IBN AL- HAITHAM J. FO R PURE & APPL. SC I.          VO L.23 (2) 2010 

  

On Almo st Bounded Submodules 
 
 

B. N.Shihab  
Department of Mathematics , College of EducationIbn-Al- Haitham , Uni vers ity of 
Baghdad 
 
 

Abstract 
        Let R be a commutative ring with identity , and let M  be a unitary  R-module. We 
introduce a concept of almost bounded submodules as follows: A submodule N of an R-
module M  is called an almost bounded submodule if there exists xM , xN such that 
annR(N)=annR(x). 
        In this p aper, some p rop erties of almost bounded submodules are given. Also, various 
basic results about almost  bounded submodules are considered. 
        M oreover, some relations between almost bounded submodules and other ty p es of 
modules are considered. 

 

Introduction 

        Every  ring considered in this p aper will be assumed to be commutative with identity  and 
every  module is unitary . We introduce the following: A submodule N of an R-module M  is 
called an almost bounded submodule, if there exists xM , xN such that annR(N)=annR(x), 
where annRN={r:rR and rN=0}. 
        Our concern in this p aper is to st udy  almost bounded submodules and to look for any 
relation between almost bounded submodules and certain ty p es of well-known modules 
esp ecially  with p rime modules. 
        This p aper consists of two sections. Our main concern in section one, is to define and 
st udy  almost bounded submodules. Also, we give some basic results for t his concept. 
        In section two, we st udy  the relation between almost bounded submodules and bounded 
modules. We show that the p rop er submodule of bounded module is not necessary  to be 
almost bounded submodule and we give some conditions under which a p rop er submodule of 
bounded module is an almost bounded submodule. Next we invest igate the relationship s 
between almost bounded submodules, p rime and fully  st able module. 
1- Basi c Propertie s of Almost Bounded S ubmodul es  

        In this section, we introduce the concept of almost bounded submodule.     We 
establishe some basic p rop erties of this concept. 
        First , we introduce the following definition. 
1.1 De fini tion: 
        A p rop er submodule N of an R-module M  is called almost bounded submodule if there 
exists xM , xN such that annR(N)=annR(x). 
        An ideal I of a ring R is an almost bounded ideal if I is an almost bounded R-
submodule. 
1.2 Remarks and Examples: 
1. Let M =ZZ as a Z-module and N=2Z0 be a submodule of M . Then N is an almost 

bounded submodule. 
2. Every  submodule of the Z-module Z is an almost  bounded submodule. 
 
Key words: almost bounded submodule, bounded module, p rime module, quasi-prime 
module, fully st able module. 

IHJPAS



IBN AL- HAITHAM J. FO R PURE & APPL. SC I.          VO L.23 (2) 2010 
 

3. Consider the Z-module M =ZZp, where p  is a p rime number and the Z-suubmodule 
N=qZZp, where q is any  p rime number. Then N an almost bounded sumodule. 

4. For each p ositive integer n and n is not p rime number, every  p rop er submodule of a Zn-
module Zn is not almost bounded submodule. 

5. 2  as a Z-submodule of Z12 is not almost bounded. In general, let n be a p ositive 
integer, then the Z-module Zn has no p rop er almost  bounded submodule. 

6. Let p  be a prime number. The Z-module Zp dose not contain any p rop er almost bounded 
submodule. 
The following remark ensures that t he almost  boundedness p rop erty  is not hereditary . 

