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2nd Reading
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Asian-European Journal of Mathematics
Vol. 8, No. 4 (2015) 1550064 (12 pages)
c
World Scientific Publishing Company
DOI: 10.1142/S1793557115500643
On 2-absorbing primary submodules of modules
over commutative ring with unity
Manish Kant Dubey
SAG, DRDO, Metcalf House, Delhi 110054, India
kantmanish@yahoo.com
Pakhi Aggarwal
Department of Mathematics
University of Delhi, Delhi 110007, India
scarlet2k 81@yahoo.com
CommunicatedbyV.S.Ngayen
Received October 7, 2014
Revised June 8, 2015
Published September 28, 2015
In this paper, we introduce the concept of a 2-absorbing primary submodule over a
commutative ring with nonzero identity which is a generalization of primary submodule.
Let Mbe an R-module and Nbe a proper submodule of M. Then Nis said to be a
2-absorbing primary submodule of Mif whenever a, b ∈R,m∈Mand abm ∈N,
then am ∈Nor bm ∈Nor ab ∈p(N:RM).Wehavegivenanexampleandproved
number of results concerning 2-absorbing primary submodules. We have also proved the
2-absorbing primary avoidance theorem for submodules.
Keywords: 2-absorbing primary submodules; 2-absorbing submodules; 2-absorbing pri-
mary avoidance theorem.
AMS Subject Classification: 13A15
1. Introduction
The prime ideal in ring theory plays crucial role in algebra. The concept of 2-
absorbing ideal which is a generalization of prime ideal of a commutative ring was
introduced in [6] and later studied in [2]. Several authors in [8,9,15] have extended
the notion of 2-absorbing ideal to 2-absorbing submodule which is a generalization
of prime submodule of an R-module M.Badawiet al. in [7] have introduced the
concept of 2-absorbing primary ideal which is a generalization of primary ideal in
a commutative ring and studied several properties. In this paper, we have intro-
duced the concept of a 2-absorbing primary submodules and proved many results.
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Throughout the paper Ris considered as a commutative ring with a nonzero identity
and module Mis unital.
An ideal Iof Ris said to be proper if I=R. A nonzero proper ideal Iof R
is said to be a 2-absorbing ideal of Rif abc ∈Ifor any a, b, c ∈R,thenab ∈Ior
ac ∈Ior bc ∈I. AproperidealIof Ris said to be a 2-absorbing primary ideal
of Rif whenever a, b, c ∈Rwith abc ∈I, then ab ∈Ior ac ∈√Ior bc ∈√I,
where √I={r∈R:thereexistsn∈Nwith rn∈I}denotes the radical of I.
Let Nbe a submodule of M.Then(N:RM)orsimply(N:M) denotes the ideal
{r∈R:rM ⊆N}.A proper submodule Nof an R-module Mis said to be a prime
submodule if for a∈R,m∈M, and am ∈N,thenm∈Nor a∈(N:RM).A
proper submodule Nof Mis said to be a 2-absorbing submodule of Mif whenever
a, b ∈R,m∈Mwith abm ∈N,thenab ∈(N:RM)oram ∈Nor bm ∈N. The
annihilator of Mdenoted by annR(M)is(0:
RM).An R-module Mis called a
multiplication module if every submodule Nof Mhas the form IM for some ideal I
of R. It is easy to see that, since I⊆(N:RM)thenN=IM ⊆(N:RM)M⊆N,
so that N=(N:RM)M[18]. For a submodule Nof M, if N=IM for some ideal
Iof R,thenwesaythatIis a presentation ideal of N. Clearly, every submodule of
Mhas a presentation ideal if and only if Mis a multiplication module. Let Nand
Kbe submodules of a multiplication R-module Mwith N=I1Mand K=I2M
for some ideals I1and I2of R. The product of Nand Kdenoted by NK is defined
by NK =I1I2M. By [1], the product of Nand Kis independent of presentations
of Nand K. Also, the term ab, for a, b ∈Mrepresents the product of Ra and Rb.
Clearly, NK is a submodule of Mand NK ⊆N∩K. A submodule Nof Mis called
pure Nif aN =N∩aM for every a∈R. A module Mis said to be cancellative
module if whenever rm =rn for elements m, n ∈Mand r∈R,thenm=n. Let
Nbe a proper submodule of a nonzero R-module M. Then the M-radical of N,
denoted by M-rad N, is defined to be the intersection of all prime submodules of
Mcontaining N. Itisshownin[18, Theorem 2.12] that if Nis a proper submodule
of a multiplication R-module M, then M-rad N=(N:M)M.
The paper is organized as follows. In Sec. 2, we introduce the notion of 2-
absorbing primary submodules and give some characterizations of the same. The
properties of 2-absorbing primary submodule are also studied in some specific
domain. In Sec. 3, we give the 2-absorbing primary avoidance theorem for sub-
modules.
