A Note on Primary and Weakly Primary Submodules

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

ISSN 1307-5543 – www.ejpam.com

A Note on Primary and Weakly Primary Submodules

Gulsen Ulucak

and Rabia Nagehan Uregen

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 andP(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 16=0. An idealIofRis called a proper ideal ifI6=R. Then the radical of a proper idealIofRis denoted by r ad(I)andr ad(I) ={x ∈R| xn∈I for some positive integern}. A proper ideal PofRis called prime (primary) if a b∈P for somea,b∈Rimplies that eithera∈P or b∈P(either a ∈ P or bn ∈ P for some positive integer n). A proper ideal P ofR is said to be a weakly prime ideal if 06=a b∈Pfor somea,b∈Rimplies that eithera∈Por b∈P, and it is called a weakly primary ideal if 06=a b∈Pfor somea,b∈Rimplies that eithera∈Por bn∈Pfor some positive integern(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 r m ∈ N for some r ∈ R and m∈M implies that eitherm∈N or r M ⊆N and it is said to be a weakly prime submodule if 0 6= r m ∈ N for some r ∈ R and m ∈ M implies that either m ∈ N or r M ⊆ N. A non empty subsetSofRis said to be multiplicative closed set if 0∈/Sand whenevera,b∈S, then a b∈S. LetSbe a multiplicative closed set inR. It can be easily seen thatM_Sis anR_S-module under the operations a_s + _ub= ua_su+s b and r_va_s = r a_vs for any r_v ∈RS and as,

b

u ∈MS [5]. A proper

submoduleN ofM is said to beS(N)-locally prime (S(N)-weakly prime) submodule ifNMis

a prime (a weakly prime) submodule of MMfor each maximal idealMwithS(N)⊆M[4]. ∗_{Corresponding author.}

Email addresses:[email protected](G. Ulucak),[email protected](R. Uregen)

A proper submoduleN ofM is said to be a primary submodule if r m∈N for somer ∈R, m∈M implies that eitherm∈Nor rnM ⊆N for some positive integernand it is said to be a weakly primary submodule if 06=r m∈Nfor somer∈R,m∈M implies that eitherm∈Nor rnM ⊆N for some positive integern. The ideal{r ∈R| r M⊆N}will be denoted by(N:M)

and(0 : N) ={r ∈R| r N =0} where N is a submodule of M. Then the annihilator of M is(0 :M)where (0 :M) ={r ∈R | r M =0}. AnR-moduleM is called a faithful module if

(0 : M) = (0). Note that ifN is a primary submodule ofM, then(N : M) is a primary ideal ofRand r ad(N:M) ={r ∈R | rnM ⊆ N for some positive integer n} is a prime ideal ofR ([1, 6, 7]).

Main aim is to obtain the two generalization on primary submodules and weakly primary submodules of an R-module M.LetN be a proper submodule ofM. An element r ∈Ris said to be primary toN if rnm∈N, where m∈M andnis a positive integer, thenm∈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 ofRthat are not primary toN is denoted by P(N). Then we getP(N) ={r∈R|rnm∈N for some positive integern, for some elementm∈M\N}. If N = (0), thenP((0)) ={r∈R|rnm=0 for some positive integern, for some 06=m∈M}. A proper submoduleN ofM is said to be anM_{-primal ifP(N)forms an ideal ofR.}

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 Mwhere

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 MMfor every maximal idealMwhere

P(N)⊆M_.

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

Proof. Let r ∈ r ad(N:M). Then rnM ⊆ N for some positive integer n. There exists m∈M\N such thatrnm∈N. Thenr∈P(N). Thusr ad(N:M)⊆P(N).

Every primary submodule N is proposed as P(N)-locally primary submodule and every weakly primary submoduleN 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 Rwhere P(N)}⊆M_{. By the previous lemma, we say}

that r ad(N:M) ⊆ P(N) ⊆ M_{. NM is a proper submodule of MM. Indeed, if NM = MM,}

then m₁ ∈ MM for any m ∈ M. Then r m ∈ N for some r ∈/ M. We get rn ∈/ r ad(N:M).

Since N is a primary submodule of M, then m ∈ N. Thus N = M, a contradiction. Since r ad(N:M)∩(R\M_{) =} ;_{, then NM is a primary submodule of MM. Consequently, N is a}

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

Proof. Suppose thatM_{is a maximal ideal ofRwhere P(N)}⊆M_{. In the same manner as}

in the proof of the previous proposition, we have thatNMis a proper submodule ofMM. Let

0M6= r_sm_p ∈NMfor some r_s ∈RMandm_p ∈MM(for somer∈R,m∈Mands,p∈R\M). Then

there is aq∈R\M_{such thatqr m}∈_{N. Assume thatqr m=0. Then} r

s m p =

q q r s

m p =

qr m qsp =0M,

this is a contradiction. So 06=qr m∈N. Asr ad(N:M)⊆ P(N)⊆M_{, thenq}∈_{/ r ad(N:M).}

Thus r m∈Nsince Nis a weakly primary submodule. It is clear that r m6=0. Hence

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

m

p ∈NM or

rn

snMM ⊆NM for some positive integernby [4, Corollary 2.9]. Then we get that

NMis a weakly primary submodule ofMM. Consequently,Nis aP(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. ThenN is a weakly primary submodule. Thus,Nis a P(N)-locally weakly primary submodule by Proposition 2.

