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On classical n-absorbing submodules

  • Osama A. NajiEmail author
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Abstract

Let R a commutative ring with identity and M be a unitary R-module. In this paper, we investigate some properties of n-absorbing submodules of M as a generalization of 2-absorbing submodules. We also define the classical n-absorbing submodule, a proper submodule N of an R-module M is called a classical n-absorbing submodule if whenever \(a_1 a_2\ldots a_{n+1} m\in N\) for \(a_1, a_2,\ldots , a_{n+1}\in R\) and \(m \in M\), there are n of \(a_i\)’s whose product with m is in N. Furthermore, we give some characterizations of n-absorbing and classical n-absorbing submodules under some conditions.

Mathematics Subject Classification

13C05 13C13 13C99 

1 Introduction

Throughout this paper, we assume that all rings are commutative with \(1 \ne 0\). Let R be a commutative ring. An ideal I of R is said to be proper if \(I \ne R\). Let M a unitary module over R and N be a submodule of M. The residual of N by M, \((N :_R M)\) or simply (N : M), denotes the ideal \(\{r \in R : rM \subseteq N\}\). For any element x of M, the ideal (N : x) is defined by \((N : x) = \{r \in R : rx \in N\}\). Let \(a \in R\). Then, \(N_a = \{x : x \in M ~~ and~~ ax \in N\}\) is a submodule of the R-module M. Let \(m \in M\), a cyclic submodule that is generated by m is a submodule of M has the form \(Rm = \{rm : r \in R\}\). A proper submodule N of M is said to be irreducible if N is not an intersection of two submodules of M that properly contain it. The set of zero divisors of M, denoted by Zd(M) is defined by \(Zd(M) = \{r \in R : ~~for~~ some~~ x \in M~~ and~~ x \ne 0, rx = 0\}\). An R-module M is called a multiplication module if every submodule N of M has the form IM for some ideal I of R. Prime ideals play a crucial role in ring theory, since they interfere with many branches of algebra and they represent an important role in understanding the structure of ring. A proper ideal I of a ring R is called a prime ideal if, whenever \(ab\in I\) for \(a, b \in R\), then \(a\in I\) or \(b\in I\). A proper submodule N of an R-module M is said to be a prime submodule if, whenever \(a \in R\), \(m \in M\), and \(am \in N\), then \(m \in N\) or \(a \in (N : M)\).

In [5], Badawi introduced a new generalization of prime ideals in a commutative ring R. He defined a nonzero proper ideal I of R to be a 2-absorbing ideal of R if, whenever \(a, b, c \in R\) and \(abc \in I\), then \(ab \in I\) or \(ac \in I\) or \(bc \in I\). The concept of 2-absorbing ideal has been transferred to modules. A proper submodule N of an R-module M is a 2-absorbing submodule of M [6] if, whenever \(abm \in N\) for \(a, b \in R\) and \(m \in M\), then \(am \in N\) or \(bm \in N\) or \(ab \in (N : M)\). The class of 2-absorbing submodules of modules was introduced as a generalization of the class of 2-absorbing ideals of rings. Then, many generalizations of 2-absorbing submodules were studied such as primary 2-absorbing [8], almost 2-absorbing [3], almost 2-absorbing primary [2], and classical 2-absorbing [9]. In this article, we investigate some properties of n-absorbing submodules of M as a generalization of 2-absorbing submodules. We also define the classical n-absorbing submodule. Furthermore, we give some characterizations of n-absorbing and classical n-absorbing submodules under some conditions. In addition, we investigate the sufficient and necessary conditions for a submodule N to be classical n-absorbing submodule of M.

2 n-Absorbing submodules

The concept of 2-absorbing has been extended to n-absorbing in ideals and submodules, where n is any positive integer. In this section, we investigate some properties of n-absorbing submodules.

Definition 2.1

[1] A proper ideal I of a ring R is said to be an n-absorbing ideal if, whenever \(a_1\ldots a_{n+1}\in I\) for \(a_1,\ldots ,a_{n+1}\in R\), then there are n of \(a_i's\) whose product is in I.

