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# On classical n-absorbing submodules

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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 , 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  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 , almost 2-absorbing , almost 2-absorbing primary , and classical 2-absorbing . 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

 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

 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

 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 . $$\square$$

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. $$\square$$

## References

1. 1.
Anderson, D.F.; Badawi, A.: On n-Absorbing Ideal of Commutative Rings. Commun. Algebra 39, 1646–1672 (2011)
2. 2.
Ashour, A.E.; Al-Ashker, M.M.; Naji, O.A.: On almost 2-absorbing primary sub-modules. IUG J. Nat. Stud. (2017)Google Scholar
3. 3.
Ashour, A.E.; Al-Ashker, M.M.; Naji, O.A.: Some results on almost 2-absorbing sub-modules. J. Al Azhar Univ.-Gaza (Nat. Sci.) 18, 1–13 (2016)Google Scholar
4. 4.
Azizi, A.: On prime and weakly prime submodules. Vietnam J. Math. 36(3), 315–325 (2008)
5. 5.
Badawi, A.: On 2-absorbing ideals of commutative rings. Bull. Aust. Math. Soc. 75, 417–429 (2007)
6. 6.
Darani, A.; Soheilnia, F.: 2-Absorbing and weakly 2-absorbing submodules. Thai J. Math. 9(3), 577–584 (2011)
7. 7.
Darani, A.; Soheilnia, F.: On n-absorbing submodules. Math. Commun. 17, 547–557 (2012)
8. 8.
Dubey, M.; Aggarwal, P.: On 2-absorbing primary submodules of modules over commutative rings. Asian-Eur. J. Math. 8(4), 335–351 (2015)Google Scholar
9. 9.
Mostafanasab, H.; Tekir, U.; Oral, K.: Classical 2-absorbing submodules of modules over commutative rings. Eur. J. Pure Math. 8(3), 417–430 (2015)

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## Authors and Affiliations

1. 1.Department of MathematicsSakarya UniversitySakaryaTurkey