎟⎟⎟⎠

If A′ has an entry ai,j, which is not divisible by a1,1, we addthe i-th row to the first row and go back to (2). Again, d (a1,1)decreases strictly.

4) If all entries of A are divisible by a1,1, then also the entries ofA′, since they are R-linear combinations of entries of A.

By induction on min (n,m) the claim follows.

For the induction start (n = 1 or m = 1) the steps (1) − (3) arethe Euclidean algorithm on the entries of the matrix.

Before discussing the general case over a principle ideal domainR and the uniqueness, we test the algorithm at an example:

Example 4.2.4 We compute the Smith normal form of

A = (6 9 66 6 7

) ∈ Z2×3

and at the same time S ∈ GL(2,Z) and T ∈ GL(3,Z) by simultane-ously doing the row and column operations on a 2 × 2 respectively3 × 3 unit matrix. Division by remainder (step 2) yields

(1 00 1

) (6 3 66 0 7

)⎛⎜⎝

1 −1 00 1 00 0 1

⎞⎟⎠


Permuting (step 1)

(1 00 1

) (3 6 60 6 7

)⎛⎜⎝

−1 1 01 0 00 0 1

⎞⎟⎠

Reducing the first row and column (step 3)

(1 00 1

) (3 0 00 6 7

)⎛⎜⎝

−1 3 21 −2 −20 0 1

⎞⎟⎠

Since 7 is not divisible by 3, we add the second to the first row

(1 10 1

) (3 6 70 6 7

)⎛⎜⎝

−1 3 21 −2 −20 0 1

⎞⎟⎠

Division by remainder (step 2)

(1 10 1

) (3 6 10 6 7

)⎛⎜⎝

−1 3 41 −2 −40 0 1

⎞⎟⎠

Permuting (step 1)

(1 10 1

) (1 6 37 6 0

)⎛⎜⎝

4 3 −1−4 −2 11 0 0

⎞⎟⎠

Reducing the first row and column (step 3)

(1 1−7 −6

) (1 0 00 −36 −21

)⎛⎜⎝

4 −21 −13−4 22 131 −6 −3

⎞⎟⎠

Applying the algorithm to the submatrix obtained by deleting (orignoring) the first row and column, we can proceed, for example, asfollows

( −36 −21 )↦ ( −15 −21 )↦ ( −15 −6 )↦ ( −3 −6 )↦ ( −3 0 )

(in this case this is just the Euclidean algorithm computing the gcd).This yields the Smith normal form

D = S ⋅A ⋅ T


with

S = (1 1−7 −6

) D = (1 0 00 −3 0

) T =⎛⎜⎝

4 2 −9−4 1 21 −3 6

⎞⎟⎠

Hence, the elementary divisors of A are

d1 = 1 d2 = 3

up to multiplication with units Z× = {1,−1}.Using the Maple package LinearAlgebra we can do the cal-

culation as follows:with(LinearAlgebra):

A := Matrix([[6,9,6],[6,6,7]]):

SmithForm(A);

(1 0 00 3 0

)

(S, T) := SmithForm(A, output=[’U’, ’V’]):

S, T;

(0 1−1 3

) ,⎛⎜⎝

−2 1 −91 −1 21 0 6

⎞⎟⎠

Note that S and T are not uniquely determined.

If, more generally, R is a principal ideal domain, we can proveTheorem 4.2.2 in a similar way, replacing the Euclidean algorithmby the following remark:

Remark 4.2.5 Let R be a principal ideal domain, and A = (a1,1, a1,2) ∈R1×2. Then there are x, y ∈ R with

gcd (a1,1, a1,2) = d = x ⋅ a1,1 + y ⋅ a1,2

(note that if R is not a principal ideal domain then in generalgcd (a1,1, a1,2) ∉ ⟨a1,1, a1,2⟩, hence such a representation of the gcddoes not exist; note also, that over a Euclidean domain R, we canfind x and y by the Euclidean algorithm). Write

a1,1 = u ⋅ d a1,2 = v ⋅ d

with u, v ∈ R. Then with

T = (x −vy u

) ∈ GL (2,R)


we haveA ⋅ T = (d, 0) .

This procedure allows us to replace the steps (1) − (3) of the ele-mentary divisor algorithm: If a1,1 ≠ 0 then we can make all otherentries in the first row and column zero. This process terminates,since the entries a1,1 appearing in the process generate an ascendingchain of ideals, which terminates, since R is Noetherian.

Note that det (T ) = 1, that is, T ∈ SL (2,R).

Remark 4.2.6 In the elementary divisor algorithm all transforma-tions have determinant ±1 (row/column transformations and per-mutation matrices). Hence, in Theorem 4.2.2 we can achieve thatT ∈ SL (m,R) and S ∈ SL (n,R) (how?).

That r and the elementary divisors di are unique up to units,follows from:

Theorem 4.2.7 For the elementary divisors d1, ..., dr of A ∈ Rn×m

it holds

d1 ⋅ ... ⋅ di = gcd (det (AI,J) ∣ ∣I ∣ = ∣J ∣ = i) =∶Di

for i ≤ r. For i > r all det (AI,J) = 0.Here, if I ⊂ {1, ..., n} and J ⊂ {1, ...,m}, we denote by

AI,J ∈ R∣I ∣×∣J ∣

the submatrix of A with the rows I and the columns J . The det (AI,J)with ∣I ∣ = ∣J ∣ = i are also called the i × i-minors of A.

The Di are also called the i-th determinantal divisors of A.Note that d1 = D1 is the greatest common divisor of all entries

of A.

Proof. We sketch the proof using some results from (multi-)linearalgebra: The entries of the representing matrix of the i-th exteriorpower ⋀iA of A (with respect to a suitable basis) are exactly thei × i-minors of A. Moreover, if S ∈ Rn×n, then

(⋀iS) ⋅ (⋀iA) = ⋀i (S ⋅A)

If S ∈ GL (n,R), then for the gcd of the entries it holds

gcd (⋀i (S ⋅A)) = gcd (⋀iA)


(up to units), as any entry of ⋀i (S ⋅A) is a linear combination of theentries of ⋀iA and hence gcd (⋀i (S ⋅A)) is divisible by gcd (⋀iA).The converse is also true since A = S−1 ⋅ (S ⋅A).

For the base change T in the source we proceed in a similarfashion.

If we now have

S ⋅A ⋅ T =D =

⎛⎜⎜⎜⎝

d1

⋱ 0dr

0 0

⎞⎟⎟⎟⎠

according to the elementary divisor theorem, then

gcd (⋀iA) = gcd (⋀iD)

= gcd (dj1 ⋅ ... ⋅ dji ∣ 1 ≤ j1 < ... < ji ≤ r)

= d1 ⋅ ... ⋅ di

since dj ∣ dk for j ≤ k.

Example 4.2.8 For

A =

⎛⎜⎜⎜⎝

1 1 1−3 1 11 −3 11 1 −3

⎞⎟⎟⎟⎠

we compute the Smith normal form:

A↦

⎛⎜⎜⎜⎝

1 0 00 4 40 −4 00 0 −4

⎞⎟⎟⎟⎠

↦

⎛⎜⎜⎜⎝

1 0 00 4 40 −4 00 0 −4

⎞⎟⎟⎟⎠

↦

⎛⎜⎜⎜⎝

1 0 00 4 00 0 40 0 −4

⎞⎟⎟⎟⎠

↦

⎛⎜⎜⎜⎝

1 0 00 4 00 0 40 0 0

⎞⎟⎟⎟⎠

=D

On the other hand,D1 = 1 = d1

the 2 × 2-minors are equal to 0, ±4 or 8, hence

D2 = 4 = d1 ⋅ d2

and the 3 × 3-minors are ±16, hence

D3 = 16 = d1 ⋅ d2 ⋅ d3.

The elementary divisor theorem 4.2.2 describes the structure offinitely generated modules over principal ideal domains:


4.3 Modules and presentations

We recall some basic facts on modules. Here R is any (not neces-sarily commutative) ring.

Definition 4.3.1 An R-(left)-module (M,+, ⋅) is a set M withmaps

+ ∶M ×M Ð→M

⋅ ∶ R ×M Ð→M

such that

1) (M,+) is an abelian group,

2) the scalar multiplication ⋅ is distributive over the addition +,that is,

r ⋅ (m1 +m2) = r ⋅m1 + r ⋅m2

(r1R+ r2) ⋅m = r1 ⋅m + r2 ⋅m

for all r, r1, r2 ∈ R and m,m1,m2 ∈M , and

3) for all r, s ∈ R and m ∈M it holds

(r R⋅ s) ⋅m = r ⋅ (s ⋅m)

If R has a multiplicative unit 1, then we require, in addition,that 1 ⋅m =m.

Usually it is clear from the context whether + refers to theaddition in R or M . If we want to be more precise, we note Rrespectively M over + (and in the same way for ⋅).

Example 4.3.2 1) Let R = K be a field. Then a K-module isnothing else than a K-vector space.

2) A Z-module G is nothing else than an abelian group (G,+).The scalar multiplication is

Z ×GÐ→ G

(n, g)z→ n ⋅ g ∶= g + ... + g´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶n-times

with (−1) ⋅ g ∶= −g.


3) Let (R,+, ⋅) be a commutative ring. Then a subset I ⊂ R isan ideal if and only if (I,+, ⋅) is an R-module.

4) If M1 and M2 are modules over R, then also the direct productM1 ×M2 is an R-module with r ⋅ (m1,m2) = (r ⋅m1, r ⋅m2), inparticular:

5) If R is a ring, then

Rn = R × ... ×R´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

n

is an R-module.

6) Let (M,+, ⋅) be an R-module. A submodule U ⊂ M is asubgroup of (M,+), on which the scalar multiplication can berestricted, that is, with

r ⋅m ∈ U

for all m ∈ U and r ∈ R. A submodule is again an R-module.

A subset U ⊂M is a submodule if and only if U ≠ ∅ and

m1 +m2 ∈ U

r ⋅m ∈ U

for all mi,m ∈ U and r ∈ R.

7) Let M be an R-module and U ⊂ M a submodule. Then thequotient group M/U is again an R-module (the quotientmodule) with the scalar multiplication

r ⋅ m = r ⋅ (m +U) = r ⋅m +U = r ⋅m

for r ∈ R and m ∈M .

8) Let V be a K-vector space and A ∈ End (V ) an endomor-phism (for example, V = Kn and A ∈ Kn×n). Then by thesubstitution homomorphism

K[x] Ð→ End (V )x z→ A

we can make the K-vector space V into a K[x]-module bydefining the scalar multiplication as

K[x] × V Ð→ V

(f, v)z→ f ⋅ v ∶= f (A) (v)


9) An R-Algebra S is a ring (S,+, ⋅) with an injective ring ho-momorphism ϕ ∶ R → S and ϕ (r) ⋅ s = s ⋅ ϕ (r) for all r ∈ Rand s ∈ S. With the scalar multiplication

∗ ∶ R × S Ð→ S(r, s) z→ r ∗ s ∶= ϕ (r) ⋅ s

S becomes an R-module.

Hence, an R-algebra is nothing else than a set S with opera-tions

+ ∶ S × S → S (addition)

⋅ ∶ S × S → S (multiplication)

∗ ∶ R × S → S (scalar multiplication)

such that

(a) (S,+,∗) is an R-module,

(b) (S,+, ⋅) is a ring, and

(c) for all r ∈ R and s1, s2 ∈ S it holds

r ∗ (s1 ⋅ s2) = (r ∗ s1) ⋅ s2 = s1 ⋅ (r ∗ s2) .

Definition 4.3.3 An R-module homomorphism is an R-lineargroup homomorphism f ∶M → N between R-modules, that is,

1) f (m1 +m2) = f (m1) + f (m2) for all m1,m2 ∈M

2) f (r ⋅m) = r ⋅ f (m) for all m ∈M and r ∈ R.

Kernel ker (f) and image im (f) are submodules, and the ho-momorphism theorem for modules holds

M/ker (f) ≅ im (f) .

We define coker(f) = N/ im (f) as the cokernel of f .

Definition 4.3.4 Let M be an R-module.

1) M is called finitely generated, if there is a surjective R-module homomorphism

ϕ ∶ Rr →M .

The images mi = ϕ (ei) ∈M of the standard basis vector ei arecalled generators. The map ϕ is surjective if and only if,


every element M can be written as an R-linear combinationof m1, ...,mr, that is,

∀m ∈M ∃a1, ..., ar ∈ R with m = a1m1 + ... + armr.

We then writeM = ⟨m1, ...,mr⟩ .

2) M is called free of rank r, if there is an isomorphism ϕ ∶Rr →M , that is

M ≅ Rr.

The above representation m = a1m1+ ...+armr is then unique,and we call m1, ...,mr a basis of M .

3) An R-module M is called finitely presented, if M is finitelygenerated and ker(ϕ) is also finitely generated.

Example 4.3.5 Considered as a Z-module, the rational numbersQ are not finitely generated: Assume Q is generated by r1, ..., rn.Then there is a d ∈ Z coprime to the denominators ri and 1

d doesnot lie in the Z-module (that is, abelian group) ⟨r1, ..., rn⟩ generatedby r1, ..., rn.

A K-vector space of dimension r is, considered as a K-module,free of rank r (as any vector space has a basis).

The ring R = K [x1, x2, ...] of polynomials in countably manyvariables is as an R-module finitely generated (by 1). However itssubmodule

M = {f ∈ R ∣ f (0) = 0}

is not, since it contains all xi, however any finite set of polynomialsinvolves only finitely many variables.

The following short hand notation is very useful:

Definition 4.3.6 A sequence of R-module homomorphisms

...→Miϕi→Mi+1

ϕi+1→ Mi+2 → ...

is called exact, if

im (ϕi) = ker (ϕi+1) ∀i.

Remark 4.3.7 A homomorphism ϕ ∶ N → M is surjective, if thesequence

Nϕ→M → 0

is exact, and it is injective, if

0→ Nϕ→M

is exact.


Remark 4.3.8 Hence, a module M is finitely generated if there isan exact sequence

Rn π→M → 0.

With the inclusion ι of the kernel, we obtain an exact sequence

0→ ker (ϕ)ι→ Rn π

→M → 0

So a module M is finitely presented if there is an exact sequence

Rm A→ Rn π

→M → 0

with A ∈ Rn×m. This matrix A is called presentation matrixof M , and describes M completely, since by the homomorphismtheorem

Rn/ im (A) = Rn/ker (π) ≅ im (π) = Mei ↦ π(ei)

That is, the image of A contains all relations between the generatorsπ (ei) of M (where ei denotes the standard basis of Rn). Or to putit in a different way, up to isomorphism we obtain M from the freemodule Rn by imposing the calculation rules

r1e1 + ... + rnen = 0

for all

⎛⎜⎝

r1

⋮rn

⎞⎟⎠∈ im (A) .

Every finitely generated module over a Noetherian ring can berepresented in this way: In the same way as for ideals we can showthe following equivalence (Exercise 4.4):

Definition and Theorem 4.3.9 Let R be a commutative ring with1. An R-module M is called Noetherian, if it satisfies the follow-ing equivalent conditions:

1) Every ascending chain of submodules terminates.

2) Every submodule of M is finitely generated.

3) Every set of submodules has a maximal element.

Using the chain condition it is any easy exercise to show thefollowing (Exercise 4.5):


Lemma 4.3.10 Let

0→ Uι→ F

π→M → 0

be an exact sequence of R-modules. Then F is Noetherian if andonly if U and M are Noetherian.

Lemma 4.3.11 Let R be a Noetherian ring. Then Rn is a Noethe-rian module.

Proof. First note, that R is a Noetherian R-module by Example4.3.2.(3). Since

0 → R → Rn → Rn−1 → 0

a1 ↦

⎛⎜⎜⎜⎝

a1

0⋮0

⎞⎟⎟⎟⎠

,

⎛⎜⎜⎜⎝

a1

a2

⋮an

⎞⎟⎟⎟⎠

↦⎛⎜⎝

a2

⋮an

⎞⎟⎠

is exact, the claim follows by induction and Lemma 4.3.10.Since for any finitely generated R-module M we have an exact

sequence0→ U → Rn →M → 0

with a submodule U ⊂ Rn, it follows by Lemma 4.3.10:

Theorem 4.3.12 Finitely generated modules over Noetherian ringsare finitely presented.

This also proves: Finitely generated modules over Noetherianrings are Noetherian.

In particular, we can apply Theorem 4.3.12 to describe Z-modules(that is, finitely generated abelian group) and finitely generatedK [x]-modules by a presentation matrix.

Example 4.3.13 We consider the abelian group G with the pre-sentation as Z-module

0→ Z3 A→ Z4 π

→ G→ 0

given by the matrix

A =

⎛⎜⎜⎜⎝

1 1 1−3 1 11 −3 11 1 −3

⎞⎟⎟⎟⎠


First compute the Smith normal form D = S ⋅A ⋅ T of A with

D =

⎛⎜⎜⎜⎝

1 0 00 4 00 0 40 0 0

⎞⎟⎟⎟⎠

S =

⎛⎜⎜⎜⎝

1 0 0 03 1 0 02 1 1 01 1 1 1

⎞⎟⎟⎟⎠

T =⎛⎜⎝

1 −1 00 1 −10 0 1

⎞⎟⎠

(see Exercise 4.2.8). So we have a commutative diagram

0 → Z3 A→ Z4 π

→ G → 0↑ T ↓ S ≅

0 → Z3 D→ Z4 → coker(D) → 0

Hence, the columns ci = S−1 (ei) of

S−1 =

⎛⎜⎜⎜⎝

1 0 0 0−3 1 0 01 −1 1 01 0 −1 1

⎞⎟⎟⎟⎠

represent generators c1, c2, c3, c4 of coker(A) = Z4/ im (A) with therelations

1 ⋅ c1 = 0 4 ⋅ c2 = 0 4 ⋅ c3 = 0

In particular, the generator c1 can be omitted. We can see thisdirectly:

⎛⎜⎜⎜⎝

1−311

⎞⎟⎟⎟⎠

= A ⋅ e1 ∈ im (A)

In the same way, since G ≅ coker(A), the elements

vi = π(S−1 (ei)) ∈ G

are generators of G with the relations

4 ⋅ v2 = 0 4 ⋅ v3 = 0

Hence we can write G as a product of cyclic groups

G ≅ coker(D) ≅ Z/4 ×Z/4 ×Z

where the factors correspond to the generators

v2 = π

⎛⎜⎜⎜⎝

01−10

⎞⎟⎟⎟⎠

v3 = π

⎛⎜⎜⎜⎝

001−1

⎞⎟⎟⎟⎠

v4 = π

⎛⎜⎜⎜⎝

0001

⎞⎟⎟⎟⎠

∈ G

In the next section we give the general result:


4.4 Finitely generated modules over prin-

cipal ideal domains

Using the elementary divisor theorem we can describe finitely gen-erated modules over principal ideal domains.

Let R be a principal ideal domain and M a finitely generatedR-module. By Theorem 4.3.12, M is finitely presented

Rm A→ Rn π

→M → 0

and A ∈ Rn×m. The elementary divisor theorem 4.2.2 yields S ∈GL (n,R) and T ∈ GL (m,R) with

Rm A→ Rn π

→ M → 0↑ T ↓ S ≅

Rm D→ Rn → M ′ → 0

and

D =

⎛⎜⎜⎜⎝

d1

⋱ 0dr

0 0

⎞⎟⎟⎟⎠

With the standard basis vectors ei of Rn

vi = π (S−1 (ei))

are generators of M with the relations

d1v1 = 0 . . . drvr = 0

If di is a unit, then divi = 0 implies vi = 0, hence we can delete thegenerator vi (and with it the i-th row of D). The last m−r columnscan be deleted, as this does not change im (A) = ker (π).

To formulate this as a theorem we use the following notation:

Definition 4.4.1 An R-module M is the direct sum

M = U1 ⊕ ...⊕Un

of submodules U1, ..., Un ⊂M , if M is generated by the elements ofthe Ui and for all ui ∈ Ui it holds

u1 + ... + un = 0 Ô⇒ u1 = ... = un = 0

Definition 4.4.2 A module is called cyclic, if it is generated by asingle element.


Theorem 4.4.3 Let R be a principal ideal domain and M a finitelygenerated R-module. Then it holds:

1) There are generators v1, ..., vn of M and d1, ..., dr ∈ R, r ≤ nwith di ∉ R× and di ∣ di+1 for i = 1, ..., r − 1, such that M isdescribed by the relations

d1v1 = 0 . . . drvr = 0

2) M is the direct sum

M = U1 ⊕ ...⊕Un

of cyclic submodules and

Ui ≅ {R/ ⟨di⟩ for i ≤ rR for i > r

3) that is,M ≅ R/ ⟨d1⟩ × . . . × R/ ⟨dr⟩ × Rn−r

The rank n − r of M is uniquely determined by M , also theelementary divisors di of M (up to units).

Corollar 4.4.4 Let R be a principal ideal domain and M a finitelygenerated R-module. Let

T = {m ∈M ∣ ∃r ∈ R with r ⋅m = 0} ⊂M

be the torsion submodule of M . There is a free submodule F ⊂Mwith

M = T ⊕ F .

The torsion submodule T is canonical as a submodule, the freepart F , on the other hand, is not (rather the quotient M/T ≅ F iscanonical).

Definition 4.4.5 The module M is called torsion-free, if T ={0}, and M is called a torsion module, if M = T .

Remark 4.4.6 Corollary 4.4.4 shows: A finitely generated moduleover a principal ideal domain is free if and only if it is torsion-free.

From this it follows:


Remark 4.4.7 Let R be a principal ideal domain. Then everysubmodule of a free R-module is again free.

Example 4.4.8 The ideal M = ⟨x, y⟩ ⊂ R = K [x, y] is not princi-pal. As an R-submodule of R the module M is torsion-free, howevernot free (see Exercise 4.6).

Example 4.4.9 In Example 4.3.13

G = ⟨v2⟩⊕ ⟨v3⟩´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

T

⊕ ⟨v4⟩°F

with the (unique) torsion submodule

T = ⟨v2, v3⟩ ≅ Z/4 ×Z/4

and the (non-unique) free part

F = ⟨v4⟩ ≅ Z.

In Section 4.6 and in Exercise 4.8 we will also see examples overK[x].

The decomposition in Theorem 4.4.3 can be refined: If d ∈ Rand d = pe11 , ..., p

ekk is a prime factorization (with all ei > 0), then by

the Chinese remainder theorem

R/ (d) ≅ R/ ⟨pe11 ⟩ × ... ×R/ ⟨pekk ⟩

as the ideals ⟨peii ⟩ are pairwise coprime. Hence it holds:

Theorem 4.4.10 Let R be a principal ideal domain and M a finitelygenerated R-module. Then there is a t ∈ N0 and prime elementsp1, ..., ps ∈ R and e1, ..., es ∈ N with

M ≅ R/ ⟨pe11 ⟩ × . . . × R/ ⟨pess ⟩ × Rt

and this product is unique up to reordering of the factors.

4.5 Fundamental theorem of finitely gen-

erated abelian groups

As a special case we consider finitely generated Z-modules, that is,finitely generated abelian groups. We summarize for this case theresults of the preceding section:


Theorem 4.5.1 Let G be a finitely generated abelian group. Thenit holds:

1) G is the direct sum of cyclic subgroups.

2) There is 0 ≤ r ≤ n and d1, ..., dr ≥ 2 with di ∣ di+1 for i =1, ..., r − 1, such that

G ≅ Z/ (d1) × . . . × Z/ (dr) × Zn−r

3) There are prime numbers p1, ..., ps (not necessarily pairwisedifferent) and e1, ..., es ∈ N with

G ≅ Z/ (pe11 ) × . . . × Z/ (pess ) × Zn−r

Here r and n and the di (up to permutation) are uniquely de-termined by G. The same is true for the peii .

Example 4.5.2 Let G be given by the presentation matrix

A =

⎛⎜⎜⎜⎝

2 0 0 00 3 0 00 0 4 00 0 0 18

⎞⎟⎟⎟⎠

∈ Z5×5

By the elementary divisor algorithm we obtain the Smith normalform D of A

A↦

⎛⎜⎜⎜⎝

1 0 0 00 6 0 00 0 4 00 0 0 18

⎞⎟⎟⎟⎠

↦

⎛⎜⎜⎜⎝

1 0 0 00 2 0 00 0 12 00 0 0 18

⎞⎟⎟⎟⎠

↦

⎛⎜⎜⎜⎝

1 0 0 00 2 0 00 0 6 00 0 0 36

⎞⎟⎟⎟⎠

=D

Hence, the elementary divisor representation in Theorem 4.5.1 is

G ≅ Z/2 ×Z/6 ×Z/36

and the prime power representation (by 6 = 2 ⋅ 3 and 36 = 22 ⋅ 32) is

G ≅ Z/2 ×Z/2 ×Z/3 ×Z/22 ×Z/32.