1.3 Remark: 
        A submodule of an almost bounded submodule need not be almost bounded in general. 
For examp le: 
M =ZZp as a Z-module, where p  any p rime number, N=qZZp be a submodule of M , where 
q is any p rime number. Then N is an almost bounded submodule of M , but K=0Zp as a 
submodule of N which is not almost bounded submodule of N. 
        We st ate and prove the following p rop osition. 
1.4 Proposi tion: 
        Let M 1 and M 2 be two R-modules, M =M 1M 2. If N1 and N2 are almost bounded R-
submodules of M 1 and M 2 resp ectively, then N1N2 is an almost  bounded R-submodule of M . 
Proof:  We have N1 and N2 are almost bounded R-submodules of M 1 and M 2 resp ectively. 
Then there exists xM 1, xN1 such that annRN1=annR(x) and also there exists yM 2, yN2 
such that annRN2=annR(y ). Therefore (x,y )M 1M 2, (x,y )  N1N2. Now, annR(x,y ) = 
annR(x)  annR(y ) = annRN1  annRN2 = annR(N1N2). Hence N1N2 is an almost bounded 
R-submodule of M . 
       The converse of prop osition (1.4) is not  true in general as the following examp le shows. 
1.5 Example: 

        Consider M =Z 6Z12 as a Z-module. Let N= N1N2= 3 2      be a              Z-
submodule of M . Then N is an almost bounded submodule of M . Since 

annZN=annZ( 3 2     )= Z Zann 3 ann 2     =2Z6Z=6Z and there exists 

(2, 2) M , (2, 2) N such that annZN=annZ (2, 2) = Z Zann ( 2) ann (2) =3Z6Z=6Z. But 

N1= 3   and N2= 2  is not almost bounded submodules of M 1 and M 2 resp ectively. Since 

for each xZ6, x=1, 2, 4, 5N1, annZ(1 )=6Z , annZ( 2 )=3Z , annZ( 4 )=3Z , annZ( 5 )=6Z . 

Therefore for each xZ6, xN1 annZ(x)  annZN1 = annZ 3   =2Z. Thus N1 is not almost 
bounded submodule of M 1. 
In the same way, N2 is not almost bounded. 
        Using the mathematical induction, we obtain the following corollary . 
1.6 Corollary: 
        Let M 1, M 2, …, M n be a finite collection of R-modules and   M = M 1M 2…M n. If N1, 
N2, … and Nn are almost bounded R-submodules of M 1, M 2, … and M n resp ectively, then N= 
N1N2…Nn is an almost  bounded submodule of M . 
        So, we have the following app lications of (1.4) 
1.7 Corollary: 
        Let N1 and N2 be two almost bounded submodules of an R-module M . Then N1N2 is an 
almost bounded submodule of MM . 
Proof:  We haveN1 and N2 are almost bounded submodules of M , means  
there exists xM , xN1 such that annRN1=annR(x) and there exists yM , yN2 such that 
annRN2=annR(y ), imp lies (x,y )N1N2. Now, we claim that annR(N1N2)=annR(x,y ). Let 
rannR(x,y ). Then r(x,y )=(0,0), implies rx=0 and ry=0. Therefore rannR(x)= annRN1 and  
 

IHJPAS



IBN AL- HAITHAM J. FO R PURE & APPL. SC I.          VO L.23 (2) 2010 
 

rannR(y )=annRN2. Thus rannRN1 annRN2. But annR(N1N2)= annRN1annRN2, so we get 
rannR(N1N2). 
Conversely , let rannR(N1N2). Then r(a,b)=(0,0) for all (a,b)N1N2 which imp lies ra=0 
for all aN1 and rb=0 for all bN2 which imp lies rN1=0 and rN2=0. Thus rannRN1 and 
rannRN2. But annRN1=annR(x) and annR(y )=annRN2. This imp lies that rannR(x) and 
rannR(y ), that is, rx=0 and ry =0. Thus (rx,ry )=(0,0), so that r(x,y )=(0,0). Hence rannR(x,y ). 
This comp letes the proof. 
        The following corollary  is a sp ecial case of p rop osition (1.4). 
1.8 Corollary: 
        Let M  be an R-module, N be an almost bounded submodule of M . Then N

2
=NN is an 

almost bounded submodule of M
2
=M M . 

