aa r X i v : . [ m a t h . A C ] F e b ECE YETKIN CELIKEL
Abstract.
Let R be a commutative ring with non-zero identity and M bea unitary R -module. The goal of this paper is to extend the concept of 1-absorbing primary ideals to 1-absorbing primary submodules. A proper sub-module N of M is said to be a 1-absorbing primary submodule if whenever non-unit elements a, b ∈ R and m ∈ M with abm ∈ N , then either ab ∈ ( N : R M )or m ∈ M − rad ( N ) . Various properties and chacterizations of this class of sub-modules are considered. Moreover, 1-absorbing primary avoidance theorem isproved. Introduction
Throughout this paper, we shall assume unless otherwise stated, that all rings arecommutative with non-zero identity and all modules are considered to be unitary.A prime (resp. primary ) submodule is a proper submodule N of M with theproperty that for a ∈ R and m ∈ M , am ∈ N implies that m ∈ N or a ∈ ( N : R M )(resp. a ∈ p ( N : R M )). Since prime and primary ideals (submodules) have animportant role in the theory of modules over commutative rings, generalizations ofthese concepts have been studied by several authors [1]-[8], [14], [15]. For a surveyarticle consisting some of generalizations see [5]. In 2007, Badawi [4] called a non-zero proper ideal I of R a of R if whenever a, b, c ∈ R and abc ∈ I ,then ab ∈ I or ac ∈ I or bc ∈ I . As an extension of 2-absorbing primary ideals, theconcept of 2-absorbing submodules are introduced by Darani and Soheilnia [8] andstudied by Payrovi, Babaei [15]. We recall from [8] that a proper submodule N of M is said to be a if whenever a, b ∈ R and m ∈ M with abm ∈ N , then ab ∈ ( N : R M ) or am ∈ N or bm ∈ N . In 2014, Badawi, Tekir andYetkin [6] introduced the concept of 2-absorbing primary ideals. A proper ideal I of R is called if whenever a, b, c ∈ R with abc ∈ I , then ab ∈ I or ac ∈ √ I or bc ∈ √ I . After that, the notion of 2-absorbing pimary submodulesis introduced and studied in [14]. According to [14], a proper submodule N of M is said to be provided that a, b ∈ R, m ∈ M and abm ∈ N imply either ab ∈ ( N : R M ) or am ∈ M - rad ( N ) or bm ∈ M - rad ( N ) . As a recentstudy, the class of was defined in [7]. According to [7],a proper ideal I of R is said to be a if whenever non-unitelements a, b, c of R and abc ∈ I , then ab ∈ I or c ∈ √ I. Our aim is to extend thenotion of 1-absorbing primary ideals to 1-absorbing primary submodules.
Date : January, 2021.2000
Mathematics Subject Classification.
Primary 13A15, Secondary 13F05.
Key words and phrases.
For the sake of thoroughness, we give some definitions which we will needthroughout this study. Let I be an ideal of a ring R . By √ I, we mean the radicalof I which is the intersection of all prime ideals containing I , that is { r ∈ R : r n ∈ I for some n } . Let M be an R -module and N be a submodule of M . We will de-note by ( N : R M ) the residual of N by M , that is, the set of all r ∈ R suchthat rM ⊆ N . The annihilator of M denoted by Ann R ( M ) is (0 : R M ). The M -radical of N , denoted by M -rad( N ), is defined to be the intersection of allprime submodules of M containing N . If M is a multiplication R -module, then M − rad ( N ) = { m ∈ M : m k ⊆ N for some k ≥ } [1, Theorem 3.13]. If there isno such a prime submodule, then M − rad ( N ) = M. For the other notations andterminologies that are used in this article, the reader is referred to [10].We summarize the content of this article as follows. We call a proper submodule N of M a 1-absorbing primary submodule if whenever non-unit elements a, b ∈ R and m ∈ M with abm ∈ N , then ab ∈ ( N : R M ) or m ∈ M − rad ( N ) . Itis clear that a prime submodule is a 1-absorbing primary submodule, and a 1-absorbing primary submodule is a 2-absorbing primary submodule. In Section 2,we start with examples (Example 1 and Example 2) showing that the inverses ofthese implications are not true in general. Various characterizations for 1-absorbingprimary submodules are given (Theorem 1, Theorem 2 and Theorem 3). Moreover,the behavior of 1-absorbing primary submodules in modules under homomorphism,module localizations and direct product of modules are investigated (Proposition3, Proposition 4 and Proposition 5). Finally, in Section 3, the 1-absorbing primaryavoidance theorem is proved.2.