From this we can recover the elementary divisor representation

G ≅ Z/2 × Z/2 × Z/22

× Z/3 × Z/32

≅ Z/2 × Z/6 × Z/36

since by the condition di ∣ di+1 it holds: If we write the elementarydivisors

di = pei,11 ⋅ ... ⋅ p

ei,tt

with prime numbers p1, ..., pt and ei,j ≥ 0, then ei,j ≤ ei+1,j. Moreoverwe can use any prime power only once. See also Exercise 4.7.


4.6 The Jordan normal form

In linear algebra over a field K the Jordan normal form of an en-domorphism A = (ai,j) ∈Kn×n is one of the key constructions.

Considering the K[x]-module structure on Kn associated to A,we can determine the Jordan normal form of A by applying theelementary divisor algorithm to the presentation matrix of Kn as aK[x]-module.

Remark 4.6.1 In Example 4.3.2.(8) we have seen, that Kn be-comes a K[x]-module with the scalar multiplication

K[x] ×Kn Ð→Kn

(f, v)z→ f (A) ⋅ v

induced by substituting x by A. From the K[x]-module structurewe can recover the K-linear map

Kn → Kn

v ↦ x ⋅ v

Hence, a K[x]-module structure on Kn is the same as an endomor-phism A ∈Kn×n.

Remark 4.6.2 A subvector space U ⊂ Kn is a K[x]-submodule ifand only if A (U) ⊂ U .

Proof. Let u ∈ U . Then f ⋅ u ∈ U ∀f ∈K[x] if and only if x ⋅ u ∈ U ,if and only if A ⋅ u ∈ U .

Remark 4.6.3 The standard basis vectors e1, ..., en of Kn are gen-erators of Kn as a K[x]-module, that is, there is an exact sequence

K[x]nπÐ→Kn Ð→ 0

Among the ei we have the obvious relations

x ⋅ ej = A ⋅ ej =⎛⎜⎝

a1,j

⋮an,j

⎞⎟⎠=

n

∑i=1

ai,j ⋅ ei ∈Kn

By these relations, any element ∑ni=1 fi ⋅ ei ∈ K [x]

nwith fi ∈ K[x]

can be reduced to a vector of constants (that is, to an element ofKn). So these relations generate ker (π) as a K [x]-module.


Hence Kn as a K[x]-module has the presentation

K[x]nxE−AÐ→ K[x]n Ð→Kn Ð→ 0

with the presentation matrix

xE −A =

⎛⎜⎜⎜⎝

x − a1,1 −a1,n ⋯ −a1,n

−a2,1 x − a2,2 ⋮⋮ ⋱ ⋮

−an,1 ⋯ ⋯ x − an,n

⎞⎟⎟⎟⎠

In linear algebra this matrix is also called the characteristic ma-trix of A.

Remark 4.6.4 The kernel of the substitution homomorphism

ϕA ∶ K[x] → Kn×n

x ↦ A

is a principal ideal in K [x]. The minimal polynomial pA of Ais defined in linear algebra as the monic generator of the kernel

ker (ϕA) = ⟨pA⟩ .

So for every v ∈Kn the scalar multiplication pA ⋅ v = 0 is zero, thatis, Kn is a (finitely generated) torsion module.

According to the theorem of Cayley-Hamilton the characteris-tic polynomial

χA = det (xE −A) ∈ ker (ϕA)

is in the kernel, so pA ∣ χA.

Lemma 4.6.5 Assume that the characteristic polynomial χA fac-tors into linear factors. Then Kn as a K [x]-module is the directsum

Kn = U1 ⊕ ...⊕Us

of cyclic submodules

Ui ≅K[x]/ ⟨(x − λi)ei⟩

and

∏si=1 (x − λi)

ei = χA


Proof. Applying the elementary divisor algorithm to xE − A weobtain S,T ∈ SL(n,K [x]) with

S ⋅ (xE −A) ⋅ T =D =⎛⎜⎝

d1

⋱dn

⎞⎟⎠

and di ≠ 0 (note, that Kn is a torsion module). Decomposing fur-ther by the Chinese remainder theorem (that is, applying Theorem4.4.10) yields a decomposition

Kn = U1 ⊕ ...⊕Us

into a direct sum of submodules Ui and isomorphisms

αi ∶ Ui≅Ð→K[x]/ ⟨peii ⟩

with irreducible polynomials pi ∈K[x]. Then it holds

∏si=1p

eii = d1 ⋅ ... ⋅ dn = det (D) = det (xE −A) = χA.

Since χA decomposes into linear factors, the pi are of the formpi = x − λi with λi ∈K.

We now describe a K-vector space basis of Ui, in which A ∣Ui

becomes a Jordan block: As a K-vector space

K[x]/ ⟨(x − λi)ei⟩

has the basis

1, x − λi, (x − λi)2, . . . , (x − λi)

ei−1

The preimages

vi,j = α−1i ((x − λi)

j)

under the isomorphism αi ∶ Ui Ð→ K[x]/ ⟨peii ⟩ give a basis Bi =(vi,j)j=0,...,ei−1 of Ui.

Then it holds

(A − λiE) ⋅ vi,j = (x − λi) ⋅ α−1i ((x − λi)

j) = α−1

i ((x − λi)j+1

) = vi,j+1

for j = 0, ..., ei − 2 and

(A − λiE) ⋅ vi,ei−1 = (x − λi) ⋅ vi,ei−1 = 0.


Hence, with respect to the basis Bi the matrix A has as representingmatrix an ei × ei-Jordan block with eigenvalue λi

MBiBi (A ∣Ui

) = J (λi, ei) ∶=

⎛⎜⎜⎜⎝

λi 1 0⋱ ⋱

⋱ 10 λi

⎞⎟⎟⎟⎠

∈Kei×ei

This proves:

Theorem 4.6.6 (Jordan normal form) Let A ∈Kn×n, and sup-pose χA (t) factors over K into linear factors (which is true, forexample, if K = C). Then there is an S ∈ GL (n,K), such that

SAS−1 = J =

⎛⎜⎜⎜⎝

J (λ1, e1) 0J (λ2, e2)

⋱0 J (λs, es)

⎞⎟⎟⎟⎠

has block diagonal form with Jordan blocks J (λi, ri) with (not nec-essarily pairwise different) eigenvalues λ1, ..., λs on the diagonal.Up to permutation of the blocks, J is uniquely determined by A.

Remark 4.6.7 The prime power factors of the last elementary di-visor dn correspond to the maximal Jordan blocks, those of the sec-ond last to blocks of size one less, and so on. In particular, dn = pAis the minimal polynomial of A.

Example 4.6.8 We try out the algorithm for

A =

⎛⎜⎜⎜⎜⎜⎜⎝

0 1 0 0 00 0 1 0 00 0 0 0 00 0 0 0 00 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎠

(which, for simplicity is already in Jordan normal form). First wecompute the Smith normal form of xE −A:

⎛⎜⎜⎜⎜⎜⎜⎝

x −1 0 0 00 x −1 0 00 0 x 0 00 0 0 x 00 0 0 0 x − 1

⎞⎟⎟⎟⎟⎟⎟⎠

↦

⎛⎜⎜⎜⎜⎜⎜⎝

−1 0 0 0 00 x2 −1 0 00 0 x 0 00 0 0 x 00 0 0 0 x − 1

⎞⎟⎟⎟⎟⎟⎟⎠

↦

⎛⎜⎜⎜⎜⎜⎜⎝

−1 0 0 0 00 −1 0 0 00 0 x3 0 00 0 0 x 00 0 0 0 x − 1

⎞⎟⎟⎟⎟⎟⎟⎠

↦

⎛⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 x 00 0 0 0 x3 (x − 1)

⎞⎟⎟⎟⎟⎟⎟⎠

=D


From the elementary divisors we can read off, that the Jordan nor-mal form of A has one Jordan block of the size

1 × 1 with eigenvalue 03 × 3 with eigenvalue 01 × 1 with eigenvalue 1

Example 4.6.9 Suppose A would have two 1 × 1 blocks and one3 × 3 block, all with eigenvalue 0, then the Smith normal form ofxE −A would be

D =

⎛⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 00 1 0 0 00 0 x 0 00 0 0 x 00 0 0 0 x3

⎞⎟⎟⎟⎟⎟⎟⎠

For another example see Exercise 4.8.

4.7 Exercises

Exercise 4.1 1) Compute the Smith normal form of

A =⎛⎜⎝

4 6 22 3 22 3 0

⎞⎟⎠∈ Z3×3

and S,T ∈ GL (3,Z) with

S ⋅A ⋅ T =D

2) Determine a basis of im(A) and describe Z3/ im(A).

Exercise 4.2 Let

g1 =⎛⎜⎝

102

⎞⎟⎠, g2 =

⎛⎜⎝

120

⎞⎟⎠, g3 =

⎛⎜⎝

−3−3−3

⎞⎟⎠, g4 =

⎛⎜⎝

122

⎞⎟⎠∈ Z3

Determine a basis of the subgroup of Z3 generated by g1, ..., g4.

Exercise 4.3 Let G be a finite abelian group, and

Zn A→ Zn → G→ 0

a presentation of G by A = (ai,j) ∈ Zn×n. Prove that for the orderof the group G it holds

∣G∣ = ∣detA∣ .


Exercise 4.4 An R-module M is called Noetherian if it satisfiesthe following equivalent conditions:

1) Every ascending chain of submodules terminates.

2) Every submodule of M is finitely generated.

3) Every non-empty set of submodules contains a maximal ele-ment.

Prove the equivalence.

Exercise 4.5 Let0→ U

a→ F

π→M → 0

be an exact sequence of R-modules. Prove that M is Noetherian ifand only if U and M are Noetherian.

Exercise 4.6 Let K be a field and R = K[x, y]. Show that M =⟨x, y⟩ ⊂ R as an R-module is torsion-free, but not free.

Exercise 4.7 Find the elementary divisors di of

G = Z/2 ×Z/2 ×Z/2 ×Z/2 ×Z/32 ×Z/32 ×Z/35 ×Z/5 ×Z/5 ×Z/7

that is, a representation G ≅ Z/d1× ...×Z/dr with di ≥ 2 and di ∣ di+1

for i = 1, ..., r − 1.

Exercise 4.8 Let

B =⎛⎜⎝

2 0 −1−1 1 11 0 0

⎞⎟⎠∈ End (C3)

1) Compute the Smith normal form D of

A = xE −B ∈ C[x]3×3

and S,T ∈ GL (3,C[x]) with D = S ⋅A ⋅ T .

2) Determine from this the Jordan normal form of B.

5

Free resolutions andinvariants

5.1 Overview

If M is a finitely generated module over a principal ideal domainR, then by Theorem 4.4.3 there is an exact presentation sequence

0→ Rr D→ Rs π

→M → 0

where

D =

⎛⎜⎜⎜⎝

d1

⋱dr

0

⎞⎟⎟⎟⎠

Note that exactness at the module Rr is achieved by deleting anyzero column in the Smith normal form. In this way, one can describethe structure of M completely.

In which way can we generalize this to the setup of modules Mover say R = K[x1, ..., xn]? What we have in mind, is to obtaininvariants encoding information about the algebraic variety X =V (f1, ..., fr) ⊂ An

K given by fi ∈ R from

M = coker (f1, ..., fr) = R1/ im (f1, ..., fr)

which is, as an R-module, isomorphic to the coordinate ringK[X] =R/I of X.

In particular, we are interested in the dimension of X, the de-gree of X and the genus of X. We already have some idea, whatthe dimension of an algebraic variety should be, for example, thedimension of a hypersurface should be n−1. The degree of a hyper-surface should be the degree of the equation, and if X is finite, the

115

5. FREE RESOLUTIONS AND INVARIANTS 116

degree should be the number of solutions counted with multiplicity.The genus of X you perhaps know from topology. Nevertheless, wewill have to make the definitions precise and see how to computethese invariants. The canonical setting to do this, are ideals gen-erated by homogeneous polynomials. So we will also give a shortprimer on the corresponding projective varieties in Section 5.3.

5.2 Free resolutions

With the notation as in the introduction, the R-module

M = coker(A) ≅ R/I

withA = (f1, ..., fr) ∈ R

1×r

has the presentation

Rr A→ R1 →M → 0.

If X is not a hypersurface, then A will have a non-trivial kernel.

Example 5.2.1 For I = ⟨x, y⟩ ⊂ R =K[x, y], we have

A = (x, y)

and ker(A) = im(B) with

B = (y−x

)

Obviously, ker(B) = 0, hence, we have an exact sequence

0→ R1 B→ R2 A

→ R1 → R/I → 0

In Theorem 5.10.7 we will formulate a general algorithm, which,given A, will determine a matrix B with

ker(A) = im(B)

using a straightforward generalization of Buchberger’s algorithmfrom ideals to modules. Recall, that we encountered the moduleof relations (syzygies) between the generators fi in the proof ofBuchberger’s criterion. In addition to the computation of B, arigorous theory of Grobner bases of modules will allow us to give amuch nicer proof of Buchberger’s criterion.

If we iterate this process of writing ker(A) = im(B), we obtainwhat is called a free resolution:


Definition 5.2.2 Let M be an R-module. An exact sequence

...→ Fiφi→ Fi−1

φi−1→ ...

φ2→ F1

φ1→ F0

φ0→M → 0

with free modules Fi is called a free resolution of M . It is calledfinite if Fi = 0 for i > l for some integer l, which is called the lengthof the resolution.

The i-th syzygy module of M is im(φi) = ker(φi−1).

A priori it is not clear that the length is finite. This is noteven true if R is Noetherian and M is finitely generated (and hencefinitely presented):

Example 5.2.3 Let R =K[x, y]/ ⟨xy⟩. The R-module ⟨x⟩ is finitelygenerated, however, it has the infinite free resolution

...→ R1 y→ R1 x

→ R1 y→ R1 x

→ ⟨x⟩→ 0

However, for modules over the polynomial ring, we will provethe following key theorem by Hilbert:

Theorem 5.2.4 (Syzygy theorem) Let R = K[x1, ..., xn]. Ev-ery finitely generated R-module has a finite free resolution. Allmodules in the resolution are finitely generated, and the length is atmost n.

Example 5.2.5 We compute the finite free resolution in Example5.2.1 using Singular:ring R=0,(x,y),lp;

ideal I=x,y;

def L=res(I,0);

1 2 3

R <-- R <-- R

0 1 2

matrix(L[1]);

[1,1] = y

[1,2] = x

matrix(L[2]);

[1,1] = -x

[2,1] = y

If the second argument of res is positive, then Singular will stopafter computing the given number of steps of the resolution.Increasing the number of generators we get similar complexes:ring R=0,(x(1..4)),lp;


ideal I=x(1..4);

def L=res(I,0);

1 4 6 4 1

R <-- R <-- R <-- R <-- R

0 1 2 3 4In general this kind of resolution is called Koszul complex. Forn independent variables,

rank(Fi) = (n

i).

We will come back to this in Exercise 5.10.

5.3 Projective algebraic sets and graded

ideals

Many results in algebraic geometry improve, if we replace the affinespace An

K by projective space PnK . For example any line in A2K gets

added its point at infinity, which leads to the desirable fact thatany two distinct lines intersect in a point: Two parallel lines in-tersect at their common point at infinity. More generally, Bezout’stheorem shows: If K =K, two curves in P2

K of degree d1 and d2 in-tersect precisely in d1 ⋅d2 points counted with multiplicity. Our mainmotivation to pass to projective varieties is to give algorithms forcomputing the above mentioned invariants like dimension, degreeand genus.

In this section we will discuss the basics of graded rings, ho-mogeneous ideals and the associated projective algebraic sets. Weinvestigate how to associate to a projective algebraic set and affinealgebraic set, and vice versa.

Definition 5.3.1 The n-dimensional projective space over K is

Pn(K) = {1-dimensional vector subspaces of Kn+1}

If ⟨(p0, .., pn)⟩ ∈ Pn(K), so in particular (p0, .., pn) ≠ 0, then wewrite

(p0 ∶ ... ∶ pn) ∶= ⟨(p0, .., pn)⟩ .

The quotient symbol ∶ indicates that the generator of the 1-dimension vector space is only unique up to multiplication with anon-zero constant:

(p0 ∶ ... ∶ pn) = (q0 ∶ ... ∶ qn)⇐⇒ ∃λ ≠ 0 ∶ λ ⋅ (p0, ..., pn) = (q0, ..., qn)


With the group action

K× ×Kn+1 →Kn+1, (λ, (p0, ..., pn))↦ λ ⋅ (p0, ..., pn)

we havePn(K) = (Kn+1 − {0}) /K×.

This quotient construction has a generalization in the setting oftoric geometry and geometric invariant theory.

For example, we can think of P2(R) as a half sphere with oppo-site boundary points identified (since they generate the 1-dimensionvector space), see Figure 5.1.

Figure 5.1: Projective space P2(R).

To specify algebraic subsets of Pn(K) we have to consider ho-mogeneous polynomials:

Definition 5.3.2 A graded ring is a ring R together with a fixeddecomposition

R =⊕α∈GRα

as a direct sum of abelian groups, where G is a monoid (that issemigroup with unit), such that

RαRβ ⊂ Rα+β.


If f ∈ Rα then f is called homogeneous of degree α.If f ∈ R then f = ∑α∈G, finite fα with the uniquely determined

homogeneous components fα ∈ Rα.An ideal I ⊂ R is called homogeneous, if it is generated by ho-

mogeneous elements.

Definition 5.3.3 The standard grading on K[x0, ..., xn] by N0

is given by setting degxi = 1∀i. Then

K[x0, ..., xn] =⊕∞d=0K[x0, ..., xn]d

whereK[x0, ..., xn]d = {∑∣α∣=dcαx

α ∈K[x0, ..., xn]}

Example 5.3.4 In the standard grading on K[x, y] the polynomi-als 3x2y and 3x2y + 5xy2 are homogeneous, whereas 3x2y + 5x isnot.

However if you consider the grading of K[x, y] by N20 with degx =

e1 and deg y = e2, then the degree of a monomial is its exponent vec-tor. Hence 3x2y is homogeneous of degree (2,1), however 3x2y +5xy2 and 3x2y+5x are not. The homogeneous polynomials are pre-cisely the constant multiples of monomials.

If not mentioned otherwise, in what follows, we will alwaysconsider the standard grading on R =K[x0, ..., xn].

If f ∈ R is homogeneous of degree d, then

f (λ ⋅ (p0, ..., pn)) = λd ⋅ f(p0, ..., pn)

hence the condition f(p0 ∶ ... ∶ pn) = 0 for (p0 ∶ ... ∶ pn) ∈ Pn(K) iswell-defined.

Definition 5.3.5 If I ⊂ R is homogeneous, then the projectivealgebraic set defined by I is

V (I) = {p ∈ Pn(K) ∣ f(p) = 0 ∀ homogeneous f ∈ I}

and if X ⊂ Pn(K)

I(X) = ⟨f ∈ R homogeneous ∣ f(p) = 0 ∀p ∈X⟩

is the homogeneous zero ideal of X.


Remark 5.3.6 Consider the hyperplane at infinity

H = {p ∈ Pn(K) ∣ p0 = 0}

and the corresponding affine chart

U = {p ∈ Pn(K) ∣ p0 ≠ 0}

so Pn(K) = U ∪H. The map

ϕ ∶ An(K) → U(p1, ..., pn) ↦ (1 ∶ p1 ∶ ... ∶ pn)

(p1p0 , ...,pnp0

) ←[ (p0 ∶ p1 ∶ ... ∶ pn)

is a well-defined bijection. Figure 5.2 depicts the identification ofA2(R) with U ⊂ P2(R). The horizontal lines correspond to thepoints of H.

Figure 5.2: Mapping A2(R) into P2(R) by stereographic projection.

If F ∈ K[x0, ..., xn] is homogeneous, then its dehomogeniza-tion is

F a = F (1, x1, ..., xn).

Given a homogeneous ideal I ⊂K[x0, ..., xn], the ideal

Ia = {F a ∣ F ∈ I} ⊂K[x1, ..., xn]

is called the dehomogenization of I. In this way we can associate tothe projective algebraic set X = V (I) ⊂ Pn(K) the affine algebraicset

ϕ−1(X ∩U) = V (Ia).


Proof. (1 ∶ x1 ∶ ... ∶ xn) ∈ V (I) ⇔ F (1, x1, ..., xn) = 0 ∀ homoge-neous F ∈ I⇔ (x1, ..., xn) ∈ V (Ia).

Note that, more generally, any choice of a homogeneous linearform l yields an affine chart U = {l(p) ≠ 0}.

Example 5.3.7 The projective parabola X = V (x21 −x2x0) ⊂ P2(R)

consists of the points (1 ∶ x1 ∶ x2) corresponding to the point (x1, x2) ∈V (x2

1−x2) ⊂ A2(R) of the affine parabola (Figure 5.3) and one pointat infinity (0 ∶ 0 ∶ 1), so

U = {(x0 ∶ x1 ∶ x2) ∈ P2(R) ∣ x0 ≠ 0} → A2(R)∪ ∪

X = V (x21 − x2x0) → V (x2

1 − x2)(1 ∶ x1 ∶ x2) ↦ (x1, x2)

In Figure 5.3 the projective parabola is depicted by identifying a1-dimensional subspaces of A3(R) with its generator on the upperhalf sphere. Projecting the upper half sphere into the (x1, x2)-planemaps the projective parabola into the unit disc, see Figure 5.5.

Figure 5.3: Parabola x − y2 = 0 in A2(R).

Similar to affine algebraic geometry, the key to the identificationof projective algebraic sets with homogeneous ideals is a projectiveversion of the strong Nullstellensatz. To formulate this theorem, weneed the following easy but fundamental fact about homogeneousideals:


Figure 5.4: Projective parabola

Lemma 5.3.8 An ideal I ⊂ R is homogeneous if and only if forevery f ∈ I also the homogeneous summands fi ∈ Ri of f = ∑i fi arein I.

Proof. Assume the latter. Then the homogeneous summands ofany set of generators are in I, hence I is generated by these sum-mands. On the other hand, assume I = ⟨f1, ..., fr⟩ with fj ∈ Rdj

homogeneous, and let f = ∑j ajfj ∈ I. Write aj = ∑i ai,j withai,j ∈ Ri, so

f =∑j∑iai,jfj =∑d∑i+dj=d

ai,jfj´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

gd

with gd ∈ Rd. Obviously all gd ∈ I. Since there is no cancella-tion between the homogeneous polynomials gd ∈ Rd, they are thehomogeneous summands of f .

Note that this generalizes Lemma 2.3.6 if we use the grading bythe exponent vector. Using Lemma 5.3.8, it is straightforward toprove (see Exercise 5.1):

Corollar 5.3.9 The radical of a homogeneous ideal is homogeneous.


Figure 5.5: Projective parabola as a subset of the unit disc.

Theorem 5.3.10 (Projective strong Nullstellensatz) Let K =K and I ⊂K[x0, ..., xn] homogeneous with V (I) ≠ ∅. Then

I(V (I)) =√I.

Proof. Let X = V (I) ⊂ Pn(K) and consider the affine cone C(X) =V (I) ⊂ An+1(K). Then by the (affine) strong Nullstellensatz 2.1.11and Exercise 5.2 we have

√I = I(C(X)) = I(X).

From this it follows immediately:

Corollar 5.3.11 If K =K then there is a 1 ∶ 1-map

{projective algebraic sets

X ⊂ Pn(K)}

I⇄V

{homogeneous ideals J ⊂ R

with J =√J and J ≠ ⟨x0, ..., xn⟩

}

Proof. Note that there are precisely two homogeneous radicalideals I with V (I) = ∅ namely I = ⟨x0, ..., xn⟩ and I = ⟨1⟩. Hence,if we exclude the irrelevant ideal ⟨x0, ..., xn⟩, by Theorem 5.3.10 weget a 1 ∶ 1 map.

