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Weak Pseudo – 2 – Absorbing  Submodules  And  Related Concepts 

Haibat  K.  Mohmmadal                                  Sada E. Radeef 

   mfloweriraq767@gmail.co                                    dr.mohammadali2013@gmail.com 

Department  Of  Mathematics , College Of  Computer Sciences  and  Mathematics, University  Of  

Tikrit, Tikrit . Iraq. 

 

 

 

 

Abstract  

      Let  R  be a commutative ring  with identity  and  E  be a unitary left  R – module .We 

introduce  and study the concept Weak Pseudo – 2 – Absorbing submodules as  generalization 

of weakle – 2 – Absorbing submodules , where a proper submodule  A of  an  R – module  E 

is  called  Weak Pseudo – 2 – Absorbing  if   0 ≠ rsx  ∈ A   for  r, s ∈ R , x ∈ E , implies that  

rx  ∈ A + soc ( E ) or  sx ∈ A + soc (E)  or   rs ∈ [ A + soc ( E ) :𝑅 E ]. Many basic  properties, 

characterizations  and  examples  of   Weak Pseudo – 2 – Absorbing  submodule  in some  

types  of  modules  are  introduced .  

Key word : weakly – 2 – Absorbing submodules , essential submodule , socal of modules , 

multiplication modules , Z – regular modules , WP – 2 – Absorbing submodules . 

1. Introduction  

            The concept of  weakly – 2 – Absorbing  submodule was first  introduced  by Darani 

and  Soheilnia as  generalization of weakly prime submodule , where a proper submodule A 

of an R – module E is called a weakly  prime submodule of  E if  0 ≠ te ∈ A , for t ∈ R , e ∈ E  

implies that  either e ∈ A or t ∈ [ A :𝑅  E ] , where [ A :𝑅 E ] = { s ∈ R : sE   A }  [1] , and a 

proper submodule  A of an R- module  E is called a weakly – 2 – Absorbing submodule  of  E 

, if 0 ≠ ste   ∈ A , for s , t ∈  R , e ∈ E implies  that  either  se ∈ A or  te ∈ A or  st ∈  [ A :𝑅 E ] 

[2] .  

Recently , several generalizations of weakly – 2 – Absorbing  submodules  have been 

introduced  [ 3,4,5 ] . In our paper , we introduce a new generalization of weakly – 2 – 

Absorbing submodule which we callWeak Pseudo – 2 – Absorbing  submodule , where a 

proper submodule A of an R – module E is said to be Weak Pseudo – 2 – Absorbing 

Ibn Al Haitham Journal for Pure and Applied Science 

Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index 

 

Doi: 10.30526/34.2.2615 

Article history: Received,19,May,2020, Accepted 23,June,2020, Published in April 2021 

 

mailto:floweriraq767@gmail.com
mailto:dr.mohammadali2013@gmail.com
mailto:dr.mohammadali2013@gmail.com


  

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submodule  if  0 ≠ ste ∈ A , for s , t  ∈ R , e ∈ E , implies that either  se ∈ A + soc ( E )  or  te 

∈ A + soc ( E )  or st ∈ [ A + soc ( E ) :𝑅  E ] . Soc ( E ) is the intersection of all essential 

submodules of E [6] . A nonzero submodule  N of an R – module E is called an essential if  N 

∩ K  ≠ ( 0 ) for  all nonzero submodules K of  E [6]. Every weakly prime submodule of an R – 

module is weakly – 2 – Absorbing [7] . Recall that an R – module E is cyclic if  E = ˂ x >  for 

x ∈ E [8]. Recall that an R – module E is a semi simple if soc ( E ) = (0) [6]. It is well known 

that an R – module E is a semi simple if and only if soc ( 
𝐸 

𝑁 
 ) = ( 

𝑠𝑜𝑐 ( 𝐸 )+ 𝑁 

𝑁 
 ) for each 

submodule N of  E [ 6 , Ex.12(c) ] . The set [ N :𝐸  I ] = [ x ∈ E : x I   N } , where N is a 

submodule of E , and  I is an ideal of  R [ N :𝐸 I ] and is a submodule of  E  containing N. [ N 

:E R ] = N  and [ I :𝐸 R ] = I [9]. Recall that an R – module E is a multiplication if every 

submodule A of E is of the form A = IE   for some ideal I of R , Equivalently  A = [ A :𝑅 E ] 

E [10]. Recall that an R – module E is a faithful if  Ann ( E ) = { r ∈ R : rE = ( 0)} [8].  

2. Basic properties of WP – 2 – Absorbing submodules . In this part of the paper , we 

introduce the definition of WP – 2 – Absorbing submodules and , thus truth some of i𝑡,s basic 

properties , examples and characterizations.    

Definition .1.  

               A proper submodule A of an R – module E  is said to be Weak Pseudo – 2 – 

Absorbing (for shorten WP – 2 – Absorbing ) submodule of  E , if  0 ≠ ste ∈ A , for  s , t ∈ R , 

e ∈ E , implies that se ∈ A + soc( E ) or te ∈ A + soc ( E ) or  st ∈ [ A + soc ( E ) :R E ] . And 

an ideal J of a ring R is said to be WP – 2 – Absorbing ideal of R, if J is a WP – 2 – 

Absorbing R – submodule of an R – module R. 

Example and Remarks .2.  

  1. In the Z – module 𝑍36 , the only essential submodules are ˂ 2̅ >, ˂ 3 ̅>, ˂ 6̅  > and Z36  

itself  thus  Soc ( Z36 )   = ˂ 6̅  > =  { 0̅ , 6̅ , 12̅̅̅̅  , 18̅̅̅̅  , 24 ̅̅ ̅̅  , 30̅̅̅̅  }  

2 . It is clear that the submodules of  the Z – module Z36  are ˂ 4̅ > ,˂ 6̅ >, ˂ 9̅ > , ˂ 12̅̅̅̅  > and 

˂ 18̅̅̅̅  > are WP – 2 – Absorbing submodules .    