Proof:  From hy p othesis N is an almost bounded submodule of M . Then there exists xM , 
xN such that annRN=annR(x). Thus (x,x)M

2
=M M  and (x,x)N

2
=NN since 

annR(x,x)=annR(x)annR(x)=annRN=annR(NN). Hence annR(x,x)=annR(NN) which is 
what we wanted. 
        Now, we have the following p rop osition: 
1.9 Proposi tion: 
        Let M = M 1M 2 be a direct sum of two R-modules M 1 and M 2. If L1 is an almost 
bounded submodule of M 1 and annR(y )=annRM 2 for some yM 2, y ≠0, then L1M 2 is an 
almost bounded submodule of M . 
Proof:  We have L1which is an almost bounded submodule of M 1, then there exists xM 1, 
xL1 such that annRL1=annR(x), yM 2. Then (x,y )M 1M 2 and (x,y )L1M 2. We claim 
that annR(L1M 2)=annR(x,y ). Now to p rove our assump tion. Let rannR(L1M 2)= 
annRL1annRM 2. Then rannRL1annRM 2, so rannRL1 and rannRM 2=annR(y ). Therefore 
rannR(x) and rannR(y ). Thus rx=0 and ry =0 means (rx,ry )=(0,0), which imp lies 
r(x,y )=(0,0) and hence rannR(x,y ). 
     Conversely , let rannR(x,y ). Then r(x,y )=(0,0), which imp lies (rx,ry )=(0,0). Therefore 
rx=0 and ry =0. Thus rannR(x)=annRL1 and rannR(y )=annRM 2. Hence r annRL1annRM 2, 
which implies rannR(L1M 1). Therefore rannR(L1M 2)= annR(x,y ). 
        Next, we have the following remark. 
1.10 Remark: 
        A direct summand of almost  bounded need not be an almost bounded.                          
For example: 
It is known that N=qZZp is an almost bounded submodule of a Z-module M , where p ,q is 
any p rime numbers and M =ZZp. But Zp is not almost bounded because Zp has no p rop er 
almost bounded submodule. 
        We have seen by the following p rop osition that t he class of almost bounded submodule 
is closed under homomorphic image and inverse image. 
1.11 Proposi tion: 
        Let M  and M ' be two R-modules and let : M   M ' be an isomorphism. Then: 
1. If N' is an almost bounded submodule of M ', then 

–1
(N') is also almost bounded 

submodule of M . 
2. If N is an almost bounded submodule of M , then (N) is an almost  bounded submodule of 

M '. 
Proof:  1. Assume that N' is an almost bounded submodule of M ', then there exists yN' such 
that annR(y )=annRN'. Since  is an ep imorp hisim, then there exists xM  such that (x)=y . It is 
clear that x

–1
(N'). We claim that     annR(

–1
(N'))=annR(x), let rannR(x). Then rx=0, which 

imp lies (rx)=0. T hus r(x)=0. T his means rannR((x))=annR(y )=annRN'. Thus rN'=0, which 
imp lies 

–1
(rN')=0. T hen r

–1
(N')=0 and imp lies rannR(

–1
(N')). 

 
 

IHJPAS



IBN AL- HAITHAM J. FO R PURE & APPL. SC I.          VO L.23 (2) 2010 
 
On t he other hand, let rannR(

–1
(N')). Then r

–1
(N')=0, which imp lies                   

–1
(rN')=0. 

This means rN'=0. Therefore rannRN'=annR(y )=annR((x)). Thus rannR((x)) and from 
this, we get r(x)=0 which imp lies (rx)=0. Then rx=0 and hence rannR(x). Thus 
annR(x)=annR(

–1
(N')) which completes the proof. 