2. 2-Absorbing Primary Submodules
In this section, we define 2-absorbing primary submodule which is illustrated with
an example and proved several results related to the same.
Definition 2.1. Let Mbe an R-module and Nbe a proper submodule of M. Then
Nis said to be a 2-absorbing primary submodule of Mif whenever a, b ∈Rand
m∈Mwith abm ∈N,thenab ∈(N:RM)oram ∈Nor bm ∈N.
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It is easy to see that prime, primary, 2-absorbing submodules and 2-absorbing
primary submodules are entirely different concepts. Every 2-absorbing submodule
is a 2-absorbing primary submodule but converse need not be true which are illus-
trated as follows.
Example 2.2. Consider R=Zand an R-module M=Z16 ={0,1,2,...,15}.
Take a submodule N={0,8}of M.Then(N:M)={a∈R:aM ⊆N}=
{0,8,16,...}and (N:M)={a∈R:anM⊆N}={0,2,4,8,...}.Now, 2.2.2∈
Nbut 2.2/∈Nand 2.2/∈(N:M).Therefore, Nis not a 2-absorbing submodule
of Mbut it is a 2-absorbing primary submodule of M,as2.2∈(N:M).
Example 2.3. Consider a submodule N=6Zof a Z-module Z. Then 6Zis a
2-absorbing primary submodule. But it is not a primary submodule as 2.3∈Nbut
neither 2 ∈Nnor 3 ∈(N:Z).
We state the following theorems which are frequently used in the paper.
Theorem 2.4 ([7, Theorem 2.2]). If Iis a 2-absorbing primary ideal of R, then
√Iis a 2-absorbing ideal of R.
Theorem 2.5 ([7, Theorem 2.3]). Suppose that Iis a 2-absorbing primary ideal
of R. Then one of the following statements must hold.
(i) √I=Pis a prime ideal,
(ii) √I=P1∩P2,where P1and P2are the only distinct prime ideals of Rthat are
minimal over I.
Theorem 2.6. Let Nbe a 2-absorbing primary submodule of an R-module M.
Then (N:M)is a 2-absorbing primary ideal of R.
Proof. Let abc ∈(N:M)forsomea, b, c ∈R.Letab /∈(N:M)andbc /∈
(N:M). This implies ab /∈(N:M)andbc /∈(N:M).So, there exist m1,m
2∈
Msuch that abm1/∈Nand bcm2/∈Nbut ac(bm1+bm2)∈N. Since Nis a
2-absorbing primary submodule, then we have either ac ∈(N:M)ora(bm1+
bm2)∈Nor c(bm1+bm2)∈N. If ac ∈(N:M), then we are done. If a(bm1+
bm2)∈N, then abm2/∈N.Considerabcm2∈N.SinceNis a 2-absorbing primary
submodule and abm2/∈N,bcm2/∈N, therefore ac ∈(N:M).Similarly, if
c(bm1+bm2)∈N,thenwehavecbm1/∈N. Consider abcm1∈N. Since Nis a 2-
absorbing primary submodule and abm1/∈N,bcm1/∈N, therefore ac ∈(N:M).
This implies, in each case, (N:M) is a 2-absorbing primary ideal of R.
But the converse of the above theorem is not true. If (N:M) is 2-absorbing
primary ideal, then Nmay not be 2-absorbing primary. Let us consider the Z-
module M=Z×Zand N=(0,6)Zbe the submodule of M.Then(N:M)=0
but Nis not a 2-absorbing primary module. As, for r1=2,r2=3and
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m=(0,1) ∈Mwe have 2.3.(0,1) ∈Nbut neither 2.(0,1) ∈Nnor 3.(0,1) ∈N
also 2.3/∈(N:M).
Theorem 2.7. Let Nbe a 2-absorbing primary submodule of an R-module M.
Then (N:M)is a 2-absorbing ideal of R.
Proof. Let Nbe a 2-absorbing primary submodule of an R-module M. Then, by
Theorem 2.6,(N:M) is a 2-absorbing primary ideal of R. Thus, by [7,Theo-
rem 2.2], we have (N:M) is a 2-absorbing ideal of R.
Theorem 2.8. Let Mbe a faithful multiplication R-module. If Nis a 2-absorbing
primary submodule of M, then M-rad Nis a 2-absorbing primary submodule of M.