Properties of 1-absorbing primary submodules
Definition 1.
Let M be a module over a commutative ring R and N be a propersubmodule of M . We call N a 1-absorbing primary submodule if whenever non-unit elements a, b ∈ R and m ∈ M with abm ∈ N , then ab ∈ ( N : R M ) or m ∈ M − rad ( N ) . It is clear that the following implication hold: prime submodule ⇒ ⇒ M needs not to be a primary(prime) submodule; and also there are 2-absorbing primary submodules which arenot 1-absorbing primary. Example 1. (1)
Let A = K [ x, y ] , where K is a field, Q = ( x, y ) A . Con-sider R = A Q and M = R as an R -module. Then N = ( x , xy ) M is a 1-absorbing primary submodule of M [7, Example 1] . Observe that p ( N : R M ) = xR . Since x · y ∈ N , but x / ∈ N and y / ∈ p ( N : R M ) , N isnot a primary submodule (so, it is not a prime submodule) of M . (2) Consider the submodule N = p q Z of Z -module Z where p and q are distinctprime integers. Then N is a 2-absorbing primary submodule of Z by [14,Corollary 2.21] . However it is not 1-absorbing primary as p · p · q ∈ N butneither p · p ∈ ( N : Z Z ) = N nor q ∈ M − rad ( N ) = pq Z . The following example shows that there are some modules of which every propersubmodule is 2-absorbing primary but which has no 1-absorbing primary submod-ule. -ABSORBING PRIMARY SUBMODULES 3
Example 2.
Let p be a fixed prime integer. Then E ( p ) = { α ∈ Q / Z : α = r/p n + Z for some r ∈ Z and n ∈ N ∪ (0) } is a non-zero submodule of Z -module Q / Z . Foreach t ∈ N , set G t = { α ∈ Q / Z : α = r/p t + Z for some r ∈ Z } . Observe thateach proper submodule of E ( p ) is equal to G i for some i ∈ N and ( G t : Z E ( p )) = 0 for every t ∈ N . It is shown in [3, Example 1] that every submodule G t is nota primary submodule of E ( p ) . Thus there is no prime submodule in E ( p ) . Thus E ( p ) − rad ( G t ) = E ( p ) . Therefore, each G t is a 1-absorbing primary submodule of R . We next give several characterizations of 1-absorbing primary submodules of an R -module. Theorem 1.
Let N be a proper submodule of an R -module M . Then the followingstatements are equivalent: (1) N is a 1-absorbing primary submodule of M. (2) If a, b are non-unit elements of R such that ab / ∈ ( N : R M ) , then ( N : M ab ) ⊆ M − rad ( N ) . (3) If a, b are non-unit elements of R , and K is a submodule of M with abK ⊆ N , then ab ∈ ( N : R M ) or K ⊆ M − rad ( N ) . (4) If I I K ⊆ N for some proper ideals I , I of R and some submodule K of M , then either I I ⊆ ( N : R M ) or K ⊆ M − rad ( N ) . Proof. (1) ⇒ (2) Suppose that a, b are non-unit elements of R such that ab / ∈ ( N : R M ) . Let m ∈ ( N : M ab ). Hence abm ∈ N . Since N is 1-absorbing primarysubmodule and ab / ∈ ( N : R M ) , we have m ∈ M − rad ( N ) , and so ( N : M ab ) ⊆ M − rad ( N ) . (2) ⇒ (3) Suppose that ab / ∈ ( N : R M ). Since abK ⊆ N, we have K ⊆ ( N : M ab ) ⊆ M − rad ( N ) by (2).(3) ⇒ (4) Assume on the contrary that neither I I ⊆ ( N : R M ) nor K ⊆ M − rad ( N ) . Then there exist non-unit elements a ∈ I , b ∈ I with ab / ∈ ( N : R M ).Thus abK ⊆ N , it contradicts with (3).(4) ⇒ (1) Let a, b ∈ R be non-unit elements, m ∈ M and abm ∈ N . Put I = aR , I = bR , K = Rm.
