On countably generated QTAG-modules

Algebra Letters, 2019, 2019:3 https://doi.org/10.28919/al/4280 ISSN: 2051-5502

ON COUNTABLY GENERATED QTAG-MODULES FAHAD SIKANDER

College of Science and Theoretical Studies, Saudi Electronic University, Jeddah-23442, Kingdom of Saudi Arabia

Copyright c2019 the author(s). This is an open access article distributed under the Creative Commons Attribution License, which permits

unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. A right moduleM over an associative ring with unity is a QTAG-module if every finitely generated submodule of any homomorphic image ofM is a direct sum of uniserial modules. The main aim of this paper is to investigate the direct sums of countably generated QTAG-modules and obtain some significant results related to these modules.

Keywords:QTAG-module; summable modules; high submodules;h-pure submodules.

2010 AMS Subject Classification:16 K 20.

1.

Different concepts for groups like purity, projectivity, injectivity, height etc. have been

gen-eralized for modules. In generalizing the results of groups for modules sometimes we find that

they are not true for modules. So, for establishing such generalizations we impose some

condi-tions either on the module or on the underlying ring. Here we impose a condition on modules

that every finitely generated submodule of any homomorphic image of the module is a direct

sum of uniserial modules while the rings are associative with unity. After these conditions many

∗_{Corresponding author}

E-mail address: [email protected] Received August 29, 2019

elegant results of groups can be proved for QTAG-modules which are not true in general. Many

results of this paper are the generalization of the paper [3].

The study of QTAG-modules was initiated by Singh [8]. Mehdi et al. [2] worked a lot

on these modules. They introduced many concepts and generalized different notions for these

modules and contributed to the development of the theory of QTAG-modules. Yet there is much

to explore.

Singh [8] proved that a QTAG-module M is a direct sum of uniserial modules if and only

if M is the union of an ascending chain of bounded submodules. This indicates that M is a

direct sum of uniserial modules if and only if Soc(M) = L

k∈ω

S_k andH(x) =kfor everyx∈S_k.

This motivates us to define summable modules. A reduced QTAG-module M of length σ is

said to be summable if Soc(M) = L ρ<σ

Sρ and the nonzero elements of Sρ are contained in

(H_ρ(M)\H_ρ+1(M)) [1]. This implies that a QTAG-module M of length ω is a direct sum of

uniserial modules if and only if it is summable.

Here we investigate the direct sums of countably generatedQTAG-modules. The following

example shows that a summable λ-module (ω ≤λ) need not be a direct sum of countably

generated modules. Let M_α be the countably generated QTAG-module of length α, having

each of its nonzero Ulm-invariant equal toN₀.Ifγ <λ,then for eachβ (α <β <γ) Mα can

be embedded inM_β such thatH_µ(M_β)∩M_α =H_µ(M_α)for allµ.For a limit ordinalγ, Mγ'

S

α<γ

M_α and H_µ(M_β)∩M_α =H_µ(M_α) for α <β ≤γ. If γ is a non limit ordinal, γ −1 is

a limit ordinal and H_γ−1(M_γ)-high submodule of M_γ is isomorphic to M_γ−1. If γ −2 exists

then we consider Mγ−1,a direct summand ofMγ i.e.Mγ =Mγ−1⊕Hγ−1(Mγ)forα <β ≤γ.

As α ranges over the countable ordinals the modules Mα are embedded in one another. Let

M= S α<λ

M_α.NowMis summable,M/H_α(M)is a direct sum of countably generated modules

for each countableα,butMis not a direct sum of countably generated modules.

This made us investigate the other conditions as summability is not enough. The condition

thatM/H_ρ(M) should be a direct sum of countably generated modules for all ρ less than the

length ofMensures thatMis a direct sum of countably generated modules. Ifρ is a countable

is a direct sum of countably generated modules. Therefore we focused on the

QTAG-modules of limit length.

In section 3 of this article we prove some significant results related to the modules which

are the direct sums of countably generated modules. Section 4 deals with some applications of

these results.

2.

Throughout this paper, all rings are associative with unity and modules M are

uni-tal QTAG-modules. An element x ∈ M is uniform, if xR is a non-zero uniform (hence

uniserial) module and for any R-module M with a unique composition series, d(M)

de-notes 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 inM,

respec-tively. H_k(M)denotes the submodule of M generated by the elements of height at least kand

Hk(M)is the submodule of M generated by the elements of exponents at most k.A

submod-ule N ofM is h-pure [5] in M if N∩H_k(M) =H_k(N),for every integer k≥0. A submodule

N of M is said to be isotype in M, if it is σ-pure for every ordinal σ [4]. M ish-divisible if

M=M1= T∞

k=0

H_k(M)and it ish-reduced if it does not contain any h-divisible submodule. In

other words it is free from the elements of infinite height.

