1 Introduction

Throughout this paper, R denotes a left near-ring and Z(R) denotes the multiplicative center of R. A near-ring R is called 3-prime if, for all \( x,y\in R\), \(xRy=\{0\}\) implies \(x=0\) or \(y=0\). A map \(d:R\rightarrow R\) is a derivation on R if d is an additive mapping and \(d(xy)=xd(y)+d(x)y\) for all \(x,y\in R\). If the map \(d:R\rightarrow R\) defined by \(d(x)=xr-rx\) or by \( d(x)=rx-xr\) for all \(x\in R\), where \(r\in R\), is a derivation (and it will be if R is a ring), then it is called an inner derivation on R induced by 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 \(yx=0\) (\(xy=0\)). A zero divisor is a right or a left zero divisor. A near-ring R is called a zero-symmetric near-ring, if \(0x=0\) for all \(x\in R\). For subsets \( X,Y\subseteq R\), the symbol [X, Y] denotes the set \(\{xy-yx|x\in X,y\in Y\}\) . We say that U is a semigroup right (left) ideal of a near-ring R if U is a non-empty subset of R satisfying \(UR\subseteq U\) (\(RU\subseteq U\)). We say that U is a semigroup ideal if it is both a semigroup right and left ideal. For further information about near-rings, see [12, 13].

Studying when a near-ring is a ring or a commutative ring is one of the most important questions in the theory of near-rings, which has been the subject of many investigations since the 70s of the foregoing century. For example, among many others, Ligh in [11] and Bell in [2,3,4] have established some conditions that force a near-ring to be a ring. Later on, Bell and Mason in [7] have studied many properties of 3-prime near-rings that make them commutative rings. Beidar, Fong, and Wang [1] have generalized some results about rings to the setting of near-rings, while Wang [15] has established some beautiful properties of near-rings. On the other hand, Bell in [5] has generalized several results of [7] using one (two) sided semigroup ideals of near-rings. Subsequently, many authors, for instance in [8, 9] and [10], have investigated various properties of near-rings that make of them commutative rings.

In this paper, we establish some new results in this direction. In Sect. 3, we begin with Theorem 3.1 which is a generalization of Theorem 2.3 of [6] about prime rings that satisfy \(d([x,y])\in Z(R)\) for all \(x,y\in R\). We show that the result remains valid when we replace R by one of its non-zero ideals. One of the well-known results about derivations on rings and near-rings asserts that if \(x\in Z(R)\), then \(d(x)\in Z(R)\). We study its converse for 3-prime near-rings, in Theorem 3.2, and we get that if \( d(x),d(x^{2})\in Z(R)\), such that \(2x\ne 0\ne d(x)\), then \(x\in Z(R)\). We use this result to prove other results such as Theorem 3.3 (which is a partial generalization of Theorem 3.1) and Theorem 3.4. Note that Theorem 3.4 is one of our main theorems in this paper, namely that it provides a full description of the case \(\{0\}\ne d(U)\subseteq Z(R)\) where U is a subsemigroup of \((R,\cdot )\) which provides an interesting extension of many known results (see [5, 7, 9, 10]), where we show that U is commutative, \( V=\{u^{2}|u\in U\}\) is a subsemigroup of R contained in Z(R) and either \( 2R=\{0\}\) or \(U\subseteq Z(R)\). Moreover, we study three cases: \(nU=\{0\}\), \( nU\ne \{0\}\), and R is of characteristic zero. We also give an example illustrating these results. Also, one of the main theorems in this paper is Theorem 3.5, where we give a generalization of the condition \( x_{o}d(R)=\{0\} \) or \(d(R)x_{o}=\{0\}\), (which is a very useful tool in proofs), on near-rings and rings. Bell in [5] has proved if R is a 3-prime near-ring with a non-zero semigroup right (left) ideal U, such that \( Ux=\{0\}\) (\(xU=\{0\}\)) for some \(x\in R\), then one should have \(x=0\). Theorem 3.6 generalizes Bell’s result to the case \(Ux\subseteq Z(R)\) (\( xU\subseteq Z(R)\)).

2 Preliminaries

In this section, we recall some well-known results on derivations and commutativity of near-rings. These will often be used, mostly without explicit mention.

Lemma 2.1

[10, Theorem 2.7] A near-ring R is zero-symmetric if and only if R admits a derivation.

Lemma 2.2

[15, Proposition 1] Let R be a near-ring with a derivation d. Then, \(xd(y)+d(x)y=d(x)y+xd(y)\) for all \(x,y\in R\).

Lemma 2.3

[7, Lemma 1] (The partial distributive law) Let R be a near-ring and d a derivation on R. For all \(x,y,z\in R\), we have \( (xd(y)+d(x)y)z=xd(y)z+d(x)yz\).

The following results are very useful in the sequel.

