Geometry of Syzygies Edward Pearce Mathematics Institute

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Geometry of Syzygies

Edward Pearce

MA4K9 Research Project

Submitted to The University of Warwick

Mathematics Institute

April, 2016 Geometry of Syzygies

Edward Pearce 1203384

30th May 2016

Contents

1 Introduction 2

2 Free Resolutions 2 2.1 Graded Rings ...... 2 2.2 Construction of a free resolution ...... 3 2.3 Application - The Hilbert Function and Polynomial ...... 4 2.4 Minimal Free Resolutions and Hilbert Syzygy Theorem ...... 6 2.5 Graded Betti Numbers and Betti Diagrams ...... 11

3 Proof of Hilbert's Syzygy Theorem 14 3.1 Monomial Ideals and Simplicial Complexes ...... 14 3.2 Syzygies of Monomial Ideals ...... 18 3.3 Taylor and Koszul complexes ...... 21 3.4 Hilbert's Syzygy Theorem - a strengthening ...... 23

4 Castelnuovo-Mumford Regularity 25 4.1 Denition and rst applications ...... 25 4.2 The Regularity of a Cohen-Macaulay Module ...... 27

5 Toric Ideals 31

6 Eisenbud-Goto Conjecture 33 6.1 A General Regularity Conjecture ...... 33 6.2 Examples using Macaulay 2 ...... 34

A Additional Macaulay2 Code 36

1 1 Introduction

In this project we introduce several important and useful concepts and techniques from commutative and homological algebra which can prove fruitful in the study of projective geometry through means of calculating numerical invariants and convey- ing other algebraic and geometric information about ideals and their corresponding varieties. There is also some interplay with combinatorics along the way. We build our toolkit in order to consider a conjectural connection which states that under particular assumptions on a projective variety we may bound the degree of the minimal homogeneous generators of its associated ideal in terms of geometric data. This may have computational consequences, and is also deeply linked to the world of cohomology. In the nal section we look for examples in a class of polynomial ideals known as toric ideals with the aid of software system Macaulay2.

2 Free Resolutions

2.1 Graded Rings

Denition 2.1. A graded ring R is a ring which splits into a direct sum of abelian M groups R = Rn such that RiRj ⊆ Ri+j for all i,j ≥ 0, and a graded (left) n∈N module M over a graded ring R is a (left) R-module which has a direct sum M decomposition M = Md as abelian groups such that RiMj ⊆ Mi+j for all i,j. d∈Z Note that this gives each graded component Md an R0-module structure.

We say an R-module M is nitely generated if there exist elements m1, m2, . . . , mn ∈ M such that every element m ∈ M can be expressed as an R-linear combination m = r1m1 + r2m2 + ... + rnmn. The example which we shall be considering throughout is the polynomial ring M S = k[x0, . . . , xr] with each variable in degree 1, so S = Sd where Sd is the d∈N k-vector space of homogeneous polynomials of degree d. Denition 2.2. For a graded module M we may dene a new module M(d), called the d-th twist of M, by `shifting' the grading of M by d steps, i.e. M(d) is isomorphic to M as a module and has grading dened by M(d)e = Md+e. Let M,N be graded modules. Suppose f : M → N is a morphism of the underlying modules, then we say f has degree d if f(Mn) ⊂ Nn+d = N(d)n.

2 Many natural maps of graded modules take the grading of one to the grading of the other with a shift of degrees and using our notation we may interpret them as degree preserving maps between one of the modules and the shift of the other. This is convenient for keeping track of graded components and so from now on we use the term graded morphism to refer to a degree 0 morphism of graded modules.

Note that in particular a graded morphism f : M → N restricts to an R0-linear map of graded components f : Md → Nd for each d. For a graded ring R we dene a graded free R-module to be a direct sum of modules of the form R(d), for various d. Notice that R(d)−d = R0, so that R(d) has its generator in degree −d, and not d.

2.2 Construction of a free resolution

M Let M = Md be a nitely generated graded S-module, where S = k[x0, . . . , xr] d∈Z is the polynomial ring in r + 1 variables over a eld k, with all variables of degree 1. Given homogeneous elements mi ∈ M of degrees ai that generate M as an M S-module we may dene a map from the graded free module F0 = S(−ai) i onto M by sending the i-th generator of F0 to mi. Let M1 ⊂ F0 be the kernel of the map φ0 : F0 → M. Then by the Hilbert Basis theorem M1 is also a nitely generated module. The elements of M1 are called syzygies on the generators mi, or simply syzygies of M. Choosing nitely many homogeneous syzygies that generate M1 we may dene a map from a graded free S-module F1 to F0 with image M1. Continuing in this way we can construct a sequence of maps of graded free modules

ϕi ϕ1 ... −→ Fi −→ Fi−1 −→ ... −→ F1 −→ F0 called a free resolution of M. Denition 2.3. Formally, we dene a free resolution of an R-module M to be a complex ϕi ϕ1 F : ... −→ Fi −→ Fi−1 −→ ... −→ F1 −→ F0 of free R-modules such that cokerϕ1 = M and F is exact. We sometimes abuse this notation and say that an exact sequence

ϕi ϕ1 ϕ0 F : ... −→ Fi −→ Fi−1 −→ ... −→ F1 −→ F0 −→ M −→ 0 is a resolution of M. A resolution F is a graded free resolution if R is a graded ring, the Fi are graded free modules, and the maps are homogeneous maps of degree 0 (i.e. graded morphisms). If for some n < ∞ we have Fn+1 = 0, but Fi 6= 0 for 0 ≤ i ≤ n, then we say that F is a nite resolution of length n.

3 Example 2.4. Let f, g be polynomials (homogeneous or not) of positive degree in S, neither of which divides the other. Let n = deg f, m = deg g. Then

(f) K(f) : 0 → S(−n) −→ S gives a graded free resolution of S/(f) since S is an integral domain. Further, if we set h = gcd(f, g), f 0 = f/h, g0 = g/h, and p = deg h then

 g0   0  −f M f g K(f, g) : 0 → S(−(n + m − p)) −−−−−−→ S(−n) S(−m) −−−−−−→ S . gives a graded free resolution of S/(f, g).

Proof. Once again, we have exactness at S(−(n + m − p)) since S is an integral domain, so it remains to show that we have exactness at S(−n) L S(−m). First, g0 the composition f g = fg0 − f 0g = (fg − fg)/h = 0 is zero, so −f 0 g0 Im ⊆ ker f g and we have a complex. −f 0 x Now suppose ∈ ker f g , so xf + yg = 0 and by dividing throughout y by h = gcd(f, g) we nd xf 0 = −yg0. Then as f 0 | −yg0 but gcd(f 0, g0) = 1 we nd f 0 | −y, so may write −y = zf 0 for some z ∈ S. Then xf 0 = −yg0 = zf 0g0, so x zg0 g0 g0 x = zg0 and thus = = (z) ∈ Im . This shows y −zf 0 −f 0 −f 0 exactness at S(−n) L S(−m), hence the above complex is a free resolution.

2.3 Application - The Hilbert Function and Polynomial

Denition 2.5. Let M be a nitely generated graded module over S = k[x0, . . . , xr]. We dene the Hilbert function as the map HM : N → N d 7→ HM (d) = dimk Md for the dimension of the d-th graded component of M as a function of d.

For suciently large d the Hilbert function HM (d) agrees with a polynomial function PM (d), called the Hilbert Polynomial, which has degree at most r. Remark 2.6. By their denitions the Hilbert function and Hilbert polynomial are invariants under graded isomorphism of S-modules, and in fact the coecients of the Hilbert polynomial can give important invariants of the module.

4 For example, if X ⊂ r is a curve, then the Hilbert polynomial of the ho- PK mogeneous coordinate ring S of is given by a linear polynomial SX = I(X) X PM (d) = (deg X)d + (1 − genusX) where the coecients deg(X) and 1 − genus(X) give a topological classication of the embedded curve. [Eis05]

More generally, the degree of the Hilbert polynomial is the dimension of X, and the leading coecient of PM (d) multiplied by (deg PM )! is equal to the degree of X - the number of points in which X meets a general plane of complementary dimension in Pr. [Eis95] Proposition 2.7. If we have a nite free resolution of a nitely generated graded module M we can compute its Hilbert function

Let S = k[x0, . . . , xr]. If M is a graded S-module with nite free resolution

ϕm ϕ1 F : 0 −→ Fm −−→ Fm−1 −→ ... −→ F1 −→ F0 with each a nitely generated free module of the form L , then Fi Fi = j S(−ai,j) the Hilbert function of M is given by Pm i P r+d−ai,j . HM (d) = i=0(−1) j r Proof. Each map in a graded free resolution is degree preserving by denition, so we may pass to an exact sequence of nite dimensional vector spaces by taking the degree d part of each module in the sequence, and using our knowledge of exact sequences of nite dimensional vectors spaces (i.e. repeated application of the rank-nullity theorem), we get

m m X i X i HM (d) = dimK Md = (−1) dimK(Fi)d = (−1) HFi (d), i=0 i=0 so it suces to show that P r+d−ai,j where L . De- HFi (d) = j r Fi = j S(−ai,j) composing Fi as a direct sum, we get M X X HFi (d) = dimK S(−ai,j) = dimK S(−ai,j) = HS(−ai,j )(d), j j j so now it suces to show that r+d−a. Shifting back, we have HS(−a)(d) = r , and so in order to show that HS(−a)(d) = dimK S(−a)d = dimK Sd−a = HS(d − a) r+(d−a) it is enough to show that r+d . HS(−a)(d) = HS(d − a) = r HS(d) = r ∀d We complete the proof by counting the number of monomials of degree d possible

5 in a polynomial ring in r + 1 variables:

r a0 a1 ar X HS(d) = dimK Sd = dimK k[x0, . . . , xr]d = #{x0 x1 . . . xr | ai ∈ Z≥0, ai = d} i=0 r X r + d = #{a ∈ | a = d} = i Z≥0 i r i=0 by a stars and bars argument.