Proof. Suppose Nis a 2-absorbing primary submodule of M. Then using Theo-
rem 2.7,(N:M) is a 2-absorbing ideal of R. Again, by Theorem 2.5,(N:M)=
por (N:M)=p∩qwhere p, q are distinct prime ideals of R.Consider,
(N:M)=p.ThenM-rad N=(N:M)M=pM, since Mis a multiplication
module. Since pis a prime ideal, pM is a prime submodule of Mby [18, Corol-
lary 2.11]. Hence M-rad Nis a 2-absorbing submodule of Mwhich shows that it is
a 2-absorbing primary submodule of M. Again, consider when (N:M)=p∩q,
where p, q are distinct prime ideals over Rthat are minimal over (N:M). Then,
we have M-rad N=(N:M)M=(p+annM)M∩(q+annM)M=pM ∩qM,
where pM, qM are prime submodules of M,by[18, Corollary 2.11, 1.7]. Conse-
quently, M-rad Nis a 2-absorbing submodule of Mwhich shows that M-rad Nis
a 2-absorbing primary submodule of M.
Theorem 2.9. Let Nbe a 2-absorbing primary submodule of an R-module Mand
Kbe a submodule of M. Then N∩Kis a 2-absorbing primary submodule of K.
Proof. Clearly, N∩Kis a proper submodule of K. Let rsx ∈N∩Kwhere r, s ∈R
and x∈K. Since rsx ∈Nand Nis a 2-absorbing primary submodule of M,rx ∈N
or sx ∈Nor rs ∈(N:M). If rx ∈Nor sx ∈N,thenrx ∈N∩Kor sx ∈N∩K.
If rs ∈(N:M),then (rs)nM⊆Nfor some positive integer n. In particular,
(rs)nK⊆Nwhich implies (rs)nK⊆N∩K. Thus rs ∈(N∩K:K).Hence
N∩Kis a 2-absorbing primary submodule of K.
Theorem 2.10. Let M1and M2be R-modules. Let M=M1⊕M2and N⊆
M1⊕M2.ThenN=Q⊕M2is a 2-absorbing primary submodule of Mif and only
if Qis a 2-absorbing primary submodule of M1.
Proof. Let Qbe a submodule of M1such that N=Q⊕M2is a 2-absorbing primary
submodule of M.ToproveQis a 2-absorbing primary submodule of M1.Let
rsm ∈Qwhere r, s ∈Rand m∈M1and rm /∈Q,sm /∈Q.Thenrs(m, 0) ∈Q⊕M2
but (rm, 0) /∈Q⊕M2and (sm, 0) /∈Q⊕M2.As Q⊕M2is a 2-absorbing primary
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submodule of M, therefore rs ∈(Q⊕M2:M1⊕M2)or(rs)n(M1⊕M2)⊆
(Q⊕M2) for some positive integer n.Thus(rs)nM1⊆Q. Hence Qis a 2-absorbing
primary submodule of M1.Conversely,letQbe a 2-absorbing primary submodule
of M1. Assume that rs(a, b)∈Q⊕M2where r, s ∈Rand (a, b)∈M. Suppose
that r(a, b)/∈Q⊕M2and s(a, b)/∈Q⊕M2.This implies ra /∈Qand sa /∈Q.
As Qis a 2-absorbing primary submodule of M1, therefore (rs)nM1⊆Q. Thus
(rs)nM⊆Q⊕M2.Hence Q⊕M2is a 2-absorbing primary submodule of M.
For r∈R, we define (N:Mr)tobe(N:Mr)={m∈M:rm ∈N}.Clearly,
(N:Mr) is a submodule of Mcontaining N. It is well known that a submodule
Nof Mis said to be irreducible if it cannot be expressed as the intersection of
two submodules of M. We now give the following characterization of 2-absorbing
primary submodule when Nis an irreducible.
Theorem 2.11. Let Mbe an R-module and Na proper submodule of Mand
(N:M)is a 2-absorbing ideal of R. If Nis an irreducible submodule of M, then
Nis a 2-absorbing primary submodule of Mif and only if (N:Mr)=(N:Mr2)
for all r∈R\(N:M).
Proof. Let Nbe a 2-absorbing primary submodule of M. Assume that r∈
R\(N:M).Then we have to show that (N:Mr)=(N:Mr2).It is clear
that (N:Mr)⊆(N:Mr2).For the reverse inclusion, let m∈(N:Mr2), so
r2m∈N. This implies that either rm ∈Nor r2∈(N:M), since Nis a 2-
absorbing primary submodule of M.Ifrm ∈N,thenm∈(N:Mr) and hence
(N:Mr2)⊆(N:Mr), we are done. Again, if r2∈(N:M), then r∈(N:M)
and we get a contradiction. Thus (N:Mr)=(N:Mr2).