Definition 2.2

[7] A proper submodule N of an R-module M is called an n-absorbing submodule if, whenever \(a_1 \ldots a_nm\in N\) for \(a_1,\ldots ,a_n\in R\) and \(m\in M\), then either \(a_1 \ldots a_n\in (N:M)\) or there are \(n-1\) of \(a_i's\) whose product with m is in N.

Proposition 2.3

If N is an n-absorbing submodule of an R-module M, then (N : m) is an n-absorbing ideal in R for all \(m\in M - N\).

Proof

For \(m\in M - N\), (N : m) is a proper ideal of R. Assume that \(a_1 \ldots a_{n+1} \in (N : m)\) for \(a_1,\ldots ,a_{n+1} \in R\). Then, \(a_1 \ldots a_{n+1}m = a_1 \ldots a_n(a_{n+1}m) \in N\). Since N is an n-absorbing submodule, then \(a_1 \ldots a_n \in (N : M) \subseteq (N : m)\) or there are \(n - 1\) of the \(a_i's\) , \(1\le i \le n\) whose product with \(a_{n+1}m\) in N, the latter case means that there are \(n - 1\) of the \(a_i's\), \(1\le i \le n\) whose product with \(a_{n+1}\) belongs to (N : m). Thus, (N : m) is an n-absorbing ideal in R. \(\square \)

Proposition 2.4

[4] Let M an R-module and N be a proper submodule of M. Then, \(Zd(M/N) = \bigcup \nolimits _{x\in M-N} (N : x)\).

Proposition 2.5

Let N be an n-absorbing submodule of M. If the set of all zero divisors of M / N, Zd(M / N), forms an ideal in R, then it is an n-absorbing ideal of R.

Proof

Let \(a_1 \ldots a_{n+1} \in Zd(M/N)\) for \(a_1,\ldots ,a_{n+1} \in R\), and then, by Proposition 2.4, \(a_1 \ldots a_{n+1} \in (N : m')\) for some \(m' \in M - N\). Since N is an n-absorbing submodule, then \((N : m')\) is an n-absorbing ideal of R. Therefore, there are n of \(a_i's\) whose product belongs to \((N : m')\), and hence, there are n of \(a_i's\) whose product belongs to Zd(M / N). \(\square \)

Remark 2.6

The set of all zero divisors may not be an ideal. For example, consider the \({\mathbb {Z}}\)-module \(M = {\mathbb {Z}}_6\), we have \(2, 3 \in Zd(M)\) but \(2 + 3 \notin Zd(M)\).

The following theorem characterizes n-absorbing submodule in terms of submodules.

Theorem 2.7

Let N be a submodule of an R-module M. The following are equivalent:
  1. (1)

    N is an n-absorbing submodule.

     
  2. (2)

    For \(a_1,\ldots ,a_n \in R\), such that \(a_1 \ldots a_n \notin (N : M)\)\(N_{a_1 \ldots a_n} = \bigcup \nolimits _{i = 1}^n N_{{\hat{a}}_i}\), where \({\hat{a}}_i = a_1 \ldots a_{i-1}a_{i+1} \ldots a_n\).

     

Proof

\((1)\Rightarrow (2)\) Let \(m \in N_{a_1 \ldots a_n}\) and assume that \(a_1 \ldots a_n \notin (N : M)\), and then, \(a_1 \ldots a_nm \in N\). Since N is an n-absorbing submodule, then there are \(n - 1\) of \(a_i's\), \(1\le i \le n\), such that \({\hat{a}}_im \in N\), \({\hat{a}}_i = a_1 \ldots a_{i-1}a_{i+1} \ldots a_n\), and hence, \(m \in N_{{\hat{a}}_i}\). For the other containment, let \(m \in \bigcup \nolimits _{i = 1}^n N_{{\hat{a}}_i}\), then \({\hat{a}}_jm \in N\) for some \(j\in \{1,\ldots ,n\}\), then \(a_j{\hat{a}}_jm = a_1 \ldots a_nm \in N\), so \(m \in N_{a_1 \ldots a_n}\).