Corollary 2.8. There is a polynomial PM (d) called the Hilbert polynomial of M such that, if M has a free resolution as above, then PM (d) = HM (d) for all d ≥ maxi,j{ai,j} − r.

Proof. If d ≥ maxi,j{ai,j} − r, then r + d − ai,j ≥ 0 ∀i, j, in which case r + d − a (d + r − a )(d + r − a − 1) ... (d − a + 1) i,j = i,j i,j i,j r r! which is a polynomial of degree r in d. Thus in the desired range all of the terms of Pm i P r+d−ai,j become polynomials. HM (d) = i=0(−1) j r Example 2.9. Returning to example 2.4, if f, g are polynomials of positive degree in S = k[x0, . . . , xr], neither of which divides the other, and we set n = deg f, m = deg g, p = deg gcd(f, g), then by proposition 2.7 above r + d r + d − n H (d) = − , and S/(f) r r r + d r + d − n r + d − m r + d − (n + m − p) H (d) = − − + S/(f,g) r r r r which agree with their corresponding Hilbert polynomials for all d ≥ n − r, and for all d ≥ n+m−p−r respectively. A sharp bound on when the Hilbert function and polynomial coincide is given in theorem 4.2.

2.4 Minimal Free Resolutions and Hilbert Syzygy Theorem

Theorem 2.10 (Hilbert Syzygy Theorem). Any nitely generated graded S-module has a nite graded free resolution

ϕm ϕ1 F : 0 −→ Fm −−→ Fm−1 −→ ... −→ F1 −→ F0 where we may take m ≤ r + 1 the number of variables in S.

6 Each nitely generated graded S-module has a minimal free resolution which is unique up to isomorphism. The degrees of the generators (i.e. the ai,j) of its free modules not only yield the Hilbert function, as is true of any nite free resolution, but form a ner invariant. In this section we give a careful statement of the denition of minimality, and of the uniqueness theorem. Naively, minimal free resolutions can be described as follows: Given a nitely generated graded module M, choose a minimal set of homogeneous generators mi. Map a graded free module F0 onto M by sending a basis for F0 to the set of mi. Let M1 be the kernel of the map F0 → M, and repeat the procedure, starting with a minimal system of homogeneous generators for M1. Most of the applications of minimal free resolutions are based on a property that characterises them in a dierent way, which we will adopt as the formal denition.

Denition 2.11. Let m denote the homogeneous maximal ideal m = (x0, . . . , xr) ⊂ S = k[x0, . . . , xr]. A complex of graded S-modules

δi F : ... −→ Fi −→ Fi−1 −→ ...

is called minimal if for each i the image of δi is contained in mFi−1. That is, Imδi ⊂ mFi−1 for all i. Informally, we may say a complex of free modules is minimal if its dierential is represented by matrices with entries in the maximal ideal m. In particular, no nonzero constant terms are allowed. The relation between the above denition and the naive idea of a minimal free resolution is is a consequence of Nakayama's lemma.

Lemma 2.12 (Technical Lemma). Let R be a commutative ring with identity 1R, I an ideal of , and a (left) -module, then N ∼ are isomorphic R M R M R R/I = M/IM as R-modules.

Proof. The map M × R/I → M/IM, (m, r + I) 7→ r.m + IM is R-bilinear by the R-module structure on M/IM, so by the universal property of tensor products we get an -linear map N given by , R ϕ : M R R/I → M/IM ϕ : m⊗r+I 7→ r.m+IM which has inverse N . In particular, ψ : M/IM → M R R/I, m + IM 7→ m ⊗ 1R + I ψ ◦ ϕ(m ⊗ r + I) = ψ(r.m + IM) = r.m ⊗ 1R + I = m ⊗ r.1R + I = m ⊗ r + I as r ∈ R, and ϕ ◦ ψ(m + IM) = ϕ(m ⊗ 1R + I) = 1R.m + IM = m + IM as M is unital. Thus N =∼ is an isomorphism of -modules. ϕ : M R R/I −→ M/IM R Remark. We use the above lemma most commonly in the case of the polynomial ring S = k[x0, . . . , xr] and its maximal homogeneous ideal m = (x0, . . . , xr) where is a graded -module. Then N ∼ . M S M S S/m = M/mM

7 Lemma 2.13 (Nakayama's Lemma). Suppose M is a nitely generated graded S- module and m1, m2, . . . , mn ∈ M generate M/mM. Then m1, m2, . . . , mn generate M.

¯ P Proof. Let M := M/ S · mi. Consider

¯ ¯ ∼ X M/mM = (M/mM)/ S · (mi + mM) = 0 as we assume the mi generate M/mM, or more precisely that their images mi+mM are the generators. Therefore mM¯ = M¯ . Suppose for contradiction that M¯ 6= 0, then since M¯ is nitely generated (as M is), we may nd a nonzero element of least degree in M¯ . However, this element could not be in mM¯ since everything here has degree at least one greater. Thus we conclude that M¯ = 0, and so M = P S · m is generated by the mi. Corollary 2.14. A graded free resolution

δi F : ... −→ Fi −→ Fi−1 −→ ...

is minimal as a complex if and only if for each i the map δi takes a basis of Fi to a minimal set of generators of the image of δi (which is equal to the kernel of δi−1 by exactness). That is to say our denition of minimal coincides with the construction of choosing a minimal set of generators at each step.

δi+1 δi Proof. Consider the right exact sequence Fi+1 −−→ Fi −→ Imδi → 0. The complex F is minimal (i.e. Imδi ⊂ mFi−1 for all i) if and only if for each i the induced map ¯ δi+1 : Fi+1/mFi+1 → Fi/mFi is zero. This is because δi+1(mFi+1) = mδi+1(Fi+1) ⊂ mFi already so the map is well dened and for x¯ ∈ Fi+1/mFi+1 with representative ¯ x ∈ Fi+1 we have that δi+1(¯x) = 0 ∈ Fi+1/mFi+1 ⇔ δi+1(x) ∈ mFi+1. ¯ So Imδi ⊂ mFi−1 ∀i i the map δi+1 : Fi+1/mFi+1 → Fi/mFi is zero ∀i, and this ˜ holds if and only if the induced map δi : Fi/mFi → Imδi/mImδi is an isomorphism, ¯ δi+1 δ˜i by considering the induced right exact sequence Fi+1/mFi+1 −−→ Fi/mFi −→ Imδi/mImδi → 0. Finally, by Nakayama's lemma, this occurs if and only if a basis of Fi maps to a minimal set of generators of Imδi (where the minimality of ˜ the set of generators corresponds to injectivity of δi). Theorem 2.15 (Uniqueness of minimal free resolution). Let M be a nitely generated graded S-module . If F and G are minimal graded free resolutions of M, then there is a graded isomorphism of complexes F → G inducing the identity map on M. Further, any free resolution of M contains the minimal free resolution as a direct summand.

8 Proof. For uniqueness, see [Eis95, Theorem 20.2]. We can construct a minimal free resolution from any resolution, proving the second statement along the way. If F is a nonminimal complex of free modules, a matrix representing some dierential of F must contain a nonzero element of degree 0.

This corresponds to a free basis element of some Fi that maps to an element of Fi−1 not contained in mFi−1, so nonzero in the quotient Fi−1/mFi−1. Then by Nakayama's lemma, this element may be taken as a basis element of Fi−1. Thus we have found a subcomplex of F of the form

G : 0 −→ S(−a) −→c S(−a) −→ 0 for a nonzero scalar c (such a thing is called a trivial complex) embedded in F in such a way that F/G is again a free complex. As G has no homology at all (the sequence is exact as multiplication by a nonzero scalar is invertible, so an isomorphism), the long exact sequence in homology corresponding to the short exact sequence of complexes 0 → G → F → F/G → 0 shows that the homology of F/G is the same as that of F. In particular, if F is a free resolution of M, then so is F/G. Continuing in this way by quotienting out by trivial subcomplexes we eventually reach a minimal complex. If F was a resolution of M, we have constructed a minimal free resolution. Example 2.16. The nonminimal free resolution

 y −1       −x −1    M 0 1 x y x + y F : 0 → S(−2) S(−1) −−−−−−−−−−→ S(−1)3 −−−−−−−−−−−−→ S can be realised as a direct sum of the complex K(x, y) (as in example 2.4) and a trivial complex G : 0 −→ S(−1) −→1 S(−1) −→ 0 via a change of basis P =  1 0 1   0 1 1  at S(−1)3. That is 0 0 1

F : 0 / S(−2) L S(−1) / S(−1)3 / S

id P id K(x, y) L G : 0 / S(−2) L S(−1) / S(−1)3 / S Both F and K(x, y) are free resolutions of S/(x, y), but K(x, y) is minimal and F is not.

9 For us, the most important aspect of uniqueness of minimal free resolutions is that if F : ... → F1 → F0 is the minimal free resolution of a nitely generated graded S-module M, the number of generators of each degree required for the free modules Fi is determined by M only. The easiest way to state a precise result is to use the functor Tor; this useful tool from homological algebra is introduced in, for example, [Eis95, Section 6.1].

