∇-prime rings and their commutativity

Abstract

Consider a ring with an (anti)-automorphism ∇ of finite order. The fundamental aim of this manuscript is to introduce the notions of ∇-(semi)prime ideal and ∇-(semi)prime ring as a generalization of the notions of (semi)prime ideal, $\ast$-(semi)prime ideal, (semi)prime ring and $\ast$-(semi)prime ring. Furthermore, we will investigate their basic properties and study the commutative property of ∇-prime rings which satisfy certain central differential identities on ∇-ideals.

Keywords:

• Derivation
• prime ring
• ∇-prime ring
• automorphism
• involution
• anti-automorphism

AMS Subject classification 2020:

• 16N60
• 16W10
• 16W25
• 16W20

1. Introduction

In the entire manuscript, $\mathrm{S}$ represents an associative ring with $\mathcal{Z}(\mathrm{S})$ as its centre. An ideal $\mathrm{K}$ of a ring $\mathrm{S}$ is called a prime (resp. semiprime) ideal if ${\mathrm{K}}_1{\mathrm{K}}_2\subseteq \mathrm{K}$ (resp. ${\mathrm{K}}_1^2\subseteq \mathrm{K}$) implies that either ${\mathrm{K}}_1\subseteq \mathrm{K}$ or ${\mathrm{K}}_2\subseteq \mathrm{K}$ (resp. ${\mathrm{K}}_1\subseteq \mathrm{K})$ for all ideals ${\mathrm{K}}_1$ and ${\mathrm{K}}_2$ of $\mathrm{S}$. An ideal $\mathrm{K}$ of $\mathrm{S}$ is prime (resp. semiprime) ideal if and only if $\mathfrak{a},\mathfrak{b}\in \mathrm{S}$, $\mathfrak{a}\mathrm{K}\mathfrak{b}\subseteq \mathrm{K}$ (resp. $\mathfrak{a}\mathrm{K}\mathfrak{a}\subseteq \mathrm{K}$) implies that either $\mathfrak{a}\in \mathrm{K}$ or $\mathfrak{b}\in \mathrm{K}$ (resp. $\mathfrak{a}\in \mathrm{K})$. A ring in which the zero ideal is a prime (resp. semiprime) ideal is called a prime (resp. semiprime) ring [for details, see Ref. [1]]. A nonzero ideal is called an essential ideal if it has a nonzero intersection with every nonzero ideal. A bijective map $\mathrm{\nabla }:\mathrm{S}\to \mathrm{S}$ is said to be an anti-automorphism if ∇ is additive and $\mathrm{\nabla }(\mathfrak{u}\mathfrak{v})=\mathrm{\nabla }(\mathfrak{v})\mathrm{\nabla }(\mathfrak{u})$ for every $\mathfrak{u},\mathfrak{v}\in \mathrm{S}$. An (anti)-automorphism ∇ of $\mathrm{S}$ is called of the first kind if it induces the identity map on $\mathcal{Z}(\mathrm{S})$, otherwise of the second kind. An ideal $\mathrm{K}$ of $\mathrm{S}$ is said to be a ∇-closed ideal if $\mathrm{\nabla }(\mathrm{K})\subseteq \mathrm{K}$ and a ∇-ideal if $\mathrm{\nabla }(\mathrm{K})=\mathrm{K}$. Observe that an ideal $\mathrm{K}$ of $\mathrm{S}$ is a ∇-ideal if and only if $\mathrm{K}$ is a ${\mathrm{\nabla }}^{-1}$-ideal. If the order of ∇ is n, then the ∇-ideal spanned by a subset $\mathrm{R}$ of $\mathrm{S}$, denoted by $[\mathrm{R}]$, is as: $[\mathrm{R}]=⟨\mathrm{R}⟩+\mathrm{\nabla }(⟨\mathrm{R}⟩)+{\mathrm{\nabla }}^2(⟨\mathrm{R}⟩)+\cdots +{\mathrm{\nabla }}^{n-1}(⟨\mathrm{R}⟩)$, where the symbol $⟨\mathrm{R}⟩$ denotes the ideal spanned by $\mathrm{R}$. When $\mathrm{R}=\{\mathfrak{t}\}$, we shall write $[\mathfrak{t}]$ instead of $[\{\mathfrak{t}\}]$.

Until the end of the text, unless otherwise stated, we say that $\mathrm{S}$ is ∇-ring if $\mathrm{S}$ admits an (anti)-automorphism ∇ of finite order n. An anti-automorphism of order 1 or 2 is known as an involution, denoted by “$\ast$” [for details, see Ref. [2]]. Let $\mathrm{S}$ be a $\ast$-ring. Recall that an ideal $\mathrm{K}$ of $\mathrm{S}$ is known as a $\ast$-prime (resp. $\ast$-semiprime) ideal, if for any $\ast$-ideals ${\mathrm{K}}_1$ and ${\mathrm{K}}_2$ of $\mathrm{S}$, ${\mathrm{K}}_1{\mathrm{K}}_2\subseteq \mathrm{K}$ (resp. ${\mathrm{K}}_1^2\subseteq \mathrm{K}$) implies either ${\mathrm{K}}_1\subseteq \mathrm{K}$ or ${\mathrm{K}}_2\subseteq \mathrm{K}$ (resp. ${\mathrm{K}}_1\subseteq \mathrm{K})$. Moreover, $\mathrm{S}$ is called a $\ast$-prime (resp. $\ast$-semiprime) ring if the zero ideal is a $\ast$-prime (resp. $\ast$-semiprime) ideal. An ideal $\mathrm{K}$ of $\mathrm{S}$ is $\ast$-prime ideal if and only if $\mathfrak{a},\mathfrak{b}\in \mathrm{S}$, $\mathfrak{a}\mathrm{K}\mathfrak{b}\subseteq \mathrm{K}$ and $\mathfrak{a}\mathrm{K}{\mathfrak{b}}^{\ast }\subseteq \mathrm{K}$ implies that either $\mathfrak{a}\in \mathrm{K}$ or $\mathfrak{b}\in \mathrm{K}$ [for details see Ref. 3, 4, 5].

The definitions of (semi)prime ideal, (semi)prime ring, $\ast$-(semi)prime ideal and $\ast$-(semi)prime ring naturally motivate one to generalize and unify these notions. For this, we introduce the notions of ∇-(semi)prime ideal and ∇-(semi)prime ring in Section 2.

For $\mathfrak{r},\mathfrak{s}\in \mathrm{S}$, the symbol $[\mathfrak{r},\mathfrak{s}]$ represents the commutator $\mathfrak{r}\mathfrak{s}-\mathfrak{s}\mathfrak{r}$ and symbol $\mathfrak{r}\circ \mathfrak{s}$ denotes the anticommutator $\mathfrak{r}\mathfrak{s}+\mathfrak{s}\mathfrak{r}$. An additive subgroup $\mathrm{J}\subseteq \mathrm{S}$ is called a Jordan (resp. Lie) ideal of $\mathrm{S}$ if $\mathfrak{r}\circ \mathfrak{s}\in \mathrm{R}$ (resp. $[\mathfrak{r},\mathfrak{s}]\in \mathrm{R}$) for every $\mathfrak{r}\in \mathrm{J}$ and $\mathfrak{s}\in \mathrm{S}$. A Jordan ideal $\mathrm{J}$ of ∇-ring $\mathrm{S}$ is known as ∇-Jordan ideal if $\mathrm{\nabla }(\mathrm{J})=\mathrm{J}$. A map $\mathrm{\Phi }:\mathrm{S}\to \mathrm{S}$ is said to be centralizing on $\mathrm{R}\subseteq \mathrm{S}$ if $[\mathrm{\Phi }(\mathfrak{s}),\mathfrak{s}]\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{s}\in \mathrm{R}$ and skewcentralizing on $\mathrm{R}\subseteq \mathrm{S}$ if $\mathrm{\Phi }(\mathfrak{s})\circ \mathfrak{s}\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{s}\in \mathrm{R}$. An additive map $\mathrm{\Psi }:\mathrm{S}\to \mathrm{S}$ is known as derivation if $\mathrm{\Psi }(\mathfrak{s}\mathfrak{r})=\mathrm{\Psi }(\mathfrak{s})\mathfrak{r}+\mathfrak{s}\mathrm{\Psi }(\mathfrak{r})$ for every $\mathfrak{s},\mathfrak{r}\in \mathrm{S}$.

The investigation of the commutative property of rings admitting certain algebraic identities with some special types of maps has been one of the most favoured research areas among ring theorists since Divinsky [6] proved the commutative property of a simple artinian ring which admits a commuting nontrivial automorphism. For example, Mayne [7] obtained that a prime ring admitting a nontrivial centralizing automorphism must be commutative. E. C. Posner [8], established the commutativity of a prime ring with a nonzero centralizing derivation. Oukhtite [[9], Theorem 2] extended this result to Jordan ideals and proved that a prime ring is necessarily commutative if it admits a nonzero derivation centralizing on a nonzero Jordan ideal. Mamouni et al. [[10], Theorem 1], proved that if $\mathrm{S}$ is a 2-torsion free noncommutative prime ring with involution “$\ast$” of the second kind, and if $\mathrm{\Phi },\mathrm{\Psi }:\mathrm{S}\to \mathrm{S}$ are generalized derivations such that $\mathrm{\Phi }(\mathfrak{s}){\mathfrak{s}}^{\ast }-{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{s})\in \mathcal{Z}(\mathrm{S})$ for all $\mathfrak{s}\in \mathrm{S}$, then $\mathrm{\Phi }=\mathrm{\Psi }=0$. Herstein [11] proved that if $\mathrm{S}$ is a 2-torsion-free prime ring with derivation Φ such that $[\mathrm{\Phi }(\mathfrak{t}),\mathrm{\Phi }(\mathfrak{s})]=0$ for all $\mathfrak{t},\mathfrak{s}\in \mathrm{S}$, then $\mathrm{S}$ must be commutative. For other results, see Refs. ([6, 10, 12–20] and the references therein). Several authors have extended some of these results to $\ast$-prime rings and obtained the commutative property of $\ast$-prime rings which satisfy certain identities equipped with derivations. For example Oukhtite and Salhi [[21], Theorem 1.1] showed that a 2-torsion free $\ast$-prime ring is commutative if it admits a nonzero centralizing derivation. Oukhtite [[9], Theorem 1] extended this result to Jordan ideals and proved the commutativity of a 2-torsion free $\ast$-prime ring which admits a nonzero derivation centralizing on a nonzero $\ast$-Jordan ideal. Lie ideal version of this result can be seen in Ref. [22]. For further details see ([22–29] and the references therein). Following the same line of investigation, in Section 4 of the manuscript, we will investigate the commutative property of ∇-prime rings with derivations which satisfy certain algebraic identities on ∇-ideals. In fact our results improve, unify and extend several known results viz., [[12], Main Theorem], [[30], Theorems 2.1 and 2.3], [[10], Theorem 1], [[31], Theorem 3.7] and [[21], Theorem 1.1].

We remark that our results on ∇-closed ideals also hold for ∇-Jordan ideals. If $\mathrm{S}$ is a 2-torsion free semiprime ring and $\mathrm{J}$ is a Jordan ideal of $\mathrm{S}$, then by Herstein [[32], Theorem 1.1] $\mathrm{J}$ contains $\mathrm{K}$, a nonzero ideal of $\mathrm{S}$. Therefore, if $\mathrm{S}$ is ∇-ring, then $\mathrm{J}$ contains $\mathrm{K}+\mathrm{\nabla }(\mathrm{K})+\cdots +{\mathrm{\nabla }}^{n-1}(\mathrm{K})$, a nonzero ∇-ideal of $\mathrm{S}$. As a consequence all the results obtained in Ref. [9] can be extended in the setting of ∇-prime rings in the light of this research work.

The paper is organized as follows: In Section 2, we introduce the notions of ∇-(semi)prime ideal and ∇-(semi)prime ring, and provide examples in support of our study. In addition, we prove some elementary properties of ∇-semiprime rings. Section 3 is dedicated to establish some characterizations of ∇-prime ideals analogous to prime ideals and to study some fundamental properties of ∇-ideals in ∇-prime rings. In Section 4, the commutative property of ∇-prime rings which satisfy certain central identities involving derivations is explored.

