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 the qualitative version of Laurent polynomial analogue of Horrocks' Theorem.

1. Introduction

Hilbert [1] introduced free resolution on iterated syzygies of a finitely generated graded module š‘€ over polynomial ring š‘…=š¾[š‘„1,ā€¦,š‘„š‘›]. A choice of š‘Ž0=dim(š‘€āŠ—š‘…š¾) homogeneous generators of š‘€ defines a surjective homomorphism š‘…š‘Ž0ā†’š‘€, and its kernel is the first syzygy module of š‘€. Hilbert proved that the syzygies are also finitely generated (Hilbert basis theorem). Hilbert's syzygy theorem states that ā€œevery finitely generated module š‘€ over š¾[š‘„1,ā€¦,š‘„š‘›] has finite free resolution, that is, the š‘™th syzygy is š‘…š‘Žš‘™ for some š‘Žš‘™.ā€ The number of generators š‘Žš‘– of the syzygies is chosen minimally, they are independent of the choice of generators. These š‘Žš‘– are called the Betti numbers of š‘€ and the minimal length of free resolution (i.e., š‘–) is dimension of a module š‘€. When š‘€ is projective, the dimension of š‘€ is called projective dimension of š‘€ (denoted by pd(š‘€)). 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 š‘… to a quotient ring š‘…/(š‘‹), where (š‘‹) 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 ring š‘… denoted by gldim(š‘…), 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) š‘…-modules. Global dimension is an important technical notion in the dimension theory of Noetherian rings. Eisenbud [3] proved that a commutative Noetherian local ring š‘… is regular if and only if it has finite global dimension, in which case the global dimension coincides with the Krull dimension of š‘…. 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 š‘™<āˆž.

Theorem 1.1 (see [3]). The following conditions on a ring š‘… are equivalent: (1)gldim(š‘…)ā‰¤š‘™, that is, pd(š‘€)ā‰¤š‘™ for every š‘…-module š‘€, (2)pd(š‘€)ā‰¤š‘™ for every finitely generated module š‘€.

Serre [4] taught a course on multiplicities at the college de France. Part of that course focussed upon the simple inequality pdš‘…(š‘€)ā‰¤pdš‘…(š‘†)+pdš‘†(š‘€) for module š‘€ over an š‘…-algebra š‘†. 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 š‘€ be a finitely generated module over a Noetherian local ring (š‘…,š‘š). If pd(š‘€)<āˆž, then depth(š‘…)=depth(š‘€)+pd(š‘€). 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, Sectionā€‰ā€‰5] motivated by Swan's work. Horrocks proved the following theorem.

Theorem 1.2 (see [9]). Let š‘… be a local ring, š“=š‘…[š‘„] and š‘† be the set of monic polynomials of š“. If š‘ƒ is a finitely generated projective š“-module such that š‘ƒš‘† is free over š“š‘†, then š‘ƒ is a free š“-module.

This theorem draw the attention of several mathematicians, as a result several analogues of Horrocksā€™ theorem can be found in the literature [10, 11]. Mandal proved the Laurent polynomial version of Horrocksā€™ theorem in [12] whereas Nashier [13] proved the qualitative version of the same theorem over polynomial ring. But it is observed that the qualitative version of Horrocksā€™ 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 Horrocksā€™ 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 polynomial š‘“ in the Laurent polynomial ring š‘…[š‘‹,š‘‹āˆ’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 š¹š›¼š‘–+1š‘–āˆ’āˆ’ā†’š¹š‘–š¹1š›¼0āˆ’āˆ’ā†’š¹0šœ–āˆ’āˆ’āˆ’ā†’š‘€ā†’0(2.1) with only free (resp., projective) modules š¹š‘–(š‘–=0,1,2,ā€¦) is called a free (resp., projective) resolution of š‘€.

Projective modules over the Laurent polynomial ring š‘…[š‘‹,š‘‹āˆ’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 š‘€.

Let š‘… be a ring and š‘€ be an š‘…-module. š‘€[š‘‹,š‘‹āˆ’1] means the analogously defined Laurent polynomial module. We identify š‘€āŠ—š‘…š‘…[š‘‹,š‘‹āˆ’1] with š‘€[š‘‹,š‘‹āˆ’1], where š‘€[š‘‹,š‘‹āˆ’1āˆ‘]={āˆžš‘›=āˆ’āˆžš‘šš‘›š‘‹š‘›āˆ£š‘šš‘›āˆˆš‘€andš‘šš‘›=0forallbutļ¬niteš‘›}. If š‘“āˆ¶š‘€ā†’š‘ is a homomorphism of š‘…-modules, then this induces a homomorphism ī€ŗšœ“āˆ¶š‘€š‘‹,š‘‹āˆ’1ī€»ī€ŗāŸ¶š‘š‘‹,š‘‹āˆ’1ī€»,ī‚€ī“š‘šdeļ¬nedbyšœ“š‘›š‘‹š‘›ī‚=ī“š‘“ī€·š‘šš‘›ī€øš‘‹š‘›,(2.2) where šœ“ can be identified with š‘“āŠ—š‘…š‘…[š‘‹,š‘‹āˆ’1]. Given any š‘…-module š‘€ and š‘“āˆˆEndš‘…(š‘€), we can make š‘€ as š‘…[š‘‹,š‘‹āˆ’1]-module whose scalar multiplication is defined as āˆ‘š‘Žš‘š(š‘›š‘‹š‘›)=š‘Žš‘›āˆ‘š‘“š‘›(š‘š). We denote this š‘…[š‘‹,š‘‹āˆ’1]-module by š‘“š‘€. There is a canonical š‘…[š‘‹,š‘‹āˆ’1]-epimorphism šœ‘š‘“āˆ¶š‘€[š‘‹,š‘‹āˆ’1]ā†’š‘“š‘€ defined by šœ‘š‘“(āˆ‘š‘šš‘›š‘‹š‘›āˆ‘š‘“)=š‘›(š‘šš‘›). The characteristic sequence of š‘“ 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]). Suppose š‘…[š‘‹,š‘‹āˆ’1] is a Laurent polynomial ring over a local Noetherian commutative ring š‘…, and š‘ƒ is a projective š‘…[š‘‹,š‘‹āˆ’1]-module. If š‘ƒš‘“ is free for some doubly monic Laurent polynomial š‘“, then š‘ƒ is free.