3 . The submodules ˂ 12̅̅̅̅  > and ˂ 18̅̅̅̅  >  of the Z – module Z36 are  not  weakly – 2 – 

Absorbing submodules  , since  0 ≠ 2 . 3 . 2̅ ∈  ˂ 12̅̅̅̅  >  for 2 , 3 ∈ Z , 2̅ ∈ Z36  but 2 . 2̅ =  4̅  

˂ 12̅̅̅̅  > and 3 . 2̅ = 6̅  ˂ 12̅̅̅̅  > and 2. 3 = 6  [˂ 12̅̅̅̅  > :Z  Z36  ] = 12 Z . 

 4 . The submodule  ˂ 2̅ >, ˂ 3 ̅>  of  the Z- module Z36  are weakly – 2 – Absorbing 

submodules of Z36 because  they are weakly prime submodules  of  Z - Z36 .  

5 . It is clear  that the submodules ˂ 4̅ > , ˂ 6̅ >  and ˂ 9̅ >  of  the Z – module Z36  are  weakly 

– 2 – Absorbing submodules .      

6 . It is clear that every weakly – 2 – Absorbing submodule of an R – module E is a WP – 2 – 

Absorbing , but    not conversely , the following example shows that : -- In the  Z – module 

Z36 , the  submodule ˂ 18̅̅̅̅  >  is a WP –   2- Absorbing  by (2) , but  ˂ 18̅̅̅̅  >  is not weakly – 2 – 

Absorbing submodule by  (3) .           



  

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7 . It is clear that every weakly prime submodule of an R – module E is a WP – 2 – Absorbing 

but not conversely. The following example explains that in the Z- module Z36 , the submodule  

˂ 4̅ >  is a WP – 2 – Absorbing by (2) . But ˂ 4̅ > is not weakly prime submodule, since 0 ≠ 2 

. 2̅  ˂ 4̅ >  for 2  Z , 2̅  Z36 , but 2̅  ˂ 4̅ >  and  2  [˂ 4̅ > : Z  Z36 ] = 4Z.  

8 . In general ,the submodule  nZ  of the  Z – module Z is weakly – 2 – Absorbing if n = 0 , P, 

P2  and pq by  [7, Rems. And Exs. ( 1.2.2 ) (3) ] . Hence the  submodules  nZ of the Z – 

module Z is a WP – 2 – Absorbing if  n = 0 , P , P2  and pq  by ( 6 )  .    

9 . The submodules 12Z  and  18Z of the Z – module Z are not WP – 2 – Absorbing because  

soc (Z) = ( 0 ) [8]. That is , 0 ≠  2 . 3 . 2  12Z   for 2 , 3 , 2  Z , but  2 . 2  12Z + soc ( Z ) 

and  3 .2 12Z + soc ( Z ) and 2 . 3  [ 12Z + soc ( Z ) :Z Z ] = 12Z . 

Also , 0 ≠ 2 . 3 . 3  18Z  for 2 , 3  Z , but 2.3  18Z + soc ( Z ) and 3.3  18Z + soc ( Z ) 

and 2. 3  [ 18Z + soc ( Z ) :Z  Z ] = 18Z .  

10 . If  A is a WP – 2 – Absorbing submodule of an R – module E , then [ A :R  E ]  need not 

to be WP – 2 – Absorbing ideal of  R . For example the submodule ˂ 18̅̅̅̅  > of  the Z – module 

Z36  is a WP – 2 – Absorbing submodule by ( 2 ) , but [˂ 18̅̅̅̅  > : Z  Z36 ] = 18Z  is not WP – 2 – 

Absorbing ideal of Z by ( 9 ) . 

11 . The intersection of two WP – 2 – Absorbing submodules of an R- module E need not to 

be WP – 2 – Absorbing submodule .For example the submodules 3Z , 4Z are WP – 2 – 

Absorbing submodule of the Z – module Z by (8) , but 3Z  4Z = 12Z  is not WP – 2 – 

Absorbing submodul by (9) . The following results are characterizations of WP – 2 – 

Absorbing submodules .   

Proposition 3.  

    A proper submodule A of an R- module E is a WP – 2 – Absorbing submodule of E if and 

only if for any t , s  R with ts  [ A + soc ( E ) : R  E ] we have [ A :E  ts ]  [ 0 : E ts ]  [ A 

+ soc ( E ) : E  t ]  [ A + soc ( E ) :E s ]  

Proof : ( )    

  tse , it follows ≠ A . If  0  E ] ,then tse  R  A + soc ( E ) : [  ts ] with ts E A :[  Let  e     

t ] or e  E [ A + soc ( E ) : A + soc ( E ) , that is either e  A + soc ( E) or se  that either te 

[ A + soc ( E  ts ]  E [0 :  ts ] . Hence  e  E [ 0 :  s ] . If  tse =  0 then  e   E[ A + soc ( E ) : 

 [ A t ]   E[ A + soc ( E ) : ts ]  E [ 0 : st ] E  s ] . Therefore [ A :E  [ A + soc ( E ) :  t ]   E) :

] . sE  soc ( E ) : +   

 () Let 0 ≠ tse  A for  t, s  R , e  E  with  ts  [ A + soc ( E ) :R  E ] . It follows by 

hypothesis e  [ A :E  ts ] and e   [ 0 : E  ts ] , implies that e  [A + soc ( E ) :E  t ]  [ A + 

soc ( E ) : E  s ]. Hence either  te   A + soc ( E )  or se   A + soc ( E ) . Therefore A is aWP 

– 2 – Absorbing submodule of  E . 

 

 



  

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Proposition 4. 

     A proper submodule A of an R – module E is a WP – 2 – Absorbing if  0 ≠ tsK   A  for t , 

s  R and K is a submodule of E , implies that either  tK  A + soc ( E ) or  sK  A + soc ( E 

) or ts  [ A + soc ( E ) :R  E ].  

