Tag modules with complement submodules H pure

(1)

TAG-MODULES WITH

COMPLEMENT SUBMODULES

H-PURE

SURJEET SINGH

Department

ofMathematics

Kuwaituniversity

PO Box5969

Safat 13060 Kuwait

"MOHD

Z.KHAN

Deparent

of Mathematics

AligarhMuslim university

Aligarh,

(U.P.)

202001

India

(Received February 5, 1996 and in revised form August Ii, 1997)

ABSTRACT

The concept ofa QTAG-module

MR

was given by Singh[8]. The structure theory of such

modules hasbeendevelopedon similarlinesas that oftorsionabeliangroups.If amodule

Ml

is such that

MM

is aQTAG-module,it is calledastrongly TAG-module.This inturnleadstothe concept of a

primary TAG-moduleand itsperiodicity. Inthe presentpapersomedecomposition theorems for those primary TAG-modules in which all h-neat submodules are h-pure are proved. Unlike torsion abelian

groups, there existprimary TAG-modulesofinfiniteperiodicities. Such modulesarestudied inthe last section. The resultsprovedin thispaperindicate thatthestructuretheory ofprimaryTAG-modules of

infiniteperiodicityisnotverysimilartothat of torsion abeliangroups.

KEY

WORDS AND PHKASES: QTAG-modules, complement submodules, h-pure submodules, h-neat

submodules,and basic submodules.

1991AMSSUBJECT CLASSIFICATION CODES: 16D70;20K10

INTRODUCTION

Amodule

Ms

satisfying thefollowingtwoconditionsiscalledaTAG-module

[2].

(I) Every

finitelygeneratedsubmodule of any homomorphie image of

M

is adirectsum of uniserial modules.

(II)GivenanytwouniserialsubmodulesUandVofahomomorphie image of

M,

for any submoduleWof

U,

anyhomomorphism f:W Vcan beextendedto a

homomorphismg: U Vprovided the compositionlength

d(U/W)

<

d(V/).

Ifamodulesatisfiescondition

CI),

it iscalled aQTAG-module

[8].

The mainpurposeofthispaper

is toprovesome decompositiontheorems foramodule

M,

such that

MM

isaQTAG-moduleand that

(2)

802 S. SINGH AND M. Z. KHAN

module,whichisnotdecomposable,isgivenatthe endof thepaper.

However,

itfollowsfron the results

in thispaperthatanytorsionreduced module overabounded (hnp)-ring,witheverycomplement submodulepure,isdecomposable.Themainresultsaregivenin Theorems

(5.5),

(5.12)

and

(5.14).

In

section3,anecessaryandsufficient conditionforaQTAG-moduletoadmitonlyone basicsubmomdule

isgiven.Insection 4the concept ofneatheight ofa uniformelementin aQTAG-moduleis discussed.

The concept ofneatheightisusedtogive,inTheorems

(4.6)

and

(4.7),

some criterians fora

QTAGmodule, such that every h-neat moduleis/-embedded in the senseof

Moore[5].

Theresultsin

sections3and 4 canbeof independentinterest.

2PRELIMINARmS

Amodulein which the latticeofitssubmodulesislinearly orderedunderinclusion is calledaserial

module;in addition if ithas finitecomposition length,it iscalledaunlserial module.Let

Ms

bea

QTAG-module.

An

x

M

iscalledauniformelement,if

xK

isanon-zerouniform

(hence

uniserial) submodule

ofM.Forany module

AR

with acomposition series,

d(A)

denotes itscomposition length. Letx Mbe

uniform.Then

e(x)

d(xK)

is calledthe exponent ofx.The equation

Ix,

y] n,willgive thatyis a uniformelementof

M,

such thatx e

yg

andd(ylVxR) n.Forbasic

definitions

of height ofanelement of

M,

thesubmodule

I-I(M)

fork_>_{0, one}_mayreferto

[6]

or

[8].

Fo"any submoduleNof

M,

and any y

N,hN(y) will denotetheheightofyin

N;

howeverwewriteh(y)forhM(y). AsubmoduleNof

M

issaid tobeh-pure in

M,

ifilk(M) n N

H(N)

foreveryk_> 0.For anymodule

K, sue(K)

denotes the socle of K.

MR

is saidtobedecomposable,if it is a direct sumof uniserialmodules.

Byusing

[8,

Lemma(2.3)],one canprovethe following:

Proposition(2.1).AsubmoduleNofaQTAG-moduleMisb-pureinMifandonlyiffor any

uniform x

sue(N), hr,(x)

h(x).

Thefollowingisoffrequentusein thispaper.

Proposition(2.2)

[8,

Lemma(3.9)].

LetNbe anyh-puresubmodule ofaQTAG-module

M.

Then for any uniformx

M,

there exists a uniform

x"

M,

such that tbr x x+

M

M/N,

e(x)

e(x’),

x

x’

andMcx’P,=0.

By

using the aboveproposition,weget that ifM/Nis

deeomlasable

for

some

h-puresubmodule

N,

thenM T(9

N,

for somedecomposablesubmodule

T

ofM. Let

KR

beany module.Forthe

definitionsof K-injective modules and K-projective modules onemayreferto

].

Lemmas

(2.2)

and

(2.4)

in

[8]

give thefollowing:

Proposition(2.3).LetAandBbetwouniserialsubmodules ofaQTAG-modules

M,

such that

AB=0.

(i) If

d(A)

<

d(B),

thenBisA-injective. (ii) If

d(A)

>

d(B),

thenBisA-projective.

(iii)If

d(A)

d(B),

thenA_=

B

_if_and_only_{if either}

sue(A)

sue(B),

_or

Ah-It(A)

mI-IlfS).

(3)

element x

soc(M)

has finite height, then foranyuniformy

M,

wkh[x,y]

h(x),

yRbeinganh-pure submodule of

M,

isasummandof M.Forgeneral properties of ringsandmodulesonemay referto

[3].

3

BASICSUBMODULES

Throughout

MR

is aQTAG-module.AsubmoduleBofMiscalledabasic submodule of

M,

ifB is adecomposable h-pure submodule of

M,

such thatM/B ish-divisitle

[7].

