/compile/output.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 9, No. 1, 2016, 48-56

ISSN 1307-5543 – www.ejpam.com

A Note on Primary and Weakly Primary Submodules

Gulsen Ulucak1,∗and Rabia Nagehan Uregen2

1 Department of Mathematics, Faculty of Science, Gebze Technical University, 141 41400, Kocaeli,

Turkey
2 Yildiz Technical University, Graduate School of Natural and Applied Sciences, 34349, Istanbul,

Turkey

Abstract. In this paper, the generalizations of primary submodules and weakly primary submodules

are proposed as P(N)-locally primary submodules and P(N)-locally weakly primary submodules, re-

spectively. The relationships of these submodules are investigated extensively.

2010 Mathematics Subject Classifications: 13A15, 13C99, 13F05

Key Words and Phrases: Primary, Weakly primary, P(N)-locally primary, P(N)-locally weakly primary.

1. Introduction

Throughout this paper, we assume that all rings are commutative with identity 1 6= 0. An

ideal I of R is called a proper ideal if I 6= R. Then the radical of a proper ideal I of R is denoted

by rad(I) and rad(I) = {x ∈ R | xn ∈ I for some positive integer n}. A proper ideal P of R is

called prime (primary) if ab ∈ P for some a, b ∈ R implies that either a ∈ P or b ∈ P (either

a ∈ P or bn ∈ P for some positive integer n). A proper ideal P of R is said to be a weakly

prime ideal if 0 6= ab ∈ P for some a, b ∈ R implies that either a ∈ P or b ∈ P, and it is called

a weakly primary ideal if 0 6= ab ∈ P for some a, b ∈ R implies that either a ∈ P or bn ∈ P for

some positive integer n (see [2, 3]).

Let M be an R-module. A submodule N of M is called a proper submodule if N 6= M .

A proper submodule N of M is called a prime submodule if rm ∈ N for some r ∈ R and

m ∈ M implies that either m ∈ N or rM ⊆ N and it is said to be a weakly prime submodule

if 0 6= rm ∈ N for some r ∈ R and m ∈ M implies that either m ∈ N or rM ⊆ N . A non

empty subset S of R is said to be multiplicative closed set if 0 /∈ S and whenever a, b ∈ S, then

ab ∈ S. Let S be a multiplicative closed set in R. It can be easily seen that MS is an RS-module

under the operations a
s +

b
u =

ua+sb
su and r

v
a
s =

ra
vs for any r

v ∈ RS and a
s , b

u ∈ MS [5]. A proper

submodule N of M is said to be S(N)-locally prime (S(N)-weakly prime) submodule if NM is

a prime (a weakly prime) submodule of MM for each maximal ideal M with S(N) ⊆M [4].

∗Corresponding author.

Email addresses: gulsenulucak@gtu.edu.tr (G. Ulucak), rnuregen@yildiz.edu.tr (R. Uregen)

http://www.ejpam.com 48 c© 2016 EJPAM All rights reserved.



G. Ulucak and R. Uregen / Eur. J. Pure Appl. Math, 9 (2016), 48-56 49

A proper submodule N of M is said to be a primary submodule if rm ∈ N for some r ∈ R,

m ∈ M implies that either m ∈ N or rnM ⊆ N for some positive integer n and it is said to be a

weakly primary submodule if 0 6= rm ∈ N for some r ∈ R, m ∈ M implies that either m ∈ N or

rnM ⊆ N for some positive integer n. The ideal {r ∈ R | rM ⊆ N} will be denoted by (N : M)

and (0 : N) = {r ∈ R | rN = 0} where N is a submodule of M . Then the annihilator of M

is (0 : M) where (0 : M) = {r ∈ R | rM = 0}. An R-module M is called a faithful module if

(0 : M) = (0). Note that if N is a primary submodule of M , then (N : M) is a primary ideal

of R and rad(N : M) = {r ∈ R | rnM ⊆ N for some positive integer n} is a prime ideal of R

([1, 6, 7]).

Main aim is to obtain the two generalization on primary submodules and weakly primary

submodules of an R-module M.Let N be a proper submodule of M . An element r ∈ R is said

to be primary to N if rnm ∈ N , where m ∈ M and n is a positive integer, then m ∈ N . Then

r ∈ R is said to be not primary to N if rnm ∈ N for some positive integer n and for some

m ∈ M \ N . The set of all elements of R that are not primary to N is denoted by P(N). Then

we get P(N) = {r ∈ R | rnm ∈ N for some positive integer n, for some element m ∈ M \N}. If

N = (0), then P((0)) = {r ∈ R | rnm = 0 for some positive integer n, for some 0 6= m ∈ M}. A

proper submodule N of M is said to be an M-primal if P(N) forms an ideal of R.

