Int. J. Anal. Appl. (2023), 21:48 On Weakly S-2-Absorbing Submodules Govindarajulu Narayanan Sudharshana∗ Department of Mathematics, Annamalai university, Chidambaram 608001, Tamil Nadu, India ∗Corresponding author: sudharshanasss3@gmail.com Abstract. Let R be a commutative ring with identity and let M be a unitary R-module. In this paper, we introduce the notion of weakly S-2-absorbing submodule. Suppose that S is a multiplicatively closed subset of R. A submodule P of M with (P :R M)∩S = ∅ is said to be a weakly S-2-absorbing submodule if there exists an element s ∈S such that whenever a, b ∈R and m ∈M with 0 6= abm ∈P , then sab ∈ (P :M) or sam ∈P or sbm ∈P . We give the characterizations, properties and examples of weakly S-2-absorbing submodules. 1. Introduction Throughout this paper, R denotes a commutative ring with non zero identity and M is an R module. Prime ideals and submodules have vital role in ring and module theory. Of course a proper submodule P of M is called prime if am ∈ P for a ∈ R and m ∈ M implies a ∈ (P :R M) or m ∈ P where (P :R M) = {r ∈ R : rM ⊆ P}. Several generalizations of these concepts have been studied exten- sively by many authors [9], [13], [6], [16], [3], [11], [14], [5]. In 2007, Atani and Farzalipour introduced the concept of weakly prime submodules as a general- ization of prime submodules. A proper submodule P of M is defined as weakly prime if for a ∈ R and m ∈ M, whenever for 0 6= am ∈ P implies a ∈ (P :R M) or m ∈ P as in [5]. A new kind of generalization of prime submodule has been introduced and studied by Sengelen sevim et. al. in 2019 in [14]. For a multiplicatively closed subset S of R, that is, S satisfies the following conditions: (i) 1 ∈ S and (ii) s1s2 ∈ S for each s1, s2 ∈ S, a proper submodule P of an R-module M with (P :R M) ∩S = ∅ is called an S-prime submodule if there exists s ∈ S such that for a ∈ R and m ∈ M, if am ∈ P then either sa ∈ (P :R M) or sm ∈ P. In particular an ideal I of R is called Received: Apr. 5, 2022. 2010 Mathematics Subject Classification. 06F25. Key words and phrases. weakly S-prime; S-2-absorbing submodule; weakly S-2-absorbing submodule. https://doi.org/10.28924/2291-8639-21-2023-48 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-48 2 Int. J. Anal. Appl. (2023), 21:48 as S-prime ideal if I is an S-prime submodule of an R-module R, [10]. After that, the concept of weakly S-prime submodule was introduced as a generalization of S-prime submodules in [11]. Here, for a multiplicatively closed subset S of R, they called a submodule P of an R-module M with (P :R M) ∩S = ∅ a weakly S-prime submodule if there exists s ∈ S such that for a ∈ R and m ∈ M, if 0 6= am ∈ P then either sa ∈ (P :R M) or sm ∈ P. In particular, a proper ideal I of R disjoint with S is said to be weakly S-prime if there exists s ∈ S such that for a, b ∈ R and 0 6= ab ∈ I then either sa ∈ I or sb ∈ I [3]. One of the important generalizations of prime submodule is the concept of 2-absorbing submod- ule. In 2011, Darani and Soheilnia [6] introduced the concepts of 2-absorbing and weakly 2-absorbing submodules of modules over commutative rings with identities. A proper submodule P of a module M over a commutative ring R with identity is said be a 2-absorbing submodule (weakly 2-absorbing submodule) of M if whenever a, b ∈ R and m ∈ M with abm ∈ P (0 6= abm ∈ P ), then abM ⊆ P or am ∈ P or bm ∈ P . Predictably, a proper ideal I of R is 2-absorbing ideal if and only if I is a 2-absorbing submodule of R-module R. Recently, the concept of S-2-absorbing submodules was introduced in [16] which is a generalization of S-prime submodules and 2-absorbing submodules. A submodule P of M is said to be an S-2- absorbing submodule if (P :R M) ∩S = ∅ and there exists a fixed s ∈ S such that for a, b ∈ R and m ∈ M, if abm ∈ P then either sab ∈ (P :R M) or sam ∈ P or sbm ∈ P. In particular, an ideal I of R is an S-2-absorbing ideal if I is an S-2-absorbing submodule of R-module R. Our objective in this paper is to define and study the concept of weakly S-2-absorbing submodule as an extension of the above concepts. A submodule P of M is said to be a weakly S-2-absorbing submodule if (P :R M)∩S = ∅ and there exists an element s ∈ S such that for a, b ∈ R and m ∈ M, if 0 6= abm ∈ P then either sab ∈ (P :R M) or sam ∈ P or sbm ∈ P. In this case, we say that P is associated to s. In particular, an ideal I of R is a weakly S-2-absorbing ideal if I is a weakly S-2-absorbing submodule of R-module R. Some characterizations of weakly S-2-absorbing submodules are obtained. Besides, we investigate relationships between S-2-absorbing submodule and weakly S-2-absorbing submodule and also between weakly S-prime and weakly S-2-absorbing submodules of modules over commutative rings. 