Keywords

2000 Mathematics Subject Classification:

1 Introduction

In all that follows, unless stated otherwise, R will be an associative ring with the center Z(R). For any \(x, y \in R\), the symbol [x, y] and \(x\circ y \) stand for the Lie commutator \(xy-yx\) and Jordan commutator \(xy+yx\), respectively. A ring R is called 2-torsion free, if whenever \(2x=0\), with \(x\in R\), then \(x=0\). If \(S\subseteq R\), then we can define the left (resp. right) annihilator of S as \(l(S)=\{x\in R\;|\;\textit{xs}=0\;\;\textit{for} \;\textit{all} \;\;s \in S\}\) (resp. \(r(S)=\{x\in R\;|\;\textit{sx}=0\;\;\textit{for} \;\textit{all} \;\;s \in S\}\)).

Recall that a ring R is prime if for any \(a, b\in R\), \(aRb=(0)\) implies \(a=0\) or \(b=0\), and is semiprime if for any \(a\in R\), \(aRa=(0)\) implies \(a=0\). An additive subgroup U of R is said to be a Lie ideal of R if \([u,r]\in U\) for all \(u\in U\) and \(r\in R\), and a Lie ideal U is called square-closed if \(u^{2}\in U\) for all \(u\in U\). By a derivation, we mean an additive mapping \(d:R\longrightarrow R\) such that \(d(xy)=d(x)y+xd(y)\) for all \(x,y\in R\). Let \(\alpha \) and \(\beta \) be endomorphisms of R, an additive mapping \(d: R\longrightarrow R\) is said to be an \((\alpha ,\beta )\)-derivation if \(d(xy)=d(x)\alpha (y)+\beta (x)d(y)\) holds for all \(x,y \in R\). Following [1], an additive mapping \(F: R \longrightarrow R\) is called a generalized \((\alpha , \beta )\)-derivation on R if there exists an \((\alpha , \beta )\)-derivation \(d: R \longrightarrow R\) such that \(F(xy)=F(x)\alpha (y)+\beta (x)d(y)\) holds for all \( x,y \in R.\) Note that for \(I_{R}\) the identity map on R, this notion includes those of \((\alpha , \beta )\)-derivation when \(F = d\), of derivation when \(F = d\) and \(\alpha = \beta = I_{R}\), and of generalized derivation, which is the case when \(\alpha = \beta = I_{R}\).

Many results indicate that the global structure of a ring R is often tightly connected to the behavior of additive mappings defined on R. A well known result of Posner [10] states that if R is a prime ring and d a nonzero derivation of R such that \([d(x), x]\in Z(R\)) for all \(x\in R\), then R must be commutative. Over the last few decades, several authors have investigated the relationship between the commutativity of the ring R and certain specific types of derivations of R (see [3–5, 7] where further references can be found).

Daif and Bell [6] showed that if in a semiprime ring R there exists a nonzero ideal I of R and a derivation d such that \(d([x,y])-[x,y]=0\) or \(d([x,y])+[x,y]=0\) for all \(x,y\in I\), then \(I\subseteq Z(R)\). In particular, if \(I=R\) then R is commutative. At this point, the natural question is what happens in case the derivation is replaced by a generalized derivation. In [11], Quadri et al., proved that if R is a prime ring, I a nonzero ideal of R and F a generalized derivation associated with a nonzero derivation d such that any one of the following holds:  (i) \(F([x,y])-[x,y]=0\)  (ii) \(F([x,y])+[x,y]=0\)  (iii) \(F(x\circ y)-x\circ y=0\)  (iv) \(F(x\circ y)+x\circ y=0\) for all \(x,y\in I\), then R is commutative. Following this line of investigation, Ali, Kumar and Miyan [2], explored the commutativity of a prime ring R admitting a generalized derivation F satisfying any one of the following conditions:  (i) \(F([x,y])-[x,y]\in Z(R)\)  (ii) \(F([x,y])+[x,y]\in Z(R)\)  (iii) \(F(x\circ y)-x\circ y\in Z(R)\)  (iv) \(F(x\circ y)+x\circ y\in Z(R)\) for all \(x,y\in I\), a nonzero right ideal of R. On the other hand, Marubayashi et al. [8], established that if a 2-torsion free prime ring R admits a nonzero generalized \((\alpha , \beta )\)-derivation F associated with an \((\alpha , \beta )\)-derivation d such that either \(F([u,v])=0\) or \(F(u\circ v)=0\) for all \(u,v\in U\), where U is a nonzero square-closed Lie ideal of R, then \(U\subseteq Z(R)\). In the present paper, our purpose is to prove the cited results for the case when the generalized \((\alpha ,\alpha )\)-derivation F acts on one sided ideal of R.

