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

Authors: Alveera Mehdi; Fahad Sikander; Firdhousi Begum
Journal: Algebra
 (2013)

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## 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. There are many fascinating properties of QTAG modules of which h-pure submodules and high submodules are significant. A submodule N is quasi-h-dense in M if M/K is h-divisible, for every h-pure submodule K of M, containing N. Here we study these submodules and obtain some interesting results. Motivated by h-neat envelope, we also define h-pure envelope of a submodule N as the h-pure submodule K⊇N if K has no direct summand containing N. We find that h-pure envelopes of N have isomorphic basic submodules, and if M is the direct sum of uniserial modules, then all h-pure envelopes of N are isomorphic.


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## Body

## 1. Introduction

All the ringsR considered here are associative with unity, and right 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 decomposition series, d(M) denotes its decomposition 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 exponent at most k.  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.The modulesHk(M), k=0,1,…,∞ form a neighbourhood system of zero giving rise to h-topology. The closure of a submodule N⊂M is defined as N-=⋂k=0∞(N+Hk(M)), and it is closed with respect to h-topology if N=N-.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.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 submodule N⊂M is nice  [1, 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.The cardinality of the minimal generating set ofM is denoted by g(M). For all ordinals α, fM(α) is the αth-Ulm invariant of M and it is equal to g(Soc(Hα(M))/Soc(Hα+1(M))).For aQTAG module M, there is a chain of submodules M0⊃M1⊃M2⊃⋯⊃Mτ=0, for some ordinal τ. Mσ+1=(Mσ)1, where Mσ is the σth-Ulm submodule of M. Singh [2] proved that the results which hold for TAG modules also hold good for QTAG modules.


## 2. Quasi-h-Dense Submodules

In [3], we studied semi-h-pure submodules which are not h-pure but contained in h-pure submodules. Now we investigate the submodules N⊆M such that M/K is h-divisible for every h-pure submodule K⊆M, containing K. These modules are called quasi-h-dense submodules.We start with the following.Definition 1.
A submoduleN of M is quasi-h-dense in M, if for every h-pure submodule K⊆M, containing N,  M/K is h-divisible.Lemma 2.
IfSoc(Hk-1(M))⊈N+Hk(M) for some integer k, then there exists a proper submodule K of M containing N and a bounded submodule Q of M such that  (i)
M=K⊕Q;(ii)
K⊇N+Hk(M);(iii)
N+Hk(M)⊇Soc(Hk-1(K)).Proof.
The socle ofHk-1(M) can be expressed as the direct sum of S1,S2 where S1=(N+Hk(M))∩Soc(Hk-1(M)) and S2∩Hk(M)=0. Now there exists a h-pure submodule Q of M such that Soc(Q)=S2. Also (Q⊕Hk(M))/Hk(M) is h-pure in M/Hk(M) and Soc((Q⊕Hk(M))/Hk(M))=(S2⊕Hk(M))/Hk(M). Now Soc(HK(M/Hk(M)))=0 which is contained in (S2⊕Hk(M))/Hk(M)⊆Soc(HK-1(M/Hk(M))); therefore (Q⊕Hk(M))/Hk(M) is an absolute direct summand of M/Hk(M) [4]. Thus there exists a submodule K of M such that
(1)MHk(M)=(KHk(M))⊕(Q⊕Hk(M)Hk(M))
and K⊃N+Hk(M); therefore M=K⊕Q and N+Hk(K)⊇Soc(Hk-1(K)).Proposition 3.
A submoduleN of a QTAG module M is quasi-h-dense in M if and only if for all integers k≥0,N+Hk+1(M)⊇Soc(Hk(M)).Proof.
Letk be the least positive integer such that N+Hk(M)⊉Soc(Hk-1(M)). By Lemma 2, there exists a proper submodule K⊆M containing N and a bounded submodule Q of M such that M/K≅Q. This contradiction proves that N+Hk+1(M)⊇Soc(Hk(M)), for all integers k≥0.