2. P(N)-Locally Primary and P(N)-Locally Weakly Primary Submodules

Definition 1. Let N be a proper submodule of an R-module M. Then N is called a P(N)-locally

primary submodule of M if NM is a primary submodule of MM for all maximal ideal M where

P(N) ⊆M.

Definition 2. A proper submodule N of an R-module M is called a P(N)-locally weakly primary

submodule of M if NM is a weakly primary submodule of MM for every maximal ideal M where

P(N) ⊆M.

Lemma 1. Let N be a proper submodule of an R-module M. Then rad(N : M) ⊆ P(N).

Proof. Let r ∈ rad(N : M). Then rnM ⊆ N for some positive integer n. There exists

m ∈ M \ N such that rnm ∈ N . Then r ∈ P(N). Thus rad(N : M) ⊆ P(N).

Every primary submodule N is proposed as P(N)-locally primary submodule and every

weakly primary submodule N is proposed as P(N)-locally weakly primary submodule in the

following propositions, respectively.

Proposition 1. A primary submodule N of an R-module M is a P(N)-locally primary submodule.

Proof. Let M be a maximal ideal of R where P(N) ⊆M. By the previous lemma, we say

that rad(N : M) ⊆ P(N) ⊆ M. NM is a proper submodule of MM. Indeed, if NM = MM,

then m
1 ∈ MM for any m ∈ M . Then rm ∈ N for some r /∈ M. We get rn /∈ rad(N : M).

Since N is a primary submodule of M , then m ∈ N . Thus N = M , a contradiction. Since

rad(N : M) ∩ (R\M) = ;, then NM is a primary submodule of MM. Consequently, N is a

P(N)-locally primary submodule.



G. Ulucak and R. Uregen / Eur. J. Pure Appl. Math, 9 (2016), 48-56 50

Proposition 2. A weakly primary submodule N of an R-module M is a P(N)-locally weakly

primary submodule.

Proof. Suppose that M is a maximal ideal of R where P(N) ⊆M. In the same manner as

in the proof of the previous proposition, we have that NM is a proper submodule of MM. Let

0M 6=
r
s

m
p ∈ NM for some r

s ∈ RM and m
p ∈ MM (for some r ∈ R, m ∈ M and s, p ∈ R\M). Then

there is a q ∈ R \M such that qrm ∈ N . Assume that qrm= 0. Then r
s

m
p =

q
q

r
s

m
p =

qrm
qsp = 0M,

this is a contradiction. So 0 6= qrm ∈ N . As rad(N : M) ⊆ P(N) ⊆M, then q /∈ rad(N : M).

Thus rm ∈ N since N is a weakly primary submodule. It is clear that rm 6= 0. Hence

0 6= rm ∈ N implies that m ∈ N or rnM ⊆ N for some positive integer n. Thus we get
m
p ∈ NM or rn

sn MM ⊆ NM for some positive integer n by [4, Corollary 2.9]. Then we get that

NM is a weakly primary submodule of MM. Consequently, N is a P(N)-locally weakly primary

submodule.

Corollary 1. Let N be a proper submodule of an R-module M. If N is primary, then N is P(N)-

locally weakly primary.

Proof. Assume that N is a primary submodule. Then N is a weakly primary submodule.

Thus, N is a P(N)-locally weakly primary submodule by Proposition 2.

Note that if N is a P(N)-locally primary submodule of M , then N is a P(N)-locally weakly

primary submodule of M .

We give an example to show the converse is not true.

Example 1. Consider R= F[X , Y, Z]−module M = F[X , Y, Z]�(X 2, Y Z) and the zero submod-

ule N = (0) of M. One can easily see that P(N) = {0, X , Y, Z , . . .}. Note that

P(N) ⊆M= (X , Y, Z) which is the unique maximal ideal of R. Then NM = (0) is weakly primary

submodule of RM− module MM. Thus N is P(N)−weakly primary submodule. But NM is not

primary submodule since, Y
1 ·

Z̄
1 =

0̄
1 ∈ NM but
�

Y
1

�n
/∈ (NM : MM),
�

Z̄
1

�

/∈ NM for all positive

integer n. Thus N is not P(N)−primary.

