On commutativity of prime rings with skew derivations

We have

for every $a,b\in\mathscr{R}$. As $\mathscr{R}$ and $\mathscr{Q}$ satisfy the same GPIs with automorphisms by Fact 2, and hence it is a GPI for $\mathscr{Q}$. We prove it by contradiction. We assume that $dim_{D}V\geq 2$. There exists a semi-linear automorphism $\Phi\in End(V)$, by [20, p.79], such that $\mathscr{T}(a)=\Phi a\Phi^{-1}$ $\forall a\in\mathscr{Q}$. Hence, $\mathscr{Q}$ satisfies

Suppose that $\Phi u\not\in span_{D}\{u,\Phi^{-1}\vartheta u\}$, then $\{u,\Phi u,\Phi^{-1}\vartheta u\}$ is linearly $D$-independent. By density theorem for $\mathscr{R}$, there exists $a,b\in\mathscr{R}$ such that

The above relation gives $[a,b]u=0$, $[a,b]\Phi^{-1}\vartheta u=2u$ and $[a,b]\Phi u=u$. This implies that

a contradiction. Now, we assume that $\Phi u\in Span_{D}\{u,\Phi^{-1}\vartheta u\}$, then $\Phi u=u\zeta+\Phi^{-1}\vartheta u\theta$ for some $\zeta,\theta\in D$. We see that $\theta\neq 0$ otherwise if $\theta=0$, then we get $\Phi u=u\zeta$ and hence this gives that $u=\Phi^{-1}u\zeta$. Again by density theorem for $\mathscr{R}$, $\exists a,b\in\mathscr{R}$, we have

The above expression again gives that a contradiction

For $u\in V$, the set $\{u,\Phi^{-1}\vartheta u\}$ is $D$-dependent. By Fact 1, $\exists\Delta\in D$ such that $\Phi^{-1}\vartheta u=u\Delta$, $\forall u\in V$ and hence we have

and

The last expression forces that $(\mathscr{T}(a)\vartheta-\vartheta a)V=(0)$ $\forall a\in\mathscr{R}$, and hence $\mathscr{T}(a)V=(0)$ $\forall a\in\mathscr{R}$ and as $V$ is faithful, it yields that $\mathscr{T}(a)=0$ $\forall a\in\mathscr{R}$. This is a contradiction.

First we assume that $\mathscr{T}$ is an identity automorphism of $\mathscr{R}$. Then we have that $([[a,b]\vartheta,[a,b]])^{m}=[[a,b]\vartheta,[a,b]]$ is a GPI of $\mathscr{R}$. On contrary we assume that $\vartheta\not\in\mathscr{C}$. Since the identity $([[a,b]\vartheta,[a,b]])^{m}=[[a,b]\vartheta,[a,b]]$ is satisfied by $\mathscr{Q}$ (Fact 2). As $\vartheta\not\in\mathscr{C}$, then the above identity is an non-trivial GPI for $\mathscr{Q}$. By Martindale’s theorem in [24], $\mathscr{Q}$ is primitive ring which is isomorphic to a dense ring of linear transformations of a vector space $V$ over $\mathscr{C}$. Assume that dim$\mathscr{C}(V)=l$, where $1<l\in\mathbb{Z}^{+}$. For this situation, we take $\mathscr{Q}=M_{l}(\mathscr{C})$ as a ring of $l\times l$ matrices over the field $\mathscr{C}$ such that $([[a,b]\vartheta,[a,b]])^{m}=[[a,b]\vartheta,[a,b]]$ for all $a,b\in M_{l}(\mathscr{C})$. Let $e_{ij}$ be the usual unit matrix with $1$ in $(i,j)$-entry and zero elsewhere. First, we claim that $\vartheta$ is a diagonal matrix. Say $\vartheta=\sum_{ij}e_{ij}\vartheta_{ij}$, where $\vartheta_{ij}\in\mathscr{C}$. Choose $a=e_{ij},b=e_{jj}$. Then by the hypothesis, we have $([e_{ij}\vartheta,e_{ij}])^{m}=[e_{ij}\vartheta,e_{ij}]$, i.e, $e_{ij}\vartheta_{ij}=0$ and so $\vartheta_{ji}=0$, for any $i\neq j$ and hence $\vartheta$ is a diagonal matrix. Since $\xi\in Aut_{\mathscr{C}}(\mathscr{Q})$, the expression