2. Preliminaries

We begin with the following definitions:

Definition 2.1

Consider the ∇-ring $\mathrm{S}$. An ideal $\mathrm{K}$ of $\mathrm{S}$ is known as ∇-prime (resp. ∇-semiprime) ideal if ${\mathrm{K}}_1{\mathrm{K}}_2\subseteq \mathrm{K}$ (resp. ${\mathrm{K}}_1^2\subseteq \mathrm{K}$) implies that either ${\mathrm{K}}_1\subseteq \mathrm{K}$ or ${\mathrm{K}}_2\subseteq \mathrm{K}$ (resp. ${\mathrm{K}}_1\subseteq \mathrm{K}$) for all ∇-ideals ${\mathrm{K}}_1$ and ${\mathrm{K}}_2$ of $\mathrm{S}$.

Definition 2.2

A ∇-ring $\mathrm{S}$ is called a ∇-prime (resp. ∇-semiprime) ring if the zero ideal is a ∇-prime (resp. ∇-semiprime) ideal.

Below, we provide some examples of ∇-prime rings. It is easy to observe that the first two examples justify that there exist ∇-prime rings which are neither prime nor $\ast$-prime.

Example 2.1

Consider $\mathbb{C}$, the ring of all complex numbers and $\mathrm{S}=\underset{k-\mathrm{times}}{\underbrace{\mathbb{C}\times \mathbb{C}\times \cdots \times \mathbb{C}}}$. Then the map $\mathrm{\nabla }:\mathrm{S}\to \mathrm{S}$ given by $\mathrm{\nabla }({\mathfrak{u}}_1,{\mathfrak{u}}_2,{\mathfrak{u}}_3,\dots ,{\mathfrak{u}}_{k-1},{\mathfrak{u}}_k)=({\overline{\mathfrak{u}}}_k,{\overline{\mathfrak{u}}}_1,{\overline{\mathfrak{u}}}_2,{\overline{\mathfrak{u}}}_3,\dots ,{\overline{\mathfrak{u}}}_{k-1})$, where $\overline{\mathfrak{u}}$ depicts the complex conjugate of $\mathfrak{u}$, is an automorphism. Clearly $\mathrm{S}$ is ∇-prime.

Example 2.2

Let $\mathbb{F}$ be a field with an automorphism ψ of finite order and $\mathrm{S}=M_2(\mathbb{F})$ be the ring of all square matrices of order 2 over $\mathbb{F}$. For $\mathrm{A}=\left[\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right],$ we let $\overline{\mathrm{A}}=\left[\begin{array}{cc}\psi (a_4)& -\psi (a_2)\\ -\psi (a_3)& \psi (a_1)\end{array}\right]$. Consider the ring $\mathrm{R}=\underset{m-\mathrm{times}}{\underbrace{\mathrm{S}\times \mathrm{S}\times \cdots \times \mathrm{S}}}$. Define a map $\mathrm{\nabla }:\mathrm{R}\to \mathrm{R}$ by $\mathrm{\nabla }({\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3,\dots ,{\mathrm{A}}_{m-1},{\mathrm{A}}_m)=({\overline{\mathrm{A}}}_m,{\overline{\mathrm{A}}}_1,{\overline{\mathrm{A}}}_2,{\overline{\mathrm{A}}}_3,\dots ,{\overline{\mathrm{A}}}_{m-1})$. Clearly, ∇ is an anti-automorphism and $\mathrm{R}$ is ∇-prime.

Example 2.3

Consider a prime ring $\mathrm{S}$ with its opposite ring ${\mathrm{S}}^{\circ }$. Let $\mathrm{R}=\mathrm{S}\times {\mathrm{S}}^{\circ }$ and define the map $\mathrm{\nabla }:\mathrm{R}\to \mathrm{R}$ by $\mathrm{\nabla }(\mathfrak{u},\mathfrak{v})=(\mathfrak{v},\mathfrak{u})$. Then ∇ is an anti-automorphism and $\mathrm{R}$ is a ∇-prime ring.

The notion of ∇-prime ring is a unifying one i.e. it includes the notion of prime ring, by taking $\mathrm{\nabla }=$ identity map, and $\ast$-prime ring, by taking $\mathrm{\nabla }=\ast$. Hence, the study of ∇-prime rings envelop the literature of prime and $\ast$-prime rings and enrich the ring structures and we get a new class of rings. Clearly, every prime ring with (anti)-automorphism ∇ is ∇-prime but not conversely as illustrated by the above examples.

Now we prove some elementary properties of ∇-semiprime ideals in the following results.

Lemma 2.1

Let $\mathrm{K}$ be an ideal of a ∇-ring $\mathrm{S}$. Then the assertions given below are equivalent:

1. $\mathrm{K}$ is a ∇-semiprime ideal of $\mathrm{S}$,

2. $\mathrm{K}\cap \mathrm{\nabla }(\mathrm{K})\cap {\mathrm{\nabla }}^2(\mathrm{K})\cap \cdots \cap {\mathrm{\nabla }}^{n-1}(\mathrm{K})$ is a ∇-semiprime ideal of $\mathrm{S}$,

3. $\mathrm{K}\cap \mathrm{\nabla }(\mathrm{K})\cap {\mathrm{\nabla }}^2(\mathrm{K})\cap \cdots \cap {\mathrm{\nabla }}^{n-1}(\mathrm{K})$ is a semiprime ideal of $\mathrm{S}$.

Proof.

Taking $\mathrm{N}=\mathrm{K}\cap \mathrm{\nabla }(\mathrm{K})\cap {\mathrm{\nabla }}^2(\mathrm{K})\cap \cdots \cap {\mathrm{\nabla }}^{n-1}(\mathrm{K})$. So $\mathrm{N}$ is a ∇-ideal of $\mathrm{S}$.

$(i)⟹(\mathrm{ii})$ Let $\mathrm{P}$ be any ∇-ideal of $\mathrm{S}$ such that ${\mathrm{P}}^2\subseteq \mathrm{N}.$ Then ${\mathrm{P}}^2\subseteq \mathrm{K}$ and hence $\mathrm{P}\subseteq \mathrm{K}.$ Consequently $\mathrm{P}\subseteq {\mathrm{\nabla }}^i(\mathrm{K})$ for each $i:1\le i\le n$. Thus $\mathrm{P}\subseteq \mathrm{N}$ and hence $\mathrm{N}$ is a ∇-semiprime ideal.

$(\mathrm{ii})⟹(\mathrm{iii})$ Let $\mathrm{P}$ be any ideal of $\mathrm{S}$ and ${\mathrm{P}}^2\subseteq \mathrm{N}.$ Then $\{\mathrm{P}\cap \mathrm{\nabla }(\mathrm{P})\cap {\mathrm{\nabla }}^2(\mathrm{P})\cap \cdots \cap {\mathrm{\nabla }}^{n-1}(\mathrm{P}){\}}^2\subseteq \mathrm{N}$, so $\mathrm{P}\cap \mathrm{\nabla }(\mathrm{P})\cap {\mathrm{\nabla }}^2(\mathrm{P})\cap \cdots \cap {\mathrm{\nabla }}^{n-1}(\mathrm{P})\subseteq \mathrm{N}$. Therefore $\{\mathrm{P}+\mathrm{\nabla }(\mathrm{P})+{\mathrm{\nabla }}^2(\mathrm{P})+\cdots +{\mathrm{\nabla }}^{n-1}(\mathrm{P}){\}}^n\subseteq \mathrm{N}$, which further implies that $\mathrm{P}+\mathrm{\nabla }(\mathrm{P})+{\mathrm{\nabla }}^2(\mathrm{P})+\cdots +{\mathrm{\nabla }}^{n-1}(\mathrm{P})\subseteq \mathrm{N}$. Hence $\mathrm{P}\subseteq \mathrm{N}$ and so $\mathrm{N}$ is a semiprime ideal.

$(\mathrm{iii})⟹(i)$ Let $\mathrm{P}$ be any ∇-ideal of $\mathrm{S}$ and ${\mathrm{P}}^2\subseteq \mathrm{K}.$ Then ${\mathrm{P}}^2\subseteq {\mathrm{\nabla }}^i(\mathrm{K})$ for each $i=1,2,\dots ,n$. Thus ${\mathrm{P}}^2\subseteq \mathrm{N}.$ Therefore $\mathrm{P}\subseteq \mathrm{N}\subseteq \mathrm{K}$. This proves that $\mathrm{N}$ is a ∇-semiprime ideal of $\mathrm{S}$.

From Lemma 2.1, it follows that a ∇-ideal $\mathrm{K}$ of a ∇-ring $\mathrm{S}$ is ∇-semiprime if and only if $\mathrm{K}$ is semiprime. So for ∇-ideals, the notions of ∇-semiprimeness and semiprimeness are same. On the other hand if $\mathrm{K}$ is any ∇-ideal of a ∇-ring $\mathrm{S}$, then $\mathrm{K}$ may be ∇-prime but not prime. This fact is illustrated by Examples 2.1–2.3.

For $\mathrm{P},\mathrm{J}\subseteq \mathrm{S}$, let $(\mathrm{P}:\mathrm{J})=\{\mathfrak{u}\in \mathrm{S}:\mathfrak{u}\mathrm{J}\subseteq \mathrm{P}\}$ and $[\mathrm{P}:\mathrm{J}]=\{\mathfrak{u}\in \mathrm{S}:\mathrm{J}\mathfrak{u}\subseteq \mathrm{P}\}$. Observe that if $\mathrm{P}$ and $\mathrm{J}$ are ideals of $\mathrm{S}$, then so are $(\mathrm{P}:\mathrm{J})$ and $[\mathrm{P}:\mathrm{J}]$.

Lemma 2.2

Let $\mathrm{P}$ and $\mathrm{J}$ be ∇-ideals of $\mathrm{S}$. Then the following assertions hold.

1. If $\mathrm{P}$ is a ∇-semiprime ideal of $\mathrm{S}$, then $(\mathrm{P}:\mathrm{J})=[\mathrm{P}:\mathrm{J}]$ is a ∇-ideal of $\mathrm{S}$

2. If $\mathrm{P}$ is a ∇-prime ideal of $\mathrm{S}$ such that $\mathrm{J}⊈\mathrm{P}$, then $(\mathrm{P}:\mathrm{J})=[\mathrm{P}:\mathrm{J}]=\mathrm{P}$.

Proof.

(i) By Lemma 2.1, $\mathrm{P}$ is semiprime, so $\{\mathrm{J}(\mathrm{P}:\mathrm{J}){\}}^2\subseteq \mathrm{P}$ gives $\mathrm{J}(\mathrm{P}:\mathrm{J})\subseteq \mathrm{P}$. Hence $(\mathrm{P}:\mathrm{J})\subseteq [\mathrm{P}:\mathrm{J}]$. Similarly $[\mathrm{P}:\mathrm{J}]\subseteq (\mathrm{P}:\mathrm{J})$. Also it is easy to verify that $[\mathrm{P}:\mathrm{J}]=(\mathrm{P}:\mathrm{J})$ is a ∇-ideal of $\mathrm{S}$.

(ii) It follows from (i) and the fact that $(\mathrm{P}:\mathrm{J})\mathrm{J}\subseteq \mathrm{P}$.

3. Properties of ∇-prime ideals

We begin with the following characterization of ∇-prime ideals which is analogous to a well-known characterization of *-prime and prime ideals.

Theorem 3.1

Let $\mathrm{K}$ be a ∇-ideal of a ∇-ring $\mathrm{S}$. Then the statements given below are equivalent.

1. $\mathrm{K}$ is a ∇-prime ideal of $\mathrm{S}$,

2. If ${\mathrm{K}}_j$, $j=1,2,\dots ,k$, where k is any arbitrary positive integer, are ∇-ideals of $\mathrm{S}$ with ${\mathrm{K}}_1{\mathrm{K}}_2\cdots {\mathrm{K}}_k\subseteq \mathrm{K}$, then ${\mathrm{K}}_j\subseteq \mathrm{K}$ for atleast one j.

3. If $\mathfrak{p},\mathfrak{q}\in \mathrm{S}$ with $\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$, then either $\mathfrak{p}\in \mathrm{K}$ or $\mathfrak{q}\in \mathrm{K}$.

