Abstract

A module over an associative ring with unity is a -module if every finitely generated submodule of any homomorphic image of is a direct sum of uniserial modules. The study of large submodules and its fascinating properties makes the theory of QTAG-modules more interesting. A fully invariant submodule of is large in if , for every basic submodule of The impetus of these efforts lies in the fact that the rings are almost restriction-free. This motivates us to find the necessary and sufficient conditions for a submodule of a QTAG-module to be large and characterize them. Also, we investigate some properties of large submodules shared by -modules, summable modules, -summable modules, and so on.

1. Introduction and Preliminaries

All the rings considered here are associative with unity and modules are unital -modules. An element is uniform, if is a nonzero uniform (hence uniserial) module and, for any -module with a unique composition series, denotes its composition 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 exponents at most For any arbitrary , if but . 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.

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 and it is an isotype if it is -pure for every ordinal A submodule is a basic submodule of , if is -pure in , , where each is the direct sum of uniserial modules of length and is -divisible. For a QTAG-module , the th invariant of , is the cardinal number [1]. Several results which hold for -modules also hold good for -modules [2].

A module is summable if , where is the set of all elements of which are not in , where is the length of A -module is called -summable if and, for every positive integer , there is an ordinal such that length of

For any uniform element , there exist uniform elements such that and Now the -sequence of is defined as . sequences are defined as . This is analogous to the -sequences defined in groups [3]. These sequences are partially ordered because if for every For the sequence of nonnegative, nondecreasing integers we may consider as the submodule of generated by the elements of for which If is an endomorphism of , then , and therefore is fully invariant. Therefore with every large submodule of we may associate a sequence

2. Some Characterizations of Large Submodules

In this section we study and characterize the properties of fully invariant and large submodules of -modules. We also discuss the properties of large submodules inherited from the containing module.

We start with the facts which are true for any module. For a fully invariant submodule of a -module and an endomorphism of , it induces an endomorphism of such that On the other hand for the endomorphism of induced by an endomorphism of and a fully invariant submodule , That is, and is fully invariant in . For a fully invariant submodule , and each is fully invariant in

For any sequence we define as the submodule of , generated by the elements for which . This submodule is a large submodule of . In fact for every large submodule there is a sequence and, for every sequence, there is a large submodule [4].

For a -module , consider the homomorphism . As , is height preserving. This implies that for all .

We conclude that is large in if and only if is large in . In a module without elements of infinite height, consider a fully invariant submodule of , and such that for every . Let , such that . Then there exists an endomorphism of such that ; therefore and

Remark 1. For any large submodule of , for some positive integer

Lemma 2. Let be submodule of such that for , where the sequence of positive integers is monotonically increasing. Then for any basic submodule of

Proof. Let be a basic submodule of and [5]. Then Now suppose, for every , implies that Consider such that ; then there exists such that Now and , where and , ensuring the existence of such that By the -purity of , there exists , such that . Now , and thus or implying that

The following remarks are significant to be stated.

Remark 3. Let be a large submodule of an unbounded -module without elements of infinite height and a proper basic submodule of . Then therefore is unbounded. Conversely for an unbounded fully invariant submodule of , is fully invariant for all . As an immediate consequence of Lemma 2, is a large submodule of . We can say that the unbounded fully invariant submodules of are exactly the large submodules of

Remark 4. If is the direct sum of uniserial modules of length and , then where and

Remark 5. Let be the direct sum of uniserial modules of length and . Then there exists an endomorphism of with , if and only if

Remark 6. Let be a fully invariant submodule of , a direct sum of uniserial modules of length Then , where . If , if , and if , then

Remark 7. If and are the direct sums of uniserial modules of length and , respectively, and , , then(i)there exists a homomorphism such that if and only if ,(ii)there exists a homomorphism such that if and only if

Theorem 8. Let , where each is the direct sum of uniserial modules of length . Then is a fully invariant submodule of if and only if , where , for every and for . A fully invariant submodule L is large in if and only if ; the above conditions hold and the sequence is unbounded if is unbounded.