2 Main Results

In the remaining part of this paper, \(\alpha \) and \(\beta \) will denote automorphisms of R. And we shall do a great deal of calculation with commutators and anti-commutators, routinely using the following basic identities: For all \(x,y,z \in R;\)

$$[xy, z] = x[y, z] + [x, z]y ~\text{ and }~[x, yz] = y[x, z] + [x, y]z$$
$$ xo(yz) = (xoy)z-y[x,z] = y(xoz)+[x,y]z$$
$$ (xy)oz = x(yoz)-[x,z]y = (xoz)y+x[y,z]. $$

Theorem 2.1

Let R be a prime ring with center Z(R) and J a nonzero left ideal of R. Suppose that R admits a generalized \((\alpha ,\alpha )\)-derivation F associated with a nonzero \((\alpha ,\alpha )\)-derivation d such that \(d(Z(R))\ne (0)\). If \(F([x,y])-\alpha ([x,y])\in Z(R)\) for all \(x,y\in J\), then R is commutative.

Proof

It is easy to check that \(d(Z(R))\subseteq Z(R)\). Since \(d(Z(R))\ne (0)\), there exists \(0\ne c\in Z(R)\) such that \(0\ne d(c)\in Z(R)\). By assumption, we have

$$\begin{aligned} F([x,y])-\alpha ([x,y])\in Z(R) ~\text{ for } \text{ all }~ x, y\in J. \end{aligned}$$
(1)

Replacing y by cy in (1), we get

$$\begin{aligned} (F([x,y])-\alpha ([x,y]))\alpha (c)+\alpha ([x,y])d(c)\in Z(R) ~\text{ for } \text{ all }~ x, y\in J. \end{aligned}$$
(2)

Combining (1) and (2) and noting that the fact \(\alpha (c)\in Z(R)\), we find that \(\alpha ([x,y])d(c)\in Z(R)\), which implies that \([\alpha ([x,y])d(c),r]=0=[\alpha ([x,y]),r]d(c)\) for all \(x,y\in J\) and \(r\in R\). Since R is prime and \(0\ne d(c)\in Z(R)\), we have

$$\begin{aligned}{}[\alpha ([x,y]),r]=0 ~\text{ for } \text{ all }~ x, y\in J; r\in R. \end{aligned}$$
(3)

Replacing y by yx in (3) and using (3), we get

$$\begin{aligned} \alpha ([x,y])[\alpha (x),r]=0 ~\text{ for } \text{ all }~ x, y\in J; r\in R. \end{aligned}$$
(4)

Replacing r by \(r\alpha (s)\) in (4) and using (4), we arrive at \(\alpha ([x,y])r[\alpha (x),\alpha (s)]=0\) for all \(x,y\in J\) and \(r,s\in R\). The primeness of R yields that for each \(x\in J\), either \(\alpha ([x,y])=0\) or \([\alpha (x),\alpha (s)]=0\). Equivalently, either \([x,J]=0\) or \([x,R]=0\). Set \(J_{1} = \{ x \in J \mid [x,J]=0 \}\) and \(J_{2} = \{ x \in J\mid [x,R]=0 \}\). Then, \(J_{1}\) and \(J_{2}\) are both additive subgroups of I such that \(J=J_{1}\cup J_{2}\). Thus, by Brauer’s trick, we have either \(J=J_{1}\) or \(J=J_{2}\). If \(J=J_{1}\), then \([J,J]=0\), and if \(J=J_{2}\), then \([J,R]=0\). In both cases, we conclude that J is commutative and so, by a result of [9], R is commutative.

Corollary 2.2

Let R be a prime ring with center Z(R) and J a nonzero left ideal of R. Suppose that R admits a generalized \((\alpha ,\alpha )\)-derivation F associated with a nonzero \((\alpha ,\alpha )\)-derivation d such that \(d(Z(R))\ne (0)\). If \(F(xy)-\alpha (xy)\in Z(R)\) for all \(x,y\in J\), then R is commutative.

Proof

For any \(x,y\in J\), we have \(F([x,y])-\alpha ([x,y])=(F(xy)-\alpha (xy))-(F(yx)-\alpha (yx))\in Z(R)\), and hence the result follows.