For the converse, consider anh-pure submodule K⊇N. If M/K is not h-divisible, then there exist submodules P and Q of M such that M/K=(P/K)⊕(Q/K), where Q/K is bounded. If Hk(Q/K)=0 and Hk-1(Q/K)≠0, then P⊇N+Hk(M)⊇Soc(Hk-1(M)). Since P is h-pure in M, we have Hk-1(M)⊆P. On repeating the process, after a finite number of steps, we get P=M and Q/K=0, which is a contradiction. Hence, M/K is h-divisible and N is quasi-h-dense in M.Remark 4.
For a submoduleN of a QTAG module M, the following are equivalent:  (i)
N is quasi-h-dense in M;(ii)
N+Hk+1(M)⊇Soc(Hk(M)),∀k>ω;(iii)
for everyh-pure submodule K containing N, M/K is h-divisible.Lemma 5.
IfK is an h-divisible submodule of a module M and N is a K-high submodule of M, then N is h-pure in M.Proof.
For anh-divisible submodule K⊆M , K-high submodules are h-neat [5]. Therefore, N∩H1(M)=H1(N). Assume that N∩Hi(M)=Hi(N) for all i≤k. Let x∈N∩Hk+1(M). Now there exists y∈M such that d(yR/xR)=k+1. If y′∈M such that d(yR/y′R)=k, then y′+z∈Soc(M) for some z∈N and y′+z∈Soc(M)=Soc(K)⊕Soc(N)=Soc(Hk(K))⊕Soc(N). Now there exists u∈Soc(Hk(K)) such that d(xR/uR)=k and y′+u∈N. By induction y∈Hk+1(N) and N is h-pure in M.Theorem 6.
LetN be a proper submodule of M. Then N is not semi-h-pure if and only if the following hold:  (i)
N is quasi-h-dense in M;(ii)
there existsm∈ℤ+ such that Soc(Hm(M))⊆N.Proof.
By Lemma2 and Proposition 3,  (i) is satisfied. Suppose (Soc(Hk(M))+N)/N≠0, for all k≥0. If there exists a nonnegative integer m such that
(2)Soc(Hm(M))+NN=Soc(Hm+t(M))+NN,fffffffffor  everyt≥0,
then we have
(3)Soc(Hm(M))=Soc(N∩Hm(M))+Soc(Hm+t(M)),fffffffffffffffffffffffffffffffffffffffffffffff∀t≥0.
Let K be the h-neat envelope [6] of N∩Hm(M) in Hm(M). Then K is h-pure in Hm(M). Since N∩Hm(M) is quasi-h-dense in Hm(M),  Hm(M)/N is h-divisible. Thus, there exists a submodule Q of M such that N⊆Q and
(4)MK=QK⊕Hm(M)K.
Now Q is h-pure in M, hence Hm(M)=K and Soc(Hm(M))=Soc(K)=N∩Soc(Hm(M)). Therefore, we may assume that there exists a chain of integers 0≤k0<k1<k2⋯ such that
(5)Soc(Hki(M))+NN≠Soc(Hki+1(M))+NN,fffffffffffffffor  everyi≥0.
Then we may choose xi∈Soc(Hki(M)) such that xi∉Soc(Hki+1(M)) for every i and we have {xi}∩N=0. Let P be an N-high submodule of M such that ΣxiR⊆P. There exists a proper unbounded basic submodule B of P and a submodule T⊆M, containing N such that M/B=(T/B)⊕(P/B). Now T is h-pure in M which is a contradiction proving (ii).
Conversely, supposeK is an h-pure submodule of M containing N. By (i),  M/K is h-divisible. Since Soc(Hm(M))⊆K and K is h-pure in M, we have K=M.Now we should mention the following notations used by Khan and Zubair [7]:
(6)Nk(M)=(N+Hk+1(M))∩Soc(Hk(M)),Nk(M)=(N∩Soc(Hk(M)))+Soc(Hk+1(M)),Qk(M,N)=Nk(M)Nk(M).