4. If $\mathrm{P}$ is any ideal of $\mathrm{S}$ and $\mathrm{J}$ a ∇-closed ideal of $\mathrm{S}$ with $\mathrm{P}\mathrm{J}\subseteq \mathrm{K}$, then either $\mathrm{P}\subseteq \mathrm{K}$ or $\mathrm{J}\subseteq \mathrm{K}$.

5. If $\mathfrak{p},\mathfrak{q}\in \mathrm{S}$ with ${\mathrm{\nabla }}^i(\mathfrak{p})\mathrm{S}\mathfrak{q}\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$, then either $\mathfrak{p}\in \mathrm{K}$ or $\mathfrak{q}\in \mathrm{K}$.

6. $\mathrm{K}$ is ${\mathrm{\nabla }}^{-1}$-prime ideal of $\mathrm{S}$.

Proof.

$(i)⟹(\mathrm{ii})$ Let ${\mathrm{K}}_j$, $j=1,2,\dots ,k$ be ∇-ideals of $\mathrm{S}$ with ${\mathrm{K}}_1{\mathrm{K}}_2\cdots {\mathrm{K}}_k\subseteq \mathrm{K}$. If k = 2, then by (i), we are through. Suppose k>2 and assume that the statement is true for k−1. We show it also stands true for k. If ${\mathrm{K}}_k\subseteq \mathrm{K}$, then we are through. So assume ${\mathrm{K}}_k⊈\mathrm{K}$. Invoking Lemma 2.2, we have ${\mathrm{K}}_1{\mathrm{K}}_2\cdots {\mathrm{K}}_{k-1}\subseteq (\mathrm{K}:{\mathrm{K}}_k)=\mathrm{K}$. By induction hypothesis it now follows that ${\mathrm{K}}_i\subseteq \mathrm{K}$ for some i.

$(\mathrm{ii})⟹(\mathrm{iii})$ Let $\mathfrak{p},\mathfrak{q}\in \mathrm{S}$ with (1) $\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\subseteq \mathrm{K}\mathrm{for}\mathrm{each}i=1,2,\dots ,n.$(1) Let $1\le j\le n$. Firstly suppose that ∇ is an automorphism. Then applying ${\mathrm{\nabla }}^j$ to the relation (1(1) $\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\subseteq \mathrm{K}\mathrm{for}\mathrm{each}i=1,2,\dots ,n.$(1) ), we obtain ${\mathrm{\nabla }}^j(\mathfrak{p})\mathrm{S}{\mathrm{\nabla }}^{i+j}(\mathfrak{q})\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$. Given that the order of ∇ is n. Hence ${\mathrm{\nabla }}^j(\mathfrak{p})\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$. Next assume that ∇ is an anti-automorphism. If j is odd, then applying ${\mathrm{\nabla }}^j$ to the relation (1(1) $\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\subseteq \mathrm{K}\mathrm{for}\mathrm{each}i=1,2,\dots ,n.$(1) ), we have ${\mathrm{\nabla }}^i(\mathfrak{q})\mathrm{S}{\mathrm{\nabla }}^j(\mathfrak{p})\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$. Therefore ${\mathrm{\nabla }}^j(\mathfrak{p})\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\mathrm{S}{\mathrm{\nabla }}^j(\mathfrak{p})\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$. By Lemma 2.1, $\mathrm{K}$ is a semiprime ideal, so ${\mathrm{\nabla }}^j(\mathfrak{p})\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$. Also if j is even, then we get the same relation. Therefore in any case, we have ${\mathrm{\nabla }}^j(\mathfrak{p})\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\subseteq \mathrm{K}$ for all positive integers i and j. Consequently $[\mathfrak{p}]\mathrm{S}[\mathfrak{q}]\subseteq \mathrm{K}$, whence by (ii), $[\mathfrak{p}]\subseteq \mathrm{K}$ or $[\mathfrak{q}]\subseteq \mathrm{K}$. Hence $\mathfrak{p}\in \mathrm{K}$ or $\mathfrak{q}\in \mathrm{K}$.

$(\mathrm{iii})⟹(\mathrm{iv})$ Let $\mathrm{P}$ be any ideal of $\mathrm{S}$ and $\mathrm{J}$ a ∇-closed ideal of $\mathrm{S}$ with $\mathrm{P}\mathrm{J}\subseteq \mathrm{K}$. If $\mathrm{J}\subseteq \mathrm{K}$, then we are through. So assume $\mathrm{J}⊈\mathrm{K}$ and let $\mathfrak{p}\in \mathrm{J}$ with $\mathfrak{p}\notin \mathrm{K}$. Then $\mathfrak{u}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{p})\subseteq \mathrm{P}\mathrm{J}\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$ and $\mathfrak{u}\in \mathrm{P}$. From (iii), it now follows that $\mathrm{P}\subseteq \mathrm{K}$.

$(\mathrm{iv})⟹(i)$ It follows from the definition of ∇-prime ideal.

$(\mathrm{iii})⟹(v)$ Suppose ${\mathrm{\nabla }}^i(\mathfrak{p})\mathrm{S}\mathfrak{q}\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$, then $\mathfrak{q}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{p})\mathrm{S}\mathfrak{q}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{p})\subseteq \mathrm{K}$. By Lemma 2.1, $\mathrm{K}$ is a semiprime ideal, so $\mathfrak{q}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{p})\subseteq \mathrm{K}$ for each $i=1,2,\dots ,n$. Hence by (iii) either $\mathfrak{p}\in \mathrm{K}$ or $\mathfrak{q}\in \mathrm{K}$. By using similar arguments, one can show that $(v)⟹(\mathrm{iii})$.

$(i)⟺(\mathrm{vi})$ It follows from the fact that an ideal $\mathrm{K}$ is ∇-ideal iff it is ${\mathrm{\nabla }}^{-1}$-ideal.

In particular, ∇-prime rings can be characterized as below:

Corollary 3.1

Let $\mathrm{S}$ be a ∇-ring. Then the given below assertions are equivalent.

1. $\mathrm{S}$ is a ∇-prime ring.

2. $\mathrm{S}$ is a ${\mathrm{\nabla }}^{-1}$-prime ring.

3. If $\mathfrak{p},\mathfrak{q}\in \mathrm{S}$ with $\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})=\{0\}$ for each $i=1,2,\dots ,n$, then either $\mathfrak{p}=0$ or $\mathfrak{q}=0$.

4. If $\mathfrak{p},\mathfrak{q}\in \mathrm{S}$ with ${\mathrm{\nabla }}^i(\mathfrak{p})\mathrm{S}\mathfrak{q}=\{0\}$ for each $i=1,2,\dots ,n$, then either $\mathfrak{p}=0$ or $\mathfrak{q}=0$.

5. If ${\mathrm{K}}_j$, $1\le j\le k,$ are ∇-ideals of $\mathrm{S}$ with ${\mathrm{K}}_1{\mathrm{K}}_2\cdots {\mathrm{K}}_k=\{0\}$, then ${\mathrm{K}}_j=\{0\}$ for some j.

6. If $\mathrm{P}$ is any ideal of $\mathrm{S}$ and $\mathrm{J}$ a ∇-closed ideal of $\mathrm{S}$ with $\mathrm{P}\mathrm{J}=\{0\}$, then $\mathrm{P}=\{0\}$ or $\mathrm{J}=\{0\}$.

Corollary 3.2

Let $\mathrm{K}$ be an ideal of a ∇- ring $\mathrm{S}$ and m be any integer. If $\mathrm{K}$ is ${\mathrm{\nabla }}^m$-prime, then $\mathrm{K}$ is ∇-prime.

Proof.

Suppose $\mathrm{K}$ is ${\mathrm{\nabla }}^m$-prime and $\mathrm{P}$, $\mathrm{J}$ be ∇-ideals of $\mathrm{S}$ with $\mathrm{P}\mathrm{J}\subseteq \mathrm{K}$. Clearly $\mathrm{P}$, $\mathrm{J}$ are ${\mathrm{\nabla }}^m$-ideals also. Therefore by ${\mathrm{\nabla }}^m$-primeness of $\mathrm{K}$, either $\mathrm{P}\subseteq \mathrm{K}$ or $\mathrm{J}\subseteq \mathrm{K}$. Hence $\mathrm{S}$ is ∇-prime.

We remark that the converse of Corollary 3.2 does not hold. To see this consider $\mathrm{S}$ and ∇ same as in Example 2.3. Then $\mathrm{S}$ is ∇-prime but not ${\mathrm{\nabla }}^2$-prime.

Lemma 3.1

Consider a ∇-ring $\mathrm{S}$ with $\mathrm{K}$ as its nonzero ∇-ideal. Suppose $\mathrm{\nabla }(\alpha )-\alpha \notin \mathrm{K}$ for some $\alpha \in \mathcal{Z}(\mathrm{S})$ and $\mathfrak{p},\mathfrak{q}\in \mathrm{S}$. Then $\mathrm{K}$ is ∇-prime if and only if $\mathfrak{p}\mathfrak{t}\mathrm{\nabla }(\mathfrak{t}){\mathrm{\nabla }}^i(\mathfrak{q})\in \mathrm{K}$ for every $\mathfrak{t}\in \mathrm{S}$ and $i=1,2,\dots ,n$ implies that either $\mathfrak{p}\in \mathrm{K}$ or $\mathfrak{q}\in \mathrm{K}$.

Proof.

Suppose $\mathrm{K}$ is ∇-prime and $\mathfrak{p}\mathfrak{t}\mathrm{\nabla }(\mathfrak{t}){\mathrm{\nabla }}^i(\mathfrak{q})\in \mathrm{K}$ for every $\mathfrak{t}\in \mathrm{S}$ and $1\le i\le n$. Then, we have (2) $\begin{array}{rl}& \mathfrak{p}\mathfrak{t}\mathrm{\nabla }(\mathfrak{s}){\mathrm{\nabla }}^i(\mathfrak{q})+\mathfrak{p}\mathfrak{s}\mathrm{\nabla }(\mathfrak{t}){\mathrm{\nabla }}^i(\mathfrak{q})\in \mathrm{K}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S}\\ & \mathrm{and}1\le i\le n.\end{array}$(2) Let $\beta \in \mathcal{Z}(\mathrm{S})$. Then replacing $\mathfrak{s}$ by $\beta \mathfrak{s}$ in (2(2) $\begin{array}{rl}& \mathfrak{p}\mathfrak{t}\mathrm{\nabla }(\mathfrak{s}){\mathrm{\nabla }}^i(\mathfrak{q})+\mathfrak{p}\mathfrak{s}\mathrm{\nabla }(\mathfrak{t}){\mathrm{\nabla }}^i(\mathfrak{q})\in \mathrm{K}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S}\\ & \mathrm{and}1\le i\le n.\end{array}$(2) ), we have (3) $\begin{array}{rl}& \mathrm{\nabla }(\beta )\mathfrak{p}\mathfrak{t}\mathrm{\nabla }(\mathfrak{s}){\mathrm{\nabla }}^i(\mathfrak{q})+\beta \mathfrak{p}\mathfrak{s}\mathrm{\nabla }(\mathfrak{t}){\mathrm{\nabla }}^i(\mathfrak{q})\in \mathrm{K}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S}\\ & \mathrm{and}1\le i\le n.\end{array}$(3) From (2(2) $\begin{array}{rl}& \mathfrak{p}\mathfrak{t}\mathrm{\nabla }(\mathfrak{s}){\mathrm{\nabla }}^i(\mathfrak{q})+\mathfrak{p}\mathfrak{s}\mathrm{\nabla }(\mathfrak{t}){\mathrm{\nabla }}^i(\mathfrak{q})\in \mathrm{K}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S}\\ & \mathrm{and}1\le i\le n.\end{array}$(2) ) and (3(3) $\begin{array}{rl}& \mathrm{\nabla }(\beta )\mathfrak{p}\mathfrak{t}\mathrm{\nabla }(\mathfrak{s}){\mathrm{\nabla }}^i(\mathfrak{q})+\beta \mathfrak{p}\mathfrak{s}\mathrm{\nabla }(\mathfrak{t}){\mathrm{\nabla }}^i(\mathfrak{q})\in \mathrm{K}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S}\\ & \mathrm{and}1\le i\le n.\end{array}$(3) ), we find that $\begin{array}{rl}& (\mathrm{\nabla }(\beta )-\beta )\mathfrak{p}\mathfrak{t}\mathrm{\nabla }(\mathfrak{s}){\mathrm{\nabla }}^i(\mathfrak{q})\in \mathrm{K}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S}\mathrm{and}\\ & 1\le i\le n.\end{array}$Hence, we have $\begin{array}{rl}& (\mathrm{\nabla }(\beta )-\beta )\mathfrak{p}\mathfrak{t}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})\in \mathrm{K}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S}\mathrm{and}\\ & 1\le i\le n.\end{array}$By ∇-primeness of $\mathrm{K}$, we conclude that either $\mathfrak{q}\in \mathrm{K}$ or $\mathfrak{p}\mathrm{S}(\mathrm{\nabla }(\beta )-\beta )\subseteq \mathrm{K}$If the latter case prevails, then by the arbitrariness of β, we have $\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^i(\mathrm{\nabla }(\beta )-\beta )\subseteq \mathrm{K}$By the given hypothesis there exists $\alpha \in \mathcal{Z}(\mathrm{S})$ such that $\mathrm{\nabla }(\alpha )-\alpha \notin \mathrm{K}$. Therefore in the light of ∇-primeness of $\mathrm{K}$, we obtain $\mathfrak{p}\in \mathrm{K}$. Therefore $\mathrm{K}$ is ∇-prime. The converse part holds trivially.