Proof : ( ) 

 R [ A + soc ( E ) : R , K is a submodule of E . Suppose that  ts  A , for  t , s  Let 0 ≠ tsK  

  1K such that  te 2  , e1 A + soc ( E ).Then,there exists e A + soc ( E ) and  sK  E ] , tK  

E ] , then  R [ A + soc ( E ) : A and  ts  1  A + soc ( E ).  Now  0 ≠ tse 2 A + soc ( E ) and se

[ A + soc ( E  t ]   E[ A + soc ( E ) :  ts ]  E[ 0 :  st ] E  [ A : 1 by proposition (3) we have e

E  [ A + soc ( E ) :  1 t ] .It follows that e  E[ A + soc ( E ) : 1ts ] and e E[ 0 : 1s ] . But e E) :

 [ A + soc  2E ] and e R [ A + soc ( E ) :  and ts   2A + soc (E). Again 0 ≠ tse 1se ,s ] , that is

[ A + soc ( E  A and  ts  ) 2 + e1 A + soc ( E ). Now, 0 ≠ ts ( e  2s ], it follow tha  teE  ( E ) : 

ts ], it follows by proposition ( 3 )  E [ 0 :   ) 2 + e 1ts ] and ( e E [ A : ) 2 + e 1E ] , then  (e R ) :

 1s ]. That is either  t(e E[ A + soc ( E ) :  )  2 + e 1t ] or (e  E[ A + soc ( E ) :  ) 2 + e 1either (e 

A + soc ( E ) ,    2 + te 1) = te2 + e 1A + soc ( E) . If  t(e ) 2 + e 1A + soc ( E ) or s(e  )  2 e+ 

. is a contradiction A + soc ( E ) which    1A + soc ( E ) , then  te   2 and  te 

A + soc ( E )   2A + soc ( E ), then  se 1A + soc ( E ) and  se  2+ se1 ) = se2 + e 1If s (e

which is a contradiction. 

Hence either  tK  A + soc ( E ) or  sK  A + soc ( E )  or  ts  [ A + soc ( E ) :R  E ] . 

(  ) Trivial , so we omitted it . 

Proposition 5. 

      A proper submodule A of a cyclic R – module E is a WP – 2 – Absorbing if and only if 

for each t , s  R with ts  [ A + soc ( E ) :R  E ] we have [ A :R  tse ]  [ 0 : R tse ]  [ A + 

soc ( E ) :R  te ]  [ A + soc ( E ) :R se ].      

Proof : ( )     

Let  t , s  R ,with ts  [ A + soc ( E ) :R  E ] and let r [ A :R  tse ], it follows that  ts( re )  

A. If  0 ≠ ts( re )  A and A is a WP – 2 – Absorbing and ts  [ A + soc ( E ) :R  E ], then 

either  tre  A + soc ( E ) or sre  A + soc ( E ) , that is either r  [ A + soc ( E ) :R  te ] or  r 

 [ A + soc ( E ) :R se ]. If  tsre = 0 , implies that  r  [ 0 : tse]. Hence  r  [ 0 :R  tse ]  [ A + 

soc ( E ) :R  te ]  [ A + soc ( E ) :R se ] . Therefore   [ A :R  tse ]  [ 0 : R tse ]  [ A + soc ( E 

) :R  te ]  [ A + soc ( E ) :R se ].       

(  )  Since E is cyclic , then E =  e1  for  some e1  E . Let 0 ≠ tse  A for t , s  R , e  E  

with  ts  [ A + soc ( E ) :R  E ].Since e  E  then e = re1 for some r  R , that is 0 ≠ ts( re1 )  

A ,it follows that  r  [ A :R  tse1 ]  [ 0 : R tse1 ]  [ A + soc ( E ) :R  te1 ]  [ A + soc ( E ) :R 

se1 ]. But r   [ 0 : R tse1 ] ( since 0 ≠ tsre1 ) , therefore , r  [ A + soc ( E ) :R  te1 ] or r [ A + 



  

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soc ( E ) :R se1 ], it follows that  tre1  A + soc (E) or sre1  A + soc (E). That is te  A + soc 

( E ) or  se  A + soc (E).  Therefore A  is a WP – 2 – Absorbing submodule of  E .  

Proposition 6.  

     A proper submodule N of an R – module E is a WP – 2 – Absorbing submodule of  E if 

and only if ( 0 ) ≠ IJL  N for some ideals I , J of  R  and some submodule L of  E  implies 

that either IL  N + soc ( E ) or JL  N + soc ( E ) or IJ   [ N + soc ( E ) :R  E ] .  

 Proof : ( )     

Let ( 0 ) ≠ IJL  N  for some ideals I,J  of R and some submodule L of E with IJ  [ N + soc ( 

E ) :R  E ]. To prove that IL  N + soc ( E ) or JL  N + soc ( E ) .Suppose that IL   N + soc 

( E ) and JL  N + soc ( E ) , that is there exist a1  I  and  a2 J such that  a1L  N + soc ( E )  

and a2L  N + soc ( E )  .Now , ( 0 ) ≠ a1a2L  N , and N is aWP – 2 – Absorbing submodule 

of E , then by proposition ( 4 ) either a1 L  N + soc ( E )  or a2L   N + soc ( E )  or  a1a2 [ 

N + soc ( E ) :R  E ]. Since IJ  [ N + soc ( E ) :R  E ] , there exists b1  I  and b2  J such that  

b1b2  [ N + soc ( E ) :R  E ].But ( 0 ) ≠ b1b2L  N and N is a WP – 2 – Absorbing submodule  

of E , and  b1b2  [ N + soc ( E ) :R  E ] , then by proposition ( 4 ) either b1L  N + soc ( E ) or 

b2L  N + soc ( E ). 

Now : -- ( 1 ) If  b1L  N + soc ( E ) and  b2L  N + soc ( E ) . Since (0) ≠ a1b2L  N and b2L 

 N + soc ( E )  and  a1L  N + soc ( E ) , then by proposition  ( 4 )  a1b2  [N + soc ( E ) :R  

E ]. Since b1L  N + soc ( E ) and a1L  N + soc ( E ) , we get  ( a1 + b1 ) L  N + soc ( E ) 

.For there more  ( 0 ) ≠ ( a1 + b1 )b2 L  N and  N  is a WP – 2 – Absorbing with ( a1 + b1 ) L 

 N + soc ( E ) , b2L  N + soc ( E ) , it follows that  by proposition ( 4 ) ( a1 + b1 ) b2 = a1b2 + 

b1b2  [ N + soc ( E ) :R  E ] , but  a1b2  [ N + soc ( E ) :R  E ] , then b1b2  [ N + soc ( E ) :R  

E ] , this is a contradiction . 

