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

  

On Almost Bounded Submodules 
 
 

B. N.Shihab  
Department of Mathematics, College of EducationIbn-Al-Haitham , Univers 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 paper, some properties of almost bounded submodules are given. Also, various 
basic results about almost bounded submodules are considered. 
        Moreover, some relations between almost bounded submodules and other types of 
modules are considered. 

 

Introduction 

        Every ring considered in this paper 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 paper is to study almost bounded submodules and to look for any 
relation between almost bounded submodules and certain types of well-known modules 
especially with prime modules. 
        This paper consists of two sections. Our main concern in section one, is to define and 
study almost bounded submodules. Also, we give some basic results for this concept. 
        In section two, we study the relation between almost bounded submodules and bounded 
modules. We show that the proper submodule of bounded module is not necessary to be 
almost bounded submodule and we give some conditions under which a proper submodule of 
bounded module is an almost bounded submodule. Next we investigate the relationships 
between almost bounded submodules, prime and fully stable module. 
1- Basic Properties of Almost Bounded Submodules  

        In this section, we introduce the concept of almost bounded submodule.     We 
establishe some basic properties of this concept. 
        First, we introduce the following definition. 
1.1 Definition: 
        A proper 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, prime module, quasi-prime 
module, fully stable module. 

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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 prime number and the Z-suubmodule 
N=qZZp, where q is any prime number. Then N an almost bounded sumodule. 

4. For each positive integer n and n is not prime number, every proper 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 positive 
integer, then the Z-module Zn has no proper almost bounded submodule. 

6. Let p be a prime number. The Z-module Zp dose not contain any proper almost bounded 
submodule. 
The following remark ensures that the almost boundedness property  is not hereditary. 

1.3 Remark: 
        A submodule of an almost bounded submodule need not be almost bounded in general. 
For example: 
M=ZZp as a Z-module, where p any prime number, N=qZZp be a submodule of M, where 
q is any prime 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 state and prove the following proposition. 
1.4 Proposition: 
        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 M2 respectively, 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 respectively. 
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 proposition (1.4) is not true in general as the following example shows. 
1.5 Example: 

        Consider M=Z6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 M2 respectively. 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 M1. 
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 Mn respectively, then N= 
N1N2…Nn is an almost bounded submodule of M. 
        So, we have the following applications 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), implies (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  
 

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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 implies ra=0 
for all aN1 and rb=0 for all bN2 which implies rN1=0 and rN2=0. Thus rannRN1 and 
rannRN2. But annRN1=annR(x) and annR(y)=annRN2. This implies 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 completes the proof. 
        The following corollary is a special case of proposition (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 M2=MM. 
Proof:  From hypothesis 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)N2=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 proposition: 
1.9 Proposition: 
        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 prove our assumption. 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 implies 
r(x,y)=(0,0) and hence rannR(x,y). 
     Conversely, let rannR(x,y). Then r(x,y)=(0,0), which implies (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 prime numbers and M=ZZp. But Zp is not almost bounded because Zp has no proper 
almost bounded submodule. 
        We have seen by the following proposition that the class of almost bounded submodule 
is closed under homomorphic image and inverse image. 
1.11 Proposition: 
        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 epimorphisim, 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 

implies (rx)=0. Thus r(x)=0. This means rannR((x))=annR(y)=annRN'. Thus rN'=0, which 
implies –1(rN')=0. Then r–1(N')=0 and implies rannR(–1(N')). 
 

 

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

–1
(N')). Then r

–1
(N')=0, which implies                   

–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 implies (rx)=0. Then rx=0 and hence rannR(x). Thus 
annR(x)=annR(–1(N')) which completes the proof. 
2. Suppose 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). Suppose that 
(x)(N). Then (x)= (n) for some nN, which implies 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 implies (rx)=0. Thus rx=0, that is rannR(x)=annRN. Then 
rannRN, which implies 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 prove the other inclusion. 
Hence annR((N))= annR((x)) which is what we wanted.  
        The condition (: M  M' is an isomorphism) in proposition (1.11) can not be 
dropped as the following example shows. 
1.12 Example: 
1. Let :ZZ4Z4 be a projection 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 examples (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 proper 
almost bounded submodule. Since (