Thus the result is clear. (cid:3) An R -module M is called a multiplication module if every submodule N of M hasthe form IM for some ideal I of R . Equivalently, N = ( N : R M ) M [9]. Let M bea multiplication R -module and let N = IM and K = JM for some ideals I and J of R . The product of N and K is denoted by N K is defined by
IJM . Clearly,
N K is a submodule of M and contained in N ∩ K . It is shown in [1, Theorem 3.4] thatthe product of N and K is independent of presentations of N and K . It is shownin [9, Theorem 2.12] that if N is a proper submodule of a multiplication R -module M , then M -rad( N ) = p ( N : R M ) M . If M is a finitely generated multiplication R -module, then ( M -rad( N ) : M ) = p ( N : R M ) by [14, Lemma 2.4]. Now, weare ready for characterizing 1-absorbing primary submodules of finitely generatedmultiplication module M in terms of submodules of M . Theorem 2.
Let M be a finitely generated multiplication R -module and N be aproper submodule of M. Then the following statements are equivalent: (1) N is a 1-absorbing primary submodule of M . ECE YETKIN CELIKEL (2) If N N N ⊆ N for some submodules N , N , N of M, then either N N ⊆ N or N ⊆ M − rad ( N ) . Proof. (1) ⇒ (2) Suppose that N is a 1-absorbing primary submodule of M , N N N ⊆ N and N * M − rad ( N ) . Since M is a finitely generated multiplication module, N = I M and N = I M for some ideals I , I of R . Hence I I N ⊆ N . Since N * M − rad ( N ) , we have I I ⊆ ( N : R M ) by Theorem 1. Thus we conclude N N ⊆ ( N : M ) M = N .(2) ⇒ (1) Let I I K ⊆ N. Then there exists an ideal I of R such that I I I M ⊆ N which gives I I M ⊆ N or I M ⊆ M − rad ( N ) . By [16, p.231 Corollary], wehave I I ⊆ ( N : R M ) + Ann R ( M ) = ( N : R M ) or K ⊆ M − rad ( N ) . Hence, weare done from Theorem 1. (cid:3)
Lemma 1. [16, Theorem 10]
Let M be a finitely generated faithful multiplication R -module, then ( IM : M ) = I .for all ideals I of R . In [3, Corollary 2], for a proper submodule N of a multiplication R -module M, it is shown that N is primary submodule of M if and only if ( N : R M ) is primaryideal of R . Analogous with this result, we have the following. Theorem 3.