2. Characterizations of weakly S-2-absorbing submodules We start with the definitions and relationships of the main concepts of the paper. Definition 2.1. Let S be a multiplicatively closed subset of R. A submodule P of an R-module M is called a weakly S-2-absorbing submodule if (P :R M) ∩S = ∅ and there exists an element s ∈ S such that, whenever a, b ∈ R and m ∈ M, 0 6= abm ∈ P implies sab ∈ (P : M) or sam ∈ P or sbm ∈ P. In this case, we say that P is associated to s. In particular, an ideal I of R is a weakly S-2-absorbing ideal if I is a weakly S-2-absorbing submodule of R-module R Int. J. Anal. Appl. (2023), 21:48 3 Example 2.1. Consider the Z-module M = Z ×Z6 and let P = 2Z× < 3̄ >. Then P is a weakly S-2-absorbing submodule of M where S = {2n : n ∈ N∪{0}}. Indeed, let (0, 0̄) 6= r1r2(r ′,m) ∈ P for r1,r2,r ′ ∈ Z and m ∈ Z6 such that 2r1r2 /∈ (P : M) = 6Z. Then r1r2m ∈< 3̄ > with r1, r2 /∈ 3Z and so m ∈< 3̄ > also r ′ ∈ 2Z. Thus, 2r1(r ′,m) ∈ P as needed. Example 2.2. Consider the submodule P =< 6 > of the Z-module Z and the multiplicatively closed subset S = {5n : n ∈N∪{0}}. Then P is a weakly S-2-absorbing submodule. It is clear that every S-2-absorbing submodule is a weakly S-2-absorbing submodule. Since the zero submodule is (by definition) a weakly S-2-absorbing submodule of any R-module, hence the converse is not true in general and the following example shows this. Example 2.3. Consider R = Z, M = Z/30Z, P = 0 and S = Z −{0}. Then 2.3(5 + 30Z) = 0 ∈ P while 1.2.3 /∈ (P : M), 1.2(5 + 30Z) /∈ P and 1.3(5 + 30Z) /∈ P. Therefore P is not S-2-absorbing while it is weakly S-2-absorbing. Every weakly 2-absorbing submodule P of an R-module M satisfying (P : M) ∩S = ∅ is a weakly S-2-absorbing submodule of M and the two concepts coincide if S ⊆ U(R) where U(R) denotes the set of units in R. The following example shows that the converse need not be true. Example 2.4. Suppose that M = Z ×Z is an R = Z ×Z-module and P = pZ ×{0} is a submodule of M where p is prime. Then P is weakly S-2-absorbing submodule of M where S = Z −{0}×{0}. Indeed, let (0, 0) 6= (r1, r2)(r3, r4)(m1,m2) ∈ P for (r1, r2), (r3, r4) ∈ Z ×Z and (m1,m2) ∈ M such that s(r1, r2)(r3, r4) /∈ (P : M) = 0. Then either r1 or r3 or m1 must be p and either r2 or r4 or m2 must be 0. Thus s(p,r2)(m1,m2) ∈ P as needed. On the other hand, P is not a weakly 2-absorbing submodule since (0, 0) 6= (p, 1)(1, 0)(1, 1) ∈ P but neither (p, 1)(1, 0) ∈ (P : M) nor (p, 1)(1, 1) ∈ P nor (1, 0)(1, 1) ∈ P. Hence P is not weakly 2-absorbing. Lemma 2.1. Let S be a multiplicatively closed subset of R and P be a submodule of M. If P is weakly S-prime, then there exists an element s ∈ S of P such that 0 6= abm ∈ P for all a, b ∈ R and m ∈ M implies sbM ⊆ P whenever sam /∈ P . Proof. Let a, b ∈ R and m ∈ M. Assume that 0 6= abm ∈ P. Then 0 6= b(am) ∈ P. Since P is weakly S-prime, there exists s ∈ S of P such that sb ∈ (P : M) or sam ∈ P. Hence if sam /∈ P, then we get sbM ⊆ P. Proposition 2.1. Let S be a multiplicatively closed subset of R and P be a submodule of M. If P is weakly S-prime, then it is weakly S-2-absorbing. Proof. Let a, b ∈ R and m ∈ M be such that 0 6= abm ∈ P . Since P is weakly S-prime, there exists s ∈ S of P such that sa ∈ (P : M) or sbm ∈ P. If sbm ∈ P , then we are done. Suppose 4 Int. J. Anal. Appl. (2023), 21:48 sbm /∈ P , then by Lemma2.1, we get saM ⊆ P and consequently sabM ⊆ P. Hence P is weakly S-2-absorbing. The converse of the previous proposition need not be true, is illustrated in the following example. Example 2.5. Suppose that M = Z × Z is an R = Z × Z-module and P = 2Z × {0} is a submodule of M. Then P is weakly S-2-absorbing where S = (2Z + 1) × {0}. Indeed, let (0, 0) 6= (r1, r2)(r3, r4)(m1,m2) ∈ P for (r1, r2), (r3, r4) ∈ Z × Z and (m1,m2) ∈ M such that s(r1, r2)(r3, r4) /∈ (P : M) = 0. Then either r1 or r3 or m1 must be in 2Z. Without loss of generality, assume that r1 ∈ 2Z. Then s(r1, r2)(m1,m2) ∈ 2Z ×{0} as needed. On the other hand, we have (0, 0) 6= (2, 0)(1, 1) ∈ P. Now neither s(2, 0) ∈ (P : M) nor s(1, 1) ∈ P. Hence P is not weakly S-prime. Let R be a ring and S ⊆ R a multiplicatively closed subset of R. The saturation S∗ of S is defined as S∗={r ∈ R: r 1 is a unit of S−1R }. Note that S∗ is a multiplicatively closed subset containing S. Proposition 2.2. If M is an R-module and S is a mltiplicatively closed subset of R. Then the following statements hold. (i) Suppose that S1 ⊆ S2 are multiplicatively closed subsets of R. If P is a weakly S1-2-absorbing