( 2 ) If  b2L  N + soc ( E )  and   b1L  N + soc ( E ) , so by similar steps of ( 1 ) we get a 

contradiction . 

( 3 ) If  b1L   N + soc ( E )  and  b2L  N + soc ( E ) , since b2L  N + soc ( E ) and  a2L  N 

+ soc ( E ) , we get (a2 + b2 )L  N + soc ( E ) . But ( 0 ) ≠ a1 (a2 + b2 ) L  N and N is a WP – 

2 – Absorbing with a1L  N + soc ( E ) and   (a2 + b2 ) L  N + soc ( E ) then , we get  a1 (a2 + 

b2 )   [ N + soc ( E ) :R  E ] . Since a1a2   [ N + soc ( E ) :R  E ]  and  a1a2 + a1b2   [ N + soc 

( E ) :R  E ] , it follows that a1b2   [ N + soc ( E ) :R  E ]. 

Now , ( 0 ) ≠ (a1 + b1 ) a2  N and  a2L  N + soc ( E )   and ( a1 + b1 ) L  N + soc ( E ) , it 

follows by proposition ( 4 )  ( a1 + b1 )a2  = a1a2  + b1a2   [ N + soc ( E ) :R  E ] and since 

a1a2  [ N + soc ( E ) :R  E ] , we get  b1a2   [ N + soc ( E ) :R  E ]. Since ( 0 ) ≠ ( a1 + b1 ) (a2 

+ b2 ) L  N and  ( a1 + b1 ) L  N + soc ( E )  and (a2 + b2 ) L  N + soc ( E ) then by 

proposition ( 4 ) we have ( a1 + b1 ) (a2 + b2 ) = a1a2  + a1b2 + b1a2 + b1b2  [ N + soc ( E ) :R  E 

]. But  a1a2 , b1a2 , a1b2   [ N + soc ( E ) :R  E ] , we get  b1b2   [ N + soc ( E ) :R  E ] which is 

a contradiction . Thus IL  N + soc ( E )  or  JL  N + soc ( E ) . 



  

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(  )  Trivial, so we omittedit  

The following corollaries are adirect consequence of proposition ( 6 ) . 

Corollary 7. 

      A proper submodule A of an R – module E is a WP – 2 – Absorbing submodule of E if 

and only if ( 0 ) ≠ IJx  A  for some ideals I , J of  R and x  E , implies that either Ix  A + 

soc ( E ) or  Jx  A + soc ( E ) or IJ  [A + soc ( E ) :R  E ] . 

Corollary 8. 

     A proper submodule A of an R – module E is a WP – 2 – Absorbing submodule of E if and 

only if ( 0 ) ≠ sIL  A for some s  R and ideal I of R and some submodule  L of  E , implies 

that either sL  A + soc ( E ) or IL  A + soc ( E ) or sI  [ A + soc ( E ) :R  E ] .  

Corollary 9.  

     A proper submodule A of an R – module E is a WP – 2 – Absorbing submodule of E if and 

only if ( 0 ) ≠ sIx  A for some s  R , ideal I of R and some x  E , implies that either sx  

A + soc ( E ) or Ix  A + soc ( E ) or sI  [A + soc ( E ) :R  E ] . 

Proposition 10.  

      Let A be a WP – 2 – Absorbing submodule of an R – module E and  B is a submodule of  

E  with  B  A then 
   A   

B   
 is a WP – 2 – Absorbing submodule of an R – module  

E   

B  
 . 

Proof : Let  0 ≠ ts ( x + B ) = stx + B  
   A   

B   
 for s , t  R , x + B  

E   

B  
 , x  E .It follows that 

tsx  A. If   tsx = 0 then ts( x + B ) = 0 which is a contradiction . thus 0 ≠ tsx  A implies that 

either tx  A + soc ( E )  or sx  A + soc ( E )  or tsE  A + soc ( E )  .It follow that either  t( 

x + B )   
A +soc ( E )

B 
   or   s(x + B )   

A +soc ( E )

B 
   or 

tsE  

B 
    

𝐴 +𝑠𝑜𝑐 ( 𝐸 )

𝐵 
 . That is either t( x + B )  

   A   

B   
 + 

A +soc  ( E )

B 
    

   A   

B   
 + soc (  

E   

B  
 )   or s( x + B )  

   A   

B   
 + 

A +soc ( E )

B 
    

   A   

B   
 + soc (  

E   

B  
 ) or  

ts 
E   

B  
  

   A   

B   
 + 

A +soc ( E )

B 
    

   A   

B   
 + soc (  

E   

B  
 ). Hence , 

   A   

B   
  is a WP – 2 – Absorbing submodule 

of an R – module  
E   

B  
 . 

Proposition .11.  

     Let A , B be submodules of semi simple R – module E with B  A. If   B and   
   A   

B   
 are WP 

– 2 – Absorbing submodules of E ,  
E   

B  
  respectively , then A is a WP – 2 – Absorbing 

submodule  of E .  

Proof : 

Let  0 ≠ tsx  A  for  t , s  R , x  E , then 0 ≠ ts ( x + B ) = tsx + B  
   A   

B   
 . If 0 ≠ tsx  B  

and B is a WP – 2 – Absorbing , implies that  either tx  B + soc ( E )  A + soc ( E )  or sx  

B + soc ( E )  A + soc ( E )  or   tsE   B + soc ( E )  A + soc ( E )  .Thus A is a WP – 2 – 



  

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Absorbing submodule of E . Assume that  tsx  B , it follows that  0 ≠ ts( x + B )  
   A   

B   
 . But 

   A   

B   
  is a WP – 2 – Absorbing submodule of  

E   

B  
  implies that either  t( x + B )  

   A   

B   
 + soc ( 

 
E   

B  
 ) or s (x + B ) 

   A   

B   
 + soc (  

E   

B  
 ) or   

𝑡 𝑠 𝐸 

𝐵 
    

   A   

B   
 + soc (  

E   

B  
 )  . Since E is a semi simple then 

soc (
E   

B  
 )  = 

soc ( E )+B 

B 
 . It follows that either  t( x + B )  