Finally we show, that to any affine variety we can associate aprojective one:


Definition 5.3.12 Given f ∈ K[x1, ..., xn], the homogenizationof f is

fh = xdeg f0 ⋅ f (

x1

x0

, ...,xnx0

) ∈K[x0, ..., xn]

which clearly is a homogeneous polynomial.If I ⊂K[x1, ..., xn], then the homogenization of I is defined as

Ih = ⟨fh ∣ f ∈ I⟩ ⊂K[x0, ..., xn],

which, by definition, is a homogeneous ideal.

Example 5.3.13 For F = x30 + x

20x1 we get F a = 1 + x1, hence

(F a)h= x0 + x1. So if we dehomogenize and then homogenize, we

remove any common x0-power of the terms. On the other hand, bydefinition it is clear that for f ∈K[x1, ..., xn] it holds (fh)

a= f .

Definition 5.3.14 Let X ⊂ An(K) be an affine algebraic set. Wedefine the projective closure of X as

pc(X) = V (I(X)h) ⊂ Pn(K),

the vanishing locus of the homogenization of the zero ideal of X.

Theorem 5.3.15 The projective closure of an affine algebraic setX is the smallest projective algebraic set containing ϕ(X).

Proof. We note that ϕ(X) ⊂ pc(X): If f ∈ I(X) then fh(1, x1, ..., xn) =0 ∀(x1, ..., xn) ∈ X. Since any (hence any homogeneous) elementof I(X)h is a linear combination of polynomials fh with f ∈ I(X),the claim follows.

Suppose ϕ(X) ⊂ V (F1, ..., Fr) with Fi ∈ K[x0, ..., xn] homoge-neous. Then fi = Fi(1, x1, ..., xn) ∈ I(X), so fh

i ∈ I(X)h, hence alsoFi ∈ I(X)h. This shows ⟨F1, ..., Fr⟩ ⊂ I(X)h, so

pc(X) = V (I(X)h) ⊂ V (F1, ..., Fr).

Remark 5.3.16 We define the Zariski topology on Pn(K) byconsidering as closed sets the projective algebraic sets. The theoremshows that the projective closure of X is the Zariski closure of ϕ(X),that is,

pc(X) = ϕ(X).


Example 5.3.17 If we homogenize the defining ideal I of an affinevariety X = V (I), then pc(X) = V (Ih), however only if K = K,see Exercise 5.3. As a counterexample in case K ≠ K, considerI = ⟨x2

1 + x42⟩ ∈ R[x1, x2]. Then

{(1 ∶ 0 ∶ 0)} = pc(V (I)) ⫋ V (Ih) = {(1 ∶ 0 ∶ 0), (0 ∶ 1 ∶ 0)}.

Algorithm 5.3.18 Let I ⊂K[x1, ..., xn] be an ideal. If G = (g1, ..., gr)is a Grobner basis of I with respect to any weighted degree ordering> with weight vector w = (1, ...,1), then

Gh = (gh1 , ..., g

hr )

is a Grobner basis of I with respect to the monomial ordering

xa0xα >h x

b0x

β ⇐⇒ xα > xβ or (xα = xβ and a > b)

Proof. By the choice of >, we have for all f ∈K[x1, ..., xn] that

degL>(f) = deg f ,

henceL>h(f

h) = L>(f).

By the Grobner basis definition, our assumption is

L>(I) = L>(G)

and we have to show

L>h(Ih) = L>h(G

h).

By L>h(ghi ) = L>(gi) and Lemma 5.3.8, this amounts to proving that

if F ∈ Ih is homogeneous then there is an i with L>(gi) ∣ L>h(F ).For this write

F =∑ili ⋅ f

hi

with li ∈K[x0, ..., xn] and fi ∈ I. Then

F a =∑ilai ⋅ fi ∈ I

so, by assumption, there is an i with

L> (gi) ∣ L>(Fa).

Moreover, for the homogeneous polynomial F there is an s ≥ 0with F = xs0 ⋅ (F

a)h. Hence, it follows that L>(F a) = L>h((F

a)h)

divides L>h(F ).


Example 5.3.19 Consider the affine twisted cubic

C = V (I) ⊂ A3(K)

defined byI = ⟨x2

1 − x2, x31 − x3⟩ ⊂K[x1, x2, x3].

Note, that the degree reverse lexicographical ordering refines thegrading by the total degree as required in Algorithm 5.3.18. Hence,we can compute the homogenization in Singular as follows:ring R = 0,(x(0..3)),dp;

ideal I = x(1)^2-x(2), x(1)^3-x(3);

std(I);

[1]=x(2)^2-x(1)*x(3)

[2]=x(1)*x(2)-x(3)

[3]=x(1)^2-x(2)

SoIh = ⟨x2

2 − x1x3, x1x2 − x0x3, x21 − x2x0⟩ .

Note that homogenizing the given set of generators of I will notyield the correct result:

V (x21 − x2x0, x

31 − x

20x3) = pc(C) ∪ V (x0, x1),

recall Exercise 3.7.

5.4 Graded modules and the Hilbert func-

tion

Suppose we are given a projective variety X ⊂ Pn(K) and I(X) =⟨f1, ..., fr⟩ generated by homogeneous polynomials fi ∈ R =K[x0, ..., xn].We will determine the invariants of X mentioned above from a finitefree resolution of the R-module

M = coker (f1 ... fr) ≅ R/I.

This module comes with an additional structure, we will have to re-spect when computing the resolution to obtain well-defined results:M is a graded module. In general, we define:

Definition 5.4.1 Let R = ⊕i∈N0Ri be a graded ring. A graded

R-module is an R-module M together with a decomposition

M =⊕i∈ZMi


as abelian groups such that

RiMj ⊂Mi+j

for all i, j.

Example 5.4.2 Considering I = ⟨x30 − x

31 − x

32⟩ ⊂ R = K[x0, x1, x2]

choosing lp with x0 > x1 > x2 we have as a K-vector space

R/I =K ⟨xα ∣ xα ∉ L(I)⟩ =K ⟨xα ∣ 0 ≤ α0 ≤ 2, α1, α2 ≥ 0⟩

so R/I is a graded R-module with the decomposition into K-vectorspaces

R/I =K ⟨1⟩⊕K ⟨x0, x1, x2⟩⊕K ⟨x20, x0x1, x0x2, x

21, x1x2, x

22⟩

⊕K ⟨x20x1, x

20x2, x0x

21, x0x1x2, x0x

22, x

31, x

21x2, x1x

22, x

32⟩⊕ ...

In general, if I ⊂ R = K[x0, ..., xn] is a homogeneous ideal, andIj is the K-vector space of homogeneous polynomials of degree j inI, then

I =⊕jIj,

since, by Lemma 5.3.8, all homogeneous summands of an elementare again in I. So also

R/I =⊕jRj/Ij

is a graded R-module, and

dimK(Rj/Ij) = dimK(Rj) − dimK(Ij).

This immediately raises the question, which information we canderive from the K-vector space dimensions of the graded pieces.

Definition and Theorem 5.4.3 Consider R = K[x0, ..., xn] withthe standard grading, and let M be a graded R-module. If M isfinitely generated, then Mi is a K-vector space of finite dimension.The Hilbert function of M is defined as

HM ∶ Z → N0

i ↦ dimKMi

Proof. If dimKMs = ∞, then the submodule ⊕i≥sMi ⊂ M is notfinitely generated, since any element in Ms can only be a K-linearcombination of R-module generators.


Example 5.4.4 The Hilbert function of R/I in Example 5.4.2 is

i 0 1 2 3 4HR/I(i) 1 3 6 9 12

so for i ≥ 1 it seems that

HR/I(i) = 3i

agrees with a polynomial. This is a general phenomenon:

The following theorem by Hilbert shows that all informationencoded in the Hilbert function is determined by only finitely manyvalues:

Definition and Theorem 5.4.5 Let R = K[x0, ..., xn] with thestandard grading, and M a finitely generated graded R-module.Then there is an i0 ∈ Z and a polynomial PM ∈ Q[t] of degree ≤ nwith

HM(i) = PM(i)

for all integers i ≥ i0.The polynomial PM is called the Hilbert polynomial of M .

Definition 5.4.6 If X ⊂ Pn(K) is a projective variety, and I =I(X) ⊂ R = K[x0, ..., xn] is the homogeneous zero ideal, then wedefine,

1) the dimension of X as the degree of PR/I

dim(X) = deg(PR/I),

2) the degree of X as dim(X)! times the lead coefficient of PR/I

deg(X) = dim(X)! ⋅ LC(PR/I),

3) the arithmetic genus of X as

pa(X) = (−1)dim(X) (PR/I(0) − 1) .

For an affine algebraic set X we define these invariants as theinvariants of the projective closure pc(X).


Example 5.4.7 Returning to Example 5.4.2, we use Singular tocompute the Hilbert polynomial PR/I and, hence, determine theseinvariants for C = V (I):ring R=0,(x(0..2)),dp;

ideal I = x(0)^3-x(1)^3 - x(2)^3;

LIB "poly.lib";

hilbPoly(I);

0, 3

This tells us thatPR/I = 3i + 0

so dim(C) = 1, deg(C) = 3 and pa(C) = 1, which makes sense foran elliptic curve. See also Exercise 5.3.

By the proof of Theorem 5.4.5 we obtain an algorithm to com-pute the Hilbert polynomial. It uses graded free resolutions:

5.5 Graded free resolutions

Definition 5.5.1 A homomorphism f ∶M → N of graded R-modulesis called homogeneous if f(Mi) ⊂ Ni for all i.

Proposition 5.5.2 If f ∶ M → N is a homogeneous homomor-phism of R-modules, the R-modules

ker(f) ⊂M

im(f) ⊂ N

coker(f) = N/ im(f)

are graded.

Considering the homomorphisms

fi = f ∣Mi∶Mi → Ni

the proof is an easy exercise.

Remark 5.5.3 If0→ U →M → Q→ 0

is an exact sequence of graded modules with homogeneous homo-morphisms then by the homomorphism theorem

HU(i) −HM(i) +HQ(i) = 0

for all i.


Consider the homomorphism of K[x]-modules

K[x]1 →K[x]1, e1 ↦ x ⋅ e1

where e1 = 1 denotes the unit basis vector. This map clearly shouldbe homogeneous. However, according to our definition it is not,since e1 of degree deg(e1) = 0 is mapped to x⋅e1 of degree deg(x⋅e1) =deg(x)+deg(e1) = 1. This problem can easily be solved by shiftingthe degrees:

Definition 5.5.4 If M is a graded R-module and a ∈ Z, then thegraded R-module M(a) is defined by

M(a)i =Ma+i

and is called the a-th twist of M .

The module K[x]1(−1) has one generator e′1 = 1 of degreedeg(e′1) = 1, which is mapped by

K[x]1(−1)→K[x]1, e′1 ↦ x ⋅ e1

to x ⋅ e1 of degree deg(x ⋅ e1) = 1+0 = 1. Hence, this homomorphismis homogeneous. More generally:

Example 5.5.5 If fi ∈ R = K[x0, ..., xn] are homogeneous and I =⟨f1, ..., fr⟩ ⊂ R then

A = (f1, ..., fr) ∶⊕ri=1R(−deg fi)→ R

is a homogeneous homomorphism. In particular, cokerA is a gradedR-module and cokerA ≅ R/I by a homogeneous isomorphism.

Definition 5.5.6 If M is a graded R-module, then a free resolution

...→ Fiφi→ Fi−1

φi−1→ ...

φ2→ F1

φ1→ F0

φ0→M → 0

is called graded if all φi are homogeneous.

In particular, R and all Fi are graded.

Theorem 5.5.7 (Graded syzygy theorem) Let R =K[x0, ..., xn].Every finitely generated graded R-module has a finite graded freeresolution. The free modules are finitely generated and the lengthis at most n + 1.

We will prove Theorem 5.5.7 (and 5.2.4) by constructing a res-olution using Grobner basis techniques. Before, we use the gradedsyzygy theorem to prove Theorem 5.4.5 by giving an explicit con-struction of the Hilbert polynomial.


5.6 Construction of the Hilbert polyno-

mial

Remark 5.6.1 Let R = K[x0, ..., xn]. Since in R there are pre-cisely

(i + n

n) =∏

nj=1

i + j

j

monomials of degree i ≥ 0 and no monomials of negative degree, wehave

HR(i) = (i + n

n)

0

∶= {(i+nn) for i ≥ −n

0 otherwise

Note that (i+nn) = 0 for i = −n, ...,−1.

Hence for i ≥ −n the Hilbert function agrees with a polynomialof degree n:

HR(i) =1

n!(i + n) ⋅ ... ⋅ (i + 1).

So, for the twist R(a) it holds

HR(a)(i) =HR(i + a) = (i + a + n

n)

0

Example 5.6.2 For the Hilbert function of R =K[x0, x1, x2] as anR-module, we get

i 0 1 2 3 4HR(i) 1 3 6 10 15

and for i ≥ −2

HR(i) =1

2(i + 2)(i + 1),

see Figure 5.6.

Remark 5.6.3 If R =K[x0, ..., xn] and F =⊕sj=1R(aj), then

HF (i) =s

∑j=1

(i + aj + n

n)

0

and for i ≥ −min{aj} − n

HF (i) =s

∑j=1

(i + aj + n

n)


–1

0

1

2

3

4

5

6

–3 –2 –1 1 2

Figure 5.6: Hilbert function of K[x0, x1, x2].

Example 5.6.4 If R = K[x0, x1, x2] and M = R(−2) ⊕ R(−3) wehave

HR(i) =1

2(i(i − 1) + (i − 1)(i − 2)) = (i − 1)2

for i ≥ 1 = −min{−2,−3} − 2, see Figure 5.7. For example, writtenin terms of the standard basis vectors e1 of degree 2 and e2 of degree3,

M2 = ⟨(10)⟩

and

M3 =K ⟨(01) ,(

x1

0) ,(

x2

0) ,(

x3

0)⟩

Note that deg(xie1) = 1 + 2 = 3.

Using these elementary observations and the graded syzygy the-orem, we now prove Theorem 5.4.5. For a different proof see Exer-cises 5.8 and 5.6.Proof. By Theorem 5.5.7, any finitely generated graded R-modulehas a finite graded free resolution

0→ Flφl→ Fl−1

φl−1→ ...

φ2→ F1

φ1→ F0

φ0→M → 0

So by Remark 5.5.3,

HM(t) =r

∑i=0

(−1)iHFi(t).


0

1

2

3

4

1 2 3

Figure 5.7: Hilbert function of R(−2)⊕R(−3) for R =K[x0, x1, x2].

WritingFi =⊕

sij=1R(ai,j)

by Remarks 5.6.1 and 5.6.3, we have

HFi(t) =

si

∑j=1

(t + ai,j + n

n)

0

so

HM(t) =r

∑i=0

(−1)isi

∑j=1

(t + ai,j + n

n)

0

Hence for i ≥ −min{ai,j} − n the Hilbert function agrees with thepolynomial

HM(t) =r

∑i=0

(−1)isi

∑j=1

(t + ai,j + n

n)

of degree ≤ n.

Algorithm 5.6.5 The constructive proof immediately yields an al-gorithm to compute both the Hilbert function and the Hilbert poly-nomial from a graded free resolution.

Example 5.6.6 For the plane cubic I = ⟨x30 − x

31 − x

32⟩ ⊂ R =K[x0, x1, x2]

from Example 5.4.2 the resolution of R/I is

0→ R(−3)1 → R → R/I → 0


and

HM(t) = dimK Rt − dimK It

= dimK Rt − dimK Rt−3

= (t + 2

2) − (

t − 1

2)

=1

2((t + 2)(t + 1) − (t − 1)(t − 2))

= 3t

Example 5.6.7 Consider the projective twisted cubic C = V (I) ⊂P3(K), where

I = ⟨x22 − x1x3, x1x2 − x0x3, x

21 − x2x0⟩ .

A finite free resolution of R/I is given by

0→ R(−3)2 → R(−2)3 → R → R/I → 0

as the following Singular calculation shows:ring R=0,(x0,x1,x2,x3),dp;

ideal I=x2^2-x1*x3, x1*x2-x0*x3, x1^2-x2*x0;

def L=res(I,0);

1 3 2

R <-- R <-- R

0 1 2

print(betti(L),"betti");

0 1 2

0: 1 - -

1: - 3 2

total: 1 3 2

In general, for a graded free resolution

0→ Fl → Fl−1 → ...→ F1 → F0 →M → 0

the column i of this so called Betti table corresponds to the freemodule Fi. The rank of Fi is given in the last row ”total”. At rowposition j and column position i the number of free generators ofFi of degree j + i is indicated.

The shift by i has the benefit of leading to a more compact table,since, homogeneous maps given by linear polynomials have sourceand target in the same row (as we will see, we can assume that themaps φi are given by matrices with entries of degree at least 1).



Example 5.6.8 From the resolution we obtain for t ≥ 0 = 3 − 3

HR/I(t) = (t + 0 + 3

3) − 3(

t − 2 + 3

3) + 2(

t − 3 + 3

3)

=1

3!((t + 3)(t + 2)(t + 1) − 3(t + 1)t(t − 1) + 2t(t − 1)(t − 2))

= 3t + 1

so in the alternating sum the t3 and t2 terms cancel, and we obtaina polynomial of degree 1. Hence,

dim(C) = 1 deg(C) = 3 pa(C) = 0

For an example of higher dimension, see Exercise 5.5.

5.7 Monomial orderings for modules

Let R = K[x1, ..., xn]. For the moment we do not consider anygrading on R. To compute free resolution we have to extend theBuchberger algorithm from ideals of R to the setup of submodulesof the free module Rt.

First of all, what are the monomials? Write ei for the i-thstandard basis vector of Rt. Then, like in the case of polynomials,any element of Rt can be written as a sum of terms of the formcαxαei. For example, in K[x, y]2

(2x + y2

x) = 2xe1 + y

2e1 + xe2.

Hence we define:

Definition 5.7.1 An element of the form xαei ∈ Rt we call a mono-mial, and a non-zero constant multiple of a monomial we call aterm.

A monomial ordering on Rt is an ordering > on the set ofmonomials in Rt such that

1) > is a total ordering

2) > respects multiplication, that is,

xαei > xβej ⇒ xαxγei > x

βxγej

for all α,β, γ, i, j.


3) xαei > xβei⇔ xαej > xβej for all α,β, i, j.

Remark 5.7.2 By setting xα > xβ ⇔ xαei > xβei the ordering >uniquely determines a monomial ordering on R, which we againdenote by >.

Definition 5.7.3 A monomial ordering on Rt is called global ifthe induced ordering on R is global.

Remark 5.7.4 The ordering > on Rt is a well ordering if and onlyif the induced ordering > on R is a well ordering.

Proof. The implication ⇒ is trivial. On the other hand, any setof monomials is a union M = M1e1 ∪ ... ∪Mses where Mi are setsof monomials in R. So if > is a well ordering on R then Mi has asmallest element mi, and there is a smallest element in the finiteset {m1e1, ...,mses}.

Example 5.7.5 Given a monomial ordering > on R there are twocanonical ways of producing a monomial ordering on Rt:

1) Priority to the monomials (>, c):

xαei > xβej ∶⇐⇒ xα > xβ or (xα = xβ and i < j)

2) Priority to the components (c,>):

xαei > xβej ∶⇐⇒ i < j or (i = j and xα > xβ)

Example 5.7.6 In Singular we can give priority to the mono-mials byring R = 0,(x,y),(lp,c);

[1,0]>[0,1];1

[x,0]>[y,0];1

[0,x]>[y,0];1

and priority to the components byring R = 0,(x,y),(c,lp);

[0,x]>[y,0];0

[y,0]>[0,x];1

If you use C instead of c, then the ordering of the unit basisvectors is reversed.


Definition 5.7.7 With respect to a given monomial ordering > onRt, for any element

v =∑α,i

cαxαei

the leading monomial is the largest monomial xαei with cα ≠ 0and is denoted by L (v). Furthermore, we denote by LC(v) = cα theleading coefficient, and by LT (v) = cαxαei the leading term.

Example 5.7.8 Using (lp, c) on K[x, y]2, for

f = (x

5x2y + xy2 ) = 5x2y ⋅ e2 + x ⋅ e1 + xy2 ⋅ e2

we get

L(f) = x2y ⋅ e2 LC(f) = 5 LT (f) = 5x2y ⋅ e2

In Singular:ring R=0,(x,y),(lp,c);

vector f = [x, 5x2y+xy2];

leadmonom(f);

x2y*gen(2)

leadcoef(f);

5

lead(f);

5x2y*gen(2)

To do one more example, for

f = (y + zx + y

) ∈K[x, y, z]2

with respect to (lp, c) we have

L(f) = x ⋅ e2

and with respect to (c, lp)

L(f) = y ⋅ e1

In Singular:ring R=0,(x,y,z),(lp,c);

vector f = [y+z, x+y];

leadmonom(f);

x*gen(2)

ring R=0,(x,y,z),(c,lp);

vector f = [y+z, x+y];


leadmonom(f);

[y]

Note, that for (lp, c), Singular writes the elements of Rt in termsof unit basis vectors, and for (c, lp) as lists. Note also, that in thelist notation, trailing zeros are skipped. You can also use both waysto input vectors.

Definition 5.7.9 The monomials of Rt come with a natural partialorder, which we call divisibility

xαei ∣ xβej ⇐⇒ i = j and xα ∣ xβ

Moreover, given two terms c1xαei and c2xβej with xαei ∣ xβejwe define their quotient as

c2xβejc1xαei

=c2xβ

c1xα∈ R.

Definition 5.7.10 A submodule U ⊂ Rt is called monomial, if itis generated by monomials.

Lemma 5.7.11 Every monomial submodule has a unique set ofminimal generators consisting of monomials.

Proof. Let U = ⟨M⟩ ⊂ Rt be generated by the set of monomialsM . By Lemma 4.3.11 the free module Rt is Noetherian, hence Uis finitely generated by f1, ..., fr ∈ U (see Definition and Theorem4.3.9). Like in the case of ideals, write fi = ∑

uj=1 ri,jmj with mj ∈M .

Then U ⊂ ⟨m1, ...,mu⟩ ⊂ ⟨M⟩ = U . Among the mj consider theminimal elements with respect to divisibility.

Note thatxαei ≤ x

βej ⇐⇒ xαei ∣ xβej

gives a partial order on the set of monomials of Rt.

5.8 Division with remainder and Grobner

bases for modules

The basic setup for the theory of Grobner bases of submodules ofRt works just like in the case of ideals (that is, t = 1) as describedin Section 2.4. To convince ourselves that this is indeed the case,let us recall the fundamental steps.


Definition 5.8.1 Given a list G = (g1, ..., gs) of elements of Rt, anormal form is a map NF(−,G) ∶ Rt → Rt with

1) NF(0,G) = 0.

2) If NF(f,G) ≠ 0 then L(NF(f,G)) ∉ L(G).

3) For all 0 ≠ f ∈ R there are ai ∈ R with

f −NF(f,G) = ∑si=1aigi

and L(f) ≥ L(aigi) for all i with aigi ≠ 0.

We also say that NF is a normal form, if NF(−,G) is a normalform for all G.

Using Definition 5.7.9, Division with remainder in R (Algorithm2.4.1) generalizes immediately to Algorithm 5.8.1 in Rt. It gives anormal form in the sense of Definition 5.8.1.