Lemma 2.4

[15, Lemma 2] Let R be a near-ring with a derivation d . If \(x\in Z(R)\), then \(d(x)\in Z(R)\).

Lemma 2.5

[7, Lemma 3(i)] Let R be a 3-prime near-ring and \(z\in Z(R)-\{0\}\). Then, z is not a zero divisor.

Lemma 2.6

[9, Corollary 3.2] Let R be a 3-prime near-ring with a non-zero semigroup right (left) ideal U and a non-zero semigroup ideal V. If R admits a non-zero derivation d, such that \( d([V,U])=\{0\}\), then R is a commutative ring.

Lemma 2.7

[5, Lemma 1.2(iii)] Let R be a 3-prime near-ring and \( x\in Z(R)-\{0\}\). If either yx or xy in Z(R), then \(y\in Z(R)\).

Lemma 2.8

[5, Lemma 1.4] Let R be a 3-prime near-ring with a non-zero semigroup ideal U and a non-zero derivation d on R. If \(x\in R \) and \(d(U)x=\{0\}\) (\(xd(U)=\{0\}\)), then \(x=0\).

Lemma 2.9

[9, Proposition 2.8] Let R be a 3-prime near-ring with a non-zero semigroup ideal U and a non-zero derivation d. Then, for every integer \(n\geqslant 2\), the following statements are equivalent:

  1. (i)

    \(d(nR)=\{0\}\).

  2. (ii)

    \(d(nU)=\{0\}\).

  3. (iii)

    \(nU=\{0\}\).

  4. (iv)

    \(nR=\{0\}\).

Lemma 2.10

[5, Theorem 2.1] Let R be a 3-prime near-ring with a non-zero derivation d and a non-zero semigroup right (left) ideal U of R. If \(d(U)\subseteq Z(R)\), then R is a commutative ring.

Lemma 2.11

[14, Lemma 2.14] Let R be a 3-prime zero-symmetric near-ring, such that:

  1. (i)

    \(x_{1}x_{2}\ldots x_{n}\in Z(R)-\{0\}\) for some \(x_{1},x_{2},\ldots ,x_{n}\in R\), where \(n\in \mathbb { \mathbb {N} }\). Then, \(x_{n}\) is not a left zero divisor and \(x_{1}\) is not a right zero divisor.

  2. (ii)

    \(x^{n}\in Z(R)-\{0\}\) for some \(x\in R,n\in \mathbb { \mathbb {N} }\). Then, x is not a zero divisor.

Lemma 2.12

[5, Lemma 1.5] Let R be a 3-prime near-ring with a non-zero semigroup right (left) ideal U. If \(U\subseteq Z(R)\), then R is a commutative ring.

3 Main results

In this section, we will study some properties on rings and near-rings. We will begin with a result which is a generalization of Theorem 2.3 of [6].

Theorem 3.1

Let R be a prime ring with a non-zero ideals U and V of R. If R admits a non-zero derivation d, such that \( d([U,V])\subseteq Z(R)\), then R is commutative or \(2R=\{0\}\).

Proof

If \(d([U,V])=\{0\}\), then R is commutative by Lemma 2.6. Therefore, suppose \( d([U,V])\ne \{0\}\). For all \(u\in U,y,w\in V\), we have that \( d([[u,y],w])\in Z(R)\). Thus:

$$\begin{aligned}{}[u,y]d(w)+d([u,y])w-wd([u,y])-d(w)[u,y]\in Z(R). \end{aligned}$$

Since \(d([u,y])\in Z(R)\), we get that:

$$\begin{aligned}{}[u,y]d(w)-d(w)[u,y]=[[u,y],d(w)]\in Z(R). \end{aligned}$$
(1)

Replacing u by \(d(v^{\prime }v)u\), w by \(v^{\prime }v\) and y by \( d(v^{\prime }v)\), where \(v^{\prime },v\in V\), we have:

$$\begin{aligned}&{[}[d(v^{\prime }v)u,d(v^{\prime }v)],d(v^{\prime }v)] \\= & {} d(v^{\prime }v)[[u,d(v^{\prime }v)],d(v^{\prime }v)]\in Z(R) \end{aligned}$$

after simple calculations. Using (1), for each \(v^{\prime \prime }\in V^{2} \), either \(d(v^{\prime \prime })\in Z(R)\) or \([[u,d(v^{\prime \prime })],d(v^{\prime \prime })]=0\) for all \(u\in U\) by Lemma 2.7. Therefore, in all cases, \([[u,d(v^{\prime \prime })],d(v^{\prime \prime })]=0\) for all \(u\in U,v^{\prime \prime }\in V^{2}\). Now, for each \(v\in V^{2}\), define \( d_{d(v)}:R\rightarrow R\) as follows:

$$\begin{aligned} d_{d(v)}(x)=xd(v)-d(v)x=[x,d(v)]. \end{aligned}$$

Then, \(d_{d(v)}\) is an inner derivation on R induced by d(v). Therefore:

$$\begin{aligned} d_{d(v)}^{2}(u)=[d_{d(v)}(u),d(v)]=0, \end{aligned}$$

and hence, \(d_{d(v)}^{2}(U)=\{0\}\). Therefore, for all \(u,u^{\prime }\in U\), we have:

$$\begin{aligned} 0=d_{d(v)}^{2}(uu^{\prime })=2d_{d(v)}(u)d_{d(v)}(u^{\prime })=d_{d(v)}(u)2d_{d(v)}(u^{\prime }). \end{aligned}$$

From Lemma 2.8, we get that \(d_{d(v)}=0\) or \(2d_{d(v)}(U)=\{0\}\). Therefore, either \(d(v)\in Z(R)\) or \(2R=\{0\}\) by Lemma 2.9. If \(2R\ne \{0\}\), then \(d(v)\in Z(R)\) for all \(v\in V^{2}\) and R is commutative by Lemma 2.10 since \(V^{2}\) is a semigroup ideal of R. Therefore, \(d([R,R])=\{0\}\), a contradiction. Therefore, \(2R=\{0\}\). \(\square \)

The following example shows that "\(2R=\{0\}\)" in the previous theorem is not redundant.

Example 1

Let \(R=M_{2}(\mathbb { \mathbb {Z} }_{2})\). Then, R is a prime ring. Define a map d on R to be the inner derivation induced by: \(\left[ \begin{array}{cc} 0 &{} 1 \\ 0 &{} 0 \end{array} \right] \). Therefore, for all \(a,b,c,e\in \mathbb {Z} _{2}\):

$$\begin{aligned} d\left( \left[ \begin{array}{cc} a &{} b \\ c &{} e \end{array} \right] \right) =\left[ \begin{array}{cc} a &{} b \\ c &{} e \end{array} \right] \left[ \begin{array}{cc} 0 &{} 1 \\ 0 &{} 0 \end{array} \right] +\left[ \begin{array}{cc} 0 &{} 1 \\ 0 &{} 0 \end{array} \right] \left[ \begin{array}{cc} a &{} b \\ c &{} e \end{array} \right] =\left[ \begin{array}{cc} c &{} a+e \\ 0 &{} c \end{array} \right] \text {.} \end{aligned}$$

Then, clearly, it follows that:

$$\begin{aligned} d(R)=\left\{ \left[ \begin{array}{cc} 0 &{} 0 \\ 0 &{} 0 \end{array} \right] ,\left[ \begin{array}{cc} 0 &{} 1 \\ 0 &{} 0 \end{array} \right] ,\left[ \begin{array}{cc} 1 &{} 0 \\ 0 &{} 1 \end{array} \right] ,\left[ \begin{array}{cc} 1 &{} 1 \\ 0 &{} 1 \end{array} \right] \right\} \text {.} \end{aligned}$$

Also, \(d^{2}(R)=\{0\}\). In fact, \(d([x,y])\in Z(R)\) for all \(x,y\in R\) as follows: for all \(a,b,c,e,x,y,z,w\in \mathbb {Z} _{2}\), we have:

$$\begin{aligned}&d\left( \left[ \begin{array}{cc} a &{} b \\ c &{} e \end{array} \right] \left[ \begin{array}{cc} x &{} y \\ z &{} w \end{array} \right] +\left[ \begin{array}{cc} x &{} y \\ z &{} w \end{array} \right] \left[ \begin{array}{cc} a &{} b \\ c &{} e \end{array} \right] \right) \\= & {} \left[ \begin{array}{cc} cx+ez+za+wc &{} 0 \\ 0 &{} cx+ez+za+wc \end{array} \right] \in \left\{ \left[ \begin{array}{cc} 0 &{} 0 \\ 0 &{} 0 \end{array} \right] ,\left[ \begin{array}{cc} 1 &{} 0 \\ 0 &{} 1 \end{array} \right] \right\} =Z(R). \end{aligned}$$

Observe that \(2R=\{0\}\) and R is not commutative as:

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

We know from Lemma 2.4 that if \(x\in Z(R)\), then \(d(x)\in Z(R)\). In the following result, we discuss its converse for 3-prime near-rings.

Theorem 3.2

Let R be a 3-prime near-ring with a derivation d, such that \(d(u^{i})\in Z(R)\), where \(i\in \{1,2\}\), and \(d(u)\ne 0\) for some \(u\in R\). Then, either \(2u=0\) or \(u\in Z(R)-\{0\}\). Moreover, if \( d(u^{3})\in Z(R)\), then \(u^{2}\in Z(R)\).