δi Proposition 2.17. If F : ... −→ Fi −→ Fi−1 −→ ... is the minimal free resolution of a nitely generated graded S-module M and K is the residue eld S/m, then any minimal set of homogeneous generators of contains exactly S Fi dimK Tori (K,M)j ∞ M βi,j generators of degree j. Thus we may express Fi = S(−j) where βi,j := j=0 S are called the graded Betti numbers of . dimK Tori (K,M)j M Proof. By denition S is the -vector space which is the degree com- Tori (K,M)j K j ponent of the graded vector space that is the -th homology of the complex N . i K S F Claim 1: As F is minimal, the maps in the complex N are all zero. K S F Proof of claim 1: Consider a general map N N in the idK ⊗ δi : K S Fi → K S Fi−1 complex N . As the simple tensors generate N we need only consider K S F K S Fi their images. Recalling the identication K = S/m, for a simple tensor (a+m)⊗f ∈ N we have . By minimality we have K S Fi idK ⊗ δi :(a + m) ⊗ f 7→ (a + m) ⊗ δi(f) 0 0 δi(f) ∈ mFi−1, so we may write δi(f) = tf for some t ∈ m ⊂ S, f ∈ Fi−1, and then

0 idK ⊗ δi((a + m) ⊗ f) = (a + m) ⊗ δi(f) = (a + m) ⊗ tf 0 0 O = (ta + m) ⊗ f = (0 + m) ⊗ f = 0 ∈ K Fi−1. S Thus the maps in N are all zero by the minimality of F. Equivalently, this K S F might more readily be seen through the identication N N ∼ K S Fi = S/m S Fi = Fi/mFi for each i, by lemma 2.12. Therefore taking homology gives S N ∼ , and then Tori (K,M) = K S Fi = Fi/mFi by Nakayama's lemma, S gives the number of dimK Tori (K,M)j = dimK(Fi/mFi)j degree j generators that Fi requires. Corollary 2.18. If M is a nitely generated graded S-module, then the projective dimension of M is equal to the length of the minimal free resolution of M.

Proof. The projective dimension pd(M) of M is dened to be the minimal length of a projective resolution of M (that is, a resolution by projective modules). The

10 minimal free resolution of M is a projective resolution, since all free modules are projective, so pd(M) ≤ the length of the minimal free resolution of M. To show that the length of the minimal free resolution is at most the projective dimension, note that S when is greater than the projective dimension of Tori (K,M) = 0 i M (Tor is actually dened in terms of projective resolutions). Then by the above proposition, if S then has no generators of any degree for the Tori (K,M) = 0 Fi minimal free resolution, so the minimal free resolution will have length less than i too. Fact. The Quillen-Suslin theorem states that every nitely generated projective module over a polynomial ring is free. [Lang 2002]

Remark. If we allow the variables to have dierent degrees, HM (t) becomes, for large t, a polynomial with coecients that are periodic in t. (See example 5.4)

2.5 Graded Betti Numbers and Betti Diagrams

We have seen in proposition 2.7 that the numerical invariants associated to free resolutions suce to describe Hilbert functions, and will soon show that the numerical invariants of minimal free resolutions contain more information. As we will be dealing with them a lot we introduce a compact way to display them, called a Betti diagram. Denition 2.19. Suppose that F is a free complex

F : 0 −→ Fs −→ ... −→ Fi → ... → F0

∞ M βi,j where Fi = S(−j) , that is to say that Fi requires βi,j minimal generators of j=0 degree j. The Betti diagram of F has the form: 0 1 ... s

i β0,i β1,i+1 ... βs,i+s i + 1 β0,i+1 β1,i+2 ... βs,i+s+1 ...... j β0,j β1,j+1 ... βs,j+s It consists of a table with s+1 columns, labeled 0, 1, . . . , s, corresponding to the free modules F0,...,Fs. It has rows labeled with consecutive integers corresponding to the degrees. We sometimes omit the row and column labels when they are clear from context. The i-th column species the degrees of the generators of Fi. Thus, for example, the row labels at the left of the diagram correspond to the possible

11 degrees of a generator of F0. For clarity we sometimes replace a 0 in the diagram by a dash - and an indenite value by an asterisk *. Note that the entry in the j-th row of the i-th column is βi,i+j rather than βi,j which will instead be found in position (j − i, i). This choice will be explained below. If F is the minimal free resolution of a module M, we refer to the Betti diagram of F as the Betti diagram of M, and the βi,j of F are called the graded Betti numbers of M, sometimes denoted as βi,j(M). Recall from proposition 2.17 when F is minimal and the graded K-vector space S is the homology of the complex N , that since F is minimal, the Tori (M, K) F F K dierentials in this complex are zero, so S . In other words, βi,j = dimK Tori (M, K)j ∞ M βi,j when F is minimal its free modules are of the form Fi = S(−j) where βi,j = j=0 S . Note that the functor is symmetric over commutative rings dimK Tori (M, K)j Tor such as S. As the graded Betti numbers of a nitely generated graded S-module M are dened in terms of the minimal free resolution of M, we see that they are invariant under graded isomorphisms of M.

For example, the number β0,j is the number of elements of degree j required among the minimal generators of M. We will often consider the case where M is the homogeneous coordinate ring SX of a (nonempty) projective variety X. As an S-module, SX is then generated by the element 1 ∈ SX , so we will have β0,0 = 1 and β0,j = 0 for j 6= 1.

On the other hand, β1,j is the number of independent forms (polynomials) of degree j needed to generate the ideal IX of X. If SX is not the zero ring (that is, X 6= ∅), there are no elements of the ideal of X in degree 0, so β1,0 = 0. This is the case i = d = 0 of the following: Proposition 2.20. Let {βi,j} be the graded Betti numbers of a nitely generated S-module. If for a given i there exists d such that βi,j = 0 for all j < d, then βi+1,j+1 = 0 for all j < d. That is to say, if Fi has no minimal generators of degree less than d, then Fi+1 has no minimal generators of degree less than d + 1 by the minimality of the resolution.

δ1 Proof. Suppose that the minimal free resolution is F : ... −→ F1 −→ F0. By corollary 2.14, any generator of Fi+1 must map to a nonzero element of the same degree in mFi. To say that βi,j = 0 for all j < d means that all generators - and thus all nonzero elements - of Fi have degree ≥ d. Thus all nonzero elements of mFi have degree ≥ d + 1, so Fi+1 can have generators only in degree ≥ d + 1 and

12 βi+1,j+1 = 0 for j < d as claimed.

The above proposition gives a rst hint as to why it is convenient to write the Betti diagram in the form we chose to, with βi,i+j in the j-th row of the i-th column: it says that if the i-th column of the Betti diagram has zeros above the j-th row, then the (i + 1)-st column also has zeros above the j-th row. This also allows a more compact display of Betti numbers than if we had written βi,j in the j-th row of the i-th column. A deeper reason for this choice will be made clear in the description of the Castelnuovo-Mumford regularity in chapter 5. The formula for the Hilbert function given previously in proposition 2.7 has a convenient expression in terms of graded Betti numbers (since the graded Betti numbers store information more concisely than the previous ai,j notation). Corollary 2.21. If {βi,j} are the graded Betti numbers of a nitely generated graded -module , then the alternating sums P∞ i determine the S M Bj := i=0(−1) βi,j Hilbert function of M via the formula ∞ X r + d − j dim M = H (d) = B . k d M j r j=0

Moreover, the values of the Bj can be deduced inductively from the Hilbert function HM (d) via the formula X r + j − k B = H (j) − B . j M k r k:k

Note that the alternating sums Bj are actually nite with the number of terms in the sum being bounded above by the length of the minimal free resolution of M.

Proof. Recall from proposition 2.7 we know that if L then Fi = j S(−ai,j) HM (d) = Pm i P r+d−ai,j . i=0(−1) j r ∞ M βi,j By grouping generators of the same degree we may write Fi = S(−j) and j=0 then the sum becomes P r+d−ai,j P∞ r+d−j for each , hence j r = j=0 βi,j r i m ∞ ∞ X X r + d − ai,j X X r + d − j H (d) = (−1)i = (−1)i β . M r i,j r i=0 j i=0 j=0

Conversely, to compute the Bj from the Hilbert function HM (d) we proceed as follows: As M is nitely generated there is a number such that M = Md j0 ∈ Z d∈Z 13 for , i.e. take where n is a minimal HM (d) = 0 d < j0 j0 = mini=1,...n deg mi {mi}i=1 M set of homogeneous generators for M, so M = Md. It follows that β0,j = 0 for

d≥j0 all j < j0, and from the above proposition 2.20 it further follows that if j < j0 then for all . Thus P∞ i for all . Initially, we βi,j = 0 i Bj = i=0(−1) βi,j = 0 j < j0 have and inductively, assuming we know the values of for , Bj0 = HM (j0) Bk k < j then as r+j−k when , only the values of with enter into the r = 0 j < k Bk k ≤ j formula for P∞ r+j−k Pj r+j−k and knowing we HM (j) = k=0 Bj r = k=0 Bj r , HM (j) can solve for Bj to obtain the required formula. Note that, conveniently, Bj occurs with coecient r in the sum. r = 1

3 Proof of Hilbert's Syzygy Theorem

In this chapter we introduce a fundamental construction of resolutions based on simplicial complexes. This construction gives free resolutions of monomial ideals, but does not always yield minimal resolutions. It includes Koszul complexes, which we use to establish basic bounds on syzygies of all modules, including the Hilbert Syzygy theorem.

3.1 Monomial Ideals and Simplicial Complexes

We now introduce a beautiful method of writing down graded free resolutions of monomial ideals due to Bayer, Peeva, and Sturmfels. So far we have used Z- gradings of modules only, but we can think of the polynomial ring S = k[x0, . . . , xr] as r+1-graded, with a monomial a0 ar having degree r+1, Z x0 . . . xr (a0, . . . , ar) ∈ Z and the free resolutions we write down will also be Zr+1-graded. We begin by introducing the basics of the theory of nite simplicial complexes. Denition 3.1. A nite simplicial complex ∆ is a nite set N , called the set of vertices (or nodes) of ∆, and a collection F of subsets of N , called the faces of ∆, such that if A ∈ F is a face and B ⊂ A, then B is also in F. Maximal faces are called facets.