For a QTAG-moduleM, there is a chain of submodulesM0⊃M1⊃M2· · · ⊃Mτ _{=0, for}

some ordinal τ. Mσ+1= (Mσ)1, whereMσ is theσth-U lm submodule of M. Several results

which hold for TAG-modules also hold good for QTAG-modules [8]. Notations follows the

standard work of Fuchs [6, 7].

3.

This section deals with some interesting results and consequences. We start with the

follow-ing:

Remark 3.1. We may observe that if M is a summable QTAG-module of countable length thenSoc(M)is the union of an ascending chain of height finite submodules. SinceSoc(M) =

L

ρ<σ

S_ρ, S_ρ ⊆(H_ρ(M)\H_ρ+1(M))∪ {0}ifρ1, ρ2, . . . ,ρn, . . .are the ordinals less thanσ,then

Soc(M) = S∞

k=1

T_k, T_k=S_ρ1⊕. . .⊕S_ρk and T_k are height finite.

We prove the following lemma:

Lemma 3.1. Let N be a submodule of a QTAG-module M such that (i) H_α(M/N) =Hα(M) +N

N ,for everyα <β; (ii) M/N is countably generated of length≤β.

If K/N is finitely generated submodule of M/N,then K also satisfies these two conditions.

Proof. Assume thatH_α(M/K) = Hα(M) +K

K ,for allα <γ <β.Considerx+K∈Hγ(M/K).

Ifγ =α+1,then there existsy+K∈Hα(M/K) =

H_α(M) +K

K such thatd

(y+K)R

(x+K)R

=1,

thereforey∈H_α(M)andx+K∈ Hγ(M) +K

K .Otherwise, ifγ is a limit ordinal, then for each

α <γ, x+K =z+K, where z∈Hα(M). Thus x−z=u+v where u∈K, v∈N. Since u

assumes only finitely many values, there is a fixedu0∈K and a setT ⊆ {α |α <γ}such that

supT=γandx−z=u0+vwherev∈N.Nowx−u0+K=z+N∈

H_α(M) +N

N .Since supT=

γ, x−uo+N =z+N∈

H_α(M) +N

N for allα ∈T andx−uo+N∈Hγ(M/N) =

H_γ(M) +N

N .

Now we may writex−u_o+N=z+N withz∈H_γ(M)andx+K=z+K∈ Hα(M) +K

K .

To prove the second condition it is sufficient to proveH_β(M/K) =0i.e.,length ofM/Kisβ.

Forx+K∈H_β(M/K)we findu∈Ksuch thatx−u+N∈H_β(M/N) =0.Thereforex−u∈N

orx+K=0.

Theorem 3.1. If N is a submodule of a QTAG-module M such that

(i) H_α(M/N) =Hα(M) +N

N for everyα <β;

(ii) Hα(M)∩N=Hα(N)for everyα <β;

(iii) M/N is countably generated of length≤β;

Proof. SinceM/N is countably generated, we have to show that ifK/N is a finitely generated

submodule of M/N, every height increasing homomorphism f :K →N (f|N = IN) can be

extended to a height increasing homomorphism ¯f :K+xR→N whenever y∈ K such that

d

xR

yR

=1.Considerx6∈K.NowH_M/K(x+K) =α,for someα<β.Thereforex+K=z+K

with z∈H_α(M). Since u∈K where d

zR uR

=1 and xR+K =zR+K, x∈H_α(M). Now

H_M(x+v)≤H_M/K(x+K) =H_M(x) =α for allv∈K.Againα+1=β,we havew∈Hα(N)

such thatw0= f(y)∈H_α+1(M)∩N=Hα+1(N)whered

wR w0R

=1.Now we may define ¯f on

K+xRsuch that ¯f(xr+u) =wr+f(u), r∈R, u∈K.Here ¯f is a well defined height increasing

homomorphism, which is an extension of f.

As a corollary to the Theorem 3.1, we have the following theorem:

Theorem 3.2. Let N be a submodule of a QTAG-module M such that

(i) M/N

H_β(M/N) is countably generated;

(ii) N+Hβ(M)

H_β(M) is a summand of M/Hβ(M); (iii) N∩H_β(M) =H_β(N)and

(iv) H_β(M) =H_β(N)⊕T;

then M=N⊕K with K⊇T.

Proof. We may put

M H_β(M) =

N+H_β(M)

H_β(M) ⊕

Q H_β(M).