Conversely, let r1,r
2∈Rand m∈Msuch that r1r2m∈Nbut r1r2/∈
(N:M).Then,wehavetoshowthatr1m∈Nor r2m∈N. On contrary,
we assume that r1m/∈Nand r2m/∈N. We first claim that r1/∈(N:M)
and r2/∈(N:M) because if r1∈(N:M)andr2∈(N:M)implythat
r1r2∈((N:M))2⊆(N:M), which is a contradiction. Therefore we may
assume that either (N:Mr1)=(N:Mr2
1)or(N:Mr2)=(N:Mr2
2).Suppose
(N:Mr1)=(N:Mr2
1).Clearly, N⊆(N+Rr1m)∩(N+Rr2m).Let
n∈(N+Rr1m)∩(N+Rr2m).Then n=n1+s1r1m=n2+s2r2mwhere n1,
n2∈Nand s1,s
2∈R.Nowr1n=r1n1+s1r2
1m=r1n2+r1r2s2mand
s2r1r2m, r1n1,r
1n2∈N, so s1r2
1m∈Nimplies s1m∈(N:Mr2
1) but (N:M
r1)=(N:Mr2
1).Therefore s1r1m∈Nand so n∈N. Hence (N+Rr1m)∩(N+
Rr2m)⊆N.Consequently,(N+Rr1m)∩(N+Rr2m)=N, a contradiction because
Nis an irreducible submodule. Hence Nis a 2-absorbing primary submodule
of M.
Theorem 2.12. Let Nbe a 2-absorbing primary submodule of an R-module M.
Then (N:Mr)is a 2-absorbing primary submodule of Mcontaining Nfor any
r∈R\(N:M).
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Proof. Suppose r∈R\(N:M).For any s, t ∈Rand m∈Msuch that stm ∈
(N:Mr).This implies strm ∈N. Since Nis a 2-absorbing primary submodule of
M, then either trm ∈Nor srm ∈Nor st ∈(N:M).Thus sm ∈(N:Mr)
or tm ∈(N:Mr)or(st)nM⊆N. The first two cases show that (N:Mr)is
a 2-absorbing primary submodule of M. Further, (st)nM⊆Nimplies (st)nM⊆
(N:Mr) which implies st ∈((N:Mr):M). Thus (N:Mr) is a 2-absorbing
primary submodule of M.
We now study the properties of 2-absorbing primary submodules using associate
ideal (N:Rm), defined as (N:m)={r∈R:rm ∈N}where m∈M.
Theorem 2.13. If Nis a 2-absorbing primary submodule of an R-module M,
then (N:m)=Rif m∈Nor (N:m)is a 2-absorbing primary ideal containing
(N:M)if m/∈N.
Proof. If m∈N, then nothing to prove. Suppose m/∈Nand abc ∈(N:m)for
some a, b, c ∈R. This implies abcm ∈N.AsNis a 2-absorbing primary submodule
of M, then we have either am ∈Nor bcm ∈Nor abc ∈(N:M). The first
two terms am ∈Nor bcm ∈Nimply ab ∈(N:m)orbc ∈(N:m)andwe
are done. If abc ∈(N:M)and(N:M) is 2-absorbing ideal of R, then either
ab ∈(N:m)orbc ∈(N:m)orca ∈(N:m). Thus, in each case, we have
(N:m) is a 2-absorbing primary ideal of R. Clearly (N:M)⊆(N:m).
Theorem 2.14. Let Mbe a n R-module and Nbe a proper submodule of Msuch
that (N:M)is a prime ideal of R.ThenNis a 2-absorbing primary submodule of
Mif and only if for any m1,m
2∈Mif (N:m1)\((N:m2)∪(N:M)),then
N=(N+Rm1)∩(N+Rm2).
Proof. Suppose Nis a 2-absorbing primary submodule of M. Let ab ∈(N:
m1)\((N:m2)∪(N:M)) where a, b ∈R. Then abm1∈Nand abm2/∈N
and ab /∈(N:M).Clearly, N⊆(N+Rm1)∩(N+Rm2).For reverse inclu-
sion, let n∈(N+Rm1)∩(N+Rm2). Then n=n1+r1m1=n2+r2m2,where
n1,n
2∈Nand r1,r
2∈R. Now, abn =abn1+abr1m1=abn2+abr2m2and
abr1m1,abn
1,abn
2∈N, so abr2m2∈N.SinceNis a 2-absorbing primary sub-
module of Mand abm2/∈N, therefore either r2m2∈Nor abr2∈(N:M).Con-
sider the case, abr2∈(N:M) implies (ab)m(r2)m∈(N:M) for some positive
integer m. Since (N:M)isprimeandab /∈(N:M) implies (ab)m/∈(N:M).
Therefore r2m∈(N:M)givesr2∈(N:M)andsor2m2∈N.Inboththe
cases, we have n=n2+r2m2∈N.So,(N+Rm1)∩(N+Rm2)⊆N. Hence
N=(N+Rm1)∩(N+Rm2).
