# On Projective Modules and Computation of Dimension of a Module over Laurent Polynomial Ring

Authors: Ratnesh Kumar Mishra; Shiv Datt Kumar; Srinivas Behara
Journal: ISRN Algebra
 (2011)

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## Abstract

We give a procedure and describe an algorithm to compute the dimension of a module over Laurent polynomial ring. We prove the cancellation theorems for projective modules and also prove thequalitative version of Laurent polynomial analogue of Horrocks' Theorem.


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## Body

## 1. Introduction

Hilbert [1] introduced free resolution on iterated syzygies of a finitely generated graded module M over polynomial ring R=K[x1,…,xn]. A choice of a0=dim(M⊗RK) homogeneous generators of M defines a surjective homomorphism Ra0→M, and its kernel is the first syzygy module of M. Hilbert proved that the syzygies are also finitely generated (Hilbert basis theorem). Hilbert's syzygy theorem states that “every finitely generated module M over K[x1,…,xn] has finite free resolution, that is, the lth syzygy is Ral for some al.” The number of generators ai of the syzygies is chosen minimally, they are independent of the choice of generators. These ai are called the Betti numbers of M and the minimal length of free resolution (i.e., i) is dimension of a module M. When M is projective, the dimension of M is called projective dimension of M (denoted by pd(M)). A projective dimension is a measure of how far the module is being projective.Kaplansky [2] described how homological dimension changes when one passes from a ring R to a quotient ring R/(X), where (X) is the ideal generated by a nonzero divisor, without using Ext or Tor. Shanuel noticed that there is an elegant relation between different projective resolutions of the same module. Using the ideas of Shanuel, Kaplansky defined the projective dimension of a module.The global dimension (or homological dimension) of a ringR denoted by gldim(R), is a non-negative integer or infinity, and is a homological invariant of the ring. It is defined to be the supremum of the set of projective dimensions of all (finite or cyclic) R-modules. Global dimension is an important technical notion in the dimension theory of Noetherian rings. Eisenbud [3] proved that a commutative Noetherian local ring R is regular if and only if it has finite global dimension, in which case the global dimension coincides with the Krull dimension of R. This theorem opened the door for application of homological methods to commutative algebra. Now we state a theorem of Eisenbud [3] which implies that every module has a finite free resolution of length at most l<∞.Theorem 1.1 (see [3]).
The following conditions on a ringR are equivalent: (1)
gldim(R)≤l, that is, pd(M)≤l for every R-module M,(2)
pd(M)≤l for every finitely generated module M.Serre [4] taught a course on multiplicities at the college de France. Part of that course focussed upon the simple inequality pdR(M)≤pdR(S)+pdS(M) for module M over an R-algebra S. Auslander and Buchsbaum [5] realized that Serre's method could be used to study the close connection between the dimension and multiplicity over a local ring. This led them to the Auslander-Buchsbaum equality [6]: let M be a finitely generated module over a Noetherian local ring (R,m). If pd(M)<∞, then depth(R)=depth(M)+pd(M). Gago-Vargas [7] computed the projective dimension of a module over Weyl algebra. Projective dimension is a main ingredient to compute the tilted algebra which nowadays plays a very important role in the representation theory of algebras. In this paper, we give a procedure and describe an algorithm to compute the projective dimension of a module which is valid over the Laurent polynomial ring.We also prove the cancellation theorems for projective modules over the Laurent polynomial ring [8, Section  5] motivated by Swan's work. Horrocks proved the following theorem.Theorem 1.2 (see [9]).
LetR be a local ring, A=R[x] and S be the set of monic polynomials of A. If P is a finitely generated projective A-module such that PS is free over AS, then P is a free A-module.This theorem draw the attention of several mathematicians, as a result several analogues of Horrocks’ theorem can be found in the literature [10, 11]. Mandal proved the Laurent polynomial version of Horrocks’ theorem in [12] whereas Nashier [13] proved the qualitative versionof the same theorem over polynomial ring. But it is observed that the qualitative version of Horrocks’ theorem over Laurent polynomial ring is not found anywhere in the literature. One of the objectives of this paper is to fill this gap. In this paper, we prove the qualitative version of Horrocks’ theorem over the Laurent polynomial ring.


