# Different Characterizations of Large Submodules of QTAG-Modules

Authors: Fahad Sikander; Alveera Mehdi; Sabah A. R. K. Naji
Journal: Journal of Mathematics
 (2017)

---

## Abstract

A moduleM over an associative ring R with unity is a QTAG-module if every finitely generated submodule of any homomorphic image of M 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 L of M is large in M if L+B=M, for every basic submodule B of M. 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 Σ-modules, summable modules, σ-summable modules, and so on.


---

## Body

## 1. Introduction and Preliminaries

All the ringsR considered here are 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 HM(x)=sup⁡dyR/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. For any arbitrary x∈M, H(x)=k if x∈Hk(M) but x∉Hk+1(M). M is h-divisible if M=M1=⋂k=0∞Hk(M) and it is h-reduced if it does not contain any h-divisible submodule. In other words it is free from the elements of infinite height.A submoduleN of M is h-pure in M if N∩Hk(M)=Hk(N), for every integer k≥0. For a limit ordinal α, Hα(M)=⋂ρ<αHρ(M), for all ordinals ρ<α, and it is α-pure in M if Hσ(N)=Hσ(M)∩N for all ordinals σ<α and it is an isotype if it is α-pure for every ordinal α. A submodule B⊆M is a basic submodule of M, if B is h-pure in M, B=⨁Bi, where each Bi is the direct sum of uniserial modules of length i and M/B is h-divisible. For a QTAG-module M, the σth-Ulm invariant of M, fM(σ) is the cardinal number g(Soc⁡HσM/Soc⁡(Hσ+1(M))) [1]. Several results which hold for TAG-modules also hold good for QTAG-modules [2].A moduleM is summable if Soc⁡(M)=⨁τ<αSα, 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 QTAG-module M is called σ-summable if Soc⁡(M)=⋃n<ωMn,Mn⊆Mn+1 and, for every positive integer n, there is an ordinal αn such that Mn∩Hαn(M)=0,αn< length of M.For any uniform elementx∈M, there exist uniform elements x1,x2,… such that xR⊇x1R⊇x2R⊇⋯ and dxiR/xi+1R=1. Now the Ulm-sequence of x is defined as U(x)=(H(x),H(x1),H(x2),…). U sequences are defined as U(x). This is analogous to the Ulm-sequences defined in groups [3]. These sequences are partially ordered because U(x)≤U(y) if H(xi)≤H(yi) for every i. For the sequence n=(n0,n1,n2,…) of nonnegative, nondecreasing integers we may consider L as the submodule of M generated by the elements x of M for which U(x)≥n. If f is an endomorphism of M, then H(x)≤H(f(x)), and therefore L is fully invariant. Therefore with every large submodule L of M we may associate a sequence n(L).


