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223 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

On 2-Absorbing Submodules  

 

Inaam M.Ali 
Abdurehman A. Harfash 

Dept. of Mathematics/ College of Education for Pure Science (Ibn Al Haitham)/                     
University of Baghdad 

Received in: 26/May/205,Accepted in:1/August/2015 

 

Abstract  
     Let R be a commutative ring with 1 0 and M is a unitary R-module . In this paper , our 
aim is to continue studying 2-absorbing submodules which are introduced by  A.Y. Darani 
and F. Soheilina . Many new properties and characterizations are given . 
 

Key words: prime submodule , 2-absorbing submodule , quasi- prime submodule, 
multiplication module , P-primary submodule , pure submodule . 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 

This paper is a part of thesis submitted by 2nd  author under supervision by 1st author  



 

224 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

Introduction 
        The concept of 2-absorbing ideal which is a generalization of prime ideal was introduced 
by Ayman Badawi  , where "a nonzero proper ideal I of R is called a 2-absorbing ideal of R if 
whenever a , b , c ∈R  , abc ∈ I then ab ∈ I or ac	∈I or bc ∈I" , [1] . This definition can 
obviously be made for any proper ideal  . A. Y. Darani and  F. Soheilnia in [2] introduced the 
concept of 2-absorbing submodule where "a proper submodule N of M is called 2-absorbing 
submodule of M if whenever  a , b ∈ R , m∈ M and abm ∈N , then am ∈N or bm ∈N  or         
ab ∈(N:M)" 
      Our concern in this paper is to give a comprehensive study of 2-absorbing submodule , 
where we give many new properties and characterizations . 

1. 2-Absorbing Submodules – Basic Properties   
"Definition 1.1:   

        Let M be an R-module. N a proper submodule  of M is called 2-absorbing submodule if 
whenever a,b ∈ R , m ∈ M and abm ∈ N, then ab ∈ (N	: M) or am ∈ N or bm ∈ N."[2] 

     Note that an 2-absorbing ideal of a ring R is a 2-absorbing submodule of the R-module R .     

Remarks and Examples.1.2 :   

(1) "The intersection of each pair of distinct prime submodules of R-module M is 2-
absorbing" [2]   

(2) It is clear that every  prime submodule  is 2-absorbing.  However the converse is not true 
in general, for example:  

Consider Z6  as Z-module , 0  not prime submodule of Z6 since    2 .3 	0	∈ 0   but  

3 ∉ 0 	 and 2 ∉		( 0 :
Z

 Z6)=6Z  but 0  = 2 ∩ 3  is 2-absorbing submodule of Z6 as Z-

module by part (1)  

(3) It is clear that every quasi-prime submodule  is 2-absorbing, where" a proper submodule N 
of M is called quasi-prime submodule  if  whenever a, b ∈ R, m∈ M  and   a b m∈ N ,  
then  a m∈ N or   b m∈ N"[3] 

However a 2-absorbing submodule may not be quasi-prime , as the following  example 
shown : 

Consider the Z-module Z . The submodule  N =4Z is a 2-absorbing submodule of Z  since  
, if a,b,c ∈Z  with abc ∈4Z=N , then at least two of a,b,c are even . Hence either  ab	∈	N  
or ac∈	N  or  bc	∈	N 

But  4Z is not quasi-prime , since 2.2.1 ∈4Z  but 2.1∉4Z.           

(4) Let N ,W be two submodules  of an R-module M and W < N .if N is 2-absorbing 
submodule of  M  then it is not necessary that W is 2- absorbing submodule of M , for 
example  in  Z24  as Z-module . Take  N= 6  , W= 12 .    Since N =	 2 ∩ 3 	 and both 
of them are prime  submodules  then N is 2-absorbing submodule by part (1) . But 
2.2.3	∈ W ,  2.3	∉ W and   2.2=4 ∉ W	: Z  =12Z  . Thus W is not 2-absorbing 
submodule in M . 



