A Note on Elongations of Summable QTAG-Modules

Abstract

A right module 𝑀 over an associative ring with unity is a QTAG-module if every finitely generated submodule of any homomorphic image of 𝑀 is a direct sum of uniserial modules. In this paper we find a suitable condition under which a special 𝜔-elongation of a summable QTAG-module by a (𝜔+𝑘)-projective QTAG-module is also a summable QTAG-module.

1. Introduction and Preliminaries

The study of QTAG-modules was initiated by Singh [1]. Mehdi et al. studied different notions and structures on QTAG-modules, developed the theory of these modules by introducing several notions and some interesting properties of these modules and characterized different submodules of QTAG-modules. Yet there is much to explore.

Throughout this paper, all rings will be associative with unity and modules 𝑀 are unital QTAG-modules. An element 𝑥∈𝑀 is uniform, if 𝑥𝑅 is a nonzero uniform (hence uniserial) module and for any 𝑅-module 𝑀 with a unique composition series, 𝑑(𝑀) denotes its composition length. For a uniform element 𝑥∈𝑀, 𝑒(𝑥)=𝑑(𝑥𝑅) and 𝐻_𝑀(𝑥)=sup{𝑑(𝑦𝑅/𝑥𝑅)∣𝑦∈𝑀,𝑥∈𝑦𝑅and𝑦uniform} are the exponent and height of 𝑥 in 𝑀, respectively. 𝐻_𝑘(𝑀) denotes the submodule of 𝑀 generated by the elements of height at least 𝑘 and 𝐻^𝑘(𝑀) is the submodule of 𝑀 generated by the elements of exponents at most 𝑘. A QTAG-module 𝑀 is called a Σ-module, if all of its high submodules are direct sums of uniserial modules. 𝑀 is ℎ-divisible if 𝑀=𝑀¹=⋂^∞_𝑘=0𝐻_𝑘(𝑀).

A module 𝑀 is summable if Soc(𝑀)=⨁_𝛼<𝜏𝑆_𝛼, where 𝑆_𝛼 is the set of all elements of 𝐻_𝛼(𝑀) which are not in 𝐻_𝛼+1(𝑀), where 𝜏 is the length of 𝑀.

A submodule 𝑁⊂𝑀 is nice [2, Definition 2.3] in 𝑀, if 𝐻_𝜎(𝑀/𝑁)=(𝐻_𝜎(𝑀)+𝑁)/𝑁 for all ordinals 𝜎; that is, every coset of 𝑀 modulo 𝑁 may be represented by an element of the same height.

Recall that a QTAG-module 𝑀 is (𝜔+1)-projective if there exists a submodule 𝑁⊂𝐻¹(𝑀) such that 𝑀/𝑁 is a direct sum of uniserial modules and a QTAG-module 𝑀 is (𝜔+𝑘)-projective if there exists a submodule 𝑁⊂𝐻^𝑘(𝑀) such that 𝑀/𝑁 is a direct sum of uniserial modules [3].

A QTAG-module is an 𝜔-elongation of a totally projective QTAG-module by a (𝜔+𝑘)-projective QTAG-module if and only if 𝐻_𝜔(𝑀) is totally projective and 𝑀/𝐻_𝜔(𝑀) is (𝜔+𝑘)-projective.

2. Main Results

In [4], we studied strong 𝜔-elongations of summable modules by (𝜔+𝑘)-projective QTAG-modules. It was found that the module 𝑀 is a special 𝜔-elongation if 𝑀¹ is summable and there exists a submodule 𝑁⊆𝐻^𝑘(𝑀) such that 𝑀/(𝑁+𝑀¹) is a direct sum of uniserial modules. We showed there that under certain additional circumstances on 𝑁 these elongations are of necessity summable modules; in fact 𝑁∩𝐻_𝑘(𝑀)⊆𝑀¹ was taken; that is, 𝑁 is a height finite submodule of 𝑀.