Thus, (N+H_α(M))∩Q=H_α(M)∩Qfor all α ≤β.Now supposeHα(M)∩N =Hα(N)for

α <γ ≤β. If γ is a limit ordinal, we have Hγ(M)∩N =Hγ(N), otherwise γ =α+1. For

x∈H_γ(M)∩N,there existy∈N, z∈Qsuch thatd

(y+z)R xR

=1 andy+z∈H_α(M).Then

z∈(N+H_α(M))∩Q=H_α(M)∩Qandy=y+z−z∈H_α(M)∩N=H_α(N).Ifd

yR y0R

=1,

thenx−y0∈N∩Q=N∩H_β(M) =H_β(N)andx= (x−y0) +y0∈H_γ(N).In order to show that

H_α(M/N) = Hα(M) +N

consider

M

N+H_α(M) =

N+H_α(M) +Q N+H_α(M)

∼

= Q

Q∩(N+H_α(M))=

Q

Q∩H_α(M)

∼

= Hα(M) +Q

Hα(M)

⊆ M Hα(M)

, for allα≤β.

Now we apply Theorem 3.1 to N⊕T T =

N+H_β(M)

T and get

H_α

(M/T) (N⊕T)/T

∼

=H_α

M N⊕T

=H_α

M

(N+H_β(M))

∼

=H_α

(M/N)

H_β(M/N)

= Hα(M/N)

H_β(M/N)

∼

=Hα(M) +N

H_β(M) +N ∼

= (Hα(M) +N)/T (H_β(M) +N)/T

=(Hα(M/T) +N⊕T)/T

(N⊕T)/T .

These isomorphisms give identity on M/T

(N⊕T)/T when combined together and we have

((H_α(M) +N)/T) ((H_β(M) +N)/T) =

(H_α(M/T) +N⊕T)/T

(N⊕T)/T , for allα <β.

Now,

H_α(M/T)∩((N⊕T)/T) = (H_α(M)/T)∩((N⊕T)/T)

= ((H_α(M)∩N)⊕T)/T = (H_α(N)⊕T)/T

Also,

(M/T) (N⊕T)/T

∼

= M

N+H_β(M)

∼

= (M/Hβ(M))

(N+H_β(M))/(H_β(M))

= M/N

H_β(M/N)

is countably generated module of length at mostβ.Again by Theorem 3.1 we have

M

T =

N⊕T

T ⊕ K T

and thereforeM=N⊕K.

In order to prove some important results, we need the following lemmas:

Lemma 3.2. For a QTAG-module M,a submodule N⊆M and a limit ordinalλ if

Soc(H_α(M/N)) = Soc(Hα(M)) +N

N , for everyα<λ, then

H_α(M/N) = Hα(M) +N

N , for allα<λ.

Proof. Assume thatβ <λ andHα(M/N) =

H_α(M) +N

N , for allα <β.Ifβ =α+1 andx+

N∈H_β(M/N),thend

(y+N)R

(x+N)R

=1 for somey∈H_α(M/N) =Hα(M) +N

N .Without loss of

generality, we may assume thaty∈H_α(M)andx+N=y0+N∈Hβ(M) +N

N ,whered

yR

y0R

=

1.Otherwise, ifβ is a limit ordinal and we assumeHk(H_β(M/N))⊆

H_β(M) +N

N andx+N∈

Hk+1(H_β(M/N)).Ifd

(x+N)R

(y+N)R

=k,theny+N∈Soc(H_β+k(M/N)) =Soc(Hβ+k(M)) +N

N

as β+k<λ. Therefore y+N=z+N such thatz0∈Hk+1(H_β(M)), d

z0R zR

=k and(x−

z0) +N ∈Hk(H_β(M/N))⊆ Hβ(M) +N

N . Therefore x−z

0_{+N =u+N with u∈H}

β(M) and

x+N= (z0+u) +N∈ Hβ(M) +N

N .

Lemma 3.3. Let N⊆K be submodules of a QTAG-module M such that K/N is h-divisible. If Soc Hα _M N

=Soc(Hα(M)) +N

N ,then

Soc H_α M K

= Soc(Hα(M)) +K

Proof. SinceK/N is h-divisible, it is a direct summand of M/N and M N =

K N ⊕

Q

N, for some Q⊇N. For x+K ∈ Soc(Hα(M/K)), x=y+z, y∈Q, z∈N and we may define a

homo-morphism f :M/K→Q/N such that f(x+K) =y+N.Thereforey+N ∈Soc(H_α(Q/N))⊆

Soc(H_α(M/N)) = (Soc(H_α(M)) +N)/Nand there exists au∈Soc(H_α(M))such thaty+N=

u+N.Therefore

x+K=y+K=u+K∈Soc(Hα(M)) +K

K .

which completes the proof.

Lemma 3.4. For k <ω,let K be a Hα+k(M)-high submodule of M.Then Hi(M)⊆Hi(K) +

H_α(M)for every i≤k+1.

Proof. Since Soc(M) = Soc(K)⊕Soc(H_α+k(M)) ⊆ Soc(K) +H_α(M), we suppose that

Hi(M)⊆Hi(K) +H_α(M)for everyi≤k.Considerx0∈Hi+1(M).Then there existsx∈Hi(M)

such that d

x0R xR

=i and x=y+z0 where y∈Soc(K) and d

zR z0R

=k with z∈H_α(M).