Algorithm 5.8.1 Division with remainder for modules

Input: f ∈ Rt, g1, ..., gs ∈ Rt, > be a global ordering on the mono-mials of Rt.


f = q + r = ∑ti=1aigi + r

such that L(r) is not divisible by any L(gi).1: q = 02: r = f3: while r ≠ 0 and L(gi) ∣ L(r) for some i do4: Cancel the lead term of r:5: a = LT (r)

LT (gi) ∈ R as given in Definition 5.7.96: q = q + a ⋅ gi7: r = r − a ⋅ gi

Example 5.8.2 Let R = K[x1, ..., x5] and consider the monomialordering (lp, c) on R5 obtained from lp by giving priority to themonomials. Let gi ∈ R5 be the i-th column of the matrix

A =

⎛⎜⎜⎜⎜⎜⎜⎝

0 x2 0 0 −x1

−x2 0 x3 0 00 −x3 0 x4 00 0 −x4 0 x5

x1 0 0 −x5 0

⎞⎟⎟⎟⎟⎟⎟⎠


and U = im(A) = ⟨g1, ..., g5⟩ ⊂ R5. We try out Algorithm 5.8.1 todivide some vectors v ∈ R5 by G = (g1, ..., g5):

v =

⎛⎜⎜⎜⎜⎜⎜⎝

000

−x2x4

x1x3

⎞⎟⎟⎟⎟⎟⎟⎠

= x3

⎛⎜⎜⎜⎜⎜⎜⎝

0−x2

00x1

⎞⎟⎟⎟⎟⎟⎟⎠

+

⎛⎜⎜⎜⎜⎜⎜⎝

0x2x3

0−x2x4

0

⎞⎟⎟⎟⎟⎟⎟⎠

= x3

⎛⎜⎜⎜⎜⎜⎜⎝

0−x2

00x1

⎞⎟⎟⎟⎟⎟⎟⎠

+ x2

⎛⎜⎜⎜⎜⎜⎜⎝

0x3

0−x4

0

⎞⎟⎟⎟⎟⎟⎟⎠

+ 0

which shows that v ∈ U .In the following example the division terminates early:

v =

⎛⎜⎜⎜⎜⎜⎜⎝

000

x1 − x2x4

x1x3

⎞⎟⎟⎟⎟⎟⎟⎠

= x3

⎛⎜⎜⎜⎜⎜⎜⎝

0−x2

00x1

⎞⎟⎟⎟⎟⎟⎟⎠

+

⎛⎜⎜⎜⎜⎜⎜⎝

0x2x3

0x1 − x2x4

0

⎞⎟⎟⎟⎟⎟⎟⎠

This can be fixed by considering reduced division, just like in Al-gorithm 2.4.2, which will put x1e4 into the remainder. Recall, thatfor testing whether a normal form is zero, this is not a problem:If we would put terms into the remainder in intermediate steps,they would never go away again (since in every step the lead termdecreases). So once Algorithm 5.8.1 stops with a non-zero remain-der, we know that also reduced division will not give zero (and viceversa).

Finally, for

v =

⎛⎜⎜⎜⎜⎜⎜⎝

00

−x1x3

x2x5

0

⎞⎟⎟⎟⎟⎟⎟⎠

= x1g2 + x2g5

Algorithm 5.8.1 will not do anything, and return v as the remainder,although v ∈ U . The reason is that G is not a Grobner basis. Wewill return to this in the next section.In Singular these calculation can be done as follows:ring R = 0,(x(1..5)),(lp,c);

module M = [0,-x(2),0,0,x(1)],

[x(2),0,-x(3),0,0],


[0,x(3),0,-x(4),0],

[0,0,x(4),0,-x(5)],

[-x(1),0,0,x(5),0];

reduce([0,0,0,-x(2)*x(4),x(1)*x(3)],M,1);

0

reduce([0,0,0,x(1)-x(2)*x(4),x(1)*x(3)],M,1);

x(1)*gen(4)+x(2)*x(3)*gen(2)-x(2)*x(4)*gen(4)

reduce([0,0,-x(1)*x(3),x(2)*x(5),0],M,1);

-x(1)*x(3)*gen(3)+x(2)*x(5)*gen(4)

Provided we divide by a Grobner basis, division with remaindersolves the submodule membership problem:

Definition 5.8.3 Given a monomial ordering > and a subset G ⊂Rt, we define the leading module of G as

L(G) = L>(G) = ⟨L(f) ∣ f ∈ G/{0}⟩ ⊂ Rt,

the monomial submodule generated by the leading monomials.

Definition 5.8.4 Let U ⊂ Rt be a submodule and > a global mono-mal ordering on Rt. A finite set 0 ∉ G ⊂ U is called Grobner basisof U with respect to >, if

L(G) = L(U).

Theorem 2.4.8, Lemma 2.4.9 and Corollary 2.4.10 generalizeimmediately to the module setup:

Theorem 5.8.5 (Submodule membership) Let U ⊂ Rt be a sub-module and f ∈ Rt. If G = (g1, ..., gs) is a Grobner basis of U andNF is a normal form, then

f ∈ U ⇐⇒ NF(f,G) = 0.

Proof. Write f = ∑i aigi + r with r = NF(f,G). If r = 0 thenf ∈ ⟨G⟩ ⊂ U . On the other hand, if r ≠ 0 then by Definition 5.8.1(2.)

L(r) ∉ L(G) = L(U).

So, r ∉ U , hence f = ∑i aigi + r ∉ U .

Lemma 5.8.6 If V ⊂ U ⊂ Rt are submodules with L(V ) = L(U)then U = V .


Proof. Let G = (g1, ..., gs) be a Grobner basis of V , NF a normalform, f ∈ U and f = ∑i aigi + r with r = NF(f,G). So r ∈ U . Ifr ≠ 0, then by Definition 5.8.1 (2.)

L(r) ∉ L(G) = L(V ) = L(U).

Hence r ∉ U , a contradiction.

Corollar 5.8.7 If G is a Grobner basis of the submodule U ⊂ Rt,then

U = ⟨G⟩ .

Proof. We have L(U) = L(G) ⊂ L(⟨G⟩) ⊂ L(U), so G is a Grobnerbasis of ⟨G⟩ ⊂ U and L(⟨G⟩) = L(U). Equality follows from Lemma5.8.6.

Remark 5.8.8 To have a uniquely determined remainder, we pro-ceed as follows: A vector f ∈ Rt is called reduced with respect to aset G ⊂ Rt, if no term of f is contained in L(G). A normal form iscalled reduced normal form, if NF(f,G) is reduced with respectto G for all f and G.

Arguing in the same way as in proof of Theorem 2.4.15 showsthat, for a reduced normal form, the remainder is unique.

Also Algorithm 2.4.2 immediately generalizes to the module case,yielding a reduced normal form. Recall that for reduced division withremainder we do not stop if the lead monomial is not divisible bysome lead monomial of G, but instead put the lead term into theremainder and continue.

Example 5.8.9 Using reduced division with remainder, we cancontinue the second calculation in Example 5.8.2:

v =

⎛⎜⎜⎜⎜⎜⎜⎝

000

x1 − x2x4

x1x3

⎞⎟⎟⎟⎟⎟⎟⎠

= x3

⎛⎜⎜⎜⎜⎜⎜⎝

0−x2

00x1

⎞⎟⎟⎟⎟⎟⎟⎠

+

⎛⎜⎜⎜⎜⎜⎜⎝

0x2x3

0x1 − x2x4

0

⎞⎟⎟⎟⎟⎟⎟⎠

= x3

⎛⎜⎜⎜⎜⎜⎜⎝

0−x2

00x1

⎞⎟⎟⎟⎟⎟⎟⎠

+

⎛⎜⎜⎜⎜⎜⎜⎝

000x1

0

⎞⎟⎟⎟⎟⎟⎟⎠

+ x2

⎛⎜⎜⎜⎜⎜⎜⎝

0x3

0−x4

0

⎞⎟⎟⎟⎟⎟⎟⎠

In Singular:


ring R = 0,(x(1..5)),(lp,c);

module M = [0,-x(2),0,0,x(1)],

[x(2),0,-x(3),0,0],

[0,x(3),0,-x(4),0],

[0,0,x(4),0,-x(5)],

[-x(1),0,0,x(5),0];

reduce([0,0,0,x(1)-x(2)*x(4),x(1)*x(3)],M);

x(1)*gen(4)

5.9 Computing Grobner bases of sub-

modules

Fix a global monomial ordering > on Rt. We follow the same strat-egy as in Section 2.5 by cancelling lead terms via S-polynomials.Clearly, lead terms can only cancel if they occur in the same com-ponent of the vectors. Hence, we define:

Definition 5.9.1 The least common multiple of two monomi-als xαei and xβej in Rt is

lcm(xαei, xβej) = {

lcm(xα, xβ) if i = j0 if i ≠ j

Definition 5.9.2 The syzygy polynomial or S-polynomial of f, g ∈Rt is defined as the vector

spoly(f, g) =lcm(L(f), L(g))

LT (f)f −

lcm(L(f), L(g))

LT (g)g ∈ Rt.

In particular, writing L(f) = xαei and L(g) = xβej, if i ≠ j thenspoly(f, g) = 0. On the other hand, if i = j then, according toDefinition 5.7.9, we have a well defined quotient

lcm(L(f), L(g))

LT (f)∈ R.

Using this notation, Buchberger’s algorithm for ideals carriesover directly to Algorithm 5.9.1. Clearly we have to consider in thealgorithm only pairs of elements f ≠ g which have the lead term inthe same component.


Algorithm 5.9.1 Buchberger

Input: U = ⟨g1, ..., gs⟩ ⊂ Rt a submodule, > a global monomialordering, and NF a normal form.

Output: A Grobner basis of U with respect to >.1: G = {g1, ..., gs}2: repeat3: H = G4: for all f, g ∈H do5: r = NF(spoly(f, g),H)6: if r ≠ 0 then7: G = G ∪ {r}8: until G =H

Proof. If r ≠ 0 then L(r) ∉ L(H) by Definition 5.8.1(2.), hence

L(H) ⫋ L(H ∪ {r}).

Since Rt is Noetherian (Lemma 4.3.11), the algorithm terminatesby the chain condition (Definition and Theorem 4.3.9). To showthat the final result is a Grobner basis, we will prove Buchberger’scriterion in the next Section 5.10.

Example 5.9.3 Continuing Example 5.8.2, we use Buchberger’salgorithm to compute a Grobner basis of U ⊂ R5 with respect to(lp, c). In each step the first column denotes the coefficients in thesyzygy polynomial and the second the division with remainder.

↱ −x1x3e3 + x2x5e4 ↱ −x1x3x5e5 + x2x4x5e4

g1 = x1e5 − x2e2 x3x5

g2 = x2e1 − x3e3 x1

g3 = x3e2 − x4e4 x2x5

g4 = x4e3 − x5e5 x1x3

g5 = −x1e1 + x5e4 x2

g6 = x1x3e3 − x2x5e4 1 −x4

For the first calculation, recall that already in Example 5.8.2 weobserved that NF(g6, (g1, ..., g5)) = g6 ≠ 0. The second calculationstep by step:

spoly(g4, g6) = −x1x3x5e5 + x2x4x5e4

= x3x5 ⋅ g1 + (−x2x3x5e2 + x2x4x5e4)

= x3x5 ⋅ g1 + x2x5 ⋅ g3 + 0

Since there are no further non-zero S-polynomials, this provesthat (g1, ..., g6) is a Grobner basis.


Dividing by (g1, ..., g6), the submodule membership test for g6

now obviously works as expected: NF(g6, (g1, ..., g6)) = 0.The Grobner basis calculation in Singular:

ring R = 0,(x(1..5)),(lp,c);

module M = [0,-x(2),0,0,x(1)],

[x(2),0,-x(3),0,0],

[0,x(3),0,-x(4),0],

[0,0,x(4),0,-x(5)],

[-x(1),0,0,x(5),0];

std(M);

[1]=x(4)*gen(3)-x(5)*gen(5)

[2]=x(3)*gen(2)-x(4)*gen(4)

[3]=x(2)*gen(1)-x(3)*gen(3)

[4]=x(1)*gen(5)-x(2)*gen(2)

[5]=x(1)*gen(1)-x(5)*gen(4)

[6]=x(1)*x(3)*gen(3)-x(2)*x(5)*gen(4)

print(std(M));0, 0, x(2), 0, x(1), 0,

0, x(3), 0, -x(2), 0, 0,

x(4), 0, -x(3), 0, 0, x(1)*x(3),

0, -x(4), 0, 0, -x(5), -x(2)*x(5),

-x(5), 0, 0, x(1), 0, 0

Remark 5.9.4 Using the notion of divisibility as introduced inDefinition 5.7.9, the definitions of minimal and reduced Grobnerbases carry over from the ideal case immediately.

Given a Grobner basis G = (g1, ..., gs), by Lemma 5.7.11, the gicorresponding to the (with respect to divisibility) minimal elementsof {L(gi) ∣ i} form a minimal Grobner basis.

Just like in the ideal case (Theorem 2.5.7 and Remark 2.5.8) wecan obtain the unique reduced Grobner basis.

5.10 Buchberger’s criterion and compu-

tation of syzygy modules

We now prove Buchberger’s criterion for Grobner bases of submod-ules U ⊂ Rt.

As we have already observed in the ideal case (Theorem 2.5.2),the strategy for the proof is closely related to syzygies. Indeed, theBuchberger test produces a Grobner basis of the module of syzy-gies with respect to an appropriately chosen ordering. By iteratingthis, we will obtain an algorithm to compute free resolutions. This


algorithm is, by far, the fastest way of computing free resolutions.It is implemented in Singular in the command sres.

Definition 5.10.1 For a matrix G ∈ Rt×s we define the syzygymodule of G as the kernel

Syz(G) = ker(G) ⊂ Rs,

and the elements are called syzygies of G.Any matrix SG ∈ Rs×g with

im(SG) = Syz(G),

that is, with generators of Syz(G) in its columns, we call a syzygymatrix of G.

We use the notation also for a list G = (g1, ..., gs) of elementsof Rt, then defining the syzygy module of G as Syz(G) = Syz(MG)for the matrix

MG = (g1 ∣ ... ∣ gs) ∶ Rs → Rt

ei ↦ gi

To summarize, the image of MG

im(MG) = ⟨g1, ..., gs⟩ ⊂ Rt

is the submodule of Rt generated by the gi, and

im(SG) = ker(g1 ∣ ... ∣ gs) ⊂ Rs

is the module of relations between the gi.Note also that, by iteratively computing syzygy matrices, we

can compute free resolutions. We will do this in more detail in thenext section.

See also Exercise 6.1 for an application of the syzygy matrix forcomputing intersections of ideals.

Remark 5.10.2 Any syzygy v = ∑si=1 viei ∈ Syz(G) ⊂ Rs corre-

sponds to a relations

∑i=1

vigi = 0.

Given a global monomial ordering >, every successful Buchbergertest

NF(spoly(gi, gj),G) = 0


gives an expression

lcm(L(gi),L(gj))LT (gi)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶m

⋅ gi −lcm(L(gi),L(gj))

LT (gj)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

w

⋅ gj − ∑sk=1 akgk = 0

and, hence, a syzygy

s(gi, gj) ∶= m ⋅ ei −w ⋅ ej −∑sk=1 akek

∈ Syz(G) ⊂ Rs

For a Grobner basis G = (g1, ..., gs) we will even prove: Withrespect to the following ordering, the s(gi, gj) form a Grobner basisof Syz(G).

Definition 5.10.3 Let G = (g1, ..., gs) be a list of elements gi ∈ Rt

and > a monomial ordering on Rt. A monomial ordering >G on Rs

is defined by

xαei >G xβej ∶⇐⇒ L(xαgi) > L(x

βgj) or (L(xαgi) = L(xβgj) and i < j)

It is called the Schreyer ordering induced by > and G.

The key property is, that we have control over the lead mono-mials of the syzygies:

Lemma 5.10.4 If i < j then

L(s(gi, gj)) =lcm(L(gi), L(gj))

L(gi)ei

where the left hand side is with respect to >G and the right handside with respect to >.

Proof. We can assume s(gi, gj) ≠ 0; otherwise the claim is trivial.Since in an S-polynomial the lead terms cancel we have (using thenotation of Remark 5.10.2)

L(m ⋅ gi) = L(w ⋅ gj)

so by i < j it holdsm ⋅ ei >G w ⋅ ej.

Moreover, since NF is a normal form, for all k

L(akgk) ≤ L(spoly(gi, gj)) < L(m ⋅ gi),

that is, alsom ⋅ ei >G L(akek)

for all k. Using both observations,

L(s(gi, gj)) =m ⋅ ei.


Remark 5.10.5 If > is global, then also >G is global (Exercise).

Example 5.10.6 We return to the resolution of the projective twistedcubic as discussed in Example 5.6.7. So let C = V (I) ⊂ P3(K),where I is generated by the Grobner basis

G = (x22 − x1x3, x1x2 − x0x3, x

21 − x0x2)

with respect to dp. For

A =⎛⎜⎝

x1 −x0

−x2 x1

x3 −x2

⎞⎟⎠

we have MG ⋅ A = 0 hence im(A) ⊂ syz(G). We verify that thecolumns of A form a Grobner basis of im(A) with respect to >G.Observe that with respect to dp

x1 ⋅ x22 = x2 ⋅ x1x2 > x3 ⋅ x

21

x0 ⋅ x22 < x1 ⋅ x1x2 = x2 ⋅ x

21,

hence the lead terms of the columns of A with respect to >G are

A =⎛⎜⎝

x1 −x0

−x2 x1

x3 −x2

⎞⎟⎠

This tells us that there is no non-zero syzygy between the columns,so, assuming that Algorithm 5.9.1 is correct, they form a Grobnerbasis of im(A).

For the moment we do not know whether the columns of A gen-erate syz(G). This question will be solved as a by-product of theproof of Buchberger’s test:

Theorem 5.10.7 (Buchberger test) Let > be a global monomialordering on Rt and NF a normal form algorithm. Let U = ⟨g1, ..., gs⟩ ⊂Rt be a submodule. If

NF(spoly(gi, gj),G) = 0

for all i < j, then

1) G = (g1, ..., gs) is a Grobner basis of U with respect to >, and

2) the non-zero syzygies s(gi, gj) for i < j form a Grobner basisG′ of Syz(G) with respect to >G, in particular, by Corollary5.8.7,

Syz(G) = ⟨G′⟩ .


Note that this finishes the proof of correctness of Algorithm5.9.1. Before proving the theorem, we apply it to Example 5.10.6:

Example 5.10.8 We do the Buchberger test for

G = (x22 − x1x3, x1x2 − x0x3, x

21 − x0x2)

and the monomial ordering dp:

↱ −x21x3 + x0x2x3 ↱ x0x

22 − x0x1x3 ↱ x0x

32 − x3

1x3

x22 − x1x3 x1 −x0 x2

1 −x0x2

x1x2 − x0x3 −x2 x1

x21 − x0x2 x3 −x2 −x2

2 x1x3

By Theorem 5.10.7, this shows that G is a Grobner basis of I = ⟨G⟩with respect to dp. Moreover, it proves that

G′ =⎛⎜⎝

⎛⎜⎝

x1

−x2

x3

⎞⎟⎠,⎛⎜⎝

−x0

x1

−x2

⎞⎟⎠,⎛⎜⎝

x21 − x0x2

0−x2

2 + x1x3

⎞⎟⎠

⎞⎟⎠

is a Grobner basis of the syzygy module Syz(G) with respect to>G. The last vector can be deleted since its lead monomial x2

1e1 isdivisible by the lead monomial x1e1 of the first vector, hence

G′ =⎛⎜⎝

⎛⎜⎝

x1

−x2

x3

⎞⎟⎠,⎛⎜⎝

−x0

x1

−x2

⎞⎟⎠

⎞⎟⎠

is a minimal Grobner basis of Syz(G). Indeed, it is the reducedGrobner basis.

In particular,ker(MG) = im(MG′)

with

MG′ =⎛⎜⎝

x1 −x0

−x2 x1

x3 −x2

⎞⎟⎠

.

We can continue now one step further by considering syzygiesof syzygies, that is, we do the Buchberger test for G′. As alreadyobserved in Example 5.10.6, there is no non-trivial syzygy betweenthe vectors in G′, that is, ker(MG′) = 0. Hence,

0→ R2 MG′

→ R3 MG→ R1 → R/I → 0

is a free resolution.


In fact, since we begin with a graded matrix of generators MG,also the syzygy matrix MG′, as produced by the theorem, is a gradedhomomorphism. So, taking the grading into account, we have proventhat the projective twisted cubic has the graded free resolution

0→ R2(−3)MG′

→ R3(−2)MG→ R1 → R/I → 0

with Betti table0 1 2

0 ∶ 1 − −1 ∶ − 3 2total ∶ 1 3 2

Proof. Fix an arbitrary element

f = ∑si=1aigi ∈ U .

Theng = ∑

si=1aiei ∈ R

s

is a preimage of f under MG, that is,

MG(g) = f .

Using >G and a reduced normal form NF′ on Rs we get anexpression

g = ∑i,jgi,j ⋅ s(gi, gj) +NF′(g,G′)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶∑sk=1rkek

with gi,j ∈ R. Then for all k with rk ≠ 0, it holds

L(rkek) ∉ ⟨L(s(gi, gj)) ∣ i, j with s(gi, gj) ≠ 0⟩ .

By Lemma 5.10.4, for k < j

L(s(gk, gj)) =lcm(L(gk), L(gj))

L(gk)ek =∶mj,kek,

solcm(L(gk), L(gj))

L(gk)does not divide L(rk) (5.1)

for all k < j with s(gk, gj) ≠ 0.By

g −NF′(g,G′) ∈ ⟨G′⟩ ⊂ Syz(G) = ker(MG)

we getf =MG(NF′(g,G′)) = ∑

sk=1rkgk.


We show that in the sum there cannot be any cancellation oflead terms: If for k < j with rk, rj ≠ 0 it holds

L(rkgk) = L(rjgj),

then L(rkgk) = L(rk)L(gk) ∈ Rt is divisible by L(gk) and L(gj),hence also by lcm(L(gk), L(gj)). So

lcm(L(gk), L(gj))

L(gk)divides L(rk),

a contradiction to Equation 5.1 (since s(gk, gj) ≠ 0 by assumptionL(rkgk) = L(rjgj)).

• In the case f ≠ 0 this proves that L(f) = L(rkgk) for some k,and hence L(f) ∈ L(G). So L(U) = L(G), which is the firstclaim.

• If we take f = 0 then g is an arbitrary syzygy. In the case f = 0the argument shows that rk = 0 for all k. Hence NF′(g,G′) = 0for all g ∈ Syz(G). This implies the second claim by thefollowing standard Lemma 5.10.9 applied to G′.

Lemma 5.10.9 Let > be a global monomial ordering on Rt, NFa normal form, U ⊂ Rt a submodule, and G = (g1, ..., gs) a list ofelements of U . Then it is equivalent:

1) G is a Grobner basis of U .

2) NF(f,G) = 0 for all f ∈ U .

Proof. The implication (1) ⇒ (2) follows from Theorem 5.8.5,since for a Grobner basis G we get NF(f,G) = 0 for all f ∈ U .

On the other hand let f ∈ U . By NF(f,G) = 0 and Definition5.8.1(3.) we obtain an expression

f = ∑si=1aigi

with L(f) ≥ L(aigi) for all i. Hence, there has to be an i withL(f) = L(aigi), which implies L(gi) ∣ L(f).

This proves that L(U) = L(G).

Remark 5.10.10 By Theorem 5.10.7 the conditions in the lemmaare equivalent to


3) U = ⟨G⟩ and NF(spoly(gi, gj),G) = 0 for all i < j.

See Theorem 2.5.2 in the ideal case.

Remark 5.10.11 (Chain condition) Clearly it is sufficient toconsider in Theorem 5.10.7 a subset of G′ which has the same leadideal.

Suppose L(gi), L(gj) and L(gl) are in the same component andi < j < l. Then s(gi, gj), s(gi, gl), s(gj, gl) ∈ G′. If

L(gj) ∣ lcm(L(gi), L(gl))

then lcm(L(gi), L(gj)) ∣ lcm(L(gi), L(gl)). Hence

L(s(gi, gj)) ∣ L(s(gi, gl)),

that is, s(gi, gl) can be deleted from G′.

Example 5.10.12 In Example 5.10.8 for

G = (x22 − x1x3, x1x2 − x0x3, x

21 − x0x2)

we havex1x2 ∣ x2

1x22,

so only s(g1, g2) and s(g2, g3) are needed in G′ (as already observedabove).

5.11 Computing free resolutions

By iterating Theorem 5.10.7 we can compute free resolutions. How-ever, we have to show that the process of computing syzygies ofsyzygies terminates. For this we show the Syzygy Theorem 5.2.4by proving:

Theorem 5.11.1 Let R = K[x1, ..., xn]. Every finitely generatedR-module M has a finite free resolution

0→ Rtlφl→ Rtl−1

φl−1→ ...

φ1→ Rt0 →M → 0

such that, if we denote the j-th column of φi by gi,j, it holds

L(gi,j) ∈K[xi, ..., xn]ti−1

for all j. The length of the resolution l ≤ n.


To prove the theorem, we develop an algorithm computing thedesired resolution. First note that by Theorem 4.3.12, any finitelygenerated R-module is finitely presented.