In particular, we have the following characterization of ∇-prime rings, when ∇ is of the second kind.

Corollary 3.3

Let $\mathrm{S}$ be a ∇-ring. Suppose ∇ is of second kind and $\mathfrak{p},\mathfrak{q}\in \mathrm{S}$. Then $\mathrm{S}$ is ∇-prime if and only if $\mathfrak{p}\mathfrak{t}\mathrm{\nabla }(\mathfrak{t}){\mathrm{\nabla }}^i(\mathfrak{q})=0$ for every $\mathfrak{t}\in \mathrm{S}$ and $1\le i\le n$ implies that either $\mathfrak{p}=0$ or $\mathfrak{q}=0$.

In case of an arbitrary anti-automorphism ∇, the result given below holds.

Lemma 3.2

Suppose $\mathrm{S}$ is a ring with an anti-automorphism ∇ and $\mathfrak{p},\mathfrak{q}\in \mathrm{K}$. If $\mathfrak{p}\mathrm{S}\mathfrak{q}=\{0\}$ and $\mathfrak{p}\mathrm{S}\mathrm{\nabla }(\mathfrak{q})=\{0\}$ imply that either $\mathfrak{p}=0$ or $\mathfrak{q}=0$, then $\mathrm{S}$ is semiprime.

Proof.

Suppose $\mathfrak{p}\mathrm{S}\mathfrak{p}=\{0\}$ for some $\mathfrak{p}\in \mathrm{S}$. Then $\mathfrak{p}\mathrm{S}\mathfrak{p}\mathrm{S}\mathrm{\nabla }(\mathfrak{p})=\{0\}$, which further implies that $\mathfrak{p}\mathrm{S}\mathrm{\nabla }(\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^{-1}(\mathfrak{p}))=\{0\}$. Also $\mathfrak{p}\mathrm{S}(\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^{-1}(\mathfrak{p}))=\{0\}$. By the given hypothesis this implies that either $\mathfrak{p}=0$ or $\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^{-1}(\mathfrak{p})=\{0\}$. If $\mathfrak{p}=0$, then we are through. So assume that $\mathfrak{p}\mathrm{S}{\mathrm{\nabla }}^{-1}(\mathfrak{p})=\{0\}$ or equivalently $\mathfrak{p}\mathrm{S}\mathrm{\nabla }(\mathfrak{p})=\{0\}$. Also $\mathfrak{p}\mathrm{S}\mathfrak{p}=\{0\}$. Again by the given assumption $\mathfrak{p}=0$. Therefore $\mathrm{S}$ is semiprime.

Now let $\mathrm{S}$ be a ∇-ring. Set ${\mathcal{S}}_{\mathrm{\nabla }}(\mathrm{S})=\{\mathfrak{t}\in \mathrm{S}\mid \mathrm{\nabla }(\mathfrak{t})=\mathfrak{t}\}$ and ${\mathcal{H}}_{\mathrm{\nabla }}(\mathrm{S})=\{\mathfrak{t}\in \mathrm{S}\mid \mathrm{\nabla }(\mathfrak{t})=-\mathfrak{t}\}$. For $\mathfrak{t}\in \mathrm{S}$, setting $\mathfrak{p}=\mathfrak{t}+\mathrm{\nabla }(\mathfrak{t})+{\mathrm{\nabla }}^2(\mathfrak{t})+\cdots +{\mathrm{\nabla }}^{n-1}(\mathfrak{t})$ and $\mathfrak{q}=\mathfrak{t}-\mathrm{\nabla }(\mathfrak{t})+{\mathrm{\nabla }}^2(\mathfrak{t})-{\mathrm{\nabla }}^3(\mathfrak{t})\cdots +{\mathrm{\nabla }}^{n-2}(\mathfrak{t})-{\mathrm{\nabla }}^{n-1}(\mathfrak{t})$. Then $\mathfrak{t}\in {\mathcal{S}}_{\mathrm{\nabla }}(\mathrm{S})$ and for even n, $\mathfrak{q}\in {\mathcal{H}}_{\mathrm{\nabla }}(\mathrm{S})$. Obviously, if $\mathrm{S}$ is ∇-prime ring, then the elements of $\mathcal{Z}(\mathrm{S})$ which are also either in ${\mathcal{S}}_{\mathrm{\nabla }}(\mathrm{S})$ or in ${\mathcal{H}}_{\mathrm{\nabla }}(\mathrm{S})$ are not zero divisors in $\mathrm{S}$.

Now assume that $\mathrm{K}$ is a ∇-ideal of a ∇-ring $\mathrm{S}$. Then it is trivial to see that the map $\mathfrak{t}+\mathrm{K}\mapsto \mathrm{\nabla }(\mathfrak{t})+\mathrm{K}$ induces an automorphism or anti-automorphism on the quotient ring $\frac{\mathrm{S}}{\mathrm{K}}$ according as ∇ is automorphism or anti-automorphism on $\mathrm{S}$, which we again denote by ∇. In this context we have the following result:

Lemma 3.3

Let $\mathrm{K}$ be a ∇-ideal of a ∇-ring $\mathrm{S}$. Then $\mathrm{K}$ is ∇-prime if and only if the quotient ring $\frac{\mathrm{S}}{\mathrm{K}}$ is ∇-prime.

Proof.

Suppose that $\mathrm{K}$ is ∇-prime and $\mathfrak{t}+\mathrm{K},\mathfrak{s}+\mathrm{K}\in \frac{\mathrm{S}}{\mathrm{K}}$ such that $(\mathfrak{t}+\mathrm{K})(\mathfrak{r}+\mathrm{K}){\mathrm{\nabla }}^i(\mathfrak{s}+\mathrm{K})=\mathrm{K}$ for every $\mathfrak{r}+\mathrm{K}\in \frac{\mathrm{S}}{\mathrm{K}}$ and $i=1,2,\dots ,n$. Then $\mathfrak{t}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{s})\subseteq \mathrm{K}$ for every $i=1,2,\dots ,n$, whence by ∇-primeness of $\mathrm{K}$ we conclude that either $\mathfrak{t}\in \mathrm{K}$ or $\mathfrak{s}\in \mathrm{K}$. Consequently $\mathfrak{t}+\mathrm{K}=\mathrm{K}$ or $\mathfrak{s}+\mathrm{K}=\mathrm{K}$, so $\frac{\mathrm{S}}{\mathrm{K}}$ is ∇-prime. Conversely, suppose that $\frac{\mathrm{S}}{\mathrm{K}}$ is ∇-prime and $\mathfrak{t},\mathfrak{s}\in \mathrm{S}$ such that $\mathfrak{t}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{s})\subseteq \mathrm{K}$, $i=1,2,\dots ,n$. Then $(\mathfrak{t}+\mathrm{K})(\mathfrak{r}+\mathrm{K}){\mathrm{\nabla }}^i(\mathfrak{s}+\mathrm{K})=\mathrm{K}$ for every $\mathfrak{r}+\mathrm{K}\in \frac{\mathrm{S}}{\mathrm{K}}$ and $i=1,2,\dots ,n$. Hence by ∇-primeness of $\frac{\mathrm{S}}{\mathrm{K}}$ we conclude that either $\mathfrak{t}+\mathrm{K}=\mathrm{K}$ or $\mathfrak{s}+\mathrm{K}=\mathrm{K}$, which further implies that either $\mathfrak{t}\in \mathrm{K}$ or $\mathfrak{s}\in \mathrm{K}$. Hence $\mathrm{K}$ is ∇-prime. This finishes the proof.

The following result describes the characteristic and torsion freeness of a ∇-prime ring.

Lemma 3.4

Consider a ∇-prime ring $\mathrm{S}$ with nonzero characteristic. Then $\mathrm{char}(\mathrm{S})=m,$ where m is a fixed prime number. Also if m is a prime number, then char$(\mathrm{S})\ne m$ if and only if $\mathrm{S}$ is m-torsion free.

Proof.

We begin with the supposition that char$(\mathrm{S})=m$, where m is a composite number. Then $m=m_1{m}_2$ and $1<m_1,m_2<m$. Therefore $m\mathrm{S}=\{0\}$ gives us that $m_1\mathfrak{t}\mathrm{S}{\mathrm{\nabla }}^i(m_2\mathfrak{s})=\{0\}$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{S}$ and $i=1,2,\dots ,n$. Invoking ∇-primeness of $\mathrm{S}$, we conclude that either $m_1\mathrm{S}=\{0\}$ or $m_2\mathrm{S}=\{0\}$, which is a contradiction. Hence m is a prime number.

Now let m be a prime number such that char$(\mathrm{S})\ne m$ and $m\mathfrak{t}=0$ for some $\mathfrak{t}\in \mathrm{S}$. Then $\mathfrak{t}\mathrm{S}{\mathrm{\nabla }}^i(m\mathfrak{s})=\{0\}$ for every $\mathfrak{s}\in \mathrm{S}$ and $i=1,2,\dots ,n$. By ∇-primeness of $\mathrm{S}$, the last expression provides us, either $\mathfrak{t}=0$ or $m\mathrm{S}=\{0\}$. If latter case prevails, then the order of each element of $\mathrm{S}$, as an additive group, divides m and hence the order of each nonzero element of additive group, $\mathrm{S}$ is m. Therefore char$(\mathrm{S})=m$, which is a contradiction. So $\mathfrak{t}=0$ and hence $\mathrm{S}$ is m-torsion free. Obviously, if $\mathrm{S}$ is m-torsion free, then char$(\mathrm{S})\ne m$.

Lemma 3.5

Suppose $\mathrm{S}$ is a ∇-prime ring with $\mathrm{K}$ a nonzero ∇-closed left (resp. right) ideal. If $\mathfrak{q}\mathrm{K}=\{0\}$ (resp. $\mathrm{K}\mathfrak{q}=\{0\}$) for some $\mathfrak{q}\in \mathrm{S}$, then $\mathfrak{q}=0$.

Proof.

Suppose $\mathfrak{q}\mathrm{K}=\{0\}$, then $\mathfrak{q}\mathrm{S}\mathrm{K}=\{0\}$. As $\mathrm{K}$ is ∇-closed, so we also have $\mathfrak{q}\mathrm{S}{\mathrm{\nabla }}^i(\mathrm{K})=\{0\}$, $i=1,2,\dots ,n$. Therefore, by ∇-primeness of $\mathrm{S}$, $\mathfrak{q}=0$. Similarly $\mathrm{K}\mathfrak{q}=\{0\}$ implies $\mathfrak{q}=0$.

Corollary 3.4

Let $\mathrm{S}$ be a ∇-prime ring and $\mathrm{K}$ be a nonzero ∇-closed ideal of $\mathrm{S}$ and $\mathrm{J}$ be any nonzero ideal of $\mathrm{S}$. Then $\mathrm{J}\mathrm{K}\ne \{0\}$ and $\mathrm{K}\mathrm{J}\ne \{0\}$. Therefore, $\mathrm{K}$ is essential if and only if $\mathrm{K}$ is nonzero.