   A   

B   
  +   

B+soc ( E )

B 
  or  s ( t + B )  

   A   

B   
  +   

B+soc ( E )

B 
  or  

t s E 

B 
    

   A   

B   
 + 

B+soc ( E )

B 
  . But B  A , implies that  B + soc ( E )  A + soc ( E ) , hence  

A 

B 
  + 

B+soc ( E )

B 
     

   A   

B   
  +   

A + soc ( E )

B 
 . Since  

   A   

B   
  

A + soc ( E )

B 
   implies  that  

   A   

B   
  +   

A + soc ( E )

B 
  =  

A +soc  ( E )

B 
  . that is either t ( x + B )  

A +soc  ( E )

B 
   or  s(x + B )    

A + soc ( E )

B 
   ,   

t s E 

B 
  

A + soc ( E )

B 
   , it follows that either  tx  A + soc ( E ) or  sx  A + soc ( E )  or  tsE  A + soc ( 

E ) . Thus A is  WP – 2 – Absorbing submodule of  E .  

Proposition 12. 

       Let A be a proper submodule of  an R – module E with soc ( E )  A. Then A is a WP – 2 

– Absorbing submodule of E so  if and only if [ A :E I ] is a WP – 2 – Absorbing submodule of  

E  for each ideal I of R .  

Proof : (  ) 

Let (0) ≠ tsB  [ A :E  I ]  for t , s  R , B is a submodule of  E , then  ( 0 ) ≠ tsIB    A , 

implies that either tIB  A + soc ( E ) or  sIB   A + soc ( E ) or tsE  A + soc ( E ) . But soc 

( E )  A , then A + soc ( E ) = A . that is either  tIB  A or  sIB  A or  tsE  A. Thus , 

either tB  [ A :E I]  or sB  [ A :E I]  or tsE  A  [ A :E I]  . It follows that either  tB  [ A 

:E I]   [ A :E I]  + soc ( E )  or  sB    [ A :E I]   [ A :E I]  + soc ( E )  or  tsE   [ A :E I]   [ 

A :E I]  + soc ( E ). Hence , [ A :E I]  is a WP – 2 – Absorbing  submodule of E . 

(  )Since [ A :E  I ]  is a WP – 2 – Absorbing subodule for every non zero ideal I of  R . Put I 

= R , we get [ A :E  R ] = A  is a WP – 2 – Absorbing submodule of E .  

We need to introduce the following definition. 

Definition 13. Let  A be a WP – 2 – Absorbing submodule of an R – module E and r, s  R , 

e  E , we say that ( r , s , e ) is WP – triple zero of A if  rse = 0 , re  A + soc ( E ) , se  A + 

soc ( E ) and rs  [A + soc ( E ) :R  E ] .  

Proposition 14. If  A is a WP – 2 – Absorbing submodule of E with ( r , s , e ) is a WP – 

trible zero of A for some r, s  R , e  E . Then rsA = ( 0 ) . 

Proof : Suppose rsA ≠ ( 0 ) , then rsa ≠ 0 for some a  A. Since ( r , s , e ) is a WP – triple 

zero of A then  rse = 0 , re  A + soc ( E ) , se  A + soc ( E ) and rs  [A + soc ( E ) :R  E ]. 

Since 0 ≠ rsa  A and A is a WP – 2 – Absorbing submodule of  E and rs  [A + soc ( E ) :R  

E ], then either  ra A + soc ( E ) or sa   A + soc ( E).  



  

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Now, 0 ≠ rs ( e + a ) = rse + rsa = rsa  A , and rs  [A + soc ( E ) :R  E ], then either  r( e + a 

) = re + ra  A + soc ( E )  or s( e + a ) = se + sa  A + soc ( E ). If  re + sa  A + soc ( E ) 

and  ra  A + soc ( E ) implies that re  A + soc ( E ) contradiction.  If  se + sa  A + soc ( E 

) and  sa  A + soc ( E ) , implies that  se A + soc ( E)  contradiction .  Hence , rsA = ( 0 ) . 

Proposition 15. If  A is a WP – 2 – Absorbing submodule of  E with ( r , s , e ) is a WP – 

triple zero of  A for some r, s  R , e  E , then [ A :R  E ]re = [ A :R  E ]se = ( 0 ) . 

Proof : Suppose that [ A :R  E ]se ≠ ( 0 ) then yse ≠ o for some y  [ A :R  E ]. Since  ( r , s , e 

) is a WP – triple zero of  A, rse = 0 and  re  A + soc ( E )  , se  A + soc ( E ) and  rs  [A 

+ soc ( E ) :R  E ]. We have  0 ≠ yse   A and A is a WP – 2 – Absorbing submodule of  E , 

then either  ye  A + soc ( E )  or se  A + soc ( E )  or ys  [A + soc ( E ) :R  E ]. Now , 0 ≠ ( 

r + y ) se = rse + yse = yse  A and  A is a WP – 2 – Absorbing submodule , then either ( r + 

y )e = re + ye  A + soc ( E )   or  se  A + soc ( E )  or ( r + y )s  [A + soc ( E ) :R  E ].Since 

ye  A + soc ( E )   and  if  re + ye  A + soc ( E ), it follows  that  re  A + soc ( E ) a 

contradiction . If ( r + y )s = rs + ys  [A + soc ( E ) :R  E ] and  ys  [A + soc ( E ) :R  E ], 

then  rs  [A + soc ( E ) :R  E ] a contradiction . Thus [ A :R  E ]se = ( 0 ) . Similarly we can 

prove  [ A :R  E ]re = ( 0 ) .  

Proposition 16. If A is a WP – 2 – Absorbing submodule of  E with ( r , s , e ) is a WP – 

triple zero of  A for some r, s  R , e  E . Then  r[ A :R  E ]e = s[ A :R  E ]e = ( 0 ).  

Proof : Suppose that   r[ A :R  E ]e ≠ ( 0 ) , then there exists  x [ A :R  E ] such that rxe ≠ 0 . 

