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Kyungpook Mathematical Journal 2024; 64(2): 197-204

Published online June 30, 2024 https://doi.org/10.5666/KMJ.2024.64.2.197

Copyright © Kyungpook Mathematical Journal.

Minimal Generators of Syzygy Modules Via Matrices

Haohao Wang* and Peter Oman

Department of Mathematics, Southeast Missouri State University, Cape Girardeau, MO 63701, USA
e-mail : hwang@semo.edu and poman@semo.edu

Received: April 20, 2023; Revised: November 18, 2023; Accepted: March 10, 2024

Let R = K[x] be a univariate polynomial ring over an algebraically closed field K of characteristic zero. Let AMm,m(R) be an m×m matrix over R with non-zero determinate det(A) ∈ R. In this paper, utilizing linear-algebraic techniques, we investigate the relationship between a basis for the syzygy module of f1, . . . , fm and a basis for the syzygy module of g1, . . . , gm, where [g1, . . . , gm] = [f1, . . . , fm]A.

Keywords: Smith Normal Form, Cauchy-Binet Formula, Basis, Modules

Systems of linear equations, vector spaces, and bases for vector spaces are widely studied in linear algebra. Modules are a generalization of vector spaces from linear algebra in which the “scalars" are allowed to be from an arbitrary ring, rather than a field. Many linear algebraic methods remain effective over principal ideal domains where the extended Euclidean algorithm may be utilized for computing with unimodular matrices. This paper is devoted to the some applications of linear algebra in studying modules over a univariate polynomial ring R=K[x] where K is an algebraically closed field of characteristic zero.

Over the ring R, given a1,,ak,bR, the problem to determine if a single equation

a1x1+akxk=b

has a solution x1,,xkR amounts to determining if b is in the ideal generated by a1,,ak; this is the ideal membership problem. In the case of a system of several equations, the ideal membership problem turns into the submodule membership problem. Furthermore, given a1,,akRn, a solution (x1,,xk)Rk to the homogeneous equation

a1x1++akxk=0

is called a syzygy of (a1,,ak). The solution set for this system of homogeneous equations provides a system of generators for the syzygy module of (a1,,ak). Hence, the computation of the null space of the matrix of a system of linear equations over R is essentially the syzygy problem, the goal is to find a generating set for the syzygy module.

A fundamental fact of linear algebra over a field that is a finitely generated vector space has a basis – the minimal generating (spanning) sets of a vector space are linearly independent and therefore form a basis. However, modules are more complicated than vector spaces; for instance, not all modules have a basis. It remains an interesting and an active research area to find a minimal generating set, or an upper bound for the size of a minimal generating set for different kinds of modules under various of conditions [2], [4], [6], and [8]. Hilbert's syzygy theorem [7] states that, if M is a finitely generated module over a multivariate polynomial ring K[x1,,xn] over a field K, then the n-th syzygy module of M is always a free module, i.e., a module that has a basis – a generating set consisting of linearly independent elements. In particular, over a univariate polynomial ring, Hilbert's syzygy theorem asserts that over a principal ideal ring, every submodule of a free module is itself free. Hence, it is an interesting problem to find the basis for syzygy modules.

This short paper investigates the minimal set of generators of a family of special modules. Our focus is centered on two finitely generated modules – the first module is Syz(f), the syzygy module of f=[f1,,fm], where f1,,fmR=K[x] with K an algebraically closed field of characteristic zero; the second module is Syz(g), the syzygy module of g=[g1,,gm], where [g1,,gm]=[f1,,fm]A with AMm,m(R) and (A)=d(x)R. Geometrically, if (f)=1, then f may be viewed as a polynomial parametrization of a rational space curve in a m-dimensional affine space. Similarly, g=[g1,,gm]=fA can be considered as another rational space curve constructed by applying a linear transformation A to the curve f over the ring R.

A natural question to ask is “how to link a basis for Syz(f) with a basis for Syz(g)"? To answer this question, we evoke the Smith normal form of the matrix A, and take advantage of Cauchy-Binet formula to relate a basis for Syz(f) with a basis for Syz(g) via a matrix BMm-1,m-1(R) such that (A) for some λK{0}.

This paper is structured as the following. We begin in Section 2 with a brief review of results concerning a matrix factorization over PIDs, and the concepts of syzygy modules. We then answer the question that "how to link a basis for Syz(f) with a basis for Syz(g)" in Section 3. The main result of this paper is Theorem 3.1, where we provide an explicit equation to describe a relationship between a basis for Syz(f) and a basis for Syz(g). We flush out our theorem by a simple and illustrative example. We end our paper by a few conclusions in Section 4.

In this section, we provide a brief review of results concerning a matrix factorization over a PID, the concepts of syzygy modules, and the Hilbert-Burch theorem. First, we recall the Smith normal form over a PID and the Cauchy-Binet formula.

Theorem 2.1. (Smith Normal Form over PID [1, Theorem 3.1]) Let R be a PID and let AMm,n(R). Then there is a UGL(m,R) and a VGL(n,R) such that

UAV=Dr000

where r=rank(A), and D is a diagonal matrix with diagonal entries d1,d2,,dr with di0 for i=1,,r, and didi+1 for i=1,,r-1. Furthermore, if R is a Euclidean domain, the matrices U,V can be taken to be a product of elementary matrices.

