Mathematics - 393 
 

 مجلة إبن الهيثم للعلوم الصرفة و التطبيقية

 2012 السنة 25 المجلد 3 العدد

Ibn Al-Haitham Journal for Pure and Applied Science  

 No. 3 Vol. 25 Year 2012 

Strongly (Comletely) Hollow Submodules I 

 
I. M. A. Hadi and Gh. A.  Humod 
Department of Mathematics, College of Education Ibn-Al-Haitham, 
University of Baghdad 
Received in: 15 March 2012  Accepted in: 21 May 2012 
 
Abstract 
        Let R be a commutative ring with unity and let M be an R-module. In this paper we 
study strongly (completely) hollow submodules and quasi-hollow submodules. We investigate 
the basic properties of these submodules and the relationships between them. Also we study 
the be behavior of these submodules under certain class of modules such as compultiplication, 
distributive, multiplication and scalar modules. In part II we shall continue the study of these 
submodules. 
 
Key Words:  Strongly (completely)-hollow submodules, distributive modules, 
multiplication (comultiplication) modules, scalar modules. 
 
Introduction 
        Throughout this paper, all rings are commutative rings with identity elements, and all 
modules are unital modules. In this article we study strongly (completely) hollow submodules 
which are introduced in [1], also we introduce quasi-hollow submodules. In section one of 
this paper we give the basic properties of these submodules. Also we give some results under 
the class of distributive modules and compultiplication modules. In section two, we 
investigate some properties of strongly, completely and quasi-hollow submodules under the 
class of multiplication modules. In section three we introduce some properties of strongly 
(completely) and quasi-hollow submodules under certain class of modules. 
 
1- Strongly (Completely) Hollow and Quasi-Hollow Submodules 
        We begin this section with the following: 
 
 Definition: [1, 4.2] 
        Let 0 ≠ L ≤ M, then L is called a strongly hollow submodule (briefly, SH-submodule) if 
for every L1, L2 ≤ M with L ≤ L1 + L2 implies L ≤ L1  or  L ≤ L2, we say that an R-module M 
is a strongly-hollow module if M is a strongly hollow submodule of itself. 
 Remark:  
        Let 0 ≠ L ≤ M, L is a SH-submodule if for each L1, …, Ln ≤ M with L ≤ L1 + L2 + … + 
Ln, implies L ≤ L1  or  L ≤ L2 … or  L ≤ Ln. 
 Definition: [1, 4.2] 
        Let 0 ≠ L ≤ M, then L is called a completely hollow submodule (briefly, CH-submodule) 
if for any collection {Lλ}λ∈Λ of R-submodules of M with L Lλ

λ∈Λ
= ∑ , implies L = Lλ                     

for some λ∈Λ. 
        We say that an R-module M is completely hollow (briefly, CH-module) if M is 
completely hollow of itself. 
 Remarks and Examples:  



 
 
 

Mathematics - 394 
 

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 No. 3 Vol. 25 Year 2012 

The Z as Z-module is not SH, not CH, and every submodule is not SH, not CH. 
1. Z6 as Z-module is not SH, and every nonzero proper submodule is SH. 
2. Q as Z-module is not SH, since there exist two proper submodules A, B of Q such that            

Q = A + B see [2, p.187, Exc.6(b)]. 
3. Let M be an R-module, and N ≤ L ≤ M. 
If L is SH then N need not be SH-submodule. 
For example, 2< >  is SH (CH)-submodule of Z4 as Z-module. But 0< >  is not SH (not CH). 
4. Let M be an R-module, and 0 ≠ L ≤ W ≤ M. 
If L is SH-submodule, then W need not be SH-submodule. 
For example 6< >  is SH(CH)-submodule of Z48 as Z-module. But 2< >  is not SH (not CH), 
since 2< >  ⊆ 8< >  + 6< > , and 2< >  ⊈ 8< > , 2< >  ⊈ 6< > . 
5. Let M be an R-module, and L1, L2 ≤ M. If L1 and L2 are SH-submodule, then L1 + L2 need 
not be SH. 
For example: In Z12 as Z-module, 3 , 4< > < >  are SH-submodules of Z12. But 

