AlgebraVolume 2013 (2013), Article ID 873193, 6 pageshttp://dx.doi.org/10.1155/2013/873193

Research Article

On Quasi--Dense Submodules and -Pure Envelopes of QTAG Modules

Alveera Mehdi, Fahad Sikander, and Firdhousi Begum

Department of Mathematics, Aligarh Muslim University, Aligarh 202 002, India

Received 27 March 2013; Accepted 4 July 2013

Academic Editor: Antonio M. Cegarra

Copyright © 2013 Alveera Mehdi et al. 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.

A module over an associative ring with unity is a QTAG module if every finitely generated submodule of any homomorphic image of is a direct sum of uniserial modules. There are many fascinating properties of QTAG modules of which -pure submodules and high submodules are significant. A submodule is quasi--dense in if is -divisible, for every -pure submodule of containing Here we study these submodules and obtain some interesting results. Motivated by -neat envelope, we also define -pure envelope of a submodule as the -pure submodule if has no direct summand containing We find that -pure envelopes of have isomorphic basic submodules, and if is the direct sum of uniserial modules, then all -pure envelopes of are isomorphic.

1. Introduction

All the rings considered here are associative with unity, and right modules are unital modules. An element is uniform, if is a nonzero uniform (hence uniserial) module and for any -module with a unique decomposition series, denotes its decomposition length. For a uniform element , , and are the exponent and height of in , respectively. denotes the submodule of generated by the elements of height at least , and is the submodule of generated by the elements of exponent at most .   is -divisible if , and it is -reduced if it does not contain any -divisible submodule. In other words, it is free from the elements of infinite height.

The modules , form a neighbourhood system of zero giving rise to -topology. The closure of a submodule is defined as , and it is closed with respect to -topology if .

A submodule of is -pure in if , for every integer . For a limit ordinal , , for all ordinals , and it is -pure in if for all ordinals .

A module is summable if , where is the set of all elements of which are not in , where is the length of . A submodule is nice [1, Definition 2.3] in , if for all ordinals ; that is, every coset of modulo may be represented by an element of the same height.

The cardinality of the minimal generating set of is denoted by . For all ordinals , is the - invariant of and it is equal to .

For a module , there is a chain of submodules , for some ordinal . , where is the submodule of . Singh [2] proved that the results which hold for TAG modules also hold good for modules.

2. Quasi--Dense Submodules

In [3], we studied semi--pure submodules which are not -pure but contained in -pure submodules. Now we investigate the submodules such that is -divisible for every -pure submodule , containing . These modules are called quasi--dense submodules.

We start with the following.

Definition 1. A submodule of is quasi--dense in , if for every -pure submodule , containing ,   is -divisible.

Lemma 2. If for some integer , then there exists a proper submodule of containing and a bounded submodule of such that (i); (ii); (iii).

Proof. The socle of can be expressed as the direct sum of where and . Now there exists a -pure submodule of such that . Also is -pure in and . Now Soc which is contained in ; therefore is an absolute direct summand of [4]. Thus there exists a submodule of such that and ; therefore and .

Proposition 3. A submodule of a module is quasi--dense in if and only if for all integers .

Proof. Let be the least positive integer such that . By Lemma 2, there exists a proper submodule containing and a bounded submodule of such that . This contradiction proves that , for all integers .For the converse, consider an -pure submodule . If is not -divisible, then there exist submodules and of such that , where is bounded. If and , then . Since is -pure in , we have . On repeating the process, after a finite number of steps, we get and , which is a contradiction. Hence, is -divisible and is quasi--dense in .

Remark 4. For a submodule of a module , the following are equivalent: (i) is quasi--dense in ; (ii); (iii)for every -pure submodule containing , is -divisible.

Lemma 5. If is an -divisible submodule of a module and is a -high submodule of , then is -pure in .

Proof. For an -divisible submodule  , -high submodules are -neat [5]. Therefore, . Assume that for all . Let . Now there exists such that . If such that , then for some and . Now there exists such that and . By induction and is -pure in .

Theorem 6. Let be a proper submodule of . Then is not semi--pure if and only if the following hold: (i) is quasi--dense in ;(ii)there exists such that .