## 2. Preliminary Notes

In this section we define some terms used in this paper and state certain standard results without proof. We hope that this will improve the readability and understanding of the proof of the paper.A polynomialf in the Laurent polynomial ring R[X,X-1] is said to be doubly monic polynomial if coefficient of the highest degree term and the lowest degree term are unit.We need the following result of Kunz [14] to compute the projective resolution in Section 3.Definition 2.1 (see [14]).
An exact sequence(2.1)Fi+1→αiFiF1→α0F0→ϵM→0
with only free (resp., projective) modules Fi(i=0,1,2,…) is called a free (resp., projective) resolution of M.Projective modules over the Laurent polynomial ringR[X,X-1] are stably free. In [15] it is proved that a projective module is stably free provided it possesses a finite free resolution. The minimum length of free resolution is called the projective dimension of a module M.LetR be a ring and M be an R-module. M[X,X-1] means the analogously defined Laurent polynomial module. We identify M⊗RR[X,X-1] with M[X,X-1], where M[X,X-1]={∑n=-∞∞mnXn∣mn∈Mandmn=0for all but finiten}. If f:M→N is a homomorphism of R-modules, then this induces a homomorphism (2.2)ψ:M[X,X-1]⟶N[X,X-1],defined byψ(∑mnXn)=∑f(mn)Xn,
where ψ can be identified with f⊗RR[X,X-1]. Given any R-module M and f∈EndR(M), we can make M as R[X,X-1]-module whose scalar multiplication is defined as m(∑anXn)=an∑fn(m). We denote this R[X,X-1]-module by Mf. There is a canonical R[X,X-1]-epimorphism φf:M[X,X-1]→Mf defined by φf(∑mnXn)=∑fn(mn). The characteristic sequence of f is an exact sequence which proved in Proposition 4.3.The following is the Laurent polynomial version of a Horrocks Theorem which we state as follows.Theorem 2.2 (see [12]).
SupposeR[X,X-1] is a Laurent polynomial ring over a local Noetherian commutative ring R, and P is a projective R[X,X-1]-module. If Pf is free for some doubly monic Laurent polynomial f, then P is free.In the proof of the Theorem5.1, we need the following theorem.Theorem 2.3 (see [12]).
Suppose thatR[X,X-1] is a Laurent polynomial ring over a local Noetherian commutative ring R. Also suppose that P and P′ are two projective R[X,X-1]-modules with RankP′<RankP. If Pf′ is a direct summand of Pf for some doubly monic polynomial f, then P′ is also a direct summand of P.