Proof

Observe that:

$$\begin{aligned} d(u^{2})=d(u)(2u)\in Z(R), \end{aligned}$$

which implies that \(2u\in Z(R)\) as \(d(u)\ne 0\). Since

$$\begin{aligned} rd(u^{2})=d(u^{2})r \end{aligned}$$

for all \(r\in R\), we have:

$$\begin{aligned} d(u)(2ru)=d(u)(2ur)\text {.} \end{aligned}$$

Thus, \(d(u)(2ru-2ur)=0\) and hence:

$$\begin{aligned} 2ru=2ur\text { .} \end{aligned}$$
(2)

Since \(2u\in Z(R)\), we get that \(r(2u)=(2u)r\) for all \(r\in R\). However, using (2), we deduce that:

$$\begin{aligned} r(2u)=2ru=2ur. \end{aligned}$$

Therefore:

$$\begin{aligned} (2u)r=2ur \end{aligned}$$
(3)

for all \(r\in R\). Replacing r by ur in (2), we get:

$$\begin{aligned} 2uru=2uur, \end{aligned}$$

and using (3) on each side of the last equation, we have:

$$\begin{aligned} (2u)ru=(2u)ur, \end{aligned}$$

and hence:

$$\begin{aligned} (2u)(ru-ur)=0\text {.} \end{aligned}$$

Therefore, either \(2u=0\) or \(u\in Z(R)-\{0\}\) by Lemma 2.5.

To complete the proof, we will discuss the two cases. If \(u\in Z(R)-\{0\}\), then \(u^{2}\in Z(R\dot{)}\). On the other hand, if \(2u=0\), then \(d(u^{2})=0\), and hence, \(d(u^{3})=d(u)u^{2}\in Z(R)\) which implies that \(u^{2}\in Z(R)\). \(\square \)

In the following result, we will prove a similar result of Theorem 3.1 for 3-prime near-rings.

Theorem 3.3

Let R be a 2-torsion-free 3-prime near-ring. If R admits a non-zero derivation d, such that \(d([x,y]),d([x,y]^{2})\in Z(R)\) for all \(x,y\in R\), then R is a commutative ring.

Proof

From Theorem 3.2, for all \(x,y\in R\), either \([x,y]\in Z(R)-\{0\}\) or \( d([x,y])=0\). If \(d([x,y])=\{0\}\) for all \(x,y\in R\), then R is a commutative ring by Lemma 2.6. Now, suppose \(d([R,R])\ne \{0\}\). Therefore, there are \(a,b\in R\), such that \(d([a,b])\ne 0\). Thus, \([a,b]\in Z(R)-\{0\}\), and for all \(x,y\in R\), such that \(d([x,y])=0\), we have:

$$\begin{aligned} d([[a,b]x,y])= & {} d([a,b][x,y]) \\= & {} d([a,b])[x,y]\in Z(R), \end{aligned}$$

and hence, \([x,y]\in Z(R)\) also by lemma 2.7. Therefore, \([R,R]\subseteq Z(R)\). Observe that \([a,ab]=a[a,b]\in Z(R)\) implies that \(a\in Z(R)\), and then, \( [a,b]=0\), a contradiction. Therefore, \(d([R,R])=\{0\}\). \(\square \)

In Theorem 2.1 of [5], Bell has showed that, if \(d(U)\subseteq Z(R)\), then R is a commutative ring, where U is a non-zero semigroup right (left) ideal of R. The following result extends his result over subsemigroups and also generalizes Theorem 2 of [7].

Theorem 3.4

Let R be a 3-prime near-ring with a non-zero derivation d and a subsemigroup U of \((R,\cdot )\), such that \(\{0\}\ne d(U)\subseteq Z(R)\). Then, U is commutative, \(S=\{u\in U|d(u)=0\}\) and \( V=\{u^{2}|u\in U\}\) are subsemigroups of R contained in Z(R) and either \( 2R=\{0\}\) or \(U\subseteq Z(R)\). Moreover, for any positive integer n, we have the following cases:

  1. (i)

    If \(nU=\{0\}\), then \(nR=\{0\}\), and in this case: (the characteristic of R is odd and \(U\subseteq Z(R)\)) or \(2R=\{0\}\).

  2. (ii)

    If \(nU\ne \{0\}\), then R is n-torsion-free and either (\(2R=\{0\}\) and n is odd) or \(U\subseteq Z(R)\).

  3. (iii)

    If R is of characteristic zero, then \(U\subseteq Z(R)\).

