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

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 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 𝑀 over polynomial ring 𝑅=𝐾[𝑥₁,…,𝑥_𝑛]. A choice of 𝑎₀=dim(𝑀⊗_𝑅𝐾) homogeneous generators of 𝑀 defines a surjective homomorphism 𝑅^𝑎0→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,…,𝑥_𝑛] has finite free resolution, that is, the 𝑙th syzygy is 𝑅^𝑎𝑙 for some 𝑎_𝑙.” The number of generators 𝑎_𝑖 of the syzygies is chosen minimally, they are independent of the choice of generators. These 𝑎_𝑖 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 ring 𝑅 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 ring 𝑅 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 pd_𝑅(𝑀)≤pd_𝑅(𝑆)+pd_𝑆(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]). Let 𝑅 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 𝑃_𝑆 is free over 𝐴_𝑆, 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 version of 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 polynomial 𝑓 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 𝐹−−→_𝛼𝑖+1^𝑖𝐹_𝑖𝐹−−→_1𝛼⁰𝐹−−−_0𝜖→𝑀→0(2.1) with only free (resp., projective) modules 𝐹_𝑖(𝑖=0,1,2,…) is called a free (resp., projective) resolution of 𝑀.

Projective modules over the Laurent polynomial ring 𝑅[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.

Let 𝑅 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=−∞}𝑚_𝑛𝑋^𝑛∣𝑚_𝑛∈𝑀and𝑚_𝑛=0forallbutfinite𝑛}. If 𝑓∶𝑀→𝑁 is a homomorphism of 𝑅-modules, then this induces a homomorphism definedb𝜓y∶𝜓𝑀𝑋,𝑚𝑋⁻¹_𝑛⟶𝑁𝑋,𝑋⁻¹,𝑋^𝑛=𝑓𝑚^𝑛𝑋^𝑛,(2.2) where 𝜓 can be identified with 𝑓⊗_𝑅𝑅[𝑋,𝑋⁻¹]. Given any 𝑅-module 𝑀 and 𝑓∈End_𝑅(𝑀), we can make 𝑀 as 𝑅[𝑋,𝑋⁻¹]-module whose scalar multiplication is defined as 𝑚(∑𝑎_𝑛𝑋^𝑛)=𝑎_𝑛∑𝑓^𝑛(𝑚). We denote this 𝑅[𝑋,𝑋⁻¹]-module by _𝑓𝑀. There is a canonical 𝑅[𝑋,𝑋⁻¹]-epimorphism 𝜑_𝑓∶𝑀[𝑋,𝑋⁻¹]→_𝑓𝑀 defined by 𝜑_𝑓(∑𝑚_𝑛𝑋^𝑛)=∑𝑓^𝑛(𝑚_𝑛). 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]). Suppose 𝑅[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 𝑃_𝑓 is free for some doubly monic Laurent polynomial 𝑓, then 𝑃 is free.

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

Theorem 2.3 (see [12]). Suppose that 𝑅[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 𝑃^𝑓 is a direct summand of 𝑃_𝑓 for some doubly monic polynomial 𝑓, then 𝑃′ is also a direct summand of 𝑃.

3. Projective Dimension of Module

Lemma 3.1. Let 𝑀₁, 𝑀₂, 𝑀₃ be 𝑅-modules and 0⟶𝑀−−→_1𝛼¹𝑀−−→_2𝛼²𝑀₃⟶0(3.1) be a split short exact sequence, and 𝛽₁ and 𝛽₂ are the splittings corresponding to 𝛼₁ and 𝛼₂, respectively. Then the following sequence 0⟶𝑀−−→_3𝛽²𝑀−−→_2𝛽¹𝑀₁⟶0(3.2) is an exact sequence.