2. Sup p ose that N is an almost bounded submodule of M . Then xM , xN such that 
annR(x)=annRN. Since xM , we get (x)M '. We claim that (x)(N). Sup p ose that 
(x)(N). Then (x)= (n) for some nN, which imp lies that (x) – (n)=0, so that (x–
n)=0. Thus x–n=

–1
(0) and hence x–n=0. Then x=nN. Therefore xN which is a 

contradiction. Hence (x)(N). To show that annR((N))=annR((x)). Let rannR((x)).  
         Then r(x)=0, which imp lies (rx)=0. Thus rx=0, that is rannR(x)=annRN. Then 
rannRN, which imp lies that rN=0, so that (rN)=0. Then r(N)=0. Hence rannR(N)). 
Therefore annR((x))annR((N)). By  using the same way , we can p rove the other inclusion. 
Hence annR((N))= annR((x)) which is what we wanted.  
        The condition (: M   M ' is an isomorphism) in p rop osition (1.11) can not be 
drop p ed as t he following examp le shows. 
1.12 Example: 
1. Let :ZZ4Z4 be a p rojection map  such that (x,y )=y  for all (x,y )ZZ4. Let 

N=<3> 2  be a submodule of ZZ4. It is easily to show that N is an almost bounded 
submodule of ZZ4. But (N) is a submodule of Z4 and it is not almost bounded submodule 
of Z4 by  (remarks and examp les (1.2) (5)). 
2. Let : Z4 Z4Z be an injection map  such that (x)=(x,0) for all xZ4, let 

N'= 2 <3> be an almost bounded submodule of Z4Z. It is know that Z4 has no p rop er 
almost bounded submodule. Since (

–1
(N') is a submodule of Z4, then (

–1
(N') is not almost 

bounded submodule of Z4 by  (remarks and examp les               (1.2) (5)). 
2- Modul es Rel ated to Almost Bounded S ubmodul es  

        In this section, we st udy  the relationship s between almost bounded submoduls and 
bounded modules, p rime and fully  st able modules. 
        We st art with t he following definition which will be needed. 
        Recall that an R-module M  is said to be bounded module, if there exists an element xM  
such that annRM =annR(x), [1]. 
        By  using this concept, we have the following. 
2.1 Remark: 
        A submodule N of a bounded R-module M  is not necessary  be an almost bounded. For 

examp le Z4 as a Z4-module is bounded module, but 2  is not almost bounded submodule. 
        Recall that an R-module M  is called a quasi-prime R-module if and only if annRN is a 
p rime ideal for each non-zero submodule N of M , [2]. 

        Recall that a submodule N of an R-module M  is called essential if NK≠0 for every  
non-zero submodule K of M , [1]. 
        The following p rop osition gives a sufficient condition under which every  submodule of a 
bounded module is an almost bounded. 
2.2 Proposi tion: 
        Let M  be a cyclic quasi-prime R-module and N be a p rop er essential submodule of M . 
Then N is an almost  bounded submodule. 
Proof:  Assume that N is p rop er submodule of an R-module M , then there exists yM , yN. 
Since N is essential submodule of M , thus there exists rR, r≠0. Thus annRry   annRN. But 
M  quasi-prime, so annRry  = annRy . Then annRy   annRN  annRM . Let tannRy . Then ty =0, 
but M  is cyclic. Thus y =cx for some cR. Therefore tcx=0 which imp lies that tcannR(x). 
Thus either cannR(x) or tannR(x). If cannR(x), then cx=y =0. This is a contradiction. Thus 
tannR(x) = annRM   annRN. Therefore annR(y ) = annRN and hence N is an almost bounded 
submodule of M . 