\((2)\Leftarrow (1)\) Let \(a_1,\ldots ,a_n \in R\) and \(m \in M\) such that \(a_1 \ldots a_nm \in N\). Assume that \(a_1 \ldots a_n \notin (N : M)\), then \(m \in N_{a_1 \ldots a_n} = \bigcup \nolimits _{i = 1}^n N_{{\hat{a}}_i}\) then \(m \in N_{{\hat{a}}_j}\) for some \(j\in \{1,\ldots ,n\}\), implies that \({\hat{a}}_jm = a_1 \ldots a_{j-1}a_{j+1} \ldots a_nm \in N\). Thus, N is an n-absorbing submodule. \(\square \)

The following example shows that if N is not an n-absorbing submodule of M, then the second statement in the previous theorem does not hold.

Example 2.8

Take \(n = 2\). Let \(M = {\mathbb {Z}}\) be a module over itself, and let \(N = 8{\mathbb {Z}}\), N is not a 2-absorbing submodule of M and \(N_{2.2} = 2{\mathbb {Z}} \ne N_2 = 4{\mathbb {Z}}\).

Now, we give a necessary and sufficient condition for capability of reducing (by 1) the index of the residual (N : M) of the proper submodule N of M.

Theorem 2.9

Let N be an n-absorbing submodule of an R-module M. Then, (N : M) is an \((n - 1)\)-absorbing ideal of R if and only if (N : m) is an \((n - 1)\)-absorbing ideal of R for all \(m \in M - N\).

Proof

\((\Rightarrow )\) Let \(a_1,\ldots , a_n \in R\), \(m \in M - N\) and \(a_1 \ldots a_n \in (N : m)\). Then, \(a_1 \ldots a_nm \in N\). Since N is an n-absorbing submodule of M, then \(a_1 \ldots a_n \in (N : M)\) or there are \(n - 1\) of the \(a_i's\) whose product with m is in N. If \(a_1 \ldots a_n \in (N : M)\), then, by assumption, there are \(n - 1\) of the \(a_i's\), \(1\le i \le n\), whose product belongs to (N : M), and hence, there are \(n - 1\) of the \(a_i's\), \(1\le i \le n\), whose product belongs to (N : m). In the other case, if there are \(n - 1\) of the \(a_i's\) whose product with m is in N, and hence, there are \(n - 1\) of the \(a_i's\), \(1\le i \le n\), whose product belongs to (N : m) and we are done.

\((\Leftarrow )\) Suppose that \(a_1 \ldots a_n \in (N : M)\) for some \(a_1,\ldots , a_n \in R\) and assume that, for every i, \(1 \le i \le n\), there exists \(m_i \in M\), such that \({\hat{a}}_im_i \notin N\), where \({\hat{a}}_i = a_1 \ldots a_{i-1}a_{i+1} \ldots a_n\). By \(a_1 \ldots a_nm_i \in N\), it follows that \({\hat{a}}_jm_i \in N\), where \(j \ne i\) and \({\hat{a}}_j = a_1 \ldots a_{j-1}a_{j+1} \ldots a_n\), since \((N : m_i)\) is \((n - 1)\)-absorbing ideal. If \(\sum _{i=1}^{n} {m_i} \in N\), then \({\hat{a}}_jm_j \in N\), since \({\hat{a}}_jm_i \in N\), \(\forall i \ne j\), which is a contradiction. Thus, \(\sum _{i=1}^{n} {m_i} \notin N\). Now, by \(a_1 \ldots a_n\sum _{i=1}^{n} {m_i} \in N\), we have \(a_1 \ldots a_n \in (N : \sum _{i=1}^{n} {m_i})\), and then, there are \(n - 1\) of the \(a_i's\) whose product is in \((N : \sum _{i=1}^{n} {m_i})\), and hence, there are \(n - 1\) of the \(a_i's\) whose product with \(\sum _{i=1}^{n} {m_i}\) belongs to N, and then, we must have \({\hat{a}}_km_k \in N\), for some \(k \in \{1,\ldots ,n\}\), which is a contradiction. Thus, there are \(n - 1\) of the \(a_i's\) whose product with M is contained in N. Therefore, (N : M) is \((n - 1)\)-absorbing ideal of R. \(\square \)

Proposition 2.10

Let N be an n-absorbing submodule of an R-module M , \(y \in M\), and \(a_1,\ldots ,a_n\in R\). If \(a_1 \ldots a_n \notin (N : M)\), then
$$\begin{aligned} (N : a_1 \ldots a_ny) = \bigcup _{i = 1}^n (N : {\hat{a}}_iy), \end{aligned}$$
where \({\hat{a}}_i = a_1 \ldots a_{i-1}a_{i+1} \ldots a_n\).

