Abstract
In this study, we aim to introduce the concepts of 1-absorbing prime submodules and weakly 1-absorbing prime submodules of a unital module over a noncommutative ring with nonzero identity. This is a new class of submodules between prime submodules (weakly prime submodules) and 2-absorbing submodules (weakly 2-absorbing submodules). Let R be a noncommutative ring with a nonzero identity \(1\ne 0\) and M an R-module. A proper submdule P of M is said to be a 1-absorbing prime submodule (weakly 1-absorbing prime submodule) if for all nonunits \(x,y\in R\) and \(m\in M\) with xRyRm \(\subseteq \) P \((\left\{ 0\right\} \ne \) xRyRm \(\subseteq P)\), then \(xy\in (M:_{R}P)\) or \(m\in P\). Various properties and characterizations of these classes of submodules are considered.
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1 Introduction
In this article, we focus only on noncommutative rings with nonzero identity and nonzero unital left modules. Let R always denote such a ring and let M denote such an R-module. The concept of prime ideals and its generalizations have a significant place in noncommutative algebra since they are used in understanding the structure of rings. Recall that in a commutative ring a proper ideal I of R is said to be a prime ideal if whenever \(xy\in I\) then \(x\in I\) or \(y\in I.\) In [1], Anderson and Smith introduced a notion of weakly prime ideal which is a generalization of prime ideals. A proper ideal I of R is called weakly prime ideal if \(0\ne xy\in I\) for some elements \(x,y\in R\) implies that \(x\in I\) or \(y\in I.\) It is clear that every prime ideal is weakly prime but the converse is not true in general. Afterwards, Badawi, in his celebrated paper [2], introduced the notion of 2-absorbing ideals and used them to characterize Dedekind domains. Recall from [2], that a nonzero proper ideal I of R is called 2-absorbing ideal if \(xyz\in I\) for some \(x,y,z\in R\) implies either \(xy\in I\) or \(xz\in I\) or \(yz\in I\). Note that every prime ideal is also a 2-absorbing ideal. After this, over the past decades, 2-absorbing version of ideals and many generalizations of 2-absorbing ideals attracted considerable attention by many researchers. Badawi and Darani in [3] defined and studied the notion of weakly 2-absorbing ideals which is a generalization of weakly prime ideals. A proper ideal I of R is called a weakly 2-absorbing ideal if for each \(x,y,z\in R\) with \(0\ne xyz\in I\), then either \(xy\in I\) or \(xz\in I\) or \(yz\in I\).
In 2010, Hirano et al. extended the notion of weakly prime ideals in rings, not necessarily commutative or with identity. According their celebrated paper [11], a proper ideal P of R is called a weakly prime ideal of a ring R if whenever \(a,b\in R\) such that \(\left\{ 0\right\} \ne aRb\subseteq P,\) then \(a\in P\) or \(b\in P.\) They also verified that P is weakly prime ideal if and only if whenever J, K are right ideals of R such that \(\left\{ 0\right\} \ne JK\subseteq P\), then \(J\subseteq P\) or \(K\subseteq P.\) An ideal I of R is said to be proper if \(I\ne R\). Recall that a proper ideal I of R is called 2-absorbing as in [6] if whenever \(aRbRc\subseteq I\) for some \(a,b,c\in R\), then \(ab\in I\) or \(bc\in I\) or \(ac\in I\). Let I be a proper ideal of R. Recall from [7] that a proper ideal I of R is said to be a weakly 2-absorbing ideal of R if whenever \(a,b,c\in R\) with \(\left\{ 0\right\} \ne aRbRc\subseteq I\), then \(ab\in I\) or \(ac\in I\) or \(bc\in I\). Note that a 2-absorbing ideal is a weakly 2-absorbing ideal. However, these are different concepts.
In 2011, Darani and Soheilnia [5] introduced the concept 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 to be a 2-absorbing (weakly 2-absorbing) submodule of M if whenever \(a,b\in R\) and \(m\in M\) with \(abm\in P\) \((0\ne abm\in P)\), then \(abM\subseteq P\) or \(am\in P\) or \(bm\in P\). One can see that 2-absorbing submodules are generalization of prime submodules. Moreover, it is obvious that 2-absorbing ideals are special cases of 2-absorbing submodules.
In [8] and [9] the notions of 2-absorbing and weakly 2-absorbing submodules of a module over a noncommutative ring were introduced. A proper submodule P of a module M over a noncommutative ring R with identity is said to be a 2-absorbing (weakly 2-absorbing) submodule of M if whenever \(a,b\in R\) and \(m\in M\) with \(aRbRm\subseteq P\) \((\left\{ 0\right\} \ne aRbRm\subseteq P)\), then \(ab\in (P:M)\subseteq P\) or \(am\in P\) or \(bm\in P\).