In the proof of the Theorem 5.1, we need the following theorem.

Theorem 2.3 (see [12]). Suppose that š‘…[š‘‹,š‘‹āˆ’1] is a Laurent polynomial ring over a local Noetherian commutative ring š‘…. Also suppose that š‘ƒ and š‘ƒā€² are two projective š‘…[š‘‹,š‘‹āˆ’1]-modules with Rankš‘ƒā€²<Rankš‘ƒ. If š‘ƒī…žš‘“ is a direct summand of š‘ƒš‘“ for some doubly monic polynomial š‘“, then š‘ƒā€² is also a direct summand of š‘ƒ.

3. Projective Dimension of Module

Lemma 3.1. Let š‘€1, š‘€2, š‘€3 be š‘…-modules and 0āŸ¶š‘€1š›¼1āˆ’āˆ’ā†’š‘€2š›¼2āˆ’āˆ’ā†’š‘€3āŸ¶0(3.1) be a split short exact sequence, and š›½1 and š›½2 are the splittings corresponding to š›¼1 and š›¼2, respectively. Then the following sequence 0āŸ¶š‘€3š›½2āˆ’āˆ’ā†’š‘€2š›½1āˆ’āˆ’ā†’š‘€1āŸ¶0(3.2) is an exact sequence.

Proof. Since š›½1 and š›½2 are the splittings corresponding to š›¼1 and š›¼2, respectively, then š›½1š‘œš›¼1=š¼š‘€1 and š›¼2š‘œš›½2=š¼š‘€3. Let š‘„āˆˆš‘€2. Then š›¼2(š‘„āˆ’š›½2(š›¼2(š‘„)))=š›¼2(š‘„)āˆ’š›¼2(š›½2(š›¼2(š‘„)))=š›¼2(š‘„)āˆ’š›¼2(š‘„)=0. Hence š‘„āˆ’š›½2(š›¼2(š‘„))āˆˆKer(š›¼2)=Im(š›¼1). Since š›¼1 is injective, there exists a unique š›½1(š‘„)āˆˆš‘€1 such that š›¼1(š›½1(š‘„))=š‘„āˆ’š›½2(š›¼2(š‘„)). We need to show that Ker(š›½1)āŠ‚Im(š›½2). Let š‘„āˆˆKer(š›½1). Then š›½1(š‘„)=0 and š›¼1(š›½1(š‘„))=š›¼1(0)=0, by injectivity of š›¼1. Hence š‘„āˆ’š›½2(š›¼2(š‘„))=0ā‡’š‘„=š›½2(š›¼2(š‘„)). Thus there exists š›¼2(š‘„)āˆˆš‘€3 such that š›½2(š›¼2(š‘„))=š‘„. Hence š‘„āˆˆIm(š›½2). Therefore Kerš›½1āŠ‚Im(š›½2). Conversely, let š‘„āˆˆIm(š›½2). Then there exists š‘¦āˆˆš‘€3 such that š›½2(š‘¦)=š‘„. Applying š›¼2 on both sides, we get š‘¦=š›¼2(š‘„). Now applying š›½2 on both sides, we get š›½2š›¼2(š‘„)=š‘„ and therefore š‘„āˆ’š›½2š›¼2(š‘„)=0. Using the defining property of š›½1 and injectivity of š›¼1, š›¼1(š›½1(š‘„))=0=š›¼1(0). Therefore š›½1(š‘„)=0. Hence š‘„āˆˆKer(š›½1). Thus Im(š›½2)āŠ‚Ker(š›½1). Therefore Im(š›½2)=Ker(š›½1)

