On the primeness of near-rings

Abstract

In this paper, we study the different kinds of the primeness on the class of near-rings and we give new characterizations for them. For that purpose, we introduce new concepts called set-divisors, ideal-divisors, etc. and we give equivalent statements for 3-primeness which make 3-primeness looks like the forms of the other kinds of primeness. Also, we introduce a new different kind of primeness in near-rings called K-primeness which lies between 3-primeness and e-primeness. After that, we study different kinds of prime ideals in near-rings and find a connection between them and new concepts called set-attractors, ideal-attractors, etc. to make new characterizations for them. Also, we introduce a new different kind of prime ideals in near-rings called K-prime ideals.

Citation

Al-Shaalan, K.H. (2020), "On the primeness of near-rings", Arab Journal of Mathematical Sciences, Vol. 26 No. 1/2, pp. 233-243. https://doi.org/10.1016/j.ajmsc.2019.12.004

1. Introduction

We say that $R$ is a right (left) near-ring if $(R,+)$ is a group, $(R,\cdot )$ is a semigroup and $R$ satisfies the right (left) distributive law. Throughout this paper, $R$ will be a left near-ring. We say that $R$ is an abelian near-ring if $x+y=y+x$ for all $x,y\in R$ and we say that $R$ is a commutative near-ring if $xy=yx$ for all $x,y\in R$. A zero-symmetric element is an element $x\in R$ satisfying $\text{0}x=\text{0}$. A near-ring $R$ is called a zero-symmetric near-ring, if $\text{0}x=\text{0}$ for all $x\in R$. A constant element is an element $y\in R$ satisfying $zy=y$ for all $z\in R$. An element $x\in R$ is called a right (left) zero divisor in $R$ if there exists a non-zero element $y\in R$ such that $yx=\text{0}$ ($xy=\text{0}$). A zero divisor is either a right or a left zero divisor. By a near-ring without zero divisors, we mean a near-ring without non-zero divisors of zero. If $A$ and $B$ are two non-empty subsets of $R$, then the product $AB$ means the set $\left\{ab|a\in A,b\in B\right\}$. We say that $U$ is a right (left) $R$-subgroup of $R$, if $U$ is a subgroup of $(R,+)$ satisfies $UR\subseteq U$ ($RU\subseteq U$). We say that $U$ is a two-sided $R$-subgroup of $R$, if $U$ is both a right and a left $R$-subgroup of $R$. We say that $I$ is a right (left) ideal of $R$, if $I$ is a normal subgroup of $(R,+)$ satisfies $(r+i)s-rs\in I$ for all $i\in I,r,s\in R$ ($RI\subseteq I$). We say that $I$ is an ideal of $R$ if it is both a right and a left ideal of $R$. We say that $U$ is a semigroup right (left) ideal of $R$, if $U$ is a non-empty subset of $R$ satisfies $UR\subseteq U$ ($RU\subseteq U$). We say that $U$ is a semigroup ideal of $R$ if it is both a semigroup right and left ideal of $R$ (some authors call $U$ a right (left, two-sided) $R$-subset of $R$ [8]). For any group $(G,+)$, $M(G)$ denotes the near-ring of all maps from $G$ to $G$ with the two operations of addition and composition of maps. $M_o(G)$ is the zero-symmetric subnear-ring of $M(G)$ consisting of all zero preserving maps from $G$ to itself (and to make them left near-rings we should write $f(g)$ by $gf$, where $f\in M(G)$ or $M_o(G)$ and $g\in G$). A trivial zero-symmetric near-ring $R$ is a zero-symmetric near-ring such that the multiplication on the group $(R,+)$ is defined by $xy=y$ and $\text{0}y=\text{0}$ for all $x\in R-\{\text{0}\},y\in R$. A near-field $N$ is a near-ring in which $(N-\{\text{0}\},\cdot )$ is a group. For further information about near-rings, see [8] and [9].