123 4 Z< > + < >=  is not SH. 
6. If M is a chained R-module, and 0 ≠ N ≤ M. Then N is SH-submodule, where M is a 
chained module if the Lattic of submodules are linearly ordered by inclusion see [3]. 
Proof: 
        Let 0 ≠ N ≤ M. Assume N ⊆ N1 + N2 where N1, N2 ≤ M. Since M is chained, either N1 ⊆ 
N2  or  N2 ⊆ N1  
If N1 ⊆ N2, then N1 + N2 = N2, so N ⊆ N2. 
If N2 ⊆ N1, then N1 + N2 = N1, so N ⊆ N1. 
Thus N is SH-submodule. 
7. Every simple R-module M is SH and CH. 
8. Every simple submodule N of an R-module is CH-submodule. 
9. Every CH-module is SH-module. 
10. The concept SH-submodule and CH-submodule are independent 
For examples: 
(a) The Z-module 

p
Z ∞  is SH-submodule of itself; that is pZ ∞  is SH-module by Remark 

1.4 (7), 
p

Z ∞  is not CH-module. Since ip
i Z

1
Z Z

p
∞

+∈
= < + >∑ , and ip

1
Z Z

p
∞ ≠< + >  for                

any i ∈ Z+. 
(b) Let M be the vector space ℝ2 over ℝ. Let N = ℝ (1,0). N is simple submodule of M. 
Since dim N = 1. So by Remark 1.4 (9), N is CH. On the other hand, N⊆ℝ (1,1)+ℝ (1, - 
1)=ℝ2=M, and N ⊈ ℝ (1,1), N ⊈ ℝ (1, - 1). That is N is not SH-submodule. 
        As we have seen by Example 1.4 (11) (b), simple submodule need not be SH. However 
under the class of distributive (or comultiplication) modules, every simple submodule is SH. 
Before proving this result, recall that the following definitions 
        An R-module M is called distributive if the Lattic of its submodules is distributive, that 
is              L ∩ (N + K) = (L ∩ N) + (L ∩ K). 
Equivalently, L + (N ∩ K) = (L + N) ∩ ( L + K) for all submodules L, N, and K of M see [4]. 
 
        An R-module M is called comultiplication if every L ≤ M is of the form L = (O

M
: I) = 

M
ann I for some I ≤ R. Equivalently, L = (O

M
: (O

R
: L)) = 

M
ann  

R
ann L, see [5  ]. 

where (O
M
: I) = {m ∈ M: Im = (0)}, (O

R
: L) = {r ∈ R: rL = (0)}. 



 
 
 

Mathematics - 395 
 

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 2012 السنة 25 المجلد 3 العدد

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 No. 3 Vol. 25 Year 2012 

 
 
 Examples:  

1. 
p

Z ∞  as Z-module is comultiplication, since for each L ≤ pZ ∞  L = i
1

Z
p

< + > , then 

p
Z Z
ann ann L L

∞

=  for some i ∈ Z+. 

2. Z as Z-module is not comultiplication, since if L = 3Z, then 
Z

ann
Z

ann 3Z = Z ≠ 3Z. 

3. Zn as Z-module is comultiplication. 
Proof: 
        Let L ≤ M. Then L = m< >  and m/n, that is n = mk for some k ∈ Z. Hence 

Z
ann m< > =<k> and 

Z
ann k m L< >=< >= . Thus L = 

nZ Z
ann ann L . 

        Recall that a non-zero submodule N of an R-module M is said to be second submodule 
of M if for each r ∈ R, the homothety  r*  on N is either zero or surjective. Equivalently, rN = 
<0>  or  rN = N for each r ∈ R, see [6]. 
where the homothety r* is an R-endomorphism on N, means r*(x) = rx for each x ∈ N. 
 
        A submodule N of an R-module M is said to be strongly irreducible (briefly, SI-
submodule) if for any L1, L2 ≤ M, L1 ∩ L2 ⊆ N, then L1 ⊆ N  or  L2 ⊆ N, see [7]. 
Examples:  
1. 6Z is not SI-submodule of Z as Z-module since 6Z ⊇ 2Z ∩ 3Z, but 6Z ⊉ 2Z, 6Z ⊉ 3Z. 
2. It is clear that every submodule of chained module is SI. 
We state the following proposition which is needed in the next two results. 
Proposition:  
        Let M be a comultiplication R-module, and N ≤ M such that 

R
ann N is prime ideal. Then 

N is a SH-submodule. 
Proof: 
        Let N ⊆ L1 + L2, where L1, L2 ≤ M. Since M is comultiplication, L1 = 