Now y=x−z0∈H_i(M)∩K=H_i(K),as K is h-pure in M. Therefored

y0R

yR

=i for some

y0∈K. Now ifd

z00R z0R

=i,then x−y0−z00 ∈Hi(M)⊆Hi(K) +H_α(M) which implies that

x∈Hi+1(K) +H_α(M).

The above discussion leads to the following:

Remark 3.2. For a submodule N of a QTAG-module M,ifSoc(H_α(M/N)) =Soc(Hα(M)) +N

N for allα ≤β,then Hα(M)∩N=Hα(N)for allα ≤β.Forβ ≤ω,the converse holds.

Lemma 3.5. If N is a direct summand of a H_α(M)-high submodule of M,thenSoc(H_γ(M/N)) =

Soc(H_γ(M)) +N

N ,for allγ ≤α.

Proof. Being a direct summand of ah-pure submodule of M, N is alsoh-pure in M. Remark

3.2 suggests that for all γ <ω, Soc(Hγ(M/N)) =

Soc(H_γ(M)) +N

N . Without loss of gener-ality, we may assume that α ≥ω. Let α =β +k, k <ω, β a limit ordinal and Q⊕N, a

H_α(M)-high submodule of M.Suppose the result holds for all ordinalsδ <α.If ω ≤δ <α

and if T is a H_δ(N)-high submodule of N, then T is a direct summand of H_δ(M)-high

Soc(H_δ(M/T)) = Soc(Hδ(M)) +T

T .As N

T ish-divisible by Theorem 3.2,

Soc(H_δ(M)) +N

N =

Soc(H_δ(M/N))and the result holds for allγ <α.

In order to prove that Soc(H_α(M/N)) = Soc(Hα(M)) +N

N , we have to show that

Hk+1(H_β(M/N)) = H k+1_(H

β(M)) +N

N . Consider x+N ∈H

k+1

H_β M N

. Since N is

h-pure, x∈Hk+1(M) and by Lemma 3.4,x=x₁+x₂+x₃ wherex₁∈Hk+1(N), x₂∈Hk+1(Q)

andx₃∈H_β(M).Asβ is a limit ordinal, for every ordinalγ<β,by Lemma 3.2,we may write

x+N=z+N,wherez∈H_γ(M).Thus for eachγ <β,we havex=x1+x2+x3=z+z1where z₁∈N.

Now,x₁−z₁+x₂∈H_γ(M)∩(N⊕Q) =H_γ(N)⊕H_γ(Q)andx₂∈ T γ<β

H_γ(Q) =H_β(M)implying

thatx+N= (x₂+x₃) +N∈ H k+1_(H

β(M)) +N

N .

We need some more lemmas to prove our main results:

Lemma 3.6. Let K=L

i∈I

K_ibe a submodule of a QTAG-module M such that each K_iis countably

generated. If N is a submodule of M,N∩K= L

i∈I0

K_i,I0⊂I and if T/N is a countably generated

submodule of M

N,then there is a submodule Q⊆M such that T ⊆Q, Q

N is countably generated

and Q∩K=L

i∈J

K_ifor some subset J of I.If M K and

N

N∩K are h-divisible, then M can be chosen

so that Q

Q∩K is also h-divisible.

Proof. Since T∩K

N∩K is countably generated, there is a countable subsetI

00 _{ofI such that} L

i∈I00

Ki

contains a complete set of representatives of T∩K

N∩K.SetQ=T+ (

L

i∈I00

K_i) =T+ (L

i∈J

K_i),where

J=I0∪I00.

Now, suppose M K and

N

N∩K are h-divisible. Consider a submodule L⊃N such that L N is

countably generated i.e., L

N =Σ(bi+N)R. Consider the elementsx1, x2, . . . , xn, . . . such that bi−yi∈Kwhered

xiR

y_iR

=1.

IfP=L+ΣxiR,then

P

N is countably generated andL⊆H1(P)+K.In this way we can choose two ascending sequences of submodulesT⊆Q1⊆Q2⊆. . .⊆Qk⊆. . .andP1⊆P2⊆. . .⊆Pk⊆

. . . ,where Mk

N and P_k

N are countably generated for everyk,Qk∩K=

L

i∈Ik

Q_k⊆H₁(P_k)+Kfor eachk.If we putQ=

∞

S

k=1 Q_k=

∞

S

k=1

P_kandJ=

∞

S

k=1

I_k,thenP_k⊆H₁(P_k+1)+K ensures that Q

K∩Q ish-divisible.

