Algebra

Volume 2014 (2014), Article ID 256020, 5 pages

http://dx.doi.org/10.1155/2014/256020

## Finitely Generated Modules over Group Rings of a Direct Product of Two Cyclic Groups

Department of Mathematics and Informatics, Faculty of Science Dhar Mahraz, Sidi Mohamed Ben Abdellah University, 30000 Fez, Morocco

Received 27 August 2014; Accepted 15 November 2014; Published 1 December 2014

Academic Editor: Zhongshan Li

Copyright © 2014 Ahmed Najim and Mohammed Elhassani Charkani. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let be a commutative field of characteristic and let , where and are two finite cyclic groups. We give some structure results of finitely generated -modules in the case where the order of is divisible by . Extensions of modules are also investigated. Based on these extensions and in the same previous case, we show that -modules satisfying some conditions have a fairly simple form.

#### 1. Introduction

Let be a field of characteristic and let be a finite group. The study of -modules in the case where the order of is divisible by is a very difficult task. When is a finite abelian -group, we find in [1] the following statement: a complete classification of finitely generated -modules is available only when is cyclic or equal to , where is the cyclic group of order 2. In [2] we find this classification in these two cases. Still more, in the case where the Sylow -subgroup of is not cyclic, the groups such that and is dihedral, semidihedral, or generalized quaternion are the only groups for which we can (in principle) classify the indecomposable -modules (see [2]). These reasons just cited show the importance of the study of -modules when is of order divisible by and equal to a direct product of two cyclic groups.

Now, let be a commutative field of characteristic and let , where and are two finite cyclic groups. Let be a finitely generated -module. When is considered as a module over a subalgebra of for a subgroup of the group , we write .

In Section 2, we show that if is a cyclic -group and the characteristic of does not divide the order of , then we can have a complete system of indecomposable pairwise nonisomorphic -modules. In the rest, we assume that and are cyclic -groups. Under conditions that is a free -module and that is a free -module, we show that is a free -module. We also show that if is of order , , and is the subgroup of generated by with , then under certain conditions is a free -module. The fact that must be a free -module is one of these conditions, and exactly in the end of this section we give a result that shows when this condition is satisfied. In Section 3 and always in the case where and are cyclic -groups, we show that under some conditions -modules have a fairly simple form. But in case , and are two cyclic groups of respective orders and , ; these modules have this simple form without any other assumptions other than that they must be finitely generated over .

#### 2. Free -Modules of Finite Rank

Throughout this paper, rings are assumed to be commutative with unity. We begin this section by giving a weak version of Nakayama’s lemma with an elementary proof.

Lemma 1 (Nakayama). *Let be a -group with p odd, a ring of characteristic where is a natural number, an -module (not necessarily finitely generated), and a submodule of and . Then, one has the following:*(1)*if , then ;*(2)*if , then ;*(3)*if , , are representatives in of a generating family of , then generate .*

* Proof. *(1) Let be the order of . We have
For , , so is a natural number. So
since has characteristic . Now .

(2) If , then , and then, by (1), and then .

(3) If is the submodule generated by , then , and then by (2) we have .

*Remark 2. *Lemma 1 remains true if and is of characteristic .

*For a ring of prime characteristic and for a cyclic group of order generated by an element , we have the following lemma.*

*Lemma 3. Let with and let be a subgroup of generated by . Then, one has (as -algebras).*

*Proof. *Define
where is a well-defined -algebra homomorphism. It is easy to see that is surjective. As and are finite free modules of the same rank over , is an isomorphism.

*Remark 4. *With the notation of Lemma 3, is simply the subgroup of generated by .

*Let be a commutative field of characteristic and let be a direct product of two finite groups and . We have , where .*

*Assume that is a cyclic group of order generated by and does not divide the order of . is a principal Artinian local ring. Indeed ; this isomorphism is induced by the homomorphism defined by . is a principal Artinian local ring with residue field (up to isomorphism) whose maximal ideal is generated by . So is a principal Artinian local ring with residue field (up to isomorphism) whose maximal ideal is generated by . We have , where is a principal Artinian local ring of residue field . The characteristic of does not divide the order of . Under these conditions, we can apply [3, Theorem 3.6] to have a complete system of indecomposable pairwise nonisomorphic -modules.*

*In the remainder of this section, we assume that and are two cyclic groups of respective orders and and are generated, respectively, by and . We have . As is a commutative ring and local and is a -group, by [4, Proposition 10, page 239], is a local ring. Therefore is a local ring. As is commutative ring and local and is a -group, is a local ring by [4, Proposition 10, page 239]. So the -projective modules are free -modules.*

*Lemma 5. Let be a -module. Then, is a -module (also is a -module).*

*Proof. *This lemma is a particular case of a more general result (see [5, page 386]). But for this particular case, we can give the following direct proof: is a -module, and we have already seen that is the unique maximal ideal of and . So is a -module.

Similarly we show that is a -module.

*Proposition 6. Let be a free -module of rank . Then, is a free -module and is a free -module of the same rank .*

*Proof. *As is a local ring, -projective modules are free -modules, and therefore this proposition is only a particular case of a more general result (see [5, Lemma 2.2.]). But for this particular case, we can give the following specific proof: we have . So . Then, we have
Hence,
As (as we have already seen), . So is a free -module of rank .

Similarly we show that is a free -module of rank .

*Proposition 7. Let be a -module. If is a free -module and is a free -module, then is a free -module.*

*Proof. * is a principal Artinian local ring with residue field and is a generator of its maximal ideal. is a free -module and is a projective -submodule of . Then, , where is a projective -module and (according to [3, Proposition 4.13]). We have
So . By Nakayama’s lemma and the remark following it, . Therefore, which is projective -module. As is a local ring, is a free -module.