We prove the following.Proposition 7.
LetK⊇N be an h-pure submodule of a QTAG module M. Then Qk(M,N)≅Qk(K,N), for all k≥0.Proof.
SinceNk(M)∩K=Nk(K) and Nk(M)∩K=Nk(K), there exists a homomorphism f:Qk(K,N)→Qk(M,N) such that f(x+Nk(K))=x+Nk(M). If z∈Nk(M) and z+Nk(M)∈Qk(M,N), then there exist u∈N∩Hk(M) and v∈Hk+1(M) such that z=u+v and e(z)=1. Moreover if u′∈uR, v′∈vR, d(uR/u′R)=1=d(vR/v′R), then u′R=v′R is contained in N∩Hk+2(M)⊆K∩Hk+2(M)=Hk+2(K) and there exists y∈K such that u′R=v′R=y′′R, where d(yR/y′′R)=k+2 and d(y′R/y′′R)=k+1=d(yR/u′R)=d(yR/v′R) for some y′,y′′∈Hk+1(K) and Hk+2(K), respectively. Since (u+y′′)R⊆Nk(K) and (v+y′′)R⊆Soc(Hk+1(M)), we have f(u+y′+Nk(K))=z+Nk(M). Also f is one to one by definition; hence f is an isomorphism.Theorem 8.
LetN be a semi-h-pure submodule in M having an h-pure hull. Then there exists an integer l∈ℤ+ such that Qk(M,N)=0, for k≥l.Proof.
LetK be an h-pure hull of N in M. By Theorem 6, there exists a nonnegative integer l such that Soc(Hl(K))⊆N and for all k≥l,
(7)Nk(K)=(N+Hk+1(K))∩Soc(Hk(K))=Soc(Hk(K))=Nk(K).
ThereforeQk(K,N)=0, for all k≥l. Hence by Proposition 7, Qk(M,N)=0 for all k≥l.Proposition 9.
LetK be an h-pure hull of a semi-h-pure submodule N of M and l a nonnegative integer. Then Soc(Hl(K))⊆N and Soc(Hl-1(K))⊈N if and only if Qk(M,N)=0 for  all k≥l and Qk-1(M,N)≠0.Proof.
SinceNk(K)=Soc(Hk(K))=Nk(K) for all k≥l, by Proposition 7,  Qk(M,N)=Qk(K,N)=0for allk≥l. If Qk-1(M,N)=0, then Qk-1(K,N)=0 and Soc(Hl-1(K))=Nl-1(K)=Nl-1(K)⊆N. This contradiction implies that Ql-1(M,N)≠0.
For the converse, consider the integerq such that Soc(Hq(K))⊆N and Soc(Hq-1(K))⊈N. If q>l, then
(8)Nl(K)=Soc(Hl(K))=Nl(K)=Soc(N∩Hl(K))+Soc(Hq(K))⊆N,
which is a contradiction. If q<l, then
(9)Nl-1(K)=Soc(Hl-1(K))=Nl-1(K).
Again it is a contradiction, hence l=q.k
V modules are defined in [3], and we can observe the following.Remark 10.
IfN is h-pure in M, then by Proposition 7,Qk(M,N)=Qk(N,N)=0 and N⊆Soc(M) implies that Nk(M)=Nk(M) for all k≥0. In other words, N is a V module in M.Proposition 11.
LetN be an h-neat submodule of M. Then N is h-pure in M if and only if N is a V module in M.Proof.
SupposeN∩Hk(M)=Hk(N). Consider x∈N such that there exists y∈M and d(yR/xR)=k+1. This implies the existence of y′ such that d(y′R/xR)=k and d(yR/y′R)=1. Now there exists z, z′∈N such that y′-z′∈Nk-1(M), where d(zR/z′R)=k-1.  As Nk-1(M)=Nk-1(M), we have y′-z′=u+v for some u∈N∩Soc(Hk-1(N)) and v∈Soc(Hk(N)). Since y′-v∈N∩Hk(M)=Hk(N), there exists ω∈N such that d(ωR/xR)=k+1 and N is h-pure in M. The converse is trivial.Proposition 12.