Lemma 3.7. Let{K_k}∞

k=1be a sequence of submodules of a QTAG-module M and Kk=

L

i∈Ik

K_k(i),

where each K_k(i) is countably generated. Let A be a set of positive integers such that M/K_k is

h-divisible, for every k∈A.If N is a submodule of M such that N∩K_k=L

i∈Ik

K_k(i)for each k and

N

N∩K_k is h-divisible, whenever k∈A and if T

N is a countably generated submodule of M N,then

there is a submodule Q of M such that T ⊆Q, Q

N is countably generated, Q∩Kk=

L

i∈Jk

K_k(i)and

Q

Q∩K_k is h-divisible, whenever k∈A.

Proof. Let n₁, n₂, . . . be a sequence of positive integers such that for every pair of positive integersk andn there is an integer j>k such thatn_j =n. By Lemma 3.6, we may construct

a chain of submodules ofM containingT,Q₁⊆Q₂⊆. . .⊆Q_k⊆. . .such that for eachk, Qk

N

is countably generated. Q_k∩K_nk = L

i∈Ik nk

K_ni

k and

Q_k

Q_k∩K_nk ish-divisible, whenevernk ∈A.Set

Q= S∞

k=1

Q_k and for eachn,letJnbe the union of allInkk for whichnk=n.Since for eachn,Qis

the union of thoseQ_k such thatn_k=n, Qis the required submodule.

Lemma 3.8. Let N be a submodule of M such thatSoc

H_α

M

N

=Soc(Hα(M)) +N N for all

α <λ.IfSoc(M) =Soc(N)⊕T,where T is a summable subsocle such that T∩H_λ(M) =0,

then M

N is a summable module of length atmostλ.

Proof. LetT = L α<λ

T_α,whereS_α ⊆H_α(M), S_α∩H_α+1(M) =0,for everyα.Then

Soc(M/N) = Soc(M) +N

N =

T⊕N N =

M

α<λ

T_α⊕N N

We have to show that forx+N(6=0)∈Tα+N

N ,HM/N(x+N) =α.Since Soc(M) =Soc(N)⊕ T, Soc(H_α(M)) = (Soc(N)∩H_α(M))⊕(T∩H_α(M)) = (Soc(N)∩H_α(M))⊕(L

β≥α

S_β).

There-fore

Soc(Hα(M/N)) = (Soc(Hα(M)) +N)/N

= ((T∩H_α(M)) +N)/N= M

β≥α

((T_β ⊕N)/N)

= ((T_α⊕N)/N)⊕( M

β≥α+1

((T_β⊕N)/N))

= ((T_α⊕N)/N)⊕Soc(H_α+1(M/N)).

and we are done.

Now we are able to prove one of our main results.

Theorem 3.3. Let M be a QTAG-module of lengthσ,whereσis a countable limit ordinal. Then

M is a direct sum of countably generated modules if M is summable and for each ρ <σ, M

contains a H_ρ(M)-high submodule which is a direct sum of countably generated modules.

Proof. For an ordinalρ such thatω ≤ρ<σ,letKρ be aHρ(M)-high submodule ofMwhich

is a direct sum of countably generated submodules. PutK_ρ = L

i∈Iρ

K_ρi such that eachK_ρi is

count-ably generated and Soc(M) =L

i∈I

Ti,where each Ti is countably generated. Now the elements

of the minimal generating set of M, {x_α} can be well ordered α <δ. Since σ is countable,

Kρ’s can be enumerated and by Lemma 3.7, we may define submodules {Nα} satisfying the

following conditions:

(i) N0=0,Nα ⊆Nβ for everyα<β andNα = S

β<α

N_β,ifα is a limit ordinal;

(ii) x_α ∈N_α+1;

(iii) Nα∩Kρ = L

i∈Iα ρ

K_ρi,for eachα;

(iv) N_α+1/Nα is countably generated; (v) N_α/(N_α∩K_ρ)ish-divisible, for eachα;

(vi) Soc(N_α) = L

i∈Iα

By Lemma 3.5,

Soc(H_ρ(M/N_α)) = Soc(Hρ(M)) +Nα

N_α , for allρ<σ.

Therefore by Remark 3.2, eachN_α is isotype and again by Lemma 3.2,

H_ρ(M/N_α) = Hρ(M) +Nα

N_α , for allρ<σ.

Now we infer that

H_ρ(N_α+1/Nα) =

H_ρ(N_α+1) +N_α

N_α , for allρ<σ.

By Lemma 3.8, Nα+1

N_α has length at mostσ and

H_ρ(N_α+1)∩Nα = (Hρ(M)∩Nα+1)∩Nα =Hρ(M)∩Nα =Hρ(Nα), for allρ≤σ.

Now using Theorem 3.1, we get N_α+1 = Nα +Qα, for each α < δ. Consequently, M = L

α

Q_α, α<δ is a direct sum of countably generated modules.

To prove some other interesting results, we need some more lemmas.