In [5], the class of 𝑛-layered module was introduced which is a proper subclass of Σ-modules as follows: 𝑀 is an 𝑛-layered module if 𝐻^𝑘(𝑀)=⋃_𝑖<𝜔𝑀_𝑖, 𝑀_𝑖⊆𝑀_𝑖+1⊆𝐻^𝑘(𝑀), and for all 𝑖≥1. 𝑀_𝑖∩𝐻_𝑖(𝑀)⊆𝑀¹. It was proved there that every 𝑛-layered module which is a strong 𝜔-elongation of a totally projective module by a (𝜔+𝑘)-projective module is totally projective and vice versa; in particular each 𝑛-layered module is (𝜔+𝑘)-projective uniquely when it is a direct sum of modules of length at most (𝜔+𝑛).

Motivated by these elongations we investigate the relationship between the classes of 𝑛-layered modules and strong 𝜔-elongations of summable modules by (𝜔+𝑘)-projective modules, that is, how 𝑛-layered modules are situated inside these special 𝜔-elongations of summable modules by (𝜔+𝑘)-projective modules and whether there is an analogue with the strong 𝜔-elongations of a totally projective modules by (𝜔+𝑘)-projective modules.

Before doing that, we need the following preliminaries.

A module 𝑀 is said to be pillared [6], if 𝑀/𝑀¹ is a direct sum of uniserial modules. Clearly, such a module is necessarily 𝑛-layered module and hence a Σ-module [5], whereas the converse is not true in general. We investigate the conditions under which the converse holds good.

A module 𝑀 is said to be a strong 𝜔-elongation (of a summable module) by a (𝜔+𝑘)-projective module if there exists a submodule 𝑁⊆𝐻^𝑘(𝑀) with 𝑀/(𝑁+𝑀¹) a direct sum of uniserial modules. The first Ulm-factor of this module is (𝜔+𝑘)-projective. There are modules with (𝜔+𝑘)-projective first Ulm-factor which are not strong 𝜔-elongations by a (𝜔+𝑘)-projective module. These modules are with strongly (𝜔+𝑘)-projective first Ulm-factor.

Now we are in the state to prove the following main result.

Theorem 1. An 𝑛-layered module is a strong 𝜔-elongation by a (𝜔+𝑘)-projective module if and only if it is a pillared module.

Proof. Suppose that 𝑀 is an 𝑛-layered module, so 𝐻^𝑘(𝑀)=⋃_𝑖<𝜔𝑀_𝑖, 𝑀_𝑖⊆𝑀_𝑖+1⊆𝐻^𝑘(𝑀), and for all 𝑖≥1. 𝑀_𝑖∩𝐻_𝑖(𝑀)⊆𝑀¹ and suppose that there exists a submodule 𝑁 of 𝑀 such that 𝑁⊆𝐻^𝑘(𝑀) with 𝑀/(𝑁+𝑀¹) is a direct sum of uniserial modules. Clearly, 𝑀/𝑀¹𝑁+𝑀¹/𝑀¹≅𝑀𝑁+𝑀¹. (1) Since 𝑁⊆⋃_𝑖<𝜔𝑀_𝑖, we infer that 𝑁=⋃_𝑖<𝜔(𝑀_𝑖∩𝑁) and thus (𝑁+𝑀¹)/𝑀¹=∪((𝑁+𝑀¹)/𝑀¹) by putting 𝑁_𝑖=𝑀_𝑖∩𝑁.
Using modular law, we compute that 𝑁_𝑖+𝑀¹𝑀¹∩𝐻_𝑖𝑀𝑀¹=𝑁_𝑖∩𝐻_𝑖(𝑀)+𝑀¹𝑀¹={0}. (2) Hence 𝑀/𝑀¹ is a direct sum of uniserial modules and therefore 𝑀 is pillared. The proof is complete as the converse is trivial.

Proposition 2. An 𝑛-layered module is a strong 𝜔-elongation of a summable module by a (𝜔+𝑘)-projective module if and only if it is a summable pillared module.

Proof. Suppose 𝑛-layered module is a strong 𝜔-elongation of a summable module by a (𝜔+𝑘)-projective module. Since 𝑀 is a Σ-module and 𝑀¹ is summable, 𝑀 has to be summable as well and Theorem 1 ensures that 𝑀¹ must be pillared.
Suppose that 𝑀 is summable pillared module, so it ensures that 𝑀¹ is summable and pillared modules are both 𝑛-layered and strong 𝜔-elongations by (𝜔+𝑘)-projective module by taking 𝑁=0, which completes the proof.