 

225 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

(5) Let N ,W be two submodules  of an R-module M and  N < W . If N is 2-absorbing 
submodule of M , then N is a 2-absorbing submodule of W  

 Proof: 

 If   W =M ,then nothing to prove  

 Let  a b x ∈N , where a ,b ∈R  , x ∈W . Since x ∈W then x ∈ M  . But  N is 2-absorbing 
submodule of M , so either :   a x ∈	N or  b x ∈	N or a b ∈ (N:M)  and since N < W implies 
(N:M)  (N:W)  , then either  a x ∈	W or  b x ∈	W or a b ∈ (N:W)   

Hence N is 2-absorbing in W		∎ 

(6) The sum of  2-absorbing submodules is not necessary 2-absorbing . for example: 
Let  N1 2Z			, N2 3Z	 , each of N1 and N2  is 2-absorbing submodule in the Z-module 
Z , but   N1 + N2 =Z  which is not 2-absorbing  . 

(7) Let  N and W be two submodules  of an R-module M such that  N ≅ W   
If N is 2-absorbing submodule , it is not necessary that W is 2-absorbing submodule as the 
following example explains this : 

Consider the Z-module Z , the submodule 2Z is 2-absorbing submodule but                 
2Z	 ≅ 	30Z		and				30Z	is	not	2-absorbing since 	2.3.5 30	 ∈ 30	but		2.5 ∉ 30Z	and 

	3.5 ∉ 30Z	and	2.3 6 ∉ 30Z 

(8) The intersection of two 2-absorbing submodule  need not be 2-absorbing submodule   for 
example : 6Z and 5Z are  2-absorbing submodule  in the Z-module Z , but 6Z ∩ 5Z = 30Z  
which is not 2-absorbing 

(9) Let N be a 2-absorbing submodule of  M , then for each A⊆M , either A⊆N or A∩N is a 
2-absorbing submodule of A  

Proof :  

Suppose  A⊈N . Then  A∩N	≨	A . 

Let  abx ∈ A∩N , and  x∈A , a,b∈R . So   abx	∈	N . Since N is 2-absorbing , either   ax  ∈
	N  or bx ∈	N  or  ab ∈ (N:M) , then  ax  ∈	A∩N or bx ∈	A∩N  or  ab ∈ (A∩N:A) .   

Proposition  .1.3 :    

Let	φ :M	⟶ M′ be an R-epimorphism . If W is 2-absorbing submodule of 	M′ , then φ (W) 
is 2-absorbing submodule of M. 

Proof : 

It is straight for word so it is omitted     ∎   

Proposition  1.4 :    

 Let   f : M	⟶ M′  be an epimorphism  , N < M such that kerf ⊆	N , then N is 2-absorbing 
submodule  of M if and only if  f(N) is 2-absorbing submodule of M′   

 



 

226 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

Proof: 	 

⟹ Let  abḿ ∈ f N    , where  ḿ ∈		M′  a,b ∈ R . ḿ f m  for some m ∈ M , since f is onto 
. 

Then   abf(m) ∈ f N  , so abf(m)=f(n)  for some  n  ∈ N and hence  f(abm)- f(n) = 0 . Thus we 
get that abm – n	∈ kerf	 ⊆ N  which implies that abm ∈ N . But N is 2-absorbing so either  am 
∈ N		or		bm	 ∈ N	or  ab ∈ (N	: M) . 

If am ∈ N		, then  f( am) ∈ f N  , that is  a f(m) ∈ f N	 	 so  a	ḿ	 ∈ f N   . 

Similarly, bm ∈ N implies that bḿ	 ∈ f N   . 

If  ab ∈ (N	: M) , then  abM ⊆ N and so f(abM) ⊆ f N  which implies that  abM′ ⊆ f N   and 
we get that  ab ∈ (f N 	:  M′) . 