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 ring š‘…[š‘„1Ā±1,š‘„2Ā±1,ā€¦,š‘„š‘›Ā±1] in terms of quotient ring as š‘…ī€ŗš‘„1Ā±1,š‘„2Ā±1,ā€¦,š‘„š‘›Ā±1ī€»ā‰…š‘…ī€ŗš‘„1,š‘¦1,š‘„2,š‘¦2,ā€¦,š‘„š‘›,š‘¦š‘›ī€»ī€·š‘„1š‘¦1āˆ’1,š‘„2š‘¦2āˆ’1,ā€¦,š‘„š‘›š‘¦š‘›ī€øāˆ’1.(3.3) The second step is to test whether the š‘…[š‘„1Ā±1,ā€¦,š‘„š‘›Ā±1]-module š‘€ is projective or not. If š‘€ is projective, then again we test whether š‘€ is stably free or not, that is, we find a matrix that defines an isomorphism š‘€āŠ•š‘…[š‘„1Ā±1,ā€¦,š‘„š‘›Ā±1]š‘Ÿā‰…š‘…[š‘„1Ā±1,ā€¦,š‘„š‘›Ā±1]š‘ , for some positive integers š‘Ÿ and š‘ . If š‘€ is stably free, then the next step is to find finite free resolution of the projective module š‘€. We denote the homomorphisms with their matrices to simplify the notation.
Given an š‘…[š‘„1Ā±1,ā€¦,š‘„š‘›Ā±1]-module š‘€ defined by a system of generators in some š‘…[š‘„1Ā±1,ā€¦,š‘„š‘›Ā±1]š‘”, choose a free š‘…[š‘„1Ā±1,ā€¦,š‘„š‘›Ā±1]-module š‘ƒ0 and a surjection šœŽāˆ¶š‘ƒ0ā†’š‘€ with kernel š¶0, then we get an exact sequence 0āŸ¶š¶0šœ“0āˆ’āˆ’āˆ’ā†’š‘ƒ0šœŽāˆ’āˆ’ā†’š‘€āŸ¶0.(3.4) Now, choose a free š‘…[š‘„1Ā±1,ā€¦,š‘„š‘›Ā±1]-module š‘ƒ1 together with a surjection map šœ™1āˆ¶š‘ƒ1ā†’š¶0 with kernel š¶1. Again we choose a free š‘…[š‘„1Ā±1,ā€¦,š‘„š‘›Ā±1]-module š‘ƒ2 together with a surjection šœ™2āˆ¶š‘ƒ2ā†’š¶1 with kernel š¶2, then we get an exact sequence 0āŸ¶š¶2šœ“2āˆ’āˆ’āˆ’ā†’š‘ƒ2šœ™2āˆ’āˆ’ā†’š¶1šœ“1āˆ’āˆ’āˆ’ā†’š‘ƒ1šœ™1āˆ’āˆ’ā†’š¶0šœ“0āˆ’āˆ’āˆ’ā†’š‘ƒ0šœŽ=š›¼0āˆ’āˆ’āˆ’ā†’š‘€āŸ¶0.(3.5) Continuing this process, we have an exact sequence ā„˜āˆ¶00š‘„021š‘’2š‘ƒš‘™š›¼š‘™āˆ’āˆ’ā†’š‘ƒš›¼š‘™āˆ’1š‘™āˆ’1āˆ’āˆ’ā†’0š‘„021š‘’2š‘ƒ1š›¼1āˆ’āˆ’ā†’š‘ƒ0šœŽ=š›¼0āˆ’āˆ’āˆ’ā†’š‘€āŸ¶0,(3.6) which is a free resolution of the module š‘€, with rank(š‘ƒš‘–)=š‘”š‘–, where š›¼š‘™=šœ“š‘™āˆ’1šœ™š‘™. For š‘™=10āŸ¶š‘ƒ1š›¼1āˆ’āˆ’ā†’š‘ƒ0š›¼0āˆ’āˆ’ā†’š‘€āŸ¶0(3.7) is a free resolution of module š‘€. Since š‘€ is a projective module, this sequence splits, so there exists š›½1āˆ¶š‘ƒ0ā†’š‘ƒ1 such that š›½1š›¼1=š¼š‘”1. We can compute the matrix š›½1 from the rows of the matrix š›¼1. We express each vector of the canonical basis of š‘ƒ1 as a linear combination of the rows of š›¼1, and with these coefficients we construct the matrix š›½1. Using Lemma 3.1, the exact sequence in (3.7) splits, giving another exact sequence 0āŸ¶š‘€š›½0āˆ’āˆ’ā†’š‘ƒ0š›½1āˆ’āˆ’ā†’š‘ƒ1āŸ¶0.(3.8) Then š‘ƒ1āŠ•š‘€ā‰…š‘ƒ0ā‰…Ker(š›½1)āŠ•š‘ƒ1. Since š‘€ is a projective module, the short exact sequence ī€·š›¼0āŸ¶Ker0ī€øāŸ¶š‘ƒ0š›¼0āˆ’āˆ’ā†’š‘€āŸ¶0(3.9) splits, so Ker(š›¼0)=Im(š›¼1) is projective. Hence pd(š‘€)=0. Suppose š‘€ is not projective, then the short exact sequence (3.9) does not split. Therefore Im(š›¼1)=Ker(š›¼0) is not projective. For š‘™=2, the exact sequence (3.6) does not split, therefore Im(š›¼2)=Ker(š›¼1) is not projective. Continuing this process Im(š›¼1),Im(š›¼2),ā€¦,Im(š›¼š‘™āˆ’1) are not projective and Im(š›¼š‘™) is projective. In this way after performing a finite number of steps, we obtain the minimum length of free resolution of š‘€ which is the projective dimension of š‘€ such that pd(š‘€)=š‘™ see Algorithm 1.