As an immediate consequence of the above, we have the following corollary.

Corollary 3. Suppose 𝑀 is a Σ-module which is a strong 𝜔-elongation of a summable module by a (𝜔+𝑘)-projective module and the (𝜔+𝑛)th Ulm-Kaplansky invariants of 𝑀 are zero for each 𝑛 such that 0≤𝑛≤𝑘−1, if 𝑘>1. Then 𝑀 is a summable pillared module.

Proof. Since (𝜔+𝑛)th Ulm-Kaplansky invariants of 𝑀 are zero, 𝐻^𝑘(𝑀)=𝐻^𝑘(𝑁)⊕𝐻^𝑘(𝑀¹) where 𝑁 is a high submodule of 𝑀. Since it is a direct sum of uniserial modules, 𝐻^𝑘(𝑁)=⋃_𝑖<𝜔𝑁_𝑖,𝑁_𝑖⊆𝑁_𝑖+1⊆𝐻^𝑘(𝑁) where 𝑁_𝑖∩𝐻_𝑖(𝑁)=0. Putting 𝑀_𝑖=𝑁_𝑖⊕𝐻^𝑘(𝑀₁), we obtain 𝐻^𝑘(𝑀)=⋃_𝑖<𝜔𝑀_𝑖. Since 𝑁 is ℎ-pure in 𝑀, we have 𝑀_𝑖∩𝐻_𝑖(𝑀)⊆𝑀¹+𝑁_𝑖∩𝐻_𝑖(𝑀)=𝑀¹+𝑁_𝑖∩𝐻_𝑖(𝑀)=𝑀¹. (3) By Proposition 2, 𝑀 is an 𝑛-layered module.

Proposition 4. A module of length not exceeding (𝜔+𝑛−1) is an 𝑛-layered module if and only if it is a direct sum of countably generated modules.

Proof. Suppose 𝑀 is an 𝑛-layered module of length not exceeding (𝜔+𝑛−1); then 𝑀¹⊆𝐻^𝑛−1(𝑀) and hence 𝑀Soc𝑀¹=_𝑖<𝜔𝐻_𝑖(𝑀)+Soc(𝑀)𝑀¹⊆𝐻^𝑛(𝑀)𝑀¹ (4) since 𝐻(⋂_𝑖<𝜔(𝐻_𝑖(𝑀)+Soc(𝑀)))⊆𝑀¹. Moreover, we write 𝐻^𝑛(𝑀)=⋃_𝑖<𝜔𝑀_𝑖, 𝑀_𝑖⊆𝑀_𝑖+1⊆𝐻^𝑛(𝑀) and 𝑀_𝑖∩𝐻_𝑖(𝑀)⊆𝑀¹. Consequently, Soc(𝑀/𝑀¹)=⋃_𝑖<𝜔𝑇_𝑖, where 𝑇_𝑖=(𝑀_𝑖+𝑀¹)/𝑀¹∩Soc(𝑀/𝑀¹). We compute 𝑇_𝑖∩𝐻_𝑖𝑀𝑀¹=𝑇_𝑖∩𝐻_𝑖(𝑀)𝑀¹=𝑀_𝑖+𝑀¹∩𝐻_𝑖(𝑀)𝑀¹=𝑀_𝑖∩𝐻_𝑖(𝑀)+𝑀¹𝑀¹=0 (5) hence 𝑀 is pillared and as 𝑀¹ is bounded, 𝑀 is a direct sum of countable modules.
The sufficiency is obvious and follows from [5].

As an immediate consequence of the above, we have the following corollary.

Corollary 5. A module 𝑀 is an 𝑛-layered module if and only if each of its 𝐻_{(𝜔+𝑛−1)}(𝑀)-high submodules is a direct sum of countable modules.

Proof. Let 𝑀 be an 𝑛-layered module and 𝑁 be its 𝐻_{(𝜔+𝑛−1)}(𝑀)-high submodule. In [5], we show that 𝑀 is an 𝑛-layered module precisely when 𝑁 is an 𝑛-layered module. Hence using the Proposition 4, result follows immediately.

