On the 2-absorbing Submodules

Vol. 10, No. 1 (2015), pp 131-137 DOI: 10.7508/ijmsi.2015.01.010

On the 2-absorbing Submodules

Sh. Payrovi∗, S. Babaei

Imam Khomieni International University, Postal Code: 34149-1-6818, Qazvin, Iran.

E-mail: [email protected] E-mail: [email protected]

Abstract. LetRbe a commutative ring andMbe anR-module. In this

paper, we investigate some properties of 2-absorbing submodules ofM. It is shown thatNis a 2-absorbing submodule ofMif and only if whenever

IJ L⊆Nfor some idealsI, Jof R and a submoduleLofM, thenIL⊆N

orJ L⊆N orIJ ⊆N :RM. Also, ifN is a 2-absorbing submodule of

M and M/N is Noetherian, then a chain of 2-absorbing submodules of

M is constructed. Furthermore, the annihilation of E(R/p) is studied

whenever 0 is a 2-absorbing submodule of E(R/p), wherepis a prime

ideal ofRandE(R/p) is an injective envelope ofR/p.

Keywords: 2-absorbing ideal, 2-absorbing submodule, A chain of 2-absorbing submodule.

2000 Mathematics subject classification: 13C99. 1. _Introduction

Throughout this paperR is a commutative ring with non-zero identity and M is an unitary R-module. We defined a submoduleN of M is 2-absorbing whenever abm ∈N for some a, b∈ R, m ∈ M, then am ∈N or bm∈N or ab ∈ N :R M, see for instance [1, 3, 4, 6, 7, 9, 10]. It is well known that, a

submodule N of M is prime if and only if IL ⊆ N for an ideal I of R and ∗_{Corresponding Author}

Received 30 October 2013; Accepted 09 September 2014 c

2015 Academic Center for Education, Culture and Research TMU

131

a submodule L of M, then either L ⊆ N or I ⊆ N :R M. This statement

persuaded us to prove that, a submoduleN ofM is 2-absorbing if and only if IJ L⊆ N for some ideals I, J of R and a submodule L of M, then IL⊆ N or J L⊆N or IJ ⊆N :RM. As a corollary of this theorem, it is shown that

L={m∈M :p⊆r(N :m)} is a 2-absorbing submodule ofM, whereN is a 2-absorbing submodule ofM withr(N :RM) =p∩qfor some prime idealsp,q

of R. Also, it is shown that if M/N is Noetherian, then there exists a chain of 2-absorbing submodules ofM that begins with N. Assume thatE(R/p) is an injective envelope ofR/p, it is shown that if 0 is a 2-absorbing submodule ofE(R/p), thenr(0 :R E(R/p)) =p and 0 :R xis determined for all nonzero

elementxofE(R/p).

Now, we define the concepts that we will use later. For a submoduleL of M letL:R M denote the ideal {r∈R :rM ⊆L}. Similarly, for an element

m∈ M let L:R m denote the ideal{r∈R : rm∈L}. IfI is an ideal of R,

then r(I) denotes the radical of I. We say thatp∈Spec(R) is an associated prime ideal ofM if there existsm∈M with 0 :Rm=p. The set of associated

prime ideals ofM is denoted by AssR(M), the set of integers is denoted byZ.

2. _{2-absorbing Submodules}

LetN be a proper submodule of M. We say that N is a 2-absorbing sub-module of M if whenever a, b ∈ R, m ∈M and abm ∈ N, then am ∈N or bm∈N or ab∈N :RM.

Lemma 2.1. Let I be an ideal ofR andN be a 2-absorbing submodule of M. If a∈R,m∈M andIam⊆N, then am∈N orIm⊆N orIa⊆N :RM. Proof. Let am 6∈ N and Ia 6⊆ N :R M. Then there exists b ∈ I such that

ba6∈N :RM. Now, bam∈N implies thatbm∈N, sinceN is a 2-absorbing

submodule ofM. We have to show thatIm⊆N. Letcbe an arbitrary element ofI. Thus (b+c)am∈N. Hence, either (b+c)m∈N or (b+c)a∈N :RM.

If (b+c)m∈N, then bybm∈N it follows thatcm∈N. If (b+c)a∈N :RM,

then ca 6∈ N :R M, but cam ∈N. Thus cm∈ N. Hence, we conclude that

Im⊆N.

Lemma 2.2. LetI, J be ideals ofR andN be a 2-absorbing submodule of M. If m∈M andIJ m⊆N, thenIm⊆N orJ m⊆N orIJ ⊆N :RM.

Proof. LetI6⊆N:RmandJ6⊆N :Rm. We have to show thatIJ⊆N :RM.