ā€ƒā€ƒā€‚ š“ = š‘… [ š‘„ 1 Ā± 1 , š‘„ 2 Ā± 1 , ā€¦ , š‘„ š‘› Ā± 1 ] ā‰… š‘… [ š‘„ 1 , š‘¦ 1 , š‘„ 2 , š‘¦ 2 , ā€¦ , š‘„ š‘› , š‘¦ š‘› ] ( š‘„ 1 š‘¦ 1 āˆ’ 1 , š‘„ 2 š‘¦ 2 āˆ’ 1 , ā€¦ , š‘„ š‘› š‘¦ š‘› āˆ’ 1 ) .
Objective: Computation of the projective dimension of a module.
Input: A finitely generated š“ -module š‘€ = < š‘“ 1 , š‘“ 2 , ā€¦ , š‘“ š‘š > āŠ‚ š“ š‘” and
a positive integer š‘™ .
Output: A projective dimension of š‘€ and a list of matrices š›¼ 1 , š›¼ 2 , ā€¦ , š›¼ š‘™
with š›¼ š‘– āˆˆ M a t ( š‘™ š‘– āˆ’ 1 Ɨ š‘™ š‘– , š“ ) , where š‘– = 1 , 2 , ā€¦ , š‘™ , such that
ā€ƒā€ƒā€ƒ ā„˜ āˆ¶ 0 0 š‘„ 0 2 1 š‘’ 2 š‘ƒ l š›¼ š‘™ āˆ’ ā†’ š‘ƒ š›¼ š‘™ āˆ’ 1 š‘™ āˆ’ 1 āˆ’ āˆ’ ā†’ 0 š‘„ 0 2 1 š‘’ 2 š‘ƒ 1 š›¼ 1 āˆ’ ā†’ š‘ƒ 0 š›¼ 0 āˆ’ ā†’ š‘ƒ 0 / š‘€ ā†’ 0
is a free resolution of š‘ƒ 0 / š‘€ . If p d ( š‘€ ) = 0 , then š‘€ is projective. The
algorithm returns š‘€ āŠ• š“ š‘Ÿ ā‰… š“ š‘™ .
initialize š‘– = 1 ;
if ( š›¼ 1 does not split)
ā€ƒ š›¼ 1 = m a t r i x ( š‘“ 1 , š‘“ 2 , ā€¦ , š‘“ š‘š ) āˆˆ M a t ( š‘™ Ɨ š‘š , š“ ) and
ā€ƒ p d ( š‘€ ) = 1 ;
ā€ƒLet š›½ 1 be the split of š›¼ 1 . Then p d ( š‘€ ) = 0 and
ā€ƒ š‘ƒ 1 āŠ• š‘€ ā‰… š‘ƒ 0 ā‰… k e r ( š›½ 1 ) āŠ• š‘ƒ 1 ;
end if
if ( š‘– = š‘™ )
ā€ƒreturn p d ( š‘€ ) ;
while ( š‘– < š‘™ ) do
ā€ƒ š‘– + + ;
ā€ƒ š›¼ š‘– = s y z ( š›¼ š‘– āˆ’ 1 ) ;
end loop
end if
if ( š›¼ š‘™ does not split)
ā€ƒ p d ( š‘€ ) = š‘™ ;
ā€ƒLet š›½ š‘™ be the split of š›¼ š‘™ . Then p d ( š‘€ ) = š‘™ āˆ’ 1 ;
end if
ā€ƒreturn p d ( š‘€ ) ;

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 ī€ŗš“=š‘…š‘„,š‘„āˆ’1,š‘¦,š‘¦āˆ’1ī€»ā‰…š‘…[]š‘„,š‘¦,š‘,š‘ž(š‘„š‘āˆ’1,š‘¦š‘žāˆ’1)(3.10)ringš‘…=0,(š‘„,š‘¦,š‘,š‘ž),š‘‘š‘;idealš¼=š‘„š‘āˆ’1,š‘¦š‘žāˆ’1;š¼=std(š¼); š¼; š¼[1]=š‘¦š‘žāˆ’1; š¼[2]=š‘„š‘āˆ’1; qringš“=š¼; š“;**quotient ring from ideal**[1]=š‘¦š‘žāˆ’1;[2]=š‘„š‘āˆ’1;Poly š‘“1=š‘„3š‘+(š‘„3š‘)āˆ’1+š‘¦4š‘ž;Poly š‘“2=š‘„š‘+(š‘¦š‘ž)āˆ’1+š‘¦š‘ž;Ideal š‘€=(š‘“1,š‘“2);Resolution šæ = mres(š‘€,0);We find a resolution of š‘ƒ0/š‘€ over Laurent polynomial ring:0ā†’š‘ƒ0š›¼0āˆ’ā†’š‘ƒ0/š‘€ā†’0.
Hence, projective dimension of š‘€ is 0.

