2009) 3( 22مجلة ابن الھیثم للعلوم الصرفة والتطبیقیة               المجلد

 حول المقاسات الجزئیة األولیة الضعیفة

 

 أنعام محمد علي هادي

ابن الهیثم،جامعة بغداد-قسم الریاضیات ،كلیةالتربیة   

  

  الخالصة

یكـون  Mفـي  Nُنعـرف ان مقاسـاً  جزئیـاً  فعلیـاً  . Rمقاساً  أیسـر علـى  Mحلقة ابدالیة ذا محاید ولیكن  Rلتكن         

فـي .  r  (N:M)أو    x  Nیـؤدي الـى   r x  N ≠ 0 و   ، x  Mو   ، r  Rأولیـاً  ضـعیفاً  اذا كـان لكـل 

ا كـان ، یسـمى أولیـاً  ضـعیفاً  اذRفـي  Pان مثالیـاً  فعلیـاً  اذ الحقیقة ان هذا المفهوم هو تعمیم لمفهوم مثالي أولي ضعیف، 

  . a  P أو  b  Pیؤدي الى ان     a b  P ≠ 0و  ، a, b  Rلكل 

  .خواص مختلفة عن المقاسات الجزئیة األولیة الضعیفة قد أعطیت

  

  

  

  

  

 
 
 
 
 
 
 
 
 
 
 



IBN AL- HAITHAM J. FO R PURE & APPL. SC I        VO L.22 (3) 2009 

 
On Weakly Prime Submodules 

 
 
 
 

I. M.A.Hadi 
Department of Mathematics,Ibn-Al-Haitham College of Education  
University of Baghdad  
 

Abstract 
        Let R be a commutative ring with unity  and let M be a left R-module. We define a 
proper submodule N of M to be a weakly prime if whenever  r  R,  x  M, 0  r x  N 
implies  x  N  or  r  (N:M). In fact this concept is a generalization of the concept weakly  
prime ideal, where a proper ideal P of R is called a weakly prime, if for all a, b  R, 0  a b  
P implies a  P or b  P. Various properties of weakly prime submodules are considered. 

    

 1.Introduction 
        Throughout this paper, R be a commutative ring with identity  and M be a unity  R-
module. A proper submodule N of M is said to be Prime if whenever        r  R , x  M , rx  
N implies either x  N or r  (N:M), where                   (N:M) = {r  R: r M  N}, see (1). 
        Semiprime submodules was given by Dauns in (2), as a generalization of prime 
submodules, where a proper submodule N of M is semiprime if rk

x  N, for r  R, x  M, k 
 Z+ (set of positive integers) implies rx  M. Also Eman A.A. in (3) studied these notions. 
        In 1999, quasi-prime submodules was introduced and studied by Muntaha (see (4)), as 
another generalization of prime submodules, where a proper submodule N of M is quasi-
prime if r1r2m  N, for r1, r2  R, m  M implies r1m  N or r2m  N; equivalently, N is 
quasi-prime if the ideal (N:M) is prime for all m  M. 
        In 2004, M.Behoodi and H.Koohi in (5) gave the notion of weakly prime submodules,  
where a proper submodule N of M weakly prime if (N:K) is a prime ideal, for all submodules  
K of M. Also this notion was studied by A.Azizi in (6), 2006. 
        By Th.2.14 in (4), we obtain that the two concepts weakly prime submodules and quasi-
prime submodules are equivalent. 
        In this paper, we give another generalization of prime submodules namely weakly prime 
submodules, however this concept is different from the concept of quasi-prime submodule 
(see Remarks 2.1.(5)).  
        In fact, D.D.Anderson and E.Smith in (7) gave the following: A proper ideal I of R is 
said to be a weakly prime if 0  ab  I, for a, b  R, then a  I or      b  I. We define a 
proper submodule N of M is weakly prime if whenever          r  R , x  M , 0  rx  N 
implies x  N or r  (N:M). Moreover S.E.Atani and F.Farzalipour in (8) introduced the 
notion of weakly primary submodules, where a proper submodule N of M is a weakly primary 
if whenever r  R , x  M , 0  rx  N implies x  N or r

n  (N:M) for some n  Z+. Also 
they gave that : a proper ideal of R is a weakly primary if it is a weakly prime submodule of 
the R-module R, (see (8)). 
        In this paper we study weakly prime submodules and give many basic properties related 
to this concept. 
 