In near-rings, there are five well-known kinds of primeness. We say that: $R$ is 0-prime (the usual primeness) if, for every two ideals $I$ and $J$ of $R$, $IJ=\{\text{0}\}$ implies $I=\{\text{0}\}$ or $J=\{\text{0}\}$, $R$ is 1-prime if, for every two right ideals $K$ and $L$ of $R$, $KL=\{\text{0}\}$ implies $K=\{\text{0}\}$ or $L=\{\text{0}\}$. $R$ is 2-prime if, for every two right $R$-subgroups $A$ and $B$ of $R$, $AB=\{\text{0}\}$ implies $A=\{\text{0}\}$ or $B=\{\text{0}\}$. $R$ is 3-prime if, for all $x,y\in R$, $xRy=\{\text{0}\}$ implies $x=\text{0}$ or $y=\text{0}$ and $R$ is equiprime (e-prime) if, for any $\text{0}\ne a,x,y\in R$, $xca=yca$ for all $c\in R$ implies $x=$ $y$. These five kinds of primeness are equivalent in the class of rings. But in the class of near-rings, we have: (1) $R$ is equiprime implies that $R$ is zero-symmetric 3-prime, (2) $R$ is 3-prime implies that $R$ is 2-prime, (3) $R$ is zero-symmetric 2-prime implies that $R$ is 1-prime and (4) $R$ is 1-prime implies that $R$ is 0-prime. For details about these kinds and their examples and relationships see [1–3,5–7] and [10]. A near-ring (a ring) $R$ is called 3-semiprime (semiprime) if, for all $x\in R$, $xRx=\{\text{0}\}$ implies $x=\text{0}$. An ideal $P$ of $R$ is: (i) a 0-prime ideal of $R$ if for every two ideals $A$ and $B$ of $R$, $AB\subseteq P$ implies that $A\subseteq P$ or $B\subseteq P$, (ii) a 1-prime ideal of $R$ if for every two right ideals $A$ and $B$ of $R$, $AB\subseteq P$ implies that $A\subseteq P$ or $B\subseteq P$, (iii) a 2-prime ideal of $R$ if for every two right $R$-subgroups $A$ and $B$ of $R$, $AB\subseteq P$ implies that $A\subseteq P$ or $B\subseteq P$, (iv) a 3-prime ideal of $R$ if for $a,b\in R$, $aRb\subseteq P$ implies that $a\in P$ or $b\in P$, (v) an e-prime (equiprime) ideal of $R$ if for every $a\in R-P$ and $x,y\in R$, $xca-yca\in P$ for all $c\in R$ implies that $x-y\in P$. Clearly that any near-ring is a $\upsilon$-prime ideal of itself, where $\upsilon \in \left\{\text{0},\mathrm{1,2,3},e\right\}$. It is well-known that (ii) implies (i) and (iv) implies (iii). Also, for zero-symmetric near-rings we have (iii) implies (ii). An ideal $I$ of $R$ is called completely prime if, for $a,b\in R$, $ab\in I$ implies that $a\in I$ or $b\in I$. If the zero ideal is completely prime, then we say that $R$ is completely prime. Then $R$ is completely prime if and only if $R$ is without zero divisors. For more details about prime ideals, see [2,4,5] and [10].

In [1], the authors gave us a short historical view about the primeness of near-rings. We will use it and add some information to it.

Several different generalizations of primeness for rings have been introduced for near-rings. In [6], Holcombe studied three different concepts of primeness, which he called 0-prime, 1-prime and 2-prime. In [5], Groenewald obtained further results for these and introduced further notion which he called 3-primeness. In [2], Booth, Groenewald and Veldsman gave another definition, called equiprimeness, or e-primeness. In [10], Veldsman made more studies on equiprime near-rings. In [1], Booth and Groenewald gave an element-wise characterization of the radical associated with $\nu$ -primeness for $\nu =\mathrm{1,2,3},e$.

In this paper we extend the idea of primeness that they did and give some new results for the primeness of near-rings. Firstly, we introduce new concepts called set-divisors, ideal divisors, etc. These concepts are generalizations of the concept of zero divisors and give another characterization of different kinds of the primeness in near-rings and hence in rings. Also, we study the 3-primeness and give new characterizations of 3-prime (3-semiprime) near-rings and hence for prime (semiprime) rings. These characterizations make 3-primeness looks like the forms of the other kinds of primeness. In fact, we show that a near-ring (a ring) is 3-prime (prime) if and only if $UV=\{\text{0}\}$ implies $U=\{\text{0}\}$ or $V=\{\text{0}\}$, where $U$ and $V$ are semigroup left ideals of $R$. Hence, a ring is prime if and only if it is without zero-semigroup right (left) ideal divisors. A similar result is made for 3-semiprime near-rings (semiprime rings) and we conclude that: for a near-ring $R$, if $r^2\ne \text{0}$ for all $r\in R-\{\text{0}\}$, then $R$ is 3-semiprime. We show that some kinds of near-rings are 3-prime if and only if they are 2-prime. Also, we introduce a new kind of primeness in near-rings (the sixth one) called K-primeness and we show that it is totally different from the other kinds of primeness and it lies between 3-primeness and e-primeness. Depending on that, we give two chains of primeness in the class of zero-symmetric near-rings for comparison. In the last part of the paper, we study different kinds of prime ideals. We introduce a new kind of prime ideals called K-prime ideals and we show that they are different from the other kinds of prime ideals. they lie between 3-prime ideal and e-prime ideals. Also, we give a new characterization of 3-prime ideals and show that $P$ is a 3-prime ideal of $R$ if and only if $UV\subseteq P$ implies $U\subseteq P$ or $V\subseteq P$, where $U$ and $V$ are semigroup left ideals of $R$. We introduce new concepts called set-attractors, ideal-attractors, etc. which are generalizations of the new concepts above (set-divisors, etc.). We make a connection between these concepts and different kinds of prime ideals in near-rings to give a new characterization of these prime ideals. Finally, we use these concepts to show that: $P$ is a completely prime ideal of $R$ if and only if $R$ is without external $P$ set-attractors.

2. On prime near-rings

Let $R$ be a near-ring. It is clear that $R$ is without zero divisors if and only if $AB=\{\text{0}\}$ implies $A=\{\text{0}\}$ or $B=\{\text{0}\}$, where $A$ and $B$ are non-empty subsets of $R$. This observation gives us a hint of a new definition.

Definition 2.1. Let $R$ be a non-zero near-ring.

• (1) Let $A$ be a non-empty subset of $R$. We say that $A$ is a left zero-set divisor (a right zero-set divisor) of $R$ if there exists a non-empty non-zero subset $B$ of $R$ such that $AB=\{\text{0}\}$ ($BA=\{\text{0}\}$). We say that $A$ is a zero-set divisor of $R$ if $A$ is a left or a right zero-set divisor of $R$.