IHJPAS



IBN AL- HAITHAM J. FO R PURE & APPL. SC I.          VO L.23 (2) 2010 
 
        An R-module M  is said to be uniform module if every  nonzero submodule of M  is 
essential, [1]. 
        Now, we deduce the following corollary . 
2.3 Corollary: 
        Let M  be a cyclic uniform R-module and annRM  is p rime ideal of R. Let N be a p rop er 
submodule of M . Then N is an almost  bounded submodule. 
Proof:  The result follows from the definition of a uniform module, [2,Corollary (1.2.8)] and 
p rop osition (2.2). 
        Recall that an R-module M  is said to be a multiplication module if for every  submodule 
N of M , there exists an ideal I of R such that N=IM , [3]. 
        An R-module M  is called fully st able in case each submodule N of M  is st able, where a 
submodule N is said to be stable, if f(N)N for each R-homomorp hism f:NM , [4]. 
       So, we have the following p rop osition. 
2.4 Proposi tion: 
        Let N be a prop er submodule of an R-module M  such that, 
1. M  is fully  st able and bounded R-module. 
2. [N

R
: M ]  annRM . 

3. annRM  is p rime ideal of R. 
Then N is an almost  bounded submodule of M . 
Proof:  From [1,corollary  (1.1.9)], we get M  is multiplication R-module and by  [4,corollary  

(2.7)], we obtain [annRM :annR(x)][(x)
R
: M ] for each xM . 

       Now, we have M  is bounded. Then there exists xM  such that annRM =annR(x). 

Therefore [annR(x)
R
: annR(x)][(x)

R
: M ], imp lies R[(x)

R
: M ]. Thus RM =<x> is cyclic. To 

p rove N is an almost  bounded submodule of M , we must show that annRN=annR(x). 

      In the first, we claim that xN. If xN, then [(x)
R
: M ][N

R
: M ], but [(x)

R
: M ]=R. 

Therefore [N
R
: M ]=R, imp lies that RM =[N

R
: M ]M =N. Thus N=M  which is a contradiction. 

Hence xN. 
It is easily to show that annR(x)annRN. 
      On t he other hand, let rannRN. Then rN=0 but M  is multiplication [1,corollary  (1.1.9)], 

then r[N
R
: M ]M =0 imp lies r[N:M ]annRM . But annRM  is p rime ideal and [N

R
: M ]  annRM  

by  (2). Then rannRM =annR(x) because M  is bounded module. Thus annRN=annR(x) and 
hence N is an almost  bounded submodule of M . 

        The conditions [N
R
: M ]  annRM  and annRM  is p rime ideal can not be drop p ed from 

p rop osition (2.4) as in the following examp le. 
2.5 Example: 
        Let M =Z 6 as a Z-module. Since M  is bounded Z-module, see [1] and M  is fully  st able Z-
module, see [4,examp le and remarks (3.7),(c)], but annZM =6Z is not p rime ideal of Z. Let 

N1= 2  and N2= 3  . [N1
Z
: M ]=[ 2 

Z
: Z6]= 2Z  annZM =6Z and 

[N2
Z
: M ]=[ 3 

Z
: Z6]=3Z  annZM . Therefore N1, N2 are not almost bounded submodules of 

M . 
        An R-module M  is said to be I-multip lication if each submodule of M  is of the form AM  
for some idemp otent ideal A of R, [4]. 
        As an immediate consequence of p rop osition (2.4). 
2.6 Corollary: 
        Let N be a prop er submodule N of an R-module M  such that: 
1. M  is I-multiplication bounded module 

IHJPAS



IBN AL- HAITHAM J. FO R PURE & APPL. SC I.          VO L.23 (2) 2010 
 

2. annRM  is p rime ideal of R. 
3. [N

R
: M ]  annRM . 

Then N is an almost  bounded submodule of M . 
Proof:  The result follows according to [4,theorem (2.9)] and p rop osition (2.4). 
        Recall that an R-module M  is called a p rime module if annRM = annRN for every  non-
zero submodule N of M , [5], [6]. 
 