Proof. Since 𝛽₁ and 𝛽₂ are the splittings corresponding to 𝛼₁ and 𝛼₂, respectively, then 𝛽₁𝑜𝛼₁=𝐼_𝑀1 and 𝛼₂𝑜𝛽₂=𝐼_𝑀3. 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 ring 𝑅[𝑥^1±1,𝑥^2±1,…,𝑥^𝑛±1] in terms of quotient ring as 𝑅𝑥^1±1,𝑥^2±1,…,𝑥^𝑛±1≅𝑅𝑥¹,𝑦₁,𝑥₂,𝑦₂,…,𝑥_𝑛,𝑦_𝑛𝑥₁𝑦₁−1,𝑥₂𝑦₂−1,…,𝑥_𝑛𝑦_𝑛−1.(3.3) The second step is to test whether the 𝑅[𝑥^1±1,…,𝑥^𝑛±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[𝑥^1±1,…,𝑥^𝑛±1]^𝑟≅𝑅[𝑥^1±1,…,𝑥^𝑛±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 an 𝑅[𝑥^1±1,…,𝑥^𝑛±1]-module 𝑀 defined by a system of generators in some 𝑅[𝑥^1±1,…,𝑥^𝑛±1]^𝑡, choose a free 𝑅[𝑥^1±1,…,𝑥^𝑛±1]-module 𝑃₀ and a surjection 𝜎∶𝑃₀→𝑀 with kernel 𝐶₀, then we get an exact sequence 0⟶𝐶−−−→_0𝜓⁰𝑃−−_0𝜎→𝑀⟶0.(3.4) Now, choose a free 𝑅[𝑥^1±1,…,𝑥^𝑛±1]-module 𝑃₁ together with a surjection map 𝜙₁∶𝑃₁→𝐶₀ with kernel 𝐶₁. Again we choose a free 𝑅[𝑥^1±1,…,𝑥^𝑛±1]-module 𝑃₂ together with a surjection 𝜙₂∶𝑃₂→𝐶₁ with kernel 𝐶₂, then we get an exact sequence 0⟶𝐶−−−→_2𝜓²𝑃−−→_2𝜙²𝐶−−−→_1𝜓¹𝑃−−→_1𝜙¹𝐶−−−→_0𝜓⁰𝑃−−−→_0𝜎=𝛼⁰𝑀⟶0.(3.5) Continuing this process, we have an exact sequence ℘∶00𝑥021𝑒2𝑃−−→_𝑙𝛼^𝑙𝑃−−→_𝛼𝑙−1^𝑙−10𝑥021𝑒2𝑃−−→_1𝛼¹𝑃−−−→_0𝜎=𝛼⁰𝑀⟶0,(3.6) which is a free resolution of the module 𝑀, with rank(𝑃_𝑖)=𝑡_𝑖, where 𝛼_𝑙=𝜓_𝑙−1𝜙_𝑙. For 𝑙=10⟶𝑃−−→_1𝛼¹𝑃−−→_0𝛼⁰𝑀⟶0(3.7) 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=𝐼_𝑡1. 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 0⟶𝑀−−→^𝛽0𝑃−−→_0𝛽¹𝑃₁⟶0.(3.8) Then 𝑃₁⊕𝑀≅𝑃₀≅Ker(𝛽₁)⊕𝑃₁. Since 𝑀 is a projective module, the short exact sequence 0⟶Ker𝛼₀⟶𝑃−−→_0𝛼⁰𝑀⟶0(3.9) 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(𝛼_𝑙−1) are not projective and Im(𝛼_𝑙) 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.