• (2) Let $A$ be an ideal of $R$. We say that $A$ is a left zero-ideal divisor (a right zero-ideal divisor) of $R$ if there exists a non-zero ideal $B$ of $R$ such that $AB=\{\text{0}\}$ ($BA=\{\text{0}\}$). We say that $A$ is a zero-ideal divisor of $R$ if $A$ is a left or a right zero-ideal divisor of $R$.

We can do same definitions if $A$ is a left (right) ideal, a left (right) $R$ -subgroup, a two-sided $R$-subgroup, a semigroup left (right) ideal or a semigroup ideal.

Definition 2.1 generalizes the concept of zero divisors in rings and near-rings. So, we have the following remark.

Remark 2.1. From Definition 2.1, we can rewrite the definitions of different kinds of the primeness as follows:

Let $R$ be a near-ring. Then

• (1) $R$ is completely prime if and only if $R$ is without zero divisors if and only if $R$ is without zero-set divisors.

• (2) $R$ is 0-prime if and only if $R$ is without zero-ideal divisors.

• (3) $R$ is 1-prime if and only if $R$ is without zero-right ideal divisors.

• (4) $R$ is 2-prime if and only if $R$ is without zero-right $R$-subgroup divisors.

Remark 2.1 enhances a question: Can we get a definition of 3-primeness like that mentioned in Remark 2.1? The following result answers this question.

Theorem 2.1.Let $R$ be a near-ring. Then the following statements are equivalent:

• (i)$R$ is 3-prime.

• (ii)$aU=\{\text{0}\}$ implies $a=\text{0}$ or $U=\{\text{0}\}$, where $a\in R$ and $U$ is a semigroup left ideal of $R$.

• (iii)$AU=\{\text{0}\}$ implies $A=\{\text{0}\}$ or $U=\{\text{0}\}$, where $A$ is a non-empty subset of $R$ and $U$ is a semigroup left ideal of $R$.

• (iv)$UV=\{\text{0}\}$ implies $U=\{\text{0}\}$ or $V=\{\text{0}\}$, where $U$ and $V$ are semigroup left ideals of $R$.

Proof. (i) implies (ii), (ii) implies (iii) and (iii) implies (iv) are clear.

To prove that (iv) implies (i), we will use the contradiction. For that purpose, suppose $R$ is not 3-prime. So there exist non-zero elements $x,y\in R$ such that $xRy=\{\text{0}\}$. Thus, $RxRy=\{\text{0}\}$. But $Rx$ and $Ry$ are semigroup left ideals of $R$, so $Rx=\{\text{0}\}$ or $Ry=\{\text{0}\}$ by (iv). Hence, $R\left\{\text{0},x\right\}=\{\text{0}\}$ or $R\left\{\text{0},y\right\}=\{\text{0}\}$ and either $\left\{\text{0},x\right\}$ or $\left\{\text{0},y\right\}$ is a semigroup left ideal of $R$. But $R$ is also a semigroup left ideal of $R$. Thus, $\left\{\text{0},x\right\}=\text{0}$, $\left\{\text{0},y\right\}=\{\text{0}\}$ or $R=\{\text{0}\}$ by (iv), a contradiction with that $x,y,R$ are all non-zero. So $R$ is 3-prime and (iv) implies (i). ■

For zero-symmetric near-rings, we have the following extra result.

Theorem 2.2. Let $R$ be a zero-symmetric near-ring. Then the following statements are equivalent:

• (i)$R$ is 3-prime.

• (ii)$Ua=\{\text{0}\}$ implies $a=\text{0}$ or $U=\{\text{0}\}$, where $a\in R$ and $U$ is a semigroup right ideal of $R$.

• (iii)$UA=\{\text{0}\}$ implies $U=\{\text{0}\}$ or, $A=\{\text{0}\}$ where $U$ is a semigroup right ideal of $R$ and $A$ is a non-empty subset of $R$.

• (iv)$UV=\{\text{0}\}$ implies $U=\{\text{0}\}$ or$V=\{\text{0}\}$, where $U$ and $V$ are semigroup right ideals of $R$.

• (v)$UV=\{\text{0}\}$ implies $U=\{\text{0}\}$ or$V=\{\text{0}\}$, where $U$ is a semigroup right ideal of $R$ and $V$ is a semigroup left ideal of $R$.

Now, we can add (5) to Remark 2.1:

• (5)$R$ is 3-prime if and only if $UV=\{\text{0}\}$ implies $U=\{\text{0}\}$ or $V=\{\text{0}\}$, where $U$ and $V$ are semigroup left ideals of $R$ if and only if $R$ is without zero-semigroup left ideal divisors.

Since any ring is a zero-symmetric near-ring, we have the following result:

Corollary 2.3. A ring is prime if and only if it is without zero-semigroup right (left) ideal divisors.

Using the same idea, the following result gives us a result for 3-semiprime zero-symmetric near-rings.

Theorem 2.4. Let $R$ be a zero-symmetric near-ring. Then the following statements are equivalent:

• (i)$R$ is 3-semiprime.

• (ii)$aU=\{\text{0}\}$ implies $a=\text{0}$, where $a\in U$ and $U$ is a semigroup left ideal of $R$.

• (iii)$Ua=\{\text{0}\}$ implies $a=\text{0}$, where $a\in U$ and $U$ is a semigroup right ideal of $R$.