Proof

As \(d(U)\ne \{0\}\), then there exists \(v^{\prime }\in U\), such that \( d(v^{\prime })\in Z(R)-\{0\}\). For all \(u\in U\), we have:

$$\begin{aligned} v^{\prime }d(uv^{\prime })=d(uv^{\prime })v^{\prime }, \end{aligned}$$

and hence:

$$\begin{aligned} d(v^{\prime })(v^{\prime }u-uv^{\prime })=0\text {.} \end{aligned}$$

Using Lemma 2.5, we get that \(v^{\prime }\) centralizes U. Again, for all \( u,w\in U\), we have:

$$\begin{aligned} rd(uv^{\prime })=d(uv^{\prime })r, \end{aligned}$$

and by the same way above, we will get that u centralizes U for all \(u\in U\). Hence, U is commutative. Now, for all \(u\in U\), either \(d(u)=0\) or not.

If \(d(u)=0\), then:

$$\begin{aligned} d(uv^{\prime })=ud(v^{\prime })\in Z(R), \end{aligned}$$

which implies that \(u\in Z(R)\). Therefore, \(S=\{u\in U|d(u)=0\}\) is a subsemigroup of R contained in Z(R).

If \(d(u)\ne 0\), then \(u^{2}\in Z(R)\) by Theorem 3.2. Consequently, \( V=\{u^{2}|u\in U\}\) is a subsemigroup of R contained in Z(R).

Now, either \(d((v^{\prime })^{2})=0\) or not. If \(d((v^{\prime })^{2})=0\), then for all \(r\in R\), we have that:

$$\begin{aligned} 0=d((v^{\prime })^{2})r=d(v^{\prime })(2v^{\prime }r), \end{aligned}$$

and hence:

$$\begin{aligned} 0=2v^{\prime }rs=v^{\prime }r(2s) \end{aligned}$$

for all \(r,s\in R\). Therefore, \(2R=\{0\}\).

If \(d((v^{\prime })^{2})\ne 0\), then for all \(r\in R\), we have:

$$\begin{aligned} rd((v^{\prime })^{3})=d((v^{\prime })^{3})r, \end{aligned}$$

and hence:

$$\begin{aligned} d((v^{\prime })^{2})(rv^{\prime }-v^{\prime }r)=0, \end{aligned}$$

which implies that \(v^{\prime }\in Z(R)\). Now, for all \(r\in R\) and \(u\in U\), we have:

$$\begin{aligned} wd(uv^{\prime })=d(uv^{\prime })w, \end{aligned}$$

and by the same way above, we will get that \(u\in Z(R)\). Therefore, \( U\subseteq Z(R)\).

(i) Suppose \(nU=\{0\}\). Therefore, \(d((v^{\prime })^{n})=0\), and for all \(r\in R,\) we have that:

$$\begin{aligned} d((v^{\prime })^{n})r=d(v^{\prime })(n(v^{\prime })^{n-1}r)=0, \end{aligned}$$

which implies that \((v^{\prime })^{n-1}(nr)=0\) and hence:

$$\begin{aligned} (v^{\prime })^{n-1}(nrs)=(v^{\prime })^{n-1}r(ns)=0 \end{aligned}$$

for all \(r,s\in R\). Therefore, \(nR=\{0\}\) or \((v^{\prime })^{n-1}=0\). If \( (v^{\prime })^{n-1}=0\), then:

$$\begin{aligned} d((v^{\prime })^{2})(v^{\prime })^{n-2}=0 \end{aligned}$$

and either \(d((v^{\prime })^{2})=0\) or \((v^{\prime })^{n-2}=0\). Continuing on this way, we will get that \(d((v^{\prime })^{2})=0\) in all cases, and by the same discussion above, we have that \(2R=\{0\}\). Thus, we have two cases:

Case 1: n is odd which implies that \(2U\ne \{0\}\). Therefore, \(nR=\{0\}\) and \(U\subseteq Z(R)\) from above.

Case 2: n is even and either \(2R=\{0\}\) or \(U\subseteq Z(R)\). If \( 2R=\{0\}\), then clearly that \(nR=\{0\}\). If \(U\subseteq Z(R)\), then for all \( r\in R\), we have:

$$\begin{aligned} 0=r(nv^{\prime })=v^{\prime }(nr), \end{aligned}$$

which implies that \(nR=\{0\}\). If \(2R\ne 0\), then from \(d((v^{\prime })^{2})\in Z(R)\), we get that \(2v^{\prime }\in Z(R)\). If \(2v^{\prime }=0\), then \(2R=\{0\}\) by the same way above, a contradiction. Therefore, \(2v^{\prime }\in Z(R)-\{0\}\). Suppose \(n=2m\). For all \(r\in R\), we have:

$$\begin{aligned} 0=v^{\prime }(2mr)=mr(2v^{\prime })=(2v^{\prime })(mr), \end{aligned}$$

which implies that \(mR=\{0\}\). Continuing on this way, we can show that \( m_{o}R=\{0\}\), where \(m_{o}\) is an odd number and \(n=2^{k}m_{o}\), for some positive integer k.