Proof

Let \(r \in (N : a_1 \ldots a_ny)\), and then, \(ra_1 \ldots a_ny = a_1 \ldots a_n(ry) \in N\). Since N is an n-absorbing submodule and \(a_1 \ldots a_n \notin (N : M)\), then \({\hat{a}}_i(ry) \in N\), where \({\hat{a}}_i = a_1 \ldots a_{i-1}a_{i+1} \ldots a_n\), for some i, and hence, \(r \in (N : {\hat{a}}_iy)\). For the reverse inclusion, let \(r \in \bigcup \nolimits _{i = 1}^n (N : {\hat{a}}_iy)\), and then, \(r \in (N : {\hat{a}}_jy)\) for some \(j\in \{1,\ldots ,n\}\). Then, \(ra_j{\hat{a}}_jy = ra_1 \ldots a_ny \in N\) implies \(r \in (N : a_1 \ldots a_ny)\). \(\square \)

In the following two propositions, we study the absorbing property under the homomorphism and localization.

Proposition 2.11

Let \(f : M \rightarrow M'\) be an epimorphism of R-modules.
  1. (1)

    If \(N'\) is an n-absorbing submodule of \(M'\), then \(f^{-1}(N')\) is an n-absorbing submodule of M.

     
  2. (2)

    If N is an n-absorbing submodule of M containing ker(f), then f(N) is an n-absorbing submodule of \(M'\).

     

Proof

(1) Let \(a_1,\ldots , a_n \in R\) and \(m \in M\), such that \(a_1 \ldots a_nm \in f^{-1}(N')\) then \(a_1 \ldots a_nf(m) \in N'\), but \(N'\) is n-absorbing submodule of \(M'\), so \(a_1 \ldots a_n \in (N' : M')\) or \({\hat{a}}_if(m) \in N'\), where \({\hat{a}}_i = a_1 \ldots a_{i-1}a_{i+1} \ldots a_n\). If \(a_1 \ldots a_n \in (N' : M')\), then \(a_1 \ldots a_nM' \subseteq N'\), then \(a_1 \ldots a_nM \subseteq f^{-1}(N')\), so \(a_1 \ldots a_n \in (f^{-1}(N') : M)\). If \({\hat{a}}_if(m) \in N'\), then \(f({\hat{a}}_im) \in N'\) so \({\hat{a}}_im \in f^{-1}(N')\). Thus, \(f^{-1}(N')\) is an n-absorbing submodule of M.

(2) Let \(a_1,\ldots ,a_n \in R\), \(m' \in M'\), and \(a_1 \ldots a_nm' \in f(N)\). Then, there exists \(t \in N\), such that \(a_1 \ldots a_nm' = f(t)\). Since f is an epimorphism therefore for some \(m \in M\), we have \(f(m) = m'\). Thus, \(a_1 \ldots a_nf(m) = f(t)\). This implies that \(f(a_1 \ldots a_nm - t) = 0\), so \(a_1 \ldots a_nm - t \in ker(f) \subseteq N\). Thus, \(a_1 \ldots a_nm \in N\). Now, since N is an n-absorbing, therefore, \({\hat{a}}_im \in N\) or \(a_1 \ldots a_n \in (N : M)\). Thus, \({\hat{a}}_im' \in f(N)\) or \(a_1 \ldots a_n \in (f(N) : M')\). Hence, f(N) is an n-absorbing submodule of \(M'\). \(\square \)

Proposition 2.12

Let S be a multiplicatively closed subset of R and \(S^{-1}M\) be the module of fraction of M. Then, the following statements hold.
  1. (1)

    If N is an n-absorbing submodule of M , then \(S^{-1}N\) is an n-absorbing submodule of \(S^{-1}M\).

     
  2. (2)

    If \(S^{-1}N\) is an n-absorbing submodule of \(S^{-1}M\) such that \(Zd(M/N) \cap S =\phi \), then N is an n-absorbing submodule of M.