• (iv)$U^2=\{\text{0}\}$ implies $U=\{\text{0}\}$, where $U$ is a semigroup left ideal of $R$.

• (v)$U^2=\{\text{0}\}$ implies $U=\{\text{0}\}$, where $U$ is a semigroup right ideal of $R$.

Proof. (i) implies (ii). Suppose (i) holds. Let $U$ be a semigroup left ideal of $R$ such that $aU=\{\text{0}\}$, where $a\in U$. Then for all $v\in U$, we have $aRv=\{\text{0}\}$. Thus, $aRa=\{\text{0}\}$ and $a=\text{0}$ by (i).

(i) implies (iii) can be proved by the same way.

(ii) implies (iv) and (iii) implies (v) are clear.

(iv) implies (v). Suppose that (iv) holds and $U^2=\{\text{0}\}$, where $U$ is a semigroup right ideal of $R$. So $uRu=\{\text{0}\}$ for all $u\in U$ and hence $RuRu=\{\text{0}\}$. But $Ru$ is a semigroup left ideal of $R$. So $Ru=\{\text{0}\}$ for all $u\in U$ by (iv). So $\left\{\text{0},u\right\}$ is a semigroup left ideal of $R$ and $\left\{\text{0},u\right\}\left\{\text{0},u\right\}=\{\text{0}\}$ for all $u\in U$. So $u=\text{0}$ by (iv) and hence $U=\{\text{0}\}$.

(v) implies (i). Suppose that (v) holds and that $xRx=\{\text{0}\}$ for some $x\in R$. Thus, $xRxR=\{\text{0}\}$. But $xR$ is a semigroup right ideal of $R$, so $xR=\{\text{0}\}$ by (v). Hence, $\left\{\text{0},x\right\}\left\{\text{0},x\right\}=\{\text{0}\}$. But $\left\{\text{0},x\right\}$ is a semigroup right ideal of $R$. Thus, $\left\{\text{0},x\right\}=\{\text{0}\}$ by (v) and hence $x=\text{0}$. So $R$ is 3-semiprime and (v) implies (i). ■

Corollary 2.5. A ring $R$ is semiprime if and only if $U^2=\{\text{0}\}$ implies $U=\{\text{0}\}$, where $U$ is a semigroup right (left) ideal of $R$.

But in the general case of 3-semiprime near-rings, we have only the following result.

Theorem 2.6. Let $R$ be a near-ring. Then the following statements are equivalent:

• (i)$R$ is 3-semiprime.

• (ii)$aU=\{\text{0}\}$ implies $a=\text{0}$, where $a\in U$ and $U$ is a semigroup left ideal of $R$.

• (iii)$U^2=\{\text{0}\}$ implies $U=\{\text{0}\}$, where $U$ is a semigroup left ideal of $R$.

Unfortunately, we cannot remove the word “zero-symmetric” in Theorems 2.2 and 2.4. The following example is the near-ring in [9, Appendix, E, 22] and it shows that the condition “zero-symmetric” in Theorems 2.2 and 2.4 is not redundant.

Example 1. Let $(R,+)$ be the Klein’s four group $\left\{\text{0},a,b,c\right\}$. Then it is an abelian group such that $x+x=\text{0}$ for all $x\in R$ and $x+y=z$ for all different non-zero elements $x,y,z\in R$. Define the multiplication on $R$ as follows:

$$\begin{array}{cc}\cdot & \text{mtd}\text{mtd}\text{mtd}\text{mtd}\\ 0& \text{mtd}\text{mtd}\text{mtd}\text{mtd}\\ a& \text{mtd}\text{mtd}\text{mtd}\text{mtd}\\ b& \text{mtd}\text{mtd}\text{mtd}\text{mtd}\\ c& \text{mtd}\text{mtd}\text{mtd}\text{mtd}\end{array}$$

Clearly $R$ is an abelian non-zero-symmetric near-ring. The only semigroup right ideals of $R$ are $R$, $\left\{\text{0},a\right\}$ and $\left\{\text{0},a,b\right\}$. So $R$ satisfies the conditions “$UV=\{\text{0}\}$ implies $U=\{\text{0}\}$ or $V=\{\text{0}\}$, where $U$ and $V$ are semigroup right ideals of $R$” and “$U^2=\{\text{0}\}$ implies $U=\{\text{0}\}$, where $U$ is a semigroup right ideal of $R$”. But $R$ is not 3-semiprime as $bRb=\{\text{0}\}$. From Theorem 2.6, we can deduce that there is a non-zero semigroup left ideal $V$ of $R$ such that $V^2=\{\text{0}\}$ and $\upsilon V=\{\text{0}\}$, where $\upsilon \in V-\{\text{0}\}$. It is easy to find out that $V=\left\{\text{0},b\right\}$ and $v=b$.

From the above example, observe that

$$\left\{\text{0},a,b\right\}b=\left\{\text{0},a,b\right\}\{b\}=\{\text{0}\}\text{.}$$

So, we cannot use this example for (ii) or (iii) in Theorem 2.2 and for (iii) in Theorem 2.4. In fact, removing “zero-symmetric” from those parts is an open problem.

Corollary 2.7. Let $R$ be a near-ring. If $r^2\ne \text{0}$ for all $r\in R-\{\text{0}\}$, then $R$ is 3-semiprime.