 



IBN AL- HAITHAM J. FO R PURE & APPL. SC I        VO L.22 (3) 2009 

2.Basic Properties 

     As we mentioned in the introduction, we introduce the following : 
Definition 2.0 :  A proper submodule N of an R-module Mis weakly prime if whenever rR, 
x M, 0 rxN implies  xN or  r(N:M). 
 In this section, we will give basic properties of weakly prime submodules. Some of these are 
extension of the results about weakly prime ideals, which are given in (7). 
        Let us start with the following: 
 
 Remarks 2.1:  
(1) It is clear that every prime submodule is weakly prime. However, since (0) the zero 

submodule of any module) is always weakly prime (by definition), a weakly prime 
submodule may not be prime; for example: the zero submodule of the Z-module Z4 is 
weakly prime, but it is not prime. 
Moreover it is easy  to check that in the class of torsion free modules, the concepts of 
prime submodule and weakly prime submodule are equivalent. 

(2) Every weakly prime ideal P of a ring R is a weakly prime submodule of the R-module 
R. 

(3) Every weakly prime submodule is weakly primary, but the converse is false as the 
following example shows. 

The submodule N = )4(  of the Z-module Z8 is weakly primary but it is not weakly 
prime. 

(4) It is easy  to check that :if P is a weakly primary submodule of an R-module M and 
(P:M) is a semiprime ideal, then P is weakly prime. 

(5)  (a) Weakly prime submodule need not be quasi-prime as the following example shows: 
The zero submodule of the Z-module Z12 is weakly prime, but it is not quasi-prime 

since (0 : 3)  = 4Z which is not a prime ideal of Z. 

 (b) Quasi-prime submodule need not be weakly prime submodule, as the following 
example shows 
If M is the Z-module ZZ, and N = 2Z(0), then N is a quasi-prime submodule of M 
(see (4), Rem.2.1.2(1)). But N is not weakly prime submodule, since (0,0)  2 (3,0)  
N, (3,0)  N and 2  (N:M) = (0). 

(6) If P is a proper submodule of an R-module M. Then P is a weakly prime  R-submodule 
of M iff P is a weakly prime R / I-submodule of M, where I is an ideal of R with  I  
ann M. 

        The following result gives characterizations of weakly prime submodules. 
 Theorem 2.2 :    Let M be an R-module. The following asserations are equivalent: 
1. P is a weakly prime submodule of M. 
2. (P:x) = (P:M)  (0:x) for any x  M, x  P. 
3. (P:x) = (P:M) or (P:x) = (0:x) for any x  M, x  P. 
4. If (0)  (a)N  P, then either N  P or (a)  (P:M), where a  R, N is a submodule of 

M. 
Proof.  (1)  (2)   Let r  (P:x) and x  P. Then  r x  P. Suppose  r x ≠ 0. Hence  r  (P:M) 
because P is weakly prime and x  P. If r x = 0, then               r  (0:x). Thus (P:x)  (P:M)  
(0:x). Now if r  (P:M)  (0:x), then either             r  (P:M) or r  (0:x). Hence, when r  
(0:x), r x = 0  P and so r  (P:x). If                 r  (P:M) then  r M  P, and this implies  r x 
 P. Hence  r  (P:x). Thus       (P:M)  (0:x)  (P:x) and therefore (P:M)  (0:x) = (P:x). 
 (2)  (3)   It is well-known that the union of two ideals A, B of R is an ideal if A  B or B  
A. By condition, the ideals (P:M) is the union of the ideals         (P:M), (0:x), so either (P:M) 
 (0:x) or (0:x)  (P:M). Thus either (P:x) = (0:x) or (P:x) = (P:M). 
 