If $\mathrm{S}$ is a ∇-prime ring, then by Lemma 2.1, $\mathrm{S}$ is semiprime. Therefore by Beidar et al. [33] its maximal right ring of quotients ${\mathcal{Q}}_{\mathrm{mr}}(\mathrm{S})$ exists. Now from Corollary 3.4, [[3], Theorem A] and [[34], Theorem 2], we have the following result:

Lemma 3.6

Consider a ∇-prime ring $\mathrm{S}$ with $\mathrm{K}$ as its nonzero ∇-closed ideal. Then $\mathrm{K}$, $\mathrm{S}$ and ${\mathcal{Q}}_{\mathrm{mr}}(\mathrm{S})$ satisfy the same polynomial identities. In particular if $\mathrm{K}$ is commutative, then so is $\mathrm{S}$.

Lemma 3.7

Consider a ∇-prime ring $\mathrm{S}$ with $\mathrm{K}$ as its nonzero ∇-closed ideal. Assume $\mathfrak{p},\mathfrak{q}\in \mathrm{S}$. Then the following implications hold.

1. $\mathfrak{p}\mathrm{K}\mathfrak{p}=\{0\}$ implies $\mathfrak{p}=0$

2. $\mathfrak{p}\mathrm{K}{\mathrm{\nabla }}^i(\mathfrak{q})=\{0\}$ for all $i=1,2,\dots ,n$ implies $\mathfrak{p}=0$ or $\mathfrak{q}=0$.

Proof.

(i) Suppose $\mathfrak{p}\mathrm{K}\mathfrak{p}=\{0\}$, then $\mathfrak{p}\mathrm{K}\mathrm{S}\mathfrak{p}\mathrm{K}=\{0\}$. Now by Lemma 2.1, $\mathrm{S}$ is semiprime so $\mathfrak{p}\mathrm{K}=\{0\}$, which further, by Lemma 3.5, implies that $\mathfrak{p}=0$.

(ii) Suppose $\mathfrak{p}\mathrm{K}{\mathrm{\nabla }}^i(\mathfrak{q})=\{0\}$. Then $\mathfrak{p}\mathrm{K}\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{q})=\{0\}$, for all $i=1,2,\dots ,n$. Now by ∇-primeness of $\mathrm{S}$, either $\mathfrak{q}=0$ or $\mathfrak{p}\mathrm{K}=\{0\}$. In the latter case by Lemma 3.5, $\mathfrak{p}=0$.

Lemma 3.8

Let $\mathrm{S}$ be a ring with an anti-automorphism ∇ and $\mathrm{K}$ be a nonzero ∇-closed ideal of $\mathrm{S}$. If $\mathrm{S}$ is ∇-prime and $\mathfrak{p}\in \mathrm{S}$ is such that $\mathfrak{p}\mathrm{\nabla }(\mathfrak{t})=\mathfrak{t}\mathfrak{p}$ for every $\mathfrak{t}\in \mathrm{K}$, then either ∇ is the identity map or $\mathfrak{p}=0$.

Proof.

Suppose $\mathfrak{p}\mathrm{\nabla }(\mathfrak{t})=\mathfrak{t}\mathfrak{p}$ for every $\mathfrak{t}\in \mathrm{S}$. If $\mathrm{S}$ is commutative, then $(\mathrm{\nabla }(\mathfrak{t})-\mathfrak{t})\mathrm{S}\mathfrak{p}=\{0\}$ for every $\mathfrak{t}\in \mathrm{K}.$ Since $\mathrm{K}$ is ∇-closed, hence by ∇-primeness of $\mathrm{S}$, it follows that either $\mathfrak{p}=0$ or $\mathrm{\nabla }(\mathfrak{t})=\mathfrak{t}$ for every $\mathfrak{t}\in \mathrm{K}$. In the latter case, replacing $\mathfrak{t}$ by $\mathfrak{t}\mathfrak{s}$, we arrive at $\mathfrak{t}\mathfrak{s}=\mathrm{\nabla }(\mathfrak{t}\mathfrak{s})=\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})$ for every $\mathfrak{t}\in \mathrm{K}$ and $\mathfrak{s}\in \mathrm{S}$. Thus $\mathrm{K}(\mathfrak{s}-\mathrm{\nabla }(\mathfrak{s}))=\{0\}$ for every $\mathfrak{s}\in \mathrm{S}$ and hence by Lemma 3.5, we infer that ∇ is the identity map. Next assume $\mathrm{S}$ is noncommutative. Then for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$, we have $\mathfrak{t}\mathfrak{s}\mathfrak{p}=\mathfrak{p}\mathrm{\nabla }(\mathfrak{t}\mathfrak{s})=\mathfrak{s}\mathfrak{p}\mathrm{\nabla }(\mathfrak{t})=\mathfrak{s}\mathfrak{t}\mathfrak{p}$, that is, $[\mathfrak{t},\mathfrak{s}]\mathfrak{p}=0$. Now substituting $\mathfrak{u}\mathfrak{t}$ in place of $\mathfrak{t}$ in the last expression, we get $[\mathfrak{t},\mathfrak{s}]\mathrm{K}\mathfrak{p}=\{0\}$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Since $\mathrm{K}$ is ∇-closed ideal, so invoking Lemma 3.7, we get $\mathfrak{p}=0$.

Corollary 3.5

[35], Lemma 3

Suppose $\mathrm{S}$ is a prime $\ast$-ring and $\mathfrak{p}\in \mathrm{S}$. If $\mathrm{S}$ is noncommutative and $\mathfrak{p}{\mathfrak{t}}^{\ast }=\mathfrak{t}\mathfrak{p}$ for every $\mathfrak{t}\in \mathrm{S}$, then $\mathfrak{p}=0$.

We remark that Lemma 3.8, does not hold if ∇ is an automorphism. For this let $\mathrm{S}$ be any noncommutative ring and let $\mathfrak{p}$ be a noncentral invertible element of $\mathrm{S}$. Define the map $\mathrm{\nabla }:\mathrm{S}\to \mathrm{S}$ by $\mathrm{\nabla }(\mathfrak{t})={\mathfrak{p}}^{-1}\mathfrak{t}\mathfrak{p}$. Then ∇ is a nontrivial automorphism. Also $\mathfrak{p}\mathrm{\nabla }(\mathfrak{t})=\mathfrak{t}\mathfrak{p}$ for every $\mathfrak{t}\in \mathrm{S}$.

Now if $\mathrm{S}$ is semiprime and the order of ∇ is infinite in $\mathrm{S}$, then so is the order of ∇ in ${\mathcal{Q}}_s(\mathrm{S})$. Assume order of ∇ is n in $\mathrm{S}$, so ${\mathrm{\nabla }}^n(\mathfrak{t})=\mathfrak{t}$ for every $\mathfrak{t}\in \mathrm{S}$. Let $\mathfrak{q}\in {\mathcal{Q}}_s(\mathrm{S})$. Then by Beidar et al. [[33], Prop. 2.2.3], there is an ideal $\mathrm{K}$ of $\mathrm{S}$ such that $\mathfrak{q}\mathrm{K}\subseteq \mathrm{S}$. If ∇ is an anti-automorphism, then for every $\mathfrak{t}\in \mathrm{K}$, we have $\mathrm{\nabla }(\mathfrak{q}\mathfrak{t})=\mathrm{\nabla }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{q})$. Applying ${\mathrm{\nabla }}^{n-1}$ on both sides, we get $\{\mathfrak{q}-{\mathrm{\nabla }}^n(\mathfrak{q})\}\mathrm{K}=\{0\}$. Also, if ∇ is an automorphism then we get the same relation. Again by Beidar et al. [[33], Prop. 2.2.3], it follows that ${\mathrm{\nabla }}^n(\mathfrak{q})=\mathfrak{q}$. Therefore the order of ∇ is also n in ${\mathcal{Q}}_s(\mathrm{S})$. In this context, we have the following result.

Theorem 3.2

Consider a ∇-prime ring $\mathrm{S}$. Then ${\mathcal{Q}}_s(\mathrm{S})$, the symmetric ring of quotients of $\mathrm{S}$, is also ∇-prime.

Proof.

Suppose $\mathrm{S}$ is ∇-prime, then by Lemma 2.1, $\mathrm{S}$ is semiprime. Therefore its symmetric ring of quotients ${\mathcal{Q}}_s(\mathrm{S})$ exists and hence by Beidar et al. [[33], Propositions 2.5.3 and 2.5.4] ∇ can be uniquely lifted to ${\mathcal{Q}}_s(\mathrm{S})$. Thus the (anti)-automorphism ∇ of $\mathrm{S}$ can be implicitly assumed to be defined on the whole ${\mathcal{Q}}_s(\mathrm{S})$. Suppose $\mathfrak{p}{\mathcal{Q}}_s(\mathrm{S})\mathfrak{q}=0$ and $\mathfrak{p}{\mathcal{Q}}_s(\mathrm{S}){\mathrm{\nabla }}^i(\mathfrak{q})=\{0\}$ for some nonzero $\mathfrak{p},\mathfrak{q}\in {\mathcal{Q}}_s(\mathrm{S})$ and for each $i=1,2,\dots ,n$. So there exists dense ideals $\mathrm{P}$ and $\mathrm{J}$ such that $\mathfrak{p}\mathrm{P}\cup \mathrm{P}\mathfrak{p}\subseteq \mathrm{S}$ and $\mathfrak{q}\mathrm{J}\cup \mathrm{J}\mathfrak{q}\subseteq \mathrm{S}$. Also $0\ne \mathfrak{p}\mathfrak{t}\in \mathrm{S}$ and $0\ne \mathfrak{q}\mathfrak{s}\in \mathrm{S}$ for some $\mathfrak{t}\in \mathrm{P}$ and $\mathfrak{s}\in \mathrm{J}$. Thus $\mathfrak{p}\mathfrak{t}{\mathcal{Q}}_s(\mathrm{S}){\mathrm{\nabla }}^i(\mathfrak{q}\mathfrak{s})=\{0\}$ for every $i=1,2,\dots ,n$, a contradiction. Hence ${\mathcal{Q}}_s(\mathrm{S})$ is also ∇-prime.

Corollary 3.6

If $\mathcal{A}$ is a $\ast$-prime ring, then ${\mathcal{Q}}_s(\mathrm{S})$ is also $\ast$-prime.

We remark that using similar techniques as above and [[33], Prop. 2.5.3], one can prove that if $\mathrm{S}$ is ∇-prime ring, where ∇ is automorphism, then ${\mathcal{Q}}_{\mathrm{mr}}(\mathrm{S})$, the maximal right ring of quotients of $\mathrm{S}$, is ∇-prime.

4. Commutativity of ∇-prime rings

In [[31], Lemmas 2.1 and 2.2] Nejjar et al. improved [[12], Lemma 2] and showed that if $\mathrm{S}$ is a prime ring with second kind involution “$\ast$”, then $\mathrm{S}$ is commutative if and only if “$\ast$” is centralizing or skew-centralizing on $\mathrm{S}$. We shall extend this result to ∇-prime rings as follows:

Lemma 4.1

Suppose $\mathrm{S}$ is a ∇-prime ring with $\mathrm{K}$ a nonzero ∇-ideal. If ∇ is of the second kind and (4) $\mathfrak{t}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(4) for every $\mathfrak{t}\in \mathrm{K}$, where $\eta \in \{-1,1\}\cup \mathcal{Z}(\mathrm{S})$, then $\mathrm{S}$ is commutative.

Proof.