As pointedoutin[8,

Remark(3.12)],

Mhasabasicsubmodule and anytwobasicsubmodulesofMareisomorphic.

Lemma(3.1).

LetA,A2,.

A

be any finitely manyuniserialsummands ofM,such thatd(Ai)

<d(Ai+l)andN

A,

A,.

ThenNis anh-pure submodule of M. t=l

Proof. Considerauniformelementx

sot(N).

Then x

Ex,,

xi

&.

If for any < j,xi 0 xj then bythehypothesis h(xi <h(xj).Thus

h(x)

h(xi) xi

0}.

Aseach

&

ish-pure,h(xi) hA,

(x,)=

hN(Xi).Thisgives

h(x)=

hN(x). Hence Nis-pure.

Lemma(3.2).

LetMbe such that

cI-(M)

0andlet

M

haveabasic submodule

B

M.Then for

somesimple submoduleSofsot(M),there existsanh-pure submoduleN

Ey,R

suchthatevery

I----!

yiR isuniserial, d(yiR)< d(yi/R)andS soc(yiR).Theheights of the

(non-zero)

elements of the

homogeneous components of

sot(M),

determinedby

S,

donothaveanupperbound.

Proof.Let M/B.Considerauniform in

soc(

By (2.2)

there existsauniform

z

sot(M)

suchthat

.

,.

AS

c

I-I(M)

0,

h(z0

is finite.Let

h(zt)

n.

Then there exists y

M,

suchthat[zl,yl

n

.ThenytRisanh-puresubmodule ofMandg

yR

0.However

h()

**.So

there existsauniformut

M

with

soc(u’-)

-Rand

e(a,)

>

n

By (2.2)

we get uniform

z

sot(M)

with

z-=

.,

h(z=)

nz

>

n.

_{We get yz} Msuch that[z., yz]

nz. By

continuingthis

process,weget an infinitesequenceofuniformelements{yi}i of

M,

such that each yiR isanh-pure

uniserialsubmodule,soc(yiR)

zaR

forsomezi Msatisfying

,,

[z,yi] ni

h(za)

andni< IfK

=Ey,Ris

notadirectsum,we get asmallest >__{2, such that}

z

zR.

_Then_N

k"!

Yk

R

y

R. By

(3. I)

Nisanh-puresubmoduleofM. Ft.* anyuniform v e

N,

ifv Evj,

k--I k--I

with

v

yjR,then h(v)

min{h(v)}

Thisgives

h(za)

<

max{h(z)

<k<i-1

}.

This is a

contradiction,as h(zj)<

h(za)

for <i.Hence

K

BEyiR.

By

using

(3 1)

weget that

K

isanh-pure

submodule.Clearly

sot(K)

ishomogeneous.Thelast partisobvious.

(4)

804 S. SINGH AND M. Z. KHAN

Proof.

By (2.3)0)

weget monomorphismsi"yiR-->yi/,R.Write:;i(Yi) wi.Thenwi is uniform ande(wi) e(yl).Consider

B

w,R,

andM M/B Let B.Then z

(y,-6,

(yi))r, =yR +

(y,r

6,.,(y,_,

)r,_,

6,(y,)r,,forsomeria

K

andapositiveintegers.

Hereyiri-Oiq(yi.)riq yiR and-o,(y,r,)

y,R..

Using this,itcanbeeasilyprovedthat

B

wiRand yRr 0.Now 6,

(y,),

ande(o(yt))

e(6 (y,)).

Sothatch(y,)R rn B 0.As o,(y)Rgy2R,

we gety:R B 0.Bycontinuingthisprocess,we getyiR B 0.Clearly

,

R <

x

R

< gives

Mis a serialmoduleofinfinitelength. It onlyremainstoprove that ish-pure. Inview of

(3.1)

it is

enoughtoprovethat each

wR

ish-pure. Now

yR

@ yi/Rbeinga:.ummandof

M,

ish-pure. But

y

y,.R

wR

y/R.So wiRish-pureinM.Thiscompletesthe

proo."

Theorem(3.4).

AQTAG-module

Ms

hasnobasicsubmodule other than

M

ifandonlyifM is

h-reducedandforeachhomogeneouscomponent

K

of

soc(M),

thereotistsanupperbound ontheheights

ofmembers ofK

Proof.Let

M

beitsonlybasic submodule. Thenbydefinitio,Misdecomposableand h-reduced.

Forasimple submoduleSof

M,

wegetasummand

Ms

of

M,

suchehat

soc(Ms)

isthe homogeneous

componentof

soc(M)

determinedbyS.If heights of members ofsot;(Ms)donothaveanupperbound,

wegetasummandN

y,R

of

Ms

such that each yiR is uniserial and d(yiR)<d(yi/tR).

By (3.3)

N

hasabasicsubmodule

Bt

#N.As Nisasummandof

M,

wegetabasicsubmodule

B

of

M

ofwhich

B

is

asummandandB,M.Thisis acontradiction.Conversely let the given conditions hold. Then

I-I(M)

0.Therestfollows from

(3.3).

4.H-NEAT HEIGHT

Throughout

Ms

isaQTAG-module.AsubmoduleNofMiscalledanh-neatsubmodule of

M

if

H(M)

N

Ht(N).

ASobservedin

[8],

anysubmoduleNof

M

ish:teatifand only ifitisa complement submoduleof

M,

anymaximalessential extension

K’

ofasubmodule

K

of

M,

isan

h:neat

submodule of M.AnysuchK’iscalledanh-neat hullofK. Foranyuniform x

M

theminimumofall

d(K’

xR),

where

K’

runsoverallh-neat hulls of

xlL

iscalled theh-neat height,,fx"itisdenotedby

h’(x).

If x N

M,

then

h

(x)

willdenote theneatheightofx inN.IfN isanh-neatsubmodule of

M,

thenanyh-neat submodule ofNish-neatin

M,

sothat foranyuniform x

N, h’(x)

_<

h

(x).

Weput

h’(0)

oo.Inan

h-divisibleQTAG-module

M,

everyuniformelement is of infinite h-neat height.

For anytwomodules

A

and

Bs

anyhomomorphismfromasubmoduleof

A

into

B

iscalleda

subhomomorphism fromAto

B;

thesetof all subhomomorphisms fromAto

B

isdenotedby

SH(A, B).