A submoduleN⊆M is a V module if and only if Soc(N+Hk(M))=Soc(N)+Soc(Hk(M)) for all k≥1.Proof.
SupposeN is a V module in M. Therefore Soc(N+H1(M))=Soc(N)+Soc(H1(M)). We will prove the result by induction. Assume that Soc(N+Hk(M))=Soc(N)+Soc(Hk(M)) for all k≤l. Since Soc(N+Hl+1(M))⊆Soc(N+Hl(M))⊆Soc(N)+Soc(Hl(M)), we have
(10)Soc(N+Hl+1(M))=(N+Hl+1(M))∩(Soc(N)+Soc(Hl(M)))=Soc(N)+Nl(M)=Soc(N)+Nl(M)=Soc(N)+Soc(Hl+1(M)).
The converse is trivial.Following results are the immediate consequences of the previous discussion.Remark 13.
(i)N is a V module in M if and only if H1(N∩Hk(M))=H1(N)∩Hk+1(M) for all k≥1.
(ii) IfHn(N+Hk(M))=Hn(Hk(M)) for all n,k≥0, then N is a V module in M.


## 3.h-Pure Envelopes

For a proper submoduleN⊆M, it is not always possible for N to have an h-pure hull in M. We study the situation when there is a proper h-pure submodule K⊇N of M, but no proper direct summand of K contains N. This motivates us to define h-pure envelopes like h-neat envelope defined earlier [6]. We find that the Ulm-Kaplansky invariants are same for all h-pure envelopes.Definition 14.
LetN be a submodule of a QTAG module M. An h-pure submodule P⊇N is an h-pure envelope of N if P has no proper direct summand containing N.Proposition 15.
In a moduleM, an h-pure submodule P⊇N is an h-pure envelope of N if and only if P contains no h-divisible summand disjoint from N and for any k, no uniserial summand of decomposition length k disjoint from N+Hk(P).Proof.
IfP is not an h-pure envelope of N, then P=K⊕L with N⊆K,L≠0. If L is not h-reduced, then it contains an h-divisible summand disjoint from N, and if L is h-reduced, then it contains a uniserial summand L′ such that d(L′)=k for some k>0. Without loss of generality, we may assume that L=L′. Now
(11)Hk(P)=(K∩Hk(P))⊕(L∩Hk(P))=K∩Hk(P),
implying that Hk(P)⊆K. Therefore, L∩(N+Hk(P))=0 and N+Hk(P)⊆K.
Conversely, ifL is h-divisible submodule of P with L∩N=0, then L is a summand of P; therefore P cannot be an h-pure envelope of N. If L≠0, it is a uniserial summand such that d(L)=k and L∩(N+Hk(P))=0, then P=K⊕L such that N+Hk(P)⊆K. This K can be chosen to be any submodule of M which is maximal with respect to the properties of containing N+Hk(P) and disjoint from L. This P can not be the h-pure envelope of N.LetxR be a uniserial summand of N such that d(xR)=k+1. If y∈xR such that d(xR/yR)=k, then xR∩(N+Hk+1(P))=0 if and only if yR⊈N+Hk+1(P). Since y∈Soc(Hk(P)), this implies that Soc(Hk(P))⊈N+Hk+1(N). This enables us to prove the following.Proposition 16.
Anh-pure submodule P⊆M containing the submodule N of a separable module M is an h-pure envelope if and only if Soc(Hk(P))⊆N+Hk+1(P), for all k∈N.Proof.
LetN be a semi-h-pure submodule of a separable module M and P the h-pure hull of N. Now we have Soc (N∩Hk(P))+Hk+1(P)=Soc(Hk(P)). This implies that fk(P,N)=0 for every k≥0. This is true for h-pure envelopes, and if M is separable, this is sufficient also.Remark 17.