Recently, in [16], Yassine et al. introduced a 1-absorbing prime ideal. This type of ideal which is a generalization of prime ideals of a commutative ring with identity. A proper ideal I of R is called 1–absorbing prime ideal if whenever \(xyz\in I\) for some nonunits \(x,y,z\in R,\) then either \(xy\in I\) or \(z\in I\). Note that every prime ideal is 1-absorbing prime and every 1-absorbing prime ideal is 2-absorbing. The converses are not true. More currently in [12] Koç et al. defined weakly 1-absorbing prime ideals which is a generalization of 1-absorbing prime ideal. A proper ideal I of R is called weakly 1-absorbing prime ideal if \(0\ne xyz\in I\) for some nonunits \(x,y,z\in R\) implies that \(xy\in I\) or \(z\in I\). Following Yassine et al. [15] and Koç [12] in [10] we introduced 1-absorbing prime ideals and weakly 1-absorbing prime ideals in noncommutative rings. For a noncommutative ring R, whenever \(xRyRz\subseteq I\) \((\left\{ 0\right\} \ne xRyRz\subseteq I)\) for some nonunits \(x,y,z\in R\), then \(xy\in I\) or \(z\in I,\) then I is a 1-absorbing prime ideal (weakly 1 absorbing prime ideal). In [14] Ugurlu introduced the concept of a 1-absorbing prime submodule of a unital module over a commutative ring with a non-zero identity. Also in [4] Celikel introduced the notion of 1-absorbing primary submodules of a unital module over a commutative ring with a non-zero identity. In this paper, after introducing the notion of 1-absorbing and weakly 1-absorbing prime submodules of a unital left module over a noncommutative ring with nonzero identity, we examine the properties of the new classes. We show that many of the results of Ugurlu in [14] for 1-absorbing prime submodules of a unital module over a commutative ring with a non-zero identity are also valid for 1-absorbing prime submodules of a unital left module over a noncommutative ring with nonzero identity. For all nonunit elements \(a,b\in R\) and \(m\in M\), if \(aRbRm\subseteq N\), \((\left\{ 0\right\} \ne aRbRm\subseteq N)\) either \(ab\in (N:_{R}M)\) or \(m\in N\), then N is called a 1-absorbing prime submodule (weakly 1-absorbing prime submodule) of M. Recall that a proper submodule N of the R-module M is a prime (weakly prime) submodule if \(aRm\subseteq N(\left\{ 0\right\} \ne aRm\subseteq N)\) for \(a\in R\) and \(m\in M\) then \(a\in (N:_{R}M)\) or \(m\in N.\)
Among many results in this paper, it is shown in Proposition 2.4 if N is a 1-absorbing prime submodule of M then \((N:_{R}M)\) is a 1-absorbing prime ideal of R. It is also proved in Corollary 2.7 that if M is an R-module and \(N_{1}\), \(N_{2}\) submodules of M with \(N_{2}\subseteq N_{1},\) then \(N_{1}\) is a 1-absorbing prime submodule of M if and only if \(N_{1}/N_{2}\) is a 1-absorbing prime submodule of \(M/N_{2}\). We also have the following charaterization of 1-absorbing prime submodules in Theorem 2.9. A proper submodule of an R-module M is a 1-absorbing prime submodule of M if \(I_{1}I_{2}K\subseteq N\) for some proper ideals \(I_{1},I_{2}\) of R and some submodule K of M, then either \(I_{1}I_{2}\subseteq (N:_{R}M)\) or \(K\subseteq N.\) If there exists a weakly 1-absorbing prime submodule N in the R-module M that is not a prime submodule, then we show in Theorem 2.10 that R is a local ring. If R is a local ring and N is a weakly 1-absorbing prime submodule that is not 1-absorbing prime, then we show in Proposition 3.10 that \((N:_{R}M)^{2}N=\left\{ 0\right\} ,\) and in particular, \((N:_{R}M)^{3}\subseteq \)Ann(M). If R is a local ring and M is a multiplication module and N is a weakly 1-absorbing prime submodule of M that is not a 1-absorbing prime submodule, then \(N^{3}=\left\{ 0\right\} \) (Proposition 3.11). It is shown in Theorem 3.14 that if N is a proper submodule of the R module M, then N is a weakly 1-absorbing prime submodule of M if for any proper ideals \(I_{1},I_{2}\) of R and a submodule K of M such that \(\left\{ 0\right\} \ne I_{1}I_{2}K\subseteq N\) and N is free triple-zero with respect to \(I_{1},I_{2},K\), we have either \(I_{1} I_{2}\subseteq \) \((N:_{R}M)\) or \(K\subseteq N.\)
We have the following diagram which clarifies the place of 1-absorbing prime submodules and weakly 1-absorbing prime submodules. Here, the arrows in the diagram are irreversible. \(\begin{array}{ccccc} \text {prime submodule} &{} \Rightarrow &{} 1\text {-absorbing prime} &{} \Rightarrow &{} 2\text {-absorbing}\\ \Downarrow &{} &{} \Downarrow &{} &{} \Downarrow \\ \text {weakly prime } &{} \Rightarrow &{} \text {weakly }1\text {-absorbing prime} &{} \Rightarrow &{} \text {weakly }2\text {-absorbing} \end{array}\)
2 1-absorbing prime submodules
Definition 2.1
Let M be an R-module and N be a proper submodule of M. For all nonunit elements \(a,b\in R\) and \(m\in M\) if \(aRbRm\subseteq N\) either \(ab\in (N:_{R}M)\) or \(m\in N\), then N is called 1-absorbing prime submodule of M.
Proposition 2.2
Prime submodules \(\Rightarrow \) 1-absorbing prime submodules \(\Rightarrow \) 2-absorbing submodules.
Proof
Let N be a prime submodule of M. Take nonunit elements \(a,b\in R\) and \(m\in M\) such that \(aRbRm\subseteq N\). Now \(abRm\subseteq N\) and since N is a prime submodule, \(ab\in (N:_{R}M)\) or \(m\in N\), as desired.
Suppose N is a 1-absorbing prime submodule of M. Take any \(a,b\in R\) and \(m\in M\) such that \(aRbRm\subseteq N\). If a and b are nonunits, we have \(ab\in (N:_{R}M)\) or \(m\in N\) and we are done. If a is a unit element, then \(abm\in N\) implies \(bm\in N.\) If b is a unit element, then there exists \(b^{\prime }\in R\) such that \(b^{\prime }b=1\) and we have \(am=ab^{\prime }bm\in N\), as desired. \(\square \)
Example 2.3
For field K the ring \(R=\left\{ \left[ \begin{array}{ccc} a &{}\quad 0 &{}\quad b\\ 0 &{}\quad a &{}\quad c\\ 0 &{}\quad 0 &{}\quad a \end{array} \right] :a,b,c\in K\right\} \) is a local ring whose unique maximal ideal M has square zero. Consider the R-module M. Then every proper submodule is a 1-absorbing prime submodule of M. To see this, choose nonunits \(x,y\in R\) and \(m\in M\) such that \(xRyRm\subseteq N\). Since \(xRy\subseteq M^{2}=\left\{ 0\right\} \), we have \(xy\in (N:_{R}M)\) which implies N is a 1-absorbing prime submodule of M.
Proposition 2.4
If N is a 1-absorbing prime submodule of M then we have the following:
-
1.
\((N:_{R}M)\) is a 1-absorbing prime ideal of R.
-
2.
