Strong Commutativity Preserving Skew Derivations in Semiprime Rings

Abstract

Let R be a semiprime ring of characteristic different from 2, C its extended centroid, Z(R) its center, F and G non-zero skew derivations of R with associated automorphism $\alpha$ and m, n positive integers such that

$$\begin{aligned}{}[F(x),G(y)]_m=[x,y]^n ~\mathrm{for \, all}~x,y \in R. \end{aligned}$$

Then R is commutative.

1 Introduction

Let R be a prime ring of characteristic different from 2 with center Z(R), extended centroid C, right Martindale quotient ring $Q_r$, and symmetric Martindale quotient ring Q.

An additive map $d:R\rightarrow R$ is a derivation on R if $d(xy)=d(x)y+xd(y)$ for all $x,y\in R$. Let $a\in R$ be a fixed element. A map $d:R\rightarrow R$ defined by $d(x)=[a,x]=ax-xa$, $x\in R$, is a derivation on R, which is called the inner derivation defined by a. Many results in the literature indicate how the global structure of a ring R is often tightly connected to the behavior of additive maps defined on R. A well-known result of Posner [18], states that if d is a derivation of R such that $[d(x),x]\in Z(R)$ for any $x\in R$, then either $d=0$ or R is commutative.

In this paper, we study the structure of prime and semiprime rings having skew derivations satisfying strong commutativity preserving conditions. Specifically, let $\alpha$ be an automorphism of a ring R. An additive map $D:R\rightarrow R$ is called an $\alpha$-derivation (or a skew derivation) on R if $D (xy)=D(x)y+\alpha (x)D(y)$ for all $x,y\in R$. In this case, $\alpha$ is called an associated automorphism of D. Basic examples of $\alpha$-derivations are the usual derivations and the map $\alpha -id$, where id denotes the identity map. Let $b\in Q$ be a fixed element. Then a map $D:R\rightarrow R$ defined by $D(x)=bx-\alpha (x)b$, $x\in R$, is an $\alpha$-derivation on R and it is called an inner $\alpha$-derivation (an inner skew derivation) defined by b. If a skew derivation D is not inner, then it is called outer.

If $\mathcal {S}\subseteq \mathcal {R}$, the map $F: \mathcal {R}\rightarrow \mathcal {R}$ is called commutativity preserving on $\mathcal {S}$ if $[x,y]=0$ implies $[F(x),F(y)]=0$; it is called strong commutativity preserving (for brevity we will always say SCP) on $\mathcal {S}$ if $[F(x),F(y)]=[x,y]$, for all $x,y \in \mathcal {S}$.

In [1], Bell and Daif proved that if R is a semiprime ring admitting a derivation d which is SCP on the right ideal I of R, then $I\subseteq Z$. The natural possibility when an additive map preserves commutativity appears in a paper by Bresar and Miers [2]. They showed that any additive map F which is SCP on a semiprime ring R is of the form $F(x)=\lambda x+\mu (x)$, where $\lambda \in C$, $\lambda ^{2}=1$, and $\mu :R \rightarrow C$ is an additive map of R into C.

Later in [15], Lin and Liu extended this result to Lie ideals, in case the ring R is prime. More precisely they proved that if L is a non-central Lie ideal of R and F is an additive map satisfying $[F(x),F(y)]-[x,y]\in C$ for all $x,y \in L$, then $F(x)=\lambda x+\mu (x)$, where $\lambda \in C$, $\lambda ^2=1$, and $\mu :R\rightarrow C$, unless when $char(R)=2$ and R satisfies the standard identity $s_4$ of degree 4.

More recently, in [16] Liu showed that if R is a semiprime ring, I a non-zero right ideal of R, F, and G non-zero skew derivations of R, with associated automorphism $\alpha$ such that $[F(x),G(y)]=[x,y]$ for all $x,y \in I$, then $[\alpha (x),x]=0$ for any $x\in I$. Moreover, if $\alpha (I)\subseteq I$ then $\sigma (I)\subseteq Z(R)$. Finally, it is proved that if R is prime then R is commutative.

Here we continue this line of investigation and we examine what happens in case F and G are skew derivations of R such that $[F(x),G(y)]_m=[x,y]^n$ for all $x,y \in I$, where I is a non-zero ideal of R and $m, n \ge 1$ are positive integers.

The results we obtained are the following:

Theorem 1

Let R be a prime ring of characteristic different from 2, C its extended centroid, Z(R) its center, I a non-zero ideal of R, F, and G non-zero skew derivations of R with associated automorphism $\alpha$ and m, n positive integers such that

$$\begin{aligned}{}[F(x),G(y)]_m=[x,y]^n \,\mathrm{{for \, all}}~ x,y \in I. \end{aligned}$$

Then R is commutative.

Theorem 2

Let R be a semiprime ring of characteristic different from 2, C its extended centroid, Z(R) its center, F and G non-zero skew derivations of R with associated automorphism $\alpha$, m, n positive integers such that

$$\begin{aligned}{}[F(x),G(y)]_m=[x,y]^n \,\mathrm{for \, all}~ x,y \in R. \end{aligned}$$

Then R is commutative.

2 Preliminaries

We denote the set of all skew derivations on Q by $\mathrm{SDer}(Q)$. By a skew-derivation word, we mean an additive map $\Delta$ of the form $\Delta = d_{1},d_{2},\dots d_{m}$, with each $d_{i}\in \mathrm{SDer}(Q)$. A skew-differential polynomial is a generalized polynomial with coefficients in Q, of the form $\Phi (\Delta _{j}(x_{i}))$ involving non-commutative indeterminates $x_{i}$ on which the derivations words $\Delta _{j}$ act as unary operations. The skew-differential polynomial $\Phi (\Delta _{j}(x_{i}))$ is called a skew-differential identity on a subset T of Q if it vanishes for any assignment of values from T to its indeterminates $x_{i}$.