Theorem 2.2. (Cauchy-Binet formula [1, Theorem 2.34, Page 210]) Suppose that A is an a×b matrix, B is an b×c matrix, I is a subset of {1,,a} with k elements, and J is a subset of {1,,c} with k elements. Let [A]I,J be the minor of A associated to the ordered sequences of indexes I and J. Then the Cauchy-Binet formula is

[AB]I,J=K[A]I,K[B]K,J,

where the sum extends over all subsets K of {1,,b} with k elements.

Next, we review a few concepts and results concerning syzygies, please refer [3, Chapters 4, 5] and [5, Chapter 20] for details. Recall that given a generating set m1,,mk of an ideal (or a module) over a ring R, a relation or first syzygy between the generators is a k-tuple (a1,,ak)Rk such that

a1m1+a2m2+akmk0.

The set of syzygies form a module Syz(m1,,mk).

Let f=[f1,,fm] with fiR. The syzygy module, Syz(f)=Syz(f1,,fm), is the kernel of the map Rmf1,,fm that takes the standard basis elements of Rm to the given set of generators. Hilbert–Burch theorem yields that Syz(f) is a free module – a module that has a basis; and is minimally generated by m-1 elements – rank(Syz(f))=m-1. In addition, if the minimal set of generators of Syz(f), namely a μ-basis, are expressed as the columns of a matrix SMm,m-1(R), then the ideal f is generated by the minors of size m-1 of the matrix of S. That is, if {1,,i^,,m} are subsets of {1,2,,m} without the element i, and [S]{1,,i^,,m},{1,,m-1} denotes the minor of size m-1 of the matrix of S without the i-th row, then the columns of S form a μ-basis if and only if

fi=α(-1)i+1[S]{1,,i^,,m},{1,,m-1}

for some αK{0},i=1,,m.

Equation (2.2) is formulated from the following Hilbert–Burch theorem by applying M=S and N=f.

Theorem 2.3.(Hilbert–Burch Theorem [3, Proposition 2.6, Chater 6])

Suppose that an ideal I in R=K[x1,,xn] has a free resolution of the form

0Rm1MRmNI0, for some m.

Then there exists a non-zero element gR such that N=[gh1,,ghm], where hi is the determinant of the (m-1)×(m-1) submatrix of M obtained by deleting row i. If K is algebraically closed and the variety of I has dimension n-2, then we can take g=1.

In this section, we investigate the relationship between a μ-basis for Syz(f) and a μ-basis for Syz(g), and prove Theorem 3.1, our main result.

We continue our notation – set R=K[x] with K an algebraically closed field of characteristic zero, and let AMm,m(R) where (A)=d(x)R. In addition, we define f=[f1,,fm] where f1,,fmR, and g=[g1,,gm]=[f1,,fm]A.

Theorem 3.1. Let S,TMm,m-1(R) such that the columns of S and T are μ-bases for Syz(f) and Syz(g) respectively. Then there exists a BMm-1,m-1(R) such that AT=SB with (A) for some λK{0}.

Proof. By Theorem 2.1,

A=Ud10000d20000dm10000dmV,

where (A)=i=1mdi=d(x)0.

Set f'=fU=[f1',,fm'], g'=gV-1=[g1',,gm'], and let the columns of S',T'Mm,m-1(R) be μ-bases for Syz(f') and Syz(g') respectively.

First, we claim that there exists a BMm-1,m-1(R) such that DT'=S'B and (A), where α,βK{0} are such that for i=1,,m-1

f'i=α(-1)i+1[S']{1,,i^,,m},{1,,m-1},g'i=β(-1)i+1[T']{1,,i^,,m},{1,,m-1}.

To prove our claim, we note that the columns of T' form a μ-basis for Syz(g') implies

g'T'=gV-1T'=fAV-1T'=f(UDV)V-1T'=(fU)DT'=f'(DT')=0.

By the definition of Syz(f'), the columns of the matrix DT' are elements of Syz(f'). Hence, they are generated by the columns of S', i.e.,

DT=SB  for some BMm1,m1(R)
(T)tDt=(DT)t=(SB)t=Bt(S)t,

where (T')t,(S')tMm-1,m(R), Dt,BtMm,m(R).

Now, we apply Cauchy-Binet formula in Theorem 2.2 to both sides of Equation (3.3). First, we compute all (m-1)×(m-1) minors of Bt(S')t obtained by deleting the i-th row for i=1,2,,m:

[Bt(S)t]{1,,m1},{1,,i^,m}=det(Bt)[(S)t]{1,,m1},{1,,i^,m}=det(B)α(1)i+1f i.

The last equality holds since the columns of S' form a μ-basis for Syz(f'), and by Equation (2.2),

[(S')t]{1,,m-1},{1,,i^,,m}=α(-1)i+1f'i, form some αK{0},i=1,2,,m.