(N : Rm) is a 1-absorbing prime ideal of R for every \(m\in M\backslash N\).
Proof
Let N be a 1-absorbing prime submodule of M.
1. Choose nonunits \(a,b,c\in R\) such that \(aRbRc\subseteq (N:_{R}M).\) For all \(m\in M,\) then \(aRbRcm\subseteq N\). By our hypothesis, \(ab\in (N:_{R}M)\) or \(cm\in N\). If \(ab\in (N:_{R}M),\) then we done. So suppose \(ab\notin (N:_{R}M).\) Hence \(cm\in N\) for all \(m\in M.\) This implies that \(c\in (N:_{R}M)\). Consequently, \((N:_{R}M)\) is 1-absorbing prime ideal of R.
2. Choose nonunits \(a,b,c\in R\) such that \(aRbRc\subseteq (N:Rm).\) Hence \(aRbRcRm\subseteq N\) and therefore \(aRbRcrm\subseteq N\) for all \(r\in R.\) By our hypothesis or \(ab\in (N:_{R}M)\) or \(crm\in N\) for all \(r\in R.\) Thus \(ab\in (N:_{R}M)\subseteq (N:Rm)\) or \(c\in (N:Rm).\) Consequently, (N : Rm) is 1-absorbing prime ideal of R. \(\square \)
The converse of the above proposition is not true in general.
Example 2.5
Let p be a fixed prime integer. Then \(\mathbb {Z}(p^{\infty })=\{a\in Q/\mathbb {Z}:a=r/p^{n}+\mathbb {Z}\) for some \(r\in \mathbb {Z}\) and \(n\ge 0\}\) is a nonzero submodule of \(Q/\mathbb {Z}\). Let \(G_{t}=\{a\in Q/\mathbb {Z}:a=r/p^{t}+\mathbb {Z}\) for some \(r\in \mathbb {Z}\}\) for all \(t\ge 0\). It is well known that each proper submodule of \(\mathbb {Z}(p^{\infty })\) is equal to \(G_{t}\) for some \(t\ge 0\). \(G_{t}\) is not a 1-absorbing prime submodule of \(\mathbb {Z}(p^{\infty })\) since for \(p^{2}(1/p^{t+2}+\mathbb {Z})\in G_{t}\) we have \((1/p^{t+2}+Z)\notin G_{t}\) and \(p^{2}\notin (G_{t}:_{\mathbb {Z} }\mathbb {Z}(p^{\infty }))=\left\{ 0\right\} \). We can see that \((G_{t}:_{\mathbb {Z}}\mathbb {Z}(p^{\infty }))=\left\{ 0\right\} \) is a 1-absorbing prime ideal of \(\mathbb {Z}\) for all \(t\ge 0\).
Note that from the above remark we have that some modules do not have any 1-absorbing prime submodules. Since each proper submodule of \(\mathbb {Z} (p^{\infty })\) is equal to \(G_{t}\) for some \(t\ge 0\), so \(\mathbb {Z}(p^{\infty })\) does not have any 1-absorbing prime submodule.
Proposition 2.6
Let \(M_{1}\) and \(M_{2}\) be R-modules and \(f:M_{1} \rightarrow M_{2}\) be a module homomorphism. Then the following statements hold:
-
1.
If \(N_{2}\) is a 1-absorbing prime submodule of \(M_{2}\), then \(f^{-1}(N_{2})\) is a 1-absorbing prime submodule of \(M_{1}\).
-
2.
Let f be an epimorphism. If \(N_{1}\) is a 1-absorbing prime submodule of \(M_{1}\) containing \(\ker (f)\), then \(f(N_{1})\) is a 1-absorbing prime submodule of \(M_{2}\).
Proof
1. Suppose that a, b are nonunit elements of R, \(m_{1}\in M_{1}\) and \(aRbRm_{1}\subseteq f^{-1}(N_{2})\). Then \(aRbRf(m_{1})\subseteq N_{2}.\) Since \(N_{2}\) is a 1-absorbing prime submodule, we have either \(ab\in (N_{2}:_{R}M_{2})\) or \(f(m_{1})\in N_{2}\). Here, we show that \((N_{2}:_{R} M_{2})\subseteq (f^{-1}(N_{2}):M_{1})\). Let \(r\in (N_{2}:M_{2})\). Then \(rM_{2}\subseteq N_{2}\) which implies that \(rf^{-1}(M_{2})\subseteq f^{-1}(N_{2})\), i.e., \(rM_{1}\subseteq f^{-1}(N_{2})\). Thus \(r\in (f^{-1} (N_{2}):M_{1})\). Hence \(ab\in \) \((f^{-1}(N_{2}):M_{1})\) or \(m_{1}\in f^{-1}(N_{2})\). Hence \(f^{-1}(N_{2})\) is a 1-absorbing prime submodule of \(M_{1}\).
2. Suppose that are nonunit elements a and b of R, \(m_{2}\in M_{2}\) and \(aRbRm_{2}\in f(N_{1})\). Since f is an epimorphism, there exists \(m_{1}\in M_{1}\) such that \(f(m_{1})=m_{2}\). Since \(\ker (f)\subseteq N_{1}\), \(aRbRm_{1}\subseteq N_{1}\). Hence \(ab\in (N_{1}:M_{1})\) or \(m_{1}\in N_{1}\). Here, we show that \((N_{1}:M_{1})\subseteq (f(N_{1}):M_{2})\). Let \(r\in (N_{1}:M_{1})\). Then \(rM_{1}\subseteq N_{1}\) which implies that \(rf(M_{1} )\subseteq f(N_{1})\). Since f is onto, we conclude that \(rM_{2}\subseteq f(N_{1})\), that is, \(r\in (f(N_{1}):M_{2})\). Thus \(ab\in (f(N_{1}):M_{2})\) or \(m_{2}\in f(N_{1})\), as desired \(\square \)
As a consequence of Proposition 2.6, we have the following result.
Corollary 2.7
Let M be an R-module and \(N_{1}\), \(N_{2}\) be submodules of M with \(N_{2}\subseteq N_{1}\). Then \(N_{1}\) is a 1-absorbing prime submodule of M if and only if \(N_{1}/N_{2}\) is a 1-absorbing prime submodule of \(M/N_{2}\).