Let     𝐴=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 generated 𝐴-module 𝑀=<𝑓1,𝑓2,…,𝑓𝑚>⊂𝐴𝑡 and a positive integer 𝑙. Output: A projective dimension of 𝑀 and a list of matrices 𝛼1,𝛼2,…,𝛼𝑙 with 𝛼𝑖∈Mat(𝑙𝑖−1×𝑙𝑖,𝐴), where 𝑖=1,2,…,𝑙, such that    ℘∶00𝑥021𝑒2𝑃−→l𝛼𝑙𝑃−−→𝛼𝑙−1𝑙−10𝑥021𝑒2𝑃−→1𝛼1𝑃−→0𝛼0𝑃0/𝑀→0 is a free resolution of 𝑃0/𝑀. If pd(𝑀)=0, then 𝑀 is projective. The algorithm returns 𝑀⊕𝐴𝑟≅𝐴𝑙. START: initialize 𝑖=1; if (𝛼1 does not split)  𝛼1=matrix(𝑓1,𝑓2,…,𝑓𝑚)∈Mat(𝑙×𝑚,𝐴) and  pd(𝑀)=1; else  Let 𝛽1 be the split of 𝛼1. Then pd(𝑀)=0 and  𝑃1⊕𝑀≅𝑃0≅ker(𝛽1)⊕𝑃1; end if if (𝑖=𝑙)  return pd(𝑀); else while (𝑖<𝑙) do  𝑖++;  𝛼𝑖=syz(𝛼𝑖−1); end loop end if if (𝛼𝑙 does not split)  pd(𝑀)=𝑙; else  Let 𝛽𝑙 be the split of 𝛼𝑙. Then pd(𝑀)=𝑙−1; end if  return pd(𝑀); STOP.

Algorithm 1 

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 𝐴=𝑅𝑥,𝑥⁻¹,𝑦,𝑦⁻¹≅𝑅[𝑥,𝑦,𝑝,𝑞](𝑥𝑝−1,𝑦𝑞−1)(3.10)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;Poly 𝑓₁=𝑥³𝑝+(𝑥³𝑝)⁻¹+𝑦⁴𝑞;Poly 𝑓₂=𝑥𝑝+(𝑦𝑞)⁻¹+𝑦𝑞;Ideal 𝑀=(𝑓₁,𝑓₂);Resolution 𝐿 = mres(𝑀,0);We find a resolution of 𝑃₀/𝑀 over Laurent polynomial ring:0→𝑃−→_0𝛼⁰𝑃₀/𝑀→0.
Hence, projective dimension of 𝑀 is 0.

Example 3.4. Let 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)(3.11)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−1poly 𝑓=𝑥³𝑝+(𝑥³𝑝)⁻¹+𝑦⁴𝑞+𝑧𝑢²; poly 𝑔=𝑥⁶𝑝²+(𝑦𝑞)⁻¹+𝑦𝑞+𝑠³𝑡+𝑦𝑞;ideal𝑀=(𝑓,𝑔);resolution 𝐿=mres(𝑀,0).
We find a resolution of 𝑃₀/𝑀 over Laurent polynomial ring:
0→𝑃³³²⊕𝑃−−→^6𝛼3311𝑃₃₀⊕𝑃−−→^5𝛼3110𝑃²²⁸⊕𝑃−→^3𝛼299𝑃²²⁶⊕𝑃−→^2𝛼278𝑃⁻¹²⁰⊕𝑃₂₁⊕𝑃³²²⊕𝑃₂₃⊕𝑃₂₄⊕𝑃₂₅−→^𝛼7𝑃²¹⁵⊕𝑃²¹⁶⊕𝑃²¹⁷⊕𝑃₁₈⊕𝑃−→_𝛼19⁶𝑃³¹¹⊕𝑃²¹²⊕𝑃²¹³⊕𝑃−→_𝛼14⁵𝑃⁴⁸⊕𝑃²⁹⊕𝑃−→^2𝛼104𝑃³⁵⊕𝑃₆⊕𝑃₇−→^𝛼3𝑃²⁻¹⊕𝑃₃⊕𝑃−→_4𝛼²𝑃−→_1𝛼¹𝑃−→^00𝛼0𝑃₀/𝑀→0.
Hence, projective dimension of 𝑀 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 𝑃 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. Since 𝑃 and 𝑃^ are 𝑅[𝑋,𝑋⁻¹]-projective and 𝑅[𝑋,𝑋⁻¹] is 𝑅-projective, then we have 𝑃⊕𝑃₁≅𝑅𝑋,𝑋⁻¹^𝑛𝑃^⊕𝑃^1≅𝑅𝑋,𝑋⁻¹^𝑚𝑅𝑅𝑋𝑋,𝑋,𝑋⁻¹₋₁⋮⊕𝑄≅𝑅𝑃⊕𝑃₁^𝑛⊕𝑄^𝑛≅𝑅𝑃⊕𝑄⊕_𝑛1𝑄^𝑛≅𝑅≅𝑅,(4.1) where 𝑃₁⊕𝑄^𝑛=𝑄^𝑛1. Therefore, 𝑃 is a 𝑅-projective. Similarly 𝑃′ is also a 𝑅-projective. Since 0⟶𝑃−−^𝛽→𝑃^⟶𝑃^/𝜓𝑃⟶0,(4.2)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≅𝑅^𝑛 follows from the definition of 𝑅-projective module, where 𝑃₀ is an 𝑅-module. Let P=𝑃^⊕𝑃₀ be an 𝑅-module. Then 𝑃/𝜑𝑃′⊕P≅𝑅^𝑛. Hence 𝑃/𝜑𝑃^ is an 𝑅-projective. Similarly 𝑃^/𝜓𝑃 is also an 𝑅-projective.