⟸  Let  abm ∈ N  then  f(abm)	∈ f N 		so   abf(m)	∈ f N 	 . Since f(N) is 2-absorbing  either 
af(m) ∈ f N	 	 or  bf(m) ∈ f N	 	or ab ∈ (f N 	:  M′). 

1) If af(m) ∈ f N	 	    then  f(am) f n	 	 for some   n	∈ f N  ,  henc  am-n	∈ kerf	 ⊆ N  so 
am∈ N 

2)If  bf(m)	∈ f N  then Similarly that  bm	 ∈ N	 

3) If ab ∈ (f N 	: M′) then  abM′ ⊆ f N  so abf(x)	∈ f N  for each  x ∈ M  

so that  f(abx)=f(n) for some n ∈ N and hence abx ∈ N for each x	∈ M. Thus ab ∈ (N	: M) . ∎ 

By using Proposition 1.4 we can get the following result which is given in [ 2] as a direct 
consequence   

"Corollary  1.5 : 

      Let R be a ring , M an R- module and N , K submodules of M with K ⊆N . Then  N is a    
2-absorbing submodule of M if and only if   is a 2-absorbing submodule of   "  

       A.Y. Darani and F. Soheilina in [2] introduced the following: 

"Proposition  1.6 : 

    Let R be a commutative ring , M is a cyclic R-module and N is a submodule of M . Then N 
is a 2-absorbing submodule of M if and only if (	N	: M) is a 2-absorbing ideal of R". 

          Sh. Payrovi and S. Babaei in [4] and S. moradi  and  A. Azizi in [5]  introduced the 
following: 

Theorem 1.7 :  

"If N is a 2-absorbing submodule of M , then (	N	: M) is a 2-absorbing ideal of R" .  

          Sh . Payrovi , Babaei in [4] proved the following " Let R be a Noetherian ring , M a 
finitely generated multiplication R-module and N a proper submodule of M such that 
AssR(M/N) is a totally ordered set  .  If (N :R M) is a 2-absorbing ideal of R, then N is a 2-
absorbing submodule of M "  , where R is a commutative ring with non zero identity . 



 

227 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

         However we get the same conclusion under the class of multiplication modules  . Before 
giving our result , recall that "An R-module is called a multiplication module if  for every 
submodule N of M, there exists an ideal I of R such that IM = N  .  Equivalently , M is a 
multiplication module if  for every submodule N of M ,N=(N :RM)M" . [6] 

Theorem  1.8  : 

Let M be a multiplication R-module and N is a proper submodule  of M , If N	: M 	is a 2-
absorbing ideal of R , then N is 2-absorbing submodule  of M   

Proof: 

Let  abm ∈ N where a,b ∈ R    , m ∈ M  then  ab(m)⊆ N . 

But (m)= IM for some I	 R since M is a multiplication R-module , so abIM	⊆ N .   

Hence abI	⊆ N	: M  ,so we get that  ( a)(b)I	⊆ N	: M  . Since N	: M  is a 2-absorbing ideal , 
therefore  (a)I	⊆ N	: M  or (b)I	⊆ N	: M  or (a)(b)	⊆ N	: M  by [1] 

1) If (a)I	⊆ N	: M  ,  then (a)IM	⊆ N so (a)(m)⊆ N thus  am	∈ N 
2) If (b)I	⊆ N	: M  , then similarly  bm	∈ N. 
3) If (a)(b) ⊆ N	: M  , then  ab ∈ N	: M 	∎ 

Corollary 1.9 : 

Let M be a multiplication R-module and N is a proper submodule  of M , then N is 2-
absorbing submodule  of M if and only if  N	: M 	is 2-absorbing ideal. 

Remark 1.10 : 

The condition M is a multiplication R-module can't be dropped from Theorem 1.8  .  