Proof. By Lemma 2 and Proposition 3,   is satisfied. Suppose , for all . If there exists a nonnegative integer such that then we have Let be the -neat envelope [6] of in . Then is -pure in . Since is quasi--dense in ,   is -divisible. Thus, there exists a submodule of such that and Now is -pure in , hence and . Therefore, we may assume that there exists a chain of integers such that Then we may choose such that for every and we have . Let be an -high submodule of such that . There exists a proper unbounded basic submodule of and a submodule , containing such that . Now is -pure in which is a contradiction proving (ii).Conversely, suppose is an -pure submodule of containing . By (i),   is -divisible. Since and is -pure in , we have .

Now we should mention the following notations used by Khan and Zubair [7]: We prove the following.

Proposition 7. Let be an -pure submodule of a QTAG module . Then , for all .

Proof. Since and , there exists a homomorphism such that . If and , then there exist and such that and . Moreover if , , , then is contained in and there exists such that , where and for some and , respectively. Since and , we have . Also is one to one by definition; hence is an isomorphism.

Theorem 8. Let be a semi--pure submodule in having an -pure hull. Then there exists an integer such that , for .

Proof. Let be an -pure hull of in . By Theorem 6, there exists a nonnegative integer such that and for all , Therefore , for all . Hence by Proposition 7, for all .

Proposition 9. Let be an -pure hull of a semi--pure submodule of and a nonnegative integer. Then and if and only if for  all and .

Proof. Since for all , by Proposition 7,  . If , then and . This contradiction implies that .For the converse, consider the integer such that and . If , then which is a contradiction. If , then Again it is a contradiction, hence .

modules are defined in [3], and we can observe the following.

Remark 10. If is -pure in , then by Proposition 7, and implies that for all . In other words, is a module in .

Proposition 11. Let be an -neat submodule of . Then is -pure in if and only if is a module in .

Proof. Suppose . Consider such that there exists and . This implies the existence of such that and . Now there exists , such that , where . As , we have for some and . Since , there exists such that and is -pure in . The converse is trivial.

Proposition 12. A submodule is a module if and only if for all .

Proof. Suppose is a module in . Therefore . We will prove the result by induction. Assume that for all . Since , we have The converse is trivial.

Following results are the immediate consequences of the previous discussion.

Remark 13. (i) is a module in if and only if for all .(ii) If for all , then is a module in .

3. -Pure Envelopes

For a proper submodule , it is not always possible for to have an -pure hull in . We study the situation when there is a proper -pure submodule of , but no proper direct summand of contains . This motivates us to define -pure envelopes like -neat envelope defined earlier [6]. We find that the -Kaplansky invariants are same for all -pure envelopes.

Definition 14. Let be a submodule of a QTAG module . An -pure submodule is an -pure envelope of if has no proper direct summand containing .

Proposition 15. In a module , an -pure submodule is an -pure envelope of if and only if contains no -divisible summand disjoint from and for any , no uniserial summand of decomposition length disjoint from .

Proof. If is not an -pure envelope of , then with . If is not -reduced, then it contains an -divisible summand disjoint from , and if is -reduced, then it contains a uniserial summand such that for some . Without loss of generality, we may assume that . Now implying that . Therefore, and .Conversely, if is -divisible submodule of with , then is a summand of ; therefore cannot be an -pure envelope of . If , it is a uniserial summand such that and , then such that . This can be chosen to be any submodule of which is maximal with respect to the properties of containing and disjoint from . This can not be the -pure envelope of .

Let be a uniserial summand of such that . If such that , then if and only if . Since , this implies that . This enables us to prove the following.

Proposition 16. An -pure submodule containing the submodule of a separable module is an -pure envelope if and only if , for all .

Proof. Let be a semi--pure submodule of a separable module and the -pure hull of . Now we have Soc . This implies that for every . This is true for -pure envelopes, and if is separable, this is sufficient also.

Remark 17. Since for all , if and only if Soc for every , an -pure submodule is an -pure envelope of if and only if , for every .

Remark 18. Since the union of a chain of -pure envelopes of may not contain any -divisible direct summand disjoint from or a uniserial summand of decomposable length disjoint from , every -pure envelope of is contained in a maximal -pure envelope of .

Now we investigate -pure envelopes of containing other -pure envelopes of .

Theorem 19. Let be a submodule of a separable module and , -pure submodules of containing . Then (i)for every , the natural embedding induces a monomorphism (ii)the map is onto for every if and only if is an -dense submodule of .