In order to prove our result, we need to recall the following known facts:

Fact 1

In [9], Chuang and Lee investigate polynomial identities with skew derivations. More precisely as a consequence of in [9, Theorem 1], we have that if D is an outer skew derivation of R which satisfies the generalized polynomial identity $\Phi (x_i,D(x_j))$, then $\Phi (x_i,y_j)$ is also a generalized polynomial identity for R, where $x_i$ and $y_j$ are distinct indeterminates.

Fact 2

Let R be a prime ring and I a two-sided ideal of R. Then I, R, and Q satisfy the same generalized polynomial identities with coefficients in Q (see [5]). Furthermore, I, R, and Q satisfy the same generalized polynomial identities with automorphisms [7, Theorem 1].

Fact 3

Recall that, in case $char(R)=0$, an automorphism $\alpha$ of Q is called Frobenius if $\alpha (x)=x$ for all $x\in C$. Moreover, in case $char(R)=p\ge 2$, an automorphism $\alpha$ is Frobenius if there exists a fixed integer t such that $\alpha (x)=x^{p^t}$ for all $x\in C$. In [7, Theorem 2], Chuang proves that if $\Phi (x_i,\alpha (x_i))$ is a generalized polynomial identity for R, where R is a prime ring and $\alpha \in Aut(R)$ an automorphism of R which is not Frobenius, then R also satisfies the non-trivial generalized polynomial identity $\Phi (x_i,y_i)$, where $x_i$ and $y_i$ are distinct indeterminates.

Fact 4

Let R be a domain and $\alpha \in Aut(R)$ an automorphism of R which is outer. In [13], Kharchenko proves that if $\Phi (x_i,\alpha (x_i))$ is a generalized polynomial identity for R, then R also satisfies the non-trivial generalized polynomial identity $\Phi (x_i,y_i)$, where $x_i$ and $y_i$ are distinct indeterminates.

Finally, let us mention that if R is a prime ring satisfying a non-trivial generalized polynomial identity and $\alpha$ an automorphism of R such that $\alpha (x)=x$ for all $x\in C$, then $\alpha$ is an inner automorphism of R [3, Theorem 4.7.4].

Lemma 1

Let R be a semiprime ring of characteristic different from 2, Q its symmetric Martindale quotient ring, F and G non-zero skew derivations of R, m, n positive integers such that $n\ge 2$ and

$$\begin{aligned} \left[ F(x),G(y)\right] _m=[x,y]^n~ \mathrm{{for \, all}}~ x,y \in R. \end{aligned}$$

(2.1)

Then R is commutative.

Proof

Assume first that R is prime. From the relation (2.1), we have both

$$\begin{aligned} \left[ F(x+z),G(y)\right] _m=[x+z,y]^n \end{aligned}$$

(2.2)

and

$$\begin{aligned} \left[ F(x+z),G(y)\right] _m=[F(x),G(y)]_m+[F(z),G(y)]_m=[x,y]^n+[z,y]^n. \end{aligned}$$

(2.3)

From (2.2) and (2.3), it follows that R satisfies the polynomial identity $[x+z,y]^n-[x,y]^n-[z,y]^n$. By Posner’s theorem [11, Theorem 2 p. 57, Lemma 1 p. 89], $Q\subseteq M_k(E)$, the ring of $k\times k$ matrices over a field E. Moreover, Q and $M_k(E)$ satisfy the same polynomial identities. If $k\ge 2$ and for $x=e_{21}$, $y=e_{11}$, $z=e_{11}$ we have the contradiction

$$\begin{aligned} 0=\left[ e_{21}+e_{12},e_{11}\right] ^n- \left[ e_{21},e_{11}\right] ^n-[e_{12},e_{11}]^n=(e_{21}-e_{12})^n\ne 0. \end{aligned}$$

Thus, $k=1$ and we have that Q is commutative, as well as R.

Let now R be semiprime. Since R is a semiprime ring for which $[x+z,y]^n-[x,y]^n-[z,y]^n$ is a polynomial identity, R is a subdirect product of prime rings $R_\alpha$, each of which still satisfies the identity $[x+z,y]^n-[x,y]^n-[z,y]^n$. In this case, by the above argument, any $R_\alpha$ is commutative. Thus, we conclude that R must be commutative. $\square$

Lemma 2

Let R be a non-commutative prime ring of characteristic different from 2, F and G non-zero skew derivations of R, m a positive integer such that

$$\begin{aligned}{}[F(x),G(y)]_m=[x,y]~\mathrm{{for \, all}}~x,y \in R. \end{aligned}$$

Then $char(R)=p > 0$ and m is odd.

Proof

For any $x,y \in R$ we have

$$\begin{aligned}{}[F(x),G(y+y)]_m=[x,y+y]=2[x,y] \end{aligned}$$

(2.4)

and also, by computing the m-th commutator

$$\begin{aligned}{}[F(x),G(2y)]_m=[F(x),2G(y)]_m=2^m[F(x),G(y)]_m=2^m[x,y]. \end{aligned}$$

(2.5)

Comparing (2.4) with (2.5) it follows $(2^m-2)[x,y]=0$ for all $x,y \in Q$. Hence, since R is not commutative and $char(R)\ne 2$, $char(R)=p\ne 0$ (since $2^m-2\equiv 0$, modulo p).

Moreover,

$$\begin{aligned} -[x,y]=[x,-y]=[F(x),G(-y)]_m=(-1)^m[F(x),G(y)]_m=(-1)^m[x,y] \end{aligned}$$

which implies that m must be an odd integer. $\square$

3 Results

Lemma 3

Let R be a non-commutative prime ring of characteristic different from 2, which is isomorphic to a dense subring of the ring of linear transformations of a vector space V over a division ring D, $\alpha :R\rightarrow R$ an automorphism of R and $q, u\in R$ such that

$$\begin{aligned}{}[qx,\alpha (y)u-uy]_m=[x,y] \end{aligned}$$

for all $x,y\in R$. Then $dim_DV=1$.

Proof

By Theorem 1 in [7], R and Q satisfy the same generalized polynomial identities with automorphisms and hence $[qx,\alpha (y)u-uy]_m-[x,y]$ is a generalized polynomial identity for Q. We assume that $dim_DV\ge 2$ and prove that a contradiction follows.