Algorithm 5.11.1 Free resolution

Input: An R-module M with a finite presentation

Rt′1φ′1→ Rt0 →M → 0

and a monomial ordering > on Rt0 .Output: A finite free resolution of M as in Theorem 5.11.1.1: Put a Grobner basis of im(φ′1) with respect to > into the

columns of the matrix φ1 (apply Algorithm 5.9.1).2: a = 13: while φa ≠ 0 do4: Applying Remark 5.9.4, replace φa by a matrix

φa ∶ Rta → Rta−1

with a minimal Grobner basis in its columns.5: Sort the columns of φa = (ga,1 ∣ ... ∣ ga,ta) such that the L(ga,i)

involving the same unit vector are in descending order withrespect to lp.

6: Compute a syzygy matrix φa+1 ∶ Rta+1 → Rta withim(φa+1) = ker(φa) such that the columns of φa+1 form aGrobner basis of im(φa+1) with respect to >φa (apply Theo-rem 5.10.7).

7: a = a + 1

Proof. First observe: If the L(ga,i) involve only xv, ..., xn then theL(s(ga,i, ga,j)) with respect to >φa involve only xv+1, ..., xn.

Proof: Suppose i < j, and

L(ga,i) = xαvv ⋅ ... ⋅ xαn

n ⋅ eq

L(ga,j) = xβvv ⋅ ... ⋅ xβnn ⋅ eq.

By Lemma 5.10.4

L(s(ga,i, ga,j)) = (∏nl=vx

max{αl,βl}−αl

l ) ⋅ ei.

If the leftmost nonzero entry of α − β is positive (lexicographicordering), then αv ≥ βv, hence

max{αv, βv} − αv = 0,


which proves the claim.Applying this observation inductively, there is some a ≤ n such

that L(s(ga,i, ga,j)) does not involve x1, ..., xn for all s(ga,i, ga,j) ≠ 0.Suppose that there is some s(ga,i, ga,j) ≠ 0. Then, in the above

notation,αi ≥ βi for all i,

which implies thatL(ga,j) ∣ L(ga,i).

This contradicts the assumption that the Grobner basis (ga,1, ..., ga,ta)was chosen minimal, hence φa+1 must be zero.

Remark 5.11.2 Let U ⊂ Rt be a graded R-module and g1, ..., gs ∈ Uhomogeneous of degree deg(ga) = da. Since monomials can onlycancel if they have the same degree, also any spoly(gi, gj) ≠ 0 is ho-mogeneous of some degree di,j. By the same argument, the equation

lcm(L(gi),L(gj))LT (gi) ⋅ gi −

lcm(L(gi),L(gj))LT (gj) ⋅ gj − ∑

sk=1 akgk = 0

involves only monomials of degree di,j. If we now define a gradingon Rs by

deg(ea) = da

also s(gi, gj) is homogeneous of degree di,j. If we put the s(gi, gj)into the columns of φ, we obtain a graded homomorphism

φ ∶⊕i,jR(−di,j)→⊕aR(−da).

Together with Theorem 5.11.1, this proves the graded Syzygy Theo-rem 5.5.7.

Example 5.11.3 Let R =K[x0, ..., x6]. We compute a graded freeresolution of R/I for the ideal I generated by the entries of

φ1 = (x1x2x5, x1x2x6, x3x4x6, x3x4x7, x5x7) .

This is a graded map

φ1 ∶ R(−3)4 ⊕R(−2)1 → R1.

The generators are already sorted by lexicographically. Using dp,the Buchberger test gives:

↱ 0 ↱ 0 ↱ 0 ↱ 0 ↱ 0 ↱ 03 x1x2x5 −x6 −x7

3 x1x2x6 x5 −x3x4 −x5x7

3 x3x4x6 −x7 x1x2

3 x3x4x7 x6 −x5

2 x5x7 x3x4 x1x2 x1x2x6

deg 4 4 4 4 5 5


In the table, we also indicate (in blue) the degrees of the generatorsand the syzygies, according to Remark 5.11.2.

Note, that the remaining syzygies

↱ 0 ↱ 0 ↱ 0 ↱ 0x1x2x5 −x3x4x6 −x3x4x7

x1x2x6 −x3x4x7

x3x4x6 x1x2x5 −x7

x3x4x7 x1x2x5 x1x2x6

x5x7 x3x4

can be skipped, since they do not give any new lead terms. Hence

φ2 =

⎛⎜⎜⎜⎜⎜⎜⎝

−x6 0 0 −x7 0 0x5 0 0 0 −x3x4 −x5x7

0 −x7 0 0 x1x2 00 x6 −x5 0 0 00 0 x3x4 x1x2 0 x1x2x6

⎞⎟⎟⎟⎟⎟⎟⎠

Observer that the lead terms of the syzygies do not involve x1.As a graded map

φ2 ∶ R(−4)4 ⊕R(−5)2 → R(−3)4 ⊕R(−2)1.

We now proceed with the next step: The syzygies g2,i are alreadysorted lexicographically by the lead monomials.

↱ −g2,6 ↱ −x1x2x5x7e3 + x1x2x3x4x6e54 g2,1 = −x6e1 + x5e2 x7

4 g2,2 = −x7e3 + x5e4 −x1x2x5

4 g2,3 = −x5e4 + x3x4e5 −x1x2x6

4 g2,4 = −x7e1 + x1x2e5 −x6

5 g2,5 = −x3x4e2 + x1x2e3 −x5x7

5 g2,6 = −x5x7e2 + x1x2x6e5 1 x3x4

deg 5 7

The division with remainder for the second syzygy in detail:

−x1 x2x5x7e3 + x1x2x3x4x6e5

= x1x2x5(−x7e3 + x6e4) −x1x2x5x6e4 + x1x2x3x4x6e5

= x1x2x5(−x7e3 + x6e4) + x1x2x6(−x5e4 + x3x4e5)

Hence

φ3 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

x7 00 −x1x2x5

0 −x1x2x6

−x6 00 −x5x7

1 x3x4

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠


As a graded homomorphism,

φ3 ∶ R(−7)1 ⊕R(−5)1 → R(−4)4 ⊕R(−5)2.

Since there are no syzygies between the columns of φ3, we havefound a free resolution

0→ R2 φ3→ R6 φ2

→ R5 φ1→ R1 → R/I → 0.

which is graded as

0→ R(−7)1⊕R(−5)1 φ3→ R(−4)4⊕R(−5)2 φ2

→ R(−3)4⊕R(−2)1 φ1→ R1 → R/I → 0.

Written as a Betti diagram

0 1 2 30 ∶ 11 ∶ 12 ∶ 4 4 13 ∶ 24 ∶ 1total ∶ 1 5 6 2

In Singular we can compute the resolution as follows:ring R = 0,(x(1..7)),dp;

ideal I = x(5)*x(7), x(3)*x(4)*x(7),

x(3)*x(4)*x(6), x(1)*x(2)*x(6), x(1)*x(2)*x(5);

resolution r = sres(I,0);1 5 6 2

R <-- R <-- R <-- R

0 1 2 3resolution not minimized yet

The last output refers to the fact that we can simplify the resolutionat the expense of loosing the Grobner basis property for the columnsof the φi. Essentially, the constant entry of φ3 allows us to removethe last column of φ2.r = minres(list(r));

1 5 5 1

R <-- R <-- R <-- R

0 1 2 3print(betti(r),"betti");

0 1 2 2

0: 1 - - -

1: - 1 - -

2: - 4 4 -

3: - - 1 -

1: - - - 1

total: 1 5 5 1


In Exercise 5.5 we have already derived the numerical invariantsof X = V (I) ⊂ P6 from this resolution. Note that the Hilbert poly-nomial, by Theorem 5.4.5, is an invariant of R/I, hence does notdepend on the choice of the graded free resolution.

In general we proceed as follows:

Definition 5.11.4 Let R =K[x0, ..., xn]. A free resolution

0→ Flφl→ Fl−1

φl−1→ ...

φ1→ F0 →M → 0

of a graded R-module M is called minimal if

im(φi) ⊂ ⟨x0, ..., xn⟩ ⋅ Fi−1

for all i.

This means that all entries of the maps φi are contained in theideal ⟨x0, ..., xn⟩, that is, they do not have any constant non-zeroentry.

Remark 5.11.5 One can show that the minimal free resolution isunique up to isomorphism. See Exercise 5.12.

We can make a resolution minimal as follows:

Remark 5.11.6 Consider a graded exact sequence

...Ð→ Rti+1φi+1Ð→ Rti

φiÐ→ Rti−1 Ð→ ...

Note that a change of basis

Rti+1φi+1 - Rti

φi - Rti−1

Rti

T

?�-

with graded T ∈ GL(ti,R), results in a graded exact sequence

...Ð→ Rti+1T○φi+1Ð→ Rti

φi○T−1Ð→ Rti−1 Ð→ ...

Suppose φi+1 has a non-zero constant entry at position (a, b). Wecan make all other entries of column b and row a zero by gradedrow and column operations.


Deleting row a and column b in φi+1, and column a in φi, weobtain an exact sequence

...Ð→ Rti+1−1φ′i+1Ð→ Rti−1

φ′iÐ→ Rti−1 Ð→ ...

For more than one non-zero constant entry, first do row and columnreductions to φi+1 such that φi+1 mod ⟨x0, ..., xn⟩ is of the form

⎛⎜⎜⎜⎝

0 01 0

⋱ 00 1

⎞⎟⎟⎟⎠

Note that R/ ⟨x0, ..., xn⟩ ≅K is a field. Then proceed as above. Notethat, by graded operations, constant entries stay constant.

In Exercise 5.13 we turn these observations into an explicit al-gorithm.

Example 5.11.7 By row operations we get

T ○ φ3 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 −x3x4x7

0 −x1x2x5

0 −x1x2x6

0 x3x4x6

0 −x5x7

1 x3x4

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

with

T =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 −x7

11

1 x6

11

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

T −1 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 x7

11

1 −x6

11

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

Then, necessarily, in

φ2 ○ T−1 =

⎛⎜⎜⎜⎜⎜⎜⎝

−x6 0 0 −x7 0 0x5 0 0 0 −x3x4 00 −x7 0 0 x1x2 00 x6 −x5 0 0 00 0 x3x4 x1x2 0 0

⎞⎟⎟⎟⎟⎟⎟⎠

the last column vanishes. Denoting by φ′2 and φ′3 the submatricesas marked above we obtain the minimal graded free resolution

0→ R(−7)1φ′3→ R(−4)4⊕R(−5)1

φ′2→ R(−3)4⊕R(−2)1 φ1

→ R1 → R/I → 0.


The Betti diagram is obtained from that of the non-minimial res-olution by a diagonal cancellation between a position (i, j) and(i − 1, j + 1):

0 1 2 30 ∶ 11 ∶ 12 ∶ 4 4 13 ∶ 24 ∶ 1total ∶ 1 5 6 2

z→

0 1 2 30 ∶ 11 ∶ 12 ∶ 4 43 ∶ 14 ∶ 1total ∶ 1 5 5 1

Note that any diagonal cancellation does not change the Hilbertpolynomial.

5.12 Exercises

Exercise 5.1 Prove that the radical of a homogeneous ideal is ho-mogeneous.Hint: Use Lemma 5.3.8 considering the homogeneous summand ofmaximal degree.

Exercise 5.2 Let R =K[x1, ..., xn] and I ⊂ R a homogeneous ideal.Consider X = V (I) ⊂ Pn(K) and let

C(X) = V (I) ⊂ An+1(K)

be the affine cone of X. Show that if X ≠ ∅, then

I(C(X)) = I(X).

Hints: Recall that

I(C(X)) = {f ∈ R ∣ f(p) = 0 ∀p ∈ C(X)}

andI(X) = ⟨f ∈ R homogeneous ∣ f(p) = 0 ∀p ∈X⟩ .

Use Lemma 5.3.8 considering the summand of degree zero.

Exercise 5.3 Let K = K, I ⊂ K[x1, ..., xn] an ideal and X =V (I) ⊂ An(K). Show that the projective closure of X is the van-ishing locus of the homogenization of I, that is,

pc(X) = V (Ih).


Hint: Recall thatpc(X) = V (I(X)h).

Prove thatf ∈

√I Ô⇒ fh ∈

√Ih

and use the strong Nullstellensatz.

Exercise 5.4 Consider the affine twisted cubic curve C = V (I)with

I = ⟨x21 − x2, x

31 − x3⟩ ⊂K[x1, x2, x3].

1) Compute the reduced Grobner basis of Ih ⊂ R =K[x0, x1, x2, x3]with respect to dp.

2) Compute a free resolution of R/Ih as an R-module.

3) Determine the dimension, degree and arithmetic genus of theprojective twisted cubic pc(C) = V (Ih) ⊂ P3(K).

Exercise 5.5 Compute for the projective variety X ⊂ P6, I = I(X) ⊂R = C[x0, ..., x6] with graded free resolution

0← R/I ← R1 ← R(−2)⊕R(−3)4 ← R(−5)1⊕R(−4)4 ← R(−7)1 ← 0.

1) the Hilbert polynomial PR/I ,

2) dimension dim(X), degree deg(X), and arithmetic genus pa(X),and

3) the Betti table of the resolution.

Exercise 5.6 Let R = K[x0, ..., xn], let p1, ..., pd ∈ Pn(K) be dis-tinct points and Ij = I(pj) ⊂ R.

1) Determine the Hilbert polynomial PR/Ij of one point.

2) Show that the sequence

0 → Ii ∩ Ij → Ii ⊕ Ij → Ii + Ij → 0(f, g) ↦ f − g

is exact.

3) Considering I = I(p1)∩ ...∩I(pd), prove by induction that theHilbert polynomial of d distinct points satisfies

PR/I = d.

Hint: Ii + Ij =?


Exercise 5.7 Let hi ∈ Z be a sequence of integers with i ∈ N, andlet

h′i = hi − hi−1

be its first difference. Assume that h′i agrees for all i ≥ i0 with apolynomial of degree ≤ n − 1 with rational coefficients.

Show that hi agrees with a polynomial of degree ≤ n with rationalcoefficients for all i ≥ i0.

Exercise 5.8 Let M be a finitely generated graded R =K[x0, ..., xn]-module.

1) Consider the multiplication by xn

f ∶ M(−1) → Mm ↦ xn ⋅m

Show that coker(f) and ker(f) are finitely generated gradedK[x0, ..., xn−1]-modules.

2) Using induction on n and Exercise 5.7, deduce that the Hilbertfunction HM(i) agrees for large i with a polynomial of degree≤ n with rational coefficients (reproving Theorem 5.4.5 by dif-ferent means).

Hints: What do you get if you multiply coker(f) and ker(f) by xn?Derive from f an exact sequence of graded K-vector spaces.

Exercise 5.9 Let R =K[x, y] and

f =⎛⎜⎝

xy + y2

xy2 − 1x2y + x2 + xy2 + xy

⎞⎟⎠, g1 =

⎛⎜⎝

y0

xy + x

⎞⎟⎠, g2 =

⎛⎜⎝

x + 1y2

0

⎞⎟⎠∈ R3

Divide f by (g1, g2) using the monomial ordering extending lp toR3 by giving

1) priority to the monomials in R,

2) priority to the components.

Exercise 5.10 Let R = K[x0, x1, x2, x3] and let > be the orderingdp. Consider the ideal

I = ⟨x0, x1, x2, x3⟩ ⊂ R,

with Grobner basis G = (x0, x1, x2, x3).


1) By verifying Buchberger’s criterion for G, determine for thesyzygy module Syz(G) a Grobner basis with respect to >G.

2) Iterate this to compute a graded free resolution of R/I.

3) Determine PR/I .

Exercise 5.11 In R =K[x1, ..., x5], consider the ideal

I = ⟨x1x3, x2x4, x3x5, x4x1, x5x2⟩ ,

and let > be the ordering dp.

1) Using the Singular commands

resolution A = res(I,0);

resolution B = sres(I,0);

compute graded free resolutions of R/I.

2) From the matrices A[i] and B[i] determine the respectiveBetti tables.

3) From each of the Betti tables derive the Hilbert polynomialPR/I . What do you observe?

4) For the projective variety X = V (I) ⊂ P4, compute dim(X),deg(X), and pa(X).

Exercise 5.12 Let R =K[x1, ..., xn], and

...f2→ F1

f1→ F0

f0→M → 0

...g2→ G1

g1→ G0

g0→ N → 0

be exact complexes of R-modules with all Fi and Gi free.

1) Show that every homomorphism α ∶ M → N extends to acommutative diagram

...f2 - F1

f1 - F0

f0 - M - 0

...g2 - G1

α1

? g1 - G0

α0

? g0 - N

α

?- 0

with homomorphisms αi ∶ Fi → Gi.


2) The homomorphisms αi are unique up to homotopy, that is,if α′i ∶ Fi → Gi also extend α then there are hi ∶ Fi → Gi+1 with

αi − α′i = hi−1 ○ fi + gi+1 ○ hi

for all i.

3) Considering two minimal free resolutions

...f2→ F1

f1→ F0

f0→M → 0

...g2→ G1

g1→ G0

g0→M → 0

of a graded module M and f = idM , show that all αi areisomorphisms.

Exercise 5.13 Describe an algorithm to transform a graded freeresolution into a minimal one.

Exercise 5.14 Let R =K[x0, ..., x6] and

I = ⟨x0x1x4, x0x3x4, x1x2x5, x1x4x5, x0x3x6, x2x3x6, x2x5x6⟩

1) Using Algorithm 5.11.1, compute a graded free resolution ofR/I.

2) Determine from this a minimal free resolution.

3) For both resolutions, write down the Betti tables and mark thediagonal cancellations.

4) Compute dimension, degree and genus of V (I) ⊂ P6.

6

Computing with modules

6.1 Overview

Based on Theorem 5.10.7 to compute the syzygy module of a Grobnerbasis, we can derive algorithms to compute with finitely presentedmodules over the polynomial ring R = K[x1, ..., xn]. We start outwith an algorithm computing generators of the kernel of an arbi-trary matrix over R. We can apply this, for example, to derive anideal intersection algorithm, see Exercise 6.1. The second basic taskis to find a presentation of the subquotient of two submodules ofRt, which are given by generators. Based on these basic tasks, wewill derive an algorithm for computing the module HomR(M,N) ofhomomorphisms M → N between two R-modules M and N . Thisis a fundamental algorithm for many tasks in algebraic geometryand commutative algebra. One application, we will have a quicklook at, are deformations of algebraic varieties. Here, given a vari-ety X = V (I) by an ideal I ⊂ R, one can obtain from HomR(I,R/I)information about deforming X in a way which keeps key invariantslike the Hilbert polynomial constant. Figure 6.1 shows the defor-mation of the union of three lines V (x0x1x2) ⊂ P2 into an ellipticcurve, that is, a smooth plane cubic.

6.2 Computing kernels

Theorem 5.10.7 only provides an algorithm to compute the kernel ofa matrix (g1 ∣ ... ∣ ga) ∈ Rt×a whose columns gi ∈ Rt form a Grobnerbasis with respect to some ordering. However, we can easily derivean algorithm for the more general task of computing the kernel(syzygy module) of an arbitrary matrix (that is, homomorphism offree R-modules of finite rank).

165

6. COMPUTING WITH MODULES 166

Figure 6.1: Deformation of three lines into an elliptic curve (realpicture).

Algorithm 6.2.1 Kernel

Input: Matrix M = (g1 ∣ ... ∣ ga) ∈ Rt×a.Output: A matrix S with im(S) = ker(M).1: Compute a Grobner basis g1, ..., ga, ga+1, ..., ga+b.2: From the syzygies s(gi, gj) build a matrix

T =

⎛⎜⎜⎜⎝

∗ ∗

1 ∗⋱ ∗

0 1

⎞⎟⎟⎟⎠

∈ R(a+b)×s

that is, put the Buchberger tests leading to a new Grobner basiselement first.

3: Apply column operations to T to obtain

T ′ =

⎛⎜⎜⎜⎝

∗ S1 ∗

⋱ 00 1

⎞⎟⎟⎟⎠

∈ R(a+b)×s

4: Return S.

Proof. By Theorem 5.10.7, ker(g1, ..., ga+b) = im(T ). Since thefirst a columns of T ′ are linearly independent, any syzygy whichinvolves only g1, ..., ga is a linear combination of the last b columns.


Example 6.2.1 In Example 5.11.3, by column operations

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

x7 00 −x1x2x5

0 −x1x2x6

−x6 00 −x5x7

1 x3x4

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

↦

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

x7 −x3x4x7

0 −x1x2x5

0 −x1x2x6

−x6 x3x4x6

0 −x5x7

1 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

hence

Syz

⎛⎜⎜⎜⎜⎜⎜⎝

−x6 0 0 −x7 0x5 0 0 0 −x3x4

0 −x7 0 0 x1x2

0 x6 −x5 0 00 0 x3x4 x1x2 0

⎞⎟⎟⎟⎟⎟⎟⎠

= ⟨

⎛⎜⎜⎜⎜⎜⎜⎝

−x3x4x7

−x1x2x5

−x1x2x6

x3x4x6

−x5x7

⎞⎟⎟⎟⎟⎟⎟⎠

⟩

In Singular we can compute a syzygy matrix using the com-mand syz:ring R = 0, (x(1..7)),dp;

matrix M[5][5] = -x(6), 0, 0, -x(7), 0,

x(5), 0, 0, 0, -x(3)*x(4), 0,

0, -x(7), 0, 0, x(1)*x(2),

0, x(6), -x(5), 0, 0,

0, 0, x(3)*x(4), x(1)*x(2), 0;

print(syz(M));

x(3)*x(4)*x(7),

x(1)*x(2)*x(5),

x(1)*x(2)*x(6),

-x(3)*x(4)*x(6),

x(5)*x(7)

For another example, see Exercise 6.2.

6.3 Computing subquotients

Definition 6.3.1 Given two matrices A ∈ Rt×a and B ∈ Rt×b thesubquotient with generators A and relations B is the R-module

subquot(A,B) =im(A) + im(B)

im(B).


Here we add im(B), before dividing it out again, to make thequotient well-defined.

Remark 6.3.2 In particular,

im(A) = subquot(A,0)

coker(B) = subquot(Et,B)

where Et denotes the t × t unit matrix.

The following algorithm determines a presentation

Rc C→ Ra → subquot(A,B)→ 0

of the subquotient.

Algorithm 6.3.1 Subquotient

Input: A ∈ Rt×a and B ∈ Rt×b.Output: C ∈ Ra×c with

subquot(A,B) ≅ coker(C)

1: By computing a syzygy matrix of ( A B ) ∈ Rt×(a+b) deter-mine C ∈ Ra×c and D ∈ Rb×c with

ker ( A B ) = im(CD

)

2: Return C.

Proof. We have a presentation

Rc

⎛⎜⎜⎝

CD

⎞⎟⎟⎠

Ð→ Ra+b( A B )

Ð→ im(A) + im(B)→ 0

Dividing by im(B), yields an exact sequence

Rc CÐ→ Ra A

Ð→im(A) + im(B)

im(B)→ 0

where A denotes the map

A ∶ Ra → Rt/ im(B)ei ↦ Aei + im(B)

induced by A.See also Exercise 6.3.


Example 6.3.3 For

A = (x, y) B = (x2, y2)

we have

ker(x, y, x2, y2) = im

⎛⎜⎜⎜⎝

−y x 0x 0 y0 −1 00 0 −1

⎞⎟⎟⎟⎠

hencesubquot(A,B) ≅ coker(C)

with

C = (−y x 0x 0 y

)

The presentation sequence of subquot(A,B) is

R3 CÐ→ R2 Ð→ ⟨x, y⟩ / ⟨x2, y2⟩ Ð→ 0

e1 z→ xe2 z→ y

In Singular we can obtain C by applying the modulo commandto the images of A and B:ring R=0,(x,y),dp;

module A = x,y;

module B = x2,y2;

print(modulo(A,B));

0,x,-y,

y,0,x

Note that the sorting of the columns of C is irrelevant.

Remark 6.3.4 Suppose M is a finitely presented module

Rc C→ Ra π

→M → 0

and π(ei) = A ⋅ ei. Then

M = subquot(A,B) ≅ coker(C).

Writing M as the subquotient of A and B, keeps track of the gen-erators of M , that is, of the map π. Writing M as the cokernel ofC looses the information about π.


6.4 Hom and the Hilbert scheme

Definition 6.4.1 For R-modules M and N let

HomR(M,N) = {f ∶M → N ∣ f is an R-module homomorphism}

be the module of homomorphisms. It has the structure of anR-module by

(f + g)(m) = f(m) + g(m)

(r ⋅ f)(m) = r ⋅ f(m)

for m ∈M and r ∈ R.