Anf SH(A,

B)

issaid tobe maximal, ifithas noextension in

SH(A, B

.

Now

(2.3)

gives the following:

Lemma(4.1).

Let xRandyRbe anytwouniserial submodules of

M,

such thatxR

yR

0.Then

(a)

For anymaximalf

SH(xR,

yR),eitherdomain(f) xRorrange(f) yR.

(5)

hold:

(i). zR_=x’R.

(ii)Given anyu

v+w,

v

xR,

w

yR

such that z

uR,

(z)

if

y’

0, then

Ix’,

v] [y;

w]

([3)

if

y’

0, then

e(w)

<_

Ix’,

v]

Lemma

(4.2).

Let

xR

and

yR

betwounisefialsubmodulesof

M

such that

xRc

yR

0.Letz

x’+ y’, x’

xR, y’

yR,

be uniform such thatd(y’R)_<

d(x’R). For T

xR B

yR,

thefollowing hold:

(i). For y’ ; 0,

hr(Z)

isthe minimumof

Ix’,

x]

and[y’,y].

(ii). For

y’

0, letf

SH(xR,

yR)be maximalwith s

d(ker 0),

minimalunder the

conditionthat

x’R ;

kerf. Ifdomain(0

uR,

then

hr

(z)

[x’, u]

minimum of

Ix’,

x]ande(y) +s-

e(x’).

Proof.g" x’R -->

y’R.

such thatg(x’r)

y’r

is anR-epimorphism. Ifw a+b,a e b

yR,

isuniformand za

wR,

then f"aR -->bR such that

f(ar)

br,isan .*tension ofg;further[z, w] [x’,

a]. Any

extension h"a’R -->

yR,

a’

xR,

ofggivesuniform w" a" +

h(a’)

such thatz w’R. Consequently wRis anh-neat hull ofzRif andonly if f ismaximal.Inthatcaseby

(4.1)

eitherdomain(f)

xRorrange(f)

yR.

Thus for

domain(f)

aR,

anduR kerf,

e(a)

stheminimumof

e(x)

and

e(y)+e(u). Tominimize

e(a),

weneedtominimises

e(u).

Sothat

r.minimal

e(u),

h

(z)

Ix’,

a]

e(a) e(x’)

min{e(x), e(y)+e(u)}

e(x’)

min{[x’, x], e(y)+e(u)

x’) },

as

e(x) e(x’)

Ix’,

x] If

y’

; 0, then

e(x’)

e(u)+e(y’),sothate(y)+e(u)

e(x’)

e(y) ety’) [y; y]. For

y’

0,it is

obviousthat

x’R

g7kerf. Thisprovesthe result.

Lemma(4.3).

Let M A

B

Bandxa

M

be uniform.Ifx a+b,a

A,

baB and

d(aR)

>

d(bR),then the following hold:

(i). Forb 0,

h’(x)

min{h, (a),

h

(b)}.

(ii).Ifb 0, and

B

ish-divisible,then

h’(x)

h, (a)

ProofNowg"aR -->bR givenby g(ar) br,is anepimorplsm. Let,t,and2betheprojections

A(9_{B -->}

A,

andA B -->

B

respectively. Consideranh-neat hull

K

ofxR.ThenKisserial.LetK,

ri(K). As

d(bR)

<_

d(aR),

wegetanepimorphismts"

K

-->

K2

such that for any

x

K, c(x)

x2if and

onlyifx+x2 K.FurtheraR

=

K,

bR

K2

and

d(K/xR)

=d(KdaR)’

By

using

(2.3)

weget thateither

Kt

ish-neator

K2

ish-neatinM.

Case I"b 0.Theneither

K

isanh-neat hullofaRor

K2

is anh-neat hull of hR.Sothat

h’(x)

_>

min{hr(a),h(b)}. Let

min{h, (a),

h(b)}

<

h’(x).

Tobe definite let

h

(a).

Thenwe

getanh-neat hull

atR

ofaR with

[a, a]

t,andauniform

b

in

M

with[b,

b]

_>t.

By (2.3)

gextendsto a

homomorphismf:air -->

blR.

Then

(a+f(at))R

isanh-neathullof;tRwith

Ix,

at+f(a0]

<

h’(x)

Thisis a

contradiction.Similarargumentsholdif

h

(b)

Thisproves (i).

Case IIb 0andBish-divisible.

Any

h-neatserialsubmodule of

B

is eitherzero orof infinite

(6)

806 S. SINGH AND M. Z. KHAN

Thus forx a,

h’(x)

h_A

(a)

Lemma(4.4).

Let

KR

$x,R

beaQTAG-modulewitheachxiRuniserial.

Let

z

Z

zq,

z

xiR,beuniform.Letz,besuchthat

e(z)

e(z,).

Then

h’(z)

istheminimumof the following numbers"

(i).All [zi,xi].with

z

#0.

(ii).The neatheights of

z

in variousx,,R@xjR,withzj 0.

Proof. Thehypothesison z, gives that for any i,oi"

zR

-->

zR

such that

oi(zr)

zr

is an

epimorphims.Let y Zyi,ye

xR,

be any uniform element in

K

such that za

yR.

Thenl

yuR

yR

givenbyrli(yur) yr isanextensionof

o.

Clearlyif a

z

#0, then [z,, yu] [z,yi]. Sothate(y)is notmorethans,theminimumof all those

[z,

xi]for which

z

#0. Thush’(z) <s.

However,

ifevery

z

#

0,thenby

(2.3),

it isimmediatethatfor

yR

tobe an h-neat hullofzR,itisnecessarythat [z, y] s,i.e

h’(z) s.Supposethat forsomej,

z

0andthatforT

x

(B_xjR,

hr(zu)

<s.Wehaveamaximal f SH(xjL,xjR)withkerfofsmallestlength amongthose containing

z.

Letwd{ domain(f),then

s’

hr(Zu)

[z,,

wu]. By

using

(2.3),

weobtain auniformy

Zi

y,withze

yR,

y,--’wuandyj

f(w,).

ThenyRis anh-neathullofzRsuch that[z, y] s’.Thus

h’(z)

< so,theminimum of the numberslisted in(i)and(ii).