Let I be an ideal of a ring R and N be a submodule of a finitelygenerated faithful multiplication R -module M . Then (1) I is a 1-absorbing primary ideal of R if and only if IM is a 1-absorbingprimary submodule of M .(2) N is a 1-absorbing primary submodule of M if and only if ( N : M ) is a1-absorbing primary ideal of R .(3) N is a 1-absorbing primary submodule of M if and only if N = IM forsome 1-absorbing primary ideal of R . Proof. (1) Suppose I is a 1-absorbing primary ideal of R . If IM = M , then I = ( IM : M ) = R by Lemma 1, a contradiction. Thus, IM is proper in M . Now,let a, b ∈ R be non-unit elements and m ∈ M such that abm ∈ IM and ab / ∈ ( IM : M ) = I . Then ab (( m ) : M ) = (( abm ) : M ) ⊆ ( IM : M ) ⊆ ( √ IM : M ) = √ I .Since I is a 1-absorbing primary ideal, we conclude that (( m ) : M ) ⊆ √ I . Thus, m ∈ (( m ) : M ) M ⊆ √ IM = M - rad ( IM ). Conversely, suppose IM is 1-absorbingprimary submodule of M . Then clearly I is proper in R . Let a, b, c ∈ R be non-unit elements with abc ∈ I and ab / ∈ ( IM : M ) = I . Since abM ∈ IM and IM is a 1-absorbing primary submodule, then cM ⊆ M - rad ( IM ) = √ IM . Therefore, c ∈ ( √ IM : M ) = √ I and I is a 1-absorbing primary ideal of R .(2) Since N = ( N : M ) M , it follows by (1).(3) Putting I = ( N : M ) in (2), the claim is clear. (cid:3) The following example shows that if ( N : R M ) is a 1-absorbing primary ideal of R, then N is not needed to be a 1-absorbing primary submodule in general. Example 3.
Let R = Z and M = Z × Z be an R -module and p a prime integer.Consider the submodule N = p n Z × { } of M for n ≥ . Then ( N : R M ) = { } is a 1-absorbing primary ideal of R . However, N is not a 1-absorbing primarysubmodule of M since p · p n − · (1 , ∈ N but neither p · p n − = p n ∈ ( N : R M ) = { } nor (1 , ∈ M − rad ( N ) = p Z × { } . In view of Theorem 3, we conclude the following result. -ABSORBING PRIMARY SUBMODULES 5
Proposition 1.
Let M be a finitely generated multiplication R -module and N bea 1-absorbing primary submodule of M . Then the following are satisfied: (1) p ( N : R M ) is a prime ideal of R .(2) p ( N : R m ) is a prime ideal of R containing p ( N : R M ) = P for every m / ∈ M − rad ( N ).(3) M − rad ( N ) is a prime submodule of M . Proof. (1) Let N be a 1-absorbing primary submodule of M . Then ( N : R M ) is1-absorbing primary ideal of R by Theorem 3. From [7, Theorem 2], we concludethat p ( N : R M ) is a prime ideal of R .(2) Since N is a 1-absorbing primary ideal, p ( N : R M ) = P is a prime idealof R by (1). Suppose that a, b ∈ R such that ab ∈ p ( N : R m ). Without loss ofgenerality we may assume that a and b are non-unit elements of R . Then thereexists a positive integer n such that a n b n m ∈ N . Since N is 1-absorbing primarysubmodule, and m / ∈ M − rad ( N ), it implies that either ( ab ) n ∈ ( N : R M ). Since P is prime and ab ∈ P , we conclude either a ∈ P = p ( N : R M ) ⊆ p ( N : R m ) or b ∈ P = p ( N : R M ) ⊆ p ( N : R m ) . (3) Suppose thet N is a 1-absorbing primary submodule. Since p ( N : R M ) is aprime ideal of R by (1), we conclude that M − rad ( N ) = p ( N : R M ) M is a primesubmodule of M by [9, Corollary 2.11]. (cid:3) Note that the intersection of two distinct non-zero 1-absorbing primary submod-ules need not be a 1-absorbing primary submodule. Consider Z -module Z . Then2 Z and 3 Z are clearly 1-absorbing primary submodules but 2 Z ∩ Z = 6 Z is not. In-deed, 2 · · ∈ Z but neither 2 · ∈ ( Z : 6 Z ) = 6 Z nor 3 ∈ Z − rad (6 Z ) = 6 Z . We calla proper submodule N of M a P -1-absorbing submodule of M if p ( N : R M ) = P is a prime submodule of R. In the next theorem, we show that if N i ’s are P -1-absorbing primary submodules of a multiplication module M , then the intersectionof these submodules is a P -1-absorbing primary submodule of M . Proposition 2.