M
ann I1, L2 = 

M
ann I2 

for some ideals I1 and I2 of R. Then N ⊆ 
M

ann I1 + 
M

ann I2 ⊆ 
M

ann (I1 ∩ I2), that is N ⊆ 
M

ann (I1 ∩ 

I2). So 
R

ann N ⊇ 
R

ann
M

ann (I1 ∩ I2) ⊇ I1 ∩ I2. But 
R

ann N is prime so 
R

ann N is SI-ideal, hence 

R
ann N ⊇ I1  or  

R
ann N ⊇ I2. Then 

M
ann

R
ann N ⊆ 

M
ann I1 = L1  or  

M
ann

R
ann N ⊆ 

M
ann I2 = L2. So N 

⊆ L1  or               N ⊆ L2, that is N is SH. 
        The following result is given in [1]. However we get it directly by Proposition 1.7. 
Corollary:  
        Let M be a comultiplication R-module, and N ≤ M. Then  
1. N is a second submodule implies N is SH. 
2. N is a finitely generated second submodule, implies N is CH. 
Proof: 
(1) Since N is second, then 

R
ann N is a prime ideal by [6]. Hence the result is obtained by 

Proposition 1.7. 

(2) By part (1) N is SH. But N is finitely generated, so 
n

i
i 1

N Rx
=

= ∑  for some x1, …, xn. Hence          

N ⊆ Rxi for some i = 1, …, n. But Rxi ⊆ N. Thus N = Rxi. 



 
 
 

Mathematics - 396 
 

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 2012 السنة 25 المجلد 3 العدد

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 No. 3 Vol. 25 Year 2012 

 Corollary:  
        Let M be a comultiplication R-module, and let N be a simple submodule. Then N is SH. 
Proof: 
       It is clear that every simple submodule is second, hence the result follows by                
corollary 1.8 (1). 
        Recall that an R-module M is said to be prime if 

R
ann M = 

R
ann N for every non-zero 

submodule N of M., see [8]. 
If M is a prime R-module, then 

R
ann M is prime by [8]. 

An R-module M is called a quasi-prime if 
R

ann N is a prime for each non-zero submodule N of 

M, see [9, Definition 1.2.1]. 
Notice that every prime R-module M is quasi-prime by [9, Remark 1.2.2]. 
 Corollary:  
        Let M be a comultiplication prime (or quasi-prime) R-module. Then every non-zero 
submodule is SH. 
Proof: 
       Since M is prime (or quasi-prime) implies 

R
ann N is prime ideal for each non-zero 

submodule N of M. Hence the result follows from Proposition 1.7. 
Proposition:  
        Let M be a distributive R-module, and <0> ≠ N ≤ M. If N is a simple submodule of M, 
then N is SH. 
Proof: 
       Assume N is simple, N ≤ L1 + L2 where L1, L2 ≤ M. Hence  
N = N ∩ (L1 + L2) 
    = (N ∩ L1) + (N ∩ L2), since M is distributive. 
Then (N ∩ L1 = <0>  or  N ∩ L1 = N) and (N ∩ L2 = <0>  or  N ∩ L2 = N). But N ≠ 0. So we 
have only three possible cases 
(1) N ∩ L1 = <0>, N ⊆ L2. 
(2) N ∩ L2 = <0>, N ⊆ L1. 
(3) N ⊆ L1, N ⊆ L2. 
Thus either N ⊆ L1  or  N ⊆ L2; that is N is SH. 
 Remark:  
        The condition M is distributive or comultiplication is necessary condition in Proposition 
1.11 and Corollary 1.9. 
As we have seen in Remark 1.4(11)(b), N = ℝ(1,0) and N is simple but not SH. Moreover the 
vector space ℝ2 over ℝ is not distributive since ℝ2 = ℝ(1,1) + ℝ(1,–1) and N ∩ ℝ2= N, but                  
(N ∩ ℝ(1,1)) + (N ∩ ℝ(1, –1)) = {(0,0)}. Thus ℝ2 is not distributive. 
Also ℝ2 is not comultiplication R-module. For if L = ℝ(1,1), then ann


L = {0} and 

2

2ann{0} =


 , thus L ≠ 
2

ann ann L


. 

        Now we introduce the following concept. 
 Definition:  
        Let <0> ≠ L ≤ M, L is called a quasi-hollow submodule (briefly qH-submodule) if for 
each L1, L2 ≤ M with L = L1 + L2, then L = L1  or  L = L2. 
An R-module M is said a quasi-hollow module if M is a quasi-hollow submodule. 
 Remark:  



 
 
 

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 No. 3 Vol. 25 Year 2012 

        Let <0> ≠ L ≤ M, L is a quasi-hollow submodule if for each L1, …,Ln with L = L1 + 
…+Ln, then L = L1  or  … or L = Ln. 
 