Suppose h’(z)

<_{so. We}getauniformw Zwi,wi x,P, such thatwRisanh-neat hull of

zRand[z,w] h’(z).Then for some j,wjR xjR.Forthisj,

z

0 .rod

(w,+w)R

is anh-neathull of

z,R. Consequentlyfor

T

x

@

x,R,

hr

(zo)

<

h’(z).

This isacontradiction.Thiscompletestheproof Wenowgiveacriterianintermsof h-neat heights, foraQTAG-module,in whicheveryh-neat

submoduleish-pure. We shall giveamoregeneralresult. Analogoustothe definitionofan/-embedded

subgroup of

an.abelian

p-groupgivenby Moore

[5],

wedefinean/-embeddedsubmodule ofa

QTAG-module.Let

Z*

bethesetof all non-negativeintegersand1"

Z

-->

Z’

beanyfunction such thatn _<

l(n),

n

Z

+"

AsubmoduleNofaQTAG-moduleMis saidtobel-embeddd ifH,(M) N H,

fN)

forevery

Z/

ne ThusifIisthe identitymapon

Z,

asubmoduleNof

M

ish-purein

M

if andonlyifN is

I-embedded.Givenl"

Z4-

---)

Z

_satisfying

l(n)

>__{n, we define}

l

Z

such’that

for_any_ne

Z

/,/(n)

is the minimumofall

l(k),

k>_ n.Then

l

ismonotonic. FurtheranysubmoduleNof

M

is/-embeddedifand onlyif it is

l-embedded.

Sowithoutloss of generalityweassumeth is monotic.Furtherdefine

Proposition4.5.Let

M

beanh-reducedQTAG-moduleand l"

Z’

Z4-

beamonotonicfunction

such thatn _<l(n),ne Z/. Then every h-neat submodule ofMis/-embeddedif andonly ifh(y) < l(h’(y)+1)-1foreveryuniformye M.

Proof.Let everyh-neatsubmoduleof

M

be/-embedded.Consider auniformyeM. As

M

is

h-reduced,every h-neat hullofyRisoffinitelength.Let zRbean h-neathullofyRsuch that[3’, z] h’(y)

t.Then

Ht(zR)

yR

and

I-It+(zR)

<

yR.

Thenbythehypothesis,

I-Itet)(M)

zR

Ht(zR)

yR,

but

Ho/)(M)

czR<

yR.

Consequentlyh(y)

</(t+l)-I

=/(h’(y)+

1)-

Converselylettheinequality hold.

(7)

l-embedded. We get smallest positiveintegernsuch that

Ht(,)0VI’)

N

:

H,(N).

Then

I-Itt.,)(M)

Nc

Hn.

(N).

Thereexistsauniformy

I.it,)0Vl)

cNsuch thaty

I-I,(N).

As

l(n)

_>/(n-l), y

I-I.t(N).

Sothat h(y) n-1.Consequently h’(y) <n-1.Bythe hypothesish(y) </(h’(y)

+ 1)-1

<

l(n)-1.

Howeverasy

Ht,)(M),

h(y)>

l(n).

This is a contradiction.Thisprovesthe result.

Theorem(4.6).LetMbe anyQTAG-moduleandl’Z

--

Z beamonotonicfunction such that n

Z

/"

_<fin),n Theneveryh-neatsubmoduleof

M

is/-embeddedif andonlyif foranyuniformye

M,

h(y) _</(h’(y)

+ 1)-

1.

Proof. Let everyh-neatsubmodule ofMbe/-embedded. WriteM L

D,

whereDisthe largest

h-divisiblesubmoduleofM.NowLish-reduced and every h-neat submodule ofLis/-embeddedinL.

Consider auniform y M.Writey y,+y2,y,

L,

y2 D.

Suppose

y 0. Thenh(y) h(y).

By (4.3),

h’(y)

h[(y) By

using

(4.5),

we geth(y) h(y) _</(h’(y)+

1)-

I.Suppose y--0.theny=y2

D,

henceandh(y) LetKbeanyh-neathullofyR.Consideranyn>_0.ThenI-Itc,(M)

I-I4t.)(L)

D. As

K

rD 0,

Ht,)(M)

K

H,(K),

weget

H,(K)

0.Sothat

d(K)

’, h’(y) h(y). Onceagain h(y)=/(h’(y)

+ l)-

1.Converselylet the given condition be satisfied.

By

essentiallyfollowingthe

arguments in(4.5),wecompletetheproof.

Theorem(4.7).

Let

M

L

DbeaQTAG-modulesuch that

L

ish-reducedand

D

ish-divisible.

Foramonotonic function l"

Z

-

Z satisfyingn_</(n), everyh-neatsubmodule ofMis/-embeddedif

andonlyif

(i) everyh-neat submodule ofLis/-embedded in

L;

and

(ii)for any serial submodueWof

D,

anynon-zerohomomorphism f"W

-

Lisa monomorphism.

Proof Letevery h-neat submodule ofMbe/-embedded.Then obviously(i)hold. Considera non-zerohomomorphism f"W Lwith kerf 0.thenbR

soc(W.).c

kerf. Consider

soe(f(W))

b,R. As h(bt)<

oo,

by using

(2.3)

we canchooseWtobesuch thatqW)is h-neat inL.Then

L

{x+f(x) x

W}

is anh-neat hullofbR. Sothat

h’(b)

<

oo.

By (4.6)

h’(b)

**.This givesa

contradiction.

Conversely,let the conditions be satisfied. Considerauniform yeM. Let y

y+y,

ye

L,

ye

D.

Suppose

y 0.Thenby

(4.3)

hfy) hL(y)

<-

l(h[

(y)

+ 1)

1.

Suppose

y 0.Thimy ye D. Let

K

be any h-neat hull of

yR.

Let

K

and

K

beprojectionsof

K

in

L

and

D

respectively.Then

K

_=

K

andwe

getanepimorphism f"

K

withye kerf.

By

(ii), f=0.Consequently

K

D

and hence

d(K)

Soonceagainh(y) =/(h’(y)+ 1)-1. Hence

(4.6)

completestheproof.

By

taking

I,

weget thefollowing:

Corollary

(4.8).