Conversely, suppose r1r2m∈Nwhere r1,r
2∈R, m ∈Mand r1m/∈Nand
r1r2/∈(N:M). Then, we have to show that r2m∈N. Now, we have r1∈(N:
r2m)\(N:m)∪(N:M). Put m1=r2m, m2=m, in given assumption we have
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N=(N+Rr2m)∩(N+Rm).This gives r2m∈(N+Rm)∩(N+Rr2m)=N.
Therefore, Nis a 2-absorbing primary submodule of M.
Theorem 2.15. Let Nbe a 2-absorbing primary submodule of Mand m∈M\N.
Then (N:m)is a prime ideal of Ror there exists an element a∈Rsuch that
(N:anm)is a prime ideal for some positive integer n.
Proof. Let Nbe a 2-absorbing primary submodule of M.Then(N:M)isa
2-absorbing ideal of R.Since(N:M) is a 2-absorbing ideal of R, therefore by
Theorem 2.5,wehaveeither(N:M)=por (N:M)=p∩q,wherepand
qare distinct prime ideals of R. Suppose (N:M)=p.Thenp=(N:M)⊆
(N:m). We show that (N:m) is a prime ideal of R.Letrs ∈(N:m)for
some r, s ∈R.Then(rs)n∈(N:m) implies rnsnm∈N. As Nis 2-absorbing
primary submodule of M, then either rnm∈Nor snm∈Nor rs ∈(N:M).If
rnm∈Nor snm∈N,thenr∈(N:m)ors∈(N:m). If rs ∈(N:M),
then rs ∈p.Sincepis prime ideal of R, then either r∈p⊆(N:m)ors∈p⊆
(N:m).Therefore, (N:m) is a prime ideal of R.
Again, consider (N:M)=p∩q.Ifp⊆(N:m), then by previous
argument, we have (N:m) is a prime ideal of R.Ifp(N:Rm)then
there exists a∈Rsuch that a∈pbut a/∈(N:m)oranm/∈N. Also,
pq ⊆√pq ⊆√p∩q=(N:M)⊆(N:m).Thus pq ⊆(N:M), which
implies anqn⊆(N:m)andsoq⊆(N:anm).Again, on similar manner, we can
easily show that (N:anm) is a prime ideal of R.
Theorem 2.16. Let Nbe an arbitrary submodule of an R-module Mand
N1,N
2,...,N
nbe 2-absorbing primary submodules of M.Suppose(Nj:M)
(Ni:m)for all m∈M\Niwith i=jand 1≤i≤n. If NNifor al l i, then
there exists an element a∈Nsuch that a/∈∪Ni;hence N∪Ni.
Proof. Since NNi,thereexistsmi∈Nsuch that mi/∈Nifor all 1 ≤i≤n.
Since each Niis a 2-absorbing primary submodule of M,byTheorem2.15,wehave
either (Ni:mi) is a prime ideal or there exists pi∈Rsuch that (Ni:pitimi)
is a prime ideal of R.Let
(Ni:mi) be a prime ideal. Then, by given assumption
there exist rj∈(Nj:M)andrj/∈(Ni:mi). Thus rjnj∈(Nj:M)andrjnj/∈
(Ni:mi). Let xi=r1r2...r
i−1ri+1 ...r
n=j=irj.Then xi∈(Nj:M)and
xi/∈(Ni:mi). If li=max(n1,n
2,...,n
i−1,n
i+1,...,n
n),then ai=mixili∈Nj
for all j=i.Butai/∈Nibecause if ai∈Ni,thenmixili∈Ni, this implies
xi∈(Ni:mi), a contradiction. Let a=a1+a2+···+an.Then ai=a−
j=iaj.Since j=iaj∈Ni, therefore a/∈Niotherwise we would have ai∈Ni,
a contradiction. So a/∈∪Ni.Therefore a∈Nand a/∈∪Ni. Consider the case
when (Ni:pitimi) is a prime ideal of Rfor some pi∈R. Then there is sj∈
(Nj:M)andsj/∈(Ni:pitimi),which implies sjnj
∈(Nj:M)andsjnj
/∈
(Ni:pitimi),Let yi=s1s2...s
i−1si+1 ...s
n=j=isj.Then yi∈(Nj:M) but
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yi/∈(Ni:mi).If ki=max(n
1,n
2,...,n
i−1,n
i+1,...,n
n),then bi=pitimiyiki∈
Njfor all j=i.Butbi/∈Nibecause, if bi∈Ni,thenpitimiyiki∈Nithis
implies yiki∈(Ni:pitimi), a contradiction. Let b=b1+b2+···+bn.Then
bi=b−j=ibj.Sincej=ibj∈Ni, therefore b/∈Niotherwise we would have
bi∈Ni, a contradiction. So b/∈∪Ni. Hence Nn
i=1 Ni.