 



IBN AL- HAITHAM J. FO R PURE & APPL. SC I        VO L.22 (3) 2009 
 

 (3)  (4)   If 0  (a) N  P. Suppose that N   P and (a)   (P:M). N   P implies that 

there exists x  N and x  P, hence ax  aN  P; that is a  (P:x). By condition (3), either 
(P:x) = (P:M) or (P:x) = (0:x). If (P:x) = (P:M), we get        a  (P:M) which is a contradiction. 
Thus (P:x) = (0:x) and so ax = 0. 
On the other hand, 0  (a) N  P implies that there exists y  N such that             0  ay  P 
and so a  (P:y). Moreover we can see that y  P, for if we assume that y  P, then by 
condition 3, either (P:y) = (P:M) or (P:y) = (0:y).                    If (P:y) = (P:M), then a  (P:M) 
which is a contradiction. If (P:y) = (0,y), we get ay = 0 which is a contradiction. Moreover 0  
ay = ay + ax = a(y + x)  P; that is a  (P:y + x). But y + x  P because x  P, y  P, hence 
by condition 3 , either (P: y + x) = (P:M) or (P: y + x) = (0: y + x). If (P: y + x) = (P:M) then          
a  (P:M) which is a contradiction. If (P: y + x) = (0: y + x), then a(y + x) = 0 and hence ay + 
ax = ay + 0 = 0 which is a contradiction. Therefore our assumption is false and so either N  
P or (a)  (P:M). 
 (4)  (1)   Let r  R, x  M, such that 0  r x  P. Then 0  (r) (x)  P. By condition (4), (x) 
 P or (r)  (P:M) and hence either x  P or r  (P:M); that is, P is weakly prime. 
 Remark 2.3 
        It is known that if P is a prime submodule of an R-module M, then (P:M) is a prime ideal 
of R. However the "weak" analogs of this statement is not true in general, for example: 

The zero submodule of the Z-module Z4, is weakly prime, but (0

:  Z4) = 4 Z is not a weakly 

prime ideal of Z. 
        We give the following: 
Proposition 2.4:    If P is a weakly prime submodule of a faithful R-module M, then (P

R
: M) 

is a weakly prime ideal of R. 
Proof.  Let a, b  R. If 0  a b  (P:M); then a b M  P. Since M is faithful,  a b M  (0), 
hence 0  (a) (b M)  P and so by  Th.2.2 either (a)  (P:M) or          b M  P; that is, either a 
 (P:M) or b  (P:M). Thus (P:M) is a weakly prime ideal of R. 
     The converse of prop.2.4 is not true as the following example shows: 

        Let M be the Z-module Z  Z, let P = (0)  2Z. Then (P

:  M) = (0) which is a weakly 

prime ideal in Z, however P is not a weakly prime submodule of M because (0,0)  2(0,1)  

P, but (0,1)  P and 2  (P

:  M) = (0). 

        Also, we have the following:- 
  Proposition 2.5 :     Let P be a weakly prime submodule of an R-module M. Then (P

R
:  M) is 

a weakly prime ideal of R , where  annRR / . 

Proof.  By remark 2.1 (6), P is a weakly prime R -submodule of M. But M is a faithful R -

module, so by prop.2.6, (P
R
:  M) is a weakly prime ideal of R . 

       Recall that an R-module M is called multiplication module if for each submodule N of M, 
N = I M for some ideal I of R, equivalently N = (N:M)M (see (9)). 
        In the class of finitely generated faithful multiplication modules, we have the following:   
Theorem 2.6:     Let M be a faithful finitely generated multiplication R-module, let N be a 
proper submodule of M. Then the following statements are equivalent 

1. N is a weakly prime submodule of M. 

2. (N
R
: M) is a weakly prime ideal of R. 