LetM L DbeaQTAG-modulesuch thatLish-reducedand

D

ish-divisible Then thefollowingareequivalent:

(8)

808 S. SINGH AND M. Z. K}IAN

(iii)

Evey

h-neat

submoduie ofl

ish-pureand foranyuniserialsubmodule WofD anynon-zero homomorphism f"W -->

L

isamonomorphism

5.H-NEAT

SUBMODULES

Amodule

MR

is calledastrongly TAG-modue,ifM Mis a

OTAG-module.We

start with the

following:

Lemma(5.1).

Let

Mz

beastronglyTAG-module,Aand

B

betwouniserial submodulesof some

homomorphicimages ofM. Then the following hold:

(i) Ifd(A)<_d(B),thenBisA-injective.

(ii)

If

d(A)

>

d(B),

thenBisA-projective.

(iii)If

d(A)

d(B),

thenA_-.-B,whenever

sot(A)

_=

sot(B)

or

A/H(A)

=

B/H(B).

(iv) Mis aTAG-module.

Proof.Now AandBaresubmodules ofM/KandM/Lforsondesubmodules

K

and

L

ofM. As N

M/K M/Lisahomorphicimage ofM

M,

Ax0,0>d3aresubmadules ofNwithzero intersection, (i),(ii),and(iii)followfrom

(2.3).

Finally(iv)follows from(i).

Let

MR

be astronglyTAG-module. Let spec(M)be thesetofallsimple R-moduleswhichoccur

ascomposition factors ofsomefinitely generated submodules ofM. Let

S,

S’e spec(M).ThenS’is called an immediatepredecessorofS

(and

Siscalledan immediatesuccessor ofS’)iffor some uniserial submodule A of

M, A/H(A)=

S’and

H(A)/H2(A)

S.

By

using

(5.1)

weget that anySe spec0Vl)does

nothave more than one immediate successorandmore thanoneimmediatepredecessor.(seealso

[9]).

LetS, S’espec(M), S’ iscalledak-th successor of

S,

if there existsasequence S

So,

$1,.

Sk

S’

of k+l distinct membersSiofspec(M),suchthat for <k,S/ is an immediate successor of Si; in this

situationSis called ak-thpredecessorofS’. Siscalledits own0-th

sueeessor(0-th

predecessor).S’is

called asuccessorof

S,

ifS’is ak-th successorofSforsomepositiveinteger k.DefiiaeS S’if for somek>_0,S’is ak-th successorork-th predecessorofS.Thisisan

equivilence

relation.

Any

equivalence classCdeterminedby thisrelation iscalledaprimaryela;,s.Fora torsion abeliangroup,each

suchCisasingleton.Howeverforatorsionmodule overabounded{hnp)-ring,eachCis finite.Forany primary classCinspec(M),the submodule1Vl,of allthose x e Msuchthatevery composition factor of

xRis in

C,

iscalled the C-primary submodule ofM.

By

using

(5.1)

onecaneasilysee that

M

isadirect

sumofitsC-primary submodules.Amodule

M

iscalledaprimaryTAG-J,oduleifM Mis a

TAG-modulesuchthatspec(M)is aprimary class. Consideraprimary TAG-modueM.Letspee(M)havek members, then either kisfiniteoreoumable.Thisk is called theperiodicityofM.Inthissectionwestudy primary TAG-modules.

Lemma(5.2).

Let

Ms

beanh-reducedprimaryTAG-module offiniteperiodicity.If thereexists a

function f"

Z

--->

Z

such thatforany uniformx e

M, h(x)

<

f(h’(x)),

thenMisbounded.

Proof.Let

M

be of periodicity k.Foranyuniform x e

M, h’(x)

<

oo.

Thisgivesh(x)<

f(h’(x))

<

(9)

writeM

xR

@

x2R

@

M’,

withxiRnon-zerounisedal,

zR

soc(xiR), h(z2)>

max{

f(j) < _<

k+d(xR) }and

e(x=)

>k.Now

h(zz)

[z=, x=].

As

M

isofperiodicitykand

e(x2)

>k,wegety2

x=R

such that[zz,y=] < k-1andsoc(x=R/y2R) soc(xR).Thisgivesamaximalg SH(x=R,

xR)

withd(ker g)_<_kand

z2R

_

kerg.Consequentlyd(domain(g)) <

k+d(xlR),

h’(z:)

<

k+d(x,R).

As h(z=)<

f[h’(z:

)),

wegeth(z=)<

max{

f(j) 0 _< <

k+d(xR)}.

This isacontradiction.Hence Misbounded.

Lemma(5.3).Let

Ms

be any primaryTAGfmoduleoffiniteperiodicity. If everyh-neat submodule

ofMish-pure, theneitherMis h-divisibleorh-reduced.

Proof.LetMbeneitherh-reduced nor h-divisible. ThenM xR@AB

MI

for someuniform element x and a serialmoduleAof infinite length.LetzR

soc(A).

Thenh(z)

oo.

If the periodicity of

Misk, then forsomeu, _< u<k,wegetasubmodule ofAoflengthusatisfyingsoc(A/yR)

-=

soc(xR).

By (2.3),wegetamaximalf

SH(A, x.R)

withd(domain(f))<

e(x)+u.

Thisgivesanh-neat hull

K

of

zR

lengthe(x)+u.AsKish-pure,we geth(z) d(K)-1<

oo.

Thisiscontradiction.Hence theresultfollows.

Lemma(5.4).

Let

MR

beaprimary TAG-module of finite periodicityk.Let T xR B Abe a

submoduleofM,withxRunisedal, such thateveryh-neatsubmodul oftish-pureinT.Then the followinghold

(i) If

soc(xR)

sot(A),

then

d(A)

<

d(xR)+k.

(ii).If

soc(xR)

is theu-thpredecessorof

soc(A)

forsome u _>1, then

d(A)

<

d(xR)+u.

Proof.Let

soc(A)

zR. Let soc(xR) zR.Foramaximalf;e0 inSH(A,

xR)

withzR

=kerf

and

d(ker f)minimal,wehave

d(ker f)

k, domain(f) yRc:A;furtherh’(z) e(y)-I [z, y]_<e(x)+k-1.