A simplex is a simplicial complex in which every subset of N is a face, i.e. F = P(N ). For any vertex set N we may form the void simplicial complex, which has no faces at all, i.e F = ∅. However, if ∆ has any faces at all, then the empty set is necessarily a face of ∆, that is ∅ ∈ F, so F 6= ∅.

14 By contrast, we call the simplicial complex whose only face is ∅ the irrelevant simplicial complex on N , i.e. where F = {∅}. Remark (Stanley-Reisner correspondence). We may associate to any simplicial complex ∆ with vertex set N = {x0, . . . , xr} the squarefree monomial ideal in S = k[x0, . . . , xr] whose generators are the monomials with support equal to a non-face of ∆. [Miller and Sturmfels 2005, Chapter 1] Under this correspondence, the irrelevant simplicial complex corresponds to the irrelevant ideal (x0, . . . , xr), while the void simplicial complex corresponds to the unit ideal (1) ⊂ S. Remark 3.2 (Geometric realisation). Suppose ∆ = (N , F) is a simplicial complex, we consider L . Any simplicial complex has a geometric realisation, that is, #N R ∆ a topological space that is a union of simplexes corresponding to the faces of ∆. It may be constructed by realising the set of vertices of ∆ as a linearly independent set in a suciently large real vector space, and realising each face of ∆ as the convex hull of its vertex points; the realisation of ∆ is then the union of these faces.

An orientation of a simplicial complex consists of an ordering of the vertices of ∆. Thus a simplicial complex may have many orientations - this is not the same as an orientation of the underlying topological space. Denition 3.3 (Labeling by monomials). We will say that ∆ is labeled (by monomials of S) if there is a monomial of S associated to each vertex of ∆. We then label each face A of ∆ by the least common multiple of the labels of the vertices in A, so mA := lcma∈A(ma). By convention, the label of the empty face is m∅ = 1. Let ∆ be an oriented labeled simplicial complex, and write I ⊂ S for the ideal αj generated by the monomials mj = x labeling the vertices of ∆. That is, I = (mj : j ∈ N ) ⊂ S. We will associate to ∆ a graded complex of free S-modules δ δ C(∆) = C(∆; S): ... −→ Fi −→ Fi−1 −→ ... −→ F1 −→ F0 where Fi is the free S-module whose basis consists of the set of faces of ∆ having i elements, so F = L S(−a ). The complex will thus be nite with i fj ∈F,|fj |=i i,j Fi = 0 for i > |N |. The complex C(∆) is sometimes a resolution of S/I. The dierential δ for C(∆) is given by its action on the basis elements, which correspond to faces A ∈ F. X mA δ(A) := (−1)pos(n,A) (A \ n) m n∈A A\n where pos(n, A), the position of the vertex n in A, is the number of elements preceding n in the ordering of A, and A \ n denotes the face obtained from A by removing n. So for n ∈ A, pos(n, A) may start at zero and be at most #A − 1.

15 If ∆ is not void (i.e. F 6= ∅) then F0 = S, with generator corresponding to the face of ∆ which is the empty set. Further the generators of F1 correspond to the vertices of ∆, and each of these generators maps by δ to its labeling monomial (remembering the convention that m∅ = 1). That is δ(n) = mn · ∅ for vertices n δ of ∆. Therefore H0(C(∆)) = coker(F1 −→ S) = S/I since imδ1 = (mn : n ∈ N∆) = I∆. We set the degree of the basis element corresponding to a face A equal to the αA exponent vector of the monomial that is the label of A, so if mA = x , then #N . deg A := αA ∈ N Claim. With respect to this grading, C(∆) is a Zr+1-graded free complex.

Proof. First, we prove that δ2 = 0 in order to show that C(∆) is a complex. Let A be a face of ∆ and i, j ∈ A vertices. If j precedes i in A in the orientation on ∆ then pos(j, A \ i) = pos(j, A), and if i precedes j in the ordering of A then pos(j, A \ i) = pos(j, A) − 1. Thus

X mA δ (δA) = (−1)pos(i,A) δ(A \ i) m i∈A A\i   X pos(i,A) mA X pos(j,A\i) mA\i = (−1)  (−1) A \ i, j mA\i mA\i,j i∈A j∈A\i X mA = (−1)pos(i,A)+pos(j,A) A \ i, j m i,j∈A,j

For the dierential to be degree-preserving we require m deg A = deg A A \ i mA\i for each vertex i ∈ A. As we dened the degree of a face A in terms of the degree of its labeling monomial mA we nd that m m deg A A \ i = deg A + deg A \ i mA\i mA\i

= deg mA − deg mA\i + deg mA\i = deg mA = deg A as required.

16 If we take S = K and label all the vertices of ∆ with 1 ∈ K then C(∆; K) is, up to a shift in homological degree, the usual reduced chain complex of ∆ with coecients in K. [Hatcher 2001]

δ δ C(∆; K): ... −→ Fi −→ Fi−1 −→ ... −→ F1 −→ F0, ⊕#{i-faces of ∆} X pos(n,A) Fi = K , δA = (−1) A \ n. n∈A The homology of is written and is called the reduced homology C(∆; K) Hi(∆; K) of ∆ with coecients in K. The shift in homological degree comes about as follows: the homological degree of a simplex in C(∆) is the number of vertices in the simplex, which is one more than the dimension of the simplex, so that is the -st homology of Hi(∆; K) (i + 1) C(∆; K). If for all (that is, for all ), then we say that is -acyclic, Hi(∆; K) = 0 i ≥ −1 i ∆ K and since is a free module over , this is the same as saying that S K Hi(∆; S) = 0 for all i ≥ −1. The homology and the homology are independent of the Hi(∆; K) Hi(C(∆; S)) orientation of ∆ - in fact they depend only on the homotopy type of the geometric realisation of ∆ and the ring K or S. Thus we often ignore orientations. Roughly speaking we may say that the complex C(∆; S), for an arbitrary labeling, is obtained by extending scalars from K to S and `homogenizing' the formula for the dierential of with respect to the degrees of the generators of the C(∆; K) Fi dened for the S-labeling of ∆. Example 3.4. Suppose that ∆ is the labeled simplicial complex

x0x1x2 x0x1x2 x0x1 x0x2 x1x2 with orientation obtained by ordering the vertices from left to right. The complex C(∆) is

  −x2 0      x1 −x1    0 x0 x0x1 x0x2 x1x2 0 → S(−3)2 −−−−−−−−−−−→ S(−2)3 −−−−−−−−−−−−−−−−→ S.

This complex is represented by the Betti diagram

17 0 1 2 0 1 - - 1 - 3 2

We will soon show that the complex C(∆) is a minimal free resolution of the S-module S/(x0x1, x0x2, x1x2). Denition 3.5. Let ∆ be a labeled simplicial complex. If m is any monomial, we write ∆m for the subcomplex of ∆ consisting of those faces of ∆ whose labels divide m.

For example, if m is not divisible by any of the vertex labels , then ∆m is the empty simplicial complex, with no vertices and the single face ∅. On the other hand, if m is divisible by all the labels of ∆, then ∆m = ∆. 0 Moreover, is equal to 0 for some subset of the vertex set of ∆m ∆lcm{mi|i∈N } N 0 ∆. More precisely, N = {i ∈ N : mi|m}, so by the denition of least common 0 multiple, divides , and so 0 . Conversely, if lcm{mi|i ∈ N } m ∆lcm{mi|i∈N } ⊂ ∆m A ∈ ∆m is a face, then mA|m by denition of ∆m, then as mA = lcm{mi|i ∈ 0 0 A} for all i ∈ A we have mi|mA|m which means i ∈ N , so A ⊂ N and thus 0 and therefore 0 . mA|lcm{mi|i ∈ N } A ∈ ∆lcm{mi|i∈N } Denition 3.6. A full subcomplex of ∆ is a subcomplex consisting of all the faces of ∆ that involve a particular set of vertices. Note that we have shown that all the subcomplexes ∆m ⊂ ∆ are full in ∆.

3.2 Syzygies of Monomial Ideals

Theorem 3.7 (Bayer, Peeva, and Sturmfels). Let ∆ be a simplicial complex labeled by monomials m1, . . . , mt ∈ S, and let I = (m1, . . . , mt) ⊂ S be the ideal in S generated by the vertex labels. The complex C(∆) = C(∆; S) is a free resolution of if and only if the reduced simplicial homology vanishes for every S/I Hi(∆m; K) monomial m and every i ≥ 0. Moreover, C(∆) is a minimal free resolution if and 0 only if mA 6= mA0 for every proper subface A of a face A for all faces A ∈ ∆. By the above remarks, using this theorem we can determine whether C(∆) is a resolution just by checking the vanishing condition for monomials that are least common multiples of sets of vertex labels.