is also a GPI for $\mathscr{Q}$, therefore $\xi(\vartheta)$ is also diagonal. The automorphism, in particular $\xi(\vartheta)=(1+e_{ij})\vartheta(1-e_{ij})$, for any $i\neq j$ and say $\vartheta^{\xi}=\sum_{ij}e_{ij}\vartheta^{\prime}_{ij}$, where $\vartheta^{\prime}_{ij}\in\mathscr{C}$. Since $\vartheta^{\prime}_{ij}=0$, then we get $0=\vartheta^{\prime}_{ij}=\vartheta_{jj}-\vartheta_{ii}$, by easy computation. So that $\vartheta_{jj}=\vartheta_{ii}$ hold for any $i\neq j$, and we get a contradiction that $\vartheta\in\mathscr{C}$. Assume that $dim_{\mathscr{C}}V=\infty$.

By Martindale’s theorem [24], it observes that $Soc(\mathscr{Q})=F\neq(0)$ and $eFe$ is finite dimensional simple central algebra over $\mathscr{C}$, for any minimal idempotent element $e\in F$. We can also suppose that $F$ is non-commutative, because else $\mathscr{Q}$ must be commutative. Clearly, $F$ satisfies $([[a,b]\vartheta,[a,b]])^{m}=[[a,b]\vartheta,[a,b]]$ (see, for example, the proof of [23, Theorem 1]). As $F$ is a simple ring, either $F$ does not contain any non-trivial idempotent element or $F$ is generated by its idempotents. In this last case, assume that $F$ contains two minimal orthogonal idempotent elements $e$ and $f$. Using the assumption, one can see that, for $[a,b]=[ea,f]=eaf$, we have

in this case we get $f\vartheta eaf\vartheta eaf\vartheta e=0$, and primeness of $\mathscr{R}$, we get $f\vartheta e=0$ for any rank $1$ orthogonal idempotent element $e$ and $f$. Notably, for any rank $1$ idempotent element $e$, we have $e\vartheta(1-e)=0$ and $(1-e)\vartheta e=0$, that is, $e\vartheta=e\vartheta e=\vartheta e$. Hence, $[\vartheta,e]=0$ gives that $F$ is commutative or $\vartheta\in\mathscr{C}$. We get a contradiction, in this case. Now, we consider when $F$ cannot contain two minimal orthogonal idempotent elements and so, $F=D$ for suitable finite dimensional division ring $D$ over its center which implies that $\mathscr{Q}=F$ and $\vartheta\in F$. By [20, Theorem 2.3.29] (see also [23, Lemma 2]), there exists a field $\mathbb{K}$ such that $F\subseteq M_{n}(\mathbb{K})$ and $M_{n}(\mathbb{K})$ satisfies $([[a,b]\vartheta,[a,b]])^{m}=[[a,b]\vartheta,[a,b]]$. If $n=1$ then $F\subseteq\mathbb{K}$ and we have also a contradiction. Moreover, as we have just seen, if $n\geq 2$, then $\vartheta\in Z(M_{n}(\mathbb{K}))$. Finally, if $F$ does not contain any non-trivial idempotent element, then $F$ is finite dimensional division algebra over $\mathscr{C}$ and $\vartheta\in F=\mathscr{R}\mathscr{C}=\mathscr{Q}$. If $\mathscr{C}$ is finite, then $F$ is finite division ring, that is, $F$ is a commutative field and so $\mathscr{R}$ is commutative too. If $\mathscr{C}$ is infinite, then $F\bigotimes_{\mathscr{C}}\mathbb{K}\cong M_{n}(\mathbb{K})$, where $\mathbb{K}$ is a splitting field of $F$. We get the conclusion. Henceforward, $\mathscr{T}$ is non-identity automorphism of $\mathscr{R}$. Now, we have two cases: Case I: If $\mathscr{T}$ is $\mathscr{Q}-$inner, then there exists an invertible element $\alpha$ of $\mathscr{Q}$ such that $\mathscr{T}(a)=\alpha a\alpha^{-1}$ for every $a\in\mathscr{R}$. Thus, $([\alpha[a,b]\alpha^{-1}\vartheta,[a,b]])^{m}=[\alpha[a,b]\alpha^{-1}\vartheta,[a,b]]$ for every $a,b\in\mathscr{R}$. If $\alpha^{-1}\vartheta\in\mathscr{C}$, then $\mathscr{R}$ satisfies $([\alpha[a,b],[a,b]])^{m}=[\vartheta[a,b],[a,b]]$ and we get the conclusion as above. Now we assume that $\alpha^{-1}\vartheta\not\in\mathscr{C}$, therefore