 
2.7 Proposi tion: 
        Let M  be a p rime R-module and N, K be two submodules of M  such that NKM , K is 
an almost  bounded submodule of M . Then N is an almost  bounded submodule of M . 
Proof:  Assume that K is almost bounded submodule of M , that is there exists xM , xK 
such that annRK=annR(x). Since, xK, NK. Then we obtain xN. To p rove annRN= 
annR(x). annRKannRN (since NKM ), imp lies annR(x) annRN. Hence annR(x)annRN. 
      Now, let rR, rannRN= annRM  for each submodule N of M  (since M  is p rime module), 
but annRM annRK= annR(x). Therefore r annR(x). Thus annRN annR(x). Hence N is an 
almost bounded submodule of M . 
        So, we have the following app lication of (2.7). 
2.8 Corollary: 
        Let M  be a p rime R-module and N, K be two submodules of M  such that N is an almost 
bounded submodule of M . Then NK is also almost bounded submodule of M . 
Proof:  It is know that NKN. So according to p rop osition (2.7), NK is an almost 
bounded submodule of M . 
        As a generalization of corollary  (2.8), we give the following corollary . 
2.9 Corollary: 

        Let M  be a p rime R-module and ni i 1{N }   be a finite collection of submodules of M  such 

that Ni is an almost bounded submodule of M  for some i, i=1,2,…,n. Then 
n

i
i 1

N

  is also 

almost bounded submodule f M . 
Proof:  The proof is by  induction on n and corollary  (2.8). 
        The following examp le shows that the intersection of an infinite collection of almost 
bounded submodules of M  need not be almost  bounded submodule of M . 
2.10 Example: 
        Consider Z as a Z-module, Z is p rime Z-module. Since pZ is an almost  bounded of Z, 
for each p where p is a prime number. However 

p isp rime
pZ =0 is not almost bounded submodule 

of Z. 

 
Re ferences: 
1. Ammen, Sh.A., (2002), Bounded M odules, M .D.T hesis, University  of Baghdad. 
2. Abdul-Razak, H.M ., (1999), Quasi-Prime M odules and Quasi-Prime Submodules, M .D. 

Thesis, University  of Baghdad. 
3. Ansari-Toroghy , H. and Farshadifar, F., (2008), On Endomorp hisims of M ultip lication 

and Comultip lication M odules, Archivum M athematicum (BRNO), T omus 44, 9-15. 
4. Abass, M .S., (1990), On Fully  Stable M odules, Ph.D. T hesis University  of Baghdad. 
5. Desale, G.,and Nicholson, K.W., (1981), Endomorphisim Rings, J. Algebra,.70: 548-560. 
6. Ebrahimi, At ani, S., (2008), On Generalized Dist inguished Prime Submodules, Thai 

Journal of M athematics,.6(2): 369-376. 
  
  

IHJPAS



  2010) 2( 23مجلة ابن الھیثم للعلوم الصرفة والتطبیقیة               المجلد

  

  حول المقاسات الجزئیة المقیدة تقریبا ً 
  
  

  بثینة نجاد شھاب
جامعة بغداد ،ابن الھیثم  - كلیة التربیة،قسم الریاضیات   

 
 
 

  الخالصة
فـي هــذا البحـث قــدمنا . Rمقاسـاً  احادیــاً  أیسـراً  علــى الحلقـة  Mعنصـر محایــد، ولـیكن  يحلقـة ابدالیــة ذ Rلـتكن         

مقیـد تقریبــاً  اذا وجـد عنصــر  Mمـن المقــاس  Nیطلــق علـى المقــاس الجزئـي : اتيمفهـوم مقـاس جزئــي مقیـد تقریبــاً  كمـا یـ

xM  وxN  بحیث انannR(N)=annR(x) .ئج في هذا البحث، اعطیت بعض الخـواص وكـذلك ُدرسـت العدیـد مـن النتـا

الـى هـذا ُدرســت بعـض العالقــات بینـه وبـین انــواع اخـرى مــن  فضـال عــن. االساسـیة حـول المقاســات الجزئیـة المقیـدة تقریبــا ً 

  .المقاسات

  
  

 

IHJPAS