Proof. The maps send the coset upon the coset . Since , there exists such that and ; therefore If , then or ; that is, there exists such that . Now . Therefore or and is a monomorphism, which completes part (i).(ii) Suppose is -dense in , therefore for . For such that , there exists such that . Since is -dense in ,  , and is surjective.Conversely, suppose each is surjective, therefore Soc, for every . Let . Now there exists such that . Now with , and such that there exists with . Also and . We will prove that implying that . For , if , then , and if and implies that . Therefore , and we are done.

Now we investigate the relation between -Kaplansky invariants of the submodules and their -pure envelopes.

Theorem 20. Let be an -pure envelope of the submodule of a separable module . Then

Proof. We have Since is separable, by Remark 17, we have therefore, and we are able to write This implies that is an extension of by , and the result follows.

Following are the immediate consequences of the previous discussion.

Corollary 21. Every -pure envelope of the submodule in a separable module has same Ulm-Kaplansky invariants. Therefore -pure envelopes of have isomorphic basic submodules.

Corollary 22. Let be the direct sum of uniserial modules. Then all -pure envelopes of a submodule are isomorphic, if they exist.

Proposition 23. If is -pure submodule of a separable module such that its closure is not -pure in , then has no -pure envelope in .

Proof. An -pure envelope of must have a larger basic submodule than , and any uniserial summand in this larger basic submodule not in contradicts the Proposition 15, and the result follows.

Theorem 24. Every submodule of a separable module admits -pure envelopes if and only if is quasicomplete; that is, the closure of every -pure submodule of is -pure in .

Proof. Suppose on the contrary that is not quasicomplete. Thus it contains an -pure submodule such that is not -pure in . By Proposition 23, the submodule does not have an -pure envelope. This contradiction proves that is quasicomplete.For the converse, consider a submodule of a quasicomplete module . We construct a countable sequence of subsocles of with another sequence, of -pure submodules of such that and , for every . Now we may say that is the maximal submodule such that and . Put such that the nonzero elements of are of height and . Now , and is bounded by ; therefore there exists a minimal -pure submodule containing , which is here. Now is also bounded, hence a summand of and by Proposition 16, . By the same argument, we get , an -pure submodule of . Now is an -dense subsocle of , and we have to show that does not have a uniserial summand of decomposition length , which is disjoint from . Suppose is a uniserial summand of such that and . Let be the least positive integer such that . Since and there are elements , , , such that . Here , because . Therefore implies that . As , we have ; therefore there exist such that . Also there are elements such that and . Since is -pure, there is a uniserial summand and and , which is a contradiction implying that .Suppose that is the closure of with respect to -topology. Then, is -pure in and .Case i. When is closed [8] with respect to -topology [4], by the structure of ,   is a summand of . Consider the decomposition . If is a submodule of such that Soc, then If has height in , then and is -pure in . Thus, and , and is an -pure envelope of in the closed module M.Case ii. When is an arbitrary QTAG module, again consider the decomposition . If is bounded, then we are done; otherwise is a closed module. By Case , has an -pure envelope . Assume on the contrary that is a summand of such that . If Soc, then is a uniserial summand of and implying that which is a contradiction. Therefore Soc, but then contains a summand such that Soc and implies that , again a contradiction. Thus, no summand exists, and is an -pure envelope of .

Now we investigate the conditions under which every submodule of a QTAG module has an -pure envelope.

Proposition 25. has an -pure envelope in if and only if it is -divisible.

Proof. If is not -divisible, then the basic submodule of any -pure submodule of containing is nontrivial and it has a nonzero uniserial summand disjoint from . By Proposition 15, can not be an -pure envelope of . The converse is trivial.

The following theorem characterizes the module whose every submodule has an -pure envelope.

Theorem 26. In a module , every submodule has an -pure envelope if and only if is the direct sum of an -divisible and a quasicomplete module.

Proof. Suppose that every submodule of has an -pure envelope. By Proposition 25, is -divisible. Now , where is the maximal -divisible submodule of and is separable. Let and , the -pure envelope of in . Now if we project into , then the projection of is an -pure envelope of in . Therefore, all the submodules of have -pure envelopes and Theorem 24 implies that is a quasicomplete module.Conversely, suppose , where is -divisible and a quasicomplete module. For , where may be chosen. By Theorem 24, has an -pure envelope in . Now there exists an -divisible submodule that contains as an essential submodule. If we put , then which is -pure in . cannot have a uniserial or -divisible summand disjoint from , because such a summand would be disjoint from and . Therefore, is an -pure envelope of .

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