3. N = I M for some weakly prime ideal I of R. 
Proof.  (1)  (2)   It holds by prop. 2.4 
 

 



IBN AL- HAITHAM J. FO R PURE & APPL. SC I        VO L.22 (3) 2009 
 
(2)  (1)   Let r  R, x  M, such that 0  r x  N. (x) is a submodule of M, hence (x) = J M  
for some ideal J of R. Thus 0  r J M   N = (N:M) M. But M is a faithful finitely generated 

multiplication R-module, so by (10, Th.3.1)                r J  (N
R
: M). Moreover r J  (0) and 

since (N
R
: M) is weakly prime ideal, either  r  (N

R
: M) or J  (N

R
: M) (see Th.3 in (7)). 

Hence either r  (N
R
: M) or (x) = J M  (N

R
: M) M = N, that is r  (N

R
: M)  or  x  N. Thus N 

is weakly prime. 

 (2)  (3)   Since (N
R
: M) is weakly prime and N = (N

R
: M) M, so condition (3) hold. 

 (3)  (2)   By (3), N = I M for some weakly prime ideal I of R. But M is a multiplication 

module, so N = (N
R
: M) M. Hence I M = (N:M) M and so by      (10, Th.3.1)  I = (N

R
: M). 

    Proposition 2.7 :     Let P be a weakly prime submodule of an R-module M. If P is not 
prime, then (P:M) P = (0). 
Proof.  Suppose (P:M) P  0. We will show that P is prime. Let r x  P. If         r x  0, then 
either x  P  or  r  (P:M), since P is weakly prime. Now assume   r x = 0. First suppose  rP  
(0), so there exists  y  P such that 0  r y  P. Hence  0  r y = r (x + y)  P  which implies 
that either  x + y  P or  r  (P:M). Hence either  x  P  or  r  (P:M). Now we can assume 
that  r P = 0 and     (P:M) x = 0. Since (P:M) P  (0), there exists  s  (P:M), y  P such that            
0  s y  P. Thus 
(r + s) (x + y) = r x + s x + r y + s y 
                      = 0 + 0 + 0 + s y 
                      = s y 
That is 0  (r + s) (x + y)  P. Then P is weakly prime gives  x + y  P  or           r + s  
(P:M). Hence  x  P  or  r  (P:M). 
         Now we have the following: 
  Proposition 2.8:  Let M and M  be R-modules and let  f : M  M  be an R-epimorphism. If 
N is a weakly prime submodule of M such that  ker f  N, then f (N) is a weakly prime 
submodule of M . 
Proof.  Let r  R, y  M , such that 0  r y  f (N). Then there exists  x  N  such that 0  r y 
= f (x), and since f  is an epimorphism, y = f (x1) for some      x1 M. Thus  f (r x1 – x) = 0 and 
so r x1 – x  ker f  N. It follows that   0  r x1  N and since N is weakly prime either x1  
N  or  r (N

R
: M). Thus  y = f (x1)  f (N)  or  r  ( f (N): M ); that is f (N) is weakly prime. 

        As a particular case of prop.(2.8), we have the following: if N, W are submodules of an 
R-module M such that N  W and N is weakly prime, then   N / W is a weakly prime R-
submodule of M / N. 
        The following result discussos the localization of weakly prime submodules. 
 Proposition 2.9 :     Let P be a weakly prime R-submodule and S be a multiplicative subset 

of R with (P
R
:  M)  S =  . Then PS is weakly prime            RS-submodule of M S. 

Proof.  Let 
b

a
 RS and 

c

x
 M S such that 0s  

b

a

c

x
  PS. Hence 0s  

cb

xa
  PS and so there 

exists  y  P and  d  S such that 
d

y

cb

xa
  , and this implies that there exists  t  S  such that  

t a d x = t b c y. On the other hand 
1

0


cb

xa
= (0s) which implies that  f a x  0  for all  f  S. 

Hence 0  t a d x  P. But P is a weakly prime R-submodule of M, so either  t d x  P  or  a   
 
 
 



IBN AL- HAITHAM J. FO R PURE & APPL. SC I        VO L.22 (3) 2009 

 (P:M) and hence either 
cdt

xdt
  PS  or  

b

a
  (P:M)S. Because (P

R
:  M)S  (PS

SR
: M S), we have 

either  
c

x
 PS  or  

b

a
  (PS

SR
: M S). 