Since(12)Soc(Hk(P))Soc(Hk+1(P)+N)∩Soc(Hk(P))≅Soc(Hk(P))+Soc(Hk+1(P)+N)Soc(Hk+1(P)+N)=0,
for all k<ω, if and only if  Soc(Hk(P))⊆N+Hk+1(P) for every k<ω, an h-pure submodule P⊇N is an h-pure envelope of N if and only if fk(N,P)=0, for every k<ω.Remark 18.
Since the union of a chain ofh-pure envelopes of N may not contain any h-divisible direct summand disjoint from N or a uniserial summand of decomposable length k disjoint from N+Hk(P), every h-pure envelope of N⊆M is contained in a maximal h-pure envelope of N.Now we investigateh-pure envelopes of N containing other h-pure envelopes of N.Theorem 19.
LetN be a submodule of a separable module M and P⊆Q, h-pure submodules of M containing N. Then  (i)
for everyk<ω, the natural embedding P→Q induces a monomorphism
(13)fk:Soc(Hk(P))+Soc(N+Hk+1(P))Soc(N+Hk+1(P))⟶Soc(Hk(Q))+Soc(N+Hk+1(Q))Soc(N+Hk+1(Q)),(ii)
the mapfk is onto for every k>ω if and only if P is an h-dense submodule of Q.Proof.
The mapsfk send the coset x+Soc(N+Hk+1(P)) upon the coset x+Soc(N+Hk+1(Q)). Since x∈Hk(P), there exists y∈P such that d(yR/xR)=k and e(x)=1; therefore
(14)fk(Soc(Hk(P))+Soc(N+Hk+1(P))Soc(N+Hk+1(P)))=Soc(Hk(P))+Soc(N+Hk+1(Q))Soc(N+Hk+1(Q)).
If x+Soc(N+Hk+1(P))∈kerfk, then fk(x+Soc(N+Hk+1(P)))=Soc(N+Hk+1(Q)) or x=u+z,u∈N,z∈Hk+1(Q); that is, there exists v∈Q such that d(vR/zR)=k+1. Now x-u=z∈N∩Hk+1(Q)=Hk+1(P). Therefore x∈N+Hk+1(P) or x∈Soc(N+Hk+1(P)) and fk is a monomorphism, which completes part (i).
(ii) SupposeP is h-dense in Q, therefore Q=P+Hk(Q) for k<ω. For x∈Soc(Hk(Q)) such that x∉Soc(Hk+1(Q)), there exists y∈Q such that d(yR/xR)=k. Since P is h-dense in Q,x=u+z,  u∈P,z∈H1(Q),x∈Soc(Hk(P))+Soc(N+Hk+1(Q)), and fk is surjective.
Conversely, suppose eachfk is surjective, therefore Soc(Hk(Q))⊆Soc(Hk(P))+Soc(N+Hk+1(Q)), for every k<ω. Let x∈Soc(Hk(Q)). Now there exists y∈Q such that d(yR/xR)=k. Now x=z+(u+v) with z∈Soc(Hk(P)),u∈N, and v∈Q such that there exists v′∈Q with d(v′R/vR)=k+1. Also u∈Hk(N) and z+u∈Hk(P). We will prove that Hk(Q)⊆P+H1(Q) implying that Q=P+H1(Q). For k=1, if y∈Soc(Q), then y∈P+H1(Q), and if y∈Hk+1(Q),x∈Soc(Q) and x=z+u+v implies that y-z-u-v∈P+H1(Q). Therefore y∈P+H1(Q), and we are done.Now we investigate the relation betweenUlm-Kaplansky invariants of the submodules N and their h-pure envelopes.Theorem 20.
LetP be an h-pure envelope of the submodule N of a separable module M. Then
(15)Soc(P∩Hk(M))Soc(P∩Hk+1(M))≅Soc(N∩Hk(P))Soc(N∩Hk+1(P))⊕Soc(Hk+1(P)+(N∩Hk(P)))Soc(Hk+1(P))+Soc(N∩Hk(P)).Proof.