## 3. Projective Dimension of Module

Lemma 3.1.
LetM1, M2, M3 be R-modules and
(3.1)0⟶M1→α1M2→α2M3⟶0
be a split short exact sequence, and β1 and β2 are the splittings corresponding to α1 and α2, respectively. Then the following sequence
(3.2)0⟶M3→β2M2→β1M1⟶0
is an exact sequence.Proof.
Sinceβ1 and β2 are the splittings corresponding to α1 and α2, respectively, then β1oα1=IM1 and α2oβ2=IM3. Let x∈M2. Then α2(x-β2(α2(x)))=α2(x)-α2(β2(α2(x)))=α2(x)-α2(x)=0. Hence x-β2(α2(x))∈Ker(α2)=Im(α1). Since α1 is injective, there exists a unique β1(x)∈M1 such that α1(β1(x))=x-β2(α2(x)). We need to show that Ker(β1)⊂Im(β2). Let x∈Ker(β1). Then β1(x)=0 and α1(β1(x))=α1(0)=0, by injectivity of α1. Hence x-β2(α2(x))=0⇒x=β2(α2(x)). Thus there exists α2(x)∈M3 such that β2(α2(x))=x. Hence x∈Im(β2). Therefore Kerβ1⊂Im(β2). Conversely, let x∈Im(β2). Then there exists y∈M3 such that β2(y)=x. Applying α2 on both sides, we get y=α2(x). Now applying β2 on both sides, we get β2α2(x)=x and therefore x-β2α2(x)=0. Using the defining property of β1 and injectivity of α1, α1(β1(x))=0=α1(0). Therefore β1(x)=0. Hence x∈Ker(β1). Thus Im(β2)⊂Ker(β1). Therefore Im(β2)=Ker(β1)By applying the technique discussed in [7] for Weyl algebra, we give a procedure to calculate the projective dimension of a module over the Laurent polynomial ring. We also describe an algorithm to compute the projective dimension of a module.Procedure.
The first step is to define the Laurent polynomial ringR[x1±1,x2±1,…,xn±1] in terms of quotient ring as
(3.3)R[x1±1,x2±1,…,xn±1]≅R[x1,y1,x2,y2,…,xn,yn](x1y1-1,x2y2-1,…,xnyn-1).
The second step is to test whether the R[x1±1,…,xn±1]-module M is projective or not. If M is projective, then again we test whether M is stably free or not, that is, we find a matrix that defines an isomorphism M⊕R[x1±1,…,xn±1]r≅R[x1±1,…,xn±1]s, for some positive integers r and s. If M is stably free, then the next step is to find finite free resolution of the projective module M. We denote the homomorphisms with their matrices to simplify the notation.
Given anR[x1±1,…,xn±1]-module M defined by a system of generators in some R[x1±1,…,xn±1]t, choose a free R[x1±1,…,xn±1]-module P0 and a surjection σ:P0→M with kernel C0, then we get an exact sequence
(3.4)0⟶C0→ψ0P0→σM⟶0.
Now, choose a free R[x1±1,…,xn±1]-module P1 together with a surjection map ϕ1:P1→C0 with kernel C1. Again we choose a free R[x1±1,…,xn±1]-module P2 together with a surjection ϕ2:P2→C1 with kernel C2, then we get an exact sequence
(3.5)0⟶C2→ψ2P2→ϕ2C1→ψ1P1→ϕ1C0→ψ0P0→σ=α0M⟶0.
Continuing this process, we have an exact sequence
(3.6)℘:0⇢Pl→αlPl-1→αl-1⇢P1→α1P0→σ=α0M⟶0,
which is a free resolution of the module M, with rank(Pi)=ti, where αl=ψl-1ϕl. For l=1(3.7)0⟶P1→α1P0→α0M⟶0
is a free resolution of module M. Since M is a projective module, this sequence splits, so there exists β1:P0→P1 such that β1α1=It1. We can compute the matrix β1 from the rows of the matrix α1. We express each vector of the canonical basis of P1 as a linear combination of the rows of α1, and with these coefficients we construct the matrix β1. Using Lemma 3.1, the exact sequence in (3.7) splits, giving another exact sequence