δ δ Proof. Let C(∆) be the complex C(∆) : ... −→ Fi −→ Fi−1 −→ ... −→ F1 −→ F0. Then S/I = coker(δ1 : F1 → F0) = F0/imδ1 = S/imδ1 since imδ1 = I. We will identify the homology of C(∆) at Fi with a direct sum of copies of vector spaces and then the complex is a resolution if and only if it is exact if Hi(∆m; K) C(∆)

18 and only if its homology vanishes, i.e. Hi(C(∆; S)) = 0, which will be if and only if all . The identication is as follows: Hi(∆m; K) = 0 For each α ∈ Zr+1 we will compute the homology of the complex of vector spaces δ δ C(∆)α : ... −→ (Fi)α −→ (Fi−1)α −→ ... −→ (F0)α, formed from the degree-α components of each free module Fi in C(∆). If any of the components of α are negative then C(∆)α = 0 as a complex, so the homology vanishes in this degree. That is to say, we may suppose α ∈ Nr+1 as we are dealing with monomials and an r+1-grading. Now x some α α0 αr . For each face of , N m = x = x0 . . . xr ∈ S A ∆ the complex C(∆) has a rank-one free summand S.A which, as a vector space, has a -basis is a monomial . The degree of is the exponent of K {n.A|n ∈ S } n.A n.mA where mA is the label of the face A. Thus for the degree α part of S.A we have

( xα . .A if mA|m K mA (S.A)α = 0 otherwise It follows that the complex has a -basis corresponding bijectively to the C(∆)α K α faces of ∆m where m = x .

Using this correspondence we identify the terms of the complex C(∆)α with the terms of the reduced chain complex of having coecients in (up to a shift ∆m K in homological degree as for the case where the vertex labels are all 1). Further the dierentials of the complexes agree:

δ For C(∆)α : ... −→ (Fi)α −→ (Fi−1)α −→ ... we have ! X m X m δ βA A = βA δA mA mA A∈∆m,#A=i A∈∆m,#A=i X m X pos(n,A) mA = βA (−1) A \ n mA mA\n A∈∆m,#A=i n∈A X pos(n,A) m = (−1) βA A \ n mA\n A∈∆m,#A=i,n∈A and correspondingly in we have C(∆m; K) ! X X δ βAA = βAδA

A∈∆m,#A=i A∈∆m,#A=i X X pos(n,A) = βA (−1) A \ n

A∈∆m,#A=i n∈A X pos(n,A) = (−1) βAA \ n

A∈∆m,#A=i,n∈A

19 Thus, having identied r+1 with the reduced chain complex of C(∆)α, α ∈ N ∆m where m = xα, we see that the complex C(∆) is a resolution of S/I if and only if for all , as required for the rst statement. Hi(∆m; K) = 0 i ≥ 0 δ In summary, C(∆; S): ... −→ Fi −→ Fi−1 −→ ... is a resolution of S/I i it is exact at each Fi, ∀i ≥ 1 i the homology Hi(C(∆; S)) = 0, ∀i ≥ 0 i the homology of vector spaces for all graded components r+1, Hi(C(∆)α) = 0, ∀i ≥ 0 C(∆)α, α ∈ N since the degree of the dierential is zero so preserves graded pieces and restricts to an exact sequence of graded pieces/vector spaces; and ∼ Hi(C(∆; S)α) = Hi(∆m; K) where m = xα. For minimality, note that if A is an (i + 1)-face and A0 is an i-face of ∆, then the component of the dierential of C(∆) that maps S.A to S.A0 is zero unless A0 ⊂ A, in which case it is ± mA by denition. Thus C(∆) is minimal if and only mA0 0 if mA 6= mA0 for all A ⊂ A, as required.

Example 3.8. We continue with the ideal (x0x1, x0x2, x1x2) as above in example 3.4. For the labeled simplicial complex ∆

x0x1x2 x0x1x2 x0x1 x0x2 x1x2

0 the distinct subcomplexes ∆ of the form ∆m are the empty complex ∆1, the complexes ∆ , ∆ , ∆ , each of which consists of a single point, and the x0x1 x0x2 x1x2 whole complex ∆ itself. As all the subcomplexes of ∆ of the form ∆m are contractible, they have no higher reduced homology (with any coecients) than H0, and thus we can conclude that the complex C(∆) is the minimal free resolution of S/(x0x1, x0x2, x1x2). Example 3.9. The ideal I = (x3, x2y, x2z, z3) ⊂ S = K[x, y, z] has minimal free resolution C(∆) where ∆ is the labeled simplicial complex

x2y x2yz x2z3 x3y x3yz x2z z3 x3z x3

The distinct subcomplexes of the form ∆m are the empty complex ∆1, the vertices ∆x3 , ∆x2y, ∆x2z, ∆z3 , the edges ∆x3y, ∆x3z, ∆x2yz, ∆x2z3 , two paths in ∆ described

20 by ∆x3z3 and ∆x2yz3 , the 2-face ∆x3yz, and the whole complex ∆ = ∆x3yz3 , all of which are contractible. The Betti diagram for S/I is 0 1 2 3 0 1 - - - 1 - - - - 2 - 4 3 1 3 - - 1 - Then by corollary 2.21 we have Hilbert function

d + 2 d − 1 d − 2 H (d) = − 4 + 3 S/I 2 2 2 and Hilbert polynomial PS/I (d) ≡ 6. By directly computing small values of the Hilbert function we nd that in this case the bound given in corollary 2.8 is not sharp. That is, the Hilbert function and polynomial coincide for all d ≥ 2 where 2 < 3 = maxi,j{ai,j} − r. Example 3.10. Noting that any full subcomplex of a simplex is itself a simplex, and the above examples, we come to an idea which gives a result rst proved (in a dierent way) by Diana Taylor [Eis95].

3.3 Taylor and Koszul complexes

Denition 3.11. Let I = (m1, . . . , mn) ⊂ S be any monomial ideal, and let ∆ be a simplex with n vertices, labeled m1, . . . , mn. The complex C(∆), called the Taylor complex of m1, . . . , mn is a free resolution of S/I. This is because a simplex is contractible, so has no higher reduced homology. Example 3.12. The Taylor complex is rarely minimal. For instance, taking

(m1, m2, m3) = (x0x1, x0x2, x1x2) as before, we get a Taylor complex

    1 0 −x2 −x2         −1 −x1 0 x1      1 x0 x0 0 x0x1 x0x2 x1x2 0 → S(−3) −−−−→ S(−3)3 −−−−−−−−−−−−−−→ S(−2)3 −−−−−−−−−−−−−−→ S

This Taylor complex is a nonminimal resolution with Betti diagram: 0 1 2 3 0 1 - - 1 1 - 3 3 -

21  0 1 1  Further, via the change of basis P =  1 1 0  at S(−3)3 we may decompose 0 −1 0 the above resolution into a direct sum of the resolution seen in example 3.4 and a trivial complex.

Denition 3.13. We may dene the Koszul complex K(x0, . . . , xr) of x0, . . . , xr to be the Taylor complex in the special case where the mi = xi are variables. The smallest cases were exhibited in example 2.4. By theorem 3.7 the Koszul complex is a minimal free resolution of the residue class eld since each K = S/(x0, . . . , xr) face has unique label which is the product (as least common multiple) of its vertex labels.

Example. K(x0, x1, x2):

    x0 0 x2 −x1          x1   −x2 0 x0      x2 x1 −x0 0 x0 x1 x2 0 → S(−3) −−−−−→ S3(−2) −−−−−−−−−−−−−−−−→ S3(−1) −−−−−−−−−−−→ S

We can replace the variables x0, . . . , xr by any polynomials f0, . . . , fr to obtain a complex which we will write as K(f0, . . . , fr), the Koszul complex of the sequence . In fact, since the dierentials have only -coecients, we could even f0, . . . , fr Z take the fi to be elements of an arbitrary commutative ring. Under nice circum- stances, for example when the fi are homogeneous elements of positive degree in a graded ring, this complex is a resolution if and only if the fi form a regular sequence. [Eis95, Theorem 17.6]

Example 3.14. By theorem 3.7, the Koszul complex K(x0, . . . , xn−1) of x0, . . . , xn−1 in S = k[x0, . . . , xr], n ≤ r+1 is a minimal free resolution of SX = S/(x0, . . . , xn−1) where the associated variety X ⊂ r is a complete intersection of hyperplanes. PK Each i-face of this simplex on n-vertices has (unique) label which is the product of its vertex labels, so is of degree , and there are n such -faces in a simplex on i i i n vertices. Therefore

( n n if ( ) i j = i Fi = S(−i) i , βi,j = 0 otherwise and we have Betti diagram: 0 ... i ... n n 0 1 ... i ... 1

22 and in the notation of corollary 2.21 we have

∞ X n B = (−1)iβ = (−1)j , j i,j j i=0 thus we may deduce the combinatorial formula

∞ n X r + d − j X nr + d − j r + d − n H (d) = B = (−1)j = SX j r j r r − n j=0 j=0 ∼ ∼ by identifying SX = S/(x0, . . . , xn−1) = k[xn, . . . , xr] = k[x0, . . . , xr−n] which may otherwise follow inductively from the combinatorial relation n n−1 n−1. k = k + k−1 Note that is given by the associated polynomial for all . HSX (d) d ≥ 0

3.4 Hilbert's Syzygy Theorem - a strengthening

We can use the Koszul complex and theorem 3.7 to prove a sharpening of Hilbert's Syzygy theorem (theorem 2.10), which is the vanishing statement in the following proposition. We also get an alternate way to compute the graded Betti numbers.