Proof. Let be a fully invariant submodule of . Then by the facts mentioned above and Remark 6. Now for and the first condition holds. If , then for every ; therefore for every and the second condition holds. If , then there exists a least positive integer such that . Then for all , where . Since , this implies that Again for ; we have Now suppose and If , then and and the second condition holds. We assume that . Consider such that and such that . Now, by Remark 7, there exists an endomorphism of mapping onto Hence and ; thus
Now suppose . Then so . If and such that , we may choose such that . Then and . By Remark 7, there exists an endomorphism of with . Thus and we have ; therefore
If , then but if , we may define so that this inequality holds for all . Thus all fully invariant submodules of are the direct sums of . If is a large submodule of and is unbounded, then, by Lemma 2, is also unbounded. Therefore must be unbounded.
For the converse, suppose , where for all and for all . To establish the full invariance of , we consider any and We have to show that for any endomorphism of , . Consider , such that , where and If , then so , because ; hence . If , then because Thus
This implies that is a fully invariant submodule of If is unbounded and is also unbounded, then is unbounded and is therefore a large submodule of by Remark 3.

Corollary 9. If is a large submodule of a -module , then is a direct sum of uniserial modules.

Proof. For any basic submodule of , and the result follows.

Corollary 10. For any large submodule of , .

Proof. Since is a direct sum of uniserial modules, or

Theorem 11. Let be -pure submodule of a -module and a large submodule of Then there exists a large submodule of such that . If is -divisible, then is the closure of in and is therefore uniquely determined by and .

Proof. Let and . Since is -pure in and is a -sequence for , we have that is a -sequence for . Thus is a large submodule of
If , then ; therefore and . Conversely if , then implies that or . Thus .
Let be -divisible and a large submodule of with Then That is, is -divisible. But , where is a direct summand of ; we have and is a direct sum of uniserial modules [6]. Now , thus

Now we characterize large submodules in terms of invariants.

Theorem 12. Let be a submodule of a -module . Then is a large submodule of if and only if , where (i),(ii),(iii)the sequence is unbounded if is unbounded and the Ulm-invariants of are given by ,for all .

Proof. Suppose . Since ’s are fully invariant submodules, their sum is again fully invariant submodule of If is bounded, then is large. If is unbounded, then, by the third condition, for each , there exists a positive integer such that or .
Since, and , If , there exists such that , where . Now , where or ; thus and .
If , then and because and . Now, by Lemma 2, , for every basic submodule of , and is a large submodule of .
Conversely suppose is a large submodule of Then for any basic submodule of is a large submodule of and, by Theorem 8, , where and ’s satisfy the given conditions.
Now, and, for each , This implies that
For the converse, consider , where Then for and for all . Now, for , we have and
If for , then . (by the given condition). Therefore and Let Now . Since is -pure in and is -divisible, , by Theorem 11. Again is a basic submodule of ; thus , for all
If , where is the direct sum of uniserial modules of length , then , where is a direct sum of uniserial modules of length .
Again,And the proof is complete.

3. Properties of Large Submodules of QTAG-Modules

In this section we compare the structures of -modules and their large submodules. We investigate the characteristics of -modules which are preserved by their large submodules. We start with the -modules, that is, the modules whose high submodules are direct sums of uniserial modules [7]. Then we study summable, -summable, -projective, and -pure complete -modules.

Singh [8] proved that a -module is a direct sum of uniserial submodules if and only if is the union of an ascending sequence of submodules , such that, for every , there exists and for all .

This helps us to prove the following.

Theorem 13. A -module is a -module if and only if , where and for every

Proof. Since is a -module, it contains a high submodule such that is a direct sum of uniserial modules.
Again is a high submodule [9] of if and only if is -pure in and . Therefore by the above result [8], , , and , and we deduce . If we put , then and , because is a high submodule of
For the converse if , where we put , then Also Therefore is a direct sum of uniserial modules and is a -module.

Now we may prove the following.

Theorem 14. A -module is a - module if and only if its large submodule is a -module.