But ( r , s , e ) is a WP – triple zero of  A , rse = 0 , re  A + soc ( E )  or  se  A + soc ( E )  

and   rs  [A + soc ( E ) :R  E ].  

For 0 ≠ rxe  A and  A is a WP – 2 – Absorbing submodule of  E, then either  re  A + soc ( 

E )   or  xe  A + soc ( E )  or  rx  [A + soc ( E ) :R  E ]. Now , 0 ≠ r( s + x )e = rse + rxe = 

rxe  A , and  A is a WP – 2 – Absorbing submodule , then either  re  A + soc ( E )  or ( s + 

x ) e = se + xe  A + soc ( E )  or   r( s + x ) = rs + rx  [A + soc ( E ) :R  E ].  That is  re  A 

+ soc ( E )  a contradiction. If ( s + x )e = se + xe  A + soc ( E ), implies that  se  A + soc ( 

E )  a contradiction. If  rs + rx [A + soc ( E ) :R  E ], implies that  rs  [A + soc ( E ) :R  E ] a 

contradiction. Thus r[ A :R  E ]e = ( 0 ) . In similary way  s[ A :R  E ]e = ( 0 ) . 

As direct consequence of  proposition ( 16 ) , we get the following corollary : 

Corollary 17. If A is a WP – 2 – Absorbing submodule of an R – module E with ( r , s , e ) is 

a WP – triple zero of  A for some r, s  R , e  E , then  r [ A :R  E ] A =  s [ A :R  E ]A  = ( 0 

) .  

Proposition 18. If A is a WP – 2 – Absorbing submodule of  E with ( r , s , e ) is a WP – 

triple zero of  A for some r, s  R , e  E , then  [ A :R  E ] sA  = [ A :R  E ] rA = ( 0 ).  

Proof : Suppose that [ A :R  E ] sA  ≠ ( 0 ) , then xsa ≠ ( 0 ) for some x  [ A :R  E ], a  A . 

Since ( r , s , e ) is a WP – triple zero of  A, then  rse = 0 , re  A + soc ( E ) , se   A + soc ( 

E ) and  rs  [ A + soc ( E ) :R E ]. For 0 ≠ xsa  A , it follows that either xa  A + soc ( E ) 

or  sa  A + soc ( E ) or  xs [ A + soc ( E ) :R E ]. We have ( r + x ) s ( a + e ) = rsa + rse + 



  

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xsa + xse = xsa  A         ( since rse = 0 , and  rsa = 0 for proposition ( 14 ) and  xsa = 0 from 

proposition ( 16 )). That is 0 ≠   ( r + x )( a + e ) = ra + re + xa + xe  A , implies that  re  A 

+ soc ( E) a contradiction  or s( a + e ) = sa + se  A + soc ( E ) implies that se  A + soc ( E ) 

a contradiction  or ( r + x )s = rs + xs [ A + soc ( E ) :R E ], implies that  rs [ A + soc ( E ) 

:R E ] a contradiction .Thus [ A :R  E ] sA  = ( 0 ) .  

In similar steps , we can show that [ A :R E ]rA = ( 0 ) . 

 Proposition 19. If A is a WP – 2 – Absorbing submodule of  E with ( r , s , e ) is a WP – 

triple zero of  A for some r, s  R , e  E , then  [ A :R  E ] [ A :R  E ] e = ( 0 ) . 

Proof : Suppose that [ A :R  E ] [ A :R  E ] e  ≠ ( 0 ), then 0 ≠ xye  A for  some x , y  [ A :R  

E ]. For ( r , s , e ) is a WP – triple zero of  A, then  rse = 0, re  A + soc ( E ) , se   A + soc ( 

E ) and  rs  [ A + soc ( E ) :R E ].  

Now, 0 ≠ xye  A , implies that  either xe  A + soc ( E ) or ye   A + soc ( E ) or xy   [ A 

+ soc ( E ) :R  E ].  

Now , 0 ≠ ( r + x )( s + y )e = rse + rye + xse + xye = xye  A (since rse = 0 , rye = 0 , xse = 0  

by proposition (16)).It follows that either  ( r + x )e = re + xe  A + soc ( E ), implies that  re 

 A + soc ( E ) a contradiction . or ( s + y )e = se + ye  A + soc ( E ), implies that  se  A + 

soc ( E ) a contradiction, or ( r + x )( s + y ) = rs + ry + xs + xy [ A + soc ( E ) :R E ], implies  

that  rs [ A + soc ( E ) :R E ] a contradiction. 

 Hence  [ A :R  E ] [ A :R  E ] e = ( 0 ) . 

Proposition 20.  If A is a WP – 2 – Absorbing submodule of  E with ( r , s , e ) is a WP – 

triple zero of  A for some r, s  R , e  E , then  [ A :R  E ] [ A :R  E ]A = ( 0 ) . 

Proof :  By proposition ( 14 ) and  proposition ( 19 ) .  

Proposition 21 .  Let A be a WP – 2 – Absorbing submodule of  E and  rsB  A for some r, s 

 R, and some submodule  B of  E with  ( r , s , x ) is not WP – triple zero of  A for every  x  

B . If  rs  [A + soc ( E ) :R  E ], then  rx  A + soc ( E )  or   sx  A + soc ( E ) . 

Proof : Suppose that  ( r , s , x ) is not WP – triple zero of  A for every  x  B and suppose 

that  rB  A + soc ( E ) and sB  A + soc ( E ), then ry1  A + soc ( E ) or  sy2  A + soc ( E 

) for some  y1, y2  B. If  0 ≠ rsy1 A with  rs  [A + soc ( E ) :R  E ] and since ry1  A + soc 

( E ) then sy1  A + soc ( E ) ( for A is a WP – 2 – Absorbing submodule ) . If  rsy1 = 0 and  

ry1  A + soc ( E ) , rs  [A + soc ( E ) :R  E ] and ( r , s , y1 ) is not WP – triple zero of  A , 

we get  sy1 A + soc ( E ). By similar arguments since ( r , s , y2 ) is not WP – triple zero of  

A , we get ry2  A + soc ( E ). Now , rs (y1 + y2  )  A and ( r 1 s , y1 + y2  ) is not WP – triple 

zero of  A and  rs  [A + soc ( E ) :R  E ], we get  r (y1 + y2  )  A + soc ( E ) or  s(y1 + y2  )  

A + soc ( E ) . 