 Consider the following example : 

Let M be the Z-module Z  and N = 0  . N is not 2-absorbing since : 

p.p. Z  =	0  but   p. Z  	0 and P 	 ∉	 	Z	 	
∶ Z 0   ,     Also notice that  

(N:	Z )= 0    ,  and 0   is a prime ideal in Z , so  0    is 2-absorbing ideal in Z .  

       Recall that " A proper submodule N of an R-module M is called a P-primary  submodule 
of M if whenever  a ∈ R  and  m ∈M  and  am ∈ N , then m ∈ N  or a ∈ √N ∶ 	M=P " [7]  

      Darani in [7 ] proved the following : 

" Let N be a P-primary submodule of a cyclic R-module. Then N is 2-absorbing if and only if 
(pM)2 ⊆ N " 

   However we improve this Theorem as follows . 

Proposition 1.11  : 

Let N be a p-primary submodule of a multiplication R-module M . Then N is 2-absorbing if 
and only if (pM)2 ⊆ N  (where (pM)2 = p2M) . 

 



 

228 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

Proof : 

⟹  Since M is a multiplication R-module , M-radN=√N:M M by[7,Th 2.10] 

But N is P-primary , so  P =√N:M is a prime ideal . ; that is M-radN=PM .  

It follows that  P2 ⊆ (N:M) by [1] Th. 2.4  . Thus P2M ⊆N . 

Hence  ( PM)2 ⊆N 

⟸  Let  abm∈N where a , b ∈R , m∈M  

Assume that   am∉N and  bm∉N . As N is primary with  abm∈N and bm∉N , we get              
a	∈ √N:M=P 

Also  abm∈N and  am∉N , we get   b	∈ √N:M=P 

Thus  ab ∈ P  ⊆ (N:M) and consequently N is 2-absorbing submodule of M ∎ 

       Recall that " A proper submodule N of an R-module M is called a 2-absorbing primary  
submodule of M if whenever a ; b ∈ R and m ∈M  and  abm ∈ N , then am ∈ M-rad(N ) or bm 
∈ M-rad(N ) or ab ∈ (N :R M)" [8] 

    It is clear that 2-absorbing submodule implies 2-absorbing primary , but the converse may 
be not hold , as the example shows  

Let M be the Z-module Z , let N=8Z , so N is 2-absorbing primary ,but it is not 2-absorbing 
submodule . 

        Babaei in [4] proved the following : 

"Let R be a Noetherian ring , I a 2-absorbing ideal of R , and M a faithful multiplication R-
module such that AssR(M /IM) is a non-empty totally ordered set . Then abm ∈ IM implies 
that am ∈IM or bm ∈ IM or ab ∈I whenever a; b ∈ R and m ∈ M " [4] , that IM is 2-absorbing 
. where R is a commutative ring with nonzero identity . 

   However we get the same conclusion of this theorem but with less conditions and  also we 
give a simple proof . 

Proposition  1.12 :  

Suppose M is a finitely generated multiplication R-module . If  I is 2-absorbing ideal of R 
such that annM ⊆I , then IM is 2-absorbing submodule of M.  

Proof:  

Let  abm	∈ IM where a , b ∈ R , m	∈ M ,  hence ab(m) ∈ IM  . Since M is multiplication , then 
(m)=JM  for some ideal  J of R . Thus   abJM	⊆ IM and so abJ	⊆ I+annM = I . But I is a 2-
absorbing ideal of R , so either ab	∈ I	 or aJ ⊆ I or bJ ⊆ I , it follows that ab	∈ IM	: M   or  
aJM ⊆ IM or   bJM ⊆ IM ; that is either  ab	∈ IM	: M 	 or  a(m) ⊆ IM or b(m) ⊆ IM Thus  
ab	∈ IM	: M  or        am ∈ IM or bm ∈ IM and so  IM is a 2-absorbing submodule  of M 	∎ 

 

 



 

229 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

Corollary 1.13 :  

Suppose M is a faithful finitely generated multiplication R-module . If  I is a 2-absorbing 
ideal of R, then IM is a 2-absorbing submodule of M.  