## 2. Some Characterizations of Large Submodules

In this section we study and characterize the properties of fully invariant and large submodules ofQTAG-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 submoduleN of a QTAG-module M and an endomorphism f of M, it induces an endomorphism f¯ of M/N such that f¯(x+N)=f(x)+N. On the other hand for the endomorphism f¯ of M/N induced by an endomorphism f of M and a fully invariant submodule K/N⊆M/N, f¯(x+N)=f(x)+N∈K/N. That is, f(x)∈K and K is fully invariant in M. For a fully invariant submodule A⊆M=⨁Mi, A=⨁(A∩Mi) and each A∩Mi is fully invariant in Mi.For any sequencen=(n1,n2,…) we define M(n) as the submodule of M, generated by the elements x for which U(x)≥n. This submodule is a large submodule of M. In fact for every large submodule there is a sequence and, for every sequence, there is a large submodule [4].For aQTAG-module M, consider the homomorphism f:M→M/M1. As M1=⋂k=0∞Hk(M), f is height preserving. This implies that H(x)=H(f(x))andU(x)=U(f(x)) for all x∈M.We conclude thatL/M1 is large in M/M1 if and only if L is large in M. In a module M without elements of infinite height, consider a fully invariant submodule K of M, and x∈Soc⁡(K) such that n=H(x)≤H(y) for every y∈Soc⁡(K). Let z∈Soc⁡(M), such that H(z)≥n. Then there exists an endomorphism f of M such that f(x)=z; therefore z∈K and Soc⁡(K)=Soc⁡(Hn(M)).Remark 1.
For any large submoduleL of M, Soc⁡(L)=Soc⁡(Hn(M)) for some positive integer n.Lemma 2.
LetN be submodule of M such that Soc⁡(Hnk(M))⊆Soc⁡(Hk(N)) for k=0,1,2,…, where the sequence of positive integers n0,n1,n2,… is monotonically increasing. Then M=N+B for any basic submodule B of M.Proof.
LetB=⨁Bi be a basic submodule of M and M=B1⊕⋯⊕Bk⊕(Bk∗,Hk(M)) [5]. Then (1)Soc⁡M=Soc⁡B1+⋯+Bk⊕Soc⁡HkM=Soc⁡B1⊕⋯⊕Bnk⊕Soc⁡HkN.
Now suppose, for everyx∈M, e(x)≤k implies that x∈(B+N). Consider x∈M such that e(x)=k+1; then there exists y∈M such that d(xR/yR)=k. Now y∈Soc⁡(M) and y=b+z, where b∈B and z∈Soc⁡(Hk(N)), ensuring the existence of z′ such that d(z′R/zR)=k. By the h-purity of B, there exists b′∈B, such that d(b′R/bR)=k. Now ex-b′-z′≤k, and thus x-b′-z′∈B+N or x∈B+N implying that M=B+N.The following remarks are significant to be stated.Remark 3.
LetL be a large submodule of an unbounded QTAG-module M without elements of infinite height M and B a proper basic submodule of M. Then (2)MB=B+LB≅LB∩L; therefore L is unbounded. Conversely for an unbounded fully invariant submodule L of M, Hk(L) is fully invariant for all k∈Z+. As an immediate consequence of Lemma 2, L is a large submodule of M. We can say that the unbounded fully invariant submodules of M are exactly the large submodules of M.Remark 4.
IfBi is the direct sum of uniserial modules of length i and x(≠0)∈Bi, then (3)Ux=n0,n1,…,ni-n0-1,…,∞,where n0=H(x) and nk=n0+k,0≤k≤i-n0-1.Remark 5.
LetBi be the direct sum of uniserial modules of length i and x,y∈Bi. Then there exists an endomorphism f of Bi with f(x)=y, if and only if H(x)≤H(y).Remark 6.
LetA be a fully invariant submodule of Bi, a direct sum of uniserial modules of length i. Then A=Hni(Bi), where ni≤i. If A=0, if ni=i, and if A≠0, then ni=min⁡Hx,x∈A.Remark 7.