(ii) Suppose that \(nU\ne \{0\}\). Therefore, there exists \(v_{o}\in U\), such that \( nv_{o}\ne 0\). Let

$$\begin{aligned} W=\{u\in U|d(u)\ne 0\}\text {.} \end{aligned}$$

If \(nw=0\) for some \(w\in W\), then for all \(u^{\prime }\in S\), we have:

$$\begin{aligned} 0=d(u^{\prime }(nw))=d(w(nu^{\prime }))=d(w)nu^{\prime }\text {.} \end{aligned}$$

Thus, \(nu^{\prime }=0\). Now, for all \(u\in U\) , we get that:

$$\begin{aligned} 0=u(nu^{\prime })=nuu^{\prime }=u^{\prime }(nu)\text {.} \end{aligned}$$

If \(u^{\prime }\ne 0\), then \(nU=\{0\}\), a contradiction.

Therefore, we have two cases: \(S=\{0\}\) or \(nw\ne 0\) for all \(w\in W\).

Case 1: \(S=\{0\}\). For all \(w\in W=U-\{0\}\), observe that \(d(w^{2})\in Z(R)\) implies that \(2w\in Z(R)\), and hence, \(2U\subseteq Z(R) \). If \(2U=\{0\}\), then \(2R=\{0\}\) from (i) and n is odd which implies that R is n-torsion-free. If \(2U\ne \{0\}\), then there exists \(u_{o}\in U\), such that \(2u_{o}\in Z(R)-\{0\}\). For all \(u\in U\), we get that:

$$\begin{aligned} u(2u_{o})=2uu_{o}\in Z(R), \end{aligned}$$

and hence, \(u\in Z(R)\). Therefore, \(U\subseteq Z(R)\). Suppose that \(nr=0\) for some \( r\in R\). From \(d(w^{n})\in Z(R)\), we deduce that \(nw^{n-1}\in Z(R)\), where \( w\in W\). If \(nw^{n-1}\ne 0\) for some \(w\in W\), then:

$$\begin{aligned} 0=w^{n-1}(nr)=r(nw^{n-1}), \end{aligned}$$

and then, \(r=0\). Therefore, R is n-torsion-free.

On the other hand, if \(nw^{n-1}=0\) for all \(w\in W\), then \(nw=0\) as \(w\in Z(R)-\{0\}\). Therefore, \(nW=\{0\}\), and hence, \(nU=\{0\}\), a contradiction.

Case 2: There exists \(u_{1}\in S\), such that \(u_{1}\ne 0\ne nw\) for all \(w\in W\). If \(2w=0\) for some \(w\in W\), then n is odd, and for all \( u\in U-W\), we have:

$$\begin{aligned} 0=d(u(2w))=d(w)2u, \end{aligned}$$

and hence, \(2u=0\). Now, for all \(v\in U\), observe that:

$$\begin{aligned} 0=v(2u_{1})=u_{1}(2v), \end{aligned}$$

and then, \(2U=0\). That implies that \(2R=\{0\}\) from (i) and hence R is n-torsion-free. If \(2w\ne 0\) for all \(w\in W\), then \(w\in Z(R)\) by Theorem 3.2. Therefore, \(U\subseteq Z(R)\), and by the same way in case 1, we can prove that R is n-torsion-free.

(iii) Since \(2R\ne \{0\}\), then \(U\subseteq Z(R)\). \(\square \)

The following example shows that:

  1. (i)

    It is possible in Theorem 3.2 to find a prime ring with a derivation d, such that \(d(u^{i})\in Z(R)\), where \(i\in \{1,2,3\}\), and \(d(u)\ne 0\) for some \(u\in R\), and \(2u=0\), but \(u\notin Z(R)\).

  2. (ii)

    \(2R=\{0\}\) is not redundant in Theorem 3.4.

Example 2

Let \(R=M_{2}({ \mathbb {Z} }_{2})\). Then, R is a prime ring. Define the same inner derivation d on R as defined in Example 1. Take \(u=\left[ \begin{array}{cc} 0 &{} 0 \\ 1 &{} 0 \end{array} \right] \). Then, \(d(u^{i})\in Z(R)\), where \(i\in \{1,2,3\}\), \(d(u)\ne 0\) and \(2u=0\) but \(u\notin Z(R)\), since:

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

This shows that (i) is true. For (ii), observe that:

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

is a subsemigroup and \(\{0\}\ne d(U)\subseteq Z(R)\). However, \(U\nsubseteq Z(R)\).

Bell and Mason have proved in [7] that, when R is a 3-prime near-ring, then \(x_{o}d(R)=\{0\}\) (\(d(R)x_{o}=\{0\}\)) implies that \(x_{o}=0\) or \(d=0\), which is a very useful tool in proofs. The following result generalizes their result.