In the following example, it is illustrated that a submodule N can be both P(N)-locally

primary submodule of M and P(N)-locally weakly primary submodule of M but it is neither

primary submodule of M nor weakly primary submodule of M .

Example 2. Let R = Z and consider the R − module M=Z12. Let N be the submodule of Z12

generated by 6. It is easily seen that 0 6= 23(= 32) ∈ N but 2 /∈ (N : M) and 3 /∈ N (3 /∈ (N : M)

and 2 /∈ N), that is, N is not a weakly primary submodule of M, hence N is not a primary

submodule of M. Assume that N is not a P(N)-locally primary submodule of M. Then there

exists a maximal ideal M of R with P(N) ⊆ M where NM is not a primary submodule of MM.

Note that 2,3 ∈ P(N). Thus 1 ∈ M, a contradiction. Therefore, N is a P(N)-locally primary

submodule of M. Hence N is a P(N)-locally weakly primary submodule of M.

Theorem 1. Let N be a proper submodule of an R-module M. Then the following statements

hold:



G. Ulucak and R. Uregen / Eur. J. Pure Appl. Math, 9 (2016), 48-56 51

(i) N is a primary submodule if and only if P(N) = rad(N : M).

(ii) Let P(0) ⊆ rad(N : M). Then N is a primary submodule if and only if N is a weakly

primary submodule.

Proof. (i) Assume that N is a primary submodule. Let r ∈ P(N). Then rnm ∈ N for some

positive integer n and for some m ∈ M \ N . Since N is a primary submodule, then

(rn)kM = rnkM ⊆ N for some positive integer k, that is, r ∈ rad(N : M). Hence

P(N) ⊆ rad(N : M). By Lemma 1, we get P(N) = rad(N : M).

Suppose that P(N) = rad(N : M). Let rm ∈ N and m ∈ M \ N where r ∈ R, m ∈ M . Then

r ∈ P(N). Thus r ∈ rad(N : M), that is, rkM ⊆ N for some positive integer k. Consequently,

N is a primary submodule.

(ii) It is clear that every primary submodule is a weakly primary submodule.

Now, assume that N is a weakly primary submodule. Let r ∈ P(N). Then rnm ∈ N for

some positive integer n and for some m ∈ M \N . Suppose that rnm= 0. Since m ∈ M \N , then

we get m 6= 0. So r ∈ P(0). Thus r ∈ rad(N : M), by assumption. Hence P(N) = rad(N : M)

by Lemma 1. Suppose that 0 6= rnm ∈ N . Since m ∈ M \ N and N is a weakly primary

submodule, then (rn)kM ⊆ N for some positive integer k, that is, r ∈ rad(N : M) and so

P(N) = rad(N : M). By (i), N is a primary submodule.

Corollary 2. Let N be a proper submodule of an R-module M with P(N) = rad(N : M). Then

N is a P(N)-locally primary submodule and P(N)-locally weakly primary submodule.

Proof. We get that N is a primary submodule by Theorem 1(i). Then N is a P(N)-locally

primary submodule by Proposition 1. Since N is primary submodule, then N is weakly primary

submodule. Therefore, N is P(N)-locally weakly primary submodule by Proposition 2.

Note that, by [4, Lemma 2.19], if M is a maximal ideal of R, then (N : M)M ⊆ (NM : MM).

Now, we explain that rad((N : M)M) = rad(NM : MM) when M is a maximal ideal of R with

P(N) ⊆M.

Proposition 3. Let N be a proper submodule of an R-module M. Then

rad((N : M)M) = rad(NM : MM) for any maximal ideal M of R with P(N) ⊆M.

Proof. Since S(N) ⊆ P(N) for any proper submodule N of M , it is clear from [4, Lemma

2.19 and Lemma 2.20].

Lemma 2. Let N be a proper submodule of an R-module M. Then

rad((N : M)M) = (rad(N : M))M for any maximal ideal M of R with P(N) ⊆M.