 
 
 Remarks and Examples:  
1. It is clear that every CH-submodule is qH-submodule. The converse is not true. For 
example the Z-module 

p
Z ∞  is qH-module (qH-submodule of itself) since there is no L1≨M 

and L2≨M such that pZ ∞  = L1 + L2. But by Remark 1.4(11)(b) pZ ∞  is not CH. 

2. Every simple submodule of an R-module is qH-submodule. 
3. Every SH-submodule is qH-submodule. 
The converse is not true in general, for example in the vector space ℝ2 over ℝ,                          
N = ℝ(1,0) is simple submodule, so by Remark 1.15(2), N is qH, but it is not SH by            
Remark 1.4(11)(b). 
4. If M is chained, then every submodule is qH. 
Proof:  
It follows by Remark 1.4(7) and Remark 1.15(3). 
5. Let M be an R-module. Then M is a qH-module if and only if M is SH if and only if M is 
hollow. 
where M is hollow if every proper submodule N of M is small. 
That is there is no proper submodule W of M such that N + W = M. 
Equivalently, for every submodules N, W such that N ≨ M, W ≨ M implies N + W ≨ M. 
6. Let M be CH (qH or SH) R-module, then there is no submodules N, W of M such that           
M = N ⊕W. 
7. Consider Z48 as Z-module. Each of 2 , 4 , 8< > < > < >  and Z48 is not qH, not SH. Each of 

3 , 6 , 12 , 24< > < > < > < >  is qH and SH. 
8. Consider M = Z4⊕Z2 as Z-module. 
Each of 2 40 Z , Z 0 , 2 0< > ⊕ ⊕ < > < > ⊕ < >  is qH and SH, and each of 2 4 2Z Z , 2 Z⊕ < > ⊕  
is not qH, not SH. 
9. Let <0> ≠ L ≤ M as R-module. Let N ≤ L. If L is qH-submodule, then N need not be qH. 
For example, Z-module Z48 where 3< >  is qH, but 0< >  is not qH. 
10. Let <0> ≠ L ≤ W ≤ M as R-module. If L is qH, then W need not be qH. For example,               
M = Z4⊕Z2 as Z-module, where 20 Z< > ⊕  is qH and 2 20 Z 2 Z< > ⊕ ⊆< > ⊕ . But 

22 Z< > ⊕  is not qH. 
11. If L1, L2 are qH of an R-module M, then L1 + L2 need not be qH. For example, 

1 2 2 4L 0 Z , L Z 0=< > ⊕ = ⊕ < >  are qH of M = Z4⊕Z2 as Z-module. But L1 + L2 = M is        
not qH. 
12. Let R be a ring. If A and B are qH(SH)-ideals. Then AB need not be qH(SH)-ideals of R. 
For example 2< >  and 3< >  are qH(SH)-ideals of the ring Z6. But 2 3 0< > ⋅ < >=< >  is not 
SH, not qH. 
        Now we find that under the class of distributive of modules, the concepts, qH-
submodules and SH-submodules are equivalent, as the following proposition shows: 
 
 Proposition:  
        Let M be distributive R-module, and 0 ≠ N ≨ M. Then N is SH-submodule if and only if 
N is qH-submodule. 



 
 
 

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Proof:  
(⇒) Clear by Remark 1.15(3). 
(⇐) Assume N is qH-submodule. Let N ⊆ L1 + L2 where L1, L2 ≤ M. Then N = N ∩ (L1 + 
L2), so N = (N ∩ L1) + (N ∩ L2), since M is distributive. Then N = N ∩ L1  or  N = N ∩ L2 
since N is qH. It follows that either N ⊆ L1  or  N ⊆ L2. Hence N is a SH-submodule. 
 