     

Proof

(1) Assume that \(a_1,\ldots ,a_n \in R\), \(s_1,\ldots ,s_n, l \in S\), \(m \in M\) and \(\frac{a_1 \ldots a_nm}{s_1 \ldots s_nl} \in S^{-1}N\). Then, there exists \(s' \in S\), such that \(s'a_1 \ldots a_nm = a_1 \ldots a_n(s'm) \in N\). By assumption, N is an n-absorbing submodule of M, and thus, \(a_1 \ldots a_n \in (N : M)\) or \({\hat{a}}_is'm \in N\), where \({\hat{a}}_i = a_1 \ldots a_{i-1}a_{i+1} \ldots a_n\) for some \(1 \le i \le n\). If \({\hat{a}}_is'm \in N\), then \(\frac{{\hat{a}}_is'm}{s_1 \ldots s_{i-1}s_{i+1} \ldots s_ns'l} = \frac{{\hat{a}}_im}{{\hat{s}}_il} \in S^{-1}N\), and if \(a_1 \ldots a_n \in (N : M)\), then \(\frac{a_1 \ldots a_n}{s_1 \ldots s_n} \in S^{-1}(N : M) \subseteq (S^{-1}N : S^{-1}M)\) . Therefore, \(S^{-1}N\) is an n-absorbing submodule of \(S^{-1}M\).

(2) Let \(a_1,\ldots ,a_n \in R\) and \(m \in M\) be such that \(a_1 \ldots a_nm \in N\). Then, \(\frac{a_1 \ldots a_nm}{1} \in S^{-1}N\). Since \(S^{-1}N\) is an n-absorbing submodule of \(S^{-1}M\), either \(\frac{a_1 \ldots a_n}{1} \in (S^{-1}N :_{S^{-1}R} S^{-1}M)\) or \(\frac{{\hat{a}}_im}{1} \in S^{-1}N\), where \({\hat{a}}_i = a_1 \ldots a_{i-1}a_{a+1}..a_n\) for some \(1 \le i \le n\). Therefore, there exists \(s \in S\), such that \(s{\hat{a}}_im \in N\). This implies \({\hat{a}}_im \in N\), since \(S \cap Zd(M/N) = \phi \). Now, consider the case when \(\frac{a_1 \ldots a_n}{1} \in (S^{-1}N :_{S^{-1}R} S^{-1}M)\), then \(a_1 \ldots a_nS^{-1}M\subseteq S^{-1}N\). Now, we have to show \(a_1 \ldots a_nM\subseteq N\). Assume that \(m'\in M\), and then, \(\frac{a_1 \ldots a_nm'}{1} \in a_1 \ldots a_nS^{-1}M \subseteq S^{-1}N\), so there exists \(t\in S\), such that \(ta_1 \ldots a_nm \in N\). Since \(S \cap Zd(M/N) = \phi \), then \(a_1 \ldots a_nm' \in N\), and therefore, \(a_1 \ldots a_nM \subseteq N\). Hence, N is an n-absorbing submodule of M. \(\square \)

3 Classical n-absorbing submodules

In this section, we introduce and study the concept of classical n-absorbing submodules as a generalization of n-absorbing submodules.

Definition 3.1

A proper submodule N of an R-module M is called a classical n-absorbing submodule if, whenever \(a_1 a_2\dots a_{n+1} m\in N\) for \(a_1, a_2,\dots , a_{n+1}\in R\) and \(m \in M\), there are n of \(a_i\)’s whose product with m is in N.

Example 3.2

  1. (1)

    Let \(R = {\mathbb {Z}}\) and \(M = R \times R\). The submodule \(N = \{(k,k): k\in R\}\) is a classical n-absorbing submodule of M.

     
  2. (2)