Proof. Let r
p ∈ rad((N : M)M) for some r ∈ R and p ∈ R \M. Then ( r

p )
n = rn

pn ∈ (N : M)M
for some positive integer n. There exists an element q ∈ R \M such that qrn ∈ (N : M),

that is, qrnm ∈ N for every m ∈ M . Then rnm ∈ N for every m ∈ M since q /∈ P(N). Thus

r ∈ rad(N : M). Then r
p ∈ rad((N : M)M). Conversely, assume that r

p ∈ rad((N : M)M).

There is an u ∈ R \M such that ur ∈ rad(N : M). Then (ur)n = unrn ∈ (N : M). Hence
un

un
rn

pn ∈ (N : M)M. Consequently, rn

pn = (
r
p )

n ∈ (N : M)M and so r
p ∈ rad((N : M)M).



G. Ulucak and R. Uregen / Eur. J. Pure Appl. Math, 9 (2016), 48-56 52

Corollary 3. Let N be a proper submodule of an R-module. If M is any maximal ideal of R with

P(N) ⊆M, then rad((N : M)M) = rad(NM : MM).

Proof. It is clear from Proposition 3 and Lemma 2.

Proposition 4. Let N be a proper submodule of an R-module M and m ∈ M. Then

rad((N : Rm)M) = rad(NM : (Rm)M) for any maximal ideal M of R with P(N) ⊆M.

Proof. It is clear.

If we put N = 0 in Proposition 4, we have the following corollary.

Corollary 4. Let M be an R-module and m ∈ M. Then rad((0 : Rm)M) = rad(0M : (Rm)M) for

any maximal ideal M of R with P(0) ⊆M.

Proposition 5. Let N be a proper submodule of an R-module M and M be a maximal ideal of R

with P(N) ⊆M. Then the following statements hold:

(i) Let P(0) ⊆ P(N). Then rad(N : M) is a weakly prime ideal of R if and only if rad((N : M)M)

is a weakly prime ideal of RM.

(ii) rad(N : M) is a prime ideal of R if and only if rad((N : M)M) is a prime ideal of RM.

Proof. (i) Suppose that rad(N : M) is a weakly prime ideal of R. If rad(N : M)M = RM,

then 1
1 ∈ rad((N : M)M) = (rad(N : M))M and so q1 = q ∈ rad(N : M) for some q ∈ R \M.

But by Lemma 1, rad(N : M) ⊆ P(N) ⊆M, which is a contradiction. So rad((N : M)M) 6= RM,

that is, rad((N : M)M) is a proper ideal of RM. Let 0 6= r
p

s
q ∈ rad((N : M)M), where r, s ∈ R

and p,q ∈ R \M. Then we have r
p

s
q =

rs
pq ∈ (rad(N : M))M, then there exists an u ∈ R \M

such that urs ∈ rad(N : M). If urs = 0, then r
p

s
q =

u
u

r
p

s
q =

urs
upq = 0, this is a contradiction.

So urs 6= 0. Since 0 6= urs ∈ rad(N : M) and rad(N : M) is a weakly prime ideal of R, then

ur ∈ rad(N : M) or s ∈ rad(N : M). Hence r
p =

u
u

r
p ∈ (rad(N : M))M or s

q ∈ (rad(N : M))M,

that is, r
p ∈ rad((N : M)M) or s

q ∈ rad((N : M)M).

Assume that rad((N : M)M) is a weakly prime ideal of RM. If rad(N : M) = R, then

rad((N : M)M) = RM, a contradiction. So rad(N : M) is a proper ideal of R. Let

0 6= ab ∈ rad(N : M) for some a, b ∈ R. Then ab
1 =

a
1

b
1 ∈ rad((N : M)M). If a

1
b
1 = 0, then

qab = 0 for some q ∈ R \M. As 0 6= ab, then q ∈ P(0). Thus q ∈M, which is a contradiction.

So 0 6= a
1

b
1 ∈ rad((N : M)M). Since rad((N : M)M) is a weakly prime ideal of RM, then

a
1 ∈ rad((N : M)M) or b

1 ∈ rad((N : M)M). Therefore pa ∈ rad(N : M) for some p ∈ R\M or

sb ∈ rad(N : M) for some s /∈M. As p ∈ R\M and s ∈ R\M, then p, s /∈ P(N). Consequently,

a ∈ rad(N : M) or b ∈ rad(N : M).

(ii) Assume that rad(N : M) is a prime ideal of R. In a similar way, we get rad((N : M)M)

is a proper ideal of RM. Now, let r
p

s
q ∈ rad((N : M)M), where r, s ∈ R and p,q ∈ R \M.