 
 Corollary:  
        Let M be distributive R-module, and <0> ≠ N ≨ M. If N is CH-submodule, then N is SH. 
Proof: 
        It follows by Remark 1.15(1) and previous proposition. 
 Remark:  
        Let M be an R-module, N ⊆ K ⊆ M. If N is SH(qH)-submodule in M, then N is SH(qH)           
in K. 
Proof:  It is clear 
        The converse of this remark is true under the class of distributive module as follows: 
 Proposition:  
        Let M be a distributive R-module. Let N ⊆ K ⊆ M. Then N is SH(qH)-submodule in M 
if and only if N is SH(qH) in K. 
Proof:   
(⇒) It follows by previous remark. 
(⇐) Assume N is SH-submodule in K. Let N ⊆ L1 + L2 where L1, L2 ≤ M. Since N ⊆ K then        
N = N ∩ K ⊆ (L1 + L2) ∩ K 
                   = (L1 ∩ K) + (L2 ∩ K), since M is distributive 
So N ⊆ (L1 ∩ K) + (L2 ∩ K). Then N ⊆ L1 ∩ K  or  N ⊆ L2 ∩ K, since N is SH in K. 
Then N ⊆ L1  or  N ⊆ L2. Thus N is SH in M.  
By a similar proof, if N is qH in K, then N is qH in M. 
        Now we turn our attention to image and inverse image of SH, qH and CH-submodules. 
 Proposition:  
        Let M and M' be R-modules and N be a SH-submodule of M. If  f : M → M' be an R-
epimorphism, then f (N) is SH-submodule of M'. 
Proof: 
        Let f (N) ⊆ L1 + L2 where L1, L2 ≤ M'. Then N ⊆ f  - 1f (N) ⊆ f  - 1(L1 + L2). But  
f  - 1(L1 + L2) = f  - 1(L1) + f  - 1(L2)  see [2,3.1.10(c)], so N ⊆ f  - 1(L1) + f  - 1(L2), then N ⊆ f  - 
1(L1)  or N ⊆ f  - 1(L2). It follows f (N) ⊆ f f – 1(L1) = L1  or  f (N) ⊆ f f – 1(L2) = L2. Hence f (N) 
⊆ L1  or             f (N) ⊆ L2. 
        The condition f is an epimorphism is necessary in Proposition 1.20, for example, Let              
f : Z12 → Z12, f (x) = 4x fo each x ∈ Z12, where Z12 considered as Z-module. It is clear that f 
is not epimorphism. Let N = 3< > , N is a SH submodule of Z12. But f (N) = 0< >  is not SH. 
1.21 Corollary:  
        Let N be a SH-submodule of an R-module M. Let L ≨ N, then N/L is SH-submodule of 
M/L. 
 Corollary:  
        Let M ≅ M' be R-module, if N ≤ M. Then N is a SH-submodule of M iff f (N) is a SH-
submodule of M'. 
 Proposition:  
        Let M and M' be R-modules and f : M → M' be an isomorphism, Let <0> ≠ N ≤ M. If 
N is qH(CH)-sbmodule of M, then f (N) is qH (CH)-submodule of M'. 



 
 
 

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Proof: 
        If N is qH-submodule of M. Assume f (N) = W1 + W2 for some W1, W2 ≤ M'. Since f is 
isomorphism, so W1 = f (L1), W2 = f (L2) for some L1, L2 ≤ M. Thus f (N) = f (L1) + f (L2). But           
f (L1 + L2) = f (L1) + f (L2), see [2, 3.1.10(a)]. Then f (N) = f (L1 + L2). Since f is 
monomorphism, we get   N = L1 + L2. It follows that N = L1  or  N = L2. Hence f (N) = f (L1) 
= W1  or                        f (N) = f (L2) = W2. By a similar proof, N is CH-submodule of M 
implies f (N) is CH of M'. 
 