Linearizing (4(4) $\mathfrak{t}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(4) ), we have (5) $\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})+\mathfrak{s}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathfrak{s}+\eta \mathrm{\nabla }(\mathfrak{s})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(5) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Let $\alpha \in \mathcal{Z}(\mathrm{S})$. Then replacing $\mathfrak{s}$ by $\alpha \mathfrak{s}$ in (5(5) $\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})+\mathfrak{s}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathfrak{s}+\eta \mathrm{\nabla }(\mathfrak{s})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(5) ), we find that (6) $\mathrm{\nabla }(\alpha )\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})+\alpha \mathfrak{s}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\alpha )\mathrm{\nabla }(\mathfrak{s})\mathfrak{t}+\mathrm{\alpha \eta }\mathrm{\nabla }(\mathfrak{s})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(6) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. From (5(5) $\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})+\mathfrak{s}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathfrak{s}+\eta \mathrm{\nabla }(\mathfrak{s})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(5) ) and (6(6) $\mathrm{\nabla }(\alpha )\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})+\alpha \mathfrak{s}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\alpha )\mathrm{\nabla }(\mathfrak{s})\mathfrak{t}+\mathrm{\alpha \eta }\mathrm{\nabla }(\mathfrak{s})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(6) ), we find that $(\mathrm{\nabla }(\alpha )-\alpha )(\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})+\eta \mathrm{\nabla }(\mathfrak{s})\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Since $\mathrm{K}$ is ∇-ideal, we have $(\mathrm{\nabla }(\alpha )-\alpha )(\mathfrak{t}\mathfrak{s}+\eta \mathfrak{s}\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Now α is an arbitrary element of $\mathcal{Z}(\mathrm{S})$, therefore from the last relation, we have (7) ${\mathrm{\nabla }}^i(\mathrm{\nabla }(\alpha )-\alpha )(\mathfrak{t}\mathfrak{s}+\eta \mathfrak{s}\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(7) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$ and $i=1,2,\dots ,n$. Hence (8) ${\mathrm{\nabla }}^i(\mathrm{\nabla }(\alpha )-\alpha )\mathrm{S}[\mathfrak{t}\mathfrak{s}+\eta \mathfrak{s}\mathfrak{t},\mathfrak{r}]=\{0\}$(8) for every $\mathfrak{t},\mathfrak{s},\mathfrak{r}\in \mathrm{K}$, $\mathfrak{r}\in \mathrm{S}$ and $i=1,2,\dots ,n$. By the given hypothesis ∇ is of the second kind, so $\mathrm{\nabla }(\alpha )-\alpha$ is nonzero for some $\alpha \in \mathcal{Z}(\mathrm{S})$. Therefore, in view of the ∇-primeness of $\mathrm{S}$ from (8(8) ${\mathrm{\nabla }}^i(\mathrm{\nabla }(\alpha )-\alpha )\mathrm{S}[\mathfrak{t}\mathfrak{s}+\eta \mathfrak{s}\mathfrak{t},\mathfrak{r}]=\{0\}$(8) ), we deduce that (9) $\mathfrak{t}\mathfrak{s}+\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(9) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Assume that $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}=\{0\}$. Then from (9(9) $\mathfrak{t}\mathfrak{s}+\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(9) ), we have $\mathfrak{t}\mathfrak{s}=\eta \mathfrak{s}\mathfrak{t}$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Hence for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$, we have $\mathfrak{t}\mathfrak{s}\mathfrak{u}=\eta \mathfrak{s}\mathfrak{t}\mathfrak{u}=\mathfrak{s}(\eta \mathfrak{t}\mathfrak{u})=\mathfrak{s}\mathfrak{u}\mathfrak{t}$, that is, $\mathfrak{t}\mathfrak{s}\mathfrak{u}=\mathfrak{s}\mathfrak{u}\mathfrak{t}$. Applying Lemma 3.6, it follows that $\mathfrak{t}\mathfrak{s}\mathfrak{u}=\mathfrak{s}\mathfrak{u}\mathfrak{t}$ for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{S}$. Taking $\mathfrak{u}={\mathrm{\nabla }}^i(\alpha )$, we get $(\mathfrak{t}\mathfrak{s}-\mathfrak{s}\mathfrak{t})\mathrm{S}{\mathrm{\nabla }}^i(\alpha )=\{0\}$. By ∇-primeness of $\mathrm{S}$, we infer that $\mathrm{S}$ is commutative. Thus $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}\ne \{0\}$, a contradiction.

Therefore $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}\ne \{0\}$. Now from (9(9) $\mathfrak{t}\mathfrak{s}+\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(9) ), we have $[\mathfrak{t}\mathfrak{s}\mathfrak{u},\mathfrak{r}]=[-\eta \mathfrak{t}\mathfrak{u}\mathfrak{s},\mathfrak{r}]=[\mathfrak{u}\mathfrak{t}\mathfrak{s},\mathfrak{r}]$ for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$ and $\mathfrak{r}\in \mathrm{S}$. Therefore $\mathfrak{t}\mathfrak{s}\mathfrak{u}-\mathfrak{u}\mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$. Let ${\alpha }_1\in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. Then replacing $\mathfrak{u}$ by ${\alpha }_1$, we find that ${\alpha }_1[\mathfrak{t},\mathfrak{s}]\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Since $\mathrm{\nabla }(\mathcal{Z}(\mathrm{S})\cap \mathrm{K})\subseteq \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$, therefore by ∇-primeness of $\mathrm{S}$, we infer that $[\mathfrak{t},\mathfrak{s}]\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Now using $\mathfrak{s}\mathfrak{t}$ instead of $\mathfrak{t}$ in the previous relation and commuting it with $\mathfrak{t}$, we get $\mathrm{K}$ is commutative. Invoking Lemma 3.6, we infer that $\mathrm{S}$ is commutative.

Lemma 4.2

Suppose $\mathrm{S}$ is a ∇-prime ring with $\mathrm{K}$ a nonzero ∇-closed ideal. Let $\mathrm{\Phi }:\mathrm{S}\to \mathrm{S}$ be a derivation such that $\mathrm{\Phi }(\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$. Then $\mathrm{\Phi }=0$.

Proof.

Suppose $\mathrm{\Phi }(\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$. Then $\mathrm{\Phi }(\mathfrak{u})\mathfrak{t}=\mathrm{\Phi }(\mathfrak{u}\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$ and $\mathfrak{u}\in \mathrm{S}$. Hence $\mathrm{\Phi }(\mathfrak{u})\mathrm{S}\mathfrak{t}=\{0\}$ for every $\mathfrak{t}\in \mathrm{K}$ and $\mathfrak{u}\in \mathrm{S}$. In particular, $\mathrm{\Phi }(\mathfrak{u})\mathrm{S}{\mathrm{\nabla }}^i(\mathfrak{a})=\{0\}$ for every $\mathfrak{u}\in \mathrm{S}$ and $i=1,2\cdots ,n$, where $0\ne \mathfrak{a}\in \mathrm{K}$. By ∇-primeness of $\mathrm{S}$, we get $\mathrm{\Phi }(\mathfrak{u})=0$ for every $\mathfrak{u}\in \mathrm{S}$, that is, $\mathrm{\Phi }=0$.

In Ref. [8], Posner showed that a prime ring is necessarily commutative if it possesses a nonzero centralizing derivation. Bresar [[16], Theorem 4.1], generalized this result and obtained that a prime ring $\mathrm{S}$ is commutative if it possesses derivations Φ and Ψ such that $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$ holds for every $\mathfrak{t}\in \mathrm{K}$, where $\mathrm{K}$ is a nonzero left ideal of $\mathrm{S}$ and $\mathrm{\Psi }\ne 0$. Motivated by this we obtained the following theorem which partially extends [[16], Theorem 4.1] to ∇-prime rings.

Theorem 4.1

Suppose $\mathrm{S}$ is a ∇-prime ring with $\mathrm{K}$ a nonzero ∇-closed ideal. Let Φ and $\mathrm{\Psi }:\mathrm{S}\to \mathrm{S}$ be derivations, not both zero, such that (10) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(10) holds for every $\mathfrak{t}\in \mathrm{K}$, then $\mathrm{S}$ is commutative. Moreover if $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$, then $\mathrm{\Phi }=\mathrm{\Psi }$.

Proof.

First we prove that $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}\ne \{0\}.$ On the contrary suppose that $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}=\{0\}$. Then from (10(10) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(10) ), we have (11) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}=\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(11) Replacing $\mathfrak{t}$ by $\mathfrak{t}+\mathfrak{s}$ in (11(11) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}=\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(11) ), we have (12) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s})\mathfrak{t}=\mathfrak{t}\mathrm{\Psi }(\mathfrak{s})+\mathfrak{s}\mathrm{\Psi }(\mathfrak{t})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(12) Substituting $\mathfrak{s}\mathfrak{t}$ in place of $\mathfrak{s}$ in the previous relation, we obtain $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}\mathfrak{t}+\mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^2+\mathfrak{s}\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}=\mathfrak{t}\mathrm{\Psi }(\mathfrak{s})\mathfrak{t}+\mathfrak{t}\mathfrak{s}\mathrm{\Psi }(\mathfrak{t})+\mathfrak{s}\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})$$\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$ Using (11(11) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}=\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(11) ) and (12(12) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s})\mathfrak{t}=\mathfrak{t}\mathrm{\Psi }(\mathfrak{s})+\mathfrak{s}\mathrm{\Psi }(\mathfrak{t})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(12) ), we get $[\mathfrak{s}\mathrm{\Psi }(\mathfrak{t}),\mathfrak{t}]=0$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Now replacing $\mathfrak{s}$ by $\mathfrak{u}\mathfrak{s}$ in the last relation and utilizing it again, we have $[\mathfrak{u},\mathfrak{t}]\mathfrak{s}\mathrm{\Psi }(\mathfrak{t})=0$ for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$. Therefore $\mathrm{\Psi }(\mathfrak{t})[\mathfrak{u},\mathfrak{t}]\mathrm{K}\mathrm{\Psi }(\mathfrak{t})[\mathfrak{u},\mathfrak{t}]=\{0\}$, whence by Lemma 3.7, it follows that (13) $\mathrm{\Psi }(\mathfrak{t})[\mathfrak{u},\mathfrak{t}]=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{u}\in \mathrm{K}$(13) Linearizing (14(13) $\mathrm{\Psi }(\mathfrak{t})[\mathfrak{u},\mathfrak{t}]=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{u}\in \mathrm{K}$(13) ) in $\mathfrak{t}$, we arrive at (14) $\mathrm{\Psi }(\mathfrak{t})[\mathfrak{u},\mathfrak{s}]+\mathrm{\Psi }(\mathfrak{s})[\mathfrak{u},\mathfrak{t}]=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$(14) Putting $[\mathfrak{u},\mathfrak{t}]\mathfrak{u}$ in place of $\mathfrak{u}$ in the above relation and using it again and (14(13) $\mathrm{\Psi }(\mathfrak{t})[\mathfrak{u},\mathfrak{t}]=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{u}\in \mathrm{K}$(13) ), we infer that $\mathrm{\Psi }(\mathfrak{s})[\mathfrak{u},\mathfrak{t}{]}^2=0.$ Replacing $\mathfrak{s}$ by $\mathfrak{s}\mathfrak{v}$ in the previous expression, we have $\mathrm{\Psi }(\mathfrak{s})\mathrm{K}[\mathfrak{u},\mathfrak{t}{]}^2=\{0\}$ for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$ As $\mathrm{K}$ is ∇-closed, so we conclude that either $\mathrm{\Psi }(\mathfrak{s})=0$ for every $\mathfrak{s}\in \mathrm{K}$ or $[\mathfrak{u},\mathfrak{t}{]}^2=0$ for every $\mathfrak{t},\mathfrak{u}\in \mathrm{K}$. In the latter case $\mathrm{K}$ is a PI-ring and hence by Herstein [[2],Theorem 1.4.2], $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}\ne \{0\}$ which is absurd.

Next if $\mathrm{\Psi }(\mathfrak{u})=0$ for every $\mathfrak{u}\in \mathrm{K}$, then by Lemma 4.2, $\mathrm{\Psi }=0$. Hence from (11(11) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}=\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(11) ), we have (15) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}=0\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(15) Linearizing this, we get (16) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s})\mathfrak{t}=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(16) Putting ${\mathfrak{t}}^2$ for $\mathfrak{t}$ in (17(16) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s})\mathfrak{t}=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(16) ) and using (16(15) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}=0\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(15) ), we arrive at (17) $\mathfrak{t}\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^2=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(17) Upon right multiplying (17(16) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s})\mathfrak{t}=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(16) ) by $\mathfrak{t}$ and then subtracting it from (18(17) $\mathfrak{t}\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^2=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(17) ), we have (18) $\mathfrak{t}\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}-\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}\mathfrak{t}=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(18) Substituting $\mathfrak{s}\mathrm{\Phi }(\mathfrak{t})$ for $\mathfrak{s}$ in the last relation, we have $\mathfrak{t}\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}\mathrm{\Phi }(\mathfrak{t})=0$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Thus by Lemma 3.7, it follows that $\mathfrak{t}\mathrm{\Phi }(\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$. So in view of the last expression, right multiplication of (17(16) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s})\mathfrak{t}=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(16) ) by $\mathrm{\Phi }(\mathfrak{t})$ provides us $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}\mathrm{\Phi }(\mathfrak{t})=0$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Hence by Lemma 4.2, $\mathrm{\Phi }=0$ which is a contradiction. Thus $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}\ne \{0\}$.