Note that ifN is aP(N)-locally primary submodule of M, thenN is aP(N)-locally weakly primary submodule ofM.

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](X2,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, Y₁ · Z₁¯ = ₁¯0 ∈NM but Y₁

n /

∈(NM : MM),

€_¯Z 1 Š

/

∈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 ofM nor weakly primary submodule ofM.

Example 2. Let R= Zand consider the R−modul e M=Z12. Let N be the submodule ofZ12 generated by6. It is easily seen that06=23(=32)∈N but2∈/(N:M)and3∈/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 idealM_{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 .

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

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

Proof. (i)Assume thatN is a primary submodule. Let r∈P(N). Thenrnm∈N for some positive integernand for somem∈M\N. SinceN is a primary submodule, then

(rn)kM=rnkM ⊆N for some positive integerk, that is,r∈r ad(N:M). Hence P(N)⊆r ad(N:M). By Lemma 1, we getP(N) =r ad(N :M).

Suppose that P(N) =r ad(N:M). Letr m∈N andm∈M\N wherer ∈R,m∈M. Then r∈P(N). Thus r∈r ad(N :M), that is, rkM ⊆N for some positive integerk. 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 integernand for somem∈M\N. Suppose thatrnm=0. Sincem∈M\N, then we getm6=0. Sor ∈P(0). Thusr∈r ad(N:M), by assumption. Hence P(N) =r ad(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 ∈ r ad(N:M) and so P(N) =r ad(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) = r ad(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). ThenN is aP(N)-locally primary submodule by Proposition 1. SinceNis primary submodule, thenNis weakly primary submodule. Therefore,NisP(N)-locally weakly primary submodule by Proposition 2.

Note that, by[4, Lemma 2.19], ifM_{is a maximal ideal ofR, then(N:M)M}⊆_(NM:MM).

Now, we explain thatr ad((N:M)M) =r ad(NM:MM)whenMis a maximal ideal ofRwith

P(N)⊆M_.

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

r ad((N:M)M) =r ad(NM:MM)for any maximal idealMof R with P(N)⊆M.

Proof. SinceS(N)⊆ P(N)for any proper submodule N ofM, 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

r ad((N:M)M) = (r ad(N:M))Mfor any maximal idealMof R with P(N)⊆M.

Proof. Let r_p ∈r ad((N:M)M)for somer∈Randp∈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 qr}n ∈ _{(N : M),}

that is,qrnm∈N for everym∈ M. Then rnm∈N for everym∈M since q∈/ P(N). Thus r ∈ r ad(N:M). Then _pr ∈ r ad((N:M)M). Conversely, assume that

r

p ∈ r ad((N:M)M).

There is an u∈ R\M_{such that ur} ∈ _{r ad(N:M). Then (ur)}n _{= u}n_rn ∈ _{(N : M). Hence}

un unr

n

pn ∈(N:M)M. Consequently, r

n

Corollary 3. Let N be a proper submodule of an R-module. IfM_{is any maximal ideal of R with}

P(N)⊆M_{, then r ad((N :M)M) =r ad(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 r ad((N:Rm)M) =r ad(NM:(Rm)M)for any maximal idealMof R with P(N)⊆M.

Proof. It is clear.

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

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

any maximal idealM_{of R with P(0)}⊆M_.

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

with P(N)⊆M_{. Then the following statements hold:}

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

is a weakly prime ideal of RM.

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

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

then 1₁ ∈r ad((N:M)M) = (r ad(N :M))M and soq1=q∈r ad(N:M)for someq∈R\M.

But by Lemma 1,r ad(N:M)⊆P(N)⊆M_{, which is a contradiction. Sor ad((N:M)M)}6_=RM,

that is, r ad((N:M)M) is a proper ideal ofRM. Let 06= _pr_qs ∈r ad((N:M)M), where r,s∈R

and p,q∈R\M_{. Then we have} r

p s q =

rs

pq ∈(r ad(N:M))M, then there exists an u∈R\M

such thaturs∈ r ad(N:M). If urs= 0, then _pr_qs = u_ur_pqs = _upqurs =0, this is a contradiction. Sours6=0. Since 06=urs∈r ad(N:M)and r ad(N:M)is a weakly prime ideal ofR, then ur∈r ad(N:M)ors∈r ad(N:M). Hence _pr = u_ur_p∈(r ad(N:M))M or

s

q ∈(r ad(N:M))M,

that is, r_p∈r ad((N:M)M)or

s

q∈r ad((N:M)M).

