# A Note on Elongations of SummableQTAG-Modules

Authors: Alveera Mehdi; Fahad Sikander; Sabah A. R. K. Naji
Journal: The Scientific World Journal
 (2013)

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## Abstract

A right moduleM 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.


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## Body

## 1. Introduction and Preliminaries

The study ofQTAG-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 modulesM 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 HM(x)=sup{d(yR/xR)∣y∈M,x∈yRandyuniform} are the exponent and height of x in M, respectively. Hk(M) denotes the submodule of M generated by the elements of height at least k and Hk(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=0∞Hk(M).A moduleM is summable if Soc(M)=⨁α<τSα, where Sα is the set of all elements of Hα(M) which are not in Hα+1(M), where τ is the length of M.A submoduleN⊂M is nice [2, Definition 2.3] in M, if Hσ(M/N)=(Hσ(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 aQTAG-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⊂Hk(M) such that M/N is a direct sum of uniserial modules [3].AQTAG-module is an ω-elongation of a totally projective QTAG-module by a (ω+k)-projective QTAG-module if and only if Hω(M) is totally projective and M/Hω(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⊆Hk(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∩Hk(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 Hk(M)=⋃i<ωMi, Mi⊆Mi+1⊆Hk(M), and for all i≥1. Mi∩Hi(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 ofn-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 moduleM 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 moduleM is said to be a strong ω-elongation (of a summable module) by a (ω+k)-projective module if there exists a submodule N⊆Hk(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.
Ann-layered module is a strong ω-elongation by a (ω+k)-projective module if and only if it is a pillared module.Proof.
Suppose thatM is an n-layered module, so Hk(M)=⋃i<ωMi, Mi⊆Mi+1⊆Hk(M), and for all i≥1. Mi∩Hi(M)⊆M1 and suppose that there exists a submodule N of M such that N⊆Hk(M) with M/(N+M1) is a direct sum of uniserial modules. Clearly,
(1)M/M1(N+M1)/M1≅MN+M1.
Since N⊆⋃i<ωMi, we infer that N=⋃i<ω(Mi∩N) and thus (N+M1)/M1=∪((N+M1)/M1) by putting Ni=Mi∩N.
Using modular law, we compute that(2)Ni+M1M1∩Hi(MM1)=(Ni∩Hi(M))+M1M1={0}.
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.
Ann-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.
Supposen-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 thatM 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.
SupposeM 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, Hk(M)=Hk(N)⊕Hk(M1) where N is a high submodule of M. Since it is a direct sum of uniserial modules, Hk(N)=⋃i<ωNi,Ni⊆Ni+1⊆Hk(N) where Ni∩Hi(N)=0. Putting Mi=Ni⊕Hk(M1), we obtain Hk(M)=⋃i<ωMi. Since N is h-pure in M, we have
(3)Mi∩Hi(M)⊆M1+Ni∩Hi(M)=M1+Ni∩Hi(M)=M1.
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.
SupposeM is an n-layered module of length not exceeding (ω+n-1); then M1⊆Hn-1(M) and hence
(4)Soc(MM1)=⋂i<ω‍Hi(M)+Soc(M)M1⊆Hn(M)M1
since H(⋂i<ω(Hi(M)+Soc(M)))⊆M1. Moreover, we write Hn(M)=⋃i<ωMi, Mi⊆Mi+1⊆Hn(M) and Mi∩Hi(M)⊆M1. Consequently, Soc(M/M1)=⋃i<ωTi, where Ti=(Mi+M1)/M1∩Soc(M/M1). We compute
(5)Ti∩Hi(MM1)=Ti∩(Hi(M)M1)=(Mi+M1)∩Hi(M)M1=(Mi∩Hi(M))+M1M1=0
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 moduleM 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.
LetM 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 Corollary3 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,
(6)Soc(Hω(M))=⋯=Soc(Hω+n-1(M)).A moduleM is said to be strong (ω+n-1)-elongation of a summable module by a totally projective module if Hω+n-1(M) is summable and there is a nice submodule K of M such that K∩Hω+n-1(M)=0 and M/(K⊕Hω+n-1(M)) is totally projective [3].Now we are in the state to prove our final result.Theorem 6.
Ann-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(7)MK⊕Hω+n-1(M)≅M/Hω+n-1(M)(K⊕Hω+n-1(M))/Hω+n-1(M)
is totally projective. Since K∩Hω+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<ωNi, where Ni⊆Ni+1⊆N and all Ni’s are height finite in N and hence in M as N is isotype in M. Therefore
(8)K⊕Hω+n-1(M)Hω+n-1(M)=⋃i<ω‍[(Ni+Hω+n-1(M)Hω+n-1(M))∩(K⊕Hω+n-1(M)Hω+n-1(M))].
In the same way, we can show that (Ni+Hω+n-1(M))/Hω+n-1(M) are height finite in M/Hω+n-1(M) and by [3], (K⊕Hω+n-1(M))/Hω+n-1(M) is nice in M/Hω+n-1(M) and hence M/Hω+n-1(M) is totally projective. Then
(9)MHω(M)≅M/Hω+n-1(M)Hω(M)/Hω+n-1(M)=M/Hω+n-1(M)Hω(M/Hω+n-1(M))
should be a direct sum of uniserial modules and hence M is pillared.
On the other hand,Hω+n-1(M) being summable implies that Hω(M) is summable and hence M has to be summable; thus it is summable pillared, which completes the proof.As an immediate consequence forn=1, we have the following.Corollary 7.
AΣ-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.



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*Source: 384131-2013-12-26.xml*