Proposition 3.15. Let M be a nitely generated graded module over S = k[x0, . . . , xr]. The graded Betti number βi,j(M) is the dimension of the homology, at the term NVi r+1, of the complex: Mj−i K

r+1 O ^ r+1 0 → Mj−(r+1) K → ... i+1 i−1 O ^ r+1 NVi r+1 O ^ r+1 → Mj−i−1 K → Mj−i K → Mj−i+1 K → 0 O ^ r+1 ... → Mj K → 0

In particular, we have r+1, so if . βi,j(M) ≤ HM (j − i) i βi,j(M) = 0 i > r + 1

Proof. By proposition 2.17 and the symmetry of the Tor operator, βi,j(M) = . Then as is the minimal free dimK Tori(K,M)j = dimK Tori(M, K)j K(x0, . . . , xr) resolution of we may compute S as the degree- K = S/(x0, . . . , xr) Tori (M, K)j j part of the graded K-vector space which is the i-th homology of the complex N . Now is the free -module with basis con- M S K(x0, . . . , xr) (K(x0, . . . , xr))i S sisting of faces with i elements of the simplex with r + 1 vertices, labeled by x0, . . . , xr. Faces (i.e. subsets) of {x0, . . . , xr} of size i can be identied with multi-indices on of length , and so with tuples {0, . . . , r} i ek1 ∧ ... ∧ eki , 0 ≤ k1 <

23 . . . < ki ≤ r. So the basis of (K(x0, . . . , xr))i by i-faces corresponds to a basis of r+1 V Sr+1(−i) = S(−i)( i ). Thus we need the homology at the term

! i i O O ^ r+1 O ^ r+1 M K(x0, . . . , xr) = M S (−i) = M K (−i). S i S K

L Decomposing M into its homogeneous components M = Mk, we see that the degree-j part of M N Vi r+1(−i) is K K

i ! i O ^ r+1 O ^ r+1 M K (−i) = Mj−i K . K j K

The dierentials of N preserves degrees, so the complex decom- M S K(x0, . . . , xr) poses as a direct sum of complexes of K-vector spaces of the form:

i+1 i i−1 O ^ r+1 O ^ r+1 O ^ r+1 ... → Mj−i−1 K → Mj−i K → Mj−i+1 K → ... K K K

Now homology and graded components commute since quotient and grading (i.e. direct sum) commute and the maps are degree preserving. Thus βi,j(M) is the dimension of the homology at the term N Vi r+1 of the above complex, Mj−i K thus proving the rst statement. K Then   ker M N Vi r+1 → M N Vi−1 r+1 j−i K K j−i+1 K K βi,j(M) = dimK   Im M N Vi+1 r+1 → M N Vi r+1 j−i−1 K K j−i K K i i O ^ r+1 ^ r+1 ≤ dimK Mj−i K = dimK Mj−i. dimK K K r + 1 = H (j − i) M i and so if as then r+1 . βi,j(M) = 0 i > r + 1 i = 0 The (unique) minimal free resolution of M has length at most r + 1 as there are no generators for Fi in any degree when i > r + 1 in the minimal free resolution as βi,j(M) = 0 for i > r + 1. This proves Hilbert's syzygy theorem 2.10.

24 4 Castelnuovo-Mumford Regularity

4.1 Denition and rst applications

The Castelnuovo-Mumford regularity of an ideal in S is an important measure of how complicated the ideal is. A rst approximation is the maximum degree of a generator the ideal requires; the actual denition involves the syzygies (relations between generators) as well.

Denition 4.1. Let S = k[x0, . . . , xr] and let

ϕi ϕ1 F : ... −→ Fi −→ F −→ ... −→ F1 −→ F0. be a graded complex of free -modules, with P . We dene S Fi = j S(−ai,j) the Castelnuovo-Mumford regularity, or simply regularity, of F to be regF = . The regularity of a nitely generated graded -module is the supi,j(ai,j − i) S M regularity of a minimal graded free resolution of M, which we denote by regM. In case X ⊂ r is a projective variety and I is its ideal, then regI is called the PK X regularity of X, or regX. Suppose is free, so we have ∼ L , then the regularity of M M = F0 = j S(−a0,j) M is the supremum of the degrees of a set of homogeneous minimal generators of M, since . regM = supj{a0,j − 0} = supj{a0,j} In general, the regularity of M is an upper bound for the largest degree of a minimal generator of M, which is the supremum of the numbers a0,j − 0. Assuming that M is generated by elements of degree 0, the regularity of M is the index of the last nonzero row in the Betti diagram of M: Recall that the entry in the j-th row of the i-th column of the Betti diagram of M is βi,i+j and using alternative expressions

ni ∞ M M βi,j Fi = S(−ai,k) = S(−j) k=1 j=0 for the modules in the minimal free resolution we nd that

regM = sup ai,k − i = sup j − i = sup j i,k i,j:βi,j >0 i,j:βi,i+j >0

The power of the notion of regularity comes from an alternate description in terms of cohomology, which might seem to have little to do with free resolutions. His- torically, the cohomological interpretation came rst. David Mumford dened the

25 regularity of a coherent sheaf on projective space in order to generalise a clas- sical argument of Castelnuovo. Mumford's denition is given in terms of sheaf cohomology. The denition for modules, which extends that for sheaves, and the equivalence with the condition on the resolution used as a denition above came from [Eisenbud and Goto, 1984]. In most cases the regularity of a sheaf, in the (cohomological) sense of Mumford, is equal to the regularity of the graded module of its twisted global sections.

Recall the Hilbert function of a nitely generated graded S-module M is HM (d) = , and it is equal to a polynomial function for large . The regularity dimK Md PM (d) d of M provides a bound, which is sharp in the case of a Cohen-Macaulay module for when HM (d) = PM (d). Theorem 4.2. Let M be a nitely generated graded module over the polynomial ring S = k[x0, . . . , xr]. Then,

(1) The Hilbert function HM (d) agrees with the Hilbert polynomial PM (d) for all d ≥ regM + 1;

(2) More precisely, if M is a module of projective dimension δ, then HM (d) = PM (d) for all d ≥ regM + δ − r; (3) If X ⊂ r is a nonempty set of points and M = S = S/I(X), then H (d) = PK X M PM (d) if and only if d ≥ regM. In general, if M is a Cohen-Macaulay module, then the bound in part (2) is sharp.

Proof. Part (1) follows from part (2), the Hilbert Syzygy theorem, and the result that the projective dimension δ is equal to the length of the minimal free resolution (Corollary 2.18), which is less than the number of variables in S. That is to say, δ ≤ r + 1, so regM + δ − r ≤ regM + 1.

Part (2): To prove that if M is a module of projective dimension δ, then HM (d) = PM (d) for all d ≥ regM + δ − r. Consider the minimal graded free resolution of M, which is of the form

i i Mδ M0 F : 0 −→ S(−aδ,j) −→ ... −→ S(−a0,j) → M → 0 j=1 j=1 by corollary 2.18. In these terms regM = maxi,j(ai,j − i). We can compute the Hilbert function or polynomial of M by taking the alternating sum of the Hilbert functions or polynomials of free modules in the resolution of M. In this way we obtain expressions as in chapter 2:

26 δ X X r + d − ai,j H (d) = (−1)i M r i=0 j δ X X (d − ai,j + r)(d − ai,j + r − 1) ... (d − ai,j + 1) P (d) = (−1)i M r! i=0 j

This expansion for PM is the expression for HM with each binomial coecient replaced by the polynomial to which it is eventually equal. In fact the binomial coecient d−a+r has the same value as the degree poly- r r nomial (d−a+r)(d−a+r−1)...(d−a+1) for all . Thus from we r! d ≥ a − r d ≥ regM + δ − r get d ≥ ai,j − i + δ − r ≥ ai,j − r for each ai,j with i ≤ δ. For such d, each term in the expression of the Hilbert function is equal to the corresponding term in the expression of the Hilbert polynomial, and this proves part (2). The proof of part (3) and the nal remark can be found in [Eis05, Theorem 4.2].

4.2 The Regularity of a Cohen-Macaulay Module

Denition 4.3. The (Krull) dimension of a ring R, written dim R, is the supremum of lengths of chains of primes ideals of . That is R P0 $ P1 $ ... $ Pn has length n. If P is a prime ideal, the codimension of P , written codimP , is the maximum of the lengths of chains of prime ideals descending from P % ... % P0 P . If I is any ideal, the codimension of I is the minimum of the codimension of primes containing I. See [Eis95, Chapter 8] for a discussion linking these rather algebraic notions with geometry.

The dimension dim M of an R-module M is dened dim M = dim R/annM. Denition 4.4. A sequence x = x1, . . . , xn of elements of R is called a regular sequence if x1, . . . , xn generate a proper ideal of R, and for each i the element xi is a nonzerodivisor modulo (x1, . . . , xi−1). Similarly, if M is an R-module, then x is a regular sequence on M (or M- sequence) if (x1, . . . , xn)M 6= M and for each i the element xi is a nonzerodivisor on M/(x1, . . . , xi−1)M. An ideal that can be generated by a regular sequence (or, in the geometric case, the variety it denes) is called a complete intersection.

27 Denition 4.5. If I is an ideal of R and M is a nitely generated module such that IM 6= M, then the depth of I on M, written depth(I,M), is the maximal length of a regular sequence on M contained in I. If IM = M we set depth(I,M) = ∞. In the case where R is a local or homogeneous algebra and I is the maximal homogeneous ideal we write depthM in place of depth(I,M). We dene the grade of I to be gradeI = depth(I,R).

In particular, for S = k[x0, . . . , xr] and a nitely generated S-module M we have depthM = depth(m,M) is the maximal length of a regular sequence on M contained in m = (x0, . . . , xr). It follows that depthS = r + 1 from the maximal regular sequence x0, . . . , xr. Denition 4.6. Recall that the projective dimension pdM of an R-module M is the minimum length of a projective resolution of M (or ∞ if there is no nite projective resolution), and that in corollary 2.18 we proved that when M is a nitely generated S-module its projective dimension is equal to the length of the minimal free resolution of M. Denition 4.7. A local ring R is Cohen-Macaulay if depthR = dim R. More generally, an R-module M is Cohen-Macaulay if depthM = dim M. If X ⊂ r PK is a projective variety, we say that X is arithmetically Cohen-Macaulay if the homogeneous coordinate ring is Cohen-Macaulay. SX = K[x0, . . . , xr]/I(X) Fact. a1 ak is a Cohen-Macaulay ring for any positive inte- K[x1, . . . , xn]/(x1 , . . . , xk ) gers k ≤ n and a1, . . . , ak. Lemma 4.8. Suppose that M is a nitely generated graded S-module. If x is a linear form in S (i.e. a degree 1 element) that is a nonzerodivisor on M (that is, xm = 0 =⇒ m = 0 ∈ M), then regM = regM/xM.