Corollary 4.2. Let 𝑃 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. Take 𝜑 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. Let 𝑀 be an 𝑅-module and 𝑓∈𝐸𝑛𝑑_R(𝑀). The characteristic sequence of 𝑓0⟶𝑀𝑋,𝑋⁻¹−−−−−−−−−−−−−−−→^{𝑋.1]𝑀[𝑋,𝑋−1−𝑓𝑋,𝑋−1}𝑀𝑋,𝑋⁻¹−−→^𝜑𝑓_𝑓𝑀⟶0(4.4) is an exact sequence in 𝑅[𝑋,𝑋⁻¹]-module.

Proof. Clearly 𝜑_𝑓 is surjective, and we have 𝜑_𝑓𝑋.1_{𝑀[𝑋,𝑋}⁻¹_]−𝑓𝑋,𝑋⁻¹𝑚_𝑛𝑋^𝑛==𝜑_𝑓𝑓_𝑛+1𝑚^𝑛𝑋^𝑛+1−𝑓𝑚_𝑛𝑋^𝑛=𝑓𝑚_𝑛−𝑓^𝑛𝑓𝑚_𝑛_𝑛+1=0.𝑚_𝑛−𝑓^𝑛+1𝑚_𝑛(4.5)Im(𝑋.1_{𝑀[𝑋,𝑋}⁻¹_]−𝑓[𝑋,𝑋⁻¹])⊆Ker(𝜑_𝑓). Since, 𝑋.1_{𝑀[𝑋,𝑋}⁻¹_]−𝑓[𝑋,𝑋⁻¹] raises degree by one, and preserves leading coefficients, it is monomorphism. Finally, suppose 𝑧=𝑚_𝑛𝑋^𝑛∈Ker𝜑_𝑓⟹𝜑_𝑓𝑚^𝑛𝑋^𝑛=0,thatis,𝑓^𝑛𝑚_𝑛=0.(4.6) Then 𝑧==𝑧−𝑚_𝑛𝑓^𝑛𝑚_𝑛=𝑋𝑋^𝑛−𝑓^𝑛𝑚_𝑛_𝑛=𝑋.1.1_{𝑀[𝑋,𝑋}⁻¹_{]𝑀[𝑋,𝑋−1]}−𝑓^𝑛𝑚_𝑛−𝑓𝑋,𝑋⁻¹⋯−1𝑋−𝑓𝑋^𝑛−𝑓^𝑛𝑋^𝑛𝑓^𝑛−1𝑋−𝑓𝑋^𝑛−1−𝑓^𝑛−1𝑋^𝑛−1𝑓^𝑛−1−1⋯𝑋−𝑓𝑋𝑋−𝑓𝑓+0+𝑋𝑋−−𝑓𝑓+𝑋²−𝑓²+𝑋𝑋−𝑓+⋯_𝑛−𝑓^𝑛=𝑋.1_{M[𝑋,𝑋−1]}𝑋−𝑓+⋯𝑚^𝑛⟹𝑧∈Im𝑋.1−𝑓_{𝑀[𝑋,𝑋−1]}𝑋,𝑋⁻¹ℎ_𝑛𝑚_𝑛−𝑓𝑋,𝑋⁻¹.(4.7) Therefore Ker(𝜑_𝑓)⊆Im(𝑋.1_{𝑀[𝑋,𝑋}⁻¹_]−𝑓[𝑋,𝑋⁻¹]).
Hence Im(𝑋.1_{𝑀[𝑋,𝑋}⁻¹_]−𝑓[𝑋,𝑋⁻¹])=Ker(𝜑_𝑓).