By [12, p. 79], there exists a semilinear automorphism $T\in End(V)$ such that $\alpha (x)=TxT^{-1}$ for all $x\in Q$. Hence, Q satisfies

$$\begin{aligned} \left[ qx,TyT^{-1}u-uy\right] _m=[x,y]. \end{aligned}$$

We notice that, if for any $v\in V$ there exists $\lambda _v \in D$ such that $T^{-1}uv=v\lambda _v$, then, by a standard argument, it follows that there exists a unique $\lambda \in D$ such that $T^{-1}uv=v \lambda$, for all $v\in V$ (see for example [8, Lemma 1]). In this case,

$$\begin{aligned} \alpha (x)uv=\left( TxT^{-1}\right) uv=Txv\lambda \end{aligned}$$

and

$$\begin{aligned} \biggl (\alpha (x)u-ux\biggr )v=T(xv\lambda )-uxv=T\left( T^{-1}uxv\right) -uxv=0 \end{aligned}$$

which implies that $\biggl (\alpha (x)u-ux\biggr )V=(0)$. Thus $\alpha (x)u-ux=0$ for any $x\in R$, since V is faithful, and by our assumption $0=[qx,\alpha (y)u-uy]_m=[x,y]$ for any $x,y \in R$, which is a contradiction, since R is not commutative.

Therefore, there exists $v\in V$ such that $\{v,T^{-1}uv\}$ are linearly D-independent. By the density of Q, there exist $r,s\in Q$ such that

$$\begin{aligned} rv=0,\quad rT^{-1}uv=v,\quad sv=T^{-1}uv,\quad sT^{-1}uv=\left( T^{-1}u\right) ^2v. \end{aligned}$$

Hence,

$$\begin{aligned} qrv=0,\quad \left( TsT^{-1}u-us\right) v=0,\quad [r,s]v=v \end{aligned}$$

and, by the main assumption, we get the contradiction

$$\begin{aligned} 0=\biggl ( \left[ qr,TsT^{-1}u-us\right] _m-[r,s]\biggr )v=-v\ne 0. \end{aligned}$$

$\square$

Lemma 4

Let R be a non-commutative prime ring of characteristic different from 2, C its extended centroid, Z(R) its center, I a non-zero ideal of R, F, and G non-zero skew derivations of R with associated automorphism $\alpha$, m a positive integer such that

$$\begin{aligned}{}[F(x),G(y)]_m=[x,y]\quad \text{ for } \text{ all } \quad x,y \in I. \end{aligned}$$

Then there exist $p,q,u \in Q$, with p invertible such that $F(x)=pxp^{-1}q-qx$ and $G(x)=pxp^{-1}u-ux$, for all $x\in R$.

Proof

It is known that I, R, and Q satisfy the same generalized polynomial identities with skew derivations and automorphisms, so that $[F(x),G(y)]_m=[x,y]$, for all $x,y \in Q$. Notice that if $m=1$ then the result follows by [15]. Thus we consider $m\ge 2$.

Fix $y_0 \in R$ and denote $z_0=G(y_0)$. Then for any $x\in Q$ we get $[F(x),z_0]_m=[x,y_0]$. In case F is an outer skew derivation of R, it is known that Q satisfies $[t,z_0]_m-[x,y_0]$, and in particular for $t=0$, we have $[x,y_0]=0$, for all $x\in Q$. This implies $y_0 \in C$. We may repeat this argument for any $y_0\in Q$ and conclude that Q is commutative, a contradiction. Therefore, in the sequel, we always assume that F is an inner skew derivation of R. Thus, there exists $0\ne q \in Q$ such that $F(x)=\alpha (x)q-qx$, for all $x\in R$.

Fix now $x_0 \in R$ and denote $z_0=F(x_0)$. Then for any $y\in Q$ we get $[z_0,G(y)]_m=[x_0,y]$. In case G is an outer skew derivation of R, it is known that Q satisfies $[z_0,t]_m-[x_0,y]$, and in particular for $t=0$, we have $[x_0,y]=0$ for all $y\in Q$. This implies $x_0 \in C$. We may repeat this argument for any $x_0\in Q$ and conclude that Q is commutative, a contradiction. Therefore, in the sequel, we always assume that G is an inner skew derivation of R. Thus, there exists $0\ne u \in Q$ such that $G(x)=\alpha (x)u-ux$, for all $x\in R$. Hence we have that

$$\begin{aligned}{}[\alpha (x)q-qx,\alpha (y)u-uy]_m=[x,y]~\mathrm{for\, all}~ x,y \in Q. \end{aligned}$$

(3.1)

We assume that $\alpha$ is not inner. In this case

$$\begin{aligned}{}[tq-qx,\alpha (y)u-uy]_m-[x,y] \end{aligned}$$

(3.2)

is satisfied by Q.

If $\alpha$ is not Frobenius, then by (3.2) it follows that Q satisfies the generalized identity

$$\begin{aligned}{}[tq-qx,zu-uy]_m-[x,y] \end{aligned}$$

(3.3)

and in particular for $x=y=0$ in (3.3), $[tq,zu]_m=0$, for all $t,z \in Q$. By applying the result in [14], and since $q\ne 0$ and $u\ne 0$, it follows that either Qq is a non-zero central left ideal of Q or Qu is a non-zero central left ideal of Q. In any case, Q is commutative, a contradiction.

Thus, we consider the case when $\alpha$ is Frobenius. Again by (3.2), Q satisfies

$$\begin{aligned}{}[-qx,\alpha (y)u-uy]_m-[x,y]. \end{aligned}$$

(3.4)

In case both $q\in C$ and $u\in C$, by (3.4) we have that

$$\begin{aligned} -qu^m[x,\alpha (y)-y]_m-[x,y] \end{aligned}$$

(3.5)

is satisfied by Q. Then by the main Theorem in [6] and since $qu^m\ne 0$, Q satisfies a non-trivial generalized polynomial identity. On the other hand, if either $q\notin C$ or $u\notin C$, then by (3.4) and again by the main Theorem in [6], Q satisfies a non-trivial generalized polynomial identity (hence Q is a GPI-ring in any case). Therefore, by [17, Theorem 3], Q is a primitive ring and it is a dense subring of the ring of linear transformations of a vector space V over a division ring D. Moreover, Q contains non-zero linear transformations of finite rank. By Lemma 3, it follows $dim_DV=1$, that is Q is a division algebra which is finite dimensional over C. If C is finite, then Q is finite, so that Q is a commutative field, which is a contradiction. So, we assume in all that follows that C is infinite and $char(Q)=p > 0$.