Let M be a multiplication R -module. If { N i } ki =1 is a family of P -1-absorbing primary submodules of M , then so is k \ i =1 N i .Proof. Suppose that abm ∈ k \ i =1 N i but ab / ∈ k \ i =1 N i : R M ! for non-unit elements a, b ∈ R and m ∈ M . Then ab / ∈ ( N j : R M ) for some j ∈ { , ..., k } . Since N j is 1-absorbing primary and abm ∈ N j , we have m ∈ M − rad ( N j ). Now, since M − rad k \ i =1 N i ! = k \ i =1 M − rad ( N i ) = P M by [14, Proposition 2.14 (3)], we aredone. (cid:3)
Lemma 2. [12]
Let ϕ : M −→ M be an R -module epimorphism. Then (1) If N is a submodule of M and ker ( ϕ ) ⊆ N , then ϕ ( M - rad ( N )) = M - rad ( ϕ ( N )) . (2) If K is a submodule of M , then ϕ − ( M - rad ( K )) = M - rad ( ϕ − ( K )) . Proposition 3.
Let M and M be R -modules and f : M → M be a modulehomomorphism. Then the following statements hold: ECE YETKIN CELIKEL (1) If N is a 1-absorbing primary submodule of M , then f − ( N ) is a 1-absorbing primary submodule of M .(2) Let f be an epimorphism. If N is a 1-absorbing primary submodule of M containing Ker ( f ), then f ( N ) is a 1-absorbing primary submodule of M . Proof. (1) Suppose that a, b are non-unit elements of R , m ∈ M and abm ∈ f − ( N ). Then abf ( m ) ∈ N . Since N is 1-absorbing primary, we have either ab ∈ ( N : R M ) or f ( m ) ∈ M − rad ( N ) . Here, we show that ( N : R M ) ⊆ ( f − ( N ) : R M ). Let r ∈ ( N : R M ). Then rM ⊆ N which implies that rf − ( M ) ⊆ f − ( N ); i.e. rM ⊆ f − ( N ) . Thus r ∈ ( f − ( N ) : R M ) . Hence ab ∈ ( f − ( N ) : R M ) or m ∈ f − ( M − rad ( N )) . Since f − ( M − rad ( N )) = M − rad ( f − ( N )) by Lemma 2 (2) and f − ( N ) is a 1-absorbing primary submoduleof M .(2) Suppose that a , b are non-unit elements of R , m ∈ M and abm ∈ f ( N ).Since f is an epimorphism, there exists m ∈ M such that f ( m ) = m . Since
Kerf ⊆ N , abm ∈ N . Hence ab ∈ ( N : R M ) or m ∈ M − rad ( N ). Here,we show that ( N : R M ) ⊆ ( f ( N ) : R M ). Let r ∈ ( N : R M ). Then rM ⊆ N which implies that rf ( M ) ⊆ f ( N ). Since f is onto, we conclude that rM ⊆ f ( N ) , that is, r ∈ ( f ( N ) : R M ) . Thus ab ∈ ( f ( N ) : R M ) or m = f ( m ) ∈ f ( M − rad ( N )) = M − rad ( f ( N )) by Lemma 2 (1) , as desired. (cid:3) As a consequence of Theorem 3, we have the following result.
Corollary 1.
Let M be an R -module and N , N be submodules of M with N ⊆ N . Then N is a 1-absorbing primary submodule of M if and only if N /N is a1-absorbing primary submodule of M/N .Proof. Suppose that N is a 1-absorbing primary submodule of M . Considerthe canonical epimorphism f : M → M/N in Proposition 3. Then N /N isa 1-absorbing primary submodule of M/N . Conversely, let a and b are non-unit elements of R , m ∈ M such that abm ∈ N . Hence ab ( m + N ) ∈ N /N .Since N /N is a 1-absorbing primary submodule of M/N , it implies either ab ∈ ( N /N : R M/N ) or m + N ∈ M/N − rad ( N /N ) = M − rad ( N ) /N . Therefore ab ∈ ( N : R M ) or m ∈ M − rad ( N ) . Thus N is a 1-absorbing primary submoduleof M . (cid:3) Let M be R -module and M be R -module where R and R are commutativerings with identity. Let R = R × R and M = M × M . Then M is an R -moduleand every submodule of M is of the form N = N × N for some submodules N , N of M , M , respectively. Also, M − rad ( N × N ) = M − rad ( N ) × M − rad ( N )by [2, Lemma 2.3 (ii)] . Proposition 4.