We have(16)Soc(N∩Hk(P))Soc(N∩Hk+1(P))=Hk(P)∩Soc(N)Hk+1(P)∩Soc(N)≅(Soc(Hk(P))∩N)+Soc(Hk+1(P))Soc(Hk+1(P))=Soc(Hk(P))∩(Soc(N)+Soc(Hk+1(P)))Soc(Hk+1(P)).
Since M is separable, by Remark 17, we have
(17)Soc(Hk(P))Soc((N∩Soc(Hk(P)))+Hk+1(P))=0,
therefore,
(18)Soc(Hk(P))=Soc((N∩Soc(Hk(P)))+Hk+1(P)),
and we are able to write
(19)Soc((N∩Hk(P))+Hk+1(P))Soc(N∩Hk(P))+Soc(Hk+1(P))=Soc(Hk(P))Soc(Hk(P))∩(Soc(N)+Soc(Hk+1(P))).
This implies that
(20)Soc(P∩Hk(M))Soc(P∩Hk+1(M))=Soc(Hk(P))Soc(Hk+1(P))
is an extension of Soc(N∩Hk(P))/Soc(N∩Hk+1(P)) by Soc(Hk+1(P)+(N∩Hk(P)))/(Soc(Hk+1(P))+Soc(N∩Hk(P))), and the result follows.Following are the immediate consequences of the previous discussion.Corollary 21.
Everyh-pure envelope of the submodule N in a separable module M has same Ulm-Kaplansky invariants. Therefore h-pure envelopes of N have isomorphic basic submodules.Corollary 22.
LetM be the direct sum of uniserial modules. Then all h-pure envelopes of a submodule N⊆M are isomorphic, if they exist.Proposition 23.
IfN is h-pure submodule of a separable module M such that its closure N- is not h-pure in M, then N- has no h-pure envelope in M.Proof.
Anh-pure envelope Q of N- must have a larger basic submodule than N, and any uniserial summand in this larger basic submodule not in N- contradicts the Proposition 15, and the result follows.Theorem 24.
Every submodule of a separable moduleM admits h-pure envelopes if and only if M is quasicomplete; that is, the closure of every h-pure submodule of M is h-pure in M.Proof.
Suppose on the contrary thatM is not quasicomplete. Thus it contains an h-pure submodule N such that N- is not h-pure in M. By Proposition 23, the submodule N-⊆M does not have an h-pure envelope. This contradiction proves that M is quasicomplete.
For the converse, consider a submoduleN of a quasicomplete module M. We construct a countable sequence of subsocles of N,0⊆N0⊆N1⊆⋯⊆Nk⊆⋯ with another sequence, 0⊆K0⊆K1⊆⋯⊆Kk⊆⋯ of h-pure submodules of M such that Ni⊆Ki and Hi(Ki)=0, for every i. Now we may say that Nk+1 is the maximal submodule such that Nk⊆Nk+1⊆Soc(N) and Nk+1∩Hk+1(M)=0. Put Tk=Nk+1∩Hk(M) such that the nonzero elements of Tk are of height k and Nk+1=Nk⊕Tk. Now Tk∩Nk=0, and Tk⊕Kk is bounded by k; therefore there exists a minimal h-pure submodule containing Tk⊕Kk, which is Kk+1 here. Now Kk+1 is also bounded, hence a summand of M and by Proposition 16, Hk(Kk+1)=Tk. By the same argument, we get K′=⋃k∈ωKk, an h-pure submodule of M. Now
(21)N′=⋃k∈ω‍Nk=⨁k∈ωTk
is an h-dense subsocle of N, and we have to show that K′ does not have a uniserial summand of decomposition length k, which is disjoint from N+Hk(K′). Suppose xR is a uniserial summand of K′ such that d(xR)=k and xR∩(N+Hk(K′))=0. Let m be the least positive integer such that xR∩(N+Hm(K′))=0,x∈Km+1. Since xR∩(Tm+Km+Hk(Km+1))≠0 and there are elements u∈Tm, y∈Km, x′∈Soc(xR), v,v′∈Km+1 such that d(v′R/vR)=k,x′=u+y+v. Here H(u)=m, y≠0 because xR∩(N+Hk(Km+1))=0. Therefore y∈Km implies that H(y)≤m-1. As H(x′)=k-1, we have k-1=m-1; therefore there exist z,z′∈Km+1 such that d(zR/z′R)=k,uR=z′R. Also there are elements w,w′∈Km+1 such that d(wR/w′R)=k-1 and w′R=yR. Since Km is h-pure, there is a uniserial summand tR⊆Km and d(tR/yR)=k-1 and tR∩(N+Hk(K′))≠0, which is a contradiction implying that xR∩(N+Hk(K′))≠0.