By using (3.4) in (3.2) it follows that

$$\begin{aligned}{}[tq,\alpha (y)u-uy]_m \end{aligned}$$

(3.6)

is satisfied by Q. Let $s\ge 1$ be such that $p^s\ge m$, and $k=p^s$, then by (3.6) we have that Q satisfies

$$\begin{aligned} \biggl [tq,\alpha (y)u-uy\biggr ]_{k} \end{aligned}$$

that is

$$\begin{aligned} \biggl [tq,\biggl (\alpha (x)u-ux\biggr )^k\biggr ]=0~\mathrm{for \, all}~ t,x,\in Q. \end{aligned}$$

(3.7)

Since $\alpha$ is Frobenius, $\alpha (\gamma )=\gamma ^{p^h}$, for all $\gamma \in C$ and some non-zero fixed integer h. Moreover, there exists $\lambda \in C$ such that $\lambda ^{p^h}\ne \lambda$, that is $\lambda ^{p^h-1}\ne 1$.

In particular, we choose $\gamma \in C$ such that $\gamma =\lambda ^{p^h-1}\ne 0$. In the relation (3.7), we replace x by $\lambda x$ and obtain that Q satisfies

$$\begin{aligned} \biggl [tq,\biggl (\lambda ^{p^h}\alpha (x)u-\lambda ux\biggr )^k\biggr ] \end{aligned}$$

that is

$$\begin{aligned} \biggl [tq,\biggl (\gamma \alpha (x)u-ux\biggr )^k\biggr ]. \end{aligned}$$

Let $\Phi$ and $\Omega$ be maps on Q, such that $\Phi (x) = -ux$ and $\Omega (x) = \alpha (x)u$, for any element x of Q. Thus it follows that $[tq,(\Phi (x)+\gamma \Omega (x))^k]=0$ for all $x\in Q$. Expanding $(\Phi (x)+\gamma \Omega (x))^k$, we get

$$\begin{aligned} \sum _{i=0}^n \gamma ^i \left( \sum _{(i,k-i)}\varphi _1\cdot \varphi _2\cdots \varphi _k\right) =0 \end{aligned}$$

where the inside summations are taken over all permutations of $k-i$ terms of the form $\Phi (x)$ and i terms of the form $\Omega (x)$. This means that each summation inside has exactly $k-i$ terms of the form $\Phi (x)$ and i terms of the form $\Omega (x)$ but in some different order. For any $j=0,\ldots ,k$, denote $y_j=\sum _{(j,k-j)}\varphi _1\cdot \varphi _2\cdots \varphi _k$, then we can write

$$\begin{aligned} (\Phi (x)+\gamma \Omega (x))^k=y_0+\gamma y_1+\gamma ^2 y_2+\ldots +\gamma ^k y_k \end{aligned}$$

so that

$$\begin{aligned} \biggl [tq,y_0+\gamma y_1+\gamma ^2 y_2+\ldots +\gamma ^k y_k\biggr ]=0. \end{aligned}$$

that is

$$\begin{aligned}{}[tq,y_0]+\gamma [tq,y_1]+\gamma ^2[tq,y_2]+\ldots +\gamma ^k [tq,y_k]=0. \end{aligned}$$

Here, we denote by $z_i=[tq,y_i]$, for any $i=1,\ldots ,k$, then

$$\begin{aligned} z_0+\gamma z_1+\gamma ^2 z_2+\ldots +\gamma ^k z_k=0. \end{aligned}$$

(3.8)

Replacing in the previous argument $\lambda$ successively by $1,\lambda ,\lambda ^2,\ldots ,\lambda ^k$, the equation (3.8) gives the system of equations

$$\begin{aligned} \begin{aligned}&z_0+z_1+z_2+\cdots +z_k=0\\&z_0+\gamma z_1+\gamma ^2z_2+\cdots +\gamma ^kz_k=0\\&z_0+\gamma ^2z_1+\gamma ^4z_2+\cdots +\gamma ^{2k}z_k=0\\&z_0+\gamma ^3z_1+\gamma ^6z_2+\cdots +\gamma ^{3k}z_k=0\\&\cdots \cdots \cdots \cdots \\&z_0+\gamma ^kz_1+\gamma ^{2k}z_2+\cdots +\gamma ^{k^2}z_k=0. \end{aligned} \end{aligned}$$

(3.9)

Moreover, since C is infinite, there exist infinitely many $\lambda \in C$ such that $\lambda ^{i(p^h-1)}\ne 1$ for $i=1,\ldots ,k$, that is there exist infinitely many $\gamma =\lambda ^{p^h-1} \in C$ such that $\gamma ^{i}\ne 1$ for $i=1,\ldots ,k$. Hence, the Vandermonde determinant (associated with the system (3.9))

$$\begin{aligned} \left| \begin{array}{ccccc} 1 &{} 1 &{} \cdots &{} \cdots &{} 1\\ 1 &{} \gamma &{} \gamma ^2 &{} \cdots &{} \gamma ^k\\ 1 &{} \gamma ^2 &{} \gamma ^4 &{} \cdots &{} \gamma ^{2k}\\ \cdots &{} \cdots &{} \cdots &{} \cdots &{} \cdots \\ 1 &{} \gamma ^k &{} \gamma ^{2k} &{} \cdots &{} \gamma ^{k^2} \end{array} \right| =\prod _{0\le i < j \le k} (\gamma ^i - \gamma ^j) \end{aligned}$$

is not zero. Thus, we can solve the above system (3.9) and obtain $z_i=0$ ($i=0,\ldots ,k$). In particular $z_0=0$, that is Q satisfies

$$\begin{aligned} \biggl [tq,(-ux)^k\biggr ]. \end{aligned}$$

(3.10)

Since $u\ne 0$ and $q\ne 0$, then, by (3.10) and [14], either uQ is a non-zero central right ideal of Q or Qq is a non-zero central left ideal of Q. In any case, we get the contradiction that Q is commutative.

