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

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

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

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