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

r ad((N:M)M) =RM, a contradiction. Sor ad(N:M)is a proper ideal ofR. Let

06= a b∈r ad(N:M)for some a,b∈R. Then a b₁ = a₁₁b ∈r ad((N:M)M). If

a

1

b

1 =0, then qa b=0 for someq∈R\M_{. As 0}6_{=a b, thenq}∈_{P(0). Thusq}∈M_{, which is a contradiction.}

So 0 6= a₁₁b ∈ r ad((N:M)M). Since r ad((N:M)M) is a weakly prime ideal of RM, then

a

1 ∈r ad((N:M)M)or ₁b∈r ad((N:M)M). Thereforepa∈r ad(N :M)for somep∈R\Mor

s b∈r ad(N:M)for somes∈/M_{. Asp}∈_R\M_ands∈_R\M_{, thenp,s}∈_{/P(N). Consequently,}

a∈r ad(N:M)or b∈r ad(N :M).

(ii)Assume thatr ad(N:M)is a prime ideal ofR. In a similar way, we getr ad((N:M)M)

is a proper ideal of RM. Now, let r_pqs ∈ r ad((N:M)M), where r,s ∈ R and p,q ∈ R\M.

Then we have _pqrs ∈(r ad(N:M))M and so we haveurs ∈r ad(N:M)for someu∈R\M.

Since r ad(N:M) is a prime ideal of R, then ur ∈ r ad(N:M) or s ∈ r ad(N:M). Conse-quently, r_p = u_upr ∈ (r ad(N:M))M or _qs ∈ (r ad(N:M))M, that is, r_p ∈ r ad((N:M)M) or

s

Suppose thatr ad((N:M)M)is a prime ideal ofRM. From(i), it is clear thatr ad(N:M)is

a proper ideal ofR. Then a b₁ = a₁b₁ ∈r ad((N:M)M)for somea,b∈Rand sincer ad(N:M)M

is a prime ideal ofRM, then₁a∈r ad((N:M)M)or ₁b∈r ad((N:M)M). Thuspa∈r ad(N:M)

for somep∈R\M_{ors b}∈_{r ad(N:M)for somes}∈_R\M_{. Asp}∈_R\M_ands∈_R\M_{, then}

p,s∈/ P(N). Therefore,a∈r ad(N:M)or b∈r ad(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 r ad(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], r ad(NM:MM)is a weakly prime submodule of MM. By Proposition 3,

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

weakly prime ideal ofR.

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⊆r ad(N:M),then I N=0. Particularly, r ad(N :M)N =0.

Proof. Suppose thatP(0)⊆P(N)andIis an ideal ofRsuch thatI⊆r ad(N :M). SinceNis

M_{-primal, thenP(N)is an ideal ofR. As 1}∈_{/P(N), thenP(N)is a proper ideal. Hence there is}

a maximal idealM_{ofRsuch thatP(N)}⊆M_{. Then,NMis a weakly primary submodule ofMM}

becauseN is a P(N)-locally weakly primary submodule of M. Our aim is to show thatNM is

not a primary submodule ofMM. Assume thatNMis a primary submodule ofMM. Letr m∈N

for some r ∈R, m∈M. Then r m₁ = ₁rm₁ ∈NM. By assumption, m₁ ∈NM or (r₁)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⊆r ad((N :M)M)⊆r ad(NM:MM).

By [1, Corollary 3.4], IMNM = 0. We get ₁rm₁ = r m₁ = 0 for every r ∈ I and every m∈ N.

Thereforeqr m =0 for some q ∈R\M_{. If r m=}6 _{0, thenq} ∈ _{P(0)and so q} ∈M_{, which is}

a contradiction. Hencer m=0, that is, I N=0. Particularly, by putting I= r ad(N:M), we haver ad(N:M)N=0.

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

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

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

(ii) N={m∈M | m₁ ∈N}is a submodule of M and N =NM.