Proof. Let F be the minimal free resolution of M. From the free resolution K(x): 0 → S(−1) −→x S of S/(x) we compute

O idM ⊗x O Tor∗(M, S/(x)) : 0 → M S(−1) −−−−→ M S S S x ∼= 0 → M(−1) −→ M

As x is a nonzerodivisor on M, the sequence is exact at M(−1) and so Tor0(M, S/(x)) = M/xM and Tori(M,S/(x)) = 0 for i > 0. Consider the free double complex F N K(x),

28 N id⊗δi N id⊗δi−1 N ... / S(−1) S Fi / S(−1) S Fi−1 / S(−1) S Fi−2 / ...

x⊗id x⊗id x⊗id N id⊗δi N id⊗δi N ... / S S Fi / S S Fi−1 / S S Fi−2 / ... which (as above) is isomorphic to

δi δi−1 ... / Fi(−1) / Fi−1(−1) / Fi−2(−1) / ...

x x x

 δi δi ... / Fi / Fi−1 / Fi−2 / ... We can also compute Tor as the total homology of F N K(x), [Weibel 1995]

  δi x   M 0 δi−1 M ... → Fi Fi−1(−1) −−−−−−−−−→ Fi−1 Fi−2(−1) → ... and we see that F N K(x) is the minimal free resolution of M/xM. The i-th free N L module in F K(x) is Fi Fi−1(−1) so regM/xM = regM by the denition of regularity. Corollary 4.9. Let M be a nitely generated Cohen-Macaulay graded S-module, and let y1, . . . , yt be a maximal M-regular sequence of linear forms. The regularity of is the largest such that . M d (M/ (y1, . . . , yt) M)d 6= 0

Proof. As M is Cohen-Macaulay we have dim M/ (y1, . . . , yt) M = 0, and by lemma 4.8 above we have regM = regM/ (y1, . . . , yt) M, so we have reduced to the zero dimensional case. In the case dim M = 0 we complete the proof by applying the following lemma. Lemma 4.10. If M is a graded S-module of nite length, then

regM = max{d | Md 6= 0}.

Proof. This is proven using cohomological techniques. See [Eis05, Theorem 4.3].

Theorem 4.11. Suppose that X ⊂ Pr is a projective variety not contained in any hyperplane. If SX is Cohen-Macaulay, then regSX ≤ deg X − codimX.

29 Proof. Let t = dim X, so that the dimension of SX as a module is t + 1. Without loss of generality we may extend the ground eld in order to assume that it is algebraically closed, and in particular innite. Thus we may assume that a suciently general sequence of linear forms y0, . . . , yt is a regular sequence in any order on ¯ SX . Set SX = SX /(y0, . . . , yt). As X is not contained in any hyperplane, then the corresponding homogeneous ideal IX does not contain any linear form, so we have . Thus ¯ . If the dimK(SX )1 = r + 1 dimK(SX )1 = (r + 1) − (t + 1) = r − t = codimX regularity of is , then by corollary 4.9 we have ¯ , so . Since SX d (SX )d 6= 0 HS¯X (d) 6= 0 ¯ is generated as an -module by in degree , this implies that for SX S 1 0 HS¯X (e) 6= 0 all 0 ≤ e ≤ d. On the other hand, deg X is the number of points in which X meets the linear space L dened by y1 = ... = yt = 0 by denition, since this linear space has complementary dimension dim L = r − t = codimX. By induction on d, and using the graded exact sequence

y0 ¯ 0 → SX /(y1, . . . , yt)(−1) −→ SX /(y1, . . . , yt) → SX → 0

d we see that P ¯ . That is, passing to the exact sequence HSX /(y1,...,yt)(d) = e=0 HSX (e) of the d-th graded components and applying rank-nullity theorem we nd that . For large the polynomials of HSX /(y1,...,yt)(d) = HS¯X (d) + HSX /(y1,...,yt)(d − 1) d degree d induce all possible functions on the set of points X ∩ L, so deg X = . [Eis05, Theorem 4.1] HSX /(y1,...,yt)(d) It follows that for large d

d d X X deg X = HSX /(y1,...,yt)(d) = HS¯X (e) = HS¯X (0) + HS¯X (1) + HS¯X (e) e=0 e=2 d X = 1 + codimX + HS¯X (e) ≥ 1 + codimX + (regX − 1) e=2 as there are at least positive values of for , since regX − 1 HS¯X (e) e = 2, . . . , d for all . This gives as required. HS¯X (e) 6= 0 0 ≤ e ≤ d regX ≤ deg X − codimX In the most general case, the regularity can be very large. Consider the case of a module of the form M = S/I. Groebner basis methods give a general bound for the regularity of M in terms of the degrees of generators of I and the number of variables, but these bounds are doubly exponential in the number of variables. [Eis05, Corollary 4.15]

30 5 Toric Ideals

In this chapter we introduce a class of polynomial ideals called toric ideals which allow interactions between algebra, geometry, and combinatorics. They are the dening ideals of toric varieties which are a rich but fairly accessible class of varieties in algebraic geometry. Denition 5.1. Fix a subset d such that the matrix A = {a1, a2,..., an} ⊂ Z \{0} d×n has rank . We have a semigroup homomorphism: A = a1 a2 ... an ∈ Z d

n n d X π : N → Z , u = (u1, . . . , un) 7→ aiui = Au i=1 i.e. The map π distributes over addition of vectors. We dene the monoid (semigroup) generated by A to be NA := π(Nn) = {Au : u ∈ Nn}. The semigroup ring of n is and the semigroup ring of d is N k[x] = k[x1, . . . , xn] Z the Laurent polynomial ring ±1 ±1 ±1 . The map lifts to the ring k[t ] := k[t1 , . . . , td ] π homomorphism:

±1 aj a1j a2j adj πˆ : k[x] → k[t ], xj 7→ t := t1 t2 . . . td from the matrix . A = a1 a2 ... an = (aij)d×n

The toric ideal of A, denoted as IA, is the kernel of the map πˆ. That is IA = kerπ ˆ C k[x].

In this setting it is natural to grade the polynomial ring k[x] by setting deg(xi) = ai for i = 1, . . . , n. Then the set of all degrees of polynomials in k[x] is NA. We say a polynomial in k[x] is A-homogeneous if it is homogeneous under this multigrading. Example 5.2. The Veronese and Segre embeddings each give rise to toric ideals. For example, the standard twisted cubic curve in 3 is determined by the matrix PC 3 2 1 0 A = . More generally, the degree-m Veronese embedding of n 0 1 2 3 P yields a toric variety determined by an -matrix where n+m whose N ×(n+1) N = n columns consist of all the distinct -tuples such that Pn (n + 1) i0, . . . , in j=0 ij = m, 0 ≤ ij ≤ m. That is, each column in the matrix A determining the variety n consists of nonnegative integers whose sum is . vm(P ) m Proposition 5.3. (1) The toric ideal IA is a prime ideal in k[x].

(2) The ring k[x]/IA has Krull dimension d.

(3) The ideal IA is generated as a k-vector space by the innitely-many binomials u v n , and hence u v . {x − x : π(u) = π(v), u, v ∈ N } IA = hx − x : π(u) = π(v)i

31 (4) For every term order ≺ the reduced Groebner basis of IA with respect to ≺ u v consists of a nite set of binomials of the form x − x ∈ IA.

∼ ˆ a1 an Proof. (1) By the rst isomorphism theorem we have k[x]/IA = π (k[x]) = k[t ,..., t ], which is an integral domain, hence IA is a prime ideal. (2) This follows from our requirement that the matrix A = a1 a2 ... an ∈ Zd×n has rank d. See [Eis95, Chapter 8]. (3), (4) See [Maclagan et al., Prop. 3.2.2]. Example 5.4. Let be a graded polynomial ring whose variables T = K[z0, . . . , zr] have degrees . Consider the graded exact sequence of -modules deg zi = αi ∈ N T

zr 0 → T (−αr) −→ T → T/(zr) → 0.

We may pass to the d-th graded components to obtain an exact sequence of K- vector spaces. Then by the rank-nullity theorem we have that HT/(zr)(d) = HT (d)− . As ∼ we may apply HT (−αr)(d) = HT (d) − HT (d − αr) T/(zr) = K[z0, . . . , zr−1] induction on r to deduce that the Hilbert function HT of T is, for large d, equal to a polynomial with periodic coecients: that is,

r r−1 HT (d) = h0(d)d + h1(d)d + ... for some periodic functions with values in , whose periods divide the least hi(d) Q common multiple of the αi. 1 1 1 1 Example 5.5. Consider the toric variety given by A = . Then 0 1 2 4

2 4 πˆ : S = k[x1, x2, x3, x4] → k[t1, t2], x1 7→ t1, x2 7→ t1t2, x3 7→ t1t2, x4 7→ t1t2 describes a projection to P3 of a rational normal quartic in P4. We have  1   1  * +  −2   0  2 2 ker A =   ,   ,IA = (x1x3 − x , x1x4 − x ).  1   −2  2 3 0 1

We note this to be of the form as we have seen in example 2.4 so we have minimal free resolution

 2  x1x4 − x3  2  2 2 x − x1x3 x1x3 − x x1x4 − x 0 → S −−−−−−−−−−−→2 S(−2)2 −−−−−−−−−−−−−−−−−−−→2 3 S.