If  r (y1 + y2  ) = ry1 + ry2  A + soc ( E ) and  ry2  A + soc ( E ), we get  ry1  A + soc ( E ) 

is a contradiction . 



  

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If  s (y1 + y2  ) = sy1 + sy2  A + soc ( E ) and  sy1 A + soc ( E )  then  sy2   A + soc ( E )is a 

contradiction . 

Hence  rB  A + soc ( E )  or  sB  A + soc ( E )  . 

Proposition 22.  Let A , B be WP – 2 – Absorbing submodule of  E with B is not contained in 

A and either soc ( E )  A or soc ( E )  B . Then A ∩ B  is a WP – 2 – Absorbing submodule 

of  E . 

Proof : It  is clear that  A ∩ B  is a proper submodule of  B  and  B  is a proper submodule of  

E , implies that A ∩ B  is a proper submodule of  E. Let ( 0 ) ≠ rsL  A ∩ B  for r , s  R , L 

is a submodule of E, it follows that ( 0 ) ≠ rsL  A and  ( 0 ) ≠ rsL  B . But A, B are WP – 2 

– Absorbing submodule of  E, then either rL  A + soc ( E )  or sL  A + soc ( E )  or rsE  

A + soc ( E )  and  rL  B+ soc ( E )  or sL  B+ soc ( E )  or rsE  B+ soc ( E ). Thus , either  

rL  ( A + soc ( E )) ∩ (B+ soc ( E ))  or  sL  ( A + soc ( E )) ∩ (B+ soc ( E ))  or  rsE  ( A 

+ soc ( E )) ∩ (B+ soc ( E )). If soc( E )  B  then B + soc( E ) = B, it follows that either rL   

(A + soc( E )) ∩ B  or  sL  (A + soc( E )) ∩ B  or   rsE  (A + soc( E )) ∩ B. Again Since 

soc( E )  B, then by Modular Law ( A + soc( E )) ∩ B = ( A ∩ B ) + soc ( E ). Thus either rL 

 ( A ∩ B ) + soc ( E ) or sL  ( A ∩ B ) + soc ( E )  or   rsE  ( A ∩ B ) + soc ( E ). Thus A 

∩ B is a WP – 2 – Absorbing submodule of  E . 

     Recall that for any submodules A , K a multiplication R – module E with A = IE , B = JE , 

for some ideals I, J of  R, the product  AB = IJE = IB. In particular AE = IEE = IE = A, and 

for any x  E , A = Ix  [2] .  

     The following propositions are characterizations of WP – 2 – Absorbing submodules is 

class of multiplication modules. 

Proposition 23.  Let E be a multiplication R – module, and A be a proper submodule of E. 

Then A is a WP – 2 – Absorbing submodule of  E if and only if  ( 0 ) ≠ L1L2L3   A  for some 

submodules  L1 , L2 , L3  of  E  implies that  either  L1L3   A + soc( E )  or   L2L3  A + soc( E 

)  or  L1 L2  A + soc( E ) .  

Proof : (  ) Let ( 0 ) ≠ L1L2L3   A  for some submodules L1 , L2 , L3  of  E . But  E  is a 

multiplication , then L1 = I1E ,  L2  = I2E , L3 = I3E   for some ideals  I1 , I2 , I3 of R . That is 

(0) ≠ L1L2L3 = I1 I2 I3 E  A. But A is a WP – 2 – Absorbing submodule of  E, then by 

proposition ( 6 ) either  I1 I3E   A + soc ( E )  or  I2 I3E  A + soc (E)  or   I1I2  [A + soc ( E 

) : R  E ] ( ie  I1 I2E  A + soc ( E ) ). It follows that either   L1L3   A + soc( E )   or   L2L3  

A + soc( E )   or   L1 L2  A + soc( E ) . 

(  ) Let ( 0 ) ≠ I1 I2L  A  for  I1 , I2  are  ideals of  R , L is submodule of  E. Since E is a 

multiplication, then L = I3E for some ideal I3 of R. That is (0) ≠ I1 I2 I3 E  A. Put  L1 = I1E  

and  L2  = I2E, then ( 0 ) ≠ L1L2L  A, it follows by hypothesis that  either  L1L  A + soc ( E 

) or  L2L  A + soc ( E ) or  L1L2  A + soc ( E ). That is either I1L  A + soc ( E )  or  I2L  

A + soc ( E )  or  I1I2E   A + soc ( E ), ( ie I1I2   [ A + soc ( E ) :R E ]. Thus , by proposition 

( 6 ) A is a WP – 2 – Absorbing submodule of  E .  



  

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  The following corollary is a direct consequence of proposition ( 23 ) . 

Corollary 24.  Let E be a multiplication R – module and A be a proper submodule of E. Then 

A is a WP – 2 – Absorbing submodule of  E if and only if  ( 0 ) ≠ L1L2 e   A  for  some 

submodules  L1 , L2  of  E and e  E, implies that either  L1e  A + soc ( E )  or  L2e   A + 

soc ( E )   or   L1 L2  A + soc ( E ) . 

   It is well known that if E is a faithful  multiplication R – module then  soc (E) = soc (R)E 

[11,coro. (2.14) (i)]. 

Proposition 25.  Let E be a faithful  multiplication R – module and A be a proper submodule 

of E. Then A is a WP – 2 – Absorbing submodule of  E  if and only if  [ A :R E ] is a WP – 2 – 

Absorbing ideal of  R .  