Lemma 3.9. If ρ is a countable ordinal and x∈Hρ(M),then there is a countably generated

submodule N of M such that x∈H_ρ(N).

Proof. The statement holds ifρ =0.We shall prove the result inductively by assuming that it

holds for all ordinals σ <ρ. Ifρ =σ+1, then there exists an element y∈Hρ(M) such that

d

yR xR

=1.By assumption, there is a countably generated submoduleNsuch thaty∈H_σ(N).

Nowx∈H_σ+1(N) =H_ρ(N)asd

yR

xR

=1.Otherwise, ifρ is a limit ordinal, thenx∈Hσ(M)

for all σ <ρ. Inductively, there exists a countable submodule Nσ for each σ <ρ, such that

x∈H_σ(N_σ).IfN=ΣNσ,thenN is countably generated andx∈ T

σ<ρ

H_σ(N_ρ)⊆ T σ<ρ

H_σ(N) =

H_ρ(N).

Lemma 3.10. Letσ be a countable ordinal and N a submodule of a QTAG-module M such that

H_σ(M)∩N=H_σ(N).If T/N is a countably generated submodule of M/N,then there exists a

Proof. As

T∩H_σ(M)

H_σ(N)

∼

= (T∩Hσ(N)) +N

N ⊆

T N,

we may select elements x1, x2, . . . ,xk, . . . such that

T∩Hσ(M)

H_σ(N) =ΣxiR. By Lemma 3.9, for

eachk,there is a countably generated submoduleN_k such thatx_k∈H_σ(N_k).PutK₁=T+ΣNk.

Now K1

N is countably generated andT∩Hσ(M)⊆Hσ(K1).On repeating this process, we may construct a chain of submodulesK₁⊆K₂⊆. . . such that Kk

N is countably generated andKk∩

H_σ(M)⊆H_σ(K_k+1).NowK= S∞

k=1

K_kis the required module.

Lemma 3.11. Let σ be a countable ordinal and T a submodule of M such that T = L

i∈I

T_i, M

T =

L

j∈J

Q_j and each T_i and Q_j is countably generated. Suppose N is a submodule of

M such that N∩T = L

i∈I0

T_i, N+T

T =

L

j∈J0

Q_j and H_σ(M)∩N=H_σ(N).If Q

N is a countably

gen-erated submodule of M

N,then there exists a submodule P of M such that Q⊆P, P

N is countably

generated, P∩T = L

i∈I00

Ti,

P+T T =

L

j∈J00

Qjand Hσ(M)∩P=Hσ(P).

Proof. Since Q/N is a countably generated module, so Q+T

N+T is also countably generated. Therefore there exists a subset ¯J ⊂J such that L

j∈J¯

Q_j contains a set of representatives of

Q+T

T

/

N+T

T

.IfLis generated by a complete set of representatives of L

j∈_J¯

Qj,thenLis

a countably generated submodule ofM such that L+T T =

L

j∈_J¯

Qj.IfP1=Q+LandJ1=J0∪J¯, thenP1

N is countably generated and P₁+T

T =

L

j∈J1

Q_j.Now by Lemma 3.10 and Lemma 3.6,there

exist ascending chains of submodules P1⊆P2⊆. . .⊆Pk⊆. . .;L1⊆L2⊆. . .⊆Lk⊆. . .and

K₁⊆K₂⊆. . .⊆K_k⊆. . .such thatN⊆P_k⊆L_k⊆K_k⊆P_k+1, P_k/Nis countably generated for

eachk, Pk+T

T =

L

j∈Jk

Qj,Lk∩T =

L

i∈Ik

TiandHσ(M)∩Kk=Hσ(Kk).NowP=

∞

S

k=1

P_k, I00= S∞

k=1 I_k

andJ00=

∞

S

k=1

J_k and we are done.

Theorem 3.4. If σ is a countable ordinal and if M is a h-reduced QTAG-module such that

M/H_σ(M)and H_σ(M)are direct sums of countably generated modules, then M is a direct sum

of countably generated modules.

Proof. Let M Hσ(M)

=L

j∈J

Q_j andH_σ(M) =L

i∈I

P_i, where each Q_j and P_i is countably generated.

We may define a well ordering relation on the set of generators ofM, {x_α}, α <β.By Lemma

3.11, we may construct a well ordered family of the submodules {Nα} of M satisfying the

following conditions:

(i) N₀=0,N_α ⊆N_β ifα <β andNα = S

β<α

N_β,ifα is a limit ordinal;

(ii) x_α ∈N_α+1;

(iii) N_α∩H_σ(M) = L

i∈Iα

T_i;

(iv) Nα+Hσ(M)

H_σ(M) = L

j∈Jα

Qj;

(v) H_σ(M)∩N_α =H_σ(N_α);

(vi) N_α+1/Nα is countably generated.

