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 over polynomial ring . A choice of homogeneous generators of defines a surjective homomorphism , and its kernel is the first syzygy module of . Hilbert proved that the syzygies are also finitely generated (Hilbert basis theorem). Hilbert's syzygy theorem states that “every finitely generated module over has finite free resolution, that is, the th syzygy is for some .” The number of generators of the syzygies is chosen minimally, they are independent of the choice of generators. These are called the Betti numbers of and the minimal length of free resolution (i.e., ) is dimension of a module . When is projective, the dimension of is called projective dimension of (denoted by ). 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 to a quotient ring , where 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 denoted by , is a nonnegative 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) modules. Global dimension is an important technical notion in the dimension theory of Noetherian rings. Eisenbud [3] proved that a commutative Noetherian local ring is regular if and only if it has finite global dimension, in which case the global dimension coincides with the Krull dimension of . 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 .
Theorem 1.1 (see [3]). The following conditions on a ring are equivalent: (1), that is, for every module , (2) for every finitely generated module .
Serre [4] taught a course on multiplicities at the college de France. Part of that course focussed upon the simple inequality for module over an algebra . 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 AuslanderBuchsbaum equality [6]: let be a finitely generated module over a Noetherian local ring . If , then . GagoVargas [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 be a local ring, and be the set of monic polynomials of . If is a finitely generated projective module such that is free over , then is a free 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 in the Laurent polynomial ring 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 with only free (resp., projective) modules is called a free (resp., projective) resolution of .
Projective modules over the Laurent polynomial ring 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 .
Let be a ring and be an module. means the analogously defined Laurent polynomial module. We identify with , where . If is a homomorphism of modules, then this induces a homomorphism where can be identified with . Given any module and , we can make as module whose scalar multiplication is defined as . We denote this module by . There is a canonical epimorphism defined by . The characteristic sequence of 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 is a Laurent polynomial ring over a local Noetherian commutative ring , and is a projective module. If is free for some doubly monic Laurent polynomial , then is free.
In the proof of the Theorem 5.1, we need the following theorem.
Theorem 2.3 (see [12]). Suppose that is a Laurent polynomial ring over a local Noetherian commutative ring . Also suppose that and are two projective modules with . If is a direct summand of for some doubly monic polynomial , then is also a direct summand of .
3. Projective Dimension of Module
Lemma 3.1. Let , , be modules and be a split short exact sequence, and and are the splittings corresponding to and , respectively. Then the following sequence is an exact sequence.
Proof. Since and are the splittings corresponding to and , respectively, then and . Let . Then . Hence . Since is injective, there exists a unique such that . We need to show that . Let . Then and , by injectivity of . Hence . Thus there exists such that . Hence . Therefore . Conversely, let . Then there exists such that . Applying on both sides, we get . Now applying on both sides, we get and therefore . Using the defining property of and injectivity of , . Therefore . Hence . Thus . Therefore
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 in terms of quotient ring as
The second step is to test whether the module is projective or not. If is projective, then again we test whether is stably free or not, that is, we find a matrix that defines an isomorphism , for some positive integers and . If is stably free, then the next step is to find finite free resolution of the projective module . We denote the homomorphisms with their matrices to simplify the notation.
Given an module defined by a system of generators in some , choose a free module and a surjection with kernel , then we get an exact sequence
Now, choose a free module together with a surjection map with kernel . Again we choose a free module together with a surjection with kernel , then we get an exact sequence
Continuing this process, we have an exact sequence
which is a free resolution of the module , with , where . For
is a free resolution of module . Since is a projective module, this sequence splits, so there exists such that . We can compute the matrix from the rows of the matrix . We express each vector of the canonical basis of as a linear combination of the rows of , and with these coefficients we construct the matrix . Using Lemma 3.1, the exact sequence in (3.7) splits, giving another exact sequence
Then . Since is a projective module, the short exact sequence
splits, so is projective. Hence . Suppose is not projective, then the short exact sequence (3.9) does not split. Therefore is not projective. For , the exact sequence (3.6) does not split, therefore is not projective. Continuing this process are not projective and is projective. In this way after performing a finite number of steps, we obtain the minimum length of free resolution of which is the projective dimension of such that see 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
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;**quotient ring from ideal**;;Poly ;Poly ;Ideal ;Resolution = mres;We find a resolution of over Laurent polynomial ring:.
Hence, projective dimension of is 0.
Example 3.4. Let
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**quotient ring from ideal**poly ; poly ;;resolution .
We find a resolution of over Laurent polynomial ring:
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Hence, projective dimension of 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 and be projective modules and let , be injective homomorphisms. If , , and are projective over , then and are also projective over .
Proof. Since and are projective and is projective, then we have where . Therefore, is a projective. Similarly is also a projective. Since and similarly . Now isomorphism between and induces . We have the exact sequences(4.3)By Schanuel's lemma, so that . Since , are projective and direct sum of projective module is also projective module, then is projective. By the isomorphism, is an projective. follows from the definition of projective module, where is an module. Let be an module. Then . Hence is an projective. Similarly is also an projective.
Corollary 4.2. Let and be projective modules with , where is a doubly monic polynomial of Laurent polynomial ring. If and are projective, then is also projective.
Proof. Take the inclusion map and the multiplication by from . Then and . Since , are projective and , are projective. Then and are projective. Therefore and are projective. Hence and are also projective. By Proposition 4.1, module is projective.
Proposition 4.3. Let be an module and . The characteristic sequence of is an exact sequence in module.
Proof. Clearly is surjective, and we have
. Since, raises degree by one, and preserves leading coefficients, it is monomorphism. Finally, suppose
Then
Therefore .
Hence .
Theorem 4.4. Let and be finitely generated projective modules over . Suppose that for some doubly monic polynomial . Then and are stably isomorphic. In particular, if , then and are stably isomorphic.
Proof. Take . Since is doubly monic polynomial, is free as an module. Therefore is projective, whence is projective by Corollary 4.2. We have exact sequences of modules(4.8)where . Since and , the inclusion map from to and the surjective map from to makes the first sequence an exact sequence. By Proposition 4.3, second sequence is also an exact sequence. Since is an projective, is an projective. Therefore by the Schanuel’s lemma, . Hence , are stably isomorphic.
Theorem 4.5. Let be a free module of , a projective module and . Then .
Proof. Let be a free basis for , and let . Then is free over of .Claim 4. We show that each element of can be written as elements of . where , , , , , and are the elements of . Thus with , (direct sum as module). By Corollary 4.2, is a projective over , so for some . Now and .
5. A Qualitative Version of Horrocks’ Theorem
Theorem 5.1. Let be a local ring and be a projective module. Then, for any doubly monic polynomial in , the module and the module have the same minimal number of generators, that is, .
Proof. Let be a doubly monic polynomial in . Since there is a natural surjective map , we have . To complete the proof it is enough to show . We have the following two cases. Case 1. is a free module over . By Theorem 2.2, is free and the result follows.Case 2. Suppose is not free over . Then . Let . We can write as a direct summand of . Since . By Theorem 2.3, there exists an module such that . Therefore .