(3.8)0⟶M→β0P0→β1P1⟶0.
Then P1⊕M≅P0≅Ker(β1)⊕P1. Since M is a projective module, the short exact sequence
(3.9)0⟶Ker⁡(α0)⟶P0→α0M⟶0
splits, so Ker(α0)=Im(α1) is projective. Hence pd(M)=0. Suppose M is not projective, then the short exact sequence (3.9) does not split. Therefore Im(α1)=Ker(α0) is not projective. For l=2, the exact sequence (3.6) does not split, therefore Im(α2)=Ker(α1) is not projective. Continuing this process Im(α1),Im(α2),…,Im(αl-1) are not projective and Im(αl) is projective. In this way after performing a finite number of steps, we obtain the minimum length of free resolution of M which is the projective dimension of M such that pd(M)=l see Algorithm 1.Algorithm 1:
LetA=R[x1±1,x2±1,…,xn±1]≅R[x1,y1,x2,y2,…,xn,yn](x1y1-1,x2y2-1,…,xnyn-1).Objective: Computation of the projective dimension of a module.Input: A finitely generatedA-module M=<f1,f2,…,fm>⊂At anda positive integerl.Output: A projective dimension ofM and a list of matrices α1,α2,…,αlwithαi∈Mat(li-1×li,A), where i=1,2,…,l, such that℘:0⇢Pl→αlPl-1→αl-1⇢P1→α1P0→α0P0/M→0is a free resolution ofP0/M. If pd(M)=0, then M is projective. Thealgorithm returnsM⊕Ar≅Al.START:initializei=1;if (α1 does not split)α1=matrix(f1,f2,…,fm)∈Mat(l×m,A) andpd(M)=1;elseLetβ1 be the split of α1. Then pd(M)=0 andP1⊕M≅P0≅ker(β1)⊕P1;end ifif (i=l)returnpd(M);elsewhile (i<l) doi++;αi=syz(αi-1);end loopend ifif (αl does not split)pd(M)=l;elseLetβl be the split of αl. Then pd(M)=l-1;end ifreturnpd(M);STOP.Remark 3.2.
Now we give an algorithm to calculate the projective dimension of a module over the Laurent polynomial ring. We follow the technique discussed in [16] over polynomial ring.Example 3.3.
Let(3.10)A=R[x,x-1,y,y-1]≅R[x,y,p,q](xp-1,yq-1)ringR=0,(x,y,p,q),dp;idealI=xp-1,yq-1;I=std(I);I;I[1]=yq-1;I[2]=xp-1;qringA=I;A;**quotient ring from ideal**[1]=yq-1;[2]=xp-1;Polyf1=x3p+(x3p)-1+y4q;Polyf2=xp+(yq)-1+yq;IdealM=(f1,f2);ResolutionL = mres(M,0);We find a resolution ofP0/M over Laurent polynomial ring:0→P0→α0P0/M→0.
Hence, projective dimension ofM is 0.Example 3.4.
Let(3.11)A=R[x,x-1,y,y-1,z,z-1,s,s-1]≅R[x,y,z,p,q,u,s,t](xp-1,yq-1,zu-1,st-1)ringR=0,(x,y,z,p,q,u,s,t),dp;idealI=xp-1,yq-1,zu-1,st-1;I=std(I);I;I[1]=st-1I[2]=zu-1I[3]=yq-1I[4]=xp-1qringA=I;A;**quotient ring from ideal**[1]=st-1[2]=zu-1[3]=yq-1[4]=xp-1polyf=x3p+(x3p)-1+y4q+zu2;polyg=x6p2+(yq)-1+yq+s3t+yq;idealM=(f,g);resolutionL=mres(M,0).
We find a resolution ofP0/M over Laurent polynomial ring:
0→P323⊕P336→α11P30⊕P315→α10P282⊕P293→α9P262⊕P272→α8P20-1⊕P21⊕P223⊕P23⊕P24⊕P25→α7P152⊕P162⊕P172⊕P18⊕P19→α6P113⊕P122⊕P132⊕P14→α5P84⊕P92⊕P102→α4P53⊕P6⊕P7→α3P2-1⊕P3⊕P4→α2P1→α1P00→α0P0/M→0.
Hence, projective dimension ofM is 11.Remark 3.5.
Examples3.3 and 3.4 are verified using the singular software [17].