Proof:  

It follows directly by Proposition 1.12  ∎ 

Corollary 1.14 :  
    Let M be a faithful finitely generated multiplication R-module . Then every proper 
submodule of  M is 2-absorbing if and only if every proper  ideal of R is 2-absorbing .   
Proof :  

⟸  It follows directly by Corollary (1.13) . 

⟹  Let I be a proper ideal of R . Then  N= IM is a proper submodule of M  

So it is 2-absorbing and hence by Theorem (1.7) , (N:M) is 2-absorbing ideal . But M is 
faithful finitely generated multiplication R-module , so (N:M)=I by ([6],Th.3.1)  ∎ 

" Proposition 1.15 : [4,Th 2.4] 
     Let M be an R-module , N a proper submodule  of M , if N is 2-absorbing then 
N	: m 	is 2-absorbing ideal for each  m ∈M-N "   

Proof:  

First N	: m R for any m	∉	N  

Let  abc	∈ N	: m  then ab(cm) ∈	N , but N is 2-absorbing submodule  then  

a(cm)	∈	N  or b(cm)	∈	N or ab ∈	 N	: M ,  so that  acm	∈	N  or  bcm 	∈	N or     abM ⊆ N  that 
is   ac	∈	 N	: m 	  or  bc ∈	 N	: m  or ab ∈	 N	: m  hence 	 N	: m  is 2-absorbing 
ideal ∎ 

       Recall that "a submodule N  of M is called a  pure submodule  of an R-module M  if    
IM	∩ N = IN for any ideal I of R" [9]  

 Proposition 1.16  : 

Let N be a proper pure submodule  of an R-module M  , If 0  is a 2-absorbing submodule  of 
M  , then N is 2-absorbing . 

Proof: 

Let  abm	∈ N  where a ,b ∈ R , m	∈ M 

Put I =(ab)  then abm	∈ IM ∩ N , but IM ∩ N = IN  , so abm = abn  for some n ∈ N , then                 
ab( m – n ) = 0 , but (o) is 2-absorbing then a ( m – n ) = 0 or   b( m – n ) = 0  or                   ab 
∈	annM ⊆	(N : M)  . So we get am=an ∈ N or bm=bn ∈N  or ab ∈	( N : M)  ,  

Thus N is a 2-absorbing submodule  ∎ 

2.   2-absorbing Submodules Characterizations . 
         In this section we give some characterizations of 2-absorbing submodules . We start with 
the following proposition:  



 

230 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

Proposition  2.1 : 

Let N a proper submodule of an R-module M . Then N is 2-absorbing submodule of M if and 
only if  abK ⊆N  for some a , b ∈R , K M . implies  ab∈(N:M) , or  aK ⊆N  or bK ⊆N  

Proof : 

Suppose that   ab	∉(N:M) and   aK	⊈	N and bK	⊈	N . Then there exist m1 , m2 in K such that  
am1  ∉ N and bm2  ∉ N . 

Since  abm1 ∈ N and  ab ∉(N:M) , am1  ∉ N , we get  bm1 ∈ N. 

Also since abm2 ∈N and  ab ∉(N:M) , bm2  ∉ N , we get  am2 ∈ N 

Now ,  since   ab(m1 + m2)	∈ N and  ab ∉(N:M)    we have     a(m1 + m2)	∈ N or b(m1 + m2)	∈ N  

If a(m1 + m2) 	∈ N ; i.e. am1 +am2 	∈ N and since  am2 	∈ N we get am1 ∈  N which is 
contradiction ! 

If b(m1 + m2)	∈ N ; i.e. bm1 +bm2	∈ N and since  bm2	∈ N we get bm1 ∈ N which is 
contradiction ! 