NowH_σ(N_α+1)∩N_α =H_σ(N_α)andH_σ(N_α) = L

i∈Iα

T_i,is a direct summand ofH_σ(N_α+1) = L

i∈Iα+1

Ti. If we define the canonical map f :

N_α+1+H_σ(M)

H_σ(M) →

N_α+1 H_σ(N_α+1)

, then f is an

iso-morphism and f

N_α+H_σ(M)

H_σ(M)

= Nα+Hσ(Nα+1)

H_σ(N_α+1)

.As Nα+Hσ(M)

H_σ(M) is a direct summand of

N_α+1+H_σ(M)

H_σ(M) ,

N_α+H_σ(N_α+1)

H_σ(N_α+1)

is a direct summand of Nα+1 H_σ(N_α+1)

and all the conditions of

Theorem 3.2 are satisfied, therefore for eachα<β we haveNα+1=Nα⊕Nα0.ThusM= L

α<β

N_α0

is a direct sum of countably generated modules.

Since h-pure, isotype and high submodules are very significant in the study of

QTAG-modules. We prove some results related to these subQTAG-modules.

Proposition 3.1. For a countable ordinalσ and a QTAG-module M, if M/Hσ(M)is a direct

sum of countably generated modules, then every H_σ(M)-high submodule of M is a direct sum

of countably generated submodules.

Proof. Suppose the statement holds for all the ordinal less than σ and σ ≥ω. Now we put

by Lemma 3.5 and Lemma 3.2,

M/N=H_α(M/N) =Hα(M) +N

N for allα <ρ.

Therefore,

N

H_α(N)

∼

= Hα(M) +N

H_α(M) =

M

H_α(M)

∼

= (M/Hα(M))

H_α(M/H_σ(M))

is a direct sum of countably generated submodules for all α < ρ. Now, by induction,

ev-ery H_α(M)-high submodule of N/H_ρ(N) is a direct sum of countably generated

submod-ules. Since N+Hρ(M) H_ρ(M)

∼

= N

H_ρ(N) and

N+H_ρ(M)

H_ρ(M) is isotype in the summable module

M/H_ρ(M), N/H_ρ(N)is also summable.

Now by Theorem 3.3, N/H_ρ(N)is a direct sum of countably generated modules, therefore

by Theorem 3.4, Nis a direct sum of countably generated modules.

Theorem 3.5. Let N be an isotype submodule of a QTAG-module M of countable length and M,a direct sum of countably generated h-reduced QTAG-modules. Then N is also a direct sum

of countably generated modules.

Proof. Letσbe the length ofN.AsN+Hσ(M)

H_σ(M) is isotype in

M

H_σ(M),without loss of generality,

we assume that the length ofM is also σ.Suppose the statement holds for all lengthsρ <σ.

Now N+Hρ(M)

H_ρ(M) is isotype in

M

H_ρ(M) for allρ <σ.Inductively,

N H_ρ(N)

∼

= N+Hρ(M)

H_ρ(M) is a

direct sum of countably generated modules for allρ<α.Ifσ=ρ+1,then N

H_ρ(N) andHρ(N)

are the direct sums of countably generated modules and the result follows from Theorem 3.4.

Otherwise, ifσ is a limit ordinal, N

Hρ(N)

is a direct sum of countably generated submodules

for allρ<σandNis summable as it is isotype in the summable moduleMof countable length.

Therefore,Nis a direct sum of countably generated modules.

4.

In this section we establish some interesting results using the results of section 3. We start

Lemma 4.1. Let N, K be H_σ(M)-high submodules of M where σ = ρ+n, n<ω. If x+

H_ρ(N)→y+H_ρ(K) if and only if x−y∈H_ρ(M), then this mapping is a height preserving

isomorphism fromSoc(N/H_ρ(N))toSoc(K/H_ρ(K)).

Proof. Let x ∈N and x0 ∈H_ρ(N) such that d

xR x0R

=1. Suppose x∈/ K, then (K+xR)∩

Hσ(M)6=0 and therefore there existsy0∈K and z∈Hρ(M) such that y0+x00 =z06=0.Here

d

xR x00R

= i, d

zR z0R

=n. Since K∩H_σ(M) = 0, i ≤n and there is a y00 ∈K such that

d

_y00_R

y0R

=i. By Lemma 3.4,y00+x+z00 ∈Hi(M)⊆Hi(K) +Hρ(M), where d

_z00_R

z0R

=i.

Now x= (u−y00) +z00+vfor some u∈Hi(K)andv∈H_ρ(M).Puty=u−y00. Now the map

x+H_ρ(N)→y+H_ρ(K)is a height preserving isomorphism.

Theorem 4.1. If a H_σ(M)-high submodule of a QTAG-module M is a direct sum of count-ably generated modules, then every Hσ(M)-high submodule of M is a direct sum of countably

generated submodules.