Example 3.4. Let ī€ŗš“=š‘…š‘„,š‘„āˆ’1,š‘¦,š‘¦āˆ’1,š‘§,š‘§āˆ’1,š‘ ,š‘ āˆ’1ī€»ā‰…š‘…[]š‘„,š‘¦,š‘§,š‘,š‘ž,š‘¢,š‘ ,š‘”(š‘„š‘āˆ’1,š‘¦š‘žāˆ’1,š‘§š‘¢āˆ’1,š‘ š‘”āˆ’1)(3.11)ringš‘…=0,(š‘„,š‘¦,š‘§,š‘,š‘ž,š‘¢,š‘ ,š‘”),š‘‘š‘; idealš¼=š‘„š‘āˆ’1,š‘¦š‘žāˆ’1,š‘§š‘¢āˆ’1,š‘ š‘”āˆ’1; š¼=std(š¼); š¼; š¼[1]=š‘ š‘”āˆ’1š¼[2]=š‘§š‘¢āˆ’1š¼[3]=š‘¦š‘žāˆ’1š¼[4]=š‘„š‘āˆ’1qringš“=š¼; š“; **quotient ring from ideal**[1]=š‘ š‘”āˆ’1[2]=š‘§š‘¢āˆ’1[3]=š‘¦š‘žāˆ’1[4]=š‘„š‘āˆ’1poly š‘“=š‘„3š‘+(š‘„3š‘)āˆ’1+š‘¦4š‘ž+š‘§š‘¢2; poly š‘”=š‘„6š‘2+(š‘¦š‘ž)āˆ’1+š‘¦š‘ž+š‘ 3š‘”+š‘¦š‘ž;idealš‘€=(š‘“,š‘”);resolution šæ=mres(š‘€,0).
We find a resolution of š‘ƒ0/š‘€ over Laurent polynomial ring:
Hence, projective dimension of š‘€ is 11.

Remark 3.5. Examples 3.3 and 3.4 are verified using the singular software [17].

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

Proposition 4.1. Let š‘ƒ and š‘ƒī…ž be projective š‘…[š‘‹,š‘‹āˆ’1]-modules and let šœ‘āˆ¶š‘ƒā€²ā†’š‘ƒ, šœ“āˆ¶š‘ƒā†’š‘ƒī…ž be injective homomorphisms. If š‘…[š‘‹,š‘‹āˆ’1], š‘ƒ/šœ‘šœ“š‘ƒ, and š‘ƒī…ž/šœ“šœ‘š‘ƒī…ž are projective over š‘…, then š‘ƒ/šœ‘š‘ƒī…ž and š‘ƒī…ž/šœ“š‘ƒ are also projective over š‘….

Proof. Since š‘ƒ and š‘ƒī…ž are š‘…[š‘‹,š‘‹āˆ’1]-projective and š‘…[š‘‹,š‘‹āˆ’1] is š‘…-projective, then we have š‘ƒāŠ•š‘ƒ1ī€ŗā‰…š‘…š‘‹,š‘‹āˆ’1ī€»š‘›š‘ƒī…žāŠ•š‘ƒī…ž1ī€ŗā‰…š‘…š‘‹,š‘‹āˆ’1ī€»š‘šš‘…ī€ŗš‘‹,š‘‹āˆ’1ī€»ī“š‘…ā‹®š‘…ī€ŗāŠ•š‘„ā‰…š‘‹,š‘‹āˆ’1ī€»š‘›āŠ•š‘„š‘›ā‰…ī“š‘…š‘ƒāŠ•š‘ƒ1āŠ•š‘„š‘›ā‰…ī“š‘…š‘ƒāŠ•š‘„š‘›1ā‰…ī“š‘…,(4.1) where š‘ƒ1āŠ•š‘„š‘›=š‘„š‘›1. Therefore, š‘ƒ is a š‘…-projective. Similarly š‘ƒā€² is also a š‘…-projective. Since 0āŸ¶š‘ƒš›½āˆ’āˆ’ā†’š‘ƒī…žāŸ¶š‘ƒī…ž/šœ“š‘ƒāŸ¶0,(4.2)pd(š‘ƒī…ž/šœ“š‘ƒ)ā‰¤1 and similarly pd(š‘ƒ/šœ‘š‘ƒī…ž)ā‰¤1. Now isomorphism between š‘ƒ and šœ“š‘ƒ induces š‘ƒ/šœ‘š‘ƒī…žā‰…šœ“š‘ƒ/šœ“šœ‘š‘ƒī…žhttps://static.hindawi.com/articles/isrn/volume-2011/926165/figures/thumbnails/926165.alg1_th.jpg . We have the exact sequences926165.fig.001(4.3)By Schanuel's lemma, š‘ƒī…ž/šœ“šœ‘š‘ƒī…žāŠ•š‘ƒā‰…šœ“š‘ƒ/šœ“šœ‘š‘ƒī…žāŠ•š‘ƒī…ž so that š‘ƒī…ž/šœ“šœ‘š‘ƒī…žāŠ•š‘ƒā‰…š‘ƒ/šœ‘š‘ƒī…žāŠ•š‘ƒī…ž. Since š‘ƒī…ž/šœ“šœ‘š‘ƒī…ž, š‘ƒ are š‘…-projective and direct sum of projective module is also projective module, then š‘ƒī…ž/šœ“šœ‘š‘ƒī…žāŠ•š‘ƒ is š‘…-projective. By the isomorphism, š‘ƒ/šœ‘š‘ƒī…žāŠ•š‘ƒī…ž is an š‘…-projective. š‘ƒ/šœ‘š‘ƒī…žāŠ•š‘ƒī…žāŠ•š‘ƒ0ā‰…š‘…š‘› follows from the definition of š‘…-projective module, where š‘ƒ0 is an š‘…-module. Let P=š‘ƒī…žāŠ•š‘ƒ0 be an š‘…-module. Then š‘ƒ/šœ‘š‘ƒā€²āŠ•Pā‰…š‘…š‘›. Hence š‘ƒ/šœ‘š‘ƒī…ž is an š‘…-projective. Similarly š‘ƒī…ž/šœ“š‘ƒ is also an š‘…-projective.