 Proposition:  
        Let f : M → M' be an isomorphism R-module. If K is SH(qH or CH)-submodule of M', 
then f -1(K) is SH(qH or CH)-submodule of M. 
Proof: 
        Assume K is SH in M'. Let f -1(K) ⊆ L1 + L2 where L1, L2 ≤ M. Then f f -1(K) ⊆ f (L1 + 
L2) = f (L1) + f (L2), see [2,3.1.10(a)]. Since f is epimorphism K = f f -1(K) ⊆ f (L1) + f (L2). So              
K ⊆ f (L1)  or   K ⊆ f (L2). Thus f -1(K) ⊆ f f -1(L1) = L1  or  f -1(K) ⊆ f f -1(L2) = L2. Since f is 
monomorphism. Hence  f -1(K) ⊆ L1  or  f -1(K) ⊆ L2. 
By a similar proof, K is qH(CH) of M', then f -1(K) is qH(CH). 
        The condition that f is an isomorphism is necessary in Proposition 1.24. For example, 
Consider the Z-module Z and let π: Z → Z/<4> ≃ Z4 be the natural projection. Let                         
K = 2< >  ⊆ Z4, K is SH(qH or CH) of Z4. But π - 1(k) = 2Z is not SH (not qH, not CH) in Z. 
        Now we give the next result of this section. 
 Proposition:  
        Let M1, M2 be R-modules. Let M = M1⊕M2, and let <0> ≠ K ⊆ M1⊕M2. If K = N ⊕ W 
for some N ≤ M1, W ≤ M2 such that K is SH(qH)-submodule. Then N is SH(qH) of M1 and W 
is SH(qH) of M2. 
Proof: 
        Assume K = N ⊕ W, K is a SH submodule in M, K = (N ⊕ <0>) + (<0> ⊕ W), so                  
K = N ⊕ <0>  or  K = <0> ⊕ W since K is SH of M. If K = N ⊕ <0>. We claim that N is 
SH of M1. Assume N ⊆ L1 + L2 where L1, L2 ≤ M1. So K ⊆ (L1 + L2) ⊕ <0>. 
Then K ⊆ (L1⊕ <0>) + (L2⊕ <0>). Then K ⊆ L1⊕ <0>  or  K ⊆ L2⊕ <0>. Thus                               
N ⊕ <0> ⊆ L1⊕ <0>  or  N ⊕ <0> ⊆ L2⊕ <0>. Hence N ≤ L1  or  N ≤ L2. Then N is SH of 
M1. 
Similarly, if K = <0> ⊕ W, then W is SH of M2. 
By a similar proof, if K is qH, then N, W are qH in M1, M2 respectively. 
        The converse of Proposition 1.25 is not true in general. 
For example in Z-module M = Z4⊕Z6. If K = 2 3< > ⊕ < > , then K is not SH (not qH). But 

2< >  is SH(qH) in Z4 and 3< >  is SH(qH) in Z6. 
2- SH and qH(CH)-Submodules and Multiplication Modules 
 
        In this section, we introduce some properties of SH and qH(CH) submodles in the class 
of multiplication modules. 
Recall that an R-module M is called multiplication if every N ≤ M, N is of the form N = IM 
for some ideal I ≤ R. Equivalently, N = (N

R
: M)⋅M, where (N

R
: M) = {r ∈ R, rM ⊆ N}, see 

[4]. 
Proposition:  
        Let M be a faithful finitely generated multiplication R-module, N ≤ M. Then the 
following statements are equivalent: 
(1) N is SH(qH)-submodule. 



 
 
 

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(2) (N
R
: M) is SH(qH)-ideal. 

(3) N = IM, I is SH(qH)-ideal for some <0> ≠ I ≤ R. 
Proof: 
(1) ⇒ (2) Assume N is a SH-submodule of M, and let (N

R
: M) ⊆ I1 + I2 where I1, I2 are ideals 

of R. Then (N
R
: M)⋅M ⊆ (I1 + I2)⋅M. But (I1 + I2)⋅M = I1M + I2M since M is finitely generated.           

It follows (N
R
: M)⋅M ⊆ I1M + I2M. But (N

R
: M)⋅M = N since M is multiplication. Then                  

N ⊆ I1M + I2M, so either N ⊆ I1M  or  N ⊆ I2M, that is (N
R
: M)⋅M ⊆ I1M  or  (N

R
: M)⋅M ⊆ 

I2M. Thus (N
R
: M) ⊆ I1  or (N

R
: M) ⊆ I2 by [10, Theorem 3.1]. Hence (N

R
: M) is a SH-ideal of 

R. 
By a similar proof, (N

R
: M) is qH-ideal if N is qH. 

(2) ⇒ (3) Assume (N
R
: M) is SH-ideal. Put I = (N

R
: M). Since M is multiplication, then               

N = (N
R
: M)⋅M. Hence N = IM, and I is a SH-ideal. Similarly I is a qH-ideal. 