However by(4.8), h’(z) h(z). So

yR

A.Consequentlye(y) d(A)_<

d(xR)+k.

Similarly(ii)follows.

Wenowprove thefirstdecompositiontheorem.

Theorem(5.5).

Let

MR

beaprimaryTAG-moduleofpedodicityk <

oo.

Thenevery h-neat submoduleofMish-pureifandonlyif eitherMish-divisible orM

BEx=,R

suchthat

A

(i).each

x=R

isuniserial;and

(ii)foranytwodistinct

z,

]3

A

thefollowinghold

(a)

if

soc(xR)

soc(xR),

then

d(xeR

<

(b)if

soc(x)

is au-thpredecessor of

soc(xJ.),

<_u< k-1, then

d(x,R)

<_

d(xR)+u.

Proof.Let everyh-neat submodule ofMbeh-pure.

By (5.2)

Miseitherh-divisibleorh-reduced LetMbeh-reduced.By

(5.2)

Mis bounded.SothatM

@Ex=R,

forsome unisedal submodules

x,R

A

By

applying

(5.4)

wecomplete the necessity.Converselylet the gi,:aconditions be satisfied. IfMis h-divisible,thenevery h-neat submodule NofMbeing h-divisible,is a summandof

M,

consequentlyNis h-pure.Let Mbeh-reduced.Considera

u.niform

z

.A

0}=

min{[z,, x]

z,

;

0}.

ConsiderT

x,R

@

xR

with

z=

;0,

z

; 0 anda;

13.

Letfa

SH(x,R,

xR)

(10)

810 S. SINGH AND M. Z. KHAN

xR.

If f 0, obviouslydomain(f) xjR..Letf 0.If

soc(xR)

soc(xR),

thend(ker

f)

,kforsome

k>0. If

soc(xR)

soc(xR),

thenforsomeu >_

soc(xR)

isthe u-thpredecessorof

soc(x,R)

andd(ker f)

u+tk

forsome_t_> 0.Thus

(a)

and(b)yielddomain(f)

x,R..

Cosequently

hr(z=)

[z,,

x,]

h(z)

By (4.4), h’(z)

h(z).

Thisproves the result.

Theperiodicity ofatorsionabclianp-groupisone.Wegetthe,following:

Corollary(5.6).Everyneatsubgroupofanabelianp-group G.paprime number,ispure

subgroupif andonlyif eitherGis a divisiblegrouporG A@

B,

suchthatforsomepositive integern,

Ais adirect sumof copies ofZ/(p")andBisadirect sumof copies

o:Z/(p"/).

Wenow discuss the case of aprimaryTAG-moduleofinfiniteperiodicity.Henceforth

MR

willbe aprimary TAG-module ofinfiniteperiodicity.

Lemma(5.7).

Let

xR

and

yR

betwoh-neat unisedal

submodtiies

ofMsuchthat

soc(xR)

soc(yR)andsoc(yR)isapredecessorof

soc(xR).

Then"

(i) SH(yR,

xR)

0.

(ii) For anyh-neathull

K

of

xR B yR

in

M, yR

is asummandof

K;

ifin addition

xR

is h-purein

M,

then

K=xR

ByR.

(iii)IfxRand

yR

bothareh-pure,then

xR

B yR

ish-pureinM.

Proof.AsMisof infiniteperiodicityandsoc(yR)isapredecessorof

soc(xR), soc(xR)

is not a

predecessor of soc(yR). Consequently SH(yR,xR) 0.LetKbe an h-neat hullofxR ByR.Asrank(K)

2,K

A

B A2

with

&

serial. Considerthe projections

f

A, @

A2

"->Ai.Therestdctiion ofoneof f, ,sayof

f

toxRia amonomorphism. Then

soc(xR)

_-- soc(A)andsot.(yR) soc(A2).Further

f2

embeds

yRinA2. By (i)

SH(A2, A1)

0.Thisyields

yR

c::

A=.

AsyR c::’

A2

annd

yR

ish-neat,wegetyR

A2.

LetxRbe h-pureinM.Sothat

xR

ish-pureinK.ConsequentlyxRis asummandofK. AsxR

yR,

we getK xR@

yR.

Thisproves(ii).Finally let bothxRand

yR

be h-pureinM.ThenM xR

B

M

for

somesubmodule

M.

ThenK xR

(K

r

M0.

This gives

yR

K:’

M.

So K r

M

ish-purein

M.

Thus

K

MI

isasummandof

M.

Hence K

isasummandofM.Thisgive/(iii).

Lemma(5.8).

LetKbe any submodule ofM with

soc(K)

homogeneous.Then

(i)GivenanytwounisedalsubmodulesAandBof

M,

eitherArB 0t,r

.

_[tyarecomparableunder

inclusion,

(ii)foranyuniformx

K, h:(x)

h: (),

and (iii)anyh-neatsubmoduleofKish-pureinK.

Proof.(i) Let A r B 0and

d(A)

_>

d(B).

ThenA+B A

B C,

withCaproperhomomorphic image

orB. Suppose

C 0, then

soc(A)

soc(B)

soc(C).

Thiscontradicts thehypothesisthatsoc(K)

ishomogeneous.Thus C 0,soBc:A.Thisproves(i).Considerauniform x e K.Thenby (i) xRhas

unique h-neat hull

D

inK.Then

h:

(x)

d(D)-l.

Theuniquenessof[)_gives

D

ish-pure. Consequently hK(X)

d0D)-l.

Thisproves(ii).Thelast’panisimmediatefrom(ii)

(11)

(i)soc(xg) soc(yR),or

(ii)

soc(xR)

soc(yR),oneof them say

soc(xR)

isa u-thpredecessorofsoc(yR)forsome u > andd(yR) <d(xR)+u.

Proof..Letevery h-neat submodule ofT be h-pure.Let soc(,P,)Z soc(yR),and

soc(xR)be au-th predecessor ofsoc(yR).Then as in(5.4),we get

d(yR)

<d(xR)+u.

Conversely,letTsatisfythegivenconditions.If

soc(xR)

soc(yR), men soc(T)ishomogeneous,soby

(5.8)everyh-neatsubmodule ofTish-pureinT. Let

(ii)

hold,soc(yR)beau-th predecessor

ofsoc(xR).