With the help of the last statement we can verify once again the validity of Corollary 3 as it is checked that a submodule 𝑁 of 𝑀 is ℎ-high in 𝑀 if and only if 𝑁 is (𝜔+𝑛−1)-high in 𝑀 whenever the (𝜔+𝑚)th Ulm-Kaplansky invariants of 𝑀 are zero for 0≤𝑚≤𝑛−1; that is, Soc𝐻_𝜔(𝑀)=⋯=Soc𝐻_{𝜔+𝑛−1}(𝑀). (6)

A module 𝑀 is said to be strong (𝜔+𝑛−1)-elongation of a summable module by a totally projective module if 𝐻_{𝜔+𝑛−1}(𝑀) is summable and there is a nice submodule 𝐾 of 𝑀 such that 𝐾∩𝐻_{𝜔+𝑛−1}(𝑀)=0 and 𝑀/(𝐾⊕𝐻_{𝜔+𝑛−1}(𝑀)) is totally projective [3].

Now we are in the state to prove our final result.

Theorem 6. An 𝑛-layered module is a strong (𝜔+𝑛−1)-elongation of a summable module by a totally projective module if and only if it is a summable pillared module.

Proof. Clearly 𝑀𝐾⊕𝐻_{𝜔+𝑛−1}(𝑀)≅𝑀/𝐻_{𝜔+𝑛−1}(𝑀)𝐾⊕𝐻_{𝜔+𝑛−1}(𝑀)/𝐻_{𝜔+𝑛−1}(𝑀) (7) is totally projective. Since 𝐾∩𝐻_{𝜔+𝑛−1}(𝑀)=0, 𝐾 is contained in some (𝜔+𝑛−1)-high submodule of 𝑀, say 𝑁. By Corollary 5, 𝑁 is totally projective of length at most (𝜔+𝑛−1), so we may write 𝑁+⋃_𝑖<𝜔𝑁_𝑖, where 𝑁_𝑖⊆𝑁_𝑖+1⊆𝑁 and all 𝑁_𝑖’s are height finite in 𝑁 and hence in 𝑀 as 𝑁 is isotype in 𝑀. Therefore 𝐾⊕𝐻_{𝜔+𝑛−1}(𝑀)𝐻_{𝜔+𝑛−1}(𝑀)=_𝑖<𝜔𝑁_𝑖+𝐻_{𝜔+𝑛−1}(𝑀)𝐻_{𝜔+𝑛−1}(𝑀)∩𝐾⊕𝐻_{𝜔+𝑛−1}(𝑀)𝐻_{𝜔+𝑛−1}(𝑀). (8) In the same way, we can show that (𝑁_𝑖+𝐻_{𝜔+𝑛−1}(𝑀))/𝐻_{𝜔+𝑛−1}(𝑀) are height finite in 𝑀/𝐻_{𝜔+𝑛−1}(𝑀) and by [3], (𝐾⊕𝐻_{𝜔+𝑛−1}(𝑀))/𝐻_{𝜔+𝑛−1}(𝑀) is nice in 𝑀/𝐻_{𝜔+𝑛−1}(𝑀) and hence 𝑀/𝐻_{𝜔+𝑛−1}(𝑀) is totally projective. Then 𝑀𝐻_𝜔(𝑀)≅𝑀/𝐻_{𝜔+𝑛−1}(𝑀)𝐻_𝜔(𝑀)/𝐻_{𝜔+𝑛−1}(𝑀)=𝑀/𝐻_{𝜔+𝑛−1}(𝑀)𝐻_𝜔𝑀/𝐻_{𝜔+𝑛−1}(𝑀) (9) should be a direct sum of uniserial modules and hence 𝑀 is pillared.
On the other hand, 𝐻_{𝜔+𝑛−1}(𝑀) being summable implies that 𝐻_𝜔(𝑀) is summable and hence 𝑀 has to be summable; thus it is summable pillared, which completes the proof.

As an immediate consequence for 𝑛=1, we have the following.

Corollary 7. A Σ-module is a strong 𝜔-elongation of a summable module by a totally projective module if and only if it is a summable pillared module.

Conflict of Interests

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