Now linearizing (10(10) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(10) ), we have (19) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{s})-\mathfrak{s}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(19) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Let $\alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. Then replacing $\mathfrak{s}$ by α in the previous expression, we obtain (20) $\alpha (\mathrm{\Phi }(\mathfrak{t})-\mathrm{\Psi }(\mathfrak{t}))+(\mathrm{\Phi }(\alpha )-\mathrm{\Psi }(\alpha ))\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(20) for every $\mathfrak{t}\in \mathrm{K}$. Next substituting ${\alpha }^2$ for $\mathfrak{s}$ in (20(19) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\mathrm{\Phi }(\mathfrak{s})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{s})-\mathfrak{s}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(19) ), we arrive at (21) $\begin{array}{rl}& \alpha \{\alpha (\mathrm{\Phi }(\mathfrak{t})-\mathrm{\Psi }(\mathfrak{t}))+(\mathrm{\Phi }(\alpha )-\mathrm{\Psi }(\alpha ))\mathfrak{t}\}\\ & +\alpha \{\mathrm{\Phi }(\alpha )-\mathrm{\Psi }(\alpha )\}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\end{array}$(21) for every $\mathfrak{t}\in \mathrm{K}$. From (21(20) $\alpha (\mathrm{\Phi }(\mathfrak{t})-\mathrm{\Psi }(\mathfrak{t}))+(\mathrm{\Phi }(\alpha )-\mathrm{\Psi }(\alpha ))\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(20) ) and (22(21) $\begin{array}{rl}& \alpha \{\alpha (\mathrm{\Phi }(\mathfrak{t})-\mathrm{\Psi }(\mathfrak{t}))+(\mathrm{\Phi }(\alpha )-\mathrm{\Psi }(\alpha ))\mathfrak{t}\}\\ & +\alpha \{\mathrm{\Phi }(\alpha )-\mathrm{\Psi }(\alpha )\}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\end{array}$(21) ), we conclude that $\alpha \{\mathrm{\Phi }(\alpha )-\mathrm{\Psi }(\alpha )\}\mathrm{K}[\mathfrak{t},\mathfrak{s}]=\{0\}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$ Invoking Lemma 3.7, we find that either $[\mathfrak{t},\mathfrak{s}]=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}$ or $\alpha \mathrm{\Phi }(\alpha )=\alpha \mathrm{\Psi }(\alpha )$ for every $\alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. In the former case by Lemma 3.6, $\mathrm{S}$ is commutative. If the latter case prevails, then (22(21) $\begin{array}{rl}& \alpha \{\alpha (\mathrm{\Phi }(\mathfrak{t})-\mathrm{\Psi }(\mathfrak{t}))+(\mathrm{\Phi }(\alpha )-\mathrm{\Psi }(\alpha ))\mathfrak{t}\}\\ & +\alpha \{\mathrm{\Phi }(\alpha )-\mathrm{\Psi }(\alpha )\}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\end{array}$(21) ) yields ${\alpha }^2(\mathrm{\Phi }-\mathrm{\Psi })(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{S}$ and $\alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. Note that $\mathrm{\nabla }(\mathcal{Z}(\mathrm{S})\cap \mathrm{K})\subseteq \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. Therefore by ∇-primeness of $\mathrm{S}$, we conclude that $(\mathrm{\Phi }-\mathrm{\Psi })(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{S}$. Now $\mathrm{\Phi }-\mathrm{\Psi }$ is also derivation on $\mathrm{S}$. Hence by Lee [[36], Corollary 2], it follows that either $\mathrm{\Phi }=\mathrm{\Psi }$ or $\mathcal{Z}(\mathrm{S})$ contains a nonzero ideal of $\mathrm{S}$. In the latter case, $\mathrm{S}$ contains a nonzero commutative ∇-ideal and hence by Lemma 3.6, $\mathrm{S}$ is commutative. If $\mathrm{\Phi }=\mathrm{\Psi }$, then from (10(10) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(10) ), we have $[\mathrm{\Phi }(\mathfrak{t}),\mathfrak{t}]\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{K}$, whence by Bell and Martindale [[15], Lemma 4], $[\mathrm{\Phi }(\mathfrak{t}),\mathfrak{t}]=0$ for every $\mathfrak{t}\in \mathrm{K}$. Again by Lee [[36], Corollary 2], we infer that either $\mathrm{\Phi }=0$ or $\mathrm{S}$ is commutative. But the former case is not possible.

Now suppose that $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$. Then by previous arguments $\mathrm{S}$ is commutative. Hence we have $\chi (\mathfrak{t})\mathfrak{t}=0$ for every $\mathfrak{t}\in \mathrm{K}$, where $\chi =\mathrm{\Phi }-\mathrm{\Psi }$ is also a derivation on $\mathrm{S}$. Linearizing the last relation, we get (22) $\mathfrak{t}\chi (\mathfrak{s})+\mathfrak{s}\chi (\mathfrak{t})=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(22) Replacing $\mathfrak{s}$ by $\mathfrak{t}\mathfrak{s}$ in (23(22) $\mathfrak{t}\chi (\mathfrak{s})+\mathfrak{s}\chi (\mathfrak{t})=0\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(22) ), we find that $\chi (\mathfrak{s}){\mathfrak{t}}^2=0$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Now using Lemma 4.2, it can be easily deduced that $\chi =0$. This finishes the proof.

Corollary 4.1

[21], Theorem 1.1

Suppose that $\mathrm{S}$ is a $\ast$-prime ring and $\mathrm{\Phi }:\mathrm{S}\to \mathrm{S}$ is a nonzero derivation. If char$(\mathrm{S})\ne 2$ and $[\mathrm{\Phi }(\mathfrak{t}),\mathfrak{t}]\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{S}$, then $\mathrm{S}$ is commutative.

The following example illustrates that the condition of ∇-primeness in Theorem 4.1 is not superfluous.

Example 4.1

Consider the ring $\mathrm{S}=\{\left[\begin{array}{cc}x& y\\ 0& 0\end{array}\right]\mid x,y\in \mathcal{R}\},$where $\mathcal{R}$ is any noncommutative ring. The map $\mathrm{\Phi }:\mathrm{S}\to \mathrm{S}$ given by $\mathrm{\Phi }\left[\begin{array}{cc}x& y\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& y\\ 0& 0\end{array}\right]$ is a nonzero derivation. Clearly for $\mathrm{\Psi }=0$, it can be easily verified that $\mathrm{\Phi }(\mathfrak{t})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$ holds for every $\mathfrak{t}\in \mathrm{S}$. Note that $\mathrm{S}$ is a noncommutative ring which is not ��-prime for any (anti)-automorphism ∇.

In [[12], Main Theorem] S. Ali et al. showed that a 2-torsion free prime $\ast$-ring $\mathrm{S}$ is commutative if there exists a derivation $\mathrm{\Psi }:\mathrm{S}\to \mathrm{S}$ such that $[\mathrm{\Psi }(\mathfrak{t}),{\mathfrak{t}}^{\ast }]\in \mathcal{Z}(\mathrm{S})$ for all $\mathfrak{t}\in \mathrm{S}$ and $\mathrm{\Psi }(H_{\ast }(R)\cap \mathcal{Z}(\mathrm{S}))\ne \{0\}$. Nejjar et al. [[31], Theorem 3.7] gave an improved version of this result and proved that a 2-torsion free prime ring $\mathrm{S}$ with second kind involution “$\ast$” is commutative if there exists a derivation $\mathrm{\Psi }:\mathrm{S}\to \mathrm{S}$ such that $[\mathrm{\Psi }(\mathfrak{t}),{\mathfrak{t}}^{\ast }]\in \mathcal{Z}(\mathrm{S})$ for all $\mathfrak{t}\in \mathrm{S}$ or $\mathrm{\Psi }(\mathfrak{t})\circ {\mathfrak{t}}^{\ast }\in \mathcal{Z}(\mathrm{S})$ for all $\mathfrak{t}\in \mathrm{S}$. Motivated by this, we will prove the following theorem which shows that the torsion restriction is superfluous and extends it to ∇-prime rings.

Theorem 4.2

Consider a ∇-prime ring $\mathrm{S}$ with $\mathrm{K}$ as its nonzero ∇-ideal. Assume that ∇ is of the second kind and there is a nonzero derivation $\mathrm{\Phi }:\mathrm{S}\to \mathrm{S}$ such that (23) $\mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(23) for every $\mathfrak{t}\in \mathrm{K}$, where $\eta \in \{1,0,-1\}$. Then $\mathrm{S}$ is commutative.

Proof.

Linearizing (24(23) $\mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(23) ), we have (24) $\begin{array}{rl}& \mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})\\ & +\eta \mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\end{array}$(24) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Let $\alpha \in \mathcal{Z}(\mathrm{S})$. Then replacing $\mathfrak{t}$ by $\alpha \mathfrak{t}$ in (25(24) $\begin{array}{rl}& \mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})\\ & +\eta \mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\end{array}$(24) ), we obtain (25) $\begin{array}{rl}& \alpha \mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\Phi }(\alpha )\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})+\mathrm{\nabla }(\alpha )\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})\\ & +\eta \mathrm{\nabla }(\alpha )\mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})+\mathrm{\alpha \eta }\mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\\ & +\mathrm{\Phi }(\alpha )\eta \mathrm{\nabla }(\mathfrak{s})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\end{array}$(25) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Now we proceed by considering the following two cases:

Case I: $\mathrm{\Phi }(\alpha )=0$ for every $\alpha \in \mathcal{Z}(\mathrm{S}).$ In this case (25(25) $\begin{array}{rl}& \alpha \mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\Phi }(\alpha )\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})+\mathrm{\nabla }(\alpha )\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})\\ & +\eta \mathrm{\nabla }(\alpha )\mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})+\mathrm{\alpha \eta }\mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\\ & +\mathrm{\Phi }(\alpha )\eta \mathrm{\nabla }(\mathfrak{s})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\end{array}$(25) ), reduces to (26) $\begin{array}{rl}& \alpha \mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\nabla }(\alpha )\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\alpha )\mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})\\ & +\mathrm{\alpha \eta }\mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S}),\end{array}$(26) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$ and $\alpha \in \mathcal{Z}(\mathrm{S})$. Also from (24(24) $\begin{array}{rl}& \mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})\\ & +\eta \mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\end{array}$(24) ), we have (27) $\begin{array}{rl}& \mathrm{\nabla }(\alpha )\mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\nabla }(\alpha )\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})+\mathrm{\nabla }(\alpha )\eta \mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})\\ & +\mathrm{\nabla }(\alpha )\eta \mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\end{array}$(27) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. From (26(26) $\begin{array}{rl}& \alpha \mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\nabla }(\alpha )\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\alpha )\mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})\\ & +\mathrm{\alpha \eta }\mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S}),\end{array}$(26) ) and (27(27) $\begin{array}{rl}& \mathrm{\nabla }(\alpha )\mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\nabla }(\alpha )\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})+\mathrm{\nabla }(\alpha )\eta \mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})\\ & +\mathrm{\nabla }(\alpha )\eta \mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\end{array}$(27) ), it follows that (28) $\begin{array}{rl}& (\mathrm{\nabla }(\alpha )-\alpha )\{\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\eta \mathfrak{s}\mathrm{\Phi }(\mathfrak{t})\}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}\\ & \mathrm{and}\alpha \in \mathcal{Z}(\mathrm{S}).\end{array}$(28) Now ∇ is given to be of second kind, so in the light of ∇-primeness of $\mathrm{S}$, (28(28) $\begin{array}{rl}& (\mathrm{\nabla }(\alpha )-\alpha )\{\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\eta \mathfrak{s}\mathrm{\Phi }(\mathfrak{t})\}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}\\ & \mathrm{and}\alpha \in \mathcal{Z}(\mathrm{S}).\end{array}$(28) ) yields that $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}+\eta \mathfrak{s}\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Therefore by Theorem 4.1, $\mathrm{S}$ is commutative.