Proof. Suppose there exists a non-zero semigroup left ideal $U$ of $R$ such that $aU=\{\text{0}\}$,

where $a\in U$. That means $a^2=\text{0}$. By hypothesis, $a=\text{0}$ and hence $R$ is 3-semiprime. ■

Example 2. Let $R={\mathrm{\mathbb{Z}}}_6$. Then $R$ is semiprime since $r^2\ne \text{0}$ for all $r\in R-\{\text{0}\}$.

Example 3. Let $R=\left\{\text{0},\mathrm{2,4,6,8,10,12}\right\}$ the subring of ${\mathrm{\mathbb{Z}}}_{14}$. Then $R$ is semiprime since $r^2\ne \text{0}$ for all $r\in R-\{\text{0}\}$.

The converse of Corollary 2.7 is not true as the following example shows.

Example 4.Let $R=M_2({\mathrm{\mathbb{Z}}}_2)$. Then $R$ is a prime ring and hence semiprime, but

$$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

For commutative near-rings, we have the converse and we get the following result.

Corollary 2.8. Let $R$ be a commutative near-ring. Then $r^2\ne \text{0}$ for all $r\in R-\{\text{0}\}$ if and only if $R$ is 3-semiprime.

We conclude this section by the following results about the relation between 2-primeness and 3-primeness. The fact that $R$ is 3-prime implies $R$ is 2-prime is well-known. The following results have the converse.

Theorem 2.9. Let $R$ be a zero-symmetric near-ring such that $2R=\{\text{0}\}$. Then $R$ is 3-prime if and only if $R$ is 2-prime.

Proof. Suppose that $xRy=\{\text{0}\}$. Thus, $xRyR=\{\text{0}\}$. But $xR$ and $yR$ are right $R$-subgroups of $R$. So $xR=\{\text{0}\}$ or $yR=\{\text{0}\}$ as $R$ is 2-prime. Hence, $\left\{\text{0},x\right\}R=\{\text{0}\}$ or $\left\{\text{0},y\right\}R=\{\text{0}\}$ and then either $\left\{\text{0},x\right\}$ or $\left\{\text{0},y\right\}$ is a right $R$-subgroup of $R$. But $R$ is also a right $R$-subgroup of $R$. Thus, $\left\{\text{0},x\right\}=\text{0}$, $\left\{\text{0},y\right\}=\{\text{0}\}$ or $R=\{\text{0}\}$. Hence, $x=\text{0}$ or $y=\text{0}$ and $R$ is 3-prime. ■

Theorem 2.10 Any distributive near-ring $R$ is 3-prime if and only if it is 2-prime.

Proof. Suppose that $R$ is 2-prime and $xRy=\{\text{0}\}$ for some $x,y\in R$. So $xRyR=\{\text{0}\}$ and hence $xR=\{\text{0}\}$ or $yR=\{\text{0}\}$. So $AR=\{\text{0}\}$ or $BR=\{\text{0}\}$, where $A=\left\{nx|n\in \mathrm{\mathbb{Z}}\right\}$ and $B=\left\{ny|n\in \mathrm{\mathbb{Z}}\right\}$. So $A$ and $B$ are right $R$-subgroups of $R$ and hence $A=\{\text{0}\}$ or $B=\{\text{0}\}$. Therefore, $x=\text{0}$ or $y=\text{0}$ and $R$ is 3-prime. ■

3. K-prime near-rings

In this section, we will introduce a new kind of primeness of near-rings called K-primeness. Firstly, we will begin with the following result.

Theorem 3.1.Let $R$ be a ring. Then the following statements are equivalent:

• (i)$R$ is prime.

• (ii)for any $\text{0}\ne a,x,y\in R$, $xsa=yra$ for all $s,r\in R-\{\text{0}\}$ implies $x=$ $y$.

Proof.A ring $R$ is prime if and only if it is equiprime, so we will use the definition of equiprimeness, i.e. for any $\text{0}\ne a,x,y\in R$, $xca=yca$ for all $c\in R$ implies $x=$ $y$.

(i) implies (ii) is clear.

(ii) implies (i). Suppose (ii) holds. If for all $c\in R$, $xca=yca$ for $\text{0}\ne a,x,y\in R$, then $(x-y)ca=\text{0}=\text{0}ra$ for all $c,r\in R$. So $x=y$ by (ii). ■

Part (ii) enhances the following definition for near-rings.

Definition 3.1.Let $R$ be a near-ring. We say that $R$ is K-prime if, for any $\text{0}\ne a,x,y\in R$, $xsa=yra$ for all $s,r\in R-\{\text{0}\}$ implies $x=$ $y$.

As we mentioned before for rings, a ring is prime if and only if it is equiprime. So we have the following result.

Corollary 3.2.A ring $R$ is prime if and only if it is K-prime.

The following result shows that every K-prime near-ring is zero-symmetric 3-prime.

Theorem 3.3.Let $R$ be a K-prime near-ring. Then $R$ is zero-symmetric 3-prime.

Proof.Firstly, we will show that $R$ is zero-symmetric. If $R$ is not zero-symmetric, then it has at least one non-zero constant element $c$ (see [8, Theorem 1.15). For different elements $x,y$ of $R$, we have that $xsc=yrc=c$ for all $s,r\in R-\{\text{0}\}$, a contradiction with the hypothesis. So $R$ is zero-symmetric. Now, suppose $xRy=\{\text{0}\}$ for some $x,y\in R$. So $xcy=\text{0}$ for all $c\in R$. If $y\ne \text{0}$, then $xcy=\text{0}ry$ for all $c,r\in R$. So $x=\text{0}$ from the hypothesis and hence $R$ is 3-prime. ■

In the case of near-rings, we have only that e-primeness implies K-primeness as shown in the proof of Theorem 3.1 (since an e-prime near-ring is zero-symmetric [10]). But the converse is not true as we will show in the next example. We will use the near-ring mentioned in [9, Appendix, F, 7] in the next example.