Assume that c ∈ I and d ∈ J. By assumption there exists a ∈ I such that am 6∈N but aJ m⊆N. Now, Lemma 2.1 shows that aJ ⊆N :R M and so

(I\N :R m)J ⊆ N :R M, similarly there exists b ∈ J \N :R m such that

Ib⊆N :RM and alsoI(J\N :Rm)⊆N :RM. Thus we haveab∈N :RM,

ad ∈ N :R M and cb ∈ N :R M. By a+c ∈ I and b+d ∈ J it follows

that (a+c)(b+d)m ∈ N. Therefore, (a+c)m ∈ N or (b+d)m ∈ N or

(a+c)(b+d)∈N :RM. If (a+c)m∈N, thencm6∈N hence,c∈I\N :Rm

which implies that cd∈N :R M. Similarly by (b+d)m∈N, we can deduce

thatcd∈N :RM. If (a+c)(b+d)∈N :RM, thenab+ad+cb+cd∈N :RM

and socd∈N :RM. Therefore,IJ⊆N :RM.

Theorem 2.3. Let N be a proper submodule ofM. The following statement are equivalent:

(i) N is a 2-absorbing submodule ofM;

(ii) If IJ L⊆N for some idealsI, J of R and a submodule L of M, then

IL⊆N orJ L⊆N orIJ ⊆N:RM.

Proof. (ii)⇒ (i) is obvious. To prove (i)⇒ (ii), assume that IJ L⊆ N for some ideals I, J of R and a submodule L of M and IJ 6⊆ N :R M. Then

by Lemma 2.2 for all m ∈ L either Im ⊆ N or J m ⊆ N. If Im ⊆ N, for all m ∈ L we are done. Similarly if J m ⊆ N, for all m ∈ L we are done. Suppose thatm, m0 ∈Lare such that Im6⊆N andJ m06⊆N. ThusJ m⊆N and Im0 ⊆ N. Since IJ(m+m0) ⊆ N we have either I(m+m0) ⊆ N or J(m+m0) ⊆ N. By I(m+m0) ⊆ N, it follows that Im ⊆ N which is a contradiction, similarly byJ(m+m0)⊆N we get a contradiction. Therefore

eitherIL⊆N orJ L⊆N.

A submoduleN ofM is called strongly 2-absorbing if it satisfies in condition (ii), see [5]. Therefore, Theorem 2.3 shows thatN is a 2-absorbing submodule ofM if and only ifN is a strongly 2-absorbing submodule ofM.

Corollary 2.4. Let M be anR-module and N be a 2-absorbing submodule of

M. ThenN :M I={m∈M :Im⊆N} is a 2-absorbing submodules ofM for all ideal I ofR. FurthermoreN :M In=N:M In+1, for alln≥2.

Proof. Let I be an ideal of R, a, b ∈ R, m ∈ M and abm ∈ N :M I. Thus

Iabm⊆N. Hence,Im⊆N or Iab⊆N :R M orabm∈N, by Lemma 2.2. If

Im⊆N we are done. If Iab⊆N :R M, then ab ∈(N :R M) :R I = (N :M

I) :R M. If abm ∈ N, then am ∈ N or bm ∈ N or ab ∈ N :R M. Thus

Iam⊆N or Ibm⊆N or Iab⊆N :RM which complete the proof.

For the second statement, it is enough to show thatN :M I2=N :M I3. It

is clear thatN :M I2⊆N :M I3. Letm∈N :M I3. ThenI3m⊆N. Now, by

Lemma 2.2, we have I2m⊆N or Im⊆N or I3 ⊆N :R M. IfI2m⊆N or

Im⊆N, we are done. IfI3_{⊆N :}

RM, thenI2⊆N :R M sinceN :RM is a

2-absorbing ideal ofR by [9, Theorem 2.3].

It is clear that,nZis a 2-absorbing ideal ofZif and only ifn= 0, p, p2, pq,

where p, q are distinct prime integers. It is easy to see that 4_Z :_Z 6_Z = 2_Z but 4_Z:_Z36_Z=_Z. Hence, the equality mentioned in the Corollary 2.4, is not necessarily true forn= 1.

Theorem 2.5. LetN be a 2-absorbing submodule ofM such thatr(N:RM) = p∩q where p and q are the only distinct prime ideals of R that are minimal overN :RM. ThenL={m∈M :p⊆r(N :Rm)}is a 2-absorbing submodule of M containingN. Also, either r(L:RM) =q orr(L:RM) =p0∩q, where p0 is a prime ideal of R containingp.