IfBi and Bi+j are the direct sums of uniserial modules of length i and i+j, respectively, and x∈Bi, y∈Bi+j, then(i)
there exists a homomorphismf:Bi→Bi+j such that f(x)=y if and only if e(x)≥e(y),(ii)
there exists a homomorphismg:Bi+j→Bi such that g(y)=x if and only if H(x)≥H(y).Theorem 8.
LetB=⨁iBi, where each Bi is the direct sum of uniserial modules of length i. Then L is a fully invariant submodule of B if and only if L=⨁iHni(Bi), where ni≤i, for every i∈Z+ and ni≤ni+j≤ni+j for i,j∈Z+. A fully invariant submodule L is large in B if and only if L=⨁iHni(Bi); the above conditions hold and the sequence 1-n1,2-n2,3-n3,… is unbounded if B is unbounded.Proof.
LetL be a fully invariant submodule of B. Then (4)L=L∩B=⨁Bi∩L=⨁HniBi by the facts mentioned above and Remark 6. Now ni≤i for i∈Z+ and the first condition holds. If L=0, then Hni(Bi)=0 for every i; therefore ni=i for every i and the second condition holds. If L≠0, then there exists a least positive integer k such that Hnk(Bk)≠0. Then Hni(Bi)≠0 for all i≥k, where Bi≠0. Since Soc⁡(Bk)=Soc⁡(Hk-1(Bk))⊆L, this implies that Soc⁡(Hk-1(B))⊆Soc⁡(L). Again Soc⁡(Bi)=Soc⁡(Hk-1(Bi)) for i≥k; we have Soc⁡(Bi)⊆L∩Bi=Hni(Bi). Now suppose L≠0 and Bi≠0≠Bi+j. If Hni+j(Bi+j)=0, then Hni(Bi)=0 and ni=i,ni+j=i+j=ni+j and the second condition holds. We assume that Hni+j(Bi+j)≠0. Consider x∈Bi such that H(x)≥ni+j and y∈Hni+j(Bi+j) such that H(y)=ni+j. Now, by Remark 7, there exists an endomorphism g of B mapping y onto x. Hence x∈L and Hni+j(Bi)⊆L∩Bi=Hni(Bi); thus ni≤ni+j.
Now supposeHni(Bi)=0. Then ni=i so ni+j≤i+j=ni+j. If Hni(Bi)≠0 and y∈Bi+j such that H(y)≥ni+j, we may choose x∈Bi such that H(x)=ni. Then e(x)=i-ni and e(y)≤i+j-(ni+j)=i-ni. By Remark 7, there exists an endomorphism f of B with f(x)=y. Thus y∈L and we have Hni+j(Bi+j)⊆L∩Bi+j=Hni+j(Bi+j); therefore ni+j≤ni+j.
IfBi≠0≠Bi+j, then ni≤ni+j≤ni+j but if Bi=0, we may define ni so that this inequality holds for all i. Thus all fully invariant submodules of B are the direct sums of Hni(Bi). If L is a large submodule of B and B is unbounded, then, by Lemma 2, L is also unbounded. Therefore 1-n1,2-n2,3-n3,… must be unbounded.
For the converse, supposeL=⨁Hni(Bi), where ni≤i for all i∈Z+ and ni≤ni+j≤ni+j for all i,j∈Z+. To establish the full invariance of L, we consider any i∈Z+ and x∈Hni(Bi). We have to show that for any endomorphism f of B, f(x)∈L. Consider x≠0, such that f(x)=x1+⋯+xl, where xr∈Br and H(x)≤H(f(x))=min(H(xk)),1≤k≤l,e(x)≥e(f(x))=max⁡{e(xk)∣1≤k≤l}. If k≤i, then H(xk)≥H(x)≥ni so xk∈Hni(Bk)⊆Hnk(Bk), because nk≤ni; hence xk∈L. If k=i+j, then e(xk)≤e(x)≤i-ni=i+j-(ni+j)≤i+j-ni+j because ni+k≤ni+k. Thus xk∈Hi+j-ni+j(Bi+j)=Hni+j(Bi+j)⊆L.
This implies thatL is a fully invariant submodule of B. If B is unbounded and 1-n1,2-n2,3-n3,… is also unbounded, then L is unbounded and is therefore a large submodule of B by Remark 3.Corollary 9.
IfL is a large submodule of a QTAG-module M, then M/L is a direct sum of uniserial modules.Proof.
For any basic submoduleB of M,(5)ML=B+LL≅BB∩L=⨁iBi⨁iHniBi≅⨁iBiHniBi and the result follows.Corollary 10.
For any large submoduleL of M, L1=M1.Proof.