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

Proof. Assume that N is a P(N)-locally weakly primary submodule ofM. Let 06=I D⊆N for some idealIofRand some submoduleDofM. SinceNisM_{-primal, thenP(N)is an ideal}

ofR. As 1∈/ P(N), then P(N) is a proper ideal. So we have P(N) ⊆ M_{for some maximal}

ideal M_{ofR. Thus NM is a weakly primary submodule of MM. Now, IM is an ideal ofRM}

andDM is a submodule ofMMwith(I D)M=IMDM⊆NM. Suppose that IMDM=0M. Then

r

1

m

1 =

r m

1 =0 for everyr∈I and everym∈D. So there exists aq∈R\Msuch thatqr m=0. If r m6=0, thenq∈P(0). Thus q∈M_{, which is a contradiction. Sor m=0, hence I D=0,}

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

submodule ofM, thenNMis a weakly primary submodule ofMM. By[1, Theorem 3.6], either

IM⊆r ad(NM:MM)or DM⊆NM. SinceP(N)⊆M, thenI⊆r ad(N:M)or D⊆N.

LetM_{be a maximal ideal ofRwithP(N)}⊆M_{. SinceN is a proper ideal ofR, then there}

is ana∈M\N, but a₁ ∈MM. If a₁ ∈NM, thenqa∈N such thatq∈R\M. Asa∈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 ofRM. Let I be an ideal ofRM and Dbe a submodule of MM with 0M6= I D⊆ NM. By [4, Proposition 2.16], I = IM, for some ideal I ofR and D= DM, for some submodule Dof

M. So 0M 6=IMDM ⊆NM, that is, 0M6= (I D)M⊆NM. SinceP(N)⊆M, then I D⊆N. Also

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

D⊆N. IfI⊆r ad(N :M), thenI=IM⊆r ad((N:M)M). IfD⊆N, thenD=DM⊆NM. From [1, Theorem 3.6], NM is a weakly primary submodule ofMM. Therefore, N is aP(N)-locally

weakly primary submodule ofM.

Corollary 5. Let N be anM_{-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 anM_{-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 , r ad(N:Rm) =r ad(N:M)∪(0 :Rm).

(iii) For any m∈M\N , r ad(N:Rm) =r ad(N:M)or r ad(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_{ofR. HenceNMis a weakly}

primary submodule of MM. As m∈M, then m₁ ∈MM, but m₁ ∈MM\NM. If m₁ ∈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],r ad(NM:RMm₁) =r ad(NM:MM)∪(0M:RMm₁)and from[4, Corollary 2.9],

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

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

r

rn

1 ∈(N :M)M for some positive integer nand thusqrn ∈(N :M)for someq∈R\M, that

is,qrnM ⊆ N. Assume thatrnM 6⊆ N. Then rnm∈/ N for some m∈M, howeverqrnm∈N. Hence q ∈ P(N). Then q ∈ M_{, which is a contradiction. So r}n_M ⊆ _{N for some positive}

integern, that is, r ∈r ad(N:M). If ₁r ∈(0 :Rm)M, then pr∈(0 :Rm)for somep∈R\M.

Thus prRm = 0. Assume that rRm 6= 0. Then rsm 6= 0 for some s ∈ R, but prsm = 0. Thereforep∈P(0). AsP(0)⊆M_{, thenp}∈M_{, which is a contradiction. So rRm=0. Then}

r∈(0 :Rm). Hencer ∈r ad(N:M)∪(0 :Rm). Conversely, letr ∈r ad(N:M)∪(0 :Rm). If r ∈r ad(N:M), thenrnM ⊆N for some positive integernand so we get rnRm⊆rnM ⊆N. Thus r∈r ad(N :Rm). Ifr ∈(0 :Rm), thenrRm=0⊆N. Thusr∈(N:Rm)⊆ r ad(N:Rm).

(ii)⇒(iii): Clear.

(iii)⇒(i): LetM_{be a maximal ideal ofRwithP(N)}⊆M_{. Let}m

p ∈MM\NMwherem∈M,

p∈R\M_{. Then m}∈_M \_{N. By the condition of the theorem, r ad(N:Rm) =r ad(N:M)}

or r ad(N:Rm) = (0 : Rm) for some m ∈ M \N. If r ad(N:Rm) = r ad(N :M), then r ad((N:Rm)M) =r ad((N:M)M)and from Proposition 3 and Proposition 4

r ad(NM:(Rm)M) =r ad(NM:MM). By[4, Proposition 2.8],r ad(NM:RMm_p) =r ad(NM:MM).

Ifr ad(N:Rm) = (0 :Rm), thenr ad((N:Rm)M) = (0 :Rm)Mand by Proposition 4 and

Corol-lary 4,r ad(NM:(Rm)M) = (0M:(Rm)M). By[4, Proposition 2.8],

r ad(NM:RM

m

p) = (0M:RM

m

p). By[2, Theorem 2.15],NMis a weakly primary submodule of

MM. ThusN is aP(N)-locally weakly primary submodule of M.

Theorem 4. Let M be an R-module and N be anM_{-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) 06=I D⊆N for any ideal I of R and any submodule D of M implies either I⊆r ad(N:M)

or D⊆N .

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

(iv) r ad(N:Rm) =r ad(N :M)or r ad(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.

[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.