Theorem 4.4. Let 𝑃 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 𝑃_𝑓≅𝑃^𝑓, then 𝑃 and 𝑃^ are stably isomorphic.

Proof. Take 𝑀=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. Let 𝐹 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 of 𝐹₁⊕𝑋𝐹⊕(1/𝑋)𝐹 can be written as elements of 𝐹.𝐹₁⊕𝑋𝐹⊕1𝑋𝐹⟺𝑎¹𝑓₁+𝑎₂𝑓₂+𝑋𝑔₁𝑓₁+𝑔₂𝑓₂+1𝑋ℎ¹⟺𝑎₁𝑓₁+ℎ₂𝑓₂+𝑋𝑔₁+1𝑋ℎ¹𝑓¹+𝑎₂+𝑋𝑔₂+1𝑋ℎ²𝑓²⟺𝑈𝑓₁+𝑉𝑓₂⟺𝐹(directsumas𝑅-module),(4.9) 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⊕𝑃^1 for some 𝑃^1. Now 𝐹=𝑃₁[𝑋,𝑋⁻¹]⊕𝑃^1[𝑋,𝑋⁻¹] and 𝑃𝑃=𝑃₁⊕𝑋𝐹⊕(1/𝑋)𝐹=𝑃₁⊕𝑋(𝑃₁[𝑋,𝑋⁻¹]⊕𝑃^1[𝑋,𝑋⁻¹])⊕(1/𝑋)(𝑃₁[𝑋,𝑋⁻¹]⊕𝑃^1[𝑋,𝑋⁻¹])=𝑃₁[𝑋,𝑋⁻¹]⊕_1[𝑋,𝑋⁻¹]≅𝐹.

5. A Qualitative Version of Horrocks’ Theorem

Theorem 5.1. Let 𝑅 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 𝑃_𝑓 have the same minimal number of generators, that is, 𝜇(𝑃)=𝜇(𝑃_𝑓).

Proof. Let 𝑓 be a doubly monic polynomial in 𝑅[𝑋,𝑋⁻¹]. Since there is a natural surjective map 𝜙∶𝑃→𝑃_𝑓, we have 𝜇(𝑃_𝑓)≤𝜇(𝑃). To complete the proof it is enough to show 𝜇(𝑃_𝑓)≥𝜇(𝑃). We have the following two cases. Case 1. 𝑃_𝑓 is a free module over 𝑅[𝑋,𝑋⁻¹]_𝑓. By Theorem 2.2, 𝑃 is free and the result follows.Case 2. Suppose 𝑃_𝑓 is not free over 𝑅[𝑋,𝑋⁻¹]_𝑓. Then Rank(𝑃_𝑓)<𝜇(𝑃_𝑓). Let 𝜇(𝑃_𝑓)=𝑛. We can write 𝑃_𝑓 as a direct summand of (𝑅[𝑋,𝑋⁻¹]_𝑓)^𝑛≅(𝑅[𝑋,𝑋⁻¹]^𝑛)_𝑓. Since Rank(𝑃)=Rank(𝑃_𝑓)<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(𝑃_𝑓).