Then either  ab	∈(N:M)  or aK	⊆	N and bK	⊆	N  

the converse is clear  ∎ 

       The following theorem give a useful characterization of 2-absorbing submodule .   

 Theorem 2.2 : 

 Let N a proper submodule of an R-module M , then the following statement are equivalent : 

(1) N is a 2-absorbing submodule of  M   
(2) If   I J K ⊆ N , for some ideal I and J of R  and some submodule K of M then  either           

I K	⊆ N or J K⊆ N or I J	⊆( N:M) . 
Proof :   (1) ⟹ (2)  

Suppose N is a 2-absorbing submodule of  M and  I J K		⊆ N for some ideals I and J of R  and 
some submodule K of M   and I J	⊈	( N:M) .  

To show that I K ⊆ N or   J K ⊆ N. 

Suppose   I K ⊈ N and    J K ⊈ N . then there exist a1	∈	I and a2	∈	J such   that a1 K ⊈ N and    
a2 K ⊈ N .          

But a1 a2 K ⊆ N and neither a1 K ⊈ N   nor  a2 K ⊈ N and N is 2-absorbing , so we have        
a1 a2	∈( N:M) by Proposition (2.1) 

Since I J	⊈	( N:M) , then there exist b1	∈	I and b2	∈	J such that  b1 b2 ∉(N:M). 

But b1 b2K ⊆ N , so we have b1K ⊆ N or b2K ⊆ N  by Proposition (2.1) 

Now we have the following cases : 

Case (1)  b1K ⊆ N and b2K ⊈ N   



 

231 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

Since   a1b2K ⊆ N and  b2K ⊈ N  and a1 K ⊈ N  so that a1 b2	∈( N:M) by Proposition (2.1) 

Since b1K ⊆ N  and a1K ⊈ N , we conclude ( a1+b1) K ⊈ N . On the other hand , ( a1+b1) b2 K 
⊆ N and neither ( a1+b1) K ⊆ N nor b2K ⊆ N , we get that  ( a1+b1) b2 ∈(N:M) by Proposition 
(2.1) 

But (( a1+b1) b2 = a1b2+b1b2  ∈(N:M) and  a1b2 ∈(N:M) , we get b1b2  ∈(N:M) which is a 
contradiction!  

Case(2) If b2K ⊆ N and  b1K ⊈ N . By a similar argument of case (1), we reach to a  
contradiction! 

Case(3) b1K ⊆ N and b2K ⊆ N   

Since b2K ⊆ N   and a2K ⊈ N , we conclude ( a2+b2) K ⊈ N . But a1( a2+b2) K ⊆ N and 
neither a1K⊆N nor ( a2+b2)K ⊆ N , hence a1( a2+b2)∈(N:M) by Proposition (2.1) 

Since  a1 a2∈(N:M) and a1 a2 + a1 b2 ∈(N:M) , we have a1 b2 ∈(N:M) . Since (a1+b1) a2 K ⊆ N 
and neither a2 K ⊆ N nor (a1+b1)K ⊆ N , we conclude (a1+b1) a2∈(N:M) by Proposition (2.1) 

But (a1+b1) a2 = a1a2+b1 a2	 , so  a1a2+b1 a2	 ∈(N:M) and since a1 a2∈(N:M) , we get  
b1a2	∈(N:M) . Now , since (a1+b1) (a2+b2)K ⊆N and neither (a1+b1)K ⊆N nor (a2+b2)K ⊆N, 

We have  (a1+b1) (a2+b2) =a1a2+ a1b2+b1a2+b1b2 ∈(N:M) by Proposition (2.1) 

But a1a2 , a1b2 , b1a2 ∈(N:M) , so b1b2 ∈(N:M) which is a contradiction!  . Consequently      I1K 
⊆N or I2K ⊆N 

(2)  ⟹ (1) It is clear  ∎ 

        Recall that ". for any two submodules N , K of a multiplication R-module M with             
N = I1M and K = I2M for some ideals I1 and I2 of R. The product N and K denoted by NK is 
defined by NK = I1I2M."[2] 

      By using this definition of product of  submodules , we give the following 
characterization of  2-absorbing submodules .    