(3) ⇒ (1) Assume that N = IM for some SH-ideal I of R, and let N ⊆ L1+L2 where L1, L2 ≤ 
M. But L1 = I1M, L2 = I2M since M is multiplication for some ideals I1, I2 of R. So                                   
IM ⊆ I1M+I2M ⊆ (I1 + I2)⋅M, since M is finitely generated. Then IM ⊆ (I1 + I2)⋅M, so I ⊆ I1 + 
I2 by [10, Theorem 3.1]. 
So that either I ⊆ I1  or  I ⊆ I2, which implies IM ⊆ I1M  or  IM ⊆ I2M. Hence N ⊆ L1  or  N ⊆ 
L2. Thus N is SH. 
Similarly N is qH. 
        The condition M is faithful is necessary in Proposition 2.1. Fot example, the Z-module 
Z6 is finitely generated multiplication, but not faithful, let N = 2< > , N is SH of Z6,but                     
(N

Z
: M) = 6

Z
( 2 : Z )< >  = 2Z is not SH of Z. 

Corollary:  
        Let M be a finitely generated faithful multiplication R-module. Then every non-zero 
submodule of M is SH(qH) if and only if every non-zero ideal of R is SH(qH). 
 Proposition:  
        Let M be a faithful finitely generated multiplication R-module. Then R satisfies acc(dcc) 
on SH-ideals if and only if M satisfies acc(dcc) on SH-submodules. 
Proof: 
⇒ We take the case of acc. 
Let L1 ⊆ L2 … be an ascending chain of SH-submodules of M. Since Li is SH-submodule, 
then (Li

R
: M) is SH-ideal for each i = 1,2,… by Proposition 2.1, and (L1

R
: M) ⊆ (L2

R
: M) ⊆ … 

by [10,Theorem 3.1]. But R satisfies acc on any ascending chain of SH-ideals. So ther exists n 
∈ Z+ such that (Ln

R
: M) = (Ln + 1

R
: M) = …. Then (Ln

R
: M)⋅M = (Ln + 1

R
: M)⋅M = ….. Thus                            

Ln = Ln + 1 = … for some n ∈ Z+. Hence M satisfies acc on SH-submodules. 
⇐ The proof is similar. 
SH, qH(CH)-Submodules and Other Related Concepts 
        Recall that an R-module M is called scalar if for each f ∈ EndR(M), there exists r ∈ R 
such that f (x) = rx for all x ∈ M, see [11]. 
Proposition:  
        Let M be a scalar R-module and R is SH-ring, then EndR(M) is SH-ring. 



 
 
 

Mathematics - 401 
 

 مجلة إبن الهيثم للعلوم الصرفة و التطبيقية

 2012 السنة 25 المجلد 3 العدد

Ibn Al-Haitham Journal for Pure and Applied Science  

 No. 3 Vol. 25 Year 2012 

Proof: 
        Since M is a scalar R-module, then EndR(M) ≃ R/

R
ann M, see [12, Lemma 3.6.1]. Since 

R is SH-ring, then R/
R

ann M is SH-ring by Corollary 1.21. Thus EndR(M) is SH-ring by                  

Corollary 1.22. 
 Corollary:  
        Let M be a finitely generated multiplication module over SH-ring. Then EndR(M) is SH-
ring. 
Proof: 
        Since M is finitely generated multiplication, then M is scalar, see [11]. Hence the result 
is obtained by Proposition 3.1. 
        Next we shall prove in the class of comultiplication prime modules, every submodule of 
M is SH (qH)-module. But first we prove the following proposition and lemma. 
Proposition:  
        Let M be a comultiplication R-module. If M is prime (quasi-prime or second). Then M is 
hollow. 
Proof: 
        Since M is prime (quasi-prime or second), then 

R
ann M is prime, see [8], [9], [6]. And 

EndR(M) is domain, see [5, Corollary 3.21]. Hence M is hollow, see [5, Theorem 3.24]. 
 Lemma:  
        Let M be a comultiplication R-module and N ≤ M. Then N is a comultiplication R-
module. 
Proof: 
        Let W ≤ N. So W is a submodule of M. Then there exists I ≤ R such that W = ann

Μ
I. We 

claim that W = 
N

ann I. To prove our assertion. Let m ∈ W (so, m ∈ N). Hence mI = 0, so               

m ∈ 
N

ann I. Now let m ∈ 
N

ann I, so m ∈ N and mI = 0, m ∈ M. Thus m ∈ ann
Μ

I. Then w = 

N
ann I. Henc N is comultiplication. 

Theorem:  
        Let M be a comultiplication prime R-module. Then every non-zero submodule of M is a 
SH(qH) R-module. 
Proof: 
        Since M is comultiplication prime, then by Proposition 3.3, M is hollow. Let N be a non-
zero submodule of M, then N is comultiplication by Lemma 3.4. But M is prime implies N is 
a prime R-module. Thus N is a hollow R-module by Proposition 3.3. Hence N is qH(SH)-R-
module, see Remark 1.15(5). 
 Corollary:  
        Let M be a comultiplication prime R-module. Then every non-zero submodule of M is 
qH-sumodule of M. 
 