Again, applying Equation (2.1), we compute all (m-1)×(m-1) minors of (T')tDt obtained by deleting the i-th row for i=1,2,,m:

[(T)tDt]{1,,m1},{1,,i^,,m}= j=1m [(T) t]{1,,m1},{1,,j^,,m} [ D t]{1,,j^,,m},{1,,i^,,m}= j=1mβ(1)j+1g j [ D t]{1,,j^,,m},{1,,i^,,m} for some βK{0},

where the last equality is due to Equation (2.2). It is easy to observe that the diagonal matrix D in Equation (3.1) has the property that

[Dt]{1,,j^,,m},{1,,i^,,m}=0, if ji.

Thus,

j=1mβ(1)j+1g j[Dt]{1,,j^,,m},{1,,i^,,m}=β(1)i+1g i[Dt]{1,,i^,,m},{1,,i^,,m}=β(1)i+1g id(x)di.

In addition, the equality

g1 g2 gmt=gV1=f(UDV)V1=(fU)D=fD= f1 f2 fmt d1 0 0 0 0 d2 0 0 0 0 0 d m1 0 0 0 0 dm(x) = d 1f1 d 2f2 d mfmt

yields that β(-1)i+1g'id(x)di=β(-1)i+1(dif'i)d(x)di=β(-1)i+1d(x)fi'. Hence,

[(T')tDt]{1,,m-1},{1,,i^,,m}=β(-1)i+1d(x)fi'.

Therefore, the Equation (3.3) yields that for each i{1,,m}

[Bt(S')t]{1,,m-1},{1,,i^,m}=[(T')tDt]{1,,m-1},{1,,i^,,m},

and the Equations (3.4) and (3.5) show that

(B)=λd(x),

where λ=βαK{0}.

Hence, we have proved that there exists a BMm-1,m-1(R) such that DT'=S'B, (B)=λd(x), where λ=βαK{0} and α,β are such that for each i=1,,m-1

f'i=α(-1)i+1[S']{1,,i^,,m},{1,,m-1},g'i=β(-1)i+1[T']{1,,i^,,m},{1,,m-1}.

Now we are ready to prove the statement of the theorem. Note that on one hand,

gT=g(V1T)=0V1T=TC for some CMm,m(R),

that is, columns of V-1T'Syz(g) and are generated by the μ-basis for Syz(g). On the other hand,

gT=gV-1VT=g'VT=0VT=T'C' for some C'Mm,m(R),

that is, columns of VTSyz(g') and are generated by the μ-basis for Syz(g'). Thus,

VT=T'C' and V-1T'=TCT=V-1T'C'=TCC'CC'=In,n

where In,n is an n×n identity matrix. This implies that C,C'GL(m,R), that is, T=V-1T'C'V-1T', equivalently T'=VTCVT. Since μ-bases are unique up-to isomorphism, without loss of generality, we may set T=V-1T', equivalently T'=VT. Applying a similar argument to Syz(f) and Syz(f'), we obtain an analogous result about S and S'. We conclude that μ-bases for Syz(g) and Syz(f) can be expressed as the columns of the matrices T and S respectively where

T=V-1T',S=US'T'=VT,S'=U-1S.

Thus, replacing T' and S' by VT and U-1S in Equation (3.2) yields

DVT=DT'=S'B=U-1SBUDVT=SBAT=SB.

Therefore, we conclude that there exists a BMm-1,m-1(R) such that AT=SB with (A) for some λK{0}.

Below, we will provide a simple example to illustrate Theorem 3.1.

Example 3.1. Consider [f1,f2,f3]=[1,x2-1,x3+1], and find a μ-basis of Syz(f1,f2,f3) formed by the columns of the matrix S=x21x+11x01, where the maximal minors of S, with an appropriate sign, give (f1,f2,f3)

Let

A=x+1x0x2x2x2xx+1x+1x=1000x2x10x+1110001000xx+1x0110001=UDV,det(A)=x,

and

[g1,g2,g3]=[f1,f2,f3]A=[2x4+x3x2+2x+2,2x4+x3x2+2x+1,2x4x3x2+2x].

A μ-basis for Syz(g1,g2,g3) is given by the columns of the matrix T=2x2+2x+12x2x12x222x+22x12x2+x1, where the maximal minors of T, with an appropriate sign, give 2(g1,g2,g3).

We have that AT=x+12x3x21xx(x1)1x+1=SB, where B=02x11x and det(B)=2x=2det(A).

Applying linear-algebraic techniques, we obtained a relationship on the minimal generators Syz(f) and Syz(g). We proved that if S,TMm,m-1(R) such that the columns of S and T are μ-bases for Syz(f) and Syz(g) respectively, then there exists a BMm-1,m-1(R) such that AT=SB with (A) for some λK{0}.

We took advantage the Smith normal form of the matrix A over R=K[x], a PID. However, R=K[x1,,xn], n2, is not a PID, and the Smith normal form will not apply in this case. It would be interesting to discover a relationship on the minimal generators Syz(f) and Syz(g) over R=K[x1,,xn] where n2. To our knowledge, this is still an open question.

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