    Let \(R = {\mathbb {Z}}\) and \(M = {\mathbb {Z}}_3\oplus {\mathbb {Q}}\oplus {\mathbb {Z}}\). Take \(n = 2\), the submodule \(N = {\bar{0}} \oplus \{0\}\oplus {\mathbb {Z}}\) is a classical 2-absorbing submodule of M. To see this, let \(a,b,c,z\in {\mathbb {Z}}\), \(w\in {\mathbb {Q}}\) and \({\bar{x}}\in {\mathbb {Z}}_3\) such that \(abc({\bar{x}},w,z)\in N\). Hence, \({\overline{abcx}} = {\overline{0}}\) and \(abcw = 0\). If \(abcz \ne 0\), then \(w = 0\). We have 3|abcx, then 3|ab or 3|cx, if 3|ab, then \(ab({\bar{x}},w,z) = ({\overline{abx}},0,abz) = (0,0,abz)\in N\). Similarly if 3|cx, then \(c({\bar{x}},w,z) = ({\overline{cx}},0,cz) = (0,0,cz)\in N\). Now, if \(abcz = 0\), then one of abcz is zero; first, we take one of the scalars which is zero, say a, then \(a({\bar{x}},w,z) = ({\bar{0}},0,0)\in N\), and hence \(ab({\bar{x}},w,z)\in N\). if \(a,b,c \ne 0\) and \(z = 0\), since \(abcw = 0\), then \(w = 0\) (this was a previous case). If \(a,b,c \ne 0 , z = 0\) and \(w \ne 0\), then \(abcw \ne 0\) so \(abc({\bar{x}},w,z) \notin N\), a contradiction. Thus, N is a classical 2-absorbing submodule of M.

     

Proposition 3.3

Let N be a proper submodule of an R-module M.
  1. (i)

    If N is an n-absorbing submodule of M, then N is a classical n-absorbing submodule of M.

     
  2. (ii)

    If N is an n-absorbing submodule of M and (N : M) is an \((n-1)\)-absorbing ideal of R, then N is a classical \((n-1)\)-absorbing submodule of M.

     

Proof

(i) Assume that N is an n-absorbing submodule of M. Let \(a_1, a_2, \dots ,a_{n+1}\in R\) and \(m\in M\), such that \(a_1a_2\dots a_na_{n+1}m = a_1a_2\dots a_n(a_{n+1}m)\in N\). Then, either there are \(n-1\) of \(a_i\)’s whose product with \(a_{n+1}m\) is in N or \(a_1a_2\dots a_n\in (N : M)\). The first case leads us to the claim. In the second case, we have that \(a_1a_2\dots a_nm\in N\). Consequently, N is a classical n-absorbing submodule.

(ii) Assume that N is an n-absorbing submodule of M and (N : M) is an \((n-1)\)-absorbing ideal of R. Let \(a_1a_2\dots a_nm\in N\) for some \(a_1, a_2, \dots , a_n \in R\) and \(m \in M\), such that there are no \(n-1\) of \(a_i\)’s whose product with m is in N. Then, \(a_1a_2\dots a_n \in (N : M)\), and so, there are \(n-1\) of \(a_i\)’s whose product is in (N : M), which is a contradiction. Hence, N is a classical \((n-1)\)-absorbing submodule of M. \(\square \)

Remark 3.4

The following example shows that the converse of Proposition 3.3(i) is not true. Take \(n = 2\), and let \(R = {\mathbb {Z}}\) and \(M = {\mathbb {Z}}_3 \oplus {\mathbb {Z}}_5 \oplus {\mathbb {Z}}\). The zero submodule of M is a classical 2-absorbing submodule, but is not 2-absorbing, since \(3.5(1, 1,0) = (0, 0,0)\), but \(3(1, 1,0) \ne (0, 0,0)\), \(5(1, 1,0) \ne (0, 0,0)\), and \(3.5\notin (0 : {\mathbb {Z}}_3 \oplus {\mathbb {Z}}_5 \oplus {\mathbb {Z}}) = 0\).

The following theorem characterizes classical n-absorbing submodule in terms of n-absorbing ideals.

Theorem 3.5

Let M an R-module and N be a proper submodule of M. Then, the followings are equivalent:
  1. (i)

    N is a classical n-absorbing submodule of M.

     
  2. (ii)

    (N : m) is a n-absorbing ideal of R for every \(m \in M-N\).

     

Proof

\((i)\Rightarrow (ii)\) Assume that N is a classical n-absorbing submodule. (N : m) is a proper ideal, since \(m\in M-N\). Let \(a_1a_2\dots a_{n+1}\in (N : m)\) for some \(a_1, a_2, \dots , a_{n+1}\in R\). Since N is a classical n-absorbing submodule and \(a_1a_2\dots a_{n+1}m\in N\), then there are n of \(a_i\)’s whose product with m is in N, and hence, there are n of \(a_i\)’s whose product is in (N : m). Thus, (N : m) is n-absorbing ideal.