SinceM/L is a direct sum of uniserial modules, M/L1=0 or M1=L1.Theorem 11.
LetN be h-pure submodule of a QTAG-module M and L a large submodule of N. Then there exists a large submodule L′ of M such that L′∩N=L. If M/N is h-divisible, then L′ is the closure of L in M and is therefore uniquely determined by L and M/L′≅N/L.Proof.
LetL=N(n) and L′=M(n). Since N is h-pure in M and n=n1,n2,… is a U-sequence for N, we have that n is a U-sequence for M. Thus L′ is a large submodule of M.
Ifx∈L, then UM(x)=UN(x)≥n; therefore x∈L′∩N and L⊆L′∩N. Conversely if y∈L′∩N, then UN(y)=UM(y)≥n implies that y∈L or L′∩N⊆L. Thus L=L′∩N.
LetM/N be h-divisible and L′ a large submodule of M with L′∩N=L. Then (6)L′L=L′L′∩N≅N+L′N=MN. That is, L′/L is h-divisible. But M/L′≅(M/L)/(L′/L), where L′/L is a direct summand of M/L; we have M/L≅(L′/L)⊕(M/L′) and M/L′ is a direct sum of uniserial modules [6]. Now M/L′≅N/L, thus (7)ML′=N+L′L′≅NN∩L′=NL.Now we characterize large submodules in terms ofUlm invariants.Theorem 12.
LetL be a submodule of a QTAG-module M. Then L is a large submodule of M if and only if L=∑k=1∞HnkHk-nkM, where (i)
nk≤k,k∈Z+,(ii)
nk≤nk+1≤nk+1,(iii)
the sequence1-n1,2-n2,3-n3,… is unbounded if M is unbounded and the Ulm-invariants of L are given by fL(n)=∑k(fM(k-1)),k-nk-1=n,for all n∈Z+.Proof.
SupposeL=∑k=1∞Hnk(Hk-nk(M)). Since Hnk(Hk-nk(M))’s are fully invariant submodules, their sum is again fully invariant submodule of M. If M is bounded, then L is large. If M is unbounded, then, by the third condition, for each j∈Z+, there exists a positive integer i such that i-ni>j or i>ni+j.
Since,i>ni+j and Soc⁡(Hi(M))⊆Soc⁡(Hj+ni(M))⊆Hj(L), If x∈Soc⁡(Hj+ni(M)), there exists y∈M such that d(yR/xR)=j+ni, where e(x)=1. Now y∈Hj+ni+1(M), where i-ni>j or i-ni≥j+1; thus i≥ni+j+1 and y∈Hi(M).
Ifd(yR/zR)=ni, then z∈Hni(Hi-ni(M))⊆L and x∈Hj(L) because d(yR/xR)=j+ni and d(zR/xR)=j. Now, by Lemma 2, L+B=M, for every basic submodule B of M, and L is a large submodule of M.
Conversely supposeL is a large submodule of M. Then for any basic submodule B of M,L∩B is a large submodule of B and, by Theorem 8, L∩B=⨁kHnk(Bk), where k∈Z+ and nk’s satisfy the given conditions.
Now,(8)HnjBj=HnjHj-njBj⊆HnjHj-njB,foreveryj∈Z+ and, for each j∈Z+, (9)HnjHj-njB⊆ΣHnkHk-nkB.
This implies that⨁Hnk(Bk)⊆ΣHnk(Hk-nk(B)).
For the converse, considerx∈Hnj(Hj-nj(B)), where x=x1+x2+⋯+xm,xi∈Bi. Then H(xi)≥H(x)≥nj for xi,1≤i≤m and e(xi)≤e(x)≤j-ni for all xi′s. Now, for i≤j, we have H(xi)≥nj≥ni and xi∈Hni(Bi).
Ifi=j+l for l∈Z+, then e(xi)≤j-nj=j+l-(nj+l)≤j+l-nj+l. (by the given condition). Therefore (10)xi∈Hj+l-nj+lBj+l=Hnj+lBj+l=HniBi and Hnj(Hj-nj(B))⊆⨁kHnk(Bk). Let L′=∑Hnk(Hk-nk(M)). Now L′∩B=∑Hnk(Hk-nk(B))=L∩B. Since B is h-pure in M and M/B is h-divisible, L′=L, by Theorem 11. Again L∩B is a basic submodule of L; thus fL(n)=fL∩B(n), for all n∈Z+.
IfL∩B=⨁L∩Bii, where L∩Bi is the direct sum of uniserial modules of length i, then f(L∩B)(n)=g((L∩B)n+1)=g(⨁Hnk(Bk)), where Hnk(Bk) is a direct sum of uniserial modules of length n+1.
Again,(11)fL∩Bn=g⨁kHnkBk=∑kgHnkBk=∑kgBk=∑kfBk-1=∑kfMk-1,wherek-nk-1=n.
And the proof is complete.