*Let be the Jacobson radical of for . Note that if is of characteristic (as here) and is a cyclic -group, then the Jacobson radical of is none other than , where is a generator of (see [5, page 122]).*

*Let be a finitely generated -module and a natural number such that . As , is a -module. So is a -module. is called of type if is a free -module (terminology of [6]).*

*Lemma 8. If is a -module of type with and and is the subgroup of generated by , then is a free -module.*

*Proof. *As is of type , is a free -module. Define
where is a well-defined -algebra homomorphism. It is not difficult to show that is an isomorphism (using an argument similar to that done in the proof of Lemma 3). So is a free -module.

*Theorem 9. Let be a -module of type , with , and let be the subgroup of generated by with . If is -free and , then is a free -module.*

* Proof. * is an -module -free. We have , so , and therefore . So is an -module -free. By Lemma 3, ; then is an -module -free. is a free -module, so by Lemma 8 this is a free -module. In conclusion is a -module such that
So by Proposition 7 is a free -module.

*In Theorem 9 we assumed that the -module satisfies the following condition: is -free. So it is useful to know when this condition is satisfied. This is the subject of the following result.*

*Theorem 10. Let be a -module and an element of of order . The following conditions are equivalent:(1) is free;(2);(3).*

* Proof. * Assume that is free. There exists a nonzero natural number such that . The endomorphism of defined by for all is nilpotent of nilpotency index , and is an indecomposable -module. Therefore has a basis in which the matrix of is a Jordan matrix. This matrix is formed of blocks of order all equal to
So . We can easily see that . Therefore .

Now, assume that . So . As is the number of blocks of the Jordan matrix of , the order of each block is less than or equal to , and is equal to the sum of the orders of these blocks, then the order of each block is . Therefore is equal to the number of Jordan blocks of . So ; that is, .

Assume that . So is equal to the number of Jordan blocks of . Therefore the order of each Jordan block is equal to . So the modules contained in a decomposition of as a direct sum of indecomposable modules are of the form ; that is, is free.

*3. Classification of Finitely Generated -Modules: Use of Module Extensions*

*3. Classification of Finitely Generated -Modules: Use of Module Extensions*

*Let be a finite group and let be a ring. Let and be two -modules. We put . has a natural structure as a -bimodule centralized by (see [7, section 25]). Explicitly, we have
A derivation is an -homomorphism satisfying
Derivations from into form an -module . For we equip with an -module structure by
This -module is denoted by as in [8].*

*An extension of by is an -exact sequence . Let and be a pair of extensions of by . These two extensions are equivalent if there exists an isomorphism of -modules such that and . These equivalence classes of extensions form an -module . The -modules sequence , where and denote, respectively, the canonical injection from to and the second projection from to , is exact. The equivalence class of this sequence is denoted by .*

*Remark 11. *With the previous notations, derivations and modules play the same role as the cocycles and modules defined in [8].

*From Proposition 25.10 of [7] we have the following result.*

*Proposition 12. The correspondence defined by is surjective whenever is finitely generated and projective as -module.*

*From Theorems 5.2 and 5.3 of [9] we have the following result.*

*Proposition 13. Let be a cyclic group of order generated by an element , a field of characteristic , and an indecomposable -module. Then, is isomorphic to , where is a natural number strictly less than .*

*Lemma 14. Let be a ring and a direct product of two finite groups. Let be an -module such that the action of on is trivial and let be an -module. If is isomorphic to as -modules and if we extend the action of on to by , , then is isomorphic to as -modules.*

* Proof. *Let be an isomorphism of -modules. We extend the action of on to by , . We easily see that the application is an isomorphism of -modules.

*Let be a commutative field of characteristic . Let , where and are two cyclic groups of respective orders and and are generated, respectively, by and , and let be the Jacobson radical of .*

*Proposition 15. Let be a finitely generated -module. If , then there exists a nonzero natural number such that , , as -modules, where the action of on is trivial.*

*Proof. *If , then the action of on is trivial since . By Proposition 13, there exists a nonzero natural number such that , , as -modules. Then, Lemma 14 allows concluding the following.

*Theorem 16. Let be a finitely generated -module. If , then there exist two nonzero natural numbers and and two -modules , , and , , where the action of on and is trivial, and there is a derivation from in such that .*

*Proof. *We have the exact sequence of -modules . As , . So by Proposition 15 there exists a nonzero natural number such that , , as -modules, where the action of on is trivial. We set . We have . So by Proposition 15 there exists a nonzero natural number such that , , as -modules, where the action of on is trivial. We set . Then, Proposition 12 shows that for a derivation from in .

*If , , and , then we have the following corollary.*

*Corollary 17. For all finitely generated -modules there exist two nonzero natural numbers and and two -modules , , and , , where the action of on and is trivial, and there is a derivation from in such that .*

*Proof. *We have and as since the field is of characteristic , . So for -module of finite type . The rest is a simple application of Theorem 16.

*Now we return to cases and , where is generated by an element , and let be the Jacobson radical of . For an integer with and for the subgroup of generated by , we have the following result.*

*Theorem 18. Let be a finitely generated -module, with . If is -free and of type , then there exist two nonzero natural numbers and and two -modules , , and , where the action of on is trivial, and there is a derivation from in such that .*

*Proof. *We have the following exact sequence:
As , by Proposition 15, there exists a nonzero natural number such that , , as -modules, where the action of on is trivial. is a -module of type with , more is -free, and with . Then, Theorem 9 shows that is a free -module. Therefore there exists a nonzero natural number such that . The rest is a simple application of Proposition 12.

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

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