Theorem 2.3 : 

Let N be a proper submodule of a multiplication R-module M , then  

N is a 2-asorbing submodule of M if and only if  N1 N2 N3 ⊆N implies that N1 N2 ⊆N or         
N1 N3 ⊆N or  N2 N3 ⊆N , where  N1 , N2 , N3 are submodules of M. 

Proof :   

⟹ Since M is multiplication , then  N1= I1M  , N2= I2M  , N3= I3M  for some ideals I1 , I2 , I3 
of  R  . It follows that : 

N1 N2 N3 = I1 I2 I3M	⊆	N . Hence  I1 I2 I3	⊆	(N:M) . But N is 2-absorbing submodule of M 
implies that (N:M) is 2-absorbing ideal  by [Theorem 1.7]. 

So by [1 ,Th 2.13] either I1 I2 ⊆	(N:M) or I1 I3	⊆	(N:M) or I2 I3	⊆	(N:M) . Then I1 I2 M ⊆	N  
or  I1 I3M 	⊆	N  or  I2 I3M		⊆	N .  Thus  N1 N2 ⊆N  or  N1 N3 ⊆N  or   N2 N3 ⊆N  



 

232 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

⟸  Let I1 I2 K ⊆	N   where I1 ,  I2  are ideals of R and K  M . Since M is multiplication 
module , K = JM  for some J of R . Thus  I1 I2J M	⊆	N  . Put    N1= I1M   N2= I2M  .  

It follows that  N1 N2K= I1 I2J M	⊆	N  . So by hypotheses either N1 K	⊆	N   or N2 K	⊆	N  or   
N1 N2 ⊆	N  then I1J M	⊆	N  or I2J M	⊆	N   or I1 I2	⊆	(N:M); that is  I1 K ⊆	N  or  I2K ⊆	N   or 
I1I2 ⊆	(N:M) 

Thus N is 2-absorbing submodule of M  	∎  

Proposition 2.4 : 

Let N be a proper submodule  of an R-module M  . The following statements are equivalent  :  

(1)  N is a 2-absorrbing submodule of M 
(2) (N	:  I) is 2-absorbing , for each ideal I of R with IM ⊈N 
(3)  (N	: (r)) is 2-absorbing submodule   for each r ∈R with rM ⊈N 
Proof:  

(1) ⟹(2) Let I be an ideal of R with IM ⊈M , then (N	:  I) is a proper submodule of M  

Let a b m ∈	(N	: 	I)  , where a , b ∈ R , m	∈ M . Then ab(Im) ⊆	N . 

But  N is 2-absorbing submodule of M , so by Proposition(2.1) , either  a(Im) ⊆	N or       
b(Im) ⊆	N or ab	∈ (N	: M)	⊆ ( (N	: I): M) . Hence  either am  ∈	(N	: I) or  bm ∈(N	: 	I) or 
ab	∈( (N	: I): M  . Thus  (N	: 	I) is a 2-absorbing submodule . 

(2) ⟹(3) It is clear . 

(3) ⟹(1) Take r =1 then (N	: 	 1 )= N , so N is 2-absorbing.	∎  

        Now we have the following : 

Theorem 2.5 : 

Let N be a proper submodule  of an R-module M . Consider the following statements:  

1) N is a 2-absorbing submodule of M  
2) For each a,b ∈ R , m ∈ M  . If abm ∉ N , then ( N :abm) = (N:am)∪(N:bm) 
3) For each a,b	∈ R , m∈M  if abm∉N  then (N:abm)=(N :am) or (N:abm) =(N:bm)  
Then (1) ⟹(2)	⟹(3)  and if M is cyclic , then (3)	⟹(1) . 