Proof. It is clear that L is a submodule of M containing N. Assume that a, b∈R, m∈M andabm ∈L. We have to show thatam∈L or bm∈L or ab∈L :R M. Sincep ⊆r(N :R abm), thus p2abm⊆N, by [9, Theorem 2.4]

and [2, Theorem 2.4]. Therefore, by Lemma 2.1, we haveabm∈N orp2m⊆N orp2ab⊆N :RM. Ifabm∈N, thenam∈Norbm∈Norab∈N :RMwhich

implies thatam∈Lorbm∈Lorab∈L:M. Ifp2_{m⊆N, thenp}2_{⊆N :}

Rm

and sop⊆r(N :Rm) thusm∈Land we are done. Ifp2ab⊆N:RM, then by

[2, Theorem 2.13], we have p2a⊆N :RM orp2b⊆N :R M or ab∈N :RM.

In the first case we conclude thatp2_{⊆N :}

Ramand soam∈L. By a similar

argument in the second case we can deduced that bm∈L. If ab ∈N :R M,

thenab∈L:RM. Therefore, the result follows.

For the second statement, first we show that r(N :R M) =r(L :RM)∩p.

It is clear r(N :R M) ⊆ r(L :R M)∩p. Assume that a ∈ (L :R M)∩p.

Thus aM ⊆ L and so, by definition of L, p ⊆ r(N :R am), for all m ∈ M.

Hence, [2, Theorem 2.4] shows thatp2⊆N :Ram, for all m∈M. Therefore,

a3∈N :R m, for all m∈M. So thata3 ∈N :RM and thena∈r(N :RM).

Thusr(L:R M)∩p ⊆r(N :RM). Now,L:R M is a 2-absorbing ideal ofR,

therefore either r(L:RM) =p0 or r(L:RM) =p0∩q0, for some prime ideals p0,q0 of R. In the first case we have r(N :R M) = p∩p0 which implies that p0 _{=q and in the second case we haver(N :}

R M) =p∩p0∩q0 which implies

that eitherp0 =q orq0 =q.

Corollary 2.6. Let N be a 2-absorbing submodule of M such that r(N :R

M) = p∩q where p and q are the only distinct prime ideals of R that are minimal overN :RM. IfM/N is a Noetherian R-module, then

(i) there exists a chain N = L0 ⊆ L1 ⊆ · · · ⊆ Ln−1 ⊆ Ln = M of 2-absorbing submodules of M. Furthermore,Ass(M)⊆Ass(M/Ln−1)∪ Ass(Ln−1/Ln−2)∪Ass(Ln−2/Ln−3)∪· · ·∪Ass(L1/N), whereAss(Li/N) is the union of at most two totally ordered set, for all i .

(ii) there exists a chain N ⊆ Ln ⊆ Ln−1 ⊆ · · · ⊆ L1 ⊆ L0 = M of

submodules ofM such thatLi is a 2-absorbing submodule ofLi+1, for

all0≤i≤n−1.

Proof. (i) LetL1={m∈M :p⊆r(N :Rm)}. Then by Corollary 2.4,L1 is a

2-absorbing submodule of M and so either r(L1 :R M) =q or r(L1:R M) = p1∩q, where p1 is a prime ideal of R containingp. Ifr(L1 :R M) =q, then

choose L2 ={m∈M :q ⊆r(L1:R m)} =M. Hence,N ⊆L1⊆L2 =M is

requested chain. Ifr(L1:RM) =p1∩q, setL2={m∈M :p1⊆r(L1:Rm)}

and so eitherr(L2:RM) =qorr(L2:RM) =p2∩q, wherep2is a prime ideal of R containing p1. Proceeding in this way, we can achieveN ⊆ L0 ⊆L1 ⊆

· · · ⊆Ln−1⊆Ln=M of 2-absorbing submodules ofM. The last statement is

obvious, by [9, Theorem 2.6].

(ii) Let L1 = {m ∈ M : p ⊆ r(N :R m)}. Then N is a 2-absorbing

submodule ofL1. So that eitherr(N :RL1) =p1 orr(N :RL1) =p1∩q1, for some prime idealsp1,q1 ofR. Ifr(N :RL1) =p1, then chooseL2={x∈L1: p1 ⊆r(N :Rx)}=N. Hence, in this case N ⊆L1⊆L0=M is the requested

chain. If r(N :R L1) =p1∩q1, then set L2 ={x∈L1:p1⊆r(N :R x)} and

continue the same way to achieve the chain N ⊆ Ln ⊆ Ln−1 ⊆ · · · ⊆ L1 ⊆

L0=M of 2-absorbing submodules ofM.

Theorem 2.7. Let N be a 2-absorbing submodule of M. Then N :R M is a prime ideal ofRif and only ifN :Rmis a prime ideal ofRfor allm∈M\N. Proof. Assume thata, b∈R, m∈M \N and ab∈N :Rm. Thenabm⊆N.