Let M be an R -module and M be an R -module, where R , R are commutative rings with identity, R = R × R and M = M × M . Supposethat N is a proper submodule of M . If N = N × M is a 1-absorbing primarysubmodule of R -module M , then N is a 1-absorbing primary submodule of R -module M . Proof.
Suppose that N = N × M is a 1-absorbing primary submodule of M .Put M ′ = M/ { } × M and N ′ = N/ { } × N . From Corollary 1, N ′ is a 1-absorbing primary submodule of M ′ . Since M ′ ∼ = M and N ′ ∼ = N , we concludethe result. (cid:3) -ABSORBING PRIMARY SUBMODULES 7 Proposition 5.
Let S be a multiplicatively closed subset of a commutative ring R and M be an R -module. If N is a 1-absorbing primary submodule of M and S − N = S − M, then S − N is a 1-absorbing primary submodule of S − R -module S − M. Proof.
Let as and bs be non-unit elements of S − R, ms ∈ S − M with as bs ms ∈ S − N . Hence tabm ∈ N for some t ∈ S . Since N is 1-absorbing primary, we haveeither tm ∈ M − rad ( N ) or ab ∈ ( N : R M ) . Thus we conclude either ms = tmts ∈ S − ( M − rad ( N )) ⊆ S − M − rad ( S − N ) or abs s ∈ S − ( N : R M ) ⊆ ( S − N : S − R S − M ). (cid:3) Let R be a ring and M be an R -module. The idealization of M is denoted by R ( M ) = R (+) M is a commutative ring with identity with coordinate-wise additionand multiplication defined by ( a, m )( b, m ) = ( ab, am + bm ) . An ideal H is calledhomogeneous if H = I (+) N for some ideal I of R and some submodule N of M such that IM ⊆ N . Proposition 6.
Let M be an R -module and I (+) N be a homogeneous ideal of R ( M ) . If I (+) N is a 1-absorbing primary ideal of R ( M ) , then I is a 1-absorbingprimary ideal of R. Proof.
Suppose that a, b, c are non-unit elements of R such that abc ∈ I and c / ∈ √ I .Then ( a, M ) · ( b, M ) · ( c, M ) ∈ I (+) N. Note that p I (+) N = √ I (+) M by [10,Theorem 25.1 (5)]. Then ( c, M ) / ∈ p I (+) N . Since I (+) N is 1-absorbing primary,we conclude that ( a, M ) · ( b, M ) ∈ I (+) N. Thus ab ∈ I , we are done. (cid:3) The 1-absorbing primary avoidance theorem
In this section, we prove the 1-absorbing primary avoidance theorem. Through-out this section, let M be a finitely generated multiplication R -module and N,N , ..., N n be submodules of M . Recall from [11] that a covering N ⊆ N ∪ N ∪· · · ∪ N n is said to be efficient if no N k is superfluous. Also, N = N ∪ N ∪ · · · ∪ N n is an efficient union if none of the N k may be excluded. A covering of a submoduleby two submodules is never efficient. Theorem 4.