The previous argument shows that the automorphism $\alpha$ must be inner, that is there exists an invertible element $p \in Q$, such that $\alpha (x)=pxp^{-1}$, for all $x\in R$, as required. $\square$

The following result is an easy consequence of [4, Theorem 1]. It will be useful in the proof of our result:

Lemma 5

Let R be a prime ring of characteristic different from 2, Z(R) its center, C its extended centroid. Let p be an invertible element of R, d the inner derivation of R, which is induced by p, that is, $d(x)=[p,x]$ for any $x\in R$, and $\beta (x)=pxp^{-1}$ for any $x\in R$, the inner automorphism of R induced by p. Assume that F is a non-zero skew derivation of R with associated automorphism $\beta$ and $0\ne a \in R$ such that

$$\begin{aligned} aF(x)-F(x)a=d(x)~\mathrm{for\, all}~ x \in R. \end{aligned}$$

Then $d=0$, $\beta$ is the identity map on R, F is an ordinary derivation of R and $a\in Z(R)$.

Proof

Firstly we notice that, since F is a skew derivation of R, $F(xy)=F(x)y+\beta (x)F(y)$. Thus, F is both a right $(1,\beta )$-generalized skew derivation and a left $(1,\beta )$-generalized skew derivation of R, in the sense of [4]. Therefore, we may apply Theorem 1 in [4], and one of the following holds:

• (1) (case (i) of Theorem 1 in [4]) $d(x)=apx-pxa$ for any $x\in R$. Hence $d(xy)=d(x)y+xd(y)=(apx-pxa)y+x(apy-pya)$, on the other hand $d(xy)=apxy-pxya$. Comparing the previous identities, we get

  $$\begin{aligned} -px[y,a]-xapy+xpya=0. \end{aligned}$$

  (3.11)

  In particular, for $x=y=p^{-1}$, it follows $ap^{-1}-p^{-1}a=0$ which implies $ap-pa=0$. Therefore $d(x)=p[a,x]$, moreover by (3.11) we get $-px[y,a]-xpay+xpya=0$, that is $[x,p][y,a]=0$. As an application of [19, Theorem 3], it follows that either $p\in C$ or $a\in C$, in any case $d=0$.

• (2) (cases (ii) and (iii) of [4, Theorem 1]) There exists $q\in Q$ such that $d(x)=p[q,x]$, for all $x\in R$. Hence $d(xy)=d(x)y+xd(y)=p[q,x]y+xp[q,y]$, on the other hand $d(xy)=p[q,xy]=p[q,x]y+px[q,y]$. Comparing the previous identities, we get $[p,x][q,y]=0$, for any $x,y\in R$, that is either $p\in C$ or $q\in C$. In any case $d=0$.

• (3) (case (iv) of [4, Theorem 1 ]) There exists $q\in Q$ such that $d(x)=px+p[q,x]$ for all $x\in R$. Hence $d(xy)=d(x)y+xd(y)=(px+p[q,x])y+x(py+p[q,y])$, on the other hand $d(xy)=pxy+p[q,xy]=pxy+p[q,x]y+px[q,y]$. Comparing the previous identities, we get $xpy+[p,x][q,y]=0$ for any $x,y\in R$. In particular, for $x=p^{-1}$, it follows the contradiction $y=0$, for all $y\in R$.

Since in any case $p\in C$, $\beta$ is the identity map on R and F is an ordinary derivation of R such that $[a,F(x)]=0$ for any $x\in R$. Hence, by the first Posner’s theorem in [18], it follows $a\in Z(R)$ or $F(R)\subseteq Z(R)$. In this last case, by second Posner’s theorem in [18], R is commutative. In any case we obtain that $a\in Z(R)$. $\square$

Lemma 6

Let $R=M_t(C)$ be the ring of $t\times t$ matrices over C, with $char(R)=l\ne 0, 2$, $m\ge 1$ be an odd integer, $p,q,u \in R$ with p invertible such that $F(x)=pxp^{-1}q-qx$, $G(x)=pxp^{-1}u-ux$ and

$$\begin{aligned} \left[ pxp^{-1}q-qx,pyp^{-1}u-uy\right] _m-[x,y] \end{aligned}$$

(3.12)

for any $x,y \in R$. Then t=1.

Proof

In (3.12) replace x, y by $p^{-1}x, p^{-1}y$, respectively, and denote $a=p^{-1}q$, $b=qp^{-1}$, $c=p^{-1}u$, $w=up^{-1}$. Therefore Q satisfies

$$\begin{aligned} \Phi (x,y)=\left[ xa-bx,yc-wy\right] _m-\left[ p^{-1}x,p^{-1}y\right] . \end{aligned}$$

(3.13)

In case $a \in C$, then $F=0$ and (3.12) implies $[x,y]=0$ for any $x,y \in R$, that is R is commutative. On the other hand, if $c \in C$, then $G=0$ and again it follows that R is commutative. Therefore, here we may assume that $t\ge 2$ and both $a\notin C$ and $c \notin C$. We prove that a number of contradiction follows.

Firstly, we notice that, for any inner automorphism $\varphi$ of $M_t(C)$, we have that

$$\begin{aligned} \left[ x\varphi (a)-\varphi (b)x,y\varphi (c)-\varphi (w)y\right] _m- \left[ \varphi (p^{-1})x,\varphi (p^{-1})y\right] \end{aligned}$$

(3.14)

is a generalized identity for R. We will make a frequent use of this fact.