Corollary 4.2. Let š‘ƒ and š‘ƒī…ž be projective š‘…[š‘‹,š‘‹āˆ’1]-modules with š‘ƒāŠƒš‘ƒī…žāŠƒš‘“š‘ƒ, where š‘“ is a doubly monic polynomial of Laurent polynomial ring. If š‘…[š‘‹,š‘‹āˆ’1] and š‘…[š‘‹,š‘‹āˆ’1]/š‘“š‘…[š‘‹,š‘‹āˆ’1] are š‘…-projective, then š‘ƒ/š‘ƒī…ž is also š‘…-projective.

Proof. Take šœ‘ the inclusion map š‘ƒā€²ā†Ŗš‘ƒ and šœ“ the multiplication by š‘“ from š‘ƒā†’š‘ƒī…ž. Then š‘ƒ/šœ‘šœ“š‘ƒ=š‘ƒ/š‘“š‘ƒ and š‘ƒī…ž/šœ“šœ‘š‘ƒī…ž=š‘ƒī…ž/š‘“š‘ƒī…ž. Since š‘ƒ, š‘ƒī…ž are š‘…[š‘‹,š‘‹āˆ’1]-projective and š‘…[š‘‹,š‘‹āˆ’1], š‘…[š‘‹,š‘‹āˆ’1]/š‘“š‘…[š‘‹,š‘‹āˆ’1] are š‘…-projective. Then š‘ƒ/š‘“š‘ƒ and š‘ƒī…ž/š‘“š‘ƒī…ž are š‘…[š‘‹,š‘‹āˆ’1]/š‘“š‘…[š‘‹,š‘‹āˆ’1]-projective. Therefore š‘ƒ/š‘“š‘ƒ and š‘ƒī…ž/š‘“š‘ƒī…ž are š‘…-projective. Hence š‘ƒ/šœ‘šœ“š‘ƒ and š‘ƒī…ž/šœ“šœ‘š‘ƒī…ž are also š‘…-projective. By Proposition 4.1, module š‘ƒ/š‘ƒī…ž(=š‘ƒ/šœ‘š‘ƒī…ž) is š‘…-projective.

Proposition 4.3. Let š‘€ be an š‘…-module and š‘“āˆˆšøš‘›š‘‘R(š‘€). The characteristic sequence of š‘“ī€ŗ0āŸ¶š‘€š‘‹,š‘‹āˆ’1ī€»š‘‹.1]š‘€[š‘‹,š‘‹āˆ’1ī€ŗāˆ’š‘“š‘‹,š‘‹āˆ’1ī€»ī€ŗāˆ’āˆ’āˆ’āˆ’āˆ’āˆ’āˆ’āˆ’āˆ’āˆ’āˆ’āˆ’āˆ’āˆ’āˆ’ā†’š‘€š‘‹,š‘‹āˆ’1ī€»šœ‘š‘“āˆ’āˆ’ā†’š‘“š‘€āŸ¶0(4.4) is an exact sequence in š‘…[š‘‹,š‘‹āˆ’1]-module.

