On the primeness of near-rings

We say that $R$ is a right (left) near-ring if $(R,+)$ is a group, $(R,⋅)$ 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 $x,y∈R$ and we say that $R$ is a commutative near-ring if $xy=yx$ for all $x,y∈R$⁠. A zero-symmetric element is an element $x∈R$ satisfying $0x=0$⁠. A near-ring $R$ is called a zero-symmetric near-ring, if $0x=0$ for all $x∈R$⁠. A constant element is an element $y∈R$ satisfying $zy=y$ for all $z∈R$⁠. An element $x∈R$ is called a right (left) zero divisor in $R$ if there exists a non-zero element $y∈R$ such that $yx=0$ (⁠$xy=0$⁠). 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 ${ab|a∈A,b∈B}$⁠. We say that $U$ is a right (left) $R$-subgroup of $R$⁠, if $U$ is a subgroup of $(R,+)$ satisfies $UR⊆U$ (⁠$RU⊆U$⁠). 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 $(r+i)s−rs∈I$ for all $i∈I,r,s∈R$ (⁠$RI⊆I$⁠). 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 $UR⊆U$ (⁠$RU⊆U$⁠). 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. $Mo(G)$ 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 $f∈M(G)$ or $Mo(G)$ and $g∈G$⁠). 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 $0y=0$ for all $x∈R−{0},y∈R$⁠. A near-field $N$ is a near-ring in which $(N−{0},⋅)$ 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 $R$⁠, $IJ={0}$ implies $I={0}$ or $J={0}$⁠, $R$ is 1-prime if, for every two right ideals $K$ and $L$ of $R$⁠, $KL={0}$ implies $K={0}$ or $L={0}$⁠. $R$ is 2-prime if, for every two right $R$-subgroups $A$ and $B$ of $R$⁠, $AB={0}$ implies $A={0}$ or $B={0}$⁠. $R$ is 3-prime if, for all $x,y∈R$⁠, $xRy={0}$ implies $x=0$ or $y=0$ and $R$ is equiprime (e-prime) if, for any $0≠a,x,y∈R$⁠, $xca=yca$ for all $c∈R$ 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 $x∈R$⁠, $xRx={0}$ implies $x=0$⁠. An ideal $P$ of $R$ is: (i) a 0-prime ideal of $R$ if for every two ideals $A$ and $B$ of $R$⁠, $AB⊆P$ implies that $A⊆P$ or $B⊆P$⁠, (ii) a 1-prime ideal of $R$ if for every two right ideals $A$ and $B$ of $R$⁠, $AB⊆P$ implies that $A⊆P$ or $B⊆P$⁠, (iii) a 2-prime ideal of $R$ if for every two right $R$-subgroups $A$ and $B$ of $R$⁠, $AB⊆P$ implies that $A⊆P$ or $B⊆P$⁠, (iv) a 3-prime ideal of $R$ if for $a,b∈R$⁠, $aRb⊆P$ implies that $a∈P$ or $b∈P$⁠, (v) an e-prime (equiprime) ideal of $R$ if for every $a∈R−P$ and $x,y∈R$⁠, $xca−yca∈P$ for all $c∈R$ implies that $x−y∈P$⁠. Clearly that any near-ring is a $υ$-prime ideal of itself, where $υ∈{0,1,2,3,e}$⁠. 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 $a,b∈R$⁠, $ab∈I$ implies that $a∈I$ or $b∈I$⁠. 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 $ν$ -primeness for $ν=1,2,3,e$⁠.

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 $UV={0}$ implies $U={0}$ or $V={0}$⁠, 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 $r2≠0$ for all $r∈R−{0}$⁠, 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 $UV⊆P$ implies $U⊆P$ or $V⊆P$⁠, 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.

Let $R$ be a near-ring. It is clear that $R$ is without zero divisors if and only if $AB={0}$ implies $A={0}$ or $B={0}$⁠, 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 $AB={0}$ (⁠$BA={0}$⁠). 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 $AB={0}$ (⁠$BA={0}$⁠). 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)$aU={0}$ implies $a=0$ or $U={0}$, where $a∈R$ and $U$ is a semigroup left ideal of $R$.

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

• (iv)$UV={0}$ implies $U={0}$ or $V={0}$, 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 $x,y∈R$ such that $xRy={0}$⁠. Thus, $RxRy={0}$⁠. But $Rx$ and $Ry$ are semigroup left ideals of $R$⁠, so $Rx={0}$ or $Ry={0}$ by (iv). Hence, $R{0,x}={0}$ or $R{0,y}={0}$ and either ${0,x}$ or ${0,y}$ is a semigroup left ideal of $R$⁠. But $R$ is also a semigroup left ideal of $R$⁠. Thus, ${0,x}=0$⁠, ${0,y}={0}$ or $R={0}$ 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)$Ua={0}$ implies $a=0$ or $U={0}$⁠, where$a∈R$ and $U$ is a semigroup right ideal of $R$⁠.

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

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

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

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

• (iv)$U2={0}$ implies $U={0}$, where $U$ is a semigroup left ideal of $R$.