## 3. Properties of Large Submodules ofQTAG-Modules

In this section we compare the structures ofQTAG-modules and their large submodules. We investigate the characteristics of QTAG-modules which are preserved by their large submodules. We start with the Σ-modules, that is, the modules whose high submodules are direct sums of uniserial modules [7]. Then we study summable, σ-summable, (ω+1)-projective, and h-pure complete QTAG-modules.Singh [8] proved that a QTAG-module M is a direct sum of uniserial submodules if and only if M is the union of an ascending sequence of submodules Mn,n=1,2,3,…, such that, for every n, there exists kn>0 and HM(x)≤kn for all x∈Mn.This helps us to prove the following.Theorem 13.
AQTAG-module is a Σ-module if and only if Soc⁡(M)=⋃k=1∞Mk, where Mk⊂Mk+1 and for every k∈N,Mk∩Hk(M)=Soc⁡(M1).Proof.
SinceM is a Σ-module, it contains a high submodule N such that N is a direct sum of uniserial modules.
AgainN is a high submodule [9] of M if and only if N is h-pure in M and Soc⁡(M)=Soc⁡(N)+Soc⁡(M1). Therefore by the above result [8], Soc⁡(N)=⋃k=1∞Nk, Nk⊆Nk+1, and Nk∩Hk(N)=0, and we deduce Soc⁡(M)=⋃k=1∞(Nk+Soc⁡(M1)). If we put Mk=Soc⁡(M1)+Nk, then Mk⊆Mk+1 and (Soc⁡M1+Nk)∩Soc⁡M1+Soc⁡(Hk(N))=Soc⁡(M1)+(Nk∩Hk(N))=Soc⁡(M1), because Hk(N) is a high submodule of Hk(M).
For the converse ifSoc⁡(N)=⋃k=1∞(Mk∩N)=⋃k=1∞Nk, where we put Nk=N∩Mk, then Nk⊆Nk+1. Also (12)Nk∩HkN=Mk∩HkM=Mk∩HkM∩N=Mk∩HkM∩N=M1∩N=0.Therefore N is a direct sum of uniserial modules and M is a Σ-module.Now we may prove the following.Theorem 14.
AQTAG-module M is a Σ- module if and only if its large submodule L is a Σ-module.Proof.
SinceL1=M1 [6], there is a natural number m such that Soc⁡(L)=Soc⁡(Hm(M)) and Soc⁡(Hn(L))=Soc⁡(Htn(M))for every n<ω and some tn such that m≤tn<ω. If M is a Σ-module, then, by Theorem 13, Soc⁡(M) is the union of ascending chain of submodules Mk such that Mk⊆Mk+1 and Mk∩Hk(M)=Soc⁡(M1) for every k∈N.
This implies thatSoc⁡(L)=⋃k<ω(Mk∩L) and Mk∩L⊆Mk+1∩L. Therefore (13)Mk∩L∩HkL=Mk∩HkL⊆Mk∩HkM=SocM1=Soc⁡L1. Now Theorem 13 indicates that L is a Σ-module.
Conversely supposeL is a Σ-module. Therefore (14)Soc⁡L=⋃n<ωLn,Ln⊆Ln+1,Ln∩HnL=Soc⁡L1. Again Soc⁡(Hm(M))=⋃n<ωLn. Now (15)Ln∩Soc⁡HtnM=Ln∩Soc⁡HnL=Soc⁡L1=Soc⁡M1. Thus, by Theorem 13, Hm(M) is a Σ-module, and so is M.To study the other relations between a moduleM and its large submodule L we need the following lemma.Lemma 15.
Isotype submodules of countable length of summableQTAG-modules are again summable.Proof.
LetN be an isotype submodule of countable length ρ in the summable module M. Now there is a Hρ(M)-high submodule K of M such that N⊆K. Since Soc⁡(M)=⨁σ<ρSσ, there is Hρ(M)-high submodule P of M such that Soc⁡(P)=⨁σ<ρSσ.
Again, for every ordinalρ, every Hρ(M)-high submodule is isotype; therefore P is isotype and it is summable. The socles of Hρ(M)-high submodules have the same images under the canonical map M→M/Hρ(M) because this maps Hρ(M)-high submodules isomorphically in a height preserving manner onto submodules of M/Hρ(M).