Let N ⊆ N ∪ N ∪ · · · ∪ N n be an efficient covering of submodules N , N , ..., N n of M where n > . If p ( N i : R M ) * p ( N j : R m ) for all m ∈ M \ M − rad ( N j ) whenever i = j , then no N i ( ≤ i ≤ n ) is a 1-absorbing primarysubmodule of M .Proof. Assume on the contrary that N k is a 1-absorbing primary submodule of M for some 1 ≤ k ≤ n . Since N ⊆ ∪ N i is an efficient covering, N ⊆ [ ( N i ∩ N )is also an efficient covering. From [11, Lemma 2.1], \ i = k N i ∩ N ⊆ N k ∩ N .Here, observe that p ( N i : R M ) is a proper ideal of R for all 1 ≤ i ≤ n . Also,from our assumption, there is a non-unit element a i ∈ p ( N i : R M ) \ p ( N k : R m )for all i = k and for all m ∈ M \ M − rad ( N k ). Then there is a positive in-teger n i such that a n i i ∈ ( N i : R M ) for each i = k . Put a = k − Y i =1 a i , b = ECE YETKIN CELIKEL n Y i = k +1 a i and n = max { n , ..., n k − , n k +1 , ..., n n } . Now, we show that a n b n m ∈ \ i = k N i ∩ N \ ( N k ∩ N ) . Suppose that a n b n m ∈ N k ∩ N . Then a n b n ∈ ( N k : R m ) ⊆ p ( N k : R m ) . By Theorem 1 (2), p ( N k : R m ) is a prime ideal. Itimplies that a ∈ p ( N k : R m ) or b ∈ p ( N k : R m ). Thus a i ∈ p ( N k : R m ) for some i = k , a contradiction. Therefore a n b n m ∈ \ i = k N i ∩ N \ ( N k ∩ N ) which isa contradiction. Thus N k is not a 1-absorbing primary submodule. (cid:3) Theorem 5. (1-absorbing Primary Avoidance Theorem) Let
N, N , N , ..., N n ( n ≥ be submodules of M such that at most two of N , N , ..., N n are not 1-absorbing primary with N ⊆ N ∪ N ∪ · · · ∪ N n . If p ( N i : R M ) * p ( N j : R m ) forall m ∈ M \ M − rad ( N j ) whenever i = j , then N ⊆ N k for some ≤ k ≤ n. Proof.
Since it is clear for n ≤ , suppose that n > . Since any cover consistingsubmodules of M can be reduced to an efficient one by deleting any unnecessaryterms, we may assume that N ⊆ N ∪ N ∪ · · · ∪ N n is an efficient covering ofsubmodules of M. From Theorem 4, it implies that no N k is a 1-absorbing primarysubmodule which contradicts with the hypothesis. Thus N ⊆ N k for some 1 ≤ k ≤ n. (cid:3) Corollary 2.
Let N be a proper submodule of M . If 1-absorbing primary avoidancetheorem holds for M , then the 1-absorbing primary avoidance theorem holds for M/N.
Proof.
Let
K/N, N /N, N /N, ..., N n /N ( n ≥
2) be submodules of
M/N such thatat most two of N /N, N /N, ..., N n /N are not 1-absorbing primary and K/N ⊆ N /N ∪ N /N ∪ · · · ∪ N n /N. Hence, K ⊆ N ∪ N ∪ · · · ∪ N n and at most twoof N , N , ..., N n are not 1-absorbing primary by Corollary 1. Suppose that p ( N i /N : R M/N ) * p ( N j /N : R m + N ) for all m + N ∈ ( M/N ) \ ( M − rad ( N j /N ))whenever i = j . It is easy to verify that if p ( N i : R M ) ⊆ p ( N j : R m ) for some m ∈ M, then p ( N i /N : R M/N ) ⊆ p ( N j /N : R m + N ) for some m + N ∈ M/N.