        As a generalization of Cohen theorem, the following was given in ((3),Prop.4.15,ch.1). 
        Let M be a finitely generated R-module, then M is Noetherian iff every prime submodule 
is finitely generated. 
        Since every prime submodule is weakly prime (by Rem.2.1.(11)), we have the following: 
 Proposition2.10 :    Let M be a finitely generated R-module. Then M is Noetherian if every 
weakly prime submodule is finitely generated. 
        Noteice that the condition M is finitely generated that cannt be dropped from Prop .2.10, 
as the following example shows: 

        The Z-module Z p
 is not finitely generated, also it is not Noetherian. However if G is 

a nonzero submodule, then 
1

G  
i

z
p

 for some i  Z+, and 0  P(
1


i +1

z
p

)  G. 

But P  (G: Z p
) = 0 and 

1


i +1
z

p
  G; that is G is not weakly prime. Thus (0) is the 

only weakly prime submodule of Z p
 and it is obviously finitely generated. 

 Theorem 2.11:    Let M 1, M 2 be R-modules and let N be a proper submodule of M1. Then W 
= N  M 2 is a weakly prime submodule of  M = M 1 M 2  N is a weakly prime submodule 
of M 1 and for  r  R, x  M 1 with r x = 0, but x  N, r  (N:M 1) implies  r  ann M 2. 
Proof. () Let r  R, x  M 1 such that 0  r x  N. Then (0,0) ≠ r (x,0)  W, but W is 

weakly prime, so either (x,0)  W  or  r  (W
R
: M). Thus either x  N or  r  (N

R
: M 1), so that 

N is weakly prime. Now, if r  R, x  M 1 such that        r x = 0, x  N and r  (N:M 1). 
Assume that r  ann M 2, so there exists m  M 2 such that r m2  0. Thus r (x,m2) = (r x,r m2) 
= (0, r m2)  (0,0) and hence      (0,0)  r (x,m2)  N  M 2 = W. Since W is weakly prime, so 

either (x,m2)  N  M 2  or  r  (N  M 2
R
: M 1  M 2). Thus either x  N  or   r  (N

R
: M 1) 

which is a contradiction with hypothesis. 
 ()  Let r  R, (x,y)  M. Assume (0,0)  r (x,y)  N  M 2, so if  r x  0, then either  x  N  
or  r  (N:M 1), since N is weakly prime. Thus either                (x,y)  N  M 2  or  r  (N  
M 2:M1  M 2). If  r x = 0. Suppose x  N and        r  (N1:M1), then by hypothesis  r  ann 

M 2  and so  r (x,y) = (0,0)  which is a contradiction. Thus either  x  N  or  r  (N1
R
: M 1)  and 

hence either   (x,y)  N  M 2  or  r  (N1  M 2
R
: M 1  M 2). 

       It is known that if Q is a primary submodule then ):(  Q  is a prime ideal, see (11, 

prop. 2.11, p.41). Sometimes Q is called P-primary, see(11, p.42). 
        Now we have the following result: 
 Corollary 2.12:    Let Q be P-primary submodules of an R-module M 1 with  Q = (0). If 
N is a weakly prime submodule of M 1 and M2 is an R-module such that P  annR M 2, then N 
 M 2 is a weakly prime submodule in   M 1  M 2.  
Proof.  Let r  R, x  M 1 with  r x = 0. If  x  N1 (so  x  0) and  r  (N:M 1). We will prove 
that  r  ann M 2 and hence the result is obtained by previous theorem. Suppose that  r  ann 
M 2. Hence  r  P. 
        On the other hand,  r x = 0 =  Q, but  Q  is a P-primary submodule by (11,prop.1.1, 
p.15), so either  x   Q = 0  or  r  P, which is a contradiction. Thus  r  ann M 2. 
 