Before discussing how to compute Hom by Grobner basis tech-niques (specifically, Algorithms 6.2.1 and 6.3.1) we give a sketch ofan important application in algebraic geometry:

It is based on the key fact that algebraic subsets of PnK are againparametrized by an algebraic set. More precisely, we have to con-sider schemes Proj(R/I), which corresponds to homogeneous idealsI, which are not necessarily radical. In general, the remainder ofthis subsection is intended to give you some impression. To be com-pletely precise, we would need much more technical background.

Example 6.4.2 Consider the set Hn,d of hypersurfaces in PnK ofdegree d. Every element X ∈ Hn,d is of the form

X = V (I)

where I = ⟨f⟩ and

f = ∑∣α∣=d

cαxα ∈K[x0, ..., xn]

is homogeneous of degree deg(f) = d. So X corresponds to anelement

(cα)α ∈ Pm−1K

where m is the dimension of the vector space K[x0, ..., xn]d of poly-nomials of degree d. Hence

Hn,d ≅ Pm−1K .

All elements X ∈ Hn,d have the same Hilbert polynomial

PX(t) ∶= PR/I(t) = (t + n

n) − (

t − d + n

n).

It turns out that this idea is the right strategy in general:


Definition 6.4.3 The Hilbert scheme Hn is the set all subschemesof PnK.

Theorem 6.4.4 The Hilbert scheme

Hn = ⋃P ∈K[t]Hn,P

is the disjoint union of projective schemes Hn,P such that PX = Pfor all X ∈ Hn,P . For fixed degree of X, there are only finitely manyP with Hn,P ≠ ∅.

We cannot prove this here. In general, the Hilbert scheme isalso difficult to construct. However, the following theorem (whichwe also cannot prove here) allows us to describe Hn,P locally at apoint X by computing a module of homomorphisms. As we willsee in the next section, this is a task which can easily be handledalgorithmically.

Theorem 6.4.5 Let X ∈ Hn,P given by the homogeneous ideal I ⊂K[x0, ..., xn]. The tangent space of Hn,P at X is

Hom(I,R/I)0

the K-vector space of graded homomorphisms I → R/I.The elements of Hom(I,R/I)0 are called the homogeneous first

order deformations of I.

Recall that the tangent space gives a linear approximation ofa variety at a given point. See Figure 6.2 for a tangent space ofa conic. Keep in mind, that in Theorem 6.4.5 we are not talking

Figure 6.2: A tangent to a conic

about a tangent space of X, but rather about the tangent space ofthe space of all possible X at a specific X.


Remark 6.4.6 Suppose X is defined by I = im(g1, ..., gs) and

Hom(I,R/I)0 = ⟨φ1, ..., φm⟩ .

Then for parameters t, the ideal

⟨gi +∑mj=1tj ⋅ φj(gi) ∣ i = 1, ..., s⟩

determines a deformation Xt of X. However PXt = PX is only truemodulo appropriate higher order terms in ⟨t0, ..., tm⟩

2(in analogy to

a Taylor series). Moreover, there can be algebraic relations betweenthe tj.

Example 6.4.7 In the setup of Example 6.4.2, consider the hyper-surface I = ⟨x0 ⋅ ... ⋅ xn⟩. Then a basis of Hom(I,R/I)0 is given bythe homomorphisms

φα ∶ I → R/Ix0 ⋅ ... ⋅ xn ↦ xα

where ∣α∣ = d and α ≠ (1, ...,1).

To do some examples, we first develop an algorithm to computeHom in general.

6.5 Computing Hom

First note thatHomR (Rs,Rt) ≅ Rt⋅s

is a free module. We choose the identification of matrices andvectors

⎛⎜⎝

a1,1 ⋯ a1,s

⋮ ⋮at,1 ⋯ at,s

⎞⎟⎠↦

⎛⎜⎜⎜⎜⎜⎜⎝

a1,1

⋮at,1⋮at,s

⎞⎟⎟⎟⎟⎟⎟⎠

For those who know about the tensor product: Denote by astar the dual vector respectively module. We have HomR (Rs,Rt) ≅Rt ⊗ (Rs)∗, where A = (ai,j) is identified with ∑

ti=1∑

sj=1ai,j ⋅ ei ⊗ e

∗j .

Hence, the above identification amounts to an ordering of the basisvectors ei ⊗ e∗j of the tensor product.


Algorithm 6.5.1 Hom

Input: R-modules M and N given by presentations

Rs1f1→ Rs0 →M → 0

Rt1g1→ Rt0 → N → 0

Output: HomR(M,N) as a subquotient.1: By Algorithm 6.2.1, compute g2 such that

Rt2g2→ Rt1

g1→ Rt0 → N → 0

is exact.2: Construct the (s1t0) × (s0t0 + s1t1) matrix

Hom (Rs0 ,Rt0)⊕Hom (Rs1 ,Rt1)δ→ Hom (Rs1 ,Rt0)

(A, B) ↦ A ⋅ f1 − g1 ⋅B

3: By Algorithm 6.2.1 compute a matrix γ with ker(δ) = im(γ).4: Construct the (s0t0 + s1t1) × (s0t1 + s1t2) matrix

Hom (Rs0 ,Rt1)⊕Hom (Rs1 ,Rt2)ρ→ Hom (Rs0 ,Rt0)⊕Hom (Rs1 ,Rt1)

(C, D) ↦ (g1 ⋅C, C ⋅ f1 − g2 ⋅D)

5: Return subquot(γ, ρ).

For the proof of correctness we make the following observations:

Remark 6.5.1 First note that δ ⋅ ρ = 0, so

subquot(γ, ρ) = ker(δ)/ im(ρ).

Lemma 6.5.2 The map

π ∶ ker(δ) → HomR(M,N)

(A, B) ↦ A = (ei ↦ A ⋅ ei)

is well-defined and surjective.

Proof. Let (A,B) ∈ ker(δ). Then we have a commutative diagram

Rs1f1 - Rs0 - M - 0

Rt1

B

? g1 - Rt0

A

?- N

A

?- 0


where A is well defined: Suppose v ∈ im(f1). So v = f1w withw ∈ Rs1 , hence A ⋅ v = g1Bw ∈ im(g1).

Moreover, any f ∶M → N , ej ↦ ∑i aijei is induced by A ∶ Rs0 →Rt0 , ej ↦ ∑i aijei, and

im(A ⋅ f1) ⊂ im(g1).

Hence there are bij with

A ⋅ f1 ⋅ ej =∑i

bijei

and we set B = (bij).

For correctness of the algorithm it remains to prove:

Lemma 6.5.3 ker(π) = im(ρ).

Proof. We can collect all maps in the diagram

Rs1f1 - Rs0 - M - 0

Rt2g2 -

D

�Rt1

B

? g1 -

C

�Rt0

A

?- N

A

?- 0

First note that (A,B) ∈ ker(π) if and only if im(A) ⊂ im(g1) andA ⋅ f1 = g1 ⋅B.

• If (A,B) = (g1 ⋅C, C ⋅ f1 − g2 ⋅D) ∈ im(ρ), then

A ⋅ f1 = g1 ⋅C ⋅ f1 = g1 ⋅ (C ⋅ f1 − g2 ⋅D) = g1 ⋅B

and im(A) = im(g1 ⋅C) ⊂ im(g1).

• On the other hand, let (A,B) ∈ ker(π). Since im(A) ⊂ im(g1)there is C ∈ HomR (Rs0 ,Rt1) with A = g1 ⋅ C. Moreover, byA ⋅ f1 = g1 ⋅B, we have

g1 ⋅ (C ⋅ f1 −B) = A ⋅ f1 − g1 ⋅B = 0,

that is,im(C ⋅ f1 −B) ⊂ ker(g1) = im(g2).

Hence there is D ∈ HomR (Rs1 ,Rt2) with

g2 ⋅D = C ⋅ f1 −B.


Example 6.5.4 For X = V (I) ⊂ P5 defined by

I = im(g1) ⊂ R1

g1 = (x0x1, x2x3, x4x5)

we have a minimal graded free resolution

0→ R(−6)1 → R(−4)3 g2→ R(−2)3 g1

→ R1 → R/I → 0

with

g2 =⎛⎜⎝

−x2x3 −x4x5 0x0x1 0 −x4x5

0 x0x1 −x2x3

⎞⎟⎠

HencePX(t) = 4t2 + 2,

so X is a genus 1 surface of degree 8. It is an example of a so calledK3-surface, which is the two-dimensional analogue of an ellipticcurve.

Our goal is to compute the tangent space Hom(I,R/I)0 of H5,4t2+2

at X.We use the opportunity to demonstrate one more computer alge-

bra system: Macaulay2 has a similar functionality as Singular.It is very well suited for the type of calculations required in Algo-rithm 6.5. Macaulay2 (like Singular and GAP) has a built-infunction for computing Hom:i1: R = QQ[x 0..x 5];

i2: I = ideal(x 0*x 1,x 2*x 3,x 4*x 5)

o2: ideal ( x 0x 1 x 2x 3 x 4x 5 )

o2: Ideal of R

A matrix with the generators of I:i3: g1 = gens I

o3: ∣ x 0x 1 x 2x 3 x 4x 5 ∣o3: Matrix R1 <--- R3

To consider I as a submodule of R1 do:i4: M = image g1

o4: image ∣ x 0x 1 x 2x 3 x 4x 5 ∣o4: R-module, submodule of R1

To consider R/I as a submodule of R1 do:i5: N= coker g1

o5: cokernel ∣ x 0x 1 x 2x 3 x 4x 5 ∣o5: R-module, quotient of R1


Construct HomR(I,R/I):i6: H = Hom(M, N)o6: cokernel ∣ x 0x 1 x 2x 3 x 4x 5 0 0 0 0 0 0 ∣

∣ 0 0 0 x 0x 1 x 2x 3 x 4x 5 0 0 0 ∣∣ 0 0 0 0 0 0 x 0x 1 x 2x 3 x 4x 5 ∣

A matrix with a K-vectorspace basis in degree 0 in its columns:i7: T = basis(0,H);

The K-vector space dimension of HomR(I,R/I)0:i8: rank source T

o8: 54

For example the 11th basis vector (indexing starts at 0):i9: T {10}o9: ∣ x 2^2 ∣

∣ 0 ∣∣ 0 ∣

o9: Matrix

We can interpret a given basis vector as a homomorphism, for ex-ample:i10: homomorphism T {10}o10: ∣ x 2^2 0 0 ∣o10: Matrix

This homomorphism maps the generators of I to R/I as follows:

x0x1 ↦ x22, x2x3 ↦ 0, x4x5 ↦ 0

From o6 it is clear that a basis of HomR(I,R/I)0 is given by the54 = 18 + 18 + 18 homomorphisms

x0x1 ↦ xα, x2x3 ↦ 0, x4x5 ↦ 0,

x0x1 ↦ 0, x2x3 ↦ xα, x4x5 ↦ 0

x0x1 ↦ 0, x2x3 ↦ 0, x4x5 ↦ xα

where deg(xα) = 2 and xα ≠ x0x1, x2x3, x4x5.

Example 6.5.5 For X = V (I) ⊂ P2 defined by

I = im(g1)

g1 = (x0x1, x0x2, x1x2)

we have a minimal graded free resolution

0→ R(−3)2 g2→ R(−2)3 g1

→ R1 → R/I → 0

with

g2 =⎛⎜⎝

−x2 0x1 −x1

0 x0

⎞⎟⎠


HencePX(t) = 3,

so X is the union of 3 points, indeed,

I = ⟨x0, x1⟩ ∩ ⟨x0, x2⟩ ∩ ⟨x1, x2⟩ .

We follow Algorithm 6.5, step-by-step, to obtain HomR(I,R/I).The usual diagram reads

R(−3)2g2- R(−2)3

g1 - I - 0

R(−3)2g2-

D

�R(−2)3

B

? g1 -

C

�R1

A

?- R/I

A

?- 0

Hence

A ∈ HomR (R(−2)3,R1) ≅ R(2)3

B ∈ HomR (R(−3)2,R(−2)3) ≅ R(1)6

δ(A,B) = A ⋅ g2 − g1 ⋅B ∈ HomR (R(−3)2,R1) ≅ R(3)2

andδ ∶ R(2)3 ⊕R(1)6 → R(3)2

is represented by

δ = (

a11 a12 a13 b11 b21 b31 b12 b22 b32

−x2 x1 0 −x0x1 −x0x2 −x1x2 0 0 00 −x1 x0 0 0 0 −x0x1 −x0x2 −x1x2

)

To obtain the first block, compute

( a11 a12 a13 ) ⋅ g2 = ( −a11x2 + a12x1 −a12x1 + a13x0 )

set one ai,j = 1 and all others zero and use again the above identi-fication of matrices with vectors. For the second block, compute

g1⋅⎛⎜⎝

b11 b12

b21 b22

b31 b32

⎞⎟⎠= ( x0x1b11 + x0x2b21 + x1x2b31 x0x1b12 + x0x2b22 + x1x2b32 )

and proceed in the same way. If you know about the tensor product,the blocks are tensor products of matrices.


Turning to ρ,

C ∈ HomR (R(−2)3,R(−2)3) ≅ R9

D ∈ HomR (R(−3)2,R(−3)2) ≅ R4

andρ(C,D) = (g1 ⋅C,C ⋅ g2 − g2 ⋅D)

is in the direct sum of

HomR (R(−2)3,R1) ≅ R(2)3

andHomR (R(−3)2,R(−2)3) ≅ R(1)6.

Henceρ ∶ R9 ⊕R4 → R(2)3 ⊕R(1)6

and, in the same way as above, one observes that

ρ =

⎛⎜⎜⎜⎜⎜⎜⎝

C D

g1 0 0 0 0A 0 g1 0 0 0

0 0 g1 0 0

B −x2E3 x1E3 0 −g2 00 −x1E3 x0E3 0 −g2

⎞⎟⎟⎟⎟⎟⎟⎠

where E3 denotes the 3 × 3 unit matrix.To compute ker(δ)/ im(ρ) by Algorithms 6.2.1 and 6.3.1, we con-tinue in Macaulay2:i1: R = QQ[x 0..x 2];

i2: I = ideal(x 0*x 1,x 0*x 2,x 1*x 2)

o2: ideal ( x 0x 1 x 0x 2 x 1x 2 )

o2: Ideal of R

Compute a graded minimal free resolution:i3: cc = res I;

i4: betti cc0 1 2

o4: total: 1 3 2

0: 1 . .

1: . 3 2i5: g1 = cc.dd 1

o5: ∣ x 0x 1 x 0x 2 x 1x 2 ∣o5: Matrix R1 <--- R3

i6: g2 = cc.dd 2


o6: {2} ∣ -x 2 0 ∣{2} ∣ x 1 -x 1 ∣{2} ∣ 0 x 0 ∣

o6: Matrix R3<--- R2

We form the matrices δ and ρ:i7: T0 = target g1; T1 = source g1; T2 = source g2;

i8: S0 = source g1; S1 = source g2;

i9: M0 = Hom(S1,T0);

i10: M1 = Hom(S0, T0) ++ Hom(S1, T1);

i11: M2 = Hom(S0, T1) ++ Hom(S1, T2);

i12: degrees M0

o12: {{-3},{-3}}i13: degrees M1

o13: {{-2},{-2},{-2},{-1},{-1},{-1},{-1},{-1},{-1}}i14: degrees M2

o14: {{0},{0},{0},{0},{0},{0},{0},{0},{0},{0},{0},{0},{0}}i15: delta11 = transpose g2;

i16: delta12 = -id (R^2)**g1;

i17: delta = map(M0,M1,matrix {{delta11, delta12}});o17: {-3} ∣ x 2 x 1 0 -x 0x 1 -x 0x 2 -x 1x 2 0 0 0 ∣

{-3} ∣ 0 -x 1 x 0 0 0 0 -x 0x 1 -x 0x 2 -x 1x 2 ∣

i18: rho11 = id (R^3)**g1;

i19: rho21 = (transpose g2)**id (R^3);

i20: rho12 = map(R^3,R^4,0);

i21: rho22 = -(id (R^2))**g2;

i22: rho = map(M1,M2,matrix{{rho11,rho12},{rho21,rho22}});o22: {-2} ∣ x 0x 1 x 0x 2 x 1x 2 0 0 0 0 0 0 0 0 0 0 ∣

{-2} ∣ 0 0 0 x 0x 1 x 0x 2 x 1x 2 0 0 0 0 0 0 0 ∣

{-2} ∣ 0 0 0 0 0 0 x 0x 1 x 0x 2 x 1x 2 0 0 0 0 ∣

{-1} ∣ -x 2 0 0 x 1 0 0 0 0 0 x 2 0 0 0 ∣

{-1} ∣ 0 -x 2 0 0 x 1 0 0 0 0 -x 1 x 1 0 0 ∣

{-1} ∣ 0 0 -x 2 0 0 x 1 0 0 0 0 -x 0 0 0 ∣

{-1} ∣ 0 0 0 -x 1 0 0 x 0 0 0 0 0 x 2 0 ∣

{-1} ∣ 0 0 0 0 -x 1 0 0 x 0 0 0 0 -x 1 x 1 ∣

{-1} ∣ 0 0 0 0 0 -x 1 0 0 x 0 0 0 0 -x 0 ∣

Now compute the Hom-module as a subquotient:i23: H =(ker delta)/(image rho);

Basis in degree 0:i24: B = basis(0,H)o24: {-1} ∣ x 0 0 0 0 0 0 ∣

{-1} ∣ 0 x 1 0 0 0 0 ∣{-1} ∣ 0 0 x 0 0 0 0 ∣{-1} ∣ 0 0 0 x 1 0 0 ∣{-1} ∣ 0 0 0 0 x 2 0 ∣{-1} ∣ 0 0 0 0 0 x 2 ∣{0} ∣ 0 0 0 0 0 0 ∣


To obtain the homomorphisms, we consider for each column, thecorresponding linear combination of the generators of H:i25: (gens H) * (matrix entries B)o25: {-2} ∣ x 0^2 x 1^2 0 0 0 0 ∣

{-2} ∣ 0 0 x 0^2 0 0 x 2^2 ∣{-2} ∣ 0 0 0 x 1^2 x 2^2 0 ∣{-1} ∣ 0 0 x 0 0 0 0 ∣{-1} ∣ -x 0 0 0 0 x 1 0 ∣{-1} ∣ 0 -x 1 0 0 0 x 1 ∣{-1} ∣ 0 0 -x 0 -x 1 0 0 ∣{-1} ∣ 0 0 0 0 -x 1 0 ∣{-1} ∣ 0 0 0 0 0 -x 1 ∣

Every column corresponds to a tuple (A,B), the first three entriesto A and the last six to B. Hence the possible A are

(x20,0,0), (x2

1,0,0),

(0, x20,0), (0, x2

2,0),

(0,0, x21), (0,0, x2

2).

6.6 Exercises

Exercise 6.1 Let R = K[x1, ..., xn]. Given ideals I = ⟨f1, ..., fa⟩ ⊂R and J = ⟨g1, ..., gb⟩ ⊂ R, consider the matrix

G = (f1 ... fa 0 ... 0 10 ... 0 g1 ... gb 1

) ∈ R2×(a+b+1),

and let S = syz(G) ∈ R(a+b+1)×g be a syzygy matrix of G.

1) Show that the entries of the last row of S generate I ∩ J .

2) Applying this algorithm, compute the intersection

⟨x1, x2⟩ ∩ ⟨x1 − x3, x2 − x3⟩ ⊂K[x1, x2, x3].

Hint: You can use the Singular command syz.

Exercise 6.2 Let R =K[x0, ..., x6], and

A =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 x1 0 0 0 0 −x6

−x1 0 x3 0 0 0 00 −x3 0 x5 0 0 00 0 −x5 0 x0 0 00 0 0 −x0 0 x2 00 0 0 0 −x2 0 x4

x6 0 0 0 0 −x4 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

∈ R7×7


1) Compute a Grobner basis of A with respect to (dp, c).

2) From the Buchberger test, determine generators of ker(A).

Exercise 6.3 Let R =K[x0, x1, x2]. For

A = (x1 x0 00 x2 x1

)

B = (x1x2 x0x1 0 0

0 0 x1x2 x0x1)

compute a presentation of

subquot(A,B) =im(A) + im(B)

im(B).


I = ⟨x0x1, x1x2, x2x3, x3x4, x4x0⟩ ⊂ R

With the help of a computer algebra system, determine:

1) HomR(I,R/I) as a subquotient,

2) a basis of HomR(I,R/I)0, and

3) interpret your basis vectors as homomorphisms I → R/I.


I = ⟨x0x1, x2x3⟩ ⊂ R

In a step-by-step calculation,

1) determine δ and ρ as needed in Algorithm 6.5 to presentHomR(I,R/I),

2) compute γ with im(γ) = ker(δ),

3) compute a presentation of HomR(I,R/I) = subquot(γ, ρ),

4) determine a basis of HomR(I,R/I)0, and

5) interpret your basis vectors as homomorphisms I → R/I.

7

Primary decomposition

7.1 Overview

As we have already seen in Section 3.1, the union of algebraic setsX1 = V (I1) and X2 = V (I2) corresponds to the intersection of thedefining ideals

X1 ∪X2 = V (I1 ∩ I2)

To decompose the problem of analyzing X1 ∪ X2 into the prob-lem of analyzing the simpler pieces Xi, we encounter the followingtask: Write a given ideal I as an intersection of ideals in analogyto prime factorization of integers. The main question is, what theappropriate building blocks should be. If I is radical, then the de-composition of X into a union of (irreducible) varieties correpondsto the decomposition of I into an intersection of prime ideals. Forexample,

I = ⟨x0x1, x0x2, x1x2⟩ = ⟨x0, x1⟩ ∩ ⟨x0, x2⟩ ∩ ⟨x1, x2⟩

is the intersection of three prime ideals, and V (I) ⊂ P2 is the unionof the corresponding 3 points. However we have seen (for examplein Section 6.4) that in many settings one also has to consider idealswhich are not radical. In this case, it turns out that right thebuilding blocks are primary ideals: An ideal I ≠ R of a ring R iscalled primary if every zerodivisor in R/I is nilpotent.

Let us first focus on decomposing monomial ideals:

7.2 Irreducible decomposition of mono-

mial ideals

Let R = K[x0, ..., xn]. First of all, we should avoid unnecessaryfactors:

182

7. PRIMARY DECOMPOSITION 183

Definition 7.2.1 A decomposition

I =⋂s

i=1Qi

of an ideal I ⊂ R into an intersection of ideals is called irredun-dant if omitting some Qj in the intersection does not result in Iany more.

Lemma 7.2.2 Let m1, ...,mr be the unique minimal generators ofthe monomial ideal I according to Corollary 2.3.5. If

m1 = u1 ⋅ u2

with deg(ui) > 0 and gcd(u1, u2) = 1, then

I = I1 ∩ I2

withIj = ⟨uj,m2, ...,mr⟩ .

Proof. The inclusion I ⊂ Ij is obvious. On the other hand, letm ∈ I1 ∩ I2. If mi ∣ m for some i ≥ 2, then m ∈ I. Otherwise u1 ∣ mand u2 ∣ m. Then, by gcd(u1, u2) = 1 we have m1 ∣ m, hence m ∈ I.

Theorem 7.2.3 Any monomial ideal I ⊂ R has a unique irredun-dant decomposition

I =⋂s

i=1Qi

into idealsQi = ⟨xα1

i1, ..., xαr

ir⟩

generated by powers of variables.

Proof. For the existence of a decomposition apply Lemma 7.2.2inductively.

Suppose I = ⋂si=1Qi and

⋂i≠jQi ⊂ Qj

then Qj can be omitted. Proceeding inductively, we obtain anirredundant decomposition.

For the uniqueness, suppose

⋂s

i=1Qi =⋂s′

j=1Q′j


are irredundant decompositions into ideals generated by powers ofvariables. Fix i and let

Qi = ⟨xα1i1, ..., xαr

ir⟩

with αl > 0 for all l. Assume that Q′j /⊂ Qi for all j. Then for every

j there is an xβjvj ∈ Q

′j with x

βjvj ∉ Qi, that is,

vj = il Ô⇒ αl > βj.