submodule and (P : M) ∩S2 = ∅, then P is a weakly S2-2-absorbing submodule. (ii) A submodule P of M is a weakly S-2-absorbing submodule if and only if it is a weakly S∗-2- absorbing submodule. (iii) If P is a weakly S-2-absorbing submodule of M, then S−1P is a weakly 2-absorbing submodule of S−1M. Proof. (i): It is clear. (ii):Let P be weakly S-2-absorbing. Suppose (P : M)∩S∗ 6= ∅. Then we have t ∈ (P : M)∩S∗ and this implies that t 1 .a s = 1 for some a ∈ R and s ∈ S as t 1 is a unit of S−1R. Thus ta = s ∈ S implies ta ∈ S and so (P : M) ∩S 6= ∅ which is a contradiction. Hence (P : M) ∩S∗ = ∅. By (i), P is a weakly S∗-2-absorbing submodule as S ⊆ S∗. Conversely, let a, b ∈ R and m ∈ M such that 0 6= abm ∈ P. Since P is weakly S∗-2-absorbing, there exists s” ∈ S∗ of P such that s”ab ∈ (P : M) or s”am ∈ P or s”bm ∈ P. Since s” ∈ S∗, we have s” 1 .t s = 1 for some t ∈ R, s ∈ S. Then s”t = s ∈ S and so s”t ∈ S. Then sab ∈ (P : M) or sam ∈ P or sbm ∈ P. Thus P is weakly S-2-absorbing. (iii) Let a s1 , b s2 ∈ S−1R and m s3 ∈ S−1M be such that 0M S 6= a s1 b s2 m s3 ∈ S−1P. Then we get 0M 6= sabm ∈ P for some s ∈ S. By assumption, there exists s4 ∈ S of P such that s4(sa)b ∈ (P : M) or s4(sa)m ∈ P or s4bm ∈ P. Then as1 b s2 = s4s s4s ab s1s2 ∈ S−1(P : M) ⊆ (S−1P : S−1M) or a s1 m s3 = s4s s4s am s1s3 ∈ S−1P or b s2 m s3 = s4 s4 bm s2s3 ∈ S−1P . Hence S−1P is weakly 2-absorbing submodule of S−1M. The converse of (iii) in the above proposition need not be true is shown by the following example. Int. J. Anal. Appl. (2023), 21:48 5 Example 2.6. Consider the Z-module M = Q3 and S = Z −{0}. Let P = {(r1, r2, 0) : r1, r2 ∈ Z}. Note that (P : M) = 0 and (P : M) ∩S = ∅. If a = 2, b = 3 and m = (1 2 , 1 3 , 0), then (0, 0, 0) 6= 2.3(1 2 , 1 3 , 0) = (3, 2, 0) ∈ P. If we take s = 5 ∈ S, then clearly 5.2.3 /∈ (P : M), 5.2(1 2 , 1 3 , 0) /∈ P, 5.3(1 2 , 1 3 , 0) /∈ P. Thus P is not weakly S-2-absorbing. From the fact that S−1M is a vectorspace over the field S−1Z that is Q and the proper subspace S−1P is 2-absorbing [16], we have S−1P is a weakly 2-absorbing submodule by [6]. Proposition 2.3. Let S be a multiplicatively closed subset of R and M be an R-module. Then the intersection of two weakly S-prime submodule is a weakly S-2-absorbing submodule. Proof. Let P1, P2 be two weakly S-prime submodules of M and P = P1 ∩ P2. Let a, b ∈ R and m ∈ M be such that 0 6= abm ∈ P. Since P1 is weakly S-prime and 0 6= a(bm) ∈ P1, there exists s1 ∈ S of P1 such that s1a ∈ (P1 : M) or s1bm ∈ P1. Again as P2 is weakly S-prime and 0 6= bam ∈ P2 there exists s2 ∈ S of P2 such that s2b ∈ (P2 : M) or s2am ∈ P2. Now consider the following four cases. Case 1: s1a ∈ (P1 : M) and s1bm /∈ P1 s2b ∈ (P2 : M) and s2am /∈ P2. Now, put s = s1s2 ∈ S. Then sab ∈ (P1 : M) and sab ∈ (P2 : M) and so sabM ⊆ P1 ∩ P2 = P. Hence sab ∈ (P : M). Case 2: s1a ∈ (P1 : M) and s1bm /∈ P1 s2am ∈ P2 and s2b /∈ (P2 : M). Then s1am ∈ s1aM ⊆ P1 and s2am ∈ P2 implies that sam ∈ P where s = s1s2 ∈ S. Case 3: s1bm ∈ P1 and s1a /∈ (P1 : M) s2am /∈ P2 and s2b ∈ (P2 : M) Then clearly sbm ∈ P where s = s1s2 ∈ S. Case 4: s1bm ∈ P1 and s1a /∈ (P1 : M) s2am ∈ P2 and s2b /∈ (P2 : M) As P1 is weakly S-prime and 0 6= abm ∈ P1 and also s1am /∈ P1 gives that s1bM ⊆ P1 by Lemma 2.1. For the same reason, we get s2aM ⊆ P2. Then clearly sab ∈ (P : M) where s = s1s2 ∈ S. Hence P is weakly S-2-absorbing. The following result provides some condition under which a weakly S-2-absorbing submodule is S-2- absorbing. Theorem 2.1. Let S be a multiplicatively closed subset of R and P be a weakly S-2-absorbing submodule of M. If P is not S-2-absorbing, then (P : M)2P = 0. Proof. By our assmption, there exists s ∈ S of P such that, whenever x, y ∈ R and m ∈ M, 0 6= xym ∈ P implies sxy ∈ (P : M) or sxm ∈ P or sym ∈ P. Suppose (P : M)2P 6= 0, we claim that P is S-2-absorbing. Let a, b ∈ R and m ∈ M be such that abm ∈ P. If abm 6= 0, then 6 Int. J. Anal. Appl. (2023), 21:48 sab ∈ (P : M) or sam ∈ P or sbm ∈ P. So assume that abm = 0. Now, first we assume that abP 6= 0. Then abp0 6= 0 for some p0 ∈ P implies 0 6= abp0 = ab(m + p0) ∈ P. Then sab ∈ (P : M) or sa(m + p0) ∈ P or sb(m + p0) ∈ P by our assumption. Hence sab ∈ (P : M) or sam ∈ P or sbm ∈ P. Hence we may assume that abP = 0. If a(P : M)m 6= 0, then aq0m 6= 0 for some q0 ∈ (P : M). Then 0 6= aq0m = a(b+q0)m ∈ P. Then, we get sa(b+q0) ∈ (P : M) or sam ∈ P or s(b+q0)m ∈ P. Hence sab ∈ (P : M) or sam ∈ P or sbm ∈ P. So we can assume that a(P : M)m = 0. In the same manner, we can assume that b(P : M)m = 0. Since (P : M)2P 6= 0, there exists x0, y0 ∈ (P : M) and m0 ∈ P with x0y0m0 6= 0. If ay0m0 6= 0, then 0 6= ay0m0 = a(b+y0)(m+m0) ∈ P since abm = 0, abm0 ∈ abP = 0 and ay0m = amy0 ∈ am(P : M) = 0. Hence, by our assumption sa(b + y0) ∈ (P : M) or sa(m + m0) ∈ P or s(b + y0)(m + m0) ∈ P and so sab ∈ (P : M) or sam ∈ P or sbm ∈ P. So we can assume that ay0m0 = 0. In the same manner, we can assume that x0y0m = 0 and x0bm0 = 0. Since x0y0m0 6= 0, we have 0 6= x0y0m0 = (a + x0)(b + y0)(m + m0) ∈ P since abm = 0, abm0 ∈ abP = 0 and ay0m = amy0 ∈ am(P : M) = 0. Then, s(a + x0)(b + y0) ∈ (P : M) or s(a + x0)(m + m0) ∈ P or s(b + y0)(m + m0) ∈ P. Hence sab ∈ (P : M) or sam ∈ P or sbm ∈ P. Hence P is S-2-absorbing. Recall that an R-module M is said to be a multiplication module if for each submodule N of M, N = IM for some ideal I of R. If N1, N2 are two submodules of M, then N1 = AM and N2 = BM for some ideals A, B of R. The product of N1 and N2 is defined as N1N2 = ABM [4]. Also note that this product is independent of the presentations of submodules N1 and N2 of M [4, Theorem 3.4]. A submodule N of an R-module M is called a nilpotent submodule if (N : M)kN = 0 for some positive integer k [1]. Corollary 2.1. Let S be a multiplicatively closed subset of R and P be a submodule of M. Assume that P is a weakly S-2-absorbing submodule of M that is not S-2-absorbing, then 1, P is nilpotent. 2, If M is a multiplication module, then P3 = 0. Proof. 1. Immediate from the definition of nilpotent submodule and by Theorem 2.1. 2. By Theorem 2.1, (P : M)2P = 0. Then (P : M)3M = (P : M)2(P : M)M = 0. Thus P3 = 0. If N is a proper submodule of a non-zero R-module M. Then the M-radical of N, denoted by M-radN is defined as the intersection of all prime submodules of M containing N [12], [8]. If A is an ideal of the ring R then the M-radical of A (considered as a submodule of the R-module R) is denoted by √ A and consists of all elements r of R such that rn ∈ A for some positive integer n [8]. Also it is shown in [8, Theorem 2.12] that if N is a proper submodule of a multiplication R-module M, then M-radN = ( √ (N : M))M. Int. J. Anal. Appl. (2023), 21:48 7 Proposition 2.4. Assume that M is a faithful multiplication R-module, S is a multiplicatively closed subset of R and P is a submodule of M. Let P be a weakly S-2-absorbing submodule of M. If P is not S-2-absorbing, then P ⊆ M-rad0. Proof. Suppose P is not S-2-absorbing. By Theorem 2.1, (P : M)2P = 0. Since (P : M)2(P : M)M ⊆ (P : M)2P, we have (P : M)3 ⊆ ((P : M)2P : M) = (0 : M) = 0. Let a ∈ (P : M), then a3 = 0 and so a ∈ √ 0. Thus (P : M) ⊆ √ 0. Hence P = (P : M)M ⊆ √ 0M = M-rad0. Proposition 2.5. If S is a multiplicatively closed subset of R and P is a submodule of a cyclic faithful R-module M, then P is a weakly S-2-absorbing submodule of M if and only if (P : M) is a weakly S-2-absorbing ideal of R. Proof. Let P be a weakly S-2-absorbing submodule of M. Assume that M = Rm for some m ∈ M and let 0 6= abc ∈ (P : M) for some a, b, c ∈ R. Then abcm ∈ P. If abcm 6= 0, then their exists an element s ∈ S of P such that sab ∈ (P : M) or sacm ∈ P or sbcm ∈ P. If sab ∈ (P : M), then we are done. If sacm ∈ P, then sac ∈ (P : m) = (P : M) as M is cyclic. Likewise, if sbcm ∈ P, then sbc ∈ (P : M). Then, assume that abcm = 0, we get abc ∈ (0 : m) = (0 : M). As M is faithful, we have abc = 0, a contradiction. Hence (P : M) is a weakly S-2-absorbing ideal of R. Conversely, let 0 6= abm′ ∈ P for some a, b ∈ R and m′ ∈ M. Then m′ = cm for some c ∈ R and we get 0 6= abcm ∈ P. This implies abc ∈ (P : m) = (P : M). If abc 6= 0, then there exists an element s′ ∈ S of (P : M) such that s′ab ∈ (P : M) or s′bc ∈ (P : M) or s′ac ∈ (P : M). If s′ab ∈ (P : M), then we are done. If s′bc ∈ (P : M), then s′bc ∈ (P : m) and so s′bm′ ∈ P. Likewise if s′ac ∈ (P : M), then s′am′ ∈ P . Now, assume that abc = 0, then abcm = 0.m = 0, a contradiction. Hence P is weakly S-2-absorbing. Proposition 2.6. If S is a multiplicatively closed subset of R and P is a submodule of a cyclic R- module M, then P is an S-2-absorbing submodule of M if and only if (P : M) is an S-2-absorbing ideal of R. After recalling the concepts of triple-zero in various papers like [9], [7], we give the following result which is an analogue of [9, Theorem 3.10]. Theorem 2.2. Let S be a multiplicatively closed subset of R and let P be a weakly S-2-absorbing submodule of M. If a, b ∈ R, m ∈ M with abm = 0 and sab /∈ (P : M), sam /∈ P , sbm /∈ P for any s ∈ S, then (1) abP = a(P : M)m = b(P : M)m = 0 (2) a(P : M)P = b(P : M)P = (P : M)2m = 0 Proof. (1). If abP 6= 0, then for some p ∈ P, abp 6= 0. Since 0 6= abp = ab(m + p) ∈ P, then by assumption there exists s ∈ S of P such that sab ∈ (P : M) or sa(m + p) ∈ P or sb(m + p) ∈ P. 8 Int. J. Anal. Appl. (2023), 21:48 Hence sab ∈ (P : M) or sam ∈ P or sbm ∈ P , which is not possible by our assumption. Hence abP = 0. If a(P : M)m 6= 0, then for some r ∈ (P : M), arm 6= 0. Since 0 6= arm = a(r + b)m ∈ P, then there exists s ∈ S of P such that sa(r + b) ∈ (P : M) or sam ∈ P or s(r + b)m ∈ P. That is sab ∈ (P : M) or sam ∈ P or sbm ∈ P, which is not possible by our assumption. Thus a(P : M)m = 0. The similar argument prove that b(P : M)m = 0. (2). Assume that a(P : M)P 6= 0. Then for some r ∈ (P : M), p ∈ P, 0 6= arp ∈ P. As 0 6= arp = a(b + r)(m + p). By (1), we get 0 6= a(b + r)(m + p) ∈ P, then there exists s ∈ S of P such that sa(b + r) ∈ (P : M) or sa(m + p) ∈ P or s(b + r)(m + p) ∈ P. Hence sab ∈ (P : M) or sam ∈ P or sbm ∈ P, a contradiction by our assumption. Hence a(P : M)P = 0. Now, if (P : M)2m 6= 0, then for some r1, r2 ∈ (P : M), 0 6= r1r2m ∈ P. Since by (1), 0 6= r1r2m = (a+r1)(b+r2)m ∈ P, then there exists s ∈ S of P such that s(a+r1)(b+r2) ∈ (P : M) or s(a + r1)m ∈ P or s(b + r2)m ∈ P and so sab ∈ (P : M) or sam ∈ P or sbm ∈ P, a contradiction by our assumption. Hence (P : M)2m = 0. We recall that if N is a submodule of an R-module M and A is an ideal of R, then the residual of N by A is the set (N :M A) = {m ∈ M : Am ⊆ N}. It is clear that (N :M A) is a submodule of M containing N. More generally, for any subset B ⊆ R, (N :M B) is a submodule of M containing N. Proposition 2.7. Let S be a multiplicatively closed subset of R. For a submodule P of an R-module M with (P : M) ∩S = ∅, the following assertions are equivalent. (1) P is a weakly S-2-absorbing submodule of M. (2) For any a, b ∈ R, there exists s ∈ S such that, if sabM * P, then (P : ab) = (0 : ab) or (P : ab) ⊆ (P : sa) or (P : ab) ⊆ (P : sb) (3) For any a, b ∈ R and for any submodule K of M, there exists s ∈ S such that, if 0 6= abK ⊆ P then sab ∈ (P : M) or saK ⊆ P or sbK ⊆ P. Proof. (1) =⇒ (2) Let a, b ∈ R. Let m ∈ (P : ab). If abm = 0, then clearly m ∈ (0 : ab). If abm 6= 0, that is if 0 6= abm ∈ P, then by (1), there exist s ∈ S of P such that sab ∈ (P : M) or sam ∈ P or sbm ∈ P. Clearly, if sabM * P, we conclude that either sam ∈ P or sbm ∈ P. As (0 : ab) ⊆ (P : ab), we get (P : ab) = (0 : ab) or (P : ab) ⊆ (P : sa) or (P : ab) ⊆ (P : sb). (2) =⇒ (3) Let a, b ∈ R and K be a submodule of M such that 0 6= abK ⊆ P and, for the element s ∈ S of (2), we have to claim that sab ∈ (P : M) or saK ⊆ P or sbK ⊆ P. If sab ∈ (P : M), then there is nothing to prove. Suppose sab /∈ (P : M). As abK ⊆ P, we have K ⊆ (P : ab) and by (2), we have K ⊆ (0 : ab) or K ⊆ (P : sa) or K ⊆ (P : sb). If K ⊆ (0 : ab), then abK = 0, a contradiction. If K ⊆ (P : sa), then saK ⊆ P as required. (3) =⇒ (1) Let a, b ∈ R and m ∈ M with 0 6= abm ∈ P. Clearly ab < m >⊆ P . If Int. J. Anal. Appl. (2023), 21:48 9 ab < m > 6= 0, by (3), sab ∈ (P : M) or sam ∈ sa < m >⊆ P or sbm ∈ sb < m >⊆ P. If ab < m >= 0, then abm ∈ ab < m >= 0, a contradiction. Theorem 2.3. Let S be a multiplicatively closed subset of R and P be a submodule of an R-module M. If P is a weakly S-2-absorbing submodule of M. Then (1) There exists an s ∈ S such that for any a, b ∈ R, if abK ⊆ P and 0 6= 2abK for some submodule K of M, then sab ∈ (P : M) or saK ⊆ P or sbK ⊆ P . (2) There exists an s ∈ S such that for an ideal I of R and a submodule K of M, if aIK ⊆ P and 0 6= 4aIK, where a ∈ R, then saI ∈ (P : M) or saK ⊆ P or sIK ⊆ P . (3) There exists an s ∈ S such that for all ideals I, J of R and submodule K of M, if 0 6= IJK ⊆ P and 0 6= 8(IJ + (I + J)(P : M))(K + P ), then sIJ ⊆ (P : M) or sIK ⊆ P or sJK ⊆ P. In particular this holds if the group (M, +) has no elements of order 2. Proof. (1) By our assmption, there exists s ∈ S of P such that, whenever x, y ∈ R and m ∈ M, 0 6= xym ∈ P implies sxy ∈ (P : M) or sxm ∈ P or sym ∈ P. Let a, b ∈ R such that abK ⊆ P and 0 6= 2abK for some submodule K of M. Now, we will show that sab ∈ (P : M) or saK ⊆ P or sbK ⊆ P . Suppose sab /∈ (P : M). Then proving that saK ⊆ P or sbK ⊆ P is enough. Let k be an arbitrary element of K. As abk ∈ abK ⊆ P , if abk 6= 0, then sab ∈ (P : M) or sak ∈ P or sbk ∈ P. Thus we have k ∈ (P : sa) or k ∈ (P : sb) since sab /∈ (P : M). Hence saK ⊆ P or sbK ⊆ P. If abk = 0. Since 0 6= 2abK, for some k1 ∈ K, we get 0 6= 2abk1 and clearly 0 6= abk1 ∈ P. Then we get sak1 ∈ P or sbk1 ∈ P since sab /∈ (P : M). Put k2 = k + k1 and so 0 6= abk2 ∈ P. Then sak2 ∈ P or sbk2 ∈ P since sab /∈ (P : M). This leads to the following cases. Case 1: sak1 ∈ P and sbk1 ∈ P Since sak2 ∈ P or sbk2 ∈ P, we have sak ∈ P or sbk ∈ P . Thus saK ∈ P or sbK ∈ P. Case 2: sak1 ∈ P and sbk1 /∈ P Suppose