As above, we denote by $e_{ij}$ the usual matrix unit, with 1 in the (i, j) entry and zero elsewhere, and say $a=\sum _{kl} a_{kl}e_{kl}$, $b=\sum _{kl}b_{kl}e_{kl}$, $c=\sum _{kl}c_{kl}e_{kl}$ and $w=\sum _{kl}w_{kl}e_{kl}$, for $a_{kl}, b_{kl}, c_{kl}, w_{kl}\in C$.

Suppose $t\ge 3$. In (3.13), we make the following choices: $x=e_{ji}$, $y=e_{kk}$, for any i, j, k different indices; moreover, we right multiply (3.13) by $e_{jj}$ and left multiply by $e_{kk}$. As a consequence we get $a_{kj}c_{ij}^m=0$, that is $a_{kj}c_{ij}=0$. Now, let $\varphi (x)=(1+e_{ki})x(1-e_{ki})$ and denote $(a_{ij}')_{t\times t}$ the entries of the matrix $\varphi (a)$, and $(c_{ij}')_{t\times t}$ the entries of the matrix $\varphi (c)$. By the above computations, we get $a_{kj}'c_{ij}'=0$, that is $(a_{kj}+a_{ij})c_{ij}=0$, which means $a_{ij}c_{ij}=0$. Thus, by Proposition 1 in [10], it follows that either $a \in C$ or $c\in C$, in any case a contradiction.

Therefore, we finally consider the case $t=2$ and $R=M_2(C)$. For $x=e_{ii}$ and $y=e_{ji}$ in (3.13), with $i\ne j$, right multiply (3.13) by $e_{jj}$ and left multiply by $e_{ii}$, it follows $a_{ij}(c_{ij}+w_{ij})^m=0$, that is

$$\begin{aligned} a_{ij}(c_{ij}+w_{ij}). \end{aligned}$$

(3.15)

Let now $x=y=e_{ij}$ in (3.13), with $i\ne j$, and right multiply (3.13) by $e_{ii}$ one has $(a_{ji}+c_{ji})^m=0$, that is

$$\begin{aligned} a_{ji}+c_{ji}~\mathrm{for \, all}~ i\ne j \end{aligned}$$

(3.16)

which means that $a+c$ is a diagonal matrix. In this case, a standard argument shows that $a+c$ is a central matrix, say $a=\lambda -c$, for $\lambda \in Z(R)$. Analogously, for $x=y=e_{ij}$ in (3.13), and left multiply (3.13) by $e_{jj}$ one has $(w_{ji}+b_{ji})^m=0$, that is

$$\begin{aligned} w_{ji}+b_{ji}~\mathrm{for \, all}~ i\ne j \end{aligned}$$

which means that $w+b$ is a diagonal matrix and, as above, $w+b$ is a central matrix, say $b=\mu -w$, for $\mu \in Z(R)$.

In other words, $p^{-1}q=\lambda -p^{-1}u$ and $qp^{-1}=\mu -up^{-1}$. Therefore, if either $\lambda =0$ or $\mu =0$ then $q=-u$, $\lambda =\mu =0$, and $F=-G$. On the other hand, if both $\lambda \ne 0$ and $\mu \ne 0$, it follows both $q=p\lambda -u$ and $q=p\mu -u$, that is $\lambda =\mu$ and easy computations show that $F=-G$ in any case.

Now, we write $v=c+w$ and let $v=\sum _{kl}v_{kl}e_{kl}$, for $v_{kl}\in C$. The next step is to prove that either v is diagonal or both a and c are diagonal matrices of R. To do this, we assume by contradiction that v is not diagonal, for example, let $v_{12}\ne 0$, and prove that a contradiction follows. In this case, by (3.15) and (3.16) we get $a_{12}=0$ and $c_{12}=0$. Moreover, if $v_{21}\ne 0$, then a and c are diagonal matrices and we are done. Thus we assume that $v_{21}=0$.

Let $\varphi (x)=(1+e_{12})x(1-e_{12})$ and $\chi (x)=(1-e_{12})x(1+e_{12})$ and denote $\varphi (a)=\sum _{kl} a_{kl}'e_{kl}$, $\varphi (c)=\sum _{kl}c_{kl}'e_{kl}$, $\varphi (v)=\sum _{kl}v'_{kl}e_{kl}$, $\chi (a)=\sum _{kl} a_{kl}''e_{kl}$, $\chi (c)=\sum _{kl}c_{kl}''e_{kl}$, $\chi (v)=\sum _{kl}v''_{kl}e_{kl}$.

We notice that, if both $v_{12}'\ne 0$ and $v_{12}''\ne 0$, then $a_{12}'=0$, $a_{12}''=0$ and also $c_{12}'=0$, $c_{12}''=0$ that is $a_{22}-a_{11}-a_{21}=0$, $-a_{22}+a_{11}-a_{21}=0$, $c_{22}-c_{11}-c_{21}=0$, $-c_{22}+c_{11}-c_{21}=0$, implying $a_{21}=0$ and $c_{21}=0$, so that a and c are diagonal and we are done.

Thus we may assume (without loss of generality) $v_{12}''=0$. We have proved that if $v_{12}\ne 0$ then $v_{12}=v_{22}-v_{11}$. Let $\theta (x)=(1+e_{21})x(1-e_{21})$ and $\theta (v)=\sum _{kl} v_{kl}'''e_{kl}$. Since $v_{12}'''=v_{12}\ne 0$, then $v_{12}'''=v_{22}'''-v_{11}'''$, that is $v_{12}=-v_{22}+v_{11}$, implying again the contradiction $v_{12}=0$.

The previous argument says that either v is diagonal or both a and c are diagonal matrices, and as above, we may conclude that either v is central or both a and c are central matrices of R.