Proof : (1) ⟹(2) 

Let c ∈ ( N : abm)  then  abcm ∈ N . Since abm ∉ N and N is 2-absorbing , so  by (1.2 ) , 
either  acm ∈	N  or  bcm ∈	N . Then   c ∈ ( N : am)  or c ∈ ( N : bm)   

Thus ( N : abm)	⊆ ( N : am)  ∪ ( N : bm) . To prove reverse inclusion: 

 Let c ∈( N : am)  ∪ ( N : bm) , then c ∈( N : am) or c ∈ ( N : bm) so that  acm ∈	N or bcm ∈
	N   

It follows that  abcm ∈ 	bN ⊆ N  or abcm ∈ 	aN ⊆ N  

Thus  abcm ∈ N and hence c ∈ N:abm  .  



 

233 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

Then   ( N:am) ∪( N:bm) ⊆ ( N:abm) . So we get (N:abm)=(N:am)∪(N:bm). 

(2)	⟹(3)  Since (N : abm) is an ideal of R , and (N:abm)=(N:am)∪(N:bm). 

So either ( N : bm)  ⊆ ( N : am)   or ( N : am) ⊆ ( N : bm) . Then  ( N : abm) = ( N : am)   or   
( N : abm) ( N : bm) 

(3)	⟹(1)   Now suppose that  M= m   for some m1 ∈M 

Let abm ∈  N . But m=rm1 for some r ∈ R  then abrm1 ∈  N and suppose that                        
a b	∉ N :M)= (N:	 m )  , abm1	∉ N and  r∈( N : abm1 ) .  

By (3) ( N : abm1) = ( N : a m1)   or ( N : ab m1) = ( N : b m1) . Since  r ∈( N : abm1 ) , so 
either r	∈( N : a m1)   or  r	∈( N : b m1)  . Then  a rm1 ∈	N  or b rm1 ∈	N , we get  a m ∈	N  or  
b m  ∈	N . Hence N is 2-absorbing	∎ 

 

 

References  

1. Badawi ,A(2007),  On 2-Absorbing Ideals of Commutative Rings, Bull. Austral. Math. 
Soc. 75 , 417–429. 

2. Yousefian Darani,A. and Soheilnia,F.(2011), On 2-Absorbing and Weakly 2-Absorbing 
      Submodules , Thai J. Math., 9, 577-584 

3. Abdul-Razak,H.M (1999), "Quasi-Prime Modules and Quasi-Prime Submodules",  
M.Sc.Thesis, College of Education Ibn Al-Haitham, University of Baghdad,  
4. Payrovi,Sh and Babaei,S.(2012), On 2-Absorbing Submodules, Algebra Colloq., 19, 913-
920. 
5. Moradi,S. and Azizi,A.(2012) , 2-Absorbing and n-Weakly Prime Submodule , Miskolc 
Mathematical Notes , 13 , 1,pp. 75-86 .  
6.El- Bast, Z.A. and Smith, P.F.(1988), Multiplication Modules , Comm. in Algebra, 16(4), 
755-779.  
7. Moore,M.E. and Smith,S.J.(2002), Prime and Radical Submodules of Modules over 
Commutative Rings, Comm. Algebra, 30 , 5073-5064.   
8. Hojjat Mostafanasab , Ece Yetkin , Unsal Tekir and A. Y. Darani(2015)  , On 2-Absorbing 
Primary Submodules of Modules over Commutative Rings   ,arXiv : 1503 . 00308V1 
[Math.AC] Mar 2015 .  
9. Anderson,E.W. and Fuller,K.R.(1992), "Rings and Categories of Modules", Springer-
Verlage, New York . 

 
 
 
 
 

          

  



 

234 | Mathematics 

 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية                                                                            المجلد  

Ibn Al-Haitham Jour. for Pure & Appl. Sci.                                           Vol. 28 (3) 2015 

     

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