Proof. Since [6], there is a natural number such that and for every and some such that . If is a -module, then, by Theorem 13, is the union of ascending chain of submodules such that and for every .
This implies that and Therefore Now Theorem 13 indicates that is a -module.
Conversely suppose is a -module. Therefore Again Now Thus, by Theorem 13, is a -module, and so is

To study the other relations between a module and its large submodule we need the following lemma.

Lemma 15. Isotype submodules of countable length of summable -modules are again summable.

Proof. Let be an isotype submodule of countable length in the summable module Now there is a -high submodule of such that . Since , there is -high submodule of such that .
Again, for every ordinal , every -high submodule is isotype; therefore is isotype and it is summable. The socles of -high submodules have the same images under the canonical map because this maps -high submodules isomorphically in a height preserving manner onto submodules of .
Now is isotype in a summable module of countable length . Therefore is the union of an ascending chain of submodules , where for every the heights of elements of assume but a finite numbers of values.
Now , and the heights of the elements of assume a finite numbers of different values. Thus is summable.

The following result shows that summability is shared by large submodules.

Theorem 16. Let be a large submodule of a -module . Then is summable if and only if is summable.

Proof. Suppose is summable; that is, , where the nonzero elements of ’s are contained in but they do not belong to , for every
Again is fully invariant submodule of and for all ordinals , where the nonzero elements of are contained in and not contained in for every Since , whenever , , but , for each . By transfinite induction and , for and so on; that is, and , for .
If we put , where , and , then Therefore is summable.
Conversely suppose is summable. So, is summable as its fully invariant submodule. Moreover, by Lemma 15, being summable implies that is a -module. Now by Theorem 14, is also a -module. For a high submodule of .
Since is a direct sum of uniserial modules, , where and because is -pure in . Again the summability of ensures that , where and Therefore, and This implies that We may infer now that is summable.

Theorem 17. Let be the large submodule of Then is -summable if and only if is -summable.

Proof. Suppose is unbounded. Then length of length of . If is -summable, then is also -summable being a submodule of equal length.
If is bounded the result holds trivially.
Conversely suppose is -summable. Therefore and for all and some length of
Now, . Since [6], for each ordinal and for and some because is large in and both. Thus , whenever length of length of or
We may define . Thus . By defining ’s we observe that . This implies that is -summable.

Theorem 18. If is a direct sum of -summable -modules, then so is .

Proof. Let , where each is -summable. Now because is fully invariant in Since all ’s are isotype in , we infer that is large in , for every By Theorem 17, are -summable. Thus is also a direct sum of -summable modules.

Let us recall the following.

Definition 19. A -module is -projective if there exists a submodule such that is a direct sum of uniserial modules.

Remark 20. The submodules of -projective modules are also -projective.

Theorem 21. A -module is -projective if and only if its large submodule is -projective.

Proof. Suppose is -projective. Therefore there exists a submodule such that , where and for each .
Now , for every Since for some , we haveTherefore the heights of the elements of are bounded in for all Now is a direct sum of uniserial modules [6]. Therefore is a direct sum of uniserial modules and is -projective. The converse is trivial.

The property of being -pure complete is also shared by the large submodules of -modules.

First we recall the definition of -pure completeness.

Definition 22. A -module is -pure complete if, for every subsocle , there is a -pure submodule of so that In other words every subsocle supports a -pure submodule of .

Theorem 23. Let be the large submodule of a -module If is -pure complete, so is

Proof. Let be a subsocle of Since , supports a -pure submodule of Now is also large in and is -pure in . Again , and therefore is -pure complete.

Corollary 24. A -module is -pure complete if and only if is -pure complete for some fixed but arbitrary positive integer .

Proof. Since is large in , it is -pure complete if is -pure complete. Conversely suppose is -pure complete. We shall use transfinite induction to prove the result.
Let be a subsocle of such that and for some -pure submodule of By [7] we can say that there is a -pure submodule of such that Now
We have to show that there exists a -pure submodule such that We define the submodule Now because Again
Now Therefore . Now, This implies that is -pure in

In the end we state the following unsolved problems.

Problem 25. Is it true that is a -module if and only if its large submodule is?

Problem 26. Is it true that is a direct sum of closed modules if and only if its large submodule is?

Competing Interests

The authors declare that they have no competing interests.