sak /∈ P and sbk /∈ P. Then sak2 = sak1 + sak /∈ P and so sbk2 ∈ P. Hence sa(k2 + k1) /∈ P and similarly sb(k2 + k1) /∈ P. As P is weakly S-2-absorbing and sab /∈ (P : M), hence ab(k2 + k1) = 0. But ab(k2 + k1) = ab(k1 + k + k1) = 2abk1, a contradiction as 2abk1 6= 0. Thus sak ∈ P or sbk ∈ P and so saK ⊆ P or sbK ⊆ P. Case 3: sak1 /∈ P and sbk1 ∈ P The proof is same as that of Case 2. (2) By our assmption, there exists s ∈ S of P such that, whenever x, y ∈ R and m ∈ M, 0 6= xym ∈ P implies sxy ∈ (P : M) or sxm ∈ P or sym ∈ P. Let I be an ideal of R and K be a submodule of M such that aIK ⊆ P and 0 6= 4aIK, where a ∈ R. We have to prove that saI ∈ (P : M) or saK ⊆ P or sIK ⊆ P. Suppose saI * (P : M), for some i ∈ I we have sai /∈ (P : M). Let us first prove that there exists b ∈ I such that 0 6= 4abK and sab /∈ (P : M). Since 0 6= 4aIK, for some i′ ∈ I, 0 6= 4ai′K. Suppose sai′ /∈ (P : M) or 0 6= 4aiK, if we put b = i′, we get sab /∈ (P : M) and 0 6= 4abK and if we put b = i, we get 0 6= 4abK and 10 Int. J. Anal. Appl. (2023), 21:48 sab /∈ (P : M). From the above, clearly by putting b = i′ or b = i, we get the result. Hence assume that sai′ ∈ (P : M) and 4aiK = 0. Hence 0 6= 4a(i + i′)K ⊆ P and sa(i + i′) /∈ (P : M). Thus we find b ∈ I such that 0 6= 4abK and sab /∈ (P : M). As 0 6= 4abK, we get 0 6= 2abK and by (1), since abK ⊆ aIK ⊆ P and sab /∈ (P : M), we get saK ⊆ P or sbK ⊆ P. If saK ⊆ P, there we are done. Thus assume that saK * P and so sbK ⊆ P. Now to exhibit that saI ∈ (P : M) or sIK ⊆ P. Let i” ∈ I. If 2ai”K 6= 0, then by (1), sai” ∈ (P : M) or si”K ⊆ P since saK * P. Thus we get i” ∈ ((P : M) : sa) or i” ∈ (P : sK). Therefore I ⊆ ((P : M) : sa) or I ⊆ (P : sK). Then we are done. If 2ai”K = 0, then clearly 0 6= 2a(b + i”)K and a(b + i”)K ⊆ P, by (1) sa(b + i”) ∈ (P : M) or s(b + i”)K ⊆ P since saK * P, (b + i”) ∈ (P : sK) or (b + i”) ∈ ((P : M) : sa). (i): If (b + i”) ∈ (P : sK), then si”K ⊆ P as sbK ⊆ P . Hence i” ∈ (P : sK). (ii): Now assume (b + i”) ∈ ((P : M) : sa) and (b + i”) /∈ (P : sK). Consider 0 6= 4abK = 2a(b+i” +b)K and a(b+i” +b)K ⊆ P . By (1), sa(b+i” +b) ∈ (P : M) or s(b+i” +b)K ⊆ P since saK * P . As sab /∈ (P : M), we have sa(b + i” + b) /∈ (P : M). Then we have s(b + i” + b)K ⊆ P. Since (b + i”) /∈ (P : sK), we have s(b + i” + b)K * P. Therefore (b + i”) ∈ (P : sK). Since sbK ⊆ P, we have si”K ⊆ P and so i” ∈ (P : sK). Consequently I ⊆ ((P : M) : sa) or I ⊆ (P : sK) and hence as saI * (P : M), we get sIK ⊆ P . (3) Let I, J be the ideals of R and K be a submodule of M such that 0 6= IJK ⊆ P and 0 6= 8(IJ + (I + J)(P : M))(K + P ). Since 0 6= 8(IJ + (I + J)(P : M))(K + P ) = 8IJK + 8I(P : M)K + 8J(P : M)K + 8IJP + 8I(P : M)P + 8J(P : M)P . As a result, one of the types listed below has been satisfied. Type 1: 0 6= 8IJK. Then for some j ∈ J, 0 6= 8jIK and so 0 6= 4jIK. As jIK ⊆ P , by (2), there exists s ∈ S such that sjI ⊆ (P : M) or sIK ⊆ P or sjK ⊆ P. If sIK ⊆ P, then we are done and so assume that sIK * P that is sjI ⊆ (P : M) or sjK ⊆ P. We claim that sIJ ⊆ (P : M) or sJK ⊆ P. Let j′ ∈ J be an arbitrary element. If 0 6= 4j′IK, by (2), sj′I ⊆ (P : M) or sj′K ⊆ P since sIK * P .Then j′ ∈ ((P : M) : sI) or j′ ∈ (P : sK). Hence we get the result. Now let 4j′IK = 0. As 0 6= 4(j + j′)IK ⊆ P , by (2), s(j + j′)I ⊆ (P : M) or s(j + j′)K ⊆ P since sIK * P. Hence we get s(j + j′)I ⊆ (P : M) or s(j + j′)K ⊆ P. Thereby we get the four cases. Case 1: sjI ⊆ (P : M) and s(j + j′)I ⊆ (P : M). Hence we get sj′I ⊆ (P : M), that is sIJ ⊆ (P : M) Case 2: sjK ⊆ P and s(j + j′)K ⊆ P Hence we get sj′K ⊆ P, that is sJK ⊆ P Case 3: sjI ⊆ (P : M) and sjK * P. s(j + j′)K ⊆ P and s(j + j′)I * (P : M). This can be represented as j ∈ ((P : M) : sI) and j /∈ (P : sK), j + j′ ∈ (P : sK) and Int. J. Anal. Appl. (2023), 21:48 11 j + j′ /∈ ((P : M) : sI). Hence j + j′ + j /∈ ((P : M) : sI) and j + j′ + j /∈ (P : sK). Now consider 0 6= 8jIK = 4(j + j′ + j)IK and by (2), s(j + j′ + j)I ⊆ (P : M) or s(j + j′ + j)K ⊆ P since sIK * P. Hence we get j + j′ + j ∈ ((P : M) : sI) or j + j′ + j ∈ (P : sK) and this is not possible. Therefore, since j ∈ ((P : M) : sI) or j ∈ (P : sK) and j + j′ ∈ (P : sK) or j + j′ ∈ ((P : M) : sI), there must be any one of the following holds. (i) j ∈ (P : sK) and j + j′ ∈ (P : sK) and j + j′ /∈ ((P : M) : sI), then j′ ∈ (P : sK). (ii) j ∈ ((P : M) : sI) and j /∈ (P : sK) and j + j′ ∈ ((P : M) : sI), then j′ ∈ ((P : M) : sI). Case 4: s(j + j′)I ⊆ (P : M) and s(j + j′)K * P sjK ⊆ P and sjI * (P : M). Similar to the above case, we have j′ ∈ ((P : M) : sI) or j′ ∈ (P : sK). Thus sIJ ⊆ (P : M) or sJK ⊆ P. Type 2: If 0 6= 8IJP and 8IJK = 0, then 0 6= 8IJ(K + P ) ⊆ P and by Type 1, sIJ ⊆ (P : M) or sJ(K + P ) ⊆ P or sI(K + P ) ⊆ P and so sIJ ⊆ (P : M) or sJK ⊆ P or sIK ⊆ P . Type 3: If 0 6= 8J(P : M)K and 8IJK = 0, then 0 6= 8J(P : M)K = 8J(I + (P : M))K and so by Type 1, sJ(I + (P : M)) ⊆ (P : M) or sJK ⊆ P or s(I + (P : M))K ⊆ P . Hence sIJ ⊆ (P : M) or sJK ⊆ P or sIK ⊆ P. Likewise if 0 6= 8I(P : M)K, we get the result. Type 4: If 0 6= 8J(P : M)P and 8IJK = 8IJP = 8J(P : M)K = 8I(P : M)K = 0. Then 0 6= 8J(P : M)P = 8J(I + (P : M))(K + P ) and by Type 1, sJ(I + (P : M)) ⊆ (P : M) or sJ(K + P ) ⊆ P or s(I + (P : M))(K + P ) ⊆ P. Hence sIJ ⊆ (P : M) or sJK ⊆ P or sIK ⊆ P. Likewise if 0 6= 8I(P : M)P, we have the result. To prove the particular case, let (M, +) be a group having no subgroups of order 2. We have to show that 0 6= 8IJK. If this happens, We get the result by Type 1. Suppose 8IJK = 0. Let 0 6= a ∈ IJK. As 8a = 0, so the group (M, +) has a subgroup of order 2, 4 or 8, which is a contradiction. Corollary 2.2. Let S be a multiplicatively closed subset of R and I be a weakly S-2-absorbing ideal of R. (1) There exists s ∈ S such that for any a, b ∈ R and for any ideal A of R, if abA ⊆ I and 0 6= 2abA, then sab ∈ I or saA ⊆ I or sbA ⊆ I. (2) There exists s ∈ S such that for any a ∈ R, ideals A, B of R, if aAB ⊆ I and 0 6= 4aAB, then saA ⊆ I or saB ⊆ I or sAB ⊆ I. (3) There exists s ∈ S such that for any ideals A, B, C of R, if 0 6= ABC ⊆ I and 0 6= 8(AB(C + I) + AC(B + I) + BC(A + I) + AI(B + C) + BI(A + C) + CI(A + B) + I2(A + B + C)), then sAB ⊆ I or sBC ⊆ I or sAC ⊆ I. In particular, this happens when the group (R, +) has no elements of order 2. Proposition 2.8. Let φ: M1 → M2 be a module homomorphism where M1 and M2 are R-modules and S be a multiplicatively closed subset of R. Then the following holds. 1. If φ is a monomorphism and K is a weakly S-2-absorbing submodule of M2 with (φ−1(K) : 12 Int. J. Anal. Appl. (2023), 21:48 M1) ∩S = ∅, then φ−1(K) is a weakly S-2-absorbing submodule of M1. 2. If φ is an epimorphism and P is a weakly S-2-absorbing submodule of M1 containing kerφ, then φ(P ) is a weakly S-2-absorbing submodule of M2. Proof. 1. Let a, b ∈ R and m1 ∈ M1 be such that 0 6= abm1 ∈ φ−1(K). Then 0 6= φ(abm1) = abφ(m1) ∈ K as φ is a monomorphism. since K is weakly S-2-absorbing, there exists s ∈ S such that sab ∈ (K : M2) or saφ(m1) ∈ K or sbφ(m1) ∈ K. If sab ∈ (K : M2), then sab ∈ (φ−1(K) : M1) since (K : M2) ⊆ (φ−1(K) : M1) and if saφ(m1) ∈ K or sbφ(m1) ∈ K, we have φ(sam1) ∈ K implies sam1 ∈ φ−1(K) or φ(sbm1) ∈ K implies sbm1 ∈ φ−1(K). Hence φ−1(K) is a weakly S-2-absorbing submodule of M1. 2. First observe that (φ(P ) : M2) ∩ S = ∅. Indeed, assume that s′ ∈ (φ(P ) : M2) ∩ S. Then φ(s′M1) = s′φ(M1) = s′M2 ⊆ φ(P ) and so s′M1 ⊆ P as kerφ ⊆ P. This shows that s′ ∈ (P : M1) and so (P : M1) ∩ S 6= ∅, a contradiction occurs since P is a weakly S-2-absorbing submodule of M1. Now, let a, b ∈ R and m2 ∈ M2 be such that 0 6= abm2 ∈ φ(P ). As we can write m2 = φ(m1) for some m1 ∈ M1 and so 0 6= abm2 = ab(φ(m1)) = φ(abm1) ∈ φ(P ). Since kerφ ⊆ P, we have 0 6= abm1 ∈ P. Then there exists s ∈ S such that sab ∈ (P : M1) or sam1 ∈ P or sbm1 ∈ P . Consequently we get sab ∈ (φ(P ) : M2) or φ(sam1) = saφ(m1) = sam2 ∈ φ(P ) or φ(sbm1) = sbφ(m1) = sbm2 ∈ φ(P ). Hence φ(P ) is weakly S-2-absorbing submodule of M2. Corollary 2.3. Let S be a multiplicatively closed subset of R. P1 and P2 are two submodules of M with P2 ⊆ P1. 1. If K is a weakly S-2-absorbing submodule of M with (K : P1) ∩S = ∅, then K ∩P1 is a weakly S-2-absorbing submodule of P1. 2. If P1 is a weakly S-2-absorbing submodule of M, then P1/P2 is a weakly S-2-absorbing submodule of M/P2. 3. If P1/P2 is a weakly S-2-absorbing submodule of M/P2 and P2 is a weakly S-2-absorbing submodule of M, then P1 is a weakly S-2-absorbing submodule of M. Proof. 1. Consider the injection i : P1 → M defined by i(p1) = p1 for all p1 ∈ P1. We have to show that (i−1(K) : P1) ∩ S = ∅. Indeed, if s ∈ (i−1(K) : P1) ∩ S, then sP1 ⊆ i−1(K). As i−1(K) = K∩P1, we have sP1 ⊆ K∩P1 ⊆ K and so s ∈ (K : P1)∩S, a contradiction as K is weakly S-2-absorbing. Thus by Proposition 2.8(1), we conclude the result. 2. Consider the canonical epimorphism π : M → M/P2 defined by π(m) = m + P2. Then π(P1) = P1/P2 is a weakly S-2-absorbing submodule of M/P2 by Proposition 2.8(2). 