NowN is isotype in a summable module K of countable length ρ. Therefore Soc⁡(K) is the union of an ascending chain of submodules Kn, where for every n the heights of elements of Kn assume but a finite numbers of values.
NowSoc⁡(N)=∪Kn∩N,n=1,2,3,…, and the heights of the elements of Kn∩N assume a finite numbers of different values. Thus N is summable.The following result shows that summability is shared by large submodules.Theorem 16.
LetL be a large submodule of a QTAG-module M. Then M is summable if and only if L is summable.Proof.
SupposeM is summable; that is, Soc⁡(M)=⨁β<αMβ, where the nonzero elements of Mβ’s are contained in Hβ(M) but they do not belong to Hβ+1(M), for every β<α.
AgainL is fully invariant submodule of M and Hρ(M)=Hρ(L) for all ordinals ρ≥ω,Soc⁡(L)=⨁β<α(Mβ∩L), where the nonzero elements of Mβ∩L are contained in Hβ(L) and not contained in Hβ+1(L) for every ω≤β<α. Since Soc⁡(Hn(L))=Soc⁡(Htn(M)), whenever 1≤n<ω,n≤tn<ω, Mβ∩L⊆L, but (Mβ∩L)∩H1(L)=0, for each β<t1. By transfinite induction Mβ∩L⊆H1(L) and (Mβ∩L)∩H2(L)=0, for t1≤β<t2 and so on; that is, Hβ∩L⊆Hn(L) and (Hβ∩L)∩Hn+1(L)=0, for tn≤β<tn+1.
If we putL0=⨁0≤β<t1Mβ∩L,L1=⨁t1≤β<t2Mβ∩L,…,andLn=⨁tn≤β<tn+1(Mβ∩L), where n<ω, and Mβ∩L=Lβifβ≥ω, then Soc⁡(L)=⨁β<αLβifLβ⊆Hβ(L)andLβ∩Hβ+1(L)=0. Therefore L is summable.
Conversely supposeL is summable. So, L1=M1 is summable as its fully invariant submodule. Moreover, by Lemma 15, L being summable implies that L is a Σ-module. Now by Theorem 14, M is also a Σ-module. For a high submodule N of M,Soc⁡(N)⊕Soc⁡M1=Soc⁡(M).
SinceN is a direct sum of uniserial modules, Soc⁡(N)=⨁k<ωNk, where Nk⊆Hk(M) and Nk∩Hk+1(M)=0 because N is h-pure in M. Again the summability of M1 ensures that Soc⁡M1=⨁β<αKβ, where Kβ⊆Hβ(M1) and Kβ∩Hβ+1(M1)=0. Therefore, Kβ⊆Hω+β(M) and Kβ∩Hω+β+1(M)=0. This implies that (16)Soc⁡M=⨁k<ωNk⨁ω≤ω+β<ω+αKβ. We may infer now that M is summable.Theorem 17.
LetL be the large submodule of M. Then M is σ-summable if and only if L is σ-summable.Proof.
SupposeM is unbounded. Then length of M= length of L≥ω. If M is σ-summable, then L is also σ-summable being a submodule of equal length.
IfM is bounded the result holds trivially.
Conversely supposeL is σ-summable. Therefore Soc⁡(L)=⋃n<ωLn,Ln⊆Ln+1 and Ln∩Hαn(L)=0 for all n≥0 and some αn< length of M.
Now,Soc⁡(Hm(M))=⋃n<ωLn. Since L1=M1 [6], Hα(M)=Hα(L) for each ordinal α≥ω and Soc⁡(Hαn(L))=Soc⁡(Hkn(M)) for αn<ω and some kn≥max⁡(αn,m) because Hαn(L) is large in M and Hαn(M) both. Thus Ln∩Hsn(M)=Ln∩Hαn(L)=0, whenever sn< length of M= length of Hm(M)≥ω,sn=αn≥ω or ω>sn=kn.
We may defineMn={x∣x∈Soc⁡(M)∩Lnandx∉Hm(M)}. Thus Soc⁡(M)=⋃n<ωMn,Mn⊆Mn+1. By defining Mn’s we observe that Mn∩Hsn(M)=0. This implies that M is σ-summable.Theorem 18.
IfM is a direct sum of σ-summable QTAG-modules, then so is L.Proof.