Alsoobserve that if m + N ∈ ( M/N ) \ ( M/N − rad ( N j /N )) = ( M/N ) \ ( M − rad ( N j ) /N ),then m ∈ M \ M − rad ( N j ) . Thus, from our assumption p ( N i /N : R M/N ) * p ( N j /N : R m + N ) for all m + N ∈ ( M/N ) \ ( M/N − rad ( N j /N )) whenever i = j ,we conclude that p ( N i : R M ) * p ( N j : R m ) for all m ∈ M \ M − rad ( N j ) whenever i = j. From our hypothesis and Theorem 5, we have K ⊆ N k for some 1 ≤ k ≤ n. .Consequently, K/N ⊆ N k /N for some 1 ≤ k ≤ n ; so we are done. (cid:3) In view of Theorem 4 and Theorem 5, we conclude 1-absorbing primary avoidancetheorem for rings.
Corollary 3.
Let I ⊆ I ∪ I ∪ · · · ∪ I n be an efficient covering of ideals I , I , ..., I n of a ring R where n > . If √ I i * p ( I j : x ) for all x ∈ R \ p I j whenever i = j ,then no I i ( ≤ i ≤ n ) is a 1-absorbing primary ideal of R . Corollary 4. (1-absorbing Primary Avoidance Theorem for Rings) Let
I, I , I , ..., I n ( n ≥ be ideals of a ring R such that at most two of I , I , ..., I n are not 1-absorbing -ABSORBING PRIMARY SUBMODULES 9 primary with I ⊆ I ∪ I ∪ · · · ∪ I n . If √ I i * p ( I j : x ) for all x ∈ R \ p I j whenever i = j , then I ⊆ I k for some ≤ k ≤ n. References [1] R. Ameri, On the prime submodules of multiplication modules. International journal of Math-ematics and mathematical Sciences, (2003). DOI: 10.1155/S0161171203202180.[2] M. Alkan, Y. Tira¸s, On prime submodules. The Rocky Mountain Journal of Mathematics, (3) (2007) 709-722.[3] S.E. Atani, F. Callıalp , U. Tekir , A Short Note on the Primary Submodules of MultiplicationModules, International Journal of Algebra, (1) (2007), 381-384.[4] A. Badawi, On 2-absorbing ideals of commutative rings, Bull. Austral. Math. Soc., (2007),417–429.[5] A. Badawi, n -absorbing ideals of commutative rings and recent progress on three conjectures:a survey, Rings, Polynomials, and Modules, (2017), 33-52. DOI: 10.1007/978-3-319-65874-2 3[6] A. Badawi, ¨U. Tekir and E. Yetkin, On 2-absorbing primary ideals in commutative rings,Bull. Korean Math. Soc., (4) (2014), 1163–1173.[7] A. Badawi, E. Yetkin Celikel, On 1-absorbing primary ideals of commutative rings, Journalof Algebra and Applications, (6) (2020), 2050111. DOI: 10.1142/S021949882050111X.[8] A. Yousefian Darani and F. Soheilnia, On 2-absorbing and weakly 2-absorbing submodules,Thai J. Math., (2011), 577-584[9] Z. A. El-Bast and P.F. Smith, Multiplication modules. Comm. in Algebra, (1988), 755-799.[10] J. Huckaba, Rings with zero-divisors, Marcel Dekker, NewYork/ Basil,1988.[11] C. P. Lu, Unions of prime submodules. Houston J. Math, (2) (1997), 203-213.[12] C. P. Lu, M-radicals of submodules in modules, Math. Japonica, 34 (2) (1989), 211-219.[13] R. L. McCasland and M. E. Moore, On radicals of submodules of finitely generated modules,Canad. Math. Bull., (1986), 37–39.[14] H. Mostafanasab, E. Yetkin, U. Tekir, A. Y. Darani, On 2-absorbing primary submodules ofmodules over comutative rings, An. S t. Univ. Ovidius Constanta, (1) (2016), 335-351.[15] S. Payrovi, B. Babaei, On 2-absorbing submodules, Algebra Colloquium, (1) (2012), 913-920.[16] P. F. Smith, Some remarks on multiplication modules, Archiv der Mathematik (3) (1988),223-235. Department of Electrical-Electronics Engineering, Faculty of Engineering, HasanKalyoncu University, Gaziantep, Turkey
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