AsMisofinfiniteperiodicity,soc(xR)isnotapredecessor ofyR. SoSH(ylLxR) 0,andf,orany0;*f

SH(xP,, yR),

d(ker f)

u.Considerauniformz

x

+ y,

x

xP,. y

yR..

Let

e(x0

>-

e(y0.Ify

0,then

e(x0

e(y0+u.So

[x, x]

_<[y, y],and hencehf

(z)

[x, x]

hT(z). Suppose

y 0. Consider

amaximalf SH(xR, yR)with z kerf.Theneither f 0 or

d(ke

f) u.As d(xR)<d(yR)+u, domain(f) xR. Onceagain

hfr(z)=

[x, x] h(z). Let e(y0>e(x0,then

x

0,asSH(yR,xR) 0. Thenf,orany uniformv

T,

withz

zR,

wehave

z

yR.

SoyRisthe only h-neat hull ofzRinT.

Thusinall cases hf

(z)=

hT(z). By

(4.$),theresultfollows.

Lemma(5.10).

LetNbe the submodule of

M

generated by those uniform elementsx

M

such

that

soc(xR)

hasnopredecessorin

soc(M).

Then:

(i)

Soc(N)

ishomogeneous.

(ii). Nisanh-puresubmodule ofM.

(iii)

Any

h-neat submodule ofNish-pureinM.

ProofLetAbe thesetof those uniformx

M

such that

soc’xR)

hasnopredecessorin

soc(M).

Forany x, y

A,

if

soc(xR)

soc(yR),thenoneof them beingasuccessorof theother,contradictsthe

hypothesis.Sothat

soc(xR)

-=

soc(yR) forallx,y A.Consider aunif,ormzE

soc(N).

Forsomey,E

A,

z

y,R

T.,Bj,

Bj’suniserial.

For

some j,

zR

soc(Bj). But Bjisahomomorphic image of some

Assoc(yiR)has nopredecessorin

soc(M),

weget yiR

----

Bj.Hence

soc(N)

ishgmogeneous. Itis now immediatethatif foranyuniform x

M,

soc(xP,)g;;

N,

thenx N.Thisfact gives(ii). Byusing

(4.8)

we get(iii).

The submodule NofM generated by those uniform element’

.

M,

such

that

soc(xR)

hasno

predecessorin

soc(M)

iscalledaterminal submodule ofM. Wedenote this submodule by

Ter(M).

Proposition(5.11). Let

MR

beaprimary TAG-module of’infiniteperiodicityandN Ter(M).

Then

(i)

Any

submoduleKofNhasuniqueh-neathull in

M,

(ii)

forany uniformx

N,

h(x)

h’(x);

and

(iii)for any decompositionM

(D,^

A,,

N

(Ai

N).

Proof.

By (5.10)

soc(N)

ishomogeneous. So givenauniformx

N,

by

(5.8)

anytwouniform submodules ofNcontainingx arecomparableunder inclusion. Thus there is unique h-neat hull

Ax

ofxR

in

M,

andby

(5.10)

Ax

N.

For K,

thesum

L

ofthose

A

forwhich

K,

istheuniqueh-neat hull of

(12)

812 S. SINGH AND M. Z. KHAN

mapping xR.-->xiR,xr-->xiris notone-to-one,then

soc(xR)

will have apredecessorinsoc0Vl).This givesacontradiction.HencexR

-=

xil,wheneverxi 0.Thusxia t,1md (iii)follows.

Theorem(5.12).

Let

Ms

beanh-reduced primary TAG-module of infinite periodicity.Thenevery h-neat submodule ofMish-pureifandonlyif

M

@

K

@

Ter(M)

satisfying the following conditions

(i) foreachj,Kjisdecomposable andsoc(Kj ishomogeneous,

(ii)for j <j,with

K,

0

Kj,, ifzR

and

zR

areunisesedalsummands of

K,

and

K

respectively, then

soc(zR)

isau-thpredecessorof

soc(zR)

forsomepositive integerudepending

upon j and j, and

d(zR)

<

d(zR)

+u and

(iii)if isthelengthofasmallest lengthuniserialsurnmand of

N,

antiSisthe simple module determiningsoc(N),thenfor anyKj 0, ifSis a

v-th

predecesso,"ofthe simple module

S

determiningsoc(Kj),wehaved(zR)< +vjfor any unisedal summand of

K.

Proof. Let everyh-neatsubmodule ofMbeh-pure. NowN

Ter(M).

AsNish-reduced,it hasa uniserialsummandxRofsmallestlength, sayt.ConsiderM M/N.LetSbeasimplesubmodule ofM.

Considerany uniform y

M

suchthatS=__sot(yR.).

By (2.2),

wechoose ytobe uniform suchthat

y

cN 0.Then

soc(xR)

soc(yR). As

soc(x_R)

hasno

predecessor"

insoc(M),itisav-thpredecessor

ofsoc(yR)

z

for somev>1.Now

h(z)

h’(z)

<

.

Wegety

M

such that

[z,

y]

h(z).

Thenin

x.R @ yl, boththesummandsareh-pureinM.By

(5.7)

[email protected] (5.9), d(yR)_<

d(xR)

+

v.Sothereis anupper boundontheheightsof elements ofaparticularhomogeneouscomponent of

soc(M).

Henceby

(3.4) M

isitsonlybasicsubmodule,soit isdecomposable. AS Nish-pure,by the

observationfollowing(2.2),wegetM

KB N,

with

K

itsonlybasicsubmodule.ASMisprimary,

spec(M)iscountable.WegetK

K

satisfying(i).Finally(ii)and

(ilia)

follow from

(5.9).

Conversely,letMsatisfy the given conditions..Then

K

satisfi,;sconditionsanalogoustothose

givenin

(’5.5).

Soonthesimalar lines as in

(5.5),

every h-neat submodule ofKish-pure.Considerany

uniformye M.Now y y,+y=.y,a

K,

y=a N.If y, 0,y Naz.d by(5.11), yRhasuniqueh-neat

hullinM obviouslythenh(y) h’(y). Lety 0.Suppose y=#0. thenby using

(4.3)

h’(y) h(y).