Theorem 2.3

Let R be a prime ring with center Z(R) and J a nonzero left ideal of R. Suppose that R admits a generalized \((\alpha ,\alpha )\)-derivation F associated with a nonzero \((\alpha ,\alpha )\)-derivation d such that \(d(Z(R))\ne (0)\). If \(F([x,y])+\alpha ([x,y])\in Z(R)\) for all \(x,y\in J\), then R is commutative.

Proof

If \(F([x,y])+\alpha ([x,y])\in Z(R)\) for all \(x,y\in J\), then the generalized \((\alpha ,\alpha )\)-derivation \(-F\) satisfies the condition \((-F)([x,y])-\alpha ([x,y])\in Z(R)\) for all \(x,y\in J\). It follows from Theorem 2.1 that R is commutative.

Theorem 2.4

Let R be a prime ring with center Z(R) and J a nonzero left ideal of R. Suppose that R admits a generalized \((\alpha ,\alpha )\)-derivation F associated with a nonzero \((\alpha ,\alpha )\)-derivation d such that \(d(Z(R))\ne (0)\). If \(F(x \circ y)-\alpha (x \circ y)\in Z(R)\) for all \(x,y\in J\), then R is commutative.

Proof

 We are given that

$$\begin{aligned} F(x \circ y)-\alpha (x \circ y)\in Z(R) ~\text{ for } \text{ all }~ x, y\in J. \end{aligned}$$
(5)

Since \(d(Z(R))\ne (0)\), there exists \(0\ne c\in Z(R)\) such that \(0\ne d(c)\in Z(R)\). Replacing y by cy in (5), we get

$$\begin{aligned} (F(x \circ y)-\alpha (x \circ y))\alpha (c)+\alpha (x \circ y)d(c)\in Z(R) ~\text{ for } \text{ all }~ x, y\in J. \end{aligned}$$
(6)

Combining (5) and (6), we find that \(\alpha (x \circ y)d(c)\in Z(R)\) and hence \(\alpha (x \circ y)\in Z(R)\). This implies that

$$\begin{aligned}{}[\alpha (x \circ y),r]=0 ~\text{ for } \text{ all }~ x, y\in J; r\in R. \end{aligned}$$
(7)

Replacing yx for y in (7) and using (7), we have

$$\begin{aligned} \alpha (x \circ y)[\alpha (x),r]=0 ~\text{ for } \text{ all }~ x, y\in J; r\in R. \end{aligned}$$
(8)

Replacing r by \(r\alpha (s)\) in (8) and using (8), we have \(\alpha (x \circ y)r[\alpha (x),\alpha (s)]=0\) for all \(x,y\in J\) and \(r,s\in R\). The primeness of R yields that for each \(x\in J\), either \(\alpha (x \circ y)=0\) or \([\alpha (x),\alpha (s)]=0\). Now applying similar arguments as used in the proof of Theorem 2.1, we have either \(x \circ y=0\) for all \(x,y\in J\); or \([J,R]=0\). In the former case, replacing x by xz and using the fact \(x \circ y=0\) we find \([x,y]z=0\) for all \(x,y,z\in J\). This implies that \([x,y]J=0\) and hence \([x,y]RJ=0\). Since J is nonzero and R is prime, we get \([J,J]=0\). Thus, J is commutative and so R. In the latter case, we have \([J,R]=0\), in particular \([J,J]=0\) and hence we get the required result.

Theorem 2.5

Let R be a prime ring with center Z(R) and J a nonzero left ideal of R. Suppose that R admits a generalized \((\alpha ,\alpha )\)-derivation F associated with a nonzero \((\alpha ,\alpha )\)-derivation d such that \(d(Z(R))\ne (0)\). If \(F(x \circ y)+\alpha (x \circ y)\in Z(R)\) for all \(x,y\in J\), then R is commutative.

Proof

 If \(F(x \circ y)+\alpha (x \circ y)\in Z(R)\) for all \(x,y\in J\), then the generalized \((\alpha ,\alpha )\)-derivation \(-F\) satisfies the condition \((-F)(x \circ y)-\alpha (x \circ y)\in Z(R)\) for all \(x,y\in J\). It follows from Theorem 2.4 that R is commutative.

Corollary 2.6

Let R be a prime ring with center Z(R) and J a nonzero left ideal of R. Suppose that R admits a generalized \((\alpha ,\alpha )\)-derivation F associated with a nonzero \((\alpha ,\alpha )\)-derivation d such that \(d(Z(R))\ne (0)\). If \(F(xy)+\alpha (xy)\in Z(R)\) for all \(x,y\in J\), then R is commutative.

Proof

 For any \(x,y\in I\), we have \(F(x \circ y)+\alpha (x \circ y)=(F(xy)+\alpha (xy))+(F(yx)+\alpha (yx))\in Z(R)\), and hence our result follows.