Example 5. Let $(R,+)$ be the cyclic group ${\mathrm{\mathbb{Z}}}_5$ and define the multiplication on $R$ as follows:

$$\begin{array}{cc}\cdot & \text{mtd}\text{mtd}\text{mtd}\text{mtd}\text{mtd}\\ 0& \text{mtd}\text{mtd}\text{mtd}\text{mtd}\text{mtd}\\ 1& \text{mtd}\text{mtd}\text{mtd}\text{mtd}\text{mtd}\\ 2& \text{mtd}\text{mtd}\text{mtd}\text{mtd}\text{mtd}\\ 3& \text{mtd}\text{mtd}\text{mtd}\text{mtd}\text{mtd}\\ 4& \text{mtd}\text{mtd}\text{mtd}\text{mtd}\text{mtd}\end{array}$$

So $R$ is an abelian near-ring which is not a ring (as $(1+1)2=3\ne 4=2+2=(1)2+(1)2$). Clearly that $R$ is without zero divisors. Hence, $R$ is 3-prime. $R$ is not equiprime. Indeed, $1c1=3c1=c1$ for all $c\in R$. But if $\text{0}\ne a,x,y\in R$ such that $xsa=yra$ for all $s,r\in R-\{\text{0}\}$, then $x=$ $y$. Clearly that is true if $x$ or $y$ is equal to zero, since $R$ is without zero divisors. That is the only possible case. In fact, if $xsa=yra$ for all $s,r\in R-\{\text{0}\}$ and $x,y,a$ are all non-zero, then from the table we can choose $s_o,r_o\in R-\{\text{0}\}$ to satisfy that $x{s}_o=1$ and $y{r}_o=2$. Hence, $a=2a$ which implies that $a=\text{0}$ (from the table), a contradiction with $\text{0}\ne a$. Therefore, K-primeness does not imply e-primeness.

Also, we can find zero-symmetric 3-prime near-rings which are not K-prime, as the following example shows.

Example 6. Let $R$ be a trivial zero-symmetric near-ring of order greater than 2. Clearly $R$ is 3-prime. Taking two non-zero elements $x$ and $y$ such that $x\ne y$, we have $xsx=yrx=x$ for all $s,r\in R-\{\text{0}\}$. So $R$ is not K-prime.

Theorem 3.1, Theorem 3.3 and the examples after them show that K-primeness is a new kind of primeness.

Observe that K-primeness lies between 3-primeness and e-primeness (equiprimeness). So we have the following chain of primeness in the class of zero-symmetric near-rings:

$$\begin{array}{l}\text{The}\text{class}\text{of}\text{e}‐\text{prime}\text{near}‐\text{rings}\\ \subseteq \text{The}\text{class}\text{of}\text{K}‐\text{prime}\text{near}‐\text{rings}\\ \subseteq \text{The}\text{class}\text{of}3‐\text{prime}\text{near}‐\text{rings}\\ \subseteq \text{The}\text{class}\text{of}2‐\text{prime}\text{near}‐\text{rings }\\ \subseteq \text{The}\text{class}\text{of}1‐\text{prime}\text{near}‐\text{rings}\\ \subseteq \text{The}\text{class}\text{of}\text{0}‐\text{prime}\text{near}‐\text{rings }\end{array}$$

Remark 3.1.Observe that:

(i)It is well-known that $M_o(G)$ is e-prime (see [10]) and hence K-prime. Observe that it has zero divisors.

(ii)Since $M(G)$ is not zero-symmetric, so it is not K-prime (and hence not e-prime), but it has zero divisors.

(iii)Let $N$ be any near-field. Then $N$ is e-prime and hence K-prime. Indeed, for any $\text{0}\ne a,x,y\in R$ such that $xca=yca$ for all $c\in R$, we have that $x=y$ by choosing $c=a^{-1}$. Observe that $N$ is without zero divisors.

(iv)Example 6 shows a 3-prime near-ring without zero divisors which is not K-prime (and hence not e-prime).

From the above parts in Remark 3.1, there is no relation between e-primeness (K-primeness) and the existence of zero divisors in near-rings. So, we have another chain of the primeness in the class of zero-symmetric near-rings:

$$\begin{array}{l}\text{The}\text{class}\text{of}\text{completely}\text{prime}\text{near}‐\text{rings}\\ \subseteq \text{The}\text{class}\text{of}3‐\text{prime}\text{near}‐\text{rings}\\ \subseteq \text{The}\text{class}\text{of}2‐\text{prime}\text{near}‐\text{rings}\\ \subseteq \text{The}\text{class}\text{of}1‐\text{prime}\text{near}‐\text{rings}\\ \subseteq \text{The}\text{class}\text{of}\text{0}‐\text{prime}\text{near}‐\text{rings}\end{array}$$

4. On prime ideals

The next definition introduces K-prime ideals.