Further, we show that if Nis a 2-absorbing primary submodule of M,and
IJL ⊆Nfor some ideals I,J of Rand a submodule Lof M,thenIJ ⊆(N:M)
or IL ⊆Nor JL ⊆N. But first we require the following lemma.
Lemma 2.17. Let Nbe a 2-absorbing primary submodule of Mover a ring R.
Suppose that abL ⊆Nand ab /∈√N:Mfor some elements a, b ∈Rand some
submodule Lof M, then aL ⊆Nor bL ⊆N.
Proof. Suppose that abL ⊆Nand ab /∈√N:Mfor some a, b ∈Rand some
submodule Lof M. Suppose aL Nand bL N. Then for some l1,l
2∈L,
al1/∈Nand bl2/∈N. Since abl1∈Nand al1/∈N,ab /∈(N:M), therefore
bl1∈N,asNis a 2-absorbing primary submodule of M. Similarly, abl2∈N
implies al2∈Nbecause bl2/∈Nand ab /∈(N:M). Consider ab(l1+l2)∈N,we
have either a(l1+l2)∈Nor b(l1+l2)∈N. If a(l1+l2)∈N,thenal2∈Nimplies
al1∈N, a contradiction. Similarly, if b(l1+l2)∈N,thenbl1∈Nimplies bl2∈N,
a contradiction. Thus, either aL ⊆Nor bL ⊆N.
Theorem 2.18. Let Lbe a submodule of an R-module M.ThenNis a 2-absorbing
primary submodule of Mif whenever IJL ⊆Nfor some ideals I,J of Rand a
submodule Lof M, then IJ ⊆(N:M)or IL ⊆Nor JL ⊆N.
Proof. Let Nbe a 2-absorbing primary submodule of Mand let IJL ⊆Nfor
some ideals I,J of Rand a submodule Lof M. Suppose IJ (N:M), then we
show that either IL ⊆Nor JL ⊆N. On contrary, we assume that IL Nand
JL N. Then there exist a1∈I,b1∈Jsuch that a1LNand b1LN.Since
a1b1L⊆Nand Nis a 2-absorbing primary submodule, a1b1∈(N:M).Next,
we have IJ (N:M), therefore for some a∈Iand b∈J,ab /∈(N:M).Since
abL ⊆Nand ab /∈(N:M),therefore aL ⊆Nor bL ⊆N(by Lemma 2.17).
Now, we have the following three cases:
Case I: Suppose aL ⊆Nbut bL N. Since a1bL ⊆Nand bL Nand a1L
N, then by Lemma 2.17,wehavea1b∈(N:M).Now, (a+a1)bL ⊆Nand
aL ⊆Nbut a1LN, therefore (a+a1)LN. Again, since (a+a1)bL ⊆Nand
bL N,then(a+a1)LNimplies (a+a1)b∈(N:M)byTheorem2.17.
Since (a+a1)b∈(N:M)anda1b∈(N:M),we have ab ∈(N:M),a
contradiction.
Case II: Suppose bL ⊆Nbut aL N. Since ab1L⊆Nbut aL Nand
b1LN, then by Lemma 2.17,ab1∈(N:M).Again, let a(b+b1)L⊆Nbut
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bL ⊆Land b1LNgives (b+b1)LN. Since aL Nand (b+b1)LN, we
have a(b+b1)∈(N:M) (using Lemma 2.17). Since a(b+b1)∈(N:M)and
ab1∈(N:M),we have ab ∈(N:M),a contradiction.
Case III: Suppose aL ⊆Nand bL ⊆N. Since bL ⊆Nand b1LNimply
(b+b1)LN. Since a1(b+b1)L⊆Nand (b+b1)LNand a1LN, then
using Lemma 2.17,wegeta1(b+b1)∈(N:M). Since a1b1∈(N:M), a1b∈
(N:M).Again, aL ⊆Nand a1LNimply (a+a1)LN. Since (a+
a1)b1L⊆Nand (a+a1)LNand b1L⊆N, then by Lemma 2.17,wehave
(a+a1)b1∈(N:M). Since a1b1∈(N:M)and(a+a1)b1∈(N:M),
then ab1∈(N:M).Since (a+a1)(b+b1)L⊆Nand (a+a1)LNand
(b+b1)LN, then again by Lemma 2.17,wehave(a+a1)(b+b1)∈(N:M).
Since ab1,a
1b, a1b1∈(N:M),we have ab ∈(N:M), a contradiction. Hence
IL ⊆LN or JL ⊆N.
Proposition 2.19. Let Mand Mbe R-modules and f:M→ Mbe an epimor-
phism. Then,the following statements hold.