Theorem 2.7

Let R be a prime ring and J a nonzero left ideal of R such that \(r(J)=0\). If R admits a generalized \((\alpha ,\beta )\)-derivation F associated with a nonzero \((\alpha ,\beta )\)-derivation d such that \(F([x,y])=0\) for all \(x,y\in J\), then R is commutative.

Proof

 By assumption, we have

$$\begin{aligned} F([x,y])=0 ~\text{ for } \text{ all }~ x, y\in J. \end{aligned}$$
(9)

Replacing y by yx in (9) and using (9), we get \(\beta ([x,y])d(x)=0\), which implies

$$\begin{aligned}{}[x,y]\beta ^{-1}(d(x))=0 ~\text{ for } \text{ all }~ x, y\in J. \end{aligned}$$
(10)

Now substituting ry for y in (10) and using (10), we obtain \([x,r]y\beta ^{-1}(d(x))=0\) for all \(x,y\in J\) and \(r\in R\). In particular, \([x,R]RJ\beta ^{-1}(d(x))=0\) for all \(x\in J\). The primeness of R yields that for each \(x\in J\), either \([x,R]=0\) or \(J\beta ^{-1}(d(x))=0\), in this case \(d(x)=0\). In view of similar arguments as used in the proof of Theorem 2.1, we have either \([J,R]=0\) or \(d(J)=0\). If \([J,R]=0\), then J is commutative and we are done. If \(d(J)=0\), then \(0=d(RJ)=d(R)\alpha (J)+\beta (R)d(J)\), which reduces to \(d(R)\alpha (J)=0\). And hence \(d(R)\alpha (RJ)=0=d(R)\alpha (R)\alpha (J)=d(R)R\alpha (I)\). Since J is nonzero and the last relation forces that \(d=0\), contradiction.

Using the same techniques with necessary variations, we can prove the following:

Theorem 2.8

Let R be a prime ring and J a nonzero left ideal of R such that \(r(J)=0\). If R admits a generalized \((\alpha ,\beta )\)-derivation F associated with a nonzero \((\alpha ,\beta )\)-derivation d such that \(F(x \circ y)=0\) for all \(x,y\in J\), then R is commutative.

The following example demonstrates that R to be prime is essential in the hypothesis of Theorems 2.1, 2.3, 2.4 and 2.5.

Example 2.9

Let S be any ring. Next, let R=\(\left\{ \left( \begin{array}{ccc}0&{}a&{}b \\ 0&{}0&{}c \\ 0&{}0&{}0\end{array} \right) |a,b,c \in S \right\} \) and J=\(\left\{ \left( \begin{array}{ccc}0&{}a&{}b \\ 0&{}0&{}0 \\ 0&{}0&{}0\end{array} \right) |a,b \in S \right\} \), a nonzero left ideal of R. Define maps \(F,~d~,\alpha :R\longrightarrow R\) as follows: \(F\left( \begin{array}{ccc}0&{}a&{}b \\ 0&{}0&{}c \\ 0&{}0&{}0\end{array} \right) =\left. \left( \begin{array}{ccc}0&{}0&{}-c \\ 0&{}0&{}0 \\ 0&{}0&{}0\end{array} \right) \right. ,\) \(~d\left( \begin{array}{ccc}0&{}a&{}b \\ 0&{}0&{}c \\ 0&{}0&{}0\end{array} \right) =\left. \left( \begin{array}{ccc}0&{}0&{}c \\ 0&{}0&{}0 \\ 0&{}0&{}0\end{array} \right) \right. ,\)

\(\alpha \left( \begin{array}{ccc}0&{}a&{}b \\ 0&{}0&{}c \\ 0&{}0&{}0\end{array} \right) =\left. \left( \begin{array}{ccc}0&{}-a&{}~b \\ 0&{}0&{}-c \\ 0&{}0&{}0\end{array} \right) \right. ,\) Then, it is straightforward to check that F is a generalized \((\alpha , \alpha )\)-derivation associated with a nonzero \((\alpha , \alpha )\)-derivation d such that \(d(Z(R))\ne (0)\). It is easy to see that (i) \(F([x,y])-\alpha ([x,y])\in Z(R)\) (ii) \(F([x,y])+\alpha ([x,y])\in Z(R)\) (iii) \(F(x \circ y)-\alpha (x \circ y)\in Z(R)\)  (iv) \(F(x \circ y)-\alpha (x \circ y)\in Z(R)\) for all \(x,y\in J\), however R is not commutative.