–1
(N') is a submodule of Z4, then (

–1
(N') is not almost 

bounded submodule of Z4 by (remarks and examples               (1.2) (5)). 
2- Modules Related to Almost Bounded Submodules  

        In this section, we study the relationships between almost bounded submoduls and 
bounded modules, p rime and fully stable modules. 
        We start with the 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 

example 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 
prime 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 proposition gives a sufficient condition under which every submodule of a 
bounded module is an almost bounded. 
2.2 Proposition: 
        Let M be a cyclic quasi-prime R-module and N be a proper essential submodule of M. 
Then N is an almost bounded submodule. 
Proof:  Assume that N is p roper 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 implies 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. 

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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 prime ideal of R. Let N be a proper 
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 
proposition (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 stable in case each submodule N of M is stable, where a 
submodule N is said to be stable, if f(N)N for each R-homomorphism f:NM, [4]. 
       So, we have the following proposition. 
2.4 Proposition: 
        Let N be a proper submodule of an R-module M such that, 
1. M is fully stable and bounded R-module. 
2. [N

R
: M]  annRM. 

3. annRM is prime 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], implies R[(x)

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

prove 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, implies 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 the other hand, let rannRN. Then rN=0 but M is multiplication [1,corollary (1.1.9)], 

then r[N
R
: M]M=0 implies r[N:M]annRM. But annRM is prime 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 prime ideal can not be dropped from 

proposition (2.4) as in the following example. 
2.5 Example: 
        Let M=Z6 as a Z-module. Since M is bounded Z-module, see [1] and M is fully stable Z-
module, see [4,example and remarks (3.7),(c)], but annZM=6Z is not prime 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-multiplication if each submodule of M is of the form AM  
for some idempotent ideal A of R, [4]. 
        As an immediate consequence of proposition (2.4). 
2.6 Corollary: 
        Let N be a proper submodule N of an R-module M such that: 
1. M is I-multiplication bounded module 

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

2. annRM is prime 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 proposition (2.4). 
        Recall that an R-module M is called a prime module if annRM= annRN for every non-
zero submodule N of M, [5], [6]. 
 
 
2.7 Proposition: 
        Let M be a prime 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 prove annRN= 
annR(x). annRKannRN (since NKM), implies annR(x) annRN. Hence annR(x)annRN. 
      Now, let rR, rannRN= annRM for each submodule N of M (since M is prime 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 application of (2.7). 
2.8 Corollary: 
        Let M be a prime 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 proposition (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 prime R-module and n
i 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 example 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 prime Z-module. Since pZ is an almost bounded of Z, 

for each p where p is a prime number. However 
pisp rime

pZ =0 is not almost bounded submodule 

of Z. 

 
References: 
1. Ammen, Sh.A., (2002), Bounded Modules, M .D.Thesis, University  of Baghdad. 
2. Abdul-Razak, H.M ., (1999), Quasi-Prime Modules and Quasi-Prime Submodules, M.D. 

Thesis, University  of Baghdad. 
3. Ansari-Toroghy, H. and Farshadifar, F., (2008), On Endomorphisims of Multiplication 

and Comultiplication Modules, Archivum Mathematicum (BRNO), Tomus 44, 9-15. 
4. Abass, M.S., (1990), On Fully Stable Modules, Ph.D. Thesis University  of Baghdad. 
5. Desale, G.,and Nicholson, K.W., (1981), Endomorphisim Rings, J. Algebra,.70: 548-560. 
6. Ebrahimi, Atani, S., (2008), On Generalized Distinguished Prime Submodules, Thai 

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

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  الخالصة
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ا یـ ـاً  كمـ ـاً  اذا وجـد عنصــر  Mمـن المقــاس  Nیطلــق علـى المقــاس الجزئـي : اتيمفهـوم مقـاس جزئــي مقیـد تقریبـ مقیـد تقریبـ

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

ـا ً  ـات الجزئیـة المقیـدة تقریبـ الـى هـذا ُدرســت بعـض العالقــات بینـه وبـین انــواع اخـرى مــن  فضـال عــن. االساسـیة حـول المقاسـ

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