LetM=⨁i∈IMi, where each Mi is σ-summable. Now L=⨁i∈I(L∩Mi) because L is fully invariant in M. Since all Mi’s are isotype in M, we infer that L∩Mi is large in Mi, for every i. By Theorem 17, L∩Mi are σ-summable. Thus L is also a direct sum of σ-summable modules.Let us recall the following.Definition 19.
AQTAG-module M is (ω+1)-projective if there exists a submodule N⊆Soc⁡(M) such that M/N is a direct sum of uniserial modules.Remark 20.
The submodules of(ω+n)-projective modules are also (ω+n)-projective.Theorem 21.
AQTAG-module M is (ω+1)-projective if and only if its large submodule L is (ω+1)-projective.Proof.
SupposeL is (ω+1)-projective. Therefore there exists a submodule N⊆Soc(L) such that Soc⁡(L/N)=⋃n<ω(Lk/N), where Lk⊆Lk+1⊆L and Lk∩Hk(L)⊆N for each k<ω.
NowLk⊆H2(M), for every k<ω. Since (17)HkH2L=HtkH2M+Soc⁡HjkM, for some k≤jk≤tk<ω, we have(18)HtkH2M⊆HkH2L⊆HjkH2M,Lk∩HtkM=Lk∩HtkH2M⊆Lk∩HkH2L=Lk∩HkL⊆N.Therefore the heights of the elements of Lk/N are bounded in M/N for all k<ω. Now (M/N)/(L/N)≅M/L is a direct sum of uniserial modules [6]. Therefore M/N is a direct sum of uniserial modules and M is (ω+1)-projective. The converse is trivial.The property of beingh-pure complete is also shared by the large submodules of QTAG-modules.First we recall the definition ofh-pure completeness.Definition 22.
AQTAG-module M is h-pure complete if, for every subsocle S⊆Soc⁡(M), there is a h-pure submodule N of M so that S=Soc⁡(N). In other words every subsocle supports a h-pure submodule of M.Theorem 23.
LetL be the large submodule of a QTAG-module M. If M is h-pure complete, so is L.Proof.
LetS be a subsocle of L. Since S⊆Soc⁡(M), S supports a h-pure submodule N of M. Now N∩L is also large in M and N∩L is h-pure in L. Again S=Soc⁡(N)∩Soc⁡L=Soc⁡(N∩L), and therefore L is h-pure complete.Corollary 24.
AQTAG-module M is h-pure complete if and only if Hk(M) is h-pure complete for some fixed but arbitrary positive integer k.Proof.
SinceHk(M) is large in M, it is h-pure complete if M is h-pure complete. Conversely suppose Hk(M) is h-pure complete. We shall use transfinite induction to prove the result.
LetS be a subsocle of M such that S∩H1(M)⊆Soc⁡(H1(M)) and S∩H1(M)=Soc⁡(N) for some h-pure submodule N of H1(M). By [7] we can say that there is a h-pure submodule K of M such that H1K=NandSoc⁡K=Soc⁡N=Soc⁡(H1(K)). Now S∩H1(M)=Soc⁡(H1(K)).
We have to show that there exists ah-pure submodule T⊆M such that S=Soc⁡(T). We define the submodule T=K+(S∩H2(M)). Now (19)Soc⁡T=Soc⁡K+S∩H2M=Soc⁡K+S∩H2Mbecause S∩H2(M)=Soc⁡(S∩H2(M)). Again Soc⁡(T)=Soc⁡(N)+(S∩H2(M))⊆S.
Now(20)S=S∩H1M∪S∩H2M=Soc⁡KSoc⁡S∩H2M=Soc⁡K∪S∩H2M⊆Soc⁡K+S∩H2M=Soc⁡T. Therefore S=Soc⁡(T). Now, (21)Soc⁡T∩HtM=S∩HtM=Soc⁡K+S∩H2M∩HtM=Soc⁡H1K+S∩H2M∩HtM=Soc⁡K∩HtM=Soc⁡HtK=Soc⁡HtT.This implies that T is h-pure in M.In the end we state the following unsolved problems.Problem 25.
Is it true thatM is a HF-module if and only if its large submodule L is?Problem 26.
Is it true thatM is a direct sum of closed modules if and only if its large submodule L is?



---

*Source: 2496246-2017-01-03.xml*