References 
1. Abuhlail, J.Y. (2011) Zariski Topologies for Coprime and Second Submodules, Deanshib 

of Scientific Research at King Fahad University of Peroleum and Minerals February 4.  
2. Kasch, F. (1982) Modules and Rings, Acadimic Press, London. 
3. Osofsky, B.L. (1991) A Contruction of Non Standard Uniserial Modules Over Valuation 

Domains, Bulletin Amer. Math.Soc., 25:89-97. 
4. Barnard, A. (1981) Multiplication Modules, J.Algebra, 71:174-178. 



 
 
 

Mathematics - 402 
 

 مجلة إبن الهيثم للعلوم الصرفة و التطبيقية

 2012 السنة 25 المجلد 3 العدد

Ibn Al-Haitham Journal for Pure and Applied Science  

 No. 3 Vol. 25 Year 2012 

5. Ansari Toroghy, H. and Farshadifar, F. (2007) The Dual Notion of Multiplication Modules, 
Taiwanese J.Math., 11(4):1189-1201. 

6. Yassemi, S. (2001) The Dual Notion of Prime Submodules, Arch.Math. (Brno) 37:273-278. 
7. Bourbaki, N. (1998) Commutative Algebra, Springer-Verlag. 
8. Desale, G. and Nicholson,W.K. (1981) Endoprimitive Rings, J.Algebra, 7:546-560. 
9. Muntaha. A.AL. (1999) Quasi-Prime Modules and Quasi-Prime Submodules, M.Sc. 

Thesis, College of Education Ibn-Al-Haitham, University of Baghdad. 
10. Abd El-Bast, Z. and Simith, P.F. (1988) Multiplication Modules, Commun.Algebra, 16 

(4):755-779. 
11. Najad, B. (2004) Scalar Reflexive Modules, Ph.D. Thesis, College of Science, University 

of Baghdad. 
12. Eman, A. (2006) On Ikeda-Nakayama Modules, Ph.D. Thesis, College of Education Ibn-

Al-Haitham, University of Baghdad. 
 
 

 
 
 
 

 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 



 
 
 

Mathematics - 403 
 

 مجلة إبن الهيثم للعلوم الصرفة و التطبيقية

 2012 السنة 25 المجلد 3 العدد

Ibn Al-Haitham Journal for Pure and Applied Science  

 No. 3 Vol. 25 Year 2012 

  Iالمقاسات الجزئية المجوفة (التامة) بقوة  
 
 

 غالب أحمد حمود هادي، انعام محمد علي
، جامعة بغداد ابن الهيثم -قسم الرياضيات ، كلية التربية  

  2012ايار  21قبل البحث في:   2012اذار 15في :استلم البحث 
 

 الخالصة
. في هذا البحث درسنا المفاهيم: المقاسات الجزئية  Rاسا ً على مق Mحلقة ابدالية ذا محايد. وليكن  Rلتكن             

المجوفة بقوة (التامة) والمقاسات الجزئية شبه المجوفة وقدمنا الخواص المتعلقة بهم والعالقات فيما بينهم. كذلك درسنا 
لمقاسات التوزيعية، والمقاسات الجدائية المضادة، سلوك هذه المقاسات الجزئية في أصناف معينة من المقاسات ، مثل ا

 والمقاسات الجدائية والمقاسات القياسية.
 

المقاس���ات الجزئي���ة المجوف���ة (التام���ة) بق���وة، المقاس���ات التوزيعي���ة، المقاس���ات الجدائي���ة (الجدائي���ة   الكلم���ات المفتاحي���ة:
 المضادة)، المقاسات القياسية.

 
 


	Received in: 15 March 2012  Accepted in: 21 May 2012
	Introduction
	قسم الرياضيات ، كلية التربية- ابن الهيثم ، جامعة بغداد
	لتكن R حلقة ابدالية ذا محايد. وليكن M مقاسا ً على R . في هذا البحث درسنا المفاهيم: المقاسات الجزئية المجوفة بقوة (التامة) والمقاسات الجزئية شبه المجوفة وقدمنا الخواص المتعلقة بهم والعلاقات فيما بينهم. كذلك درسنا سلوك هذه المقاسات الجزئية ف...