IBN AL- HAITHAM J. FO R PURE & APPL. SC I        VO L.22 (3) 2009 
 
  Remark 2.13:    Let M 1, M 2 be R-modules. If (0) is a prime submodule of M1, then (0)  M2 
is a weakly prime submodule of M = M 1  M 2. 
Proof.  Let r  R, (x,y)  M. If (0,0)  r (x,y)  (0)  M 2, then  r x = 0  and    r y  M 2. Since 
(0) is prime in M 1, either  x = 0  or  r  (0:M1). Hence either         (x,y) = (0,y)  (0)  M 2  or  
r  ((0) + M 2:M1  M 2); that is (0)  M 2 is weakly prime in M. 
        Thus we can give the following example: 
N = (0)  Z4 is a weakly prime submodule of the Z-module Z  Z4. 
        Next we have the following: 
Proposition 2.14:    Let M 1, M 2 be R-modules. If N = U  W be a weakly prime submodule 
in M = M 1  M 2, then U, W are weakly prime submodules in M1, M 2 respectively. 
Proof.  The proof is a straight forword, so it is omitted. 
       Remark 2.15:    The converse of proposition 2.14 is not true in general as the following 
example shows. 
  Example:    (0) is a weakly prime submodule of the Z-module Z, (2Z) is a prime submodule 
of the Z-module Z so it is weakly prime. But N = (0)  2Z is not weakly prime in the Z-
module Z  Z. 
        For the next results we will assume that R = R1  R2 where each Ri is a commutative ring 
with identity , M i be an Ri-module, where i = 1,2. and            M = M 1  M 2 be the R-module 
with action (r1,r2) (m1,m2) = (r1 m1, r2 m2) where ri  Ri, mi  M i, i = 1,2. 
 Proposition 2.16 :    If P is a proper R1-submodule of M 1, then the following statements are 
equivalent 
1. P is a prime R1-submodule of M 1. 
2. P  M 2 is a prime R-submodule of M = M 1  M 2. 
3. P  M 2 is a weakly prime R-submodule of M = M 1  M 2. 
Proof.  (1)  (2)  Let (r1,r2)  R, (x,y)  M such that (r1,r2) (x,y)  P  M 2. Then  r1 x  P 
and since P is prime, either x  P  or  r1  (P

1

:
R

M 1). If      x  P, then (x,y)  P  M 2. If r1  

(P
1

:
R

M 1), then (r1,r2)  (P  M 2
R
: M). Thus     P  M 2 is a prime R-submodule of M. 

 (2)  (3)  It holds by remark 2.1 (1). 
 (3)  (1)  Let r  R1, x  M 1 such that r x  P. Then for each y  M 2, y  0, (0,0)  (r,1) 
(x,y)  P  M 2. But P  M 2 is a weakly prime R-submodule of M, so either (x,y)  P  M 2  or  

(r,1)  (P  M 2
R
: M). Thus either x  P  or                r  (P

1

:
R

M 1); that is P is a prime R1-

submodule of M 1. 
        Similarly we have 
 Proposition 2.17:    If P is a proper R2-submodule of M 2, then the following statements are 
equivalent 

1. P is a prime R2-submodule of M 2. 

2. M 1 P is a prime R-submodule of M = M 1  M 2. 

3. M 1 P is a weakly prime R-submodule of M = M 1  M 2. 
Proposition 2.18:     Let M 1, M 2 be R1, R2-modules respectively. If     P =P1  P2 is a weakly 
prime R-submodule of M = M 1  M 2, then either P = 0 or P is a prime submodule of M. 
Proof.  Assume P  0, so either P1  0  or  P2  0. Suppose that P2  0, hence there exists y  
P2, y  0. Let r  (P1

1

:
R

M 1) and let x  M 1, then                       (0,0)  (r,1) (x,y) = (r x,y)  P1 

 P2 = P. Since P is weakly prime in M, either (x,y)  P  or  (r,1)  (P1  P2: M1  M 2). Hence 
if (x,y)  P, then x  P1 and so M 1 = P1 which implies P = M 1  P2. If (r,1)  (P1  P2: M1  
M 2), then M2 = P2 which implies P = P1  M 2. Hence by propositions 2.16, 2.17, P is a prime 
R-submodule of M. 
 

 



IBN AL- HAITHAM J. FO R PURE & APPL. SC I        VO L.22 (3) 2009 

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