Proof
Suppose that \(N_{1}\) is a 1-absorbing primary submodule of M. Consider the canonical epimorphism \(f:M\rightarrow M/N_{2}\) in Proposition 2.6. Then \(N_{1}/N_{2}\) is a 1-absorbing prime submodule of \(M/N_{2}\). Conversely, let a and b are nonunit elements of R, \(m\in M\) such that \(aRbRm\subseteq N_{1}\). Hence \(aRbR(m+N_{2})\subseteq N_{1}/N_{2}.\) Since \(N_{1}/N_{2}\) is a 1-absorbing prime submodule of \(M/N_{2}\), it implies either \(ab\in (N_{1}/N_{2}:_{R}M/N_{2})\) or \(m+N_{2}\in M/N_{2}\). Therefore \(ab\in (N_{1}:_{R}M)\) or \(m\in N_{1}\). Thus \(N_{1}\) is a 1-absorbing prime submodule of M. \(\square \)
Let \(M_{1}\) be \(R_{1}\)-module and \(M_{2}\) be \(R_{2}\)-module where \(R_{1}\) and \(R_{2}\) are noncommutative rings with identity. Let \(R=R_{1}\times R_{2}\) and \(M=M_{1}\times M_{2}\). Then M is an R-module and every submodule of M is of the form \(N=N_{1}\times N_{2}\) for some submodules \(N_{1},N_{2}\) of \(M_{1},M_{2}\), respectively.
Proposition 2.8
Let \(M_{1}\) be \(R_{1}\)-module and \(M_{2}\) be \(R_{2}\)-module where \(R_{1}\) and \(R_{2}\) are noncommutative rings with identity. Let \(R=R_{1}\times R_{2}\) and \(M=M_{1}\times M_{2}\). Suppose that \(N_{1}\) is a proper submodule of \(M_{1}\). If \(N=N_{1}\times M_{2}\) is a 1-absorbing prime submodule of the R-module M, then \(N_{1}\) is a 1-absorbing prime submodule of M.
Proof
Suppose that \(N=N_{1}\times M_{2}\) is a 1-absorbing prime submodule of M. Put \(M^{\prime }=M/(\{0\}\times M_{2})\) and \(N^{\prime }=N/(\{0\}\times N_{2})\). From Corollary 2.7, \(N^{\prime }\) is a 1-absorbing prime submodule of \(M^{\prime }\). Since \(M^{\prime }\cong M_{1}\) and \(N^{\prime }\cong N_{1}\), we conclude the result. \(\square \)
Next we give several characterizations of 1-absorbing prime submodules of an R-module.
Theorem 2.9
Let N be a proper submodule of an R-module M. Then the following statements are equivalent:
-
(1)
N is a 1-absorbing prime submodule of M.
-
(2)
If a, b are nonunit elements of R such that \(ab\notin (N:_{R}M)\), then \((N:_{M}aRbR)\subseteq N\).
-
(3)
If a, b are nonunit elements of R, and K is a submodule of M with \(aRbK\subseteq N\), then \(ab\in (N:_{R}M)\) or \(K\subseteq N.\)
-
(4)
If \(I_{1}I_{2}K\subseteq N\) for some proper ideals \(I_{1},I_{2}\) of R and some submodule K of M, then either \(I_{1}I_{2}\subseteq (N:_{R}M)\) or \(K\subseteq N.\)
Proof
\((1)\Rightarrow (2)\) Suppose that a, b are nonunit elements of R such that \(ab\notin (N:_{R}M)\). Let \(m\in (N:_{M}aRbR)\). Hence \(aRbRm\subseteq N\). Since N is 1-absorbing prime submodule and \(ab\notin (N:_{R}M)\), we have \(m\in N\), and so \((N:_{M}aRbR)\subseteq N\).
\((2)\Rightarrow (3)\) Suppose that \(ab\notin (N:_{R}M)\). Since \(aRbRK\subseteq aRbK\subseteq N\), we have \(K\subseteq (N:_{M}aRbR)\subseteq N\) by (2). \((3)\Rightarrow (4)\) Suppose \(I_{1}I_{2}K\subseteq N\) for some proper ideals \(I_{1},I_{2}\) of R and some submodule K of M. Assume on the contrary that neither \(I_{1}I_{2}\subseteq (N:_{R}M)\) nor \(K\subseteq N\). Then there exist nonunit elements \(a\in I_{1},b\in I_{2}\) with \(ab\notin (N:_{R}M)\). Thus \(aRbK\subseteq N\), which contradicts (3).
\((4)\Rightarrow (1)\) Let \(a,b\in R\) be nonunit elements, \(m\in M\) and \(aRbRm\subseteq N\). Put \(I_{1}=RaR,I_{2}=RbR,K=Rm\). Now \(RaRRbRRm\subseteq RaRbRm\subseteq N.\) Thus \(ab\in RaRRbR\subseteq (N:_{R}M)\) or \(m\in Rm\subseteq N\) and we are done. \(\square \)
Theorem 2.10
Let M be an R-module. If N is a 1-absorbing prime submodule of M that is not a prime submodule, then R is a local ring.
Proof
Suppose that N is a 1-absorbing prime submodule of M that is not a prime submodule. Then there exist a nonunit \(r\in R\) and \(m\in M\) such that \(rRm\subseteq N\) but \(r\notin (N:_{R}M)\) and \(m\notin N\). Choose a nonunit element \(s\in R\). Hence we have that \(rRsRm\subseteq rRm\subseteq N\) and \(m\notin N\). Since N is 1-absorbing prime, \(rs\in (N:_{R}M)\). Let us take a unit element \(u\in R\). We claim that \(s+u\) is a unit element of R. To see this, assume \(s+u\) is a nonunit. Then \(rR(s+u)Rm\subseteq rRm\subseteq N\). As N is 1-absorbing prime, \(r(s+u)\in (N:_{R}M)\). This means that \(ru\in (N:_{R}M)\), i.e., \(r\in (N:_{R}M)\), which is a contradiction. Thus for any nonunit element s and unit element u in R, we have \(s+u\) is a unit element. From [9, Lemma 4.1], we have that R is a local ring. \(\square \)
Corollary 2.11
Let M be an R-module where R is not a local ring. Then a proper submodule N of M is a 1-absorbing prime submodule if and only if N is a prime submodule of M.