Proof : (  ) Let (0) ≠  I1I2 I3  [ A :R E ] for I1 , I2, I3  are ideals of R , it follows that  (0) ≠ 

I1I2 I3 E  A. But E is a multiplication then  (0) ≠ I1I2 I3 E = L1L2L3    A   by  taking  L1 = I1E 

, L2  = I2E  and  L3 = I3E . Now since A is a WP – 2 – Absorbing , then by proposition (23)  

either   L1 L3  A + soc ( E )  or   L2L3  A + soc ( E )  or   L1L2  A + soc ( E ) . But E is a 

faithful  multiplication then soc (E) = soc (R)E . Thus either  I1 I3E   [ A :R E ]E  +  soc ( R 

)E  or   I2 I3 E    [ A :R E ]E  +  soc ( R )E   or   I1I2E   [ A :R E ]E  +  soc ( R )E . That is 

either  I1 I3   [ A :R E ] +  soc (R)  or   I2 I3    [ A :R E ]  +  soc (R)   or   I1I2   [ A :R E ]  +  

soc ( R ) =[ [A :R E ]  +  soc ( R ) :E R ]. Therefore by proposition ( 6 ) [ A :R E ] is a WP – 2 – 

Absorbing ideal of  R . 

(  ) Let ( 0 ) ≠ I1I2L  A  for I1 , I2  are  ideals of  R  and  L is submodule of  E. Since E is a 

multiplication , then L = I3E  for  some ideal  I3  of  R .  That is ( 0 ) ≠ I1 I2 I3 E  A , it 

follows  that (0) ≠ I1 I2 I3   [ A:R E ] . But [ A:R E ] is a WP – 2 – Absorbing ideal of  R , then 

by proposition (6) either  I1 I3   [ A :R E ] +  soc ( R )  or   I2 I3    [ A :R E ]  +  soc ( R ) or   

I1I2   [ A :R E ]  +  soc ( R ) .Thus either  I1 I3E   [ A :R E ]E  +  soc ( R )E  or  I2 I3 E    [ A 

:R E ]E  +  soc ( R )E  or   I1I2E   [ A :R E ]E  +  soc ( R )E . That is either  I1L  A + soc ( E )  

or   I2L  A + soc ( E )  or   I1I2E   A + soc ( E ), ( ie I1I2   [ A + soc ( E ) :R E ]. Hence by  

proposition (6)  A is a WP – 2 – Absorbing  submodule  of   E .  

It is well known that cyclic R- module is multiplication [10]. We get the following corollary: 

Corollary 26.  Let E be faithful cyclic R – module and A be a proper submodule of E. Then 

A is a WP – 2 – Absorbing  if and only if   [ A :R E ] is a WP – 2 – Absorbing ideal of  R . 

Proposition .27.  Let E  be a faithful  finitely generated  multiplication R – module and  I  be 

a WP – 2 – Absorbing ideal of  R . Then , IE  is a WP – 2 – Absorbing submodule of  E . 

Proof : Let  (0) ≠ rI1K  IE  for  r  R , I1 be an ideal of  R , K is a submodule of  E. It 

follows that 0 ≠ rI1I2E    IE f or some ideal I2 of R. Since E is a finitely generated  

multiplication , then by[2  coro. of  Theo. ( 9 )]  we have  0 ≠ r I1I2    I  + ann( E ) = I . But I 

is a WP – 2 – Absorbing , then,by corollary (8)  either  rI2     I  + soc ( R )  or   I1I2   I  + 

soc ( R )   or    rI1  [ I  + soc ( R ) :R R] = I  + soc ( R ). That is either   rI2E  IE  + soc ( R )E 

or I1I2E    IE  + soc ( R )E   or   rI1E    IE  + soc ( R )E . Thus either   rK   IE  + soc ( E ) 



  

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or I1K  IE  + soc ( E )  or    r I1  [ IE  + soc ( E ) :R E ]. Therefore ,by corollary (8) IE is a 

WP – 2 – Absorbing submodule of  E . 

It is well known that cyclic R – modules are finitely generated [8], we get the following 

corollary which is a direct consequence of proposition (27)  

Corollary 28. Let  E be a faithful  cyclic R – module , and  I be a WP – 2 – Absorbing ideal 

of R. Then , IE  is a WP – 2 – Absorbing submodule of  E . 

3 . Conclusion . …  A new generalization of weakly – 2 – Absorbing submodule was 

introduced , and  many characterizations were given. The definition of WP – triple zero of 

WP – 2 – Absorbing submodules were introduced. A lot of basic properties of these concepts 

were established. Among the main new characterizations of WP – 2 – Absorbing submodules 

are the following   

 A proper submodule A of E is a WP- 2 – Absorbing if and only if  for  any  t , s  R with  ts 

 [ A + soc( E ) :R E ] ; we have [ A :E  ts ]  [ 0 :E  ts ]  [ A + soc ( E ) :E  t ]  [ A + soc ( E 

) :E  s ]. 

 A proper submodule A of E is a WP- 2 – Absorbing if and only if  0 ≠ tsK  A  for t , s  R  

and K is a submodule of  E , implies that either  tK   A + soc ( E )  or  sK  A + soc ( E )  or   

ts  [ A + soc( E ) :R E ]. 

 A proper submodule A of a cyclic R – module E is a WP- 2 – Absorbing if and only if  for  

each   t , s  R  with  ts  [ A + soc( E ) :R E ] , we have [ A :R  tse ]  [ 0 :R  tse ]  [ A + soc 

( E ) :R  te ]  [ A + soc ( E ) :R  se ]. 

 A proper submodule N of  E is a WP- 2 – Absorbing if and only if ( 0 ) ≠ IJL  N for some 

ideal  I , J of  R and submodule  L of  E  implies that either  IL  N+ soc( E)  or  JL N+ soc( 

E)  or  IJ  [ N+ soc( E) :R  E ] .  

  If  A is a WP- 2 – Absorbing submodule of  E with ( r , s , e ) is a WP – triple zero of A for  

some t , s  R, e  E. Then ,  rsA = (0) ,  [ A :R  E ] re = (0), r [ A :R  E ]e = (0),  r [ A :R  E ] A 

= (0),             [ A :R  E ]sA = (0)  and [ A :R E ] [ A :R E ] A = (0) .  

 A proper submodule A of multiplication module E is a WP- 2 – Absorbing if and only if  ( 0 

) ≠ L1L2L3    A for  some submodules  L1 , L2 , L3 of  E  implies that either  L1L3  A + soc ( 

E )   or L2L3  A + soc ( E )    or    L1L2   A + soc ( E ) .  

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