Then we have rs
pq ∈ (rad(N : M))M and so we have urs ∈ rad(N : M) for some u ∈ R \M.

Since rad(N : M) is a prime ideal of R, then ur ∈ rad(N : M) or s ∈ rad(N : M). Conse-

quently, r
p =

u
u

r
p ∈ (rad(N : M))M or s

q ∈ (rad(N : M))M, that is, r
p ∈ rad((N : M)M) or

s
q ∈ rad((N : M)M).



G. Ulucak and R. Uregen / Eur. J. Pure Appl. Math, 9 (2016), 48-56 53

Suppose that rad((N : M)M) is a prime ideal of RM. From (i), it is clear that rad(N : M) is

a proper ideal of R. Then ab
1 =

a
1

b
1 ∈ rad((N : M)M) for some a, b ∈ R and since rad(N : M)M

is a prime ideal of RM, then a
1 ∈ rad((N : M)M) or b

1 ∈ rad((N : M)M). Thus pa ∈ rad(N : M)

for some p ∈ R \M or sb ∈ rad(N : M) for some s ∈ R \M. As p ∈ R \M and s ∈ R \M, then

p, s /∈ P(N). Therefore, a ∈ rad(N : M) or b ∈ rad(N : M).

Proposition 6. Let M be a faithful cyclic R-module and N be a proper submodule of M with

P(0) ⊆ P(N). If N is a P(N)-locally weakly primary submodule of M, then rad(N : M) is a

weakly prime ideal of R.

Proof. Let M be a maximal ideal of R with P(N) ⊆ M. By [4, Proposition 2.18], MM

is a faithful cyclic RM-module. Then NM is a weakly primary submodule of MM. Thus by

[1, Proposition 2.3], rad(NM : MM) is a weakly prime submodule of MM. By Proposition 3,

rad((N : M)M) is a weakly prime submodule of MM. By Proposition 5 i), rad(N : M) is a

weakly prime ideal of R.

Proposition 7. Let M be an R-module. Suppose that N is an M-primal and a P(N)-locally

weakly primary submodule of M not primary submodule of M. If P(0) ⊆ P(N) and I is an ideal

of R such that I ⊆ rad(N : M), then IN = 0. Particularly, rad(N : M)N = 0.

Proof. Suppose that P(0) ⊆ P(N) and I is an ideal of R such that I ⊆ rad(N : M). Since N is

M-primal, then P(N) is an ideal of R. As 1 /∈ P(N), then P(N) is a proper ideal. Hence there is

a maximal ideal M of R such that P(N) ⊆M. Then, NM is a weakly primary submodule of MM

because N is a P(N)-locally weakly primary submodule of M . Our aim is to show that NM is

not a primary submodule of MM. Assume that NM is a primary submodule of MM. Let rm ∈ N

for some r ∈ R, m ∈ M . Then rm
1 =

r
1

m
1 ∈ NM. By assumption, m

1 ∈ NM or ( r
1)

nMM ⊆ NM

for some positive integer n. By using a similar technique in the previous proofs, m ∈ N or

rnM ⊆ N for some positive integer n since P(N) ⊆ M, but this contradicts with N which is

not a primary submodule of M . By [4, Lemma 2.19], IM ⊆ rad((N : M)M) ⊆ rad(NM : MM).

By [1, Corollary 3.4], IMNM = 0. We get r
1

m
1 =

rm
1 = 0 for every r ∈ I and every m ∈ N .

Therefore qrm = 0 for some q ∈ R \M. If rm 6= 0, then q ∈ P(0) and so q ∈ M, which is

a contradiction. Hence rm = 0, that is, IN = 0. Particularly, by putting I = rad(N : M), we

have rad(N : M)N = 0.

Proposition 8 ([4, Proposition 2.16]). Let M be an R-module and M be a maximal ideal of R.

If I is an ideal of RM and N is a submodule of MM, then

(i) I = {a ∈ R | a
1 ∈ I} is an ideal of R and I = IM.

(ii) N = {m ∈ M | m
1 ∈ N} is a submodule of M and N = NM.

Theorem 2. Let N be an M-primal submodule of an R-module M with P(0) ⊆ P(N). Then N is

a P(N)-locally weakly primary submodule of M if and only if 0 6= I D ⊆ N for some ideal I of R

and some submodule D of M implies I ⊆ rad(N : M) or D ⊆ N.