Proposition 2.12
Let \(\left\{ N_{i}:i\in \Delta \right\} \) be a chain of 1-absorbing prime submodules of the R-module M. Then \( {\textstyle \bigcap \nolimits _{i\in \Delta }} N_{i}\) is a 1-absorbing prime submodule of M.
Proof
Let \(\left\{ N_{i}:i\in \Delta \right\} \) be a chain of 1-absorbing prime submodules of M. Take nonunit elements \(a,b\in R\) and \(m\in M\) such that \(aRbRm\subseteq {\textstyle \bigcap \nolimits _{i\in \Delta }} N_{i}\). Assume that \(m\notin {\textstyle \bigcap \nolimits _{i\in \Delta }} N_{i}\), so there exists \(i\in \Delta \) such that \(m\notin N_{i}\). Since \(N_{i}\) is 1-absorbing prime, we conclude \(ab\in (N_{i}:M)\). For any \(j\in \Delta \), we have \(N_{i}\subseteq N_{j}\) or \(N_{j}\subseteq N_{i}\). Without loss of generality, if \(N_{i}\subseteq N_{j}\) then \((N_{i}:M)\subseteq (N_{j}:M)\), that is, \(ab\in (N_{j}:M)\). If \(N_{j}\subseteq N_{i}\), then \(ab\in (N_{j}:M)\) since \(m\notin N_{j}\) and \(N_{j}\) is 1-absorbing prime. Hence we have \(ab\in {\textstyle \bigcap \nolimits _{i\in \Delta }} \left\{ (N_{i}:M):i\in \Delta \right\} =(( {\textstyle \bigcap \nolimits _{i\in \Delta }} N_{i}:i\in \Delta ):M).\) \(\square \)
Definition 2.13
Let M be an R-module and N be a proper submodule of M. Let P be a 1-absorbing prime submodule of M such that \(N\subseteq P\). If there does not exist a 1-absorbing prime submodule \(P^{\prime }\) such that \(N\subseteq P^{\prime }\subset P\), then P is called a minimal 1-absorbing prime submodule over N.
Proposition 2.14
Let M be an R-module and N be a proper submodule of M. If P is a 1-absorbing prime submodule of M such that \(N\subseteq P\), then there exists a minimal 1-absorbing prime submodule over N that is contained in P.
Proof
Let \(\Lambda =\left\{ P_{i}:P_{i}\text { is a submodule of }M\text { such that }N\subseteq P_{i}\subseteq P\right\} .\) Since \(N\subseteq P,\) we have \(\Lambda \ne \emptyset .\) Consider \((\Lambda ,\supseteq )\). Let us take a chain \(\left\{ N_{i}:i\in \Delta \right\} \) in \(\Lambda \). Since by Proposition 2.12, \(\cap _{i\in \Delta }N_{i}\) is a 1-absorbing prime submodule of M, there exists a maximal element \(K\in \Lambda \) by applying Zorn’s Lemma. Then K is 1-absorbing prime and \(N\subseteq K\subseteq P\). Now we will show that K is a minimal 1-absorbing prime submodule over N. On the contrary, assume that there exists a 1-absorbing prime submodule \(K^{\prime }\) such that \(N\subseteq K^{\prime }\subseteq K\). Then \(K^{\prime } \in \Lambda \) and \(K\subseteq K^{\prime }\). This implies \(K=K^{\prime }\). Consequently, K is a minimal 1-absorbing prime submodule of N. \(\square \)
Corollary 2.15
Let M be an R-module. Every 1-absorbing prime submodule of M contains at least one minimal 1-absorbing prime submodule of M.
3 Weakly 1-absorbing prime submodules
Definition 3.1
Let R be a ring and N be a proper submodule of an R-module M. Then N is a weakly 1-absorbing prime submodule of M if \(\left\{ 0\right\} \ne aRbRm\subseteq N\) implies \(abM\subseteq N\) i.e. \(ab\in (N:_{R}M)\) or \(m\in N\) for nonunits \(a,b\in R\) and \(m\in M\).
Remark 3.2
- 1.:
-
Every 1–absorbing prime submodule is weakly 1-absorbing prime but the converse does not necessarily hold. For example consider the case where \(R=\mathbb {Z},\) \(M=\mathbb {Z}/30\mathbb {Z}\) and \(N=\left\{ 0\right\} \). Then \(2\cdot 3\cdot (5+30\mathbb {Z})=0\in N\) while \(2\cdot 3\notin (N:_{R}M),\) \((5+30Z)\notin N\). Therefore N is not 1-absorbing prime while it is weakly 1-absorbing prime.
- 2.:
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Every weakly prime submodule is weakly 1 absorbing prime but the converse does not necessarily hold. Let \(M=\mathbb {Z}_{12}\) be a module over \(\mathbb {Z}\) and \(W=\{\overline{0},\overline{4},\overline{8}\}\) be a proper submodule of M. Let \(r,s\in \mathbb {Z}\) and \(m\in M.\) Now \((W:M)=4\mathbb {Z}.\) Therfore, for \(0\ne rsm\in W,\) we get \(m=\overline{4}\) or \(m=\overline{8}\) which are elements in W or \(rs\in 4\mathbb {Z}.\) So W is a weakly 1 absorbing prime submodule. W is not weakly prime since \(\overline{0} \ne 2\cdot \overline{2}\in W\) with \(\overline{2}\notin W\) and \(2\notin (W:M).\)
Question. Suppose that L is a weakly 1-absorbing prime submodule of an R-module M and \(\left\{ 0\right\} \ne IJK\subseteq L\) for some ideals I, J of R and a submodule K of M. Does it imply that \(IJ\subseteq (M:_{R}L)\) or \(K\subseteq L\)?
This section is devoted to studying the above question for modules over noncommutative rings.