Suppose thatK is the closure of K′ with respect to h-topology. Then, K is h-pure in M and K⊃Soc(N).
Case i. When M is closed [8] with respect to h-topology [4], by the structure of M,  K=K- is a summand of M. Consider the decomposition M=K⊕L. If Q is a submodule of M such that Soc(Q)=Soc(K), then
(22)Soc(N+LN)=(N+Soc(L))N.
If x+N∈(N+Soc(L))/N has height k in M/N, then HL(x)=k and (N⊕L)/N is h-pure in M/N. Thus,
(23)MN=QN⊕(L⊕NN)
and M=Q⊕L, and Q is an h-pure envelope of N in the closed module M.
Case ii. When M is an arbitrary QTAG module, again consider the decomposition M=K⊕L. If K is bounded, then we are done; otherwise M/K is a closed module. By Case i, (N+K)/K⊂M/K has an h-pure envelope L/K. Assume on the contrary that xR is a summand of L such that d(xR)=k,xR∩(N+Hk(L))=0. If Soc(xR)∩K=0, then (xR+K)/K is a uniserial summand of L/K and
(24)(xR+KK)⋂‍(N+KK+Hk(LK))≠0,
implying that xR∩(N+Hk(L))⊈N which is a contradiction. Therefore Soc(xR)⊆N, but then N contains a summand yR such that Soc(yR)=Soc(xR) and xR∩(N+Hk(L))=0 implies that yR∩(N+Hk(K))=0, again a contradiction. Thus, no summand xR exists, and L is an h-pure envelope of N.Now we investigate the conditions under which every submodule of a QTAG moduleM has an h-pure envelope.Proposition 25.
M
1 has an h-pure envelope in M if and only if it is h-divisible.Proof.
IfM1 is not h-divisible, then the basic submodule of any h-pure submodule N of M containing M1 is nontrivial and it has a nonzero uniserial summand disjoint from M1. By Proposition 15, N can not be an h-pure envelope of M1. The converse is trivial.The following theorem characterizes the moduleM whose every submodule has an h-pure envelope.Theorem 26.
In a moduleM, every submodule has an h-pure envelope if and only if M is the direct sum of an h-divisible and a quasicomplete module.Proof.
Suppose that every submoduleN of M has an h-pure envelope. By Proposition 25, M1 is h-divisible. Now M=D⊕K, where D is the maximal h-divisible submodule of M and K is separable. Let N⊆K and P, the h-pure envelope of N in M. Now if we project P into K, then the projection of P is an h-pure envelope of N in K. Therefore, all the submodules of K have h-pure envelopes and Theorem 24 implies that K is a quasicomplete module.
Conversely, supposeM=D⊕K, where D is h-divisible and K a quasicomplete module. For N⊆M,D+N=D⊕N0, where N0⊆K may be chosen. By Theorem 24, N0 has an h-pure envelope P0 in K. Now there exists an h-divisible submodule D0⊆D that contains N∩D as an essential submodule. If we put P=P0⊕N0, then N⊆P which is h-pure in M. P cannot have a uniserial or h-divisible summand disjoint from N, because such a summand would be disjoint from N0 and N∩D. Therefore, P is an h-pure envelope of N.



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*Source: 873193-2013-07-29.xml*