## 4. A Cancellation Theorem for Projective Modules over Laurent  Polynomial Ring

Proposition 4.1.
LetP and P′ be projective R[X,X-1]-modules and let φ:P′→P, ψ:P→P′ be injective homomorphisms. If R[X,X-1], P/φψP, and P′/ψφP′ are projective over R, then P/φP′ and P′/ψP are also projective over R.Proof.
SinceP and P′ are R[X,X-1]-projective and R[X,X-1] is R-projective, then we have
(4.1)P⊕P1≅R[X,X-1]nP′⊕P1′≅R[X,X-1]mR[X,X-1]⊕Q≅∑R⋮R[X,X-1]n⊕Qn≅∑RP⊕P1⊕Qn≅∑RP⊕Q1n≅∑R,
where P1⊕Qn=Q1n. Therefore, P is a R-projective. Similarly P′ is also a R-projective. Since
(4.2)0⟶P→βP′⟶P′/ψP⟶0,pd(P′/ψP)≤1 and similarly pd(P/φP′)≤1. Now isomorphism between P and ψP induces P/φP′≅ψP/ψφP′. We have the exact sequences(4.3)By Schanuel's lemma, P′/ψφP′⊕P≅ψP/ψφP′⊕P′ so that P′/ψφP′⊕P≅P/φP′⊕P′. Since P′/ψφP′, P are R-projective and direct sum of projective module is also projective module, then P′/ψφP′⊕P is R-projective. By the isomorphism, P/φP′⊕P′ is an R-projective. P/φP′⊕P′⊕P0≅Rn follows from the definition of R-projective module, where P0 is an R-module. Let P̅=P′⊕P0 be an R-module. Then P/φP′⊕P̅≅Rn. Hence P/φP′ is an R-projective. Similarly P′/ψP is also an R-projective.Corollary 4.2.
LetP and P′ be projective R[X,X-1]-modules with P⊃P′⊃fP, where f is a doubly monic polynomial of Laurent polynomial ring. If R[X,X-1] and R[X,X-1]/fR[X,X-1] are R-projective, then P/P′ is also R-projective.Proof.
Takeφ the inclusion map P′↪P and ψ the multiplication by f from P→P′. Then P/φψP=P/fP and P′/ψφP′=P′/fP′. Since P, P′ are R[X,X-1]-projective and R[X,X-1], R[X,X-1]/fR[X,X-1] are R-projective. Then P/fP and P′/fP′ are R[X,X-1]/fR[X,X-1]-projective. Therefore P/fP and P′/fP′ are R-projective. Hence P/φψP and P′/ψφP′ are also R-projective. By Proposition 4.1, module P/P′(=P/φP′) is R-projective.Proposition 4.3.
LetM be an R-module and f∈EndR(M). The characteristic sequence of f(4.4)0⟶M[X,X-1]→X.1M[X,X-1]-f[X,X-1]M[X,X-1]→φfMf⟶0
is an exact sequence in R[X,X-1]-module.Proof.
Clearlyφf is surjective, and we have
(4.5)φf(X.1M[X,X-1]-f[X,X-1])(∑mnXn)=φf(∑(mnXn+1-f(mn)Xn))=∑(fn+1(mn)-fn(f(mn)))=∑(fn+1(mn)-fn+1(mn))=0.Im(X.1M[X,X-1]-f[X,X-1])⊆Ker(φf). Since, X.1M[X,X-1]-f[X,X-1] raises degree by one, and preserves leading coefficients, it is monomorphism. Finally, suppose
(4.6)z=∑mnXn∈Ker⁡(φf)⟹φf(∑mnXn)=0,that is,∑fn(mn)=0.
Then
(4.7)z=z-∑fn(mn)=∑(mnXn-fn(mn))=∑(Xn.1M[X,X-1]-fn)(mn)=(X.1M[X,X-1]-f[X,X-1])[⋯-1X-fXn-fnXnfn-1X-fXn-1-fn-1Xn-1fn-1⋯-1X-fX-fXf+0+X-fX-f+X2-f2X-f+⋯+Xn-fnX-f+⋯1X-fXn-fnXnfn-1X-fXn-1-fn-1Xn-1fn-1](mn)=(X.1M[X,X-1]-f[X,X-1])∑hn(mn)⟹z∈Im⁡⁡(X.1M[X,X-1]-f[X,X-1]).