Proposition 3.3
Let \(x\in M\) and \(a\in R\). Then if \(\hbox {ann}_{l}(x)\subseteq (Rx:M)\), the submodule Rx is 1-absorbing prime if and only if Rx is weakly 1-absorbing prime.
Proof
If Rx is 1-absorbing prime then it is clear that Rx is weakly 1- absorbing prime. Let Rx be a weakly 1-absorbing prime submodule of M and suppose \(r,s\in R\) are nonunits and \(m\in M\) with \(rRsRm\subseteq Rx\). Since Rx is a weakly 1-absorbing prime submodule, we may assume \(rRsRm=\left\{ 0\right\} \), otherwise Rx is 1-absorbing prime. Now \(rRsR(x+m)\subseteq Rx\). If \(rRsR(x+m)\ne \left\{ 0\right\} \) then we have \(rs\in (Rx:M)\) or \((x+m)\in Rx\), as Rx is a weakly 1-absorbing prime submodule. Hence \(rs\in (Rx:M)\) or \(m\in Rx\). Now let \(rRsR(x+m)=\left\{ 0\right\} \). Then \(rRsRm=\left\{ 0\right\} \) implies \(rRsRx=\left\{ 0\right\} \). Hence \(rs\in \)ann\(_{l}(x)\subseteq (Rx:M).\) Thus Rx is 1-absorbing prime. \(\square \)
Proposition 3.4
Let R be a ring and N be a proper submodule of an R-module M.
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1.
If N is weakly prime, then it is weakly 1-absorbing prime.
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2.
If N is a weakly 1-absorbing prime submodule of M, then it is a weakly 2-absorbing submodule.
Proof
1. Assume N is a weakly prime submodule of the R-module M and \(\left\{ 0\right\} \ne aRbRm\subseteq N\) for nonunits \(a,b\in R\) and \(m\in M\). Suppose \(m\notin N.\) Since \(abRm\subseteq N\) and \(m\notin N,\) we have \(ab\in (N:_{R}M).\) Hence N is weakly 1-absorbing prime. 2. Assume N is a weakly 1-absorbing prime submodule of the R-module M and \(\left\{ 0\right\} \ne aRbRm\subseteq N\) for \(a,b\in R\) and \(m\in M\). If a is a unit, then it is easy to see that \(bm\in N.\) If b is a unit, then there exists \(b^{\prime }\in R\) such that \(ab^{\prime }bm=am\in N.\) If both a and b are nonunits, then \(ab\in (N:_{R}M)\) since N is a weakly 1-absorbing prime submodule of the R-module M. Hence N is weakly 2-absorbing. \(\square \)
Proposition 3.5
Let N be a weakly 1-absorbing prime submodule of an R-module M. Assume that K is a submodule of M with \(N\subsetneqq K\). Then N is a weakly 1-absorbing prime submodule of K.
Proof
Let \(a,b\in R\) be nonunits and \(k\in K\) with \(\left\{ 0\right\} \ne aRbRk\subseteq N\). Then \(ab\in (N:_{R}M)\) or \(k\in N\) as N is a weakly 1-absorbing prime submodule of M. Thus \(ab\in (N:_{R}K)\) or \(k\in K\) since \((N:_{R}M)\subseteq (N:_{R}K)\) and \(N\subseteq K.\) \(\square \)
Proposition 3.6
Let N, K be submodules of an R-module M with \(K\subseteq N\). If N is a weakly 1-absorbing prime submodule of M, then N/K is a weakly 1-absorbing prime submodule of M/K. The converse is true when K is a weakly 1-absorbing prime submodule.
Proof
Assume that N is a weakly 1-absorbing prime submodule of M. Let \(a,b\in R\) be nonunits and \(m+K\in M/K\) where \(\left\{ 0_{M/K}\right\} \ne aRbR(m+K)\subseteq N/K\). Since \(aRbR(m+K)\ne \left\{ 0_{M/K}\right\} \), we get \(aRbRm\subseteq N\) and \(aRbPm\nsubseteq K\). If \(aRbRm=\left\{ 0\right\} \), we obtain \(aRbRm+K=\left\{ 0_{M/K}\right\} \). So \(aRbRm\ne \left\{ 0\right\} \). Thus \(ab\in (N:_{R}M)\) or \(m\in N\) as N is weakly 1-absorbing prime. Consequently, we get \(ab\in (N/K:_{R}M/K)\) or \(m+K\in N/K\). Conversely, let K be a weakly 1-absorbing prime submodule. Assume that N/K is a weakly 1-absorbing prime submodule of M/K. Let \(a,b\in R\) be nonunits and \(m\in M\) where \(\left\{ 0\right\} \ne aRbRm\subseteq N\). Then we have \(aRbRm+K\subseteq N/K\). If \(aRbRm+K=\left\{ 0_{M/K}\right\} \), then \(aRbRm\subseteq K\). Thus \(ab\in (K:_{R}M)\) or \(m\in K\), since K is weakly 1-absorbing prime. Therefore, \(ab\in (N:_{R}M)\) or \(m\in N\), since \(K\subseteq N\). Let \(aRbRm+K=aRbR(m+K)\ne \left\{ 0_{M/K}\right\} \). Then \(ab\in (N/K:_{R}M/K)\) or \(m+K\in N/K\). Thus \(ab\in (N:_{R}M)\) or \(m\in N\). \(\square \)
Definition 3.7
Let N be a weakly 1-absorbing prime submodule of an R-module M. For nonunits \(a,b\in R\) and \(m\in M,\) (a, b, m) is called a triple-zero of N if \(aRbRm=0\), \(ab\notin (N:_{R}M)\) and \(m\notin N\).
Note that if N is a weakly 1-absorbing prime submodule of M and there is no triple-zero of N, then N is a 1-absorbing prime submodule of M.
Proposition 3.8
Let N be a weakly 1-absorbing prime submodule of M and K be a proper submodule of M with \(K\subseteq N\). Then for any nonunits \(a,b\in R\) and \(m\in M\), (a, b, m) is a triple-zero of N if and only if \((a,b,m+K)\) is a triple-zero of N/K.