(i) If Nis a 2-absorbing primary submodule of Msuch that ker f⊆N, then f(N)
is a 2-absorbing primary submodule of M.
(ii) If Nis a 2-absorbing primary submodule of Msuch that f(M)⊆N,then
f−1(N)is a 2-absorbing primary submodule of M.
Proof. (i) Let a, b ∈Rand y∈Msuch that aby ∈f(N). Then there exists n∈N
such that aby =f(n).Since fis an epimorphism therefore for some m∈Mwe
have f(m)=y.Thusabf(m)=f(n). This implies f(abm −n)=0whichgives
abm −n∈ker f⊆N.Soabm ∈N.SinceNis a 2-absorbing primary submodule
of M,am ∈Nor bm ∈Nor ab ∈(N:M). This gives ay ∈f(N)orby ∈f(N)
or ab ∈(f(N):M). Therefore f(N) is a 2-absorbing primary submodule of M.
(ii) The proof is trivial.
Theorem 2.20. Let Nbe a submodule of an R-module Mand let Kbe any s ub-
module of Mcontained in N. Then N
Kis a 2-absorbing primary submodule of M
Kif
and only if Nis a 2-absorbing primary submodule of M.
Proof. Let Nbe a 2-absorbing primary subsemimodule of M. Consider an epimor-
phism f:N→ N
K.Then(byTheorem2.19(i)) N
Kis a 2-absorbing primary submod-
ule of M
K.Conversely,letabm ∈Nfor a, b ∈Rand m∈M.Then(abm +K)∈N
K
which implies ab(m+K)∈N
K.Since N
Kis a 2-absorbing primary submodule, either
a(m+K)∈N
Kor b(m+K)∈N
Kor ab ∈N
K:M
K. This implies, either am ∈Nor
bm ∈Nor ab ∈√N:M. Hence Nis a 2-absorbing primary submodule.
Theorem 2.21. Let Mbe a cancel lat ive R-module and Nbe proper submodule of
M.ThenNis a pure submodule of Mif and only if Nis a 2-absorbing primary
submodule of Mwith (N:M)={0}.
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Proof. Suppose that Nis a pure submodule of Mand abm ∈Nsuch that ab ∈
(N:M)wherea, b ∈Rand m∈M.Thenabm ∈abM ∩N=abN ,soabm =abn
for some n∈N. This implies bm =bn ∈N(as Mis a cancellative module). Thus N
is a 2-absorbing primary submodule of M. Next, suppose that a∈(N:M)with
a=0.SinceN=M, then there exists x∈M\Nsuch that anx∈anM∩N=anN
for some natural number n.So,thereexistsy∈Nsuch that anx=any, therefore
x=y∈N, a contradiction. Thus (N:M)={0}.Conversely, assume that Nis
a 2-absorbing primary submodule of Mwith √N:M={0}.Let abz ∈abM ∩N
implies abz ∈N,wherez∈Mand a, b ∈R. We may assume that ab /∈√N:M.
Since Nis a 2-absorbing primary submodule of M, then we have either az ∈Nor
bz ∈N.Ifbz ∈N,thenwehaveabz ∈abN . Therefore abM ∩N⊆abN. Similarly,
we can prove the case for az ∈N,thatis,abM ∩N⊆abN. Now, abN ⊆abM ∩N.
Hence abM ∩N=abN and consequently Nis a pure submodule of M.
Theorem 2.22. Let Rbe a commutative ring and Mbe a multiplication R-module.
If Mis a Dedekind module and Nis a 2-absorbing primary submodule of M, then
N=Knor N=K1nK2mwhere K, K1,K
2are maximal modules and n, m are
positive integers.
Proof. Suppose Mis a Dedekind module. Then Ris a Dedekind domain by [14,
Theorem 3.5]. For any 2-absorbing primary submodule N,wehave(N:M)isa2-
absorbing primary ideal of R.Using[7, Theorem 2.11], we have either (N:M)=In
or (N:M)=I1nI2m,whereI,I1,I
2are maximal ideals of R.By[18, Theorem 2.5],
if Iis a maximal ideal of R,thenIM is a maximal module of M. This implies
N=(IM )nor N=(I1M)n(I2M)m.
Theorem 2.23. Suppose Sis a multiplicatively closed subset of Rand S−1Mis
themoduleoffractionofM. Then the following statements hold.
(i) If Nis a 2-absorbing primary submodule of Msuch that (N:M)∩S=∅,then
S−1Nis a 2-absorbing primary submodule of S−1M.
(ii) If S−1Nis a 2-absorbing primary submodule of S−1Msuch that Z(M/N)∩
S=∅,then Nis a 2-absorbing primary submodule of M.