Definition 4.1.Let $R$ be a near-ring and $P$ an ideal of $R$. Then $P$ is a K-prime ideal of $R$ if for every $a\in R-P$ and $x,y\in R$, $xra-ysa\in P$ for all $r,s\in R-P$ implies $x-y\in P$.

Clearly $R$ is K-prime if and only if $\{\text{0}\}$ is a K-prime ideal of $R$.

The relationship between K-prime ideals and other kinds of prime ideals is stated in the following result.

Theorem 4.1.Let $R$ be a near-ring with an ideal $P$.

• (i)If $P$ is a K-prime ideal of $R$, then $P$ is a 3-prime ideal of $R$.

• (ii)If $P$ is an e-prime ideal of $R$, then $P$ is a K-prime ideal of $R$.

Proof. (i) Firstly, we will show that $P$ contains all the constant elements of $R$. Let $c$ be a constant element in $R$. If $c\in R-P$, then

$$xrc-ysc=c-c=\text{0}\in P$$

for all $x,y\in R$ and $r,s\in R-P$. So $x-y\in P$ and hence $x-\text{0}=x\in P$ for all $x\in R$. Thus, $P=R$, a contradiction with $c\notin P$. So $c\in P$.

Now, suppose $aRb\subseteq P$ for some $a,b\in R$ and $b\notin P$. From above, any element $s\in R-P$ is a zero-symmetric element. So $\text{0}sb=\text{0}\in P$ for all $s\in R-P$. So $arb-\text{0}sb\in P$ for all $r,s\in R-P$. Thus, $a\in P$ by the hypothesis and $P$ is 3-prime.

(ii)Firstly, observe that if $r\in P$ and $s\in R$ is a zero-symmetric element, then

$$rs=(r+\text{0})s-\text{0}s\in P\text{.}$$

Suppose $xra-ysa\in P$ for all $r,s\in R-P$, where $a\in R-P$ and $x,y\in R$. So $xca-yca\in P$ for all $c\in R-P$. Now, suppose $c\in P$. As $a\notin P$, we have that $a$ is a zero-symmetric element (see [10]). So $ca\in P$ and hence $xca-yca\in P$. But $P$ is e-prime. So $x-y\in P$ and $P$ is a K-prime ideal of $R$. ■

The next result generalizes Theorem 2.1 for 3-prime ideals.

Theorem 4.2. Let $R$ be a near-ring and $P$ an ideal of $R$. Then the following statements are equivalent:

(i)$P$ is a 3-prime ideal of $R$.

(ii)$BU\subseteq P$ implies $B\subseteq P$ or $U\subseteq P$, where $B$ is a non-empty subset of $R$ and $U$ is a semigroup left ideal of $R$.

(iii)$UV\subseteq P$ implies $U\subseteq P$ or $V\subseteq P$, where $U$ and $V$ are semigroup left ideals of $R$.

Proof.(i) implies (ii). Suppose (i) holds. Let $U$ be a semigroup left ideal of $R$ and $B$ be a non-empty subset of $R$ such that $BU\subseteq P$. If $B⊈P$, then there exists $b\in B-P$ such that $bRu\subseteq P$ for all $u\in U$. Thus, $U\subseteq P$ by (i).

(ii) implies (iii) is clear.

(iii) implies (i). To prove it, we will use the contradiction. Suppose that (iii) holds and $P$ is not a 3-prime ideal. So there exist $x,y$ $R-P$ such that $xRy\subseteq P$. Thus, $RxRy\subseteq P$. So $Rx\subseteq P$ or $Ry\subseteq P$ by (iii). Hence, $R(P\hspace{0.17em}\cup \hspace{0.17em}\{x\})\subseteq P$ or $R(P\hspace{0.17em}\cup \hspace{0.17em}\{y\})\subseteq P$ and then $P\hspace{0.17em}\cup \hspace{0.17em}\{x\}$ or $P\hspace{0.17em}\cup \hspace{0.17em}\{y\}$ is a semigroup left ideal of $R$. But $R$ itself is also a semigroup left ideal of $R$. Thus, $P\hspace{0.17em}\cup \hspace{0.17em}\{x\}\subseteq P$, $P\hspace{0.17em}\cup \hspace{0.17em}\{y\}\subseteq P$ or $R\subseteq P$ by (iii), a contradiction with that $x,y\in R-P$. So $P$ is 3-prime and (iii) implies (i). ■

Remark 4.1.From Theorem 4.2, a new characterization of 3-prime ideals can be written as follows:

(*) $P$ is a 3-prime ideal of $R$ if for every two semigroup left ideals $A$ and $B$ of $R$, $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$.

Using Theorem 4.2 and its proof, we can prove the following result which generalizes Theorem 2.2 for 3-prime ideals.

Theorem 4.3.Let $R$ be a zero-symmetric near-ring and $P$ an ideal of $R$. Then the following statements are equivalent:

• (i)$P$ is a 3-prime ideal of $R$.

• (ii)$UB\subseteq P$ implies $U\subseteq P$ or $B\subseteq P$, where $U$ is a semigroup right ideal of $R$ and $B$ is a non-empty subset of $R$.

• (iii)$UV\subseteq P$ implies $U\subseteq P$ or $V\subseteq P$, where $U$ and $V$ are semigroup right ideals of $R$.