Proof
Let (a, b, m) be a triple-zero of N for some nonunits \(a,b\in R\) and \(m\in M\). Then \(aRbRm=\left\{ 0\right\} \), \(ab\notin (N:_{R}M)\) and \(m\notin N\). By Proposition 3.6, we get that N/K is a weakly 1-absorbing prime submodule of M/K. Thus \(aRbR(m+K)=K,ab\notin (N/K:M/K)\) and \((m+K)\notin N/K\). Hence \((a,b,m+K)\) is a triple-zero of N/K. Conversely, assume that \((a,b,m+K)\) is a triple-zero of N/K. Suppose that \(aRbRm\ne \left\{ 0\right\} \). Then \(aRbRm\subseteq N\) since \(aRbR(m+K)=K\). Thus \(ab\in (N:_{R}M)\) or \(m\in N\) as N is weakly 1-absorbing prime, a contradiction. So it must be \(aRbRm=\left\{ 0\right\} \). Consequently, (a, b, m) is a triple-zero of N. \(\square \)
Theorem 3.9
Let R be a local ring and let N be weakly 1-absorbing prime submodule of M and (a, b, m) be a triple-zero of N for some nonunits \(a,b\in R\) and \(m\in M\). Then the following hold.
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1.
\(aRbN=a(N:_{R}M)m=b(N:_{R}M)m=\left\{ 0\right\} .\)
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2.
\(a(N:_{R}M)N=b(N:_{R}M)N=(N:_{R}M)bN=(N:_{R}M)bm=(N:_{R}M)^{2}m=\left\{ 0\right\} .\)
Proof
Suppose that (a, b, m) is a triple-zero of N for some nonunits \(a,b\in R\) and \(m\in M\). 1. Assume that \(aRbN\ne \left\{ 0\right\} \). Then there is an element \(n\in N\) such that \(aRbRn\ne \left\{ 0\right\} .\) Now \(aRbR(m+n)=aRbRm+aRbRn=aRbRn\ne \left\{ 0\right\} \) since \(aRbRm=\left\{ 0\right\} .\) Since \(\left\{ 0\right\} \ne aRbR(m+n)\subseteq N\ \)and N a weakly 1-absorbing prime submodule, we have \(ab\in (N:_{R}M)\) or \((m+n)\in N\). Hence \(ab\in (N:_{R}M)\) or \(m\in N\). This is a contradiction since (a, b, m) is a triple zero of N. Hence \(aRbN=\left\{ 0\right\} .\) Now, we suppose that \(a(N:_{R}M)m\ne \left\{ 0\right\} \). Thus there exists an element \(r\in (N:_{R}M)\) such that \(arm\ne 0\). Hence \(aR(r+b)Rm=aRrRm+aRbRm=aRrRm\). Hence \(\left\{ 0\right\} \ne aR(r+b)Rm\subseteq N.\) Since R is local, the set of nonunit elements of R is an ideal of R. Therefore \((r+b)\) is a nonunit. Since N is a weakly 1-absorbing prime submodule, we have \(a(r+b)\in (N:_{R}M)\) or \(m\in N\). Consequently, \(ab\in (N:_{R}M)\) or \(m\in N\) a contradiction since (a, b, m) is a triple-zero of N. Hence \(a(N:_{R}M)m=\left\{ 0\right\} .\) Similarly, we can proof that \(b(N:_{R}M)m=\left\{ 0\right\} \). 2. Assume that \(a(N:_{R}M)N\ne \left\{ 0\right\} \). Then there are \(r\in (N:_{R}M),\) \(n\in N\) such that \(arn\ne 0\). By (1), we get \(a(b+r)(m+n)=abm+abn+arm+arn=arn\ne 0.\) Now \(\left\{ 0\right\} \ne aR(b+r)R(m+n)\subseteq N.\) Again, since R is a local ring \((b+r)\) is a nonunit and since N is 1-absorbing prime, we have \(a(b+r)\in (N:_{R}M)\) or \((m+n)\in N\). Hence we obtain \(ab\in (N:_{R}M)\) or \(m\in N\) a contradiction. Hence \(a(N:_{R}M)N=\left\{ 0\right\} .\) In a similar way we get \(b(N:_{R}M)N=\left\{ 0\right\} .\) Now, we suppose that \((N:_{R} M)bN\ne \left\{ 0\right\} .\) Then there are \(r\in (N:_{R}M)\), \(n\in N\) such that \(rbn\ne 0\). Now, from above \((a+r)b(n+m)=abn+abm+rbn+rbm=rbn\ne 0.\) Hence \(\left\{ 0\right\} \ne (a+r)RbR(n+m)\subseteq N\) and since N is weakly 1-absorbing and \((a+r)\) is a nonunit, we have \((a+r)b\in (N:_{R}M)\) or \((n+m)\in N.\) Hence \(ab\in (N:_{R}M)\) or \(m\in N\) a contradiction since (a, b, m) is a triple-zero of N. Now, we suppose that \((N:_{R} M)bm\ne \left\{ 0\right\} .\) Then there is \(r\in (N:_{R}M)\) such that \(rbm\ne 0\). Hence \(0\ne rbm=(a+r)bm\in (a+r)RbRm=rRbRm\subseteq N.\) Since N is a weakly 1-absorbing prime submodule, we have \((a+r)b\in (N:_{R}M)\) or \(m\in N\). Therefore \(ab\in (N:_{R}M)\) or \(m\in N\) a contradiction since (a, b, m) is a triple-zero of N. Hence \((N:_{R}M)bm=\left\{ 0\right\} .\) Lastly, we show that \((N:_{R}M)^{2}m=\left\{ 0\right\} .\) Let \((N:_{R} M)^{2}m\ne \left\{ 0\right\} \). Thus there exist \(r,s\in (N:_{R}M)\) where \(rsm\ne 0\). By (1), we get \((a+r)(b+s)m=rsm\ne 0\). Thus we have \(\left\{ 0\right\} \ne (a+r)R(b+s)Rm\subseteq N\). Since R is a local ring, \((a+r)\) and \((b+s)\) are nonunits. Hence \((a+r)(b+s)\in (N:_{R}M)\) or \(m\in N\). Consequently \(ab\in (N:_{R}M)\) or \(m\in N\) a contradiction since (a, b, m) is a triple-zero of N. Therefore \((N:_{R}M)^{2}m=\left\{ 0\right\} \). \(\square \)
Proposition 3.10
Let R be a local ring. Assume that N is a weakly 1-absorbing prime submodule of an R-module M that is not 1-absorbing prime. Then \((N:_{R}M)^{2}N=\left\{ 0\right\} \). In particular, \((N:_{R}M)^{3}\subseteq \)Ann(M).