Proof. (i) Assume that a, b ∈R,s, t, l ∈S,m∈Mand a
s
b
t
m
l∈S−1Nwhich
implies µabm ∈Nfor some µ∈S. Since Nis a 2-absorbing primary submodule of
M,µam ∈Nor µbm ∈Nor ab ∈(N:M).Hence a
s
m
l∈S−1Nor b
t
m
l∈S−1N
or a
s
b
t∈S−1√N:M=(S−1N:S−1M).So, S−1Nis a 2-absorbing primary
submodule of S−1M.
(ii) Let a, b ∈Rand m∈Mbe such that abm ∈N. Then abm
1∈S−1N.
Since S−1Nis a 2-absorbing primary submodule of S−1M,either am
1∈S−1Nor
bm
1∈S−1Nor ab
1∈S−1N:S−1RS−1M. Therefore, there exists s∈Ssuch that
sam ∈Nor sbm ∈N. This implies am ∈Nor bm ∈N,sinceS∩Z(M/N)=∅.
Now, consider the case when ab
1∈S−1N:S−1RS−1M=S−1√N:M, then there
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exists µ∈Ssuch that (µab)nM⊆Nimplies ab ∈(N:M). Hence Nis a
2-absorbing primary submodule of M.
3. 2-Absorbing Primary Avoidance Theorem for Submodules
In this section, we study 2-absorbing primary avoidance theorem for submodules
which is a generalization of primary submodules. We first define an efficient covering
of submodules: let N, N1,N
2,...,N
nbe submodules of an R-module M. An efficient
covering of Nis a covering N⊆N1∪N2...∪Nnin which no Nk(where 1 ≤k≤n)
satisfies N⊆Nk.In other words, a covering N⊆N1∪N2∪···∪Nnis efficient if
no Nkis superfluous. To proceed further, we require the following lemma.
Lemma 3.1 ([11, Lemma 2.1]). Let N=N1∪···∪Nnbe an efficient union of
submodules of an R-module Mfor n>1.Then j=kNj=n
j=1 Nkfor all k.
Theorem 3.2. Let N⊆N1∪N2∪N3∪···∪Nnbe an efficient covering consisting
of submodules of an R-module M.If(Nj:M)(Nk:m)where m∈M\Nk
for every j=k, then no Nkis a 2-absorbing primary submodule.
Proof. Since N⊆∪Niand Nhas an efficient covering, then NNk,sothere
exists an element mk∈N\Nk.It is clear that N=(N∩N1)∪(N∩N2)∪···
∪(N∩Nn) is an efficient union. By Lemma 3.1,wehavej=k(N∩Nj)⊆(N∩Nk).
Suppose Nkis a 2-absorbing primary submodule. Then, using Theorem 2.15,we
have either (Nk:mk) is a prime ideal or there exists a∈Rsuch that √Nk:anmk
is a prime ideal. First, suppose (Nk:mk) is a prime ideal. By the given hypothesis
(Nj:M)(Nk:mk)forj=k. So, there exists sj∈(Nj:M) but sj/∈
(Nk:mk), where j=k. This implies sjnj∈(Nj:M) but sjnj/∈(Nk:mk)where
j=kand nj∈Z+.Let s=j=ksj.Thens∈(Nj:M) but s/∈(Nk:mk)
where j=k. Therefore, sm∈(Nj:M) for all j=kbut sm/∈(Nk:mk), where
m=max(n1,n
2,...,n
k−1,n
k+1,...,n
n).Thus smmk∈j=k(N∩Nj)\(N∩Nk),
since smmk∈(N∩Nk) implies sm∈(Nk:mk), a contradiction. So, no Nkis a
2-absorbing primary submodule of M. Now, consider the case when √Nk:anmkis
a prime ideal, where nis positive integer and a∈R. Clearly, sj∈(Nj:M) but
sj/∈(Nk:anmk), where j=k. Therefore, smanmk∈j=k(N∩Nj)\(N∩Nk),
since smanmk∈(N∩Nk) implies sm∈(Nk:anmk), a contradiction. So, no Nkis
2-absorbing primary submodule of M.
Theorem 3.3 (2-Absorbing Primary Avoidance Theorem for Submod-
ules). Let N, N1,...,N
n(n≥2) be submodules of Msuch that at most two of
N1,N
2,...,N
nare not 2-absorbing primary submodules. If N⊆N1∪N2∪···∪Nn
and (Nj:M)(Nk:m),where m∈M\Nkfor every j=k, then N⊆Nifor
some 1≤i≤n.
Proof. If n= 2, then it is obvious. Now, take n>2andNNifor all 1 ≤i≤n.
Then N⊆N1∪N2∪···∪Nnis an efficient covering. Using Theorem 3.2,noNi
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is a 2-absorbing primary submodule, which is a contradiction. Hence N⊆Nifor
some 1 ≤i≤n.
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