Suppose

y= 0andh’(y) < h(y). Wegetanh-neat hull

zR

of

yR

with[y,

z]

h’(y).Letz z,+z=, z,

K, z=

e N.AS

h<

(y) h(y),

z=

#0.One

ofzR

and

z=R

is h-neat.

As.yR

=

zR

and[y,

z]

<h(y) hK(y),

z,Ris not h-neat inKandsoinM. Consequently

z=R

ish-neatin

N,

and by

(5.10)

itish-pure.Forsome

v,

soc(zR)

is a v-thpredecessor of

soc(zR).

Sothat

d(zR)

+v. Write

soc(zR)

gR.

Thenby

using condition(iii),weget[g,

z]

<h(g)<

d(zR)

+v-1 [g,z,].Conseuently

d(z,R)

h(g). So z,R

ish-pure.Thisisacontradiction. Thiscompletestheproof.

Wenowdiscuss the caseof

M

beingnotnecessarily h-reduced. Write

M

B D,

where

D

is thelargesth-divisiblesubmoduleof M.

(13)

Proof.

Suppose

N

Ter(M)

ZD.Wegetauniform x e

sot0"0,

suchthat

h(x)

< Thenx y

+ z.Now0#y M1,z D.

By (5.11)

ya N.Consideranysimple submoduleSolD.

By

thedefinition of"S,itisnotapredecessor ofsoc(yR). Sothatsoc(yR)is apredecessorof"S.AsDish-divisible,there exists auniserialsubmoduleAoldand ahomomorphism f"A -->M,withrange(f) yRandS

=

kerf. This contradicts(4.8). Hence N

=

D.As Nish-pure,itmustbe h-divisible.

Theorem(5.14)

Let

MR

beaprimaryTAG-moduleof infiniteperiodicitysuchthatMisnot

h-divisibleand letN

Ter(M).

Thenevery h-neat submodule ofMis

h-lure

ifandonlyifthe following hold:

(a)

Nis h-divisible,and

(b)

M N@

K,

whereKjsatisfy thefollowingconditions: 9=-..

(i) if" someKj;0,thensoc(Kj)is ahomogeneouscomponent of

sot(M),

(ii) eachKjis a direct sumofserialmodules,

(iii)if’forsome < j,Ki

;

#

K

and

Ki

isnoth-divisible,then thesimplesubmoduleSi determining soc(Ki)is a v-thpredecessor of’the simple submodule determiningsoc(Kj)forsomepositive integer

vdependingon and j, and foranyuniseriaisubmoduleA of’Kj,

d(A)

< + v, where isthe length of the smallest lengthuniserialsummandofK,and

(iv)if for somej, Kj; 0 and is noth-reduced,thenfor any < j,Kiis h-divisible.

Proof.Let

D

be thelargesth-divisiblesubmoduleofM.TheD 0.Let every h-neatsubmodule ofMbe h-pure.By

(5.13)

Nish-divisible.ThusM N@

M

@

M

such thatD N(B

MI

and

M2

is h-reduced.Byapplying

(5.12)

to

M2

and usingthefactthat

M

is a direct sumofserialmodules,we get

M

@

M2

@

K

satisfying(i), (ii),and

(iii).

Finally

(iv)

isanimmediateconsequenceof(iii).

Converselyletthe given conditionsbe satisfied.

By

comparing these conditions with those in

(5.12),

weget

M

D

B

LsuchthatNg;D.ThenSH(N,

L)

0.Considerauniserialsubmodule WofD

and let f W -->

L

be anon-zero homomorphism. ThenW Z N. Forsomej, W is isomorphic to a

submoduleof

Ki.

This

K

is noth-reduced,f(W)

=

L,

and for somet, f(W)isisomorphictoasubmodule

of

Kt.

If t,obviously f isamonomorphism.

Suppose

< Then I(. ish-divisible. LetxR

soc(f(W)).

As xR

=

L,

h(x)<

.

Onthe other handx

soc(Ir

yields

h(x)

.

Thisisacontradiction. Hence <t.

SoSH(Kj,

)

0. Thisonceagaincontradictsthefactthatf; 0."lbus t.Hencefis amonomorphism.

Soby

(4.8)

theresult follows.

Weendthispaperby givinganexampleofanh-reducedprimaryTAG-module

M

ofwhichevery

h-neatsubmoduleish-pure,but it isnotdecomposable.Suchamodulehastobeof infiniteperiodicity. Example. Let

F

beaGaloisfieldand

R

be the ring ofinfinitelewer triangular matrices

[ai]

over

F,

wherei, are indexed overthe setPof allpositiveintegers.Let

e

i,

P

be the usualsetofmatrix units inR.Then

M

eu,Ris auniserial R-modulewith

d(M)

k;it isannihilated bytheideal

A

ofR

(14)

814 S. SINGH AND M. Z. KHAN

Mk/under themappinge.kr-->ek/t.kr,r R. LetT

Hk

Mk.Then

MR

{xe T

xA

0forsomek}

is a primary TAG-module ofinfinite periodicity. Its socle is homogeneous. By (5 8) every h-neat submodule ofMish-pure.Mish-reduced. Considerauniform x sot(M),then x (xk),Xk Mk.Letu be thesmallest integer such thatx,,0.ThenxR_= xj.. As d(Mu) u,by using

(2.3)

itcanbeeasilyseen thath(x) u-l. Sothatfor any > 1, soc(Hiq(M))/soc(Hi)_=

soc(Mi).

SupposethatMisdecomposable,

ThenM

Nj,

whereNjis a direct sumof uniserial modules of length j. Then

soc((H,.(M))/soc(Hi(M)) -_- soc(Ni).Thussoc(Ni) -_- soc(Mi),asiml;l,module. Consequentlyeach N,is a uniserialmodule.AsFisfinite,Niis a finite set.Consequently Miscountable.But byconstructionMis

uncountable This is acontradiction.HenceMis notdecomposable. ACKNOWLEDGEMENT

The research of Surjeet Singh was supported by the Kuwait University Research Grant No.

SM075, and of Mohd.Z.Khanbyatravel grantbyThe Third WorldAcademy of Sciences, Trieste. The

authorsarehighly thanlffltothereferee forhisvaluablecomments.

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Current Address: Surjeet Singh