Proof. Clearly šœ‘š‘“ is surjective, and we have šœ‘š‘“ī€·š‘‹.1š‘€[š‘‹,š‘‹āˆ’1]ī€ŗāˆ’š‘“š‘‹,š‘‹āˆ’1ī‚€ī“š‘šī€»ī€øš‘›š‘‹š‘›ī‚=šœ‘š‘“ī‚€ī“ī€·š‘šš‘›š‘‹š‘›+1ī€·š‘šāˆ’š‘“š‘›ī€øš‘‹š‘›ī€øī‚=ī“ī€·š‘“š‘›+1ī€·š‘šš‘›ī€øāˆ’š‘“š‘›ī€·š‘“ī€·š‘šš‘›=ī“ī€·š‘“ī€øī€øī€øš‘›+1ī€·š‘šš‘›ī€øāˆ’š‘“š‘›+1ī€·š‘šš‘›ī€øī€ø=0.(4.5)Im(š‘‹.1š‘€[š‘‹,š‘‹āˆ’1]āˆ’š‘“[š‘‹,š‘‹āˆ’1])āŠ†Ker(šœ‘š‘“). Since, š‘‹.1š‘€[š‘‹,š‘‹āˆ’1]āˆ’š‘“[š‘‹,š‘‹āˆ’1] raises degree by one, and preserves leading coefficients, it is monomorphism. Finally, suppose ī“š‘šš‘§=š‘›š‘‹š‘›ī€·šœ‘āˆˆKerš‘“ī€øāŸ¹šœ‘š‘“ī‚€ī“š‘šš‘›š‘‹š‘›ī‚ī“š‘“=0,thatis,š‘›ī€·š‘šš‘›ī€ø=0.(4.6) Then ī“š‘“š‘§=š‘§āˆ’š‘›ī€·š‘šš‘›ī€ø=ī“ī€·š‘šš‘›š‘‹š‘›āˆ’š‘“š‘›ī€·š‘šš‘›=ī“ī€·š‘‹ī€øī€øš‘›.1š‘€[š‘‹,š‘‹āˆ’1]āˆ’š‘“š‘›š‘šī€øī€·š‘›ī€ø=ī€·š‘‹.1š‘€[š‘‹,š‘‹āˆ’1]ī€ŗāˆ’š‘“š‘‹,š‘‹āˆ’1ī‚ø1ī€»ī€øā‹Æāˆ’š‘‹š‘‹āˆ’š‘“š‘›āˆ’š‘“š‘›š‘‹š‘›š‘“š‘›āˆ’1š‘‹š‘‹āˆ’š‘“š‘›āˆ’1āˆ’š‘“š‘›āˆ’1š‘‹š‘›āˆ’1š‘“š‘›āˆ’1ā‹Æāˆ’1š‘‹āˆ’š‘“š‘‹āˆ’š‘“š‘‹š‘“+0+š‘‹āˆ’š‘“+š‘‹š‘‹āˆ’š‘“2āˆ’š‘“2+š‘‹š‘‹āˆ’š‘“+ā‹Æš‘›āˆ’š‘“š‘›ī‚¹ī€·š‘šš‘‹āˆ’š‘“+ā‹Æš‘›ī€ø=ī€·š‘‹.1M[š‘‹,š‘‹āˆ’1]ī€ŗāˆ’š‘“š‘‹,š‘‹āˆ’1ī“ā„Žī€»ī€øš‘›ī€·š‘šš‘›ī€øī€·āŸ¹š‘§āˆˆImš‘‹.1š‘€[š‘‹,š‘‹āˆ’1]ī€ŗāˆ’š‘“š‘‹,š‘‹āˆ’1.ī€»ī€ø(4.7) Therefore Ker(šœ‘š‘“)āŠ†Im(š‘‹.1š‘€[š‘‹,š‘‹āˆ’1]āˆ’š‘“[š‘‹,š‘‹āˆ’1]).
Hence Im(š‘‹.1š‘€[š‘‹,š‘‹āˆ’1]āˆ’š‘“[š‘‹,š‘‹āˆ’1])=Ker(šœ‘š‘“).

Theorem 4.4. Let š‘ƒ and š‘ƒī…ž be finitely generated projective modules over š‘…[š‘‹,š‘‹āˆ’1]. Suppose that š‘ƒāŠƒš‘ƒī…žāŠƒš‘“š‘ƒ for some doubly monic polynomial š‘“āˆˆš‘…[š‘‹,š‘‹āˆ’1]. Then š‘ƒ and š‘ƒī…ž are stably isomorphic. In particular, if š‘ƒš‘“ā‰…š‘ƒī…žš‘“, then š‘ƒ and š‘ƒī…ž are stably isomorphic.

Proof. Take š‘€=š‘ƒ/š‘ƒī…ž. Since š‘“ is doubly monic polynomial, š‘…[š‘‹,š‘‹āˆ’1]/(š‘“) is free as an š‘…-module. Therefore š‘ƒ/š‘“š‘ƒ is š‘…-projective, whence š‘€ is š‘…-projective by Corollary 4.2. We have exact sequences of š‘…[š‘‹,š‘‹āˆ’1]-modules926165.fig.002(4.8)where š‘€[š‘‹,š‘‹āˆ’1]=š‘€āŠ—š‘…š‘…[š‘‹,š‘‹āˆ’1]. Since š‘ƒī…žāŠ‚š‘ƒ and š‘€=š‘ƒ/š‘ƒī…ž, the inclusion map š‘– from š‘ƒī…ž to š‘ƒ and the surjective map šœ‹ from š‘ƒ to š‘€ makes the first sequence an exact sequence. By Proposition 4.3, second sequence is also an exact sequence. Since š‘€ is an š‘…-projective, š‘€[š‘‹,š‘‹āˆ’1] is an š‘…[š‘‹,š‘‹āˆ’1]-projective. Therefore by the Schanuelā€™s lemma, š‘ƒāŠ•š‘€[š‘‹,š‘‹āˆ’1]ā‰…š‘ƒā€²āŠ•š‘€[š‘‹,š‘‹āˆ’1]. Hence š‘ƒ, š‘ƒā€² are stably isomorphic.

Theorem 4.5. Let š¹ be a free š‘…[š‘‹,š‘‹āˆ’1]-module of rank2, š‘ƒ a projective š‘…[š‘‹,š‘‹āˆ’1]-module and š¹āŠƒš‘ƒāŠƒš‘‹š¹. Then š‘ƒā‰…š¹.