is a non-trivial GPI for $\mathscr{R}$ and hence for $\mathscr{Q}$ by Fact 2. In light of “Martindale’s theorem [24], $\mathscr{Q}$ is isomorphic to a dense subring of linear transformations of a vector space $V$ over $D$, where $D$ is a finite dimensional division ring over $\mathscr{C}$”. By Lemma 2.1, the result follows. Case II: If $\mathscr{T}$ is $\mathscr{Q}$-outer, and $\mathscr{Q}$ satisfies $([\mathscr{T}([a,b])\vartheta,[a,b]])^{m}=[\mathscr{T}([a,b])\vartheta,[a,b]]$, then by Lemma 2.1 we get $dim_{D}V=1$, that is $\mathscr{Q}$ is a domain. By Fact 3, $\mathscr{Q}$ satisfies $[[r,s]\vartheta,[a,b]]^{m}=[[r,s],[a,b]]$ and in particular, for $r=a$ and $s=b$, we have $[[a,b]\vartheta,[a,b]]^{m}=[[a,b]\vartheta,[a,b]]$ for every $a,b\in\mathscr{Q}$. Hence, using the same technique as above we get the required conclusion.

We have

Firstly, we assume that $S$ is $\mathscr{Q}$-inner, that is, $S(a)=\vartheta a-\mathscr{T}(a)\vartheta$ with $0\neq\vartheta\in\mathscr{Q}$. Thus, $\forall a,b\in\mathscr{R}$, we have

If $\vartheta\in\mathscr{C}$, then $\mathscr{R}$ satisfies the GPI $([\mathscr{T}([a,b])\vartheta,[a,b]])^{m}=[\mathscr{T}([a,b])\vartheta,[a,b]]$. We get the desired conclusion, by Proposition 3.1. Therefore $\vartheta\not\in\mathscr{C}$, and so

is nontrivial GPI for $\mathscr{R}$. Thus, Lemma 2.1 yields the required result. Finally, when $S$ is $\mathscr{Q}$-outer, then the above identity can be rewritten as

and hence $\mathscr{R}$ satisfies

In particular $\mathscr{R}$ satisfies $([\mathscr{T}(a)s-sa,[a,b]])^{m}=[\mathscr{T}(a)s-sa,[a,b]]$. We divide it into two cases. First, $\mathscr{T}$ be an identity map of $\mathscr{R}$. Then $([[r,s],[a,b]])^{m}=[[r,s],[a,b]]$ for every $a,b,r,s\in\mathscr{R}$, that is, $\mathscr{R}$ is a polynomial identity ring. Thus, $\mathscr{R}$ and $M_{n}(\mathbb{K})$ satisfy the same polynomial identities [23, Lemma 1], i.e.,