32 6 Eisenbud-Goto Conjecture

Recall that the regularity of M is an upper bound for the largest degree of a minimal generator of M. The following theorem of Gruson, Lazarsfeld, and Peskine gives an optimal upper bound for the regularity of a projective curve in terms of its degree, even if SX is not Cohen-Macaulay (Compare with theorem 4.11): Theorem 6.1 (Gruson-Lazarsfeld-Peskine). Suppose K is an algebraically closed eld. If X ⊂ r is a reduced and irreducible curve, not contained in a hyperplane, PK then regSX ≤ deg X − codimX = deg X − r + 1, and thus regIX ≤ deg X − r + 2. In particular, this implies that the degrees of the polynomials needed to generate are bounded by . If the eld is the complex numbers, the degree IX deg X −r +2 K of may be thought of as the homology class of in r , so the bound X X H2(P ; K) = Z given depends only on the topology of the embedding of X. [Eis05, Theorem 5.1]

6.1 A General Regularity Conjecture

We say that a variety in a projective space is nondegenerate if it is not contained in any hyperplane. Correspondingly, we say that a homogeneous ideal is nondegenerate if it does not contain a linear form. In theorem 4.11 we proved that if X ⊂ r is arithmetically Cohen-Macaulay and nondegenerate, then regS ≤ PK X deg X − codimX, which gives deg X − r + 1 in the case of curves. This result does not hold for schemes which are not arithmetically Cohen-Macaulay, even in the case of curves; the simplest example is where X is the union of two disjoint lines in P3 (see exercise 5.2), and the result can also fail when X is not reduced or the ground eld is not algebraically closed (see exercises 5.3, 5.4). Fur- ther it is not enough to assume that the scheme is reduced and connected, since the cone over a disconnected set is connected and has the same codimension and regularity.

A possible way around these examples is to insist that X be reduced, and connected in codimension 1, meaning that X is pure-dimensional and cannot be disconnected by removing any algebraic subset of codimension 2. Conjecture 6.2 (Eisenbud and Goto, 1984). If K is an algebraically closed eld and X ⊂ r is a nondegenerate algebraic set that is connected in codimension 1, PK then regSX ≤ deg X − codimX. The conjecture is known to hold in the Cohen-Macaulay and dimension-one cases, and for smooth surfaces in characteristic 0, arithmetically Buchsbaum surfaces,

33 and toric varieties of low codimension. For a discussion of the literature, see [Eis05, Conjecture 5.2].

For the conjecture to have a chance, the number deg X − codimX must at least be nonnegative. The following proposition establishes this inequality. Further, the hypotheses can be shown to be necessary [Eis05, Exercises 5.2-5.4]. Proposition 6.3. Suppose that K is an algebraically closed eld. If X is a nondegenerate algebraically closed set in r = r and X is connected in codimension P PK 1, then deg X ≥ 1 + codimX.

To understand the bound, set c = codimX and let p0, . . . , pc be general points on X. As X is nondegenerate, these points can be chosen to span a plane L of dimension c. The degree of X is the number of points in which X meets a general c-plane and, by construction, L meets X in at least c + 1 points. The problem with this argument is that L might, a priori, meet X in a set of positive dimension (i.e. L is not a suciently general c-plane), and this can indeed happen without some extra hypothesis, such as connected in codimension 1. For the proof, see [Eis05, Proposition 5.3]. Remark. One can show that most questions about free resolutions of ideals can be reduced to the nondegenerate case, just as can most questions about varieties in projective space. This is discussed in more detail in [Eis05, Exercise 4.4].

6.2 Examples using Macaulay 2

In this section we verify conjecture 6.2 for some randomly generated toric ideals using the Macaulay2 software system as the corresponding toric varieties satisfy the hypothesis for the conjecture. We use the following formula to check whether our randomly generated toric ideal is Cohen-Macaulay. Theorem 6.4 (Auslander-Buchsbaum formula). If R is a local ring and M is a nitely generated R-module such that pdM is nite, then depthM = depthR − pdM. If M is a nitely generated S-module this allows us to compute depthM when we know the minimal free resolution of M, using corollary 2.18 and that the polynomial ring S is Cohen-Macaulay with depthS = r + 1. It follows that the module M is Cohen-Macaulay if and only if pdM = codimSM. [Eis95, Chapter 19]. We use the following method for each example:

1. Choose d such that the matrix A = {a1, a2,..., an} ⊂ Z \{0} A = a1 a2 ... an ∈ Zd×n has rank d (so d ≤ n). Further to ensure the toric ideal we obtain is

34 nondegenerate, we should take care that the columns of the matrix do not satisfy any linear relation. A = transpose(matrix {{a_1} ,... ,{a_n}});

2. Initialize a ring S = k[x] = k[x1, . . . , xn], which may be graded by setting deg(xi) = ai for i = 1, . . . , n. S = QQ[ x_1 . . x_n ] ; S = QQ[x_1..x_n,Degrees=>{{a_1} ,... ,{a_n}}];

3. Compute ker A in order to determine IA. Initialize the toric ideal IA and the homogeneous coordinate ring M = S/IA. ker A I = ideal(f_1,... ,f_{n−d } ) ; M = S^1/ I ; M= coker gens I;

4. Compute the minimal free resolution of M to determine its projective dimension and hence check whether the ring S/IA is Cohen Macaulay using the Auslander-Buchsbaum formula, or use the projective dimension function directly. r e s M pdim M

5. Compute the quantities appearing in conjecture 6.2 and check whether regSX ≤ deg X − codimX. regularity M degree M codim M  7 −5 10 6 4 4  Example 6.5. The matrix gives rise to the A1 =  9 1 0 10 5 −10  −8 10 6 3 5 −5 toric ideal 6 2 3 12 31 10 54 5 10 25 20 26 24 . IA1 = (x5x6 − x1x2x3, x3 x4 − x1 x5 x6, x1 x3 − x4 x5 x6 )  1 1 2 0 1 1  Example 6.6. The matrix gives rise to the toric A2 =  1 2 1 0 2 1  3 0 1 1 0 2 ideal 4 4 . Remark: degenerate. IA2 = (x5 − x2, x4x6 − x1, x6 − x1x2x3x4)  1 1 2 1 1 1  Example 6.7. The matrix gives rise to the toric A3 =  1 2 1 1 2 1  3 1 1 1 1 2 ideal 7 2 4 2 4 2 . Remark: degenerate. IA3 = (x5 − x2, x4x5 − x1x2x3, x4x5 − x2x3x6)

35 The ideal IA is not homogeneous with the standard grading (with variables in degree 1) and so the module S/IA does not admit a traditional graded structure, but if we choose the grading by setting deg(xi) = ai for i = 1, . . . , n, then IA becomes a homogeneous ideal. We will use a dash to denote when we use the implicit toric grading. We give the outputs from Macaulay2 when computing in these settings:

pdimM dimM codimM degreeM regularityM isHomogeneous IA

A1 4 4 2 87 Failed false 0 4 4 2 Failed 187552 true A1 A2 4 3 3 11 Failed false 0 3 3 3 840 38 true A2 A3 3 4 2 1 Failed false 0 3 4 2 1 11 true A3 Remark. I am not condent in how much faith can be put in the above results due to their many seeming inconsistencies. One might deduce that the consequences of conjecture 6.2 do not hold when considering A3 with its associated grading; however proposition 6.3 also fails in this case because deg X − codimX is negative, which is also true in the standard grading. From case A2 we see that we get diering outputs for both projective dimension and degree depending on our choice of grading. Further, it's suspected that the degree computation in case 0 A1 timed out due to an expected large degree. Finally, as previously mentioned, if we use the standard grading on k[x1, . . . , xn] naively when analysing these toric ideals in Macaulay2, then the ideals are recognised as not homogeneous, so the resolution function does not return a minimal resolution and the regularity can not be computed since the expected input is a homogeneous (graded) S-module.

A Additional Macaulay2 Code

Some explicit examples of Macaulay2 code used is provided for error-checking purposes: S = QQ[ x_1 . . x_6 ]

A = transpose(matrix{{7,9, −8} ,{ −5,1,10},{10,0,6}, {6,10,3},{4,5,5},{4, −10 , −5}})

B = matrix {{ −1 , −10 ,10} ,{ −1 ,0 ,0} ,{ −3 ,12 ,25} , {0 ,31 , −20} ,{6 , −54 , −26} ,{2 , −5 , −24}}

36 S = QQ[x_1..x_6,Degrees=>{{7,9, −8} ,{ −5,1,10},{10,0,6}, {6,10,3},{4,5,5},{4, −10 , −5}}]

I = ideal(x_5^6∗x_6^2 − x_1∗x_2∗x_3^3 , x_3^12∗x_4^31−x_1^10∗x_5^54∗x_6^5 , x_1^10∗x_3^25 − x_4^20∗x_5^26∗x_6^24)

M= coker gens I

A = transpose(matrix{{1,1,3},{1,2,0},{2,1,1}, {0,0,1},{1,2,0},{1,1,2}})

S = QQ[x_1..x_6,Degrees=>{{1,1,3},{1,2,0},{2,1,1}, {0,0,1},{1,2,0},{1,1,2}}]

I = ideal(x_5−x_2 , x_4∗x_6 − x_1 , x_6^4 − x_1∗x_2∗x_3∗x_4^4)

A = transpose(matrix{{1,1,3},{1,2,1},{2,1,1}, {1,1,1},{1,2,1},{1,1,2}}) S = QQ[x_1..x_6,Degrees=>{{1,1,3},{1,2,1},{2,1,1}, {1,1,1},{1,2,1},{1,1,2}}]

I = ideal(x_5−x_2 , x_4^7∗x_5^2 − x_1∗x_2^4∗x_3^2 , x_4^4∗x_5 − x_2^2∗x_3∗x_6)

References

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