Proof
Suppose that N is a weakly 1-absorbing prime submodule of an R-module M that is not 1-absorbing prime. Then there is a triple-zero (a, b, m) of N for some nonunits \(a,b\in R\) and \(m\in M\). Assume that \((N:_{R}M)^{2} N\ne \left\{ 0\right\} \). Thus there exist \(r,s\in (N:_{R}M)\) and \(n\in N\) with \(rsn\ne 0\). By Theorem 3.9, we get \((a+r)(b+s)(n+m)=rsn\ne 0\). Then we have \(\left\{ 0\right\} \ne (a+r)R(b+s)R(n+m)\subseteq N.\) Since N is weakly 1-absorbing prime, we have \((a+r)(b+s)\in (N:_{R}M)\) or \((n+m)\in N\) and so \(ab\in (N:_{R}M)\) or \(m\in N\) which is a contradiction. Thus \((N:_{R}M)^{2}N=\left\{ 0\right\} .\) We get \((N:_{R}M)^{3}\subseteq ((N:_{R}M)^{2}N:M)=(\left\{ 0\right\} :M)=\)Ann(M). \(\square \)
Proposition 3.11
Let R be a local ring and let M be a multiplication R-module and N be a weakly 1-absorbing prime submodule of M that is not a 1-absorbing prime submodule. Then \(N^{3}=\left\{ 0\right\} \).
Proof
We have that \((N:_{R}M)M=N\) since M is a multiplication module. Then \(N^{3}=(N:_{R}M)^{3}M=(N:_{R}M)^{2}N=\left\{ 0\right\} \). Consequently, \(N^{3}=\left\{ 0\right\} \). \(\square \)
Definition 3.12
Let N be a weakly 1-absorbing prime submodule of an R-module M and let \(a,b\in R\) be nonunits. Let \(\left\{ 0\right\} \ne I_{1}I_{2}K\subseteq N\) for some ideals \(I_{1},I_{2}\) of R and some submodule K of M. N is called free triple-zero in regard to \(I_{1},I_{2},K\) if (a, b, m) is not a triple-zero of N for every \(a\in I_{1},b\in I_{2}\) and \(m\in K\).
Lemma 3.13
Let N be a weakly 1-absorbing prime submodule of M. Assume that \(aRbK\subseteq N\) for some nonunits \(a,b\in R\) and some submodule K of M where (a, b, m) is not a triple-zero of N for every \(m\in K\). If \(ab\notin (N:_{R}M),\) then \(K\subseteq N\).
Proof
Suppose that \(aRbK\subseteq N\), but \(ab\notin (N:_{R}M)\) and \(K\nsubseteq N\). Then there exists an element \(k\in K\backslash N\). But (a, b, k) is not a triple-zero of N and \(aRbRk\subseteq N\) and \(ab\notin (N:_{R}M)\) and \(k\notin N\), a contradiction. Hence \(K\subseteq N.\) \(\square \)
Let N be a weakly 1-absorbing prime submodule of an R-module M and \(I_{1}I_{2}K\subseteq N\) for some for some ideals \(I_{1},I_{2}\) of R and some submodule K of M where N is free triple-zero in regard to \(I_{1},I_{2},K.\) Note that if \(a\in I_{1},b\in I_{2}\) and \(m\in K\), then \(ab\in (N:_{R}M)\) or \(m\in N.\)
Theorem 3.14
Suppose that N is a proper submodule of the R module M. Then the following statements are equivalent.
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1.
N is a weakly 1-absorbing prime submodule of M.
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2.
For any proper ideals \(I_{1},I_{2}\) of R and a submodule K of M such that \(\left\{ 0\right\} \ne I_{1}I_{2}K\subseteq N\) and N is free triple-zero with respect to \(I_{1},I_{2},K\), we have either \(I_{1} I_{2}\subseteq \) \((N:_{R}M)\) or \(K\subseteq N.\)
Proof
\((1)\Rightarrow (2)\) Suppose that N is a weakly 1-absorbing prime submodule of M and \(\left\{ 0\right\} \ne I_{1}I_{2}K\subseteq N\) for proper ideals \(I_{1},I_{2}\) of R and a submodule K of M such that N is free triple-zero with respect to \(I_{1},I_{2},K\). Then there are nonunit elements \(a\in I_{1}\) and \(b\in I_{2}\) such that \(ab\notin (N:_{R}M)\). Since \(aRbK\subseteq N\), \(ab\notin (N:_{R}M)\) and (a, b, k) is not a triple-zero of N for every \(k\in K\), it follows from Lemma 3.13 that \(K\subseteq N\). \((2)\Rightarrow (1)\) Suppose that \(\left\{ 0\right\} \ne aRbRm\subseteq N\) for some nonunit elements \(a,b\in R\) and \(m\in M.\) Suppose \(ab\notin (N:_{R} M)\). Let \(I_{1}=RaR,\) \(I_{2}=RbR\) and \(K=Rm.\) Now, \(\left\{ 0\right\} \ne RaRRbRRm\subseteq RN\subseteq N.\) Hence \(\left\{ 0\right\} \ne I_{1} I_{2}K\subseteq N\) and \(I_{1}I_{2}\nsubseteq \) \((N:_{R}M).\) From (2), it follows that \(m\in Rm=K\subseteq N\) and we are done. \(\square \)
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Groenewald, N. 1-absorbing and weakly 1-absorbing prime submodules of a module over a noncommutative ring. Afr. Mat. 34, 54 (2023). https://doi.org/10.1007/s13370-023-01094-0
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DOI: https://doi.org/10.1007/s13370-023-01094-0