Let $n\geq 2$ and $e_{ij}$ be the usual unit matrix. Then $r=b=e_{12}$, $s=e_{21}$ and $a=e_{11}$, we get a contradiction $2e_{12}=0$. Thus, $n=1$ and we are done. Now consider $\mathscr{T}$ is not the identity map. Therefore,

is a non-trivial GPI for $\mathscr{R}$, by Main Theorem in [5]. Moreover, by Fact 2, $\mathscr{R}$ and $\mathscr{Q}$ satisfy the same GPIs with automorphisms and hence $([\mathscr{T}(a)s-sa,[a,b]])^{m}=[\mathscr{T}(a)s-sa,[a,b]]$ is also an identity for $\mathscr{Q}$. Since $\mathscr{R}$ is a GPI-ring, by [24] “$\mathscr{Q}$ is a primitive ring, which is isomorphic to a dense subring of the ring of linear transformations of a vector space $V$ over a division ring $D$”. If $\mathscr{Q}$ is a domain, then by Fact 3, we have that $\mathscr{Q}$ satisfies the equation $([ts-sa,[a,b]])^{m}=[ts-sa,[a,b]]$. In particular, $([[a,z],[a,b]])^{m}=[[a,z],[a,b]]$ for all $a,b,z\in\mathscr{Q}$, which yields that $\mathscr{Q}$ is commutative (by using the same above argument) and hence $\mathscr{R}$. Henceforth, $\mathscr{Q}$ is not a domain. We have $\mathscr{T}(a)=hah^{-1}$ $\forall a\in\mathscr{Q}$, as mentioned above. Thus, $([hah^{-1}z-za,[a,b]])^{m}=[hah^{-1}z-za,[a,b]]$ Hence, we may consider that $dim~{}D_{V}\geq 2$. By [20, p. 79], there exists a semi-linear automorphism $h\in End(V)$ such that $\mathscr{T}(a)=hah^{-1}$ $\forall a\in\mathscr{Q}$. Hence, $\mathscr{Q}$ satisfies $([hah^{-1}z-za,[a,b]])^{m}=[hah^{-1}z-za,[a,b]]$. If for any $v\in V$ $\exists~{}\Theta_{v}\in D$ such that $h^{-1}v=v\Theta_{v}$, then, it follows that there exists a unique $\Theta\in D$ such that $h^{-1}v=v\Theta$, $\forall v\in V$ (see for example Lemma 1 in [ccl]). In this case $\mathscr{T}(a)v=(hah^{-1})v=hav\Theta$ and

since $V$ is faithful, which is a contradiction that $\mathscr{T}$ is the identity map. Thus, $\exists$ $v\in V$ such that $\{v,h^{-1}v\}$ is linearly $D$-independent. In this case, first we assume that $dim~{}V_{D}\geq 3$. Thus, $\exists~{}u\in V$ such that $\{u,v,h^{-1}v\}$ is linearly $D$-independent. Hence, the density theorem for $\mathscr{Q}$, $\exists~{}a,b,z\in\mathscr{Q}$ such that

The above relation gives that

again a contradiction. Now, the case when $dim~{}V_{D}=2$ that is, $\mathscr{Q}=M_{2}(\mathbb{K})$. Thus

Since $\mathscr{T}(a)$-word of degree $2$ and Char$(\mathscr{R})>3$ by [6, Theorem 3],

Using the same technique as above its shows that $\mathscr{Q}$ is commutative and hence $\mathscr{R}$ is commutative.

Suppose that $\mathscr{L}\not\subseteq Z(\mathscr{R})$ is a Lie ideal of $\mathscr{R}$. Then by [15], there exists an ideal $I$ of $\mathscr{R}$ such that $0\neq[I,\mathscr{R}]\subseteq\mathscr{L}$ and $[\mathscr{L},\mathscr{L}]\neq(0)$. Also, $\mathscr{R}\not\subseteq Z(\mathscr{R})$ as $\mathscr{L}$ is a noncentral Lie ideal of $\mathscr{R}$. Therefore by the given hypothesis, $I$ as well as $\mathscr{R}$ (Fact 2) satisfy $[S([a,b]),[a,b]])^{m}=[S([a,b]),[a,b]]$. By Theorem 3.3, we get the required result.