Proof. Let {š‘“1,š‘“2} be a free š‘…[š‘‹,š‘‹āˆ’1]-basis for š¹, and let š¹1={š‘Ž1š‘“1+š‘Ž2š‘“2āˆ£š‘Ž1,š‘Ž2āˆˆš‘…}. Then š¹1 is free over š‘… of rank2.Claim 4. We show that each element of š¹1āŠ•š‘‹š¹āŠ•(1/š‘‹)š¹ can be written as elements of š¹.š¹11āŠ•š‘‹š¹āŠ•š‘‹š¹āŸŗš‘Ž1š‘“1+š‘Ž2š‘“2ī€·š‘”+š‘‹1š‘“1+š‘”2š‘“2ī€ø+1š‘‹ī€·ā„Ž1š‘“1+ā„Ž2š‘“2ī€øāŸŗī‚€š‘Ž1+š‘‹š‘”1+1š‘‹ā„Ž1ī‚š‘“1+ī‚€š‘Ž2+š‘‹š‘”2+1š‘‹ā„Ž2ī‚š‘“2āŸŗš‘ˆš‘“1+š‘‰š‘“2āŸŗš¹(directsumasš‘…-module),(4.9) where š‘”1, š‘”2, ā„Ž1, ā„Ž2, š‘ˆ, and š‘‰ are the elements of š‘…[š‘‹,š‘‹āˆ’1]. Thus with š‘ƒ1=š‘ƒāˆ©š¹1, š‘ƒ=š‘ƒ1āŠ•š‘‹š¹āŠ•(1/š‘‹)š¹ (direct sum as š‘…-module). By Corollary 4.2, š¹1/š‘ƒ1=š¹/š‘ƒ is a projective over š‘…, so š¹1=š‘ƒ1āŠ•š‘ƒī…ž1 for some š‘ƒī…ž1. Now š¹=š‘ƒ1[š‘‹,š‘‹āˆ’1]āŠ•š‘ƒī…ž1[š‘‹,š‘‹āˆ’1] and š‘ƒ=š‘ƒ1āŠ•š‘‹š¹āŠ•(1/š‘‹)š¹=š‘ƒ1āŠ•š‘‹(š‘ƒ1[š‘‹,š‘‹āˆ’1]āŠ•š‘ƒī…ž1[š‘‹,š‘‹āˆ’1])āŠ•(1/š‘‹)(š‘ƒ1[š‘‹,š‘‹āˆ’1]āŠ•š‘ƒī…ž1[š‘‹,š‘‹āˆ’1])=š‘ƒ1[š‘‹,š‘‹āˆ’1š‘ƒ]āŠ•ī…ž1[š‘‹,š‘‹āˆ’1]ā‰…š¹.

5. A Qualitative Version of Horrocksā€™ Theorem

Theorem 5.1. Let š‘… be a local ring and š‘ƒ be a projective š‘…[š‘‹,š‘‹āˆ’1]-module. Then, for any doubly monic polynomial š‘“ in š‘…[š‘‹,š‘‹āˆ’1], the š‘…[š‘‹,š‘‹āˆ’1]-module š‘ƒ and the š‘…[š‘‹,š‘‹āˆ’1]š‘“-module š‘ƒš‘“ have the same minimal number of generators, that is, šœ‡(š‘ƒ)=šœ‡(š‘ƒš‘“).

Proof. Let š‘“ be a doubly monic polynomial in š‘…[š‘‹,š‘‹āˆ’1]. Since there is a natural surjective map šœ™āˆ¶š‘ƒā†’š‘ƒš‘“, we have šœ‡(š‘ƒš‘“)ā‰¤šœ‡(š‘ƒ). To complete the proof it is enough to show šœ‡(š‘ƒš‘“)ā‰„šœ‡(š‘ƒ). We have the following two cases. Case 1. š‘ƒš‘“ is a free module over š‘…[š‘‹,š‘‹āˆ’1]š‘“. By Theorem 2.2, š‘ƒ is free and the result follows.Case 2. Suppose š‘ƒš‘“ is not free over š‘…[š‘‹,š‘‹āˆ’1]š‘“. Then Rank(š‘ƒš‘“)<šœ‡(š‘ƒš‘“). Let šœ‡(š‘ƒš‘“)=š‘›. We can write š‘ƒš‘“ as a direct summand of (š‘…[š‘‹,š‘‹āˆ’1]š‘“)š‘›ā‰…(š‘…[š‘‹,š‘‹āˆ’1]š‘›)š‘“. Since Rank(š‘ƒ)=Rank(š‘ƒš‘“)<Rank(š‘…[š‘‹,š‘‹āˆ’1]š‘›). By Theorem 2.3, there exists an š‘…[š‘‹,š‘‹āˆ’1]-module š‘ƒā€² such that š‘ƒāŠ•š‘ƒā€²ā‰…š‘…[š‘‹,š‘‹āˆ’1]š‘›. Therefore šœ‡(š‘ƒ)ā‰¤š‘›=šœ‡(š‘ƒš‘“).