3. Let a, b ∈ R and m ∈ M be such that 0 6= abm ∈ P1. Then ab(m + P2) = abm + P2 ∈ P1/P2. If ab(m + P2) 6= P2, then there exists s1 ∈ S of P1/P2 implies s1ab ∈ (P1/P2 : M/P2) or s1a(m + P2) ∈ P1/P2 or s1b(m + P2) ∈ P1/P2. Hence s1ab ∈ (P1 : M) or s1am ∈ P1 or s1bm ∈ P1. If abm ∈ P2, then by assumption, there exists s2 ∈ S of P2 such that s2ab ∈ (P2 : M) ⊆ (P1 : M) or Int. J. Anal. Appl. (2023), 21:48 13 s2am ∈ P2 ⊆ P1 or s2bm ∈ P2 ⊆ P1. It follows that P1 is a weakly S-2-absorbing submodule of M associated with s = s1s2 ∈ S. We need to recall the following Lemma for the next result. Lemma 2.2. [2] For an ideal Iof a ring R and a submodule N of a finitely generated faithful multi- plication R-module M, the following hold. 1. (IN :R M) = I(N :R M). 2. If I is finitely generated faithful multiplication, then (a) (IN :M I) = N. (b) Whenever N ⊆ IM, then (JN :M I) = J(N :M I) for any ideal J of R. Proposition 2.9. Let I be a finitely generated faithful multiplication ideal of a ring R, S be a mul- tiplicatively closed subset of R and P be a submodule of a finitely generated faithful multiplication cyclic R-module M. 1. If IP is a weakly S-2-absorbing submodule of M and (P : M) ∩S = ∅, then either I is a weakly S-2-absorbing ideal of R or P is a weakly S-2-absorbing submodule of M. 2. P is a weakly S-2-absorbing submodule of IM if and only if (P :M I) is a weakly S-2-absorbing submodule of M. Proof. (1) Suppose P = M, we get I = I(P :R M) = (IP :R M) by Lemma 2.2. Since IP is a weakly S-2-absorbing submodule of M, by Proposition 2.5, I is a weakly S-2-absorbing ideal of R. Now, suppose P is a proper submodule of M. By Lemma 2.2, (IP :M I) = P and so (P : M) = ((IP :M I) :R M) = (I(P :R M) :M I). Let a, b ∈ R and m ∈ M be such that 0 6= abm ∈ P. Since I is faithful, then (0 :M I) = AnnR(I)M = 0 [2], and so 0 6= abIm ⊆ IP . By Proposition 2.7, there exists s ∈ S of IP such that sab ∈ (IP : M) or saIm ⊆ IP or sbIm ⊆ IP . If sab ∈ (IP : M), then sab ∈ (P : M). If saIm ⊆ IP, then sam ∈ (IP : I) = P. Likewise if sbIm ⊆ IP, then sbm ∈ P. Hence P is a weakly S-2-absorbing submodule of M. (2) Suppose P is a weakly S-2-absorbing submodule of IM. Then (P :R IM) ∩ S = ((P :M I) :R M) ∩ S = ∅. Let a, b ∈ R and m ∈ M be such that 0 6= abm ∈ (P :M I). If abIm = 0, then abm ∈ (0 :M I) = AnnR(I)M = 0, a contradiction. Hence 0 6= abIm ⊆ P. By Proposition 2.7, there exists s ∈ S of P such that sab ∈ (P : IM) or saIm ⊆ P or sbIm ⊆ P. If sab ∈ (P :R IM), then sab ∈ ((P :M I) :R M). If saIm ⊆ P, then sam ∈ (P : I) and similarly if sbIm ⊆ P , we get sbm ∈ (P : I) as required. Conversely, suppose (P :M I) is a weakly S-2-absorbing submodule of M. Then clearly ((P :M I) :R M) ∩S = (P :R IM) ∩S = ∅. Let a, b ∈ R and x ∈ IM be such that 0 6= abx ∈ P. Clearly ab < x >⊆ P. Since x ∈ IM, by Lemma 2.2, ab(< x >:M I) = (ab < x >:M I) ⊆ (P :M I). If ab(< x >:M I) = 0, then since abx ∈ (abIx :M I) and Ix ⊆ IM, by Lemma 2.2, we have abx ∈ ab(Ix :M I) ⊆ ab(< x >:M I) = 0, a contradiction. So we have 0 6= ab(< x >:M I) ⊆ 14 Int. J. Anal. Appl. (2023), 21:48 (P :M I). By Proposition 2.7, there exists s′ ∈ S of (P :M I) such that s′ab ∈ ((P :M I) :R M) or s′a(< x >:M I) ⊆ (P :M I) or s′b(< x >:M I) ⊆ (P :M I). If s′ab ∈ ((P :M I) :R M), then s′ab ∈ (P :R IM). If s′a(< x >:M I) ⊆ (P :M I), then Is′a(< x >:M I) ⊆ P. Since s′ax ∈ s′a < x >= s′a(I < x >:M I) = s′aI(< x >:M I) ⊆ P by Lemma 2.2. Likewise if s′b(< x >:M I) ⊆ (P :M I), then s′bx ∈ P . Hence P is a weakly S-2-absorbing submodule of IM. Acknowledgment: The author thanks Prof. Dr. P. Dheena, Professor, Department of Mathematics, Annamalai University, for suggesting the problem and going through the proof. Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the publi- cation of this paper. References [1] M.M. Ali, Idempotent and Nilpotent Submodules of Multiplication Modules, Commun. Algebra. 36 (2008), 4620–4642. https://doi.org/10.1080/00927870802186805. [2] M.M. Ali, Residual Submodules of Multiplication Modules, Beitr. 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Koç, On S-2-Absorbing Submodules and Vn-Regular Modules, An. Univ. Ovidius Const. - Ser. Mat. 28 (2020), 239–257. https://doi.org/10.2478/auom-2020-0030. https://doi.org/10.1080/00927870802186805 https://doi.org/10.2478/auom-2021-0024 https://doi.org/10.1155/s0161171203202180 https://doi.org/10.5556/j.tkjm.38.2007.77 https://doi.org/10.5556/j.tkjm.38.2007.77 https://doi.org/10.4134/JKMS.J160544 https://doi.org/10.1080/00927878808823601 https://doi.org/10.1080/00927878808823601 https://doi.org/10.5666/KMJ.2021.61.1.33 https://doi.org/10.1007/s13366-019-00476-5 https://doi.org/10.1007/s13366-019-00476-5 https://doi.org/10.48550/ARXIV.2110.14639 https://doi.org/10.48550/ARXIV.2110.14639 https://doi.org/10.4153/CMB-1986-006-7 https://doi.org/10.1007/BF01187738 https://doi.org/10.1007/BF01187738 https://doi.org/10.2478/auom-2020-0030 1. Introduction 2. Characterizations of weakly S-2-absorbing submodules References