G. Ulucak and R. Uregen / Eur. J. Pure Appl. Math, 9 (2016), 48-56 54

Proof. Assume that N is a P(N)-locally weakly primary submodule of M . Let 0 6= I D ⊆ N

for some ideal I of R and some submodule D of M . Since N is M-primal, then P(N) is an ideal

of R. As 1 /∈ P(N), then P(N) is a proper ideal. So we have P(N) ⊆ M for some maximal

ideal M of R. Thus NM is a weakly primary submodule of MM. Now, IM is an ideal of RM

and DM is a submodule of MM with (I D)M = IMDM ⊆ NM. Suppose that IMDM = 0M. Then
r
1

m
1 =

rm
1 = 0 for every r ∈ I and every m ∈ D. So there exists a q ∈ R \M such that qrm= 0.

If rm 6= 0, then q ∈ P(0). Thus q ∈M, which is a contradiction. So rm = 0, hence I D = 0,

that is a contradiction. Then 0M 6= IMDM ⊆ NM. Since N is a P(N)-locally weakly primary

submodule of M , then NM is a weakly primary submodule of MM. By [1, Theorem 3.6], either

IM ⊆ rad(NM : MM) or DM ⊆ NM. Since P(N) ⊆M, then I ⊆ rad(N : M) or D ⊆ N .

Let M be a maximal ideal of R with P(N) ⊆M. Since N is a proper ideal of R, then there

is an a ∈ M \N , but a
1 ∈ MM. If a

1 ∈ NM, then qa ∈ N such that q ∈ R \M. As a ∈ M \N , then

q ∈ P(N), that is, q ∈M, which is a contradiction. So a
1 ∈ MM \ NM. Hence NM is a proper

ideal of RM. Let I be an ideal of RM and D be a submodule of MM with 0M 6= I D ⊆ NM. By

[4, Proposition 2.16], I = IM, for some ideal I of R and D = DM, for some submodule D of

M . So 0M 6= IMDM ⊆ NM, that is, 0M 6= (I D)M ⊆ NM. Since P(N) ⊆M, then I D ⊆ N . Also

0 6= I D. On the contrary, (I D)M = 0M. By the hypothesis, we have either I ⊆ rad(N : M) or

D ⊆ N . If I ⊆ rad(N : M), then I = IM ⊆ rad((N : M)M). If D ⊆ N , then D = DM ⊆ NM. From

[1, Theorem 3.6], NM is a weakly primary submodule of MM. Therefore, N is a P(N)-locally

weakly primary submodule of M .

Corollary 5. Let N be an M-primal submodule of an R-module M with P(0) ⊆ P(N). Then N is

a P(N)-locally weakly primary submodule of M if and only if N is a weakly primary submodule

of M.

Proof. It is clear from Theorem 2 and [1, Theorem 3.6].

Theorem 3. Let M be an R-module and N be an M-primal submodule of M with P(0) ⊆ P(N).

Then the following statements are equivalent:

(i) N is a P(N)-locally weakly primary submodule of M.

(ii) For any m ∈ M \ N, rad(N : Rm) = rad(N : M)∪ (0 : Rm).

(iii) For any m ∈ M \ N, rad(N : Rm) = rad(N : M) or rad(N : Rm) = (0 : Rm).

Proof. (i) =⇒ (ii): Let N be a P(N)-locally weakly primary submodule of M and let

m ∈ M \ N . Since N is M-primal, then P(N) is an ideal of R. As i /∈ P(N), then P(N) is a

proper ideal. So we have P(N) ⊆M for some maximal ideal M of R. Hence NM is a weakly

primary submodule of MM. As m ∈ M , then m
1 ∈ MM, but m

1 ∈ MM \ NM. If m
1 ∈ NM, then

pm ∈ N for some p ∈ R \M. Since p /∈ P(N), then m ∈ N , this is a contradiction. By [2,

Theorem 2.15], rad(NM : RM

m
1 ) = rad(NM : MM)∪ (0M : RM

m
1 ) and from [4, Corollary 2.9],

rad(NM : (Rm)M) = rad(NM : MM) ∪ (0M : (Rm)M). Then by Proposition 3, Proposition 4

and Corollary 4, rad((N : Rm)M) = rad((N : M)M)∪ (0 : Rm)M. Let r ∈ rad(N : Rm). Then
r
1 ∈ rad((N : Rm)M) and so r

1 ∈ rad((N : M)M) or r
1 ∈ (0 : Rm)M. If r

1 ∈ rad((N : M)M), then



REFERENCES 55

rn

1 ∈ (N : M)M for some positive integer n and thus qrn ∈ (N : M) for some q ∈ R \M, that

is, qrnM ⊆ N . Assume that rnM 6⊆ N . Then rnm /∈ N for some m ∈ M , however qrnm ∈ N .