\((ii)\Leftarrow (i)\) Assume that (N : m) is a n-absorbing ideal of R for every \(m \in M-N\). let \(a_1, a_2, \dots , a_{n+1}\in R\) and \(m\in M\) with \(a_1a_2\dots a_{n+1}m\in N\). If \(m\in N\), we are done. Assume that \(m\notin N\), since (N : m) is a n-absorbing ideal and \(a_1a_2\dots a_{n+1}\in (N : m)\), then there are n of \(a_i\)’s whose product is in (N : m), and hence, there are n of \(a_i\)’s whose product with m is in N. Therefore, N is a classical n-absorbing submodule of M. \(\square \)

Theorem 3.6

Let M a cyclic R-module and N be a submodule of M. If N is a classical n-absorbing submodule, then N is an n-absorbing submodule of M.

Proof

Let \(M = Rm\) for some \(m\in M\). Suppose that \(a_1a_2\dots a_nx\in N\) for some \(a_1, a_2,\dots , a_n \in R\) and \(x\in M\). Then, there exists an element \(a_{n+1}\in R\), such that \(x = a_{n+1}m\). Therefore, \(a_1a_2\dots a_nx = a_1a_2\dots a_na_{n+1}m \in N\), and since N is a classical n-absorbing submodule, then there are n of \(a_i\)’s whose product with m is in N. Since M is cyclic, \((N : m) = (N : M)\); hence, there are n of \(a_i\)’s whose product with m is in N or \(a_1a_2\dots a_n \in (N : M)\). Thus, N is an n-absorbing submodule of M. \(\square \)

Now, in the following two corollaries, we characterize the classical n-absorbing submodules in terms of n-absorbing submodules and n-absorbing ideal.

Corollary 3.7

Let M a cyclic R-module and N be a submodule of M. Then, the followings are equivalent:
  1. (i)

    N is a classical n-absorbing submodule of M.

     
  2. (ii)

    N is an n-absorbing submodule of M.

     

Corollary 3.8

Let M a cyclic multiplication R-module and N be a submodule of M. Then, the followings are equivalent:
  1. (i)

    N is a classical n-absorbing submodule of M.

     
  2. (ii)

    (N : M) is an n-absorbing ideal of R.

     

Proof

Directly by Corollary 3.7 and Proposition 2.4 in [7]. \(\square \)

Here, in the next theorem, we investigate a submodule to be classical n-absorbing under some conditions.

Theorem 3.9

Let M an R-module and N be a proper irreducible submodule of M, such that \(N_r= N_{r^n}\) for all \(r\in R\), and then, N is a classical n-absorbing submodule of M.

Proof

Let \(r_1, r_2, \dots , r_{n+1} \in R\) and \(m\in N\) with \(r_1r_2\dots r_{n+1}m\in N\), and assume that N is not a classical n-absorbing submodule of M, and so, there are no n of \(a_i\)’s whose product with m is in N. We have \(N \subseteq \textstyle \bigcap \nolimits _{i=1}^{n} {(N + R{\hat{r}}_im)}\), where \({\hat{r}}_i = r_1r_2\dots r_{i-1}r_{i+1}\dots r_n\). Let \(x \in \textstyle \bigcap \nolimits _{i=1}^{n} {(N + R{\hat{r}}_im)}\), then \(x = x_1 + s_1{\hat{r}}_nm = x_2 + s_2{\hat{r}}_{n-1}m =\dots = x_n + s_n{\hat{r}}_1m\) where \(x_i \in N\) and \(s_i\in R\) for every i, then \(r_1^{n-1}x = r_1^{n-1}x_1 + s_1r_1^{n-1}{\hat{r}}_nm = r_1^{n-1}x_2 + s_2r_1^{n-1}{\hat{r}}_{n-1}m =\dots = r_1^{n-1}x_n + s_nr_1^{n-1}{\hat{r}}_1m\), since \({r_1}^{n-1}x_n, s_n{r_1}^{n-1}{\hat{r}}_1m\in N\), so \(s_1{r_1}^{n-1}{\hat{r}}_nm\in N\) which implies that \(s_1(r_2r_3\dots r_{n-1})m\in N_{{r_1}^{n}}\), but \(N_{{r_1}^n} = N_{r_1}\), and hence, \(s_1{\hat{r}}_nm\in N\), and so, \(x\in N\). Therefore, \(\textstyle \bigcap \nolimits _{i=1}^{n} {(N + R{\hat{r}}_im)} \subseteq N\); consequently, \(\textstyle \bigcap \nolimits _{i=1}^{n} {(N + R{\hat{r}}_im)} = N\), a contradiction, because N is an irreducible. Hence, N is a classical n-absorbing submodule of M. \(\square \)