Since if $a\in Z(R)$ and $c\in Z(R)$, then $F=G=0$, which is a contradiction, then we assume in what follows that $c+w=v=\nu \in Z(R)$, that is $p^{-1}q+qp^{-1}=\nu$. For $y=p$ in (3.12) it follows that

$$\begin{aligned} \biggl [pxp^{-1}q-qy,[p,u]\biggr ]_m=[x,p]. \end{aligned}$$

(3.17)

Assume that [p, u] is an invertible matrix in $M_2(C)$, thus $0\ne [p,u]^2\in Z(R)$ and by computations one has

$$\begin{aligned} 2^{m-1}\biggl (\left( pxp^{-1}q-qx\right) [p,u]^m-[p,u] \left( pxp^{-1}q-qx\right) [p,u]^{m-1}\biggr )=[x,p]. \end{aligned}$$

Since [p, u] is an invertible matrix and $m-1$ is even, then $[p,u]^{m-1}\in Z(R)$ and

$$\begin{aligned} 2^{m-1}\biggl (\left( pxp^{-1}q-qx\right) [p,u]^m-[p,u]^m \left( pxp^{-1}q-qx\right) \biggr )=[x,p]. \end{aligned}$$

Since $F\ne 0$ and $2^{m-1}[p,u]^m\ne 0$, we may apply Lemma 5 and obtain the contradiction $p\in Z(R)$.

Therefore [p, u] is not an invertible matrix in $M_2(C)$, i.e., $[p,u]^2=0$. Once again, for $y=p$ in (3.12) and since $m\ge 2$, we get

$$\begin{aligned} 0=\biggl [pxp^{-1}q-qx,[p,u]\biggr ]_m=[x,p] \end{aligned}$$

(3.18)

and as above we conclude with the contradiction $p\in Z(R)$. $\square$

4 The proof of Theorem 1

In light of Lemma 1, we may assume $n=1$. Since I, R, and Q satisfy the same generalized identities with skew derivations and automorphisms, we may assume

$$\begin{aligned}{}[F(x),G(y)]_m-[x,y]=0~\mathrm{for\,\, all}~ x,y \in Q. \end{aligned}$$

(4.1)

Moreover, by Lemma 4, there exists $p,q,u \in Q$ with p invertible such that $F(x)=pxp^{-1}q-qx$ and $G(x)=pxp^{-1}u-ux$ for all $x\in R$. Notice that, if either $F=0$ or $G=0$, then $[x,y]=0$ for all $x,y \in Q$, which means that Q is commutative, as well as R. Thus we also assume both $F\ne 0$ and $G\ne 0$. Hence Q satisfies the generalized polynomial identity

$$\begin{aligned} \left[ pxp^{-1}q-qx,pyp^{-1}u-uy\right] _m-[x,y]. \end{aligned}$$

(4.2)

In (4.2) replace x, y by $p^{-1}x, p^{-1}y$, respectively, and denote $a=p^{-1}q$, $b=qp^{-1}$, $c=p^{-1}u$, $w=up^{-1}$. Therefore Q satisfies

$$\begin{aligned} \Phi (x,y)=[xa-bx,yc-wy]_m-\left[ p^{-1}x,p^{-1}y\right] . \end{aligned}$$

(4.3)

In case $a \in C$, then $F=0$. On the other hand, if $c \in C$, then $G=0$. Therefore, we may assume that both $a\notin C$ and $c \notin C$, in particular Q is not commutative. We will prove that a number of contradiction follows. Since both a and c are not central elements, then (4.3) is a non-trivial generalized identity for R as well as for Q. Hence, Q is a primitive ring dense of linear transformations over a vector space V over C.

Assume first that $dim_CV=t$ is a finite integer. Thus, $Q\cong M_k(C)$ and by Lemma 6 it follows the contradiction that Q is commutative.

Let now $dim_CV=\infty$. Let $y_1,y_2\in Q$. By Litoff’s theorem (see Theorem 4.3.11 in [3]) there exists an idempotent element $e\in Q$ such that

$$\begin{aligned} y_1,y_2, a, b, c, w, p \in eQe \cong M_t(C) \end{aligned}$$

for some integer t. Of course $\Phi (x,y)=0$ for all $x,y \in eQe$. Thus by Lemma 6, either $a \in Ce$, or $c\in Ce$ or $[y_1,y_2]=0$. This means that either $F(eQe)=0$ or $G(eQe)=0$ or $[y_1,y_2]=0$. As above, if $F(eQe)=0$ or $G(eQe)=0$ then eQe is commutative. Therefore, in any case we get $[y_1,y_2]=0$. By the arbitrariness of $y_1,y_2 \in Q$, it follows that $[y_1,y_2]=0$ for any $y_1,y_2 \in Q$, that is Q is commutative, which is a contradiction.

5 The proof of Theorem 2

Also in this case, in light of Lemma 1, we may assume $n=1$. Let P be a prime ideal of R. Set $\overline{R}=R/P$ and write $\overline{x}=x+P \in \overline{R}$, for all $x\in R$. We start from

$$\begin{aligned} \biggl [\overline{F(x)},\overline{G(y)}\biggr ]_m=\overline{[x,y]}~\mathrm{for~ all}~ \overline{x}, \overline{y}\in \overline{R}. \end{aligned}$$

(5.1)

Case 1 $F(P)\subseteq P$, $\alpha (P)\not \subseteq P$.

In this case $\overline{\alpha (P)}$ is an ideal of $\overline{R}$. Moreover, for any $x\in R$, $p\in P$, $F(px)=F(p)x+\alpha (p)F(x)\in P$, so that $\alpha (p)F(x)\in P$ and $\overline{\alpha (p)F(x)}=\overline{0}$. Since $\overline{R}$ is prime and $\overline{F(x)}$ annihilates a non-zero ideal of $\overline{R}$, $\overline{F(x)}=\overline{0}$, that is $F(R)\subseteq P$. Thus $[F(x),G(y)]_m\in P$ for any $x,y \in R$ and, by (5.1), it follows $[R,R]\subseteq P$.

Case 2 $G(P)\subseteq P$, $\alpha (P)\not \subseteq P$.

Identical computations as in CASE 1, imply $[R,R]\subseteq P$.

Case 3 $F(P)\subseteq P$, $\alpha (P)\subseteq P$.