We cannot eliminate the condition “zero-symmetric” in Theorem 4.3 as the following example shows:

Example 7.Observe that $\{\text{0}\}$ is not a 3-prime ideal in Example 1 although it satisfies the condition “If $UV\subseteq \{\text{0}\}$, then $U\subseteq \{\text{0}\}$ or $V\subseteq \{\text{0}\}$, where $U$ and $V$ are semigroup right ideals of $R$”. This shows that “zero-symmetric” in Theorem 4.3 is not redundant.

Now, we would like to generalize Definition 2.1.

Definition 4.2.Let $R$ be a near-ring with an ideal $I$.

• (i)Let $A$ be a non-empty subset of $R$. We say that $A$ is a left $I$ set-attractor (a right $I$ set-attractor) of $R$ if there exists a non-empty subset $B$ of $R$ and $B⊈I$ such that $AB\subseteq I$ ($BA\subseteq I$). We say that $A$ is an $I$ set-attractor of $R$ if $A$ is a left or a right $I$ set-attractor of $R$.

• (ii)Let $A$ be an ideal of $R$. We say that $A$ is a left $I$ ideal-attractor (a right $I$ ideal-attractor) of $R$ if there exists an ideal $B$ of $R$ and $B⊈I$ such that $AB\subseteq I$ ($BA\subseteq I$). We say that $A$ is an $I$ ideal-attractor of $R$ if $A$ is a left or a right $I$ ideal-attractor of $R$.

We can do the same definitions if $A$ is a left (right) ideal of $R$, a left (right, two-sided) $R$-subgroup of $R$, a semigroup ideal of $R$ or a semigroup left (right) ideal of $R$.

Example 8.Let $R$ be a near-ring with an ideal $I\ne R$. Any non-empty subset of $I$ is a right $I$ set-attractor of $R$ and hence an $I$ set-attractor of $R$. In particular, $I$ is an $I$ set-attractor of $R$. Also, if there exist an ideal (a left (right) ideal, a left $R$-subgroup, a semigroup left ideal) $B$ of $R$ such that $B⊈I$, then $I$ is an $I$ ideal-attractor ($I$ left (right) ideal-attractor, $I$ left $R$-subgroup-attractor, $I$ semigroup left ideal-attractor) of $R$.

Definition 4.3.Let $R$ be a near-ring with an ideal $P$. If $A$ is a $P$ set-attractor ($P$ ideal-attractor, etc.) of $R$, then we say that $A$ is an internal $P$ set-attractor ($P$ ideal-attractor, etc.) of $R$ if $A\subseteq P$. If $A⊈P$, then we say that $A$ is an external $P$ set-attractor ($P$ ideal-attractor, etc.) of $R$. If $R$ does not have any external $P$ set-attractors ($P$ ideal-attractors, etc.), then we say that $R$ is without external $P$ set-attractors ($P$ ideal-attractors, etc.), i.e. for a $P$ set-attractor ($P$ ideal-attractor, etc.) $A$ of $R$, we have that $A\subseteq P$

Example 9. (i) Any near-ring $R$ is without external (or internal) $R$-set attractors.

(ii)Any near-ring without zero divisors is without external $\{\text{0}\}$-set attractors.

(iii)Let $R$ be the ring ${\mathrm{\mathbb{Z}}}_4$. Take $P$ to be the ideal $\left\{\text{0},2\right\}$. Then $R$ is without external $P$ set-attractors.

(iv)Let $R$ be the ring ${\mathrm{\mathbb{Z}}}_6$. Take $P$ to be the ideal $\{\text{0}\}$. Then $\{2\}$, $\{3\}$ and $\{4\}$ are external $P$ set-attractors and $\{\text{0}\}$ is an internal $P$ set-attractor.

Theorem 4.4. Let $R$ be a near-ring with an ideal $P$. Then the following statements are equivalent:

• (i)$R$ is without external $P$ set-attractors.

• (ii)$P$ is a completely prime ideal of $R$.

Proof.(i) implies (ii), Suppose (i) holds and $ab\in P$ for some $a,b\in R$. So $\{a\}\{b\}\subseteq P$. If $a\notin P$, then $b\in P$ by (i) and $P$ is completely prime.

(ii) implies (i). Suppose (ii) holds and $A$ is a $P$ set-attractor of $R$. So there exists a non-empty subset $B$ of $R$ and $B⊈P$ such that $AB\subseteq P$ or $BA\subseteq P$. Suppose the case is $AB\subseteq P$. Take $y\in B-P$. So $xy\in P$ for all $x\in A$ and then $A\subseteq P$ by (ii). By the same way we can do for the other case. So $R$ is without external $P$ set-attractors. ■

Remark 4.2. (i)If $I=\{\text{0}\}$ in Definition 4.2, then we have Definition 2.1.

(ii)From the above two definitions, Theorem 4.2 and 4.4, we can rewrite the statements of different kinds of prime ideals as follows:

Let $R$ be a near-ring with an ideal $P$. Then

• (1) $P$ is completely prime if and only if $R$ is without external $P$ set-attractors if and only if for every two non-empty subsets $A$ and $B$ of $R$, $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$.

• (2) $P$ is 0-prime if and only if $R$ is without external $P$ ideal-attractors.

• (3) $R$ is 1-prime if and only if $R$ is without external $P$ right ideal-attractors.

• (4) $R$ is 2-prime if and only if $R$ is without external $P$ right $R$-subgroup-attractors.

• (5) $R$ is 3-prime if and only if $R$ is without external $P$ semigroup left ideal-attractors.