Proof. We may assume that σ is countable and σ ≥ω. Now σ =ρ+n, where ρ is a limit

ordinal and n<ω. Let N, K be two Hσ(M)-high submodules of M and N a direct sum of

countably generated submodules. NowN/H_ρ(N) is also a direct sum of countably generated

modules and by Lemma 4.1, K/H_ρ(K) is summable. As ρ is a limit ordinal, by Proposition

3.1,we haveN/H_α(N)∼=M/H_α(M)=∼K/H_α(K)for allα<ρ.ThereforeK/Hρ(K)is a direct

sum of countably generated modules. NowH_ρ(K)is bounded, therefore by Theorem 3.4, Kis

a direct sum of countably generated modules.

Proposition 4.1. For a QTAG-module M,if M/H_ρ(M)is summable for allρ≤σ,whereσ is

countable then M/H_σ(M)is a direct sum of countably generated modules.

Proof. We assume that the result holds for all ordinals less than σ. Ifσ =α+1,then by

in-duction we may say that M/H_α(M) is a direct sum of countably generated modules. Since

(M/H_σ(M))

H_α(M/H_σ(M))

∼

= M

H_α(M) and Hα(M/Hα(M))are direct sums of countably generated

mod-ules, the result follow from Theorem 3.4.Otherwise, ifσ is a limit ordinal, then by assumption

M/Hσ(M)

H_ρ(M/H_σ(M))

∼

= M

is a direct sum of countably generated modules for each ρ <σ. Since M

H_σ(M) is summable,

M

H_σ(M) is a direct sum of countably generated modules.

An immediate consequence of this proposition may be stated as follows:

Corollary 4.1. Let σ be a countable limit ordinal and let M be a QTAG-module of lengthσ.

Then the following conditions are equivalent:

(i) M is a direct sum of countably generated modules;

(ii) M is summable and M Hρ(M)

is a direct sum of countably generated modules for all

ρ <σ;

(iii) M

Hρ(M)

is summable for allρ≤σ;

(iv) M is summable and for eachρ <σ,the Hρ(M)-high submodules of M are direct sums

of countably generated modules.

U lmfactors play a very important role in the study of QTAG-modules. In order to establish

our next result involvingU lmfactors we need the following lemma:

Lemma 4.2. Let σ be a countable ordinal. Then for a QTAG-module M, M/Hρ(M) is a

direct sum of countably generated modules for allρ <ω σ if and only if M

Mα is a direct sum of

countably generated modules for allα <σ.Here Mα is theαth- Ulm submodule of M.

Proof. Ifρ<ω σ,thenρ=ω α+n, n<ω, α <σ.Now we have

M

Mα =

M

H_{ω α}(M)

∼

= (M/Hρ(M))

H_{ω α}(M)/H_ρ(M) =

M/H_ρ(M)

H_{ω α}(M/H_ρ(M)).

Therefore,H_{ω α}

M H_ρ(M)

is bounded and if M

Mα is a direct sum of countably generated

mod-ules, then by Theorem 3.4, M

H_ρ(M) is a direct sum of countably generated modules.

Theorem 4.2. Let M be a QTAG-module of countable length. Then M is a direct sum of countably generated modules if and only if

(i) all Ulm factors of M are direct sums of uniserial modules and

Proof. The second condition implies that M is summable. Now, M will be a direct sum of

countably generated modules ifM/H_τ(M)is a direct sum of countably generated modules for

all τ less than σ, the length ofM. Assume that ρ <σ and M

Mτ is a direct sum of countably

generated modules for allτ<ρ.In the light of Lemma 4.2,we have to show that M

Mρ is a direct

sum of countably generated modules for allρ <σ.Ifρ=α+1,then

(M/Mρ₎

H_{ω α}(M/Mρ₎ =

(M/Mρ₎

Mα_/Mρ

∼

= M

Mα

is a direct sum of countably generated modules by induction. Now

H_{ω α}

M

Mρ

= M

α

Mρ =

Mα

Mα+1

is a direct sum of uniserial modules by(i).Therefore by applying Theorem 3.4, M

Mρ is a direct

sum of countably generated modules. Otherwise, ifρ is a limit ordinal, then M

Mρ =

M H_{ω ρ(M)} is

summable by(ii).Inductively, M

Mα is a direct sum of countably generated modules for allα<

ρ. By Lemma 4.2,M/Hτ(M)is a direct sum of countably generated modules for allτ <ω ρ.

Thus M H_{ω ρ}(M) =

M

Mρ is a direct sum of countably generated modules.

As an immediate consequence of this result, we can say that a summableQTAG-moduleM

of lengthω is a direct sum of countably generated modules if and only if all of itsU lmfactors

are direct sums of uniserial modules.

Conflict of Interests

The author(s) declare that there is no conflict of interests.

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