Theorem 3.5

Let R be a 3-prime near-ring with a non-zero derivation d. If \(x_{o}d(R)\subseteq Z(R)\) (\(d(R)x_{o}\subseteq Z(R)\)) for some \(x_{o}\in R\), then \(x_{o}\in Z(R)\). Moreover, If \(x_{o}\ne 0\), then R is a commutative ring.

Proof

Suppose \(x_{o}d(R)\subseteq Z(R)\) (Case 2: \(d(R)x_{o}\subseteq Z(R)\)) for some \(x_{o}\in R\). If \(x_{o}d(R)=\{0\}\), then \(x_{o}=0\in Z(R)\) by Lemma 2.8. Therefore, suppose there exists \(y_{o}\in R\), such that \(x_{o}d(y_{o})\in Z(R)-\{0\}\). If there exists \(c\in Z(R)\), such that \(d(c)\ne 0\), then \( x_{o}d(c)\in Z(R)\) implies that \(x_{o}\in Z(R)\).

Now, suppose \(d(Z(R))=\{0\}\). If \(x_{o}d(x_{o})=0\), then:

$$\begin{aligned} x_{o}d(x_{o}y_{o})=x_{o}x_{o}d(y_{o})\in Z(R), \end{aligned}$$

which implies that \(x_{o}\in Z(R)\) by Lemma 2.7. Therefore, suppose that \( x_{o}d(x_{o})\ne 0\), and hence, \(d(x_{o})\) is not a left zero divisor (Case 2: is not a right zero divisor) by Lemma 2.11(i). Observe that:

$$\begin{aligned} x_{o}d(x_{o}x_{o})= & {} x_{o}d(x_{o})x_{o}+x_{o}x_{o}d(x_{o}) \nonumber \\= & {} x_{o}d(x_{o})(2x_{o})\in Z(R), \end{aligned}$$
(4)

which implies that \(2x_{o}\in Z(R)\). From above, we deduce that \(d(2x_{o})=0\) and then:

$$\begin{aligned} 0= & {} x_{o}x_{o}d(2x_{o}) \\= & {} 2x_{o}x_{o}d(x_{o}) \\= & {} x_{o}d(x_{o})(2x_{o}\dot{)}\text {.} \end{aligned}$$

Thus, \(2x_{o}=0\) by Lemma 2.5.

(Case 2:

$$\begin{aligned}&(2x_{o})d(x_{o})x_{o} \\= & {} d(x_{o})x_{o}(2x_{o}) \\= & {} 2d(x_{o})x_{o}x_{o} \\= & {} 2x_{o}d(x_{o})x_{o} \\= & {} x_{o}d(x_{o})(2x_{o}\dot{)} \\= & {} (2x_{o}\dot{)}x_{o}d(x_{o}). \end{aligned}$$

Thus:

$$\begin{aligned} (2x_{o})(d(x_{o})x_{o}-x_{o}d(x_{o}))=0, \end{aligned}$$

and then either \(2x_{o}=0\) or \(d(x_{o})x_{o}=x_{o}d(x_{o})\). If \( d(x_{o})x_{o}=x_{o}d(x_{o})\), then:

$$\begin{aligned}&(2x_{o})d(x_{o}) \\= & {} d(x_{o})(2x_{o}) \\= & {} 2d(x_{o})x_{o} \\= & {} 2x_{o}d(x_{o}) \\= & {} x_{o}(d(2x_{o})) \\= & {} 0\text {.} \end{aligned}$$

However, \(d(x_{o})\) is not a right zero divisor. That implies \(2x_{o}=0\). Therefore, in both cases, we have that \(2x_{o}=0\).)

Hence, \(x_{o}d(x_{o}x_{o})=0\) from (4). Therefore:

$$\begin{aligned} x_{o}d(x_{o}x_{o}x_{o})=x_{o}x_{o}x_{o}d(x_{o})\in Z(R), \end{aligned}$$

which gives us \(x_{o}^{2}\in Z(R)\). If \(x_{o}^{2}=0\), then \( x_{o}^{2}d(x_{o})=0\) and, hence, \(x_{o}=0\), a contradiction with \( x_{o}d(x_{o})\ne 0\). Therefore, \(x_{o}^{2}\in Z(R)-\{0\}\) and \(x_{o}\) is not a zero divisor by Lemma 2.11(ii). From \(x_{o}^{2}d(y)=x_{o}d(y)x_{o}\) for all \(y\in R\), we deduce that:

$$\begin{aligned} x_{o}(x_{o}d(y)-d(y)x_{o})=0. \end{aligned}$$

Since \(x_{o}\) is not a zero divisor, we get that:

$$\begin{aligned} x_{o}d(y)=d(y)x_{o}\in Z(R) \end{aligned}$$

for all \(y\in R\).