Clearly

m = lcm(xβ1v1 , ..., xβs′vs′ ) ∈⋂

s′

j=1Q′j ⊂ Qi

So there is some l with xαlil

∣m, which contradicts il = vj Ô⇒ αl > βj.Hence for every i there is j with Q′

j ⊂ Qi, which proves theclaim.

Definition 7.2.4 An ideal I is called irreducible if it cannot bewritten as

I = I1 ∩ I2

with ideals Ii ⫌ I.

Corollar 7.2.5 A monomial ideal Q ⊂ R is irreducible if and onlyif it is of the form

Q = ⟨xα1i1, ..., xαr

ir⟩ .

Proof. If Q is not of this form, then by Lemma 7.2.2 it is notirreducible.

On the other hand, suppose is of the given form and

Q = I1 ∩ I2

and letIj =⋂iQj,i

be decompositions as in Theorem 7.2.3 with Qj,i generated by pow-ers of variables. From this we obtain an irredundant decomposition

Q =⋂some iQ1,i ∩⋂some iQ2,i

which by Theorem 7.2.3 is unique. Since also Q is generated bypowers of variables, for some i, j

Q = Qj,i ⊃ Ij

hence Q = Ij.By Theorem 7.2.3 and Corollary 7.2.5 we have:


Corollar 7.2.6 Any monomial ideal I ⊂ R has a unique irredun-dant decomposition into irreducible ideals.

Example 7.2.7 For

I = ⟨x0x1x2, x20x2, x

31x2⟩

we follow the constructive proof of Theorem 7.2.3:

I = ⟨x0, x31x2⟩ ∩ ⟨x1x2, x

20x2⟩

= ⟨x0, x31x2⟩ ∩ ⟨x1, x

20x2⟩ ∩ ⟨x2⟩

= ⟨x0, x31⟩ ∩ ⟨x0, x2⟩ ∩ ⟨x1, x

20⟩ ∩ ⟨x1, x2⟩ ∩ ⟨x2⟩

= ⟨x0, x31⟩ ∩ ⟨x1, x

20⟩ ∩ ⟨x2⟩

In Singular we can compute this decomposition by:LIB "monomialideal.lib";

ring R = 0,(x(0..2)),dp;

ideal I = x(0)*x(1)*x(2), x(0)^2*x(2), x(1)^3*x(2);

irreddecMon(I);

[1]:

[1]=x(0)^2

[2]=x(1)

[2]:

[1]=x(0)

[2]=x(1)^3

[3]:

[1]=x(2)

We summarize our strategy for computing this decompositionin Algorithm 7.2.1.

Algorithm 7.2.1 Irreducible decomposition of monomial ideals

Input: Monomial ideal I.Output: The irredundant irreducible decomposition of I.1: Applying Lemma 7.2.2 inductively to minimal generators com-

pute an irreducible decomposition

I =⋂s

i=1Qi

2: Inductively delete any Qj with

⋂i≠jQi ⊂ Qj

What about non-monomial ideals?


Theorem 7.2.8 Any ideal I ⊂ R has a decomposition

I =⋂s

i=1Qi

into irreducible ideals Qi.

Proof. Let X be the set of ideals for which the claim is false. ByTheorem 2.1.5, X has a maximal element I. Since I cannot beirreducible, we can write I = I1 ∩ I2 with I ⫋ Ij. Since I was chosenmaximal, both I1 and I2 have a decomposition into irreducibles,hence also I has. This contradicts the assumption.

Example 7.2.9 If we leave the class of monomial ideals, an ir-redundant irreducible decomposition is by no means unique. Forexample

I = ⟨x2, xy, y2⟩ = ⟨x2, y⟩ ∩ ⟨x, y2⟩ = ⟨x2, x + y⟩ ∩ ⟨x, y2⟩ .

This problem can be fixed (partly) by combining irreduciblesummands in a suitable way. The basic philosophy is that the idealin Example 7.2.9 should be not decomposed at all.

7.3 Existence of primary decomposition

Definition 7.3.1 An ideal I ≠ R of a ring R is called primary ifevery zerodivisor in R/I is nilpotent.

Remark 7.3.2 I is primary if and only if for f, g ∈ R it holds

fg ∈ I ⇒ f ∈ I or g ∈√I.

Definition and Theorem 7.3.3 If I is primary then P ∶=√I is

prime. It is called the associated prime of I, and I is calledP -primary.

Proof. If f ⋅ g ∈√I, there is an m with (f ⋅ g)m ∈ I. Since I is

primary, fm ∈ I or gm⋅m′

∈ I, hence f ∈√I or g ∈

√I.

Lemma 7.3.4 If I1, I2 ⊂ R are ideals then√I1 ∩ I2 =

√I1 ∩

√I2.

Proof. Let f ∈√I1 ∩ I2, so fm ∈ I1 ∩ I2 for some m. This implies

f ∈√I1 and f ∈

√I2.

On the other hand, if f ∈√I1 ∩

√I2 then fm ∈ I1 and fm

′

∈ I2

for some m and m′. Hence fmax{m,m′} ∈ I1 ∩ I2.


Lemma 7.3.5 Any intersection of P -primary ideals is P -primary.

Proof. Suppose Q1,Q2 ⊂ R are primary and√Qi = P . Let fg ∈

Q1 ∩Q2. Hence

(f ∈ Q1 or g ∈ P ) and (f ∈ Q2 or g ∈ P ),

that is, f ∈ Q1 ∩Q2 or g ∈ P . By Lemma 7.3.4

√Q1 ∩Q2 = P

and hence the claim follows.

Lemma 7.3.6 Let R be a ring, I ⊂ R an ideal. If

I ∶ ⟨f⟩ = I ∶ ⟨f 2⟩

for some f ∈ R, then

I = (I ∶ ⟨f⟩) ∩ ⟨I, f⟩ .

Proof. If g ∈ (I ∶ ⟨f⟩) ∩ ⟨I, f⟩ then there are h ∈ I, r ∈ R with

g = h + rf .

Since g ∈ (I ∶ ⟨f⟩),fh + rf 2 = fg ∈ I,

hence rf 2 ∈ I, that is, r ∈ I ∶ ⟨f 2⟩ = I ∶ ⟨f⟩. So rf ∈ I, which impliesthat g = h + rf ∈ I.

The other inclusion is obvious.

Proposition 7.3.7 For any ideal I ⫋ R it holds

prime ⇒ irreducible ⇒ primary

Proof. For the first implication, suppose I = I1 ∩ I2 with I ⫋ I1.Then there is an f ∈ I1 − I. For any g ∈ I2

fg ∈ I1I2 ⊂ I1 ∩ I2 = I.

If I is prime then g ∈ I, which proves that I2 = I.For the second implication, let f, g ∈ R with fg ∈ I and f ∉ I.

By Theorem 2.1.5 the ascending chain

I ∶ ⟨g⟩ ⊂ I ∶ ⟨g2⟩ ⊂ ...


terminates, so there is an m with I ∶ ⟨gm⟩ = I ∶ ⟨gm+1⟩. By Lemma7.3.6 this implies that

I = (I ∶ ⟨gm⟩) ∩ ⟨I, gm⟩ .

By fgm ∈ I we have f ∈ I ∶ ⟨gm⟩. As f ∉ I it holds

I ⫋ I ∶ ⟨gm⟩ ,

hence, since I is irreducible,

I = ⟨I, gm⟩ ,

This implies gm ∈ I, that is, g ∈√I.

Definition 7.3.8 A primary decomposition of an ideal I is anexpression

I =⋂s

i=1Qi

with Qi primary. A primary decomposition is called minimal if itis irredundant and all

√Qi are distinct.

Theorem 7.3.9 Any ideal has a minimal primary decomposition.

Proof. Determine an irredundant irreducible decomposition byProposition 7.3.7. By Theorem 7.2.8 this is a primary decompo-sition. By Lemma 7.3.5, we can combine factors which have thesame radical.

7.4 Algorithm for primary decomposi-

tion of monomial ideals

Proposition 7.4.1 Any irreducible monomial ideal I = ⟨xα1i1, ..., xαr

ir⟩

is ⟨xi1 , ..., xir⟩-primary.

Proof. By Proposition 7.3.7 I is primary. Since ⟨xi1 , ..., xir⟩ ⊃ I isa prime ideal, the claim follows.

Using this observation, we obtain Algorithm 7.4.1 for computinga minimal primary decomposition of a monomial.

Algorithm 7.4.1 Monomial primary decomposition

Input: Monomial ideal I.Output: Minimal primary decomposition of I.1: Compute the irredundant irreducible decomposition.2: Combine factors which have the same radical.


Example 7.4.2 In Example 7.2.7 we have computed the irredun-dant irreducible decomposition

I = ⟨x0x1x2, x20x2, x

31x2⟩ = ⟨x0, x

31⟩ ∩ ⟨x1, x

20⟩ ∩ ⟨x2⟩

By Proposition 7.4.1

√⟨x0, x3

1⟩ = ⟨x0, x1⟩ =√

⟨x1, x20⟩

hence, by⟨x0, x

31⟩ ∩ ⟨x1, x

20⟩ = ⟨x2

0, x0x1, x31⟩

a minimal primary decomposition of I is

I = ⟨x20, x0x1, x

31⟩ ∩ ⟨x2⟩

In Singular we can compute a minimal primary decomposition ofany (not necessarily monomial) ideal by:LIB "primdec.lib";

ring R = 0,(x(0..2)),dp;

ideal I = x(0)*x(1)*x(2), x(0)^2*x(2), x(1)^3*x(2);

primdecGTZ(I);

[1]:

[1]:

[1]=x(2)

[2]:

[1]=x(2)

[2]:

[1]:

[1]=x(1)^3

[2]=x(0)*x(1)

[3]=x(0)^2

[2]:

[1]=x(1)

[2]=x(0)

This command returns a list of tuples of a primary ideal Qi theprime

√Qi.

Example 7.4.3 In general, a minimal primary decomposition isnot unique. For example

⟨x2, xy⟩ = ⟨x⟩ ∩ ⟨x2, y⟩ = ⟨x⟩ ∩ ⟨x2, xy, yn⟩

for all n ≥ 1.

However, we have uniqueness in the following sense:


7.5 Uniqueness of primary decomposi-

tion

Theorem 7.5.1 IfI =⋂

s

i=1Qi

is a minimal primary decomposition, then the pairwise differentprime ideals

√Qi are uniquely determined.

They are precisely the primes P ⊂ R with P =√I ∶ ⟨b⟩ for some

b ∈ R.

Remark 7.5.2 If Q is primary and b ∈ R then

b ∈ QÔ⇒ Q ∶ b = R

b ∉ QÔ⇒√Q ∶ b =

√Q

Proof. The first claim of the remark and the inclusion ⊃ in thesecond claim are trivial. Let xm ∈ Q ∶ b for some m, that is, xmb ∈ Q.Since b ∉ Q and Q is primary, we have xm⋅m

′

∈ Q for some m′.We now prove the theorem:

Proof. By the remark

I ∶ ⟨b⟩ =⋂s

i=1 (Qi ∶ ⟨b⟩)

hence √I ∶ ⟨b⟩ =⋂

s

i=1

√Qi ∶ ⟨b⟩ =⋂b∉Qi

√Qi.

If√I ∶ ⟨b⟩ is prime then by Proposition 7.3.7

√I ∶ ⟨b⟩ =

√Qi

for some i.Conversely, by minimality of the decomposition, for every j

there isbj ∈⋂i≠jQi

with bj ∉ Qj. Then√Qj =

√I ∶ ⟨bj⟩.

Example 7.5.3 Consider

I = ⟨x0x1x2, x20x2, x

31x

22⟩


We compute an irreducible decomposition

I = ⟨x0, x31x

22⟩ ∩ ⟨x1, x

20x2⟩ ∩ ⟨x2⟩

= ⟨x0, x31⟩ ∩ ⟨x0, x

22⟩ ∩ ⟨x1, x

20⟩ ∩ ⟨x1, x2⟩ ∩ ⟨x2⟩

Since ⟨x2⟩ ⊂ ⟨x1, x2⟩ we can delete ⟨x1, x2⟩. Note that ⟨x0, x22⟩

cannot be deleted. So the irredundant irreducible decomposition is

I = ⟨x0, x31⟩ ∩ ⟨x0, x

22⟩ ∩ ⟨x1, x

20⟩ ∩ ⟨x2⟩ .

Hence, like in Example 7.4.2, a minimal primary decomposition is

I = ⟨x20, x0x1, x

31⟩ ∩ ⟨x2⟩ ∩ ⟨x0, x

22⟩ .

We verify this also using Singular:LIB "primdec.lib";

ring R = 0,(x(0..2)),dp;

ideal I = x(0)*x(1)*x(2), x(0)^2*x(2), x(1)^3*x(2)^2;

primdecGTZ(I);

[1]:

[1]:

[1]=x(2)

[2]:

[1]=x(2)

[2]:

[1]:

[1]=x(1)^3

[2]=x(0)*x(1)

[3]=x(0)^2

[2]:

[1]=x(1)

[2]=x(0)

[3]:

[1]:

[1]=x(2)^2

[2]=x(0)

[2]:

[1]=x(2)

[2]=x(0)

By computing appropriate ideal quotientsradical(quotient(I,x(0)^2));

[1]=x(2)

radical(quotient(I,x(2)^2));

[1]=x(1)


[2]=x(0)

radical(quotient(I,x(1)^3*x(2)));

[1]=x(2)

[2]=x(0)

we can obtain all possible primes√I ∶ ⟨x2

0⟩ = ⟨x2⟩√I ∶ ⟨x2

2⟩ = ⟨x0, x1⟩√I ∶ ⟨x3

1x2⟩ = ⟨x0, x2⟩ .

In fact, in the case of monomial ideals, the elements b in Theorem7.5.1 can always be chosen monomial.

Definition 7.5.4 The√Qi in the theorem are called the associ-

ated primes of I. Write

Ass(I) = {√Qi ∣ i} .

The minimal elements of Ass(I) with respect to inclusion arecalled the minimal associated primes of I. Write Min(I) forthe set of minimal associated primes.

Primary ideals Qi with√Qi minimal are called isolated pri-

mary components, the others are called embedded primary compo-nents.

Example 7.5.5 In the above example

Ass(I) = {⟨x2⟩ , ⟨x0, x1⟩ , ⟨x0, x2⟩}

andMin(I) = {⟨x2⟩ , ⟨x0, x1⟩}

The primary component ⟨x0, x22⟩ is embedded in the isolated compo-

nent ⟨x2⟩.In geometric terms, V (x0, x2

2) is a point in P2 which lies on theline V (x2).

A point specified (as a scheme) by a primary ideal that is notprime is usually referred to as a fat point.

Remark 7.5.6 If I ⊂ R is an ideal then√I = ⋂P ∈Min(I)P ,

in particular, √I = ⋂P⊃I primeP .

Proof. Clear by Theorem 7.5.1 and Lemma 7.3.4.Without proof we mention:

Theorem 7.5.7 The isolated components of a minimal primarydecomposition are uniquely determined.


7.6 Computing primary decomosition in

general

7.7 Exercises

Exercise 7.1 Let S be the set of squarefree monomials in the vari-ables x0, ..., xn. A simplicial complex ∆ on the vertices x0, ..., xnis a subset ∆ ⊂ S, such that x0, ..., xn ∈ ∆, and if m ∈ ∆ and w ∣ mthen also w ∈ ∆. The elements of ∆ are called faces, the maximalones facets.

The Stanley-Reisner ideal of ∆ is the ideal

I∆ = ⟨m ∣m ∈ S, m ∉ ∆⟩ ⊂K[x0, ..., xn]

generated by the non-faces of ∆.An element m ∈ S is called coface if x1⋅...⋅xn

m ∈ ∆, that is, if itscomplement is a face.

1) Show thatI∆ = ⋂

xi1 ⋅...⋅xia minimalcoface of ∆

⟨xi1 , ..., xia⟩

2) For the simplicial complex

S = {1, x0, x1, x2, x3, x1x2, x1x3, x0x2, x0x3, x2x3, x1x2x3}

(see Figure 7.1), determine I∆, and its decomposition accord-ing to 1). Verify your result by computing the ideal intersec-tion.

Exercise 7.2 Let R =K[x0, x1, x2] and

I = ⟨x20x1, x

20x

22, x

21, x1x

22⟩ ⊂ R

Find the irredundant decomposition of I into irreducible ideals.

Exercise 7.3 Let R =K[x0, x1, x2] and

I = ⟨x30, x

31, x

20x

22, x0x1x

22, x

21x

22⟩ ⊂ R

1) Find the irredundant decomposition of I into irreducible ide-als.

2) Find a minimal primary decomposition of I.


Figure 7.1: Simplicial complex with three maximal faces.

8

Normalization

8.1 Overview

Definition 8.1.1 Let A be a Noetherian domain. The normal-ization of A, written A, is the integral closure of A in its quotientfield quot(A). We call A normal if A = A.

Recall that the integral closure is unique up to isomorphism.The key setup we have in mind is the normalization of an affinealgebra

A =K[x1, ..., xn]/I

where I is a prime ideal, and this is the setup we will consider inthe following. We will describe an algorithm which computes thenormalization as an affine algebra A = A[t1, ..., ts]/J with a primeideal J ⊂ A. For simplicity we assume that K has characteristic zero(the algorithm will also work fine if K, more generally, is perfect,for example, if K = Fp).

On the geometric side, A is the coordinate ring of an algebraicvariety X = V (I) ⊂ An

K and A is the coordinate ring of an algebraicvariety X ⊂ An+s

K called the normalization of X. The integral ringextension A ⊂ A corresponds on the geometric side to a polynomialmap X →X .

Example 8.1.2 In Figure 3.3, we can view the twisted cubic V (x2−y, x3 − z) as the normalization X of the cusp X = V (y3 − z2). Thenormalization is only determined up to isomorphism. In fact, thetwisted cubic is in bijection to the line A1 via the paramatrization

A1 → V (x2 − y, x3 − z)

t↦ (t, t2, t3)

195

8. NORMALIZATION 196

In terms of algebra, we can see the normalization map from the lineto the cusp immediately: If I = ⟨y3 − z2⟩ ⊂K[x, y] is the ideal of thecusp, then

A =K[y, z]/I ≅ K[t2, t3] ⊂ K[t] = Ay ↦ t2

z ↦ t3

Since t is a zero of the monic polynomial T 2 − t2 ∈ K[t2, t3][T ], itis integral. Moreover, K[t] is normal since it is factorial:

Example 8.1.3 Any unique factorization domain is normal.

Proof. Let R be a unique factorization domain and rs ∈ quot(R)

be integral over R and gcd(r, s) ∈ R×. Then there are ai ∈ R with

(r

s)n

=n−1

∑i=0

ai (r

s)i

,

and, cancelling the denominator,

rn = s(n−1

∑i=0

airisn−1−i) .

So if p is a prime divisor of s, then also of r, which implies thatrs ∈ R.

In terms of geometry, normalization is a technique that improvesthe singularities of X, by removing singularities of dimension d−1 ina d-dimensional variety. In particular, for a curve X, normalizationgives a desingularization.

The basic philosophy is that X is easier to study than X, butnevertheless yields equivalent information. For example, many re-sults for curves assume, in their natural formulation, that the curveis non-singular, like the Riemann-Roch theorem classifying rationalmaps between curves.

We remark that also a desingularization of varieties of higherdimension can be achieved, for example, in the case of the sin-gular surfaces in Figures 1.4, 1.6, and 1.7. However, this is morecomplicated since desingularization does not correspond to a simplealgebraic process in terms of A (like normalization of X correspondsto normalization of A).

8.2 Basics on normalization

As in the introduction, let

A =K[x1, ..., xn]/I


with a prime ideal I ⊂K[x1, ..., xn]. We first observe that normalityis a local property, that is, it can be tested by testing it at everyelement of

X = Spec(A) = {P ⊂ A ∣ P prime ideal}

for the spectrum of A. An element P ∈ X is also called a point ofthe scheme X.

Recall that, If S ⊂ A is a multiplicatively closed set, then

S−1A = {a

s∣ s ∈ S, a ∈ A} ⊂ quot(A)

is a ring called the localization of A at S. If P ∈ X, then defineAP = S−1A with S = A − P .

Theorem 8.2.1 Normalization commutes with localization, thatis,

AP = AP

for all P ∈X.

Proof. Let au ∈ AP so a ∈ A. Hence there are ai with

an +∑n

i=1aia

n−i = 0

so

(a

u)n

+∑n

i=1aiu

i (a

u)n−i

= 0.

This implies au ∈ AP .

If au ∈ AP , then there are ai

ui∈ AP with

(a

u)n

+∑n

i=1

aiui

(a

u)n−i

= 0

Hence

(a ⋅ u1 ⋅ ... ⋅ un)n+∑

n

i=1ai (u

i ⋅∏j≠iaj) (a ⋅ u1 ⋅ ... ⋅ un)n−i = 0

so a′ = a ⋅ u1 ⋅ ... ⋅ un ∈ A. This implies a ∈ AP .

Corollar 8.2.2 A is normal if and only if AP is normal for all P .

Proof. If A = A then by Theorem 8.2.1 we have AP = AP = AP forall P . On the other hand, if AP = AP for all P then, again usingthe Theorem 8.2.1, AP = AP for all P ∈X. Using [5, Corollary 2.9]this implies A = A.


Definition 8.2.3 We call

N(A) = {P ∈X ∣ AP is not normal}

the non-normal locus of A.

Remark 8.2.4 By Corollary 8.2.2 the ring A is normal if and onlyif N(A) is empty.

Our next goal is to show that the non-normal locus can bedescribed by an ideal. It will be convenient to use the followingdefinitions (inspired by the theory of schemes): For an ideal J ⊂ Awe define by

V(J) = {P ∈X ∣ P ⊃ J} ⊂X

its vanishing locus. Subsets of the form V(J) are the closed setsof a topology on X, the Zariski topology. On the other hand, forany subset Y ⊂X we write

I(Y ) = ⋂P ∈Y P

for its zero ideal. Then, using Remark 7.5.6, it follows immedi-ately:

Theorem 8.2.5 If J ⊂ A is an ideal then

I(V(J)) =√J .

Definition 8.2.6 The conductor of A in A is

CA = {a ∈ A ∣ aA ⊂ A}.

It is an ideal of both A and A.

Lemma 8.2.7 N(A) = V(CA).

Proof. N(A) ⊂ V(CA) ∶ If P ∈ N(A), that is, AP ⫋ AP then

(CA)P = CAP= {a ∈ AP ∣ aAP ⊂ AP} ⫋ AP

However if P ∉ V(CA) then (CA)P = AP .

V(CA) ⊂ N(A) ∶ By

V(CA) = ⋃h∈AV({a ∈ A ∣ ah ∈ A}),

we have to show that for all h = pq ∈ A ⊂ quot(A) it holds

V({a ∈ A ∣ ah ∈ A}) ⊂ N(A).


So for every prime P we have to prove

{a ∈ A ∣ ah ∈ A} ⊂ P Ô⇒ AP ⫋ AP .

If AP = AP then the image of h in AP is p/q1 = p

q , that is, q ∉ P .Moreover, qh = p ∈ A, which proves the claim.

In order to compute the normalization we require some knowl-edge about the non-normal locus N(A). In general, CA can only becomputed a-posteriori from A. Fortunately, the singular locus of Acan be obtained explicitly, and contains the non-normal locus:

Definition 8.2.8 For an ideal I = ⟨f1, ..., fs⟩ ⊂ K[x1, ..., xn], theJacobian ideal Jac(I) is generated by the c × c minors of theJacobian matrix

⎛⎜⎝

∂f1∂x1

⋯ ∂f1∂xn

⋮ ⋮∂fs∂x1

⋯ ∂fs∂xn

⎞⎟⎠

where c = n − dim(X) is the codimension of X = V (I). Then thesingular locus of A is

Sing(A) = V(Jac(I) + I).

Note that we can consider Jac(I) + I both as an ideal in A =K[x1, ..., xn]/I and in K[x1, ..., xn], in fact it is easy to prove:

Remark 8.2.9 There is a bijection between the ideals of A and theideals of K[x1, ..., xn] which contain I.

Example 8.2.10 Consider the coordinate ring

A =K[x, y] =K[x, y]/I

of the plane curve X = V(I) defined by

I = ⟨x4 + y2(y − 1)3⟩.