We have am ∈ N or bm ∈ N or ab ∈ N :R M since N is a 2-absorbing

submodule ofM. Ifam∈N or bm∈N we are done. If ab∈N :R M, then

by assumption eithera∈N :RM or b∈N :RM. Thus eithera∈N :R mor

b∈N:Rm. SoN :Rmis a prime ideal.

Conversely, suppose thatab ∈N :RM for some a, b∈R and assume that

there existm, m0 ∈M such that am /∈N and bm0 ∈/ N. Byabm, abm0 ∈N it follows thatbm∈N andam0∈N sinceN :RmandN :Rm0 are prime ideals

ofR. Ifm+m0 ∈N, thenam∈N which is a contradiction. Thusm+m06∈N. Now byab(m0+m00)∈N we havea(m0+m00)∈N orb(m0+m00)∈N which is a contradiction. Thus aM ⊆N or bM ⊆N which implies that N :R M is

prime.

Corollary 2.8. Let N be a 2-absorbing submodule of M. Then N :RM is a prime ideal ofR if and only ifN:RKis a prime ideal ofRfor all submodules

K of M containingN.

Proof. By Theorem 2.7 and [9, Theorem 2.6] it follows that{N :Rx:x∈K\

N}is a totally ordered set of prime ideals ofR. Hence,N :RK=∩x∈KN :Rx

is a prime ideal ofR.

Theorem 2.9. Letpbe a prime ideal ofRandE(R/p)be an injective envelop of R/p. If0 is a 2-absorbing submodule ofE(R/p), then

(i) p2_{⊆0 :}

RE(R/p)⊆pso that r(0 :RE(R/p)) =p.

(ii) p2 ⊆0 :R x⊂0 :R ax=p, for all non-zero element xof E(R/p) and alla∈p\0 :Rx.

(iii) p2_{⊆0 :}

Rx= 0 :Ranx⊆p, for alla6∈p.

Proof. (i) We haver(0 :R x) =pfor all non-zero elementxof E(R/p), by [8,

Theorem 18.4]. Also it is obvious 0 :RE(R/p)⊆0 :Rx. Thus 0 :RE(R/p)⊆p.

Now, assume thata∈p2 and xis a non-zero element of E(R/p). Since 0 is a 2-absorbing submodule ofM, 0 :Rxis a 2-absorbing ideal ofR, by [9, Theorem

2.4]. Therefore we have p2 _{is a subset of 0 :}

R x, by [2, Theorem 2.4]. Hence,

ax= 0 and thereforeaE(R/p) = 0 andp2_{⊆0 :}

RE(R/p).

(ii) Letxbe a non-zero element ofE(R/p). Then we have p2⊆0 :Rx⊆p.

Assume that a ∈ p\0 :R x. Thus ax 6= 0 but a2x = 0 which shows that

0 :Rx⊂0 :R ax. Ifb∈p, then ab∈p2 andabx= 0. Thusb∈0 :Rax and so p⊆0 :Rax.

(iii) Assume that a 6∈ p. It is obvious that 0 :R x ⊆ 0 :R anx, for all

n ∈ _N. Let b ∈ AnnR(anx). Thus banx = 0. But multiplication by an is

an automorphism on E(R/p), so that bx = 0 and b ∈ 0 :R x. Therefore,

0 :Rx= 0 :Ranx.

Corollary 2.10. Let R be a principal ideal domain and p is a prime ideal of

R. If0 is a 2-absorbing submodule ofE(R/p), then for all non-zero elementx

of E(R/p)either0 :Rx=p2 or0 :Rx=p.

Proof. Letp= (a). Then p2 _{= (a}2_{). Letxbe a non-zero element of E(R/p).} Thenp2⊆0 :R x= (b)⊆pby Theorem 2.9(ii). Thus a2 =bcand b=aefor

some c, e ∈ R. Hence, a2 _{= aec. So a= ec ∈ p. Therefore, eitherc ∈p or} e∈p. Ifc ∈p, then c=ac0 and so a=eac0 which implies that 1 = ec0 and a=bc0 thus 0 :Rx=p. Ife∈p, thene=ae0 and soa=ae0c which implies

that 1 =e0_candb=a2_e0 _{thus 0 :}

Rx=p2.

The following example shows that the condition “0 is a 2-absorbing sub-module of E(R/p)” is essential. It is well-known that E(_Z/p_Z) = _Z(p∞) =

{m/n+Z : m, n ∈ Z, n 6= 0}, where p is a prime integer. But neither

p2Z = 0 :Z 1/p

3₊

Z nor 0 :Z 1/p

3₊

Z =pZ. Hence, 0 is not a 2-absorbing

submodule of E(_Z/p_Z). Also, this example shows that if 0 is a 2-absorbing submodule ofM, then it is not necessarily a 2-absorbing submodule ofE(M).

Acknowledgments

We would like to thank the referee for a careful reading of our article.

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