Therefore Ker(φf)⊆Im(X.1M[X,X-1]-f[X,X-1]).
HenceIm(X.1M[X,X-1]-f[X,X-1])=Ker(φf).Theorem 4.4.
LetP and P′ be finitely generated projective modules over R[X,X-1]. Suppose that P⊃P′⊃fP for some doubly monic polynomial f∈R[X,X-1]. Then P and P′ are stably isomorphic. In particular, if Pf≅Pf′, then P and P′ are stably isomorphic.Proof.
TakeM=P/P′. Since f is doubly monic polynomial, R[X,X-1]/(f) is free as an R-module. Therefore P/fP is R-projective, whence M is R-projective by Corollary 4.2. We have exact sequences of R[X,X-1]-modules(4.8)where M[X,X-1]=M⊗RR[X,X-1]. Since P′⊂P and M=P/P′, the inclusion map i from P′ to P and the surjective map π from P to M makes the first sequence an exact sequence. By Proposition 4.3, second sequence is also an exact sequence. Since M is an R-projective, M[X,X-1] is an R[X,X-1]-projective. Therefore by the Schanuel’s lemma, P⊕M[X,X-1]≅P′⊕M[X,X-1]. Hence P, P′ are stably isomorphic.Theorem 4.5.
LetF be a free R[X,X-1]-module of rank2, P a projective R[X,X-1]-module and F⊃P⊃XF. Then P≅F.Proof.
Let{f1,f2} be a free R[X,X-1]-basis for F, and let F1={a1f1+a2f2∣a1,a2∈R}. Then F1 is free over R of rank2.Claim 4.
We show that each element ofF1⊕XF⊕(1/X)F can be written as elements of F.(4.9)F1⊕XF⊕1XF⟺a1f1+a2f2+X(g1f1+g2f2)+1X(h1f1+h2f2)⟺(a1+Xg1+1Xh1)f1+(a2+Xg2+1Xh2)f2⟺Uf1+Vf2⟺F(direct sum asR-module),
where g1, g2, h1, h2, U, and V are the elements of R[X,X-1]. Thus with P1=P∩F1, P=P1⊕XF⊕(1/X)F (direct sum as R-module). By Corollary 4.2, F1/P1=F/P is a projective over R, so F1=P1⊕P1′ for some P1′. Now F=P1[X,X-1]⊕P1′[X,X-1] and P=P1⊕XF⊕(1/X)F=P1⊕X(P1[X,X-1]⊕P1′[X,X-1])⊕(1/X)(P1[X,X-1]⊕P1′[X,X-1])=P1[X,X-1]⊕P1′[X,X-1]≅F.


## 5. A Qualitative Version of Horrocks’ Theorem

Theorem 5.1.
LetR be a local ring and P be a projective R[X,X-1]-module. Then, for any doubly monic polynomial f in R[X,X-1], the R[X,X-1]-module P and the R[X,X-1]f-module Pf have the same minimal number of generators, that is, μ(P)=μ(Pf).Proof.
Letf be a doubly monic polynomial in R[X,X-1]. Since there is a natural surjective map ϕ:P→Pf, we have μ(Pf)≤μ(P). To complete the proof it is enough to show μ(Pf)≥μ(P). We have the following two cases. Case 1.
Pf is a free module over R[X,X-1]f. By Theorem 2.2, P is free and the result follows.Case 2.
SupposePf is not free over R[X,X-1]f. Then Rank(Pf)<μ(Pf). Let μ(Pf)=n. We can write Pf as a direct summand of (R[X,X-1]f)n≅(R[X,X-1]n)f. Since Rank(P)=Rank(Pf)<Rank(R[X,X-1]n). By Theorem 2.3, there exists an R[X,X-1]-module P′ such that P⊕P′≅R[X,X-1]n. Therefore μ(P)≤n=μ(Pf).



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