• (v)$U2={0}$ implies $U={0}$, 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 $aU={0}$⁠, where $a∈U$⁠. Then for all $v∈U$⁠, we have $aRv={0}$⁠. Thus, $aRa={0}$ and $a=0$ 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 $U2={0}$⁠, where $U$ is a semigroup right ideal of $R$⁠. So $uRu={0}$ for all $u∈U$ and hence $RuRu={0}$⁠. But $Ru$ is a semigroup left ideal of $R$⁠. So $Ru={0}$ for all $u∈U$ by (iv). So ${0,u}$ is a semigroup left ideal of $R$ and ${0,u}{0,u}={0}$ for all $u∈U$⁠. So $u=0$ by (iv) and hence $U={0}$⁠.

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

Corollary 2.5. A ring $R$ is semiprime if and only if $U2={0}$ implies $U={0}$, 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)$aU={0}$ implies $a=0$⁠, where $a∈U$ and $U$ is a semigroup left ideal of $R$⁠.

• (iii)$U2={0}$ implies $U={0}$⁠, 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 ${0,a,b,c}$⁠. Then it is an abelian group such that $x+x=0$ for all $x∈R$ and $x+y=z$ for all different non-zero elements $x,y,z∈R$⁠. Define the multiplication on $R$ as follows:

Clearly $R$ is an abelian non-zero-symmetric near-ring. The only semigroup right ideals of $R$ are $R$⁠, ${0,a}$ and ${0,a,b}$⁠. So $R$ satisfies the conditions “$UV={0}$ implies $U={0}$ or $V={0}$⁠, where $U$ and $V$ are semigroup right ideals of $R$” and “$U2={0}$ implies $U={0}$⁠, where $U$ is a semigroup right ideal of $R$”. But $R$ is not 3-semiprime as $bRb={0}$⁠. From Theorem 2.6, we can deduce that there is a non-zero semigroup left ideal $V$ of $R$ such that $V2={0}$ and $υV={0}$⁠, where $υ∈V−{0}$⁠. It is easy to find out that $V={0,b}$ and $v=b$⁠.

From the above example, observe that

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 $r2≠0$ for all $r∈R−{0}$, then $R$ is 3-semiprime.

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

where $a∈U$⁠. That means $a2=0$⁠. By hypothesis, $a=0$ and hence $R$ is 3-semiprime. ■

Example 2. Let $R=ℤ6$⁠. Then $R$ is semiprime since $r2≠0$ for all $r∈R−{0}$⁠.

Example 3. Let $R={0,2,4,6,8,10,12}$ the subring of $ℤ14$⁠. Then $R$ is semiprime since $r2≠0$ for all $r∈R−{0}$⁠.

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

Example 4.Let $R=M2(ℤ2)$⁠. Then $R$ is a prime ring and hence semiprime, but

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 $r2≠0$ for all $r∈R−{0}$ 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 $2R={0}$. Then $R$ is 3-prime if and only if $R$ is 2-prime.

Proof. Suppose that $xRy={0}$⁠. Thus, $xRyR={0}$⁠. But $xR$ and $yR$ are right $R$-subgroups of $R$⁠. So $xR={0}$ or $yR={0}$ as $R$ is 2-prime. Hence, ${0,x}R={0}$ or ${0,y}R={0}$ and then either ${0,x}$ or ${0,y}$ is a right $R$-subgroup of $R$⁠. But $R$ is also a right $R$-subgroup of $R$⁠. Thus, ${0,x}=0$⁠, ${0,y}={0}$ or $R={0}$⁠. Hence, $x=0$ or $y=0$ 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 $xRy={0}$ for some $x,y∈R$⁠. So $xRyR={0}$ and hence $xR={0}$ or $yR={0}$⁠. So $AR={0}$ or $BR={0}$⁠, where $A={nx|n∈ℤ}$ and $B={ny|n∈ℤ}$⁠. So $A$ and $B$ are right $R$-subgroups of $R$ and hence $A={0}$ or $B={0}$⁠. Therefore, $x=0$ or $y=0$ and $R$ is 3-prime. ■

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 $0≠a,x,y∈R$⁠, $xsa=yra$ for all $s,r∈R−{0}$ 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 $0≠a,x,y∈R$⁠, $xca=yca$ for all $c∈R$ implies $x=$ $y$⁠.

(i) implies (ii) is clear.

(ii) implies (i). Suppose (ii) holds. If for all $c∈R$⁠, $xca=yca$ for $0≠a,x,y∈R$⁠, then $(x−y)ca=0=0ra$ for all $c,r∈R$⁠. 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 $0≠a,x,y∈R$⁠, $xsa=yra$ for all $s,r∈R−{0}$ 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 $s,r∈R−{0}$⁠, a contradiction with the hypothesis. So $R$ is zero-symmetric. Now, suppose $xRy={0}$ for some $x,y∈R$⁠. So $xcy=0$ for all $c∈R$⁠. If $y≠0$⁠, then $xcy=0ry$ for all $c,r∈R$⁠. So $x=0$ 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 $ℤ5$ and define the multiplication on $R$ as follows:

So $R$ is an abelian near-ring which is not a ring (as $(1+1)2=3≠4=2+2=(1)2+(1)2$⁠). Clearly that $R$ is without zero divisors. Hence, $R$ is 3-prime. $R$ is not equiprime. Indeed, $1c1=3c1=c1$ for all $c∈R$⁠. But if $0≠a,x,y∈R$ such that $xsa=yra$ for all $s,r∈R−{0}$⁠, 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 $s,r∈R−{0}$ and $x,y,a$ are all non-zero, then from the table we can choose $so,ro∈R−{0}$ to satisfy that $xso=1$ and $yro=2$⁠. Hence, $a=2a$ which implies that $a=0$ (from the table), a contradiction with $0≠a$⁠. 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 $x≠y$⁠, we have $xsx=yrx=x$ for all $s,r∈R−{0}$⁠. 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:

Remark 3.1.Observe that:

(i)It is well-known that $Mo(G)$ 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 $0≠a,x,y∈R$ such that $xca=yca$ for all $c∈R$⁠, 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:

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 $a∈R−P$ and $x,y∈R$⁠, $xra−ysa∈P$ for all $r,s∈R−P$ implies $x−y∈P$⁠.

Clearly $R$ is K-prime if and only if ${0}$ 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 $c∈R−P$⁠, then

for all $x,y∈R$ and $r,s∈R−P$⁠. So $x−y∈P$ and hence $x−0=x∈P$ for all $x∈R$⁠. Thus, $P=R$⁠, a contradiction with $c∉P$⁠. So $c∈P$⁠.

Now, suppose $aRb⊆P$ for some $a,b∈R$ and $b∉P$⁠. From above, any element $s∈R−P$ is a zero-symmetric element. So $0sb=0∈P$ for all $s∈R−P$⁠. So $arb−0sb∈P$ for all $r,s∈R−P$⁠. Thus, $a∈P$ by the hypothesis and $P$ is 3-prime.

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

Suppose $xra−ysa∈P$ for all $r,s∈R−P$⁠, where $a∈R−P$ and $x,y∈R$⁠. So $xca−yca∈P$ for all $c∈R−P$⁠. Now, suppose $c∈P$⁠. As $a∉P$⁠, we have that $a$ is a zero-symmetric element (see [10]). So $ca∈P$ and hence $xca−yca∈P$⁠. But $P$ is e-prime. So $x−y∈P$ 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)$BU⊆P$ implies $B⊆P$ or $U⊆P$, where $B$ is a non-empty subset of $R$ and $U$ is a semigroup left ideal of $R$.

(iii)$UV⊆P$ implies $U⊆P$ or $V⊆P$, 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 $BU⊆P$⁠. If $B⊈P$⁠, then there exists $b∈B−P$ such that $bRu⊆P$ for all $u∈U$⁠. Thus, $U⊆P$ 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 $xRy⊆P$⁠. Thus, $RxRy⊆P$⁠. So $Rx⊆P$ or $Ry⊆P$ by (iii). Hence, $R(P ∪ {x})⊆P$ or $R(P ∪ {y})⊆P$ and then $P ∪ {x}$ or $P ∪ {y}$ is a semigroup left ideal of $R$⁠. But $R$ itself is also a semigroup left ideal of $R$⁠. Thus, $P ∪ {x}⊆P$⁠, $P ∪ {y}⊆P$ or $R⊆P$ by (iii), a contradiction with that $x,y∈R−P$⁠. 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 $R$⁠, $AB⊆P$ implies $A⊆P$ or $B⊆P$⁠.

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)$UB⊆P$ implies $U⊆P$ or $B⊆P$⁠, where $U$ is a semigroup right ideal of $R$ and $B$ is a non-empty subset of $R$⁠.

• (iii)$UV⊆P$ implies $U⊆P$ or $V⊆P$⁠, 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 ${0}$ is not a 3-prime ideal in Example 1 although it satisfies the condition “If $UV⊆{0}$⁠, then $U⊆{0}$ or $V⊆{0}$⁠, 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 $AB⊆I$ (⁠$BA⊆I$⁠). 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 $AB⊆I$ (⁠$BA⊆I$⁠). 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 $I≠R$⁠. 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 $A⊆P$⁠. 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 $A⊆P$

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 ${0}$-set attractors.

(iii)Let $R$ be the ring $ℤ4$⁠. Take $P$ to be the ideal ${0,2}$⁠. Then $R$ is without external $P$ set-attractors.

(iv)Let $R$ be the ring $ℤ6$⁠. Take $P$ to be the ideal ${0}$⁠. Then ${2}$⁠, ${3}$ and ${4}$ are external $P$ set-attractors and ${0}$ 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 $ab∈P$ for some $a,b∈R$⁠. So ${a}{b}⊆P$⁠. If $a∉P$⁠, then $b∈P$ 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 $AB⊆P$ or $BA⊆P$⁠. Suppose the case is $AB⊆P$⁠. Take $y∈B−P$⁠. So $xy∈P$ for all $x∈A$ and then $A⊆P$ 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 $I={0}$ 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 $R$⁠, $AB⊆P$ implies $A⊆P$ or $B⊆P$⁠.

• (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.