ThenJac(I) = ⟨x3, 2y(y − 1)3 + 3y2(y − 1)2⟩

In Singular we can compute the ideal Jac(I)+ I ⊂K[x, y] by thecommand slocus:ring R = 0,(x,y),dp;

ideal I = x^4 + y^2*(y-1)^3;

ideal singI = std(slocus(I));

singI;


[1]=y3-2y2+y

[2]=x3

We compute a minimal primary decomposition of the singular locus:primdecGTZ(singI);

[1]:

[1]:

[1]=y2-2y+1

[2]=x3

[2]:

[1]=y-1

[2]=x

[2]:

[1]:

[1]=y

[2]=x3

[2]:

[1]=y

[2]=x

So √Jac(I) + I = ⟨x, y⟩ ∩ ⟨x, y − 1⟩ ,

and the singular locus of the curve X = V (I) consists of two points(0,0), (0,1).

For the non-normal and the singular locus it holds (withoutproof):

Theorem 8.2.11 N(A) ⊂ Sing(A).

8.3 Finiteness of normalization

The following result will allow us to design an algorithm computingA in a finite number of steps:

Theorem 8.3.1 A is a finitely generated A-module.

Hence A is a Noetherian A-module by Theorem 4.3.12, since Ais a Noetherian ring by Example 2.4(3).Proof. Noether normalization gives algebraically independent y1, ..., ys ∈A such that A is a finitely generated K[y1, ..., ys]-module. Then

K[y1, ..., ys] ⊂ A ⊂ A ⊂K(y1, ..., ys) ⊂ L ∶= quot(A)

Note that K[y1, ..., ys] = A, since K[y1, ..., ys] ⊂ A is finite.


Note that in characteristic zero any finite extension is separable.Hence we may assume that the finite extension K(y1, ..., ys) ⊂ L isGalois. Otherwise we pass to the normal closure K(y1, ..., ys) ⊂L ⊂ L′. Then the same arguments as below prove that the integralclosure of A in L′ is finite over A, which implies that A is finite overA.

So, with the Galois group G of K(y1, ..., ys) ⊂ L, consider thetrace

Tr ∶ L → K(y1, ..., ys)x ↦ ∑g∈G g(x)

The trace yields a nondegenerate bilinear form

L ×L→K(y1, ..., ys)

(x, y)↦ Tr(xy)

Hence, for a basis x1, ..., xr ∈ A with

L =K(y1, ..., ys)(x1, ..., xr),

there is a dual basis x∗1, ..., x∗r ∈ L with

Tr(xix∗j ) = δi,j

andL =K(y1, ..., ys)(x

∗1, ..., x

∗r).

We show thatA ⊂ Ax∗1 + ... +Ax

∗r

and, hence, A is a finitely generated A-module: Given b ∈ A, thereare ai ∈K(y1, ..., ys) with

b = a1x∗1 + ... + arx

∗r .

By bxi ∈ A we haveai = Tr(bxi) ∈ A

using thatTr(A) ⊂ A.

Indeed, if x ∈ A then Tr(x) is a coefficient of the polynomial

P =∏g∈G

(T − g(x)) .

Since P is invariant under G, we have P ∈ K(y1, ..., ys)[T ]. Sinceall g(x) for g ∈ G are integral over A all coefficients are integral,hence P ∈ A[T ].


8.4 Grauert-Remmert criterion

The first key observation is that, for any ideal J ⊂ A, the moduleHomA(J, J) lies between A and A. Moreover it can be computedas an ideal quotient. We denote by

(I1 ∶A I2) = {b ∈ A ∣ bI2 ⊂ I1}

the quotient of ideals I1, I2 ⊂ A. Note that if Ii = Ji mod I withideals Ji ⊂ K[x1, ..., xn], then by Remark 8.2.9 this ideal quotientcan be computed by Algorithm 3.6.3 as

(I1 ∶A I2) = ((J1 + I) ∶ (J2 + I))mod I.

Lemma 8.4.1 If J ⊂ A is an ideal and 0 ≠ g ∈ J , then:

1) Any ϕ ∈ HomA(J, J) is given by multiplication

ϕ ∶ J → J

b ↦ ϕ(g)g b

2) There is a natural inclusion of rings

A ↪ HomA(J, J)a ↦ ϕa ∶ J → J

b↦ ab

3) There is a natural isomorphism of rings

HomA(J, J) ≅ 1g(gJ ∶A J)

ϕ ↦ ϕ(g)g

4) Any element of 1g(gJ ∶A J) is integral, that is,

1

g(gJ ∶A J) ⊂ A ⊂ quot(A).

Proof.

1) If ϕ ∶ J → J is a homomorphism and b ∈ J is any otherelement, then by A-linearity

g ⋅ ϕ(b) = ϕ(gb) = b ⋅ ϕ(g),

hence for b ≠ 0

ϕ(g)

g=ϕ(b)

b∈ quot(A).

This implies that ϕ(g)g b = ϕ(b)

b b = ϕ(b) for b ≠ 0. For b = 0 theclaim is trivial.


2) If ϕa = 0, that is, ab = 0 for all b ∈ J . Since 0 ≠ g ∈ J is anon-zerodivisor, it follows that a = 0.

3) If ϕ ∈ HomA(J, J) then by (1) we have ϕ(g)g J ⊂ J , that is,

ϕ(g)J ⊂ gJ . This proves that ϕ(g) ∈ (gJ ∶A J) since ϕ(g) ∈J ⊂ A, so the specified map is well-defined. Moreover, by (1)it is injective, as ϕ = 0 if ϕ(g) = 0. It is surjective since for anyg′ ∈ (gJ ∶A J) a well-defined homomorphism J → J is givenby b↦ g′

g b.

4) Finally, we have to prove that for

b ∈ (gJ ∶A J)

the element bg ∈ quot(A) is integral over A: Consider the A-

linear multiplication map Lb ∶ J → J , a ↦ ab. Fix A-modulegenerators g1, ..., gs of J . By bJ ⊂ gJ , there are

bij ∈ ⟨g⟩

withbgj = ∑ibijgi

which yields an s × s-matrix (bij) representing Lb. By thetheorem of Cayley-Hamilton, Lb is a zero of the characteristicpolynomial

χ(t) = det(t ⋅E − (bij)).

Hence, there are ai ∈ A with

bs + a1bm−1 + ... + am−1b

1 + am = 0.

Since ai is a linear combination of i×i-minors of (bij) we have

ai ∈ ⟨g⟩i.

Hence

(b

g)s

+a1

g(b

g)s−1

+ ... +as−1

gs−1(b

g)

1

+asgs

= 0

is a monic equation for bg over A.

Remark 8.4.2 If J ⊂ A is an ideal and 0 ≠ g ∈ J , then

(gJ ∶A J) = (gJ ∶quot(A) J).


Proof. The inclusion ⊂ is clear. If g′ ∈ quot(A) with g′J ⊂ gJ thena well-defined homomorphism ϕ ∶ J → J is given by b ↦ g′

g b. ByLemma 8.4.1(3.) it follows that

ϕ(g)

g∈

1

g(gJ ∶A J),

that is,g′ = ϕ(g) ∈ (gJ ∶A J).

The following criterion for normality is due to Grauert and Rem-mert. It is the basis for an algorithmic approach on computing thenormalization. It tells us that, if forming HomA(J, J) does notenlarge A any more, then we have reached A. However this onlyworks if J sees all of the non-normal locus, that is N(A) ⊂ V(J).

Theorem 8.4.3 (Grauert-Remmert criterion) Let 0 ≠ J ⊂ Abe a radical ideal with

N(A) ⊂ V(J).

Then A is normal if and only if the inclusion

A ↪ HomA(J, J)a ↦ ϕa

is an isomorphism.

Proof. If A = A then, by Lemma 8.4.1, we have injective mapsA→ HomA(J, J)→ A, which proves the claim.

For the converse: By Lemma 8.2.7

V(CA) = N(A) ⊂ V(J).

Hence, by Theorem 8.2.5,

J =√J = I(V(J)) ⊂ I(V(CA)) =

√CA,

which implies that there is a smallest m ≥ 0 with Jm ⊂ CA, that is,

JmA ⊂ A.

Assume that m > 0. Then there is a ∈ Jm−1 and b ∈ A with ab ∈A − A. Moreover, abJ ⊂ A, so multiplication by ab defines an A-homomorphism

ϕab ∶ J → Ac ↦ abc


Since ab ∈ A there is a monic equation

(ab)n + a1(ab)n−1 + ... + an = 0

with ai ∈ A. If 0 ≠ c ∈ J , then

(abc)n = −a1c(abc)n−1 − ... − anc

n ∈ J

hence abc ∈√J = J , that is,

ϕab ∈ HomA(J, J).

If we assume that A→ HomA(J, J), a↦ ϕa is an isomorphism, thenab ∈ A, a contradiction. Hence m = 0, that is, A = A.

Definition 8.4.4 A pair (J, g) with J as in Theorem 8.4.3 and0 ≠ g ∈ J is called a test pair for A, and J is called a test idealfor A.

By Lemma 1.6.13 we can choose the Jacobian ideal togetherwith any non-zero element as a test pair.

8.5 Normalization algorithm

By Lemma 8.4.1 and Theorem 8.4.3, if A is not normal, computingHom as an ideal quotient yields a ring

A ⫋ HomA(J, J) ≅1

g(gJ ∶A J) ⊂ A ⊂ quot(A),

that is strictly larger than A and is contained in A.

Example 8.5.1 For A as in Example 8.2.10, the radical of theJacobian ideal is

J ∶= ⟨x, y (y − 1)⟩A ,

and we can take g ∶= x ∈ J as a non–zerodivisor of A. Using Sin-gular we can determine HomA(J, J) as an ideal quotient:LIB "normal.lib";

ring R = 0,(x,y),dp;

ideal I = x^4+y^2*(y-1)^3;

qring A = std(I);

ideal J = x, y*(y-1);

quotient(x*J, J);

[1]=x

[2]=y3-2y2+y


HenceA ⫋ A1 ⊂ A ⊂ quot(A)

with

A1 =1

x(xJ ∶A J) =

1

x⟨x, y(y − 1)2⟩

=A ⟨1,y(y − 1)2

x⟩ .

By iterating this calculation, we obtain Algorithm 8.5.1.

Algorithm 8.5.1 Normalization

Input: Prime ideal I ⊂K[x1, ..., xn].Output: Normalization A of A =K[x1, ..., xn]/I.

1: Initial test pair: J0 ∶=√

Jac(I) + I ⊂ A and select 0 ≠ g ∈ J0.2: i = 03: repeat4: i = i + 15: Ai =

1g(gJi−1 ∶Ai−1

Ji−1)

6: Ji =√JAi−1

7: until Ai = Ai−1

We prove correctness of the algorithm:Proof. Starting from A0 = A, we get a chain of extensions ofreduced Noetherian rings

A = A0 ⊂ ⋅ ⋅ ⋅ ⊂ Ai−1 ⊂ Ai ⊂ ⋅ ⋅ ⋅ ⊂ Am = A.

Since A is Noetherian, by Theorem 8.3.1 the sequence terminates,that is, for some m we have Am = Am+1. By Theorem 8.4.3 it thenholds Am = A.

Moreover, (Ji, g) is a test pair for Ai by the following lemma:

Lemma 8.5.2 Let A ⊂ A′ ⊂ A be extensions of rings and (J, g) atest pair for A. Then (

√JA′, g) is a test pair for A′.

Proof. Clearly, every non–zerodivisor g ∈ A ofA is a non–zerodivisorof quot(A) ⊃ A′. For J ′ =

√JA′ we have to show that N(A′) =

V(CA′) ⊂ V(J ′), that is, if CA′ ⊂ Q for a prime Q then J ′ ⊂ Q.Let P = Q ∩A. Then P is prime: If f ⋅ g ∈ P ⊂ Q with f, g ∈ A

then f ∈ Q or g ∈ Q. Hence f ∈ P or g ∈ P .By A′ = A we have

Q ⊃ CA′ = {a ∈ A′ ∣ aA ⊂ A′} ⊃ {a ∈ A ∣ aA ⊂ A} = CA.


Hence Q ∩ A ⊃ CA, that is, Q ∩ A ∈ V(CA) = N(A) ⊂ V(J). SoJ ⊂ Q ∩ A, which implies that JA′ ⊂ Q. Using Remark 7.5.6 itfollows that √

JA′ = ⋂Q′⊃JA′ primeQ′ ⊂ Q.

Practical computations rely on explicit representations of the Aias A–algebras. These will be obtained as an application of Lemma8.5.3 below. To formulate the lemma, we use the following notation.Let J ⊂ A be an ideal containing a non–zerodivisor g of A, andlet A–module generators u0 = g, u1, . . . , us for (gJ ∶A J) be given.Choose variables T1, . . . , Ts, and consider the epimorphism

Φ ∶ A[T1, . . . , Ts]→1

g(gJ ∶A J), Ti ↦

uig

,

hence1

g(gJ ∶A J) ≅ A[T1, . . . , Ts]/ker(Φ).

The kernel of Φ describes the A–algebra relations on the ui/g. Thesecan be computed by Algorithm 3.5.10. In fact, in this setup, theycan be described explicitly:

• Each A–module syzygy

α0u0 + α1u1 + . . . + αsus = 0, αi ∈ A,

gives an element α0 +α1T1 + . . .+αsTs ∈ ker(Φ), which we calla linear relation.

• Developing each productuig

ujg , 1 ≤ i ≤ j ≤ s, as a sum

uig

ujg =

∑k βijkukg , we get elements TiTj −∑k βijkTk in ker Φ, which we

call quadratic relations.

Lemma 8.5.3 The linear and quadratic relations generate ker(Φ).

For a proof see [9, Lemma 3.6.7].

Example 8.5.4 For A1 as in Example 8.5.1, we find

A[t1]/I1 ≅ A1

t1 ↦y(y − 1)2

x

with

I1 = ⟨−t1x + y(y − 1)2, t1y(y − 1) + x3, t21 + x2(y − 1)⟩ .

In this representation of A1 we can compute

J1 =√A1 ⟨x, y(y − 1)⟩ =A1 ⟨x, y(y − 1), t1⟩ .


Finally, the following result will allow us to find the normaliza-tion in a way such that in all steps of the algorithm the computationof Hom can be carried through in the original ring A:

Theorem 8.5.5 Let 0 ≠ J ⊂ A be an ideal and A ⊂ A′ ⊂ A anintermediate ring, and let J ′ =

√JA′. Let U and H be ideals of A

and d ∈ A such that A′ = 1dU and J ′ = 1

dH, respectively. Then

(gJ ′ ∶A′ J′) =

1

d(dgH ∶A H) ⊂ quot(A).

Proof. By Remark 8.4.2, g ∈ J ⊂ J ′ and quot(A) = quot(A′) weget

(gJ ′ ∶A′ J′) = (gJ ′ ∶quot(A) J

′)

and, in the same way, by dg ∈H

(dgH ∶A H) = (dgH ∶quot(A) H) .

Moreover

(gJ ′ ∶quot(A) J′) = {a ∈ quot(A) ∣ a

1

dH ⊂ g

1

dH}

= {a ∈ quot(A) ∣ aH ⊂ gH}

= (gH ∶quot(A) H)

and

(dgH ∶quot(A) H) = {b ∈ quot(A) ∣ bH ⊂ dgH}

= {b ∈ quot(A) ∣b

dH ⊂ gH}

= d {a ∈ quot(A) ∣ aH ⊂ gH}

= d (gH ∶quot(A) H) ,

which proves the claim.Note that, in the setup of the algorithm, in every step, d can be

chosen to be a power of g.

Example 8.5.6 Continuing Example 8.5.4 we write J1 as an A-module

J1 =1

x⟨x2, xy(y − 1), y(y − 1)2⟩ =

1

dH1

with d = x and H1 = ⟨x2, xy(y − 1), y(y − 1)2⟩ ⊂ A.


Using the test pair (J1, x) and applying Theorem 8.5.5 and Lemma8.5.3, we get

1

x(xJ1 ∶A1 J1) =

1

x2(x2H1 ∶A H1) =

1

x2⟨x2, xy(y − 1), y(y − 1)2⟩

=A ⟨1,y(y − 1)

x,y(y − 1)2

x2⟩

andA2 ∶= A[t1, t2]/I2 ≅ A ⟨1, y(y−1)

x , y(y−1)2x2 ⟩

t1 ↦ y(y−1)2x2

t2 ↦ y(y−1)x

with

I2 = ⟨t1x − t2(y − 1), − t2x + y(y − 1), t1y(y − 1) + x2, t1y2(y − 1)2 + t2x

3,

t21 + (y − 1), t1t2 + x, t22 − t2y⟩

In the final step, we find that A2 is normal, so that A = A2.

Example 8.5.7 The normalization algorithm described above, isavailable in Singular. In fact, this is the first implementation ofa normalization algorithm which is practically useable. We apply itto the coordinate ring of the curve from Example 8.5.6:LIB "normal.lib";

ring R = 0,(x,y),dp;

ideal I = x^4+y^2*(y-1)^3;

list nor = normal(I);

def R1 = nor[1][1];

setring R1;

norid;

norid[1]=T(1)*x-T(2)*y+T(2)

norid[2]=-T(2)*x+y^2-y

norid[3]=T(1)*y^2-T(1)*y+x^2

norid[4]=T(1)*y^4-2*T(1)*y^3+T(1)*y^2+T(2)*x^3

norid[5]=T(1)^2+y-1

norid[6]=T(1)*T(2)+x

norid[7]=T(2)^2-T(1)*y

norid[8]=y^5+x^4-3*y^4+3*y^3-y^2

normap;

normap[1]=x

normap[2]=y

We verify that the normalization of the given curve is indeed smooth:std(slocus(norid));

[1]=1


8.6 Exercises

Exercise 8.1 Determine the normalization of the singularity oftype Ak defined by the ideal

⟨y2 − xk+1⟩ ⊂K[x, y]

for k ≥ 2 even.

Exercise 8.2 Let R =K[x, y] and let

I = ⟨yk − xk + xk+1⟩ ⊂ R

be the ideal of an ordinary k-fold point, where k ≥ 2. For A = R/Idetermine A-module generators of A.

Exercise 8.3 Determine the normalization of the Whitney um-brella X = V (I) ⊂ A3

I = ⟨x2 − y2z⟩ ⊂K[x, y, z]

(see Figure 3.4).

Exercise 8.4 For the Steiner surface X = V (I) ⊂ A3

I = ⟨x2y2 + x2z2 + y2z2 − xyz⟩ ⊂K[x, y, z]

(see Figure 3.5) determine A-module generators of A where A =K[x, y, z]/I.

Index

Zariski topology, 198

affine algebra, 195affine algebraic set, 20affine algebraic variety, 20affine chart, 121affine space, 19Algebraic geometry, 4algebraic variety, 4arithmetic genus, 129associated prime, 186associated primes, 192Axiom, 15

Barth sextic, 5basis, 101Betti table, 135Bezout’s theorem, 23, 118Buchberger normal form, 52Buchberger’s algorithm, 2, 56, 145Buchberger’s criterion, 56

Cayley-Hamilton, theorem of, 110,203

characteristic matrix, 110characteristic polynomial, 110closure, 31codimension, 26, 199coface, 193cokernel, 100column operation, 92conductor, 198Conj. of Birch and Swinnerton-

Dyer, 22coordinate ring, 77curve, 4cyclic module, 105

degree, 129degree reverse lexicographical or-

dering, 45dehomogenization, 121desingularization, 196determinantal divisors, 96differential equation, 16differential field, 15Diffie-Hellman key exchange, 22dimension, 129direct sum, 105divisibility, 139Division with remainder, 8domain, 78double cone, 4

elementary divisor, 92Elementary divisor theorem, 91elementary divisors of a module,

106elementary extension, 15elimination ideal, 67elimination ordering, 65elliptic curve, 21elliptical arc, 31embedded, 192Euclidean Algorithm, 9Euclidean norm, 40Euclidean ring, 40exact sequence, 117exakt, 101exponent vector, 43

face, 193facet, 193fat point, 192

211

INDEX 212

Fermat’s last theorem, 11finite, 117finitely generated, 32finitely generated module, 100finitely presented, 101first order deformations, 171free module, 101free resolution, 117fundamental thm. of arithmetic,

10

Galois group, 201Gaussian elimination, 2generating system, 30generators, 100, 167geometric invariant theory, 119global ordering, 43global ordering for modules, 137Grobner bases, 2Grobner basis, 51, 142Grobner fan, 49graded free resolution, 131graded module, 127graded ring, 119Grauert-Remmert criterion, 204group of bijections, 14group table, 15

Hilbert function, 128Hilbert polynomial, 129Hilbert scheme, 171Hilbert’s basis theorem, 34Hom, 170homogeneous, 120homogeneous homomorphism, 130homogeneous zero ideal, 120homogenization, 125homomorphism theorem for mod-

ules, 100hyperboloid, 4hyperplane at infinity, 121

icosahedron, 5ideal, 30

ideal membership problem, 39ideal quotient, 81, 202implicitation, 70initial term, 49intersection of ideals, 63, 180irreducible, 20irreducible ideal, 184irredundant, 183isolated, 192

Jacobi symbol, 25Jacobian ideal, 199Jordan block, 112

K3-surface, 175Koszul complex, 118Kummer quartic surface, 4

lead coefficient, 34, 46, 138lead monomial, 34lead term, 34leading ideal, 51leading module, 142leading monomial, 46, 138leading term, 46, 138least common multiple, 144Legendre symbol, 24lexicographic ordering, 45linear system, 76local ordering, 43localization, 197

Macaulay2, 175Maple, 3Mathematica, 15Maxima, 15maximal ideal, 37minimal, 58, 188minimal associated primes, 192minimal free resolution, 158minimal generators, 44minimal generators for monomial

submodules, 139minimal Grobner basis, 42

INDEX 213

minimal polynomial, 110minors, 96module, 98module homomorphism, 100module of homomorphisms, 170monomial ideal, 44monomial ordering, 43monomial ordering for modules,

136monomial submodule, 139monomials, 136morphism, 71multi-index notation, 43multiplicatively closed set, 197

negative lexicographical ordering,45

Noether normalization, 200Noether, Emmy, 32Noetherian, 32Noetherian module, 102non-normallocus, 198normal closure, 201normal form, 51, 140normalization of a ring, 195normalization of a variety, 195

octahedron, 12ODE, 16orbit, 13ordinary k-fold point, 210

p-adic numbers, 11P-primary, 186Pari/Gp, 11polynomial parametrization, 70presentation matrix, 102primary decomposition, 38, 188primary ideal, 38, 186prime factorization, 10prime number, 10prime number theorem, 10projection, 67projective algebraic set, 120

projective closure, 125projective space, 118

Q-rational points, 22quotient module, 99

radical, 36radical ideal, 36rank, 106rational function field, 78rational map, 74, 78rational parametrization, 76Reduce, 15reduced, 53, 58, 143reduced Buchberger normal form,

53reduced Grobner basis, 42reduced normal form, 53, 143relations, 167Riemann-Roch theorem, 196Risch’s algorithm, 15row operation, 92

S-pair, 18S-polynomial, 56, 144scheme, 35Schreyer ordering, 148semigroup ordering, 43sieve of Eratosthenes, 10simplicial complex, 193Singular, 2singular locus, 199singularity, 4singularity of type Ak, 210Smith normal form, 92spectrum, 35, 197stabilizer, 13standard grading, 120Stanley-Reisner ideal, 193Steiner surface, 72subfield of constants, 15submodule, 99subquotient, 167sum of ideals, 63

INDEX 214

surface, 4Surfer, 4syzygy, 57, 147syzygy matrix, 147syzygy module, 117, 147syzygy polynomial, 56, 144syzygy theorem, 117, 153

term, 136test ideal, 205test pair, 205tetrahedron, 5Theorem of Cayley, 14Theorem of Mordell, 22tie-break, 47Togliatti quintic, 5toric geometry, 119torsion modul, 106torsion submodule, 106torsion-free, 106total ordering, 43, 136trial division, 10twist, 131

unique factorization domain, 196

vanishing ideal, 31vanishing locus, 31vanishing locus, scheme theoretic,

198

weighted degree ordering, 47well ordering, 43Whitney umbrella, 70Wiles, Andrew, 11

Zariski topology, 32, 125zero ideal, scheme theoretic, 198

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Computer Algebra - Fachbereich Mathematikboehm/lehre/1213_CA/ca.pdf · 2017. 4. 30. · Computer Algebra Lecture Notes, Summer Term 2017 Janko B ohm April 30, 2017