In this case $\overline{F}$ is a skew derivation of $\overline{R}$. If $G(P)\subseteq P$, then also $\overline{G}$ is a skew derivation of $\overline{R}$, and by the primeness of $\overline{R}$ and Theorem 1, we have that $\overline{R}$ is commutative, that is $[R,R]\subseteq P$.

Let now $G(P)\not \subseteq P$. Then $\overline{G(P)}$ is a non-zero ideal of $\overline{R}$. For any $x,y \in R$ and $q\in P$, we get

$$\begin{aligned} \biggl [F(x),G(yq)\biggr ]_m=[x,yq]~\mathrm{for\,\, all} ~x,y\in R \end{aligned}$$

that is

$$\begin{aligned} \biggl [F(x),G(y)q+\alpha (y)G(q)\biggr ]_m=[x,yq]~\mathrm{for~all}~x,y\in R \end{aligned}$$

implying that

$$\begin{aligned} \biggl [F(x),\alpha (y)G(q)\biggr ]_m\in P~\mathrm{for~all}~x,y\in R \end{aligned}$$

and

$$\begin{aligned} \biggl [\overline{F(x)},\overline{\alpha (y)G(q)}\biggr ]_m=\overline{0}~\mathrm{for\,\,all}~ \overline{x}, \overline{y}\in \overline{R}. \end{aligned}$$

(5.2)

In particular, let $h\ge 1$ be such that $p^h\ge m$, then (5.2) is

$$\begin{aligned} \biggl [\overline{F(x)},\overline{\alpha (y)G(q)}^{p^h}\biggr ]=\overline{0}~\mathrm{for\,\,all}~ \overline{x}, \overline{y}\in \overline{R}. \end{aligned}$$

(5.3)

By the primeness of $\overline{R}$ and since $\overline{\alpha (R)G(P)}$ is a non-zero ideal of $\overline{R}$, by applying the result in [14], it follows that either $\overline{F(R)}\subseteq Z(\overline{R})$ or $\overline{\alpha (R)G(P)}\subseteq Z(\overline{R})$. In the first case, by (5.1) we get $\overline{[x,y]}=\overline{0}$. In the latter case it follows that $\overline{R}$ is commutative. In any case $[R,R]\subseteq P$.

Case 4 $G(P)\subseteq P$, $\alpha (P)\subseteq P$.

In light of previous cases, we may also assume that $F(P)\not \subseteq P$. In this case, $\overline{G}$ is a skew derivation of $\overline{R}$, moreover $\overline{F(P)}$ is a non-zero ideal of $\overline{R}$. For any $x,y \in R$ and $q\in P$, we get

$$\begin{aligned} \biggl [F(qx),G(y)\biggr ]_m=[qx,y],\quad \forall ~x,y\in R \end{aligned}$$

that is

$$\begin{aligned} \biggl [F(q)x+\alpha (q)F(x),G(y)\biggr ]_m\in P,\quad \forall ~x,y\in R \end{aligned}$$

implying that

$$\begin{aligned} \biggl [F(q)x,G(y)\biggr ]_m\in P,\quad \forall ~x,y\in R \end{aligned}$$

and

$$\begin{aligned} \biggl [\overline{F(P)R},\overline{G(R)}\biggr ]_m=\overline{0}. \end{aligned}$$

(5.4)

Since $\overline{R}$ is prime and $\overline{F(P)R}$ is a non-zero ideal of $\overline{R}$, $\overline{F(P)R}$, and $\overline{R}$ satisfy the same generalized identities with skew derivation $\overline{G}$ and automorphism $\overline{\alpha }$. Therefore, by (5.4) we get

$$\begin{aligned} \biggl [\overline{R},\overline{G(R)}\biggr ]_m=\overline{0}. \end{aligned}$$

(5.5)

From (5.5) and (5.1) it follows again $\overline{[x,y]}=\overline{0}$ for any $x,y \in R$, that is $[R,R]\subseteq P$.

Case 5 $F(P)\not \subseteq P$, $G(P)\not \subseteq P$.

For any $x,y \in R$ and $q,u\in P$, by (5.1) we get

$$\begin{aligned} \biggl [F(xq),G(yu)\biggr ]_m=[xq,yu]\in P~\mathrm{for\, all}~x,y\in R \end{aligned}$$

that is

$$\begin{aligned} \biggl [F(x)q+\alpha (x)F(q),G(y)u+\alpha (y)G(u)\biggr ]_m\in P~\mathrm{for\, all}~x,y\in R \end{aligned}$$

implying that

$$\begin{aligned} \biggl [\alpha (x)F(q),\alpha (y)G(u)\biggr ]_m\in P~\text {for all}~x,y\in R \end{aligned}$$

and

$$\begin{aligned} \biggl [\overline{\alpha (R)F(P)},\overline{\alpha (R)G(P)}\biggr ]_m=\overline{0}. \end{aligned}$$

Assume first that $\alpha (P)\subseteq P$, so that both $\overline{F(P)}$ and $\overline{G(P)}$ are non-zero ideals of $\overline{R}$. On the other hand, if $\alpha (P)\not \subseteq P$ then $\overline{\alpha (R)F(P)}$ and $\overline{\alpha (R)G(P)}$ are left ideals of $\overline{R}$. In any case, for any $\overline{x}\in \overline{\alpha (R)F(P)}$ and $\overline{y}\in \overline{\alpha (R)G(P)}$, it follows $[\overline{x},\overline{y}]=\overline{0}$. Since $\overline{R}$ is prime, we apply again a reduced version of main result in [14], and conclude that either $\overline{\alpha (R)F(P)}\subseteq Z(\overline{R})$ or $\overline{\alpha (R)G(P)}\subseteq Z(\overline{R})$. In any case $\overline{R}$ contains a non-zero central ideal (either left or two-sided), so that $\overline{R}$ is commutative, i.e., $\overline{[x,y]}=\overline{0}$, for any $x,y \in R$ and $[R,R]\subseteq P$.

Therefore in any case $[R,R]\subseteq P$, for any prime ideal P of R. Then $[R,R]\subseteq \bigcap _{i} P_{i}=(0)$ (where $P_{i}$ are all prime ideals of R), that is R is commutative.