(Case 2: By the same way, we can show that \(x_{o}^{2}\in Z(R)-\{0\}\). From

$$\begin{aligned} x_{o}^{2}d(y)x_{o}=d(y)x_{o}^{3}=x_{o}d(y)x_{o}^{2}=x_{o}^{3}d(y) \end{aligned}$$

for all \(y\in R\), we deduce that:

$$\begin{aligned} x_{o}^{2}(d(y)x_{o}-x_{o}d(y))=0\text {.} \end{aligned}$$

Therefore,

$$\begin{aligned} x_{o}d(y)=d(y)x_{o}\in Z(R) \end{aligned}$$
(5)

for all \(y\in R\). Therefore, \(d(x_{o})\) is not a left zero divisor also).

Therefore:

$$\begin{aligned} 0= & {} (2x_{o})d(y) \\= & {} d(y)(2x_{o}) \\= & {} 2d(y)x_{o} \\= & {} 2x_{o}d(y) \\= & {} x_{o}(2d(y)) \end{aligned}$$

for all \(y\in R\) and, hence, \(2d(R)=\{0\}\). Therefore, \(2R=\{0\}\) by Lemma 2.9.

For all \(y\in R\) and using (5), we have:

$$\begin{aligned} x_{o}d(x_{o}y)= & {} x_{o}^{2}d(y)+x_{o}d(x_{o})y \\= & {} d(y)x_{o}^{2}+yx_{o}d(x_{o}) \\= & {} d(y)x_{o}x_{o}+yd(x_{o})x_{o} \\= & {} d(yx_{o})x_{o} \\= & {} x_{o}d(yx_{o}), \end{aligned}$$

and then, \(x_{o}(d(x_{o}y)-d(yx_{o}))=0\). Thus, \(d(x_{o}y)=d(yx_{o})\) for all \(y\in R\). Replacing y by \(x_{o}y\) and using that \(d(Z(R))=\{0\}\), we get:

$$\begin{aligned} x_{o}^{2}d(y)= & {} d(x_{o}x_{o}y)=d(x_{o}yx_{o}) \\= & {} x_{o}d(yx_{o})+d(x_{o})yx_{o} \\= & {} x_{o}d(x_{o}y)+d(x_{o})yx_{o} \\= & {} x_{o}^{2}d(y)+x_{o}d(x_{o})y+d(x_{o})yx_{o}, \end{aligned}$$

and hence:

$$\begin{aligned} 0= & {} x_{o}d(x_{o})y+d(x_{o})yx_{o} \\= & {} d(x_{o})x_{o}y+d(x_{o})yx_{o} \\= & {} d(x_{o})(x_{o}y+yx_{o})\text {.} \end{aligned}$$

However, \(d(x_{o})\) is not a left zero divisor. Therefore:

$$\begin{aligned} x_{o}y=-yx_{o}=yx_{o} \end{aligned}$$

for all \(y\in R\) (since \(2R=\{0\}\)). Therefore, \(x_{o}\in Z(R)\). Therefore, \( d(x_{o})=0\) as \(d(Z(R))=\{0\}\), a contradiction with \(x_{o}d(x_{o})\ne 0\). Therefore, the case \(x_{o}d(x_{o})\ne 0\) is impossible. \(\square \)

Bell in [5] has proved that, if R is a 3-prime near-ring with a non-zero semigroup right (left) ideal U, such that \(Ux=\{0\}\) (\(xU=\{0\}\)) for some \( x\in R\), then \(x=0\). We conclude this paper by a generalization of his result.

Theorem 3.6

Let R be a 3-prime near-ring, such that \( Ux_{o}\subseteq Z(R)\) (\(x_{o}U\subseteq Z(R)\)), where U is a non-zero semigroup right (left) ideal of R and \(x_{o}\in R\). Then, \(x_{o}\in Z(R)\) and either \(x_{o}=0\) or R is a commutative ring.

Proof

If \(Ux_{o}=\{0\}\), then \(x_{o}=0\in Z(R)\) by the primeness of R. Therefore, suppose there exists \(u_{o}\in U\), such that \(u_{o}x_{o}\in Z(R)-\{0\}\). As \( u_{o}x_{o}x_{o}\in Z(R)\), we have that \(x_{o}\in Z(R)\). However, \(x_{o}\ne 0\), so \(U\subseteq Z(R)\) and then R is a commutative ring by Lemma 2.12. Similarly, we can write the proof for the case \(x_{o}U\subseteq Z(R)\), but we need to show that if \(x_{o}u_{o}\in Z(R)-\{0\}\), then \(x_{o}\ne 0\). Indeed, if \(x_{o}=0\), then:

$$\begin{aligned} 0u_{o}=(00)u_{o}=0u_{o}0=0 \end{aligned}$$

a contradiction with \(0u_{o}\ne 0\). Therefore, \(x_{o}\ne 0\). \(\square \)