Theorem 3.10

Let M an R-module and N be a classical n-absorbing submodule of M, such that (N : y) is a prime ideal of R for \(y\in M-N\). For \(x\in M\), if \((N : x) - \bigcup \nolimits _{x_i\in M-N} (N : x_i) \ne \phi \), then \(N = (N + Rx)\cap \bigcap \nolimits _{x_i\in M-N} (N + Rx_i)\).

Proof

Suppose that N is a classical n-absorbing submodule of M. Let \(a_1a_2\dots a_n\in (N : x) - \bigcup \nolimits _{x_i\in M-N} (N : x_i)\), where \(a_1, a_2, \dots , a_n\in R\), then \(a_1a_2\dots a_nx\in N\) and \(a_1a_2\dots a_nx_i\notin N\) for every \(x_i \in M-N\). It is Clear that \(N\subseteq (N + Rx)\cap \bigcap \nolimits _{x_i\in M-N} (N + Rx_i)\). For the reverse inclusion, let \(n\in (N + Rx)\cap \bigcap \nolimits _{x_i\in M-N} (N + Rx_i)\), then \(n = n' + r'x = n_i + r_ix_i\) for every \(x_i \in M - N\), where \(n', n_i\in N\) and \(r', r_i\in R\). Now, \(a_1a_2\dots a_nn = a_1a_2\dots a_nn' + a_1a_2\dots a_nr'x = a_1a_2\dots a_nn_i + a_1a_2\dots a_nr_ix_i\) and \(a_1a_2\dots a_nr'x, a_1a_2\dots a_nn', a_1a_2\dots a_nn_i\in N\), so \(a_1a_2\dots a_nr_ix_i\in N\). Since N is a classical n-absorbing submodule and \(a_1a_2\dots a_nx_i\notin N\), then there are \(n-1\) of \(a_i\)’s whose product with \(r_ix_i\) is in N. Hence, there are \(n-1\) of \(a_i\)’s whose product with \(r_i\) is in \((N : x_i)\). If \(x_i\in N\), then \(r_ix_i\in N\), and so \(n = n_i + r_ix_i \in N\). Assume that \(x_i\notin N\), so, by hypothesis, \((N : x_i)\) is a prime, and hence, either there are \(n-1\) of \(a_i\)’s whose product is in \((N : x_i)\) or \(r_i\in (N : x_i)\). From the first case, we have \(a_1a_2\dots a_nx_i\in N\) which is a contradiction. Therefore, \(r_i\in (N : x_i)\), and hence, \(r_ix_i\in N\). Thus, we have \(n = n_i + r_ix_i \in N\), so \((N + Rx)\cap \bigcap \nolimits _{x_i\in M-N} (N + Rx_i) \subseteq N\). Hence, \(N = (N + Rx)\cap \bigcap \nolimits _{x_i\in M-N} (N + Rx_i)\). \(\square \)

Corollary 3.11

Let M an R-module and N be a classical n-absorbing submodule of M, such that (N : y) is a prime ideal of R for \(y\in M-N\). For \(x\in M-N\), if \((N : x) - \bigcup \nolimits _{x_i\in M-N} (N : x_i) \ne \phi \), then N is not irreducible.

Proof

By Theorem 3.10, \(N = (N + Rx)\cap \bigcap \nolimits _{x_i\in M-N} (N + Rx_i)\). Since \(x\in M - N\), we have \(N\subset (N + Rx)\) and \(N\subset \bigcap \nolimits _{x_i\in M-N} (N + Rx_i)\). Thus, N is not irreducible. \(\square \)

Notes

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Authors and Affiliations

  1. 1.Department of MathematicsSakarya UniversitySakaryaTurkey

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