Hence q ∈ P(N). Then q ∈ M, which is a contradiction. So rnM ⊆ N for some positive

integer n, that is, r ∈ rad(N : M). If r
1 ∈ (0 : Rm)M, then pr ∈ (0 : Rm) for some p ∈ R \M.

Thus prRm = 0. Assume that rRm 6= 0. Then rsm 6= 0 for some s ∈ R, but prsm = 0.

Therefore p ∈ P(0). As P(0) ⊆M, then p ∈M, which is a contradiction. So rRm = 0. Then

r ∈ (0 : Rm). Hence r ∈ rad(N : M)∪ (0 : Rm). Conversely, let r ∈ rad(N : M)∪ (0 : Rm). If

r ∈ rad(N : M), then rnM ⊆ N for some positive integer n and so we get rnRm ⊆ rnM ⊆ N .

Thus r ∈ rad(N : Rm). If r ∈ (0 : Rm), then rRm= 0 ⊆ N . Thus r ∈ (N : Rm) ⊆ rad(N : Rm).

(ii)⇒ (iii): Clear.

(iii)⇒ (i): Let M be a maximal ideal of R with P(N) ⊆M. Let m
p ∈ MM\NM where m ∈ M ,

p ∈ R \M. Then m ∈ M \ N . By the condition of the theorem, rad(N : Rm) = rad(N : M)

or rad(N : Rm) = (0 : Rm) for some m ∈ M \ N . If rad(N : Rm) = rad(N : M), then

rad((N : Rm)M) = rad((N : M)M) and from Proposition 3 and Proposition 4

rad(NM : (Rm)M) = rad(NM : MM). By [4, Proposition 2.8], rad(NM : RM

m
p ) = rad(NM : MM).

If rad(N : Rm) = (0 : Rm), then rad((N : Rm)M) = (0 : Rm)M and by Proposition 4 and Corol-

lary 4, rad(NM : (Rm)M) = (0M : (Rm)M). By [4, Proposition 2.8],

rad(NM : RM

m
p ) = (0M : RM

m
p ). By [2, Theorem 2.15], NM is a weakly primary submodule of

MM. Thus N is a P(N)-locally weakly primary submodule of M .

Theorem 4. Let M be an R-module and N be an M-primal submodule of M with P(0) ⊆ P(N).

Then the following statements are equivalent:

(i) N is a P(N)-locally weakly primary submodule of M.

(ii) 0 6= I D ⊆ N for any ideal I of R and any submodule D of M implies either I ⊆ rad(N : M)

or D ⊆ N.

(iii) rad(N : Rm) = rad(N : M)∪ (0 : Rm) for any m ∈ M \ N.

(iv) rad(N : Rm) = rad(N : M) or rad(N : Rm) = (0 : Rm) for any m ∈ M \ N,

Proof. It is clear from Theorem 2 and Theorem 3.

References

[1] A.E. Ashour. On Weakly Primary Submodules, Journal of Al Azhar University-

Gaza(Natural Sciences), 13, 31-40. 2011.

[2] S.E. Atani and F. Farzalipour. On Weakly Primary Ideals, Georgian Mathematical Journal,

12, 423-429. 2005.

[3] S.E. Atani and F. Farzalipour. On Weakly Prime Submodules, Tamkang Journal of Mathe-

matics, 38, 247-252. 2007.



REFERENCES 56

[4] A.K. Jabbar. A Generalization of Prime and Weakly Prime Submodules, Pure Mathematical

Sciences, 2, 1-11. 2013.

[5] R.Y. Sharp. Steps in Commutative Algebra, Cambridge University Press, Cambridge, 1990.

[6] U. Tekir. A Note on Multiplication Modules, International Journal of Pure and Applied

Mathematics, 27, 103-107. 2006.

[7] U. Tekir. On Primary Submodules, International Journal of Pure and Applied Mathematics,

27, 283-289. 2006.