Case II: $\mathrm{\Phi }(\alpha )\ne 0$ for some $\alpha \in \mathcal{Z}(\mathrm{S}).$ Substituting $\mathfrak{t}$ for $\mathfrak{s}$ in (25(25) $\begin{array}{rl}& \alpha \mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{s})+\mathrm{\Phi }(\alpha )\mathfrak{t}\mathrm{\nabla }(\mathfrak{s})+\mathrm{\nabla }(\alpha )\mathrm{\Phi }(\mathfrak{s})\mathrm{\nabla }(\mathfrak{t})\\ & +\eta \mathrm{\nabla }(\alpha )\mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})+\mathrm{\alpha \eta }\mathrm{\nabla }(\mathfrak{s})\mathrm{\Phi }(\mathfrak{t})\\ & +\mathrm{\Phi }(\alpha )\eta \mathrm{\nabla }(\mathfrak{s})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\end{array}$(25) ) and making use of (23(23) $\mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(23) ), it follows that $\mathrm{\Phi }(\alpha )\{\mathfrak{t}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathfrak{t}\}\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{K}$. Now since $\mathrm{K}$ is ∇-ideal, so $\mathrm{\Phi }(\alpha )\mathrm{S}{\mathrm{\nabla }}^i([\mathfrak{t}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathfrak{t},s])=\{0\}$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Invoking ∇-primeness of $\mathrm{S}$, we infer that $\mathfrak{t}\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{K}$, whence by Lemma 4.1, $\mathrm{S}$ is commutative.

The following example shows that the condition of ∇-primeness in Theorem 4.2 is not superfluous.

Example 4.2

Let $\mathbb{F}$ be a field with a nontrivial automorphism σ of finite order. Consider the noncommutative ring $\mathbb{S}=\mathbb{F}[x]\times {\mathcal{M}}_2(\mathbb{F})$, where $\mathbb{F}[x]$ is the ring of all polynomials over $\mathbb{F}$ and ${\mathcal{M}}_2(\mathbb{F})$ is the ring of all $2\times 2$ matrices over $\mathbb{F}$. Define the maps $\mathrm{\nabla },\mathrm{\Phi }:\mathbb{S}\to \mathbb{S}$ by $\mathrm{\nabla }(f(x),X)=(f(x),X^{\sigma })$ and $\mathrm{\Phi }(f(x),X)=(f^{\mathrm{\prime }}(x),0)$, where $X^{\sigma }$ denotes the matrix whose entries are the images of the corresponding entries of X under σ and $f^{\mathrm{\prime }}(x)$ denotes the formal derivative of $f(x)$. Then it can be easily verified that ∇ is an automorphism of the second kind and Φ is a derivation such that $\mathrm{\Phi }(\mathfrak{t})\mathrm{\nabla }(\mathfrak{t})+\eta \mathrm{\nabla }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{S}$, where $\eta \in \{1,0,-1\}$. Note that $\mathrm{S}$ is not a ∇-prime ring.

In particular, if we take $\mathrm{\nabla }=\ast$, then we have the following generalization of Theorem 4.2, which extends [[10], Theorem 1] to $\ast$-prime rings.

Theorem 4.3

Consider a *-prime ring $\mathrm{S}$ with $\mathrm{K}$ as its nonzero *-ideal. Assume that “$\ast$” is of the second kind and $\mathrm{\Phi },\mathrm{\Psi }:\mathrm{S}\to \mathrm{S}$ are derivations, not both zero, such that (29) $\mathrm{\Phi }(\mathfrak{t}){\mathfrak{t}}^{\ast }-{\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(29) for every $\mathfrak{t}\in \mathrm{K}$. Then $\mathrm{S}$ is commutative. Moreover, if $\mathrm{\Phi }(\mathfrak{t}){\mathfrak{t}}^{\ast }-{\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$, then $\mathrm{\Phi }=\mathrm{\Psi }$.

Proof.

Linearizing (29(29) $\mathrm{\Phi }(\mathfrak{t}){\mathfrak{t}}^{\ast }-{\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(29) ), we have (30) $\begin{array}{rl}& \mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }+\mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^{\ast }-{\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{s})-{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\\ & \mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.\end{array}$(30) Let $\alpha \in \mathcal{Z}(\mathrm{S})$. Then substituting $\alpha \mathfrak{s}$ for $\mathfrak{s}$ in (30(30) $\begin{array}{rl}& \mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }+\mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^{\ast }-{\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{s})-{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\\ & \mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.\end{array}$(30) ), we have (31) $\begin{array}{rl}& {\alpha }^{\ast }\mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }+\alpha \mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^{\ast }+\mathrm{\Phi }(\alpha )\mathfrak{s}{\mathfrak{t}}^{\ast }\\ & -\alpha {\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{s})-\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{s}-{\alpha }^{\ast }{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\end{array}$(31) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Replacing $\mathfrak{s}$ by $\mathfrak{t}$ in (31(31) $\begin{array}{rl}& {\alpha }^{\ast }\mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }+\alpha \mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^{\ast }+\mathrm{\Phi }(\alpha )\mathfrak{s}{\mathfrak{t}}^{\ast }\\ & -\alpha {\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{s})-\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{s}-{\alpha }^{\ast }{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\end{array}$(31) ) and using (29(29) $\mathrm{\Phi }(\mathfrak{t}){\mathfrak{t}}^{\ast }-{\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(29) ), we see that (32) $\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{t}}^{\ast }-\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(32) Now we proceed by considering the following cases:

Case I: $\mathrm{\Phi }(\alpha )\ne 0$ and $\mathrm{\Psi }(\alpha )=0$. In this case (32(32) $\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{t}}^{\ast }-\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(32) ) reduces to $\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{t}}^{\ast }\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{K}$. Clearly $\mathfrak{t}{\mathfrak{t}}^{\ast }$ is a symmetric element. Hence by $\ast$-primeness of $\mathrm{S}$, the last expression entails that $\mathfrak{t}{\mathfrak{t}}^{\ast }\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{K}$. Therefore by Lemma 4.1, $\mathrm{S}$ is commutative.

Case II: $\mathrm{\Phi }(\alpha )=0$ and $\mathrm{\Psi }(\alpha )\ne 0$. Arguing the same as in Case I, we obtain that $\mathrm{S}$ is commutative.

Case III: $\mathrm{\Phi }(\alpha )=0$ and $\mathrm{\Psi }(\alpha )=0$. From (31(31) $\begin{array}{rl}& {\alpha }^{\ast }\mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }+\alpha \mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^{\ast }+\mathrm{\Phi }(\alpha )\mathfrak{s}{\mathfrak{t}}^{\ast }\\ & -\alpha {\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{s})-\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{s}-{\alpha }^{\ast }{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\end{array}$(31) ), we have (33) ${\alpha }^{\ast }\mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }+\alpha \mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^{\ast }-\alpha {\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{s})-{\alpha }^{\ast }{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(33) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. From (33(33) ${\alpha }^{\ast }\mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }+\alpha \mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^{\ast }-\alpha {\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{s})-{\alpha }^{\ast }{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$(33) ) and (30(30) $\begin{array}{rl}& \mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }+\mathrm{\Phi }(\mathfrak{s}){\mathfrak{t}}^{\ast }-{\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{s})-{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\\ & \mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.\end{array}$(30) ), we get (34) $({\alpha }^{\ast }-\alpha )\{\mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }-{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(34) Therefore, we also have (35) $({\alpha }^{\ast }-\alpha {)}^{\ast }\{\mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }-{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(35) Since “$\ast$” is of second kind, so ${\alpha }^{\ast }-\alpha$ is nonzero for some $\alpha \in \mathcal{Z}(\mathrm{S})$. Therefore invoking $\ast$-primeness of $\mathrm{S}$, we infer from (34(34) $({\alpha }^{\ast }-\alpha )\{\mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }-{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(34) ) and (35(35) $({\alpha }^{\ast }-\alpha {)}^{\ast }\{\mathrm{\Phi }(\mathfrak{t}){\mathfrak{s}}^{\ast }-{\mathfrak{s}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(35) ), that (36) $\mathrm{\Phi }(\mathfrak{t})\mathfrak{s}-\mathfrak{s}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(36) Thus by Theorem 4.1, $\mathrm{S}$ is commutative.

Case IV : $\mathrm{\Phi }(\alpha )\ne 0$ and $\mathrm{\Psi }(\alpha )\ne 0$. Putting $\mathfrak{t}+\mathfrak{s}$ in place of $\mathfrak{t}$ in (32(32) $\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{t}}^{\ast }-\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(32) ), we get (37) $\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{s}}^{\ast }+\mathrm{\Phi }(\alpha )\mathfrak{s}{\mathfrak{t}}^{\ast }-\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{s}-\mathrm{\Psi }(\alpha ){\mathfrak{s}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(37) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Replacing $\mathfrak{s}$ by $\alpha \mathfrak{s}$ in (37(37) $\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{s}}^{\ast }+\mathrm{\Phi }(\alpha )\mathfrak{s}{\mathfrak{t}}^{\ast }-\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{s}-\mathrm{\Psi }(\alpha ){\mathfrak{s}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(37) ), we have (38) $\begin{array}{rl}& {\alpha }^{\ast }\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{s}}^{\ast }+\alpha \mathrm{\Phi }(\alpha )\mathfrak{s}{\mathfrak{t}}^{\ast }-\alpha \mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{s}\\ & -{\alpha }^{\ast }\mathrm{\Psi }(\alpha ){\mathfrak{s}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\end{array}$(38) for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. From (37(37) $\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{s}}^{\ast }+\mathrm{\Phi }(\alpha )\mathfrak{s}{\mathfrak{t}}^{\ast }-\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{s}-\mathrm{\Psi }(\alpha ){\mathfrak{s}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$(37) ) and (38(38) $\begin{array}{rl}& {\alpha }^{\ast }\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{s}}^{\ast }+\alpha \mathrm{\Phi }(\alpha )\mathfrak{s}{\mathfrak{t}}^{\ast }-\alpha \mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{s}\\ & -{\alpha }^{\ast }\mathrm{\Psi }(\alpha ){\mathfrak{s}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\end{array}$(38) ), we have (39) $({\alpha }^{\ast }-\alpha )\{\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{s}}^{\ast }-\mathrm{\Psi }(\alpha ){\mathfrak{s}}^{\ast }\mathfrak{t}\}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(39) As “$\ast$” is of the second kind. Hence by $\ast$-primeness of $\mathrm{S}$, we have (40) $\mathrm{\Phi }(\alpha )\mathfrak{t}\mathfrak{s}-\mathrm{\Psi }(\alpha )\mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(40) Subcase I: $\mathrm{\Phi }(\alpha )=\mathrm{\Psi }(\alpha )$. From (40(40) $\mathrm{\Phi }(\alpha )\mathfrak{t}\mathfrak{s}-\mathrm{\Psi }(\alpha )\mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(40) ), we have $\mathrm{\Phi }(\alpha )[\mathfrak{t},\mathfrak{s}]\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$ Since $\mathrm{K}$ is $\ast$-closed, so by $\ast$-primeness of $\mathrm{S}$, we infer that $[\mathfrak{t},\mathfrak{s}]\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$ Now it can be easily deduced that $\mathrm{S}$ is commutative.

Subcase II: $\mathrm{\Phi }(\alpha )\ne \mathrm{\Psi }(\alpha )$. Replacing $\mathfrak{t}$ by ${\mathfrak{t}}^{\ast }$ and $\mathfrak{s}$ by $\mathfrak{t}$ in (40(40) $\mathrm{\Phi }(\alpha )\mathfrak{t}\mathfrak{s}-\mathrm{\Psi }(\alpha )\mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(40) ), we get (41) $\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{t}-\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{t}}^{\ast }\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(41) Also from (40(40) $\mathrm{\Phi }(\alpha )\mathfrak{t}\mathfrak{s}-\mathrm{\Psi }(\alpha )\mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(40) ), we have (42) $\mathrm{\Psi }(\alpha )\mathfrak{t}{\mathfrak{t}}^{\ast }-\mathrm{\Phi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(42) From (41(41) $\mathrm{\Psi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{t}-\mathrm{\Phi }(\alpha )\mathfrak{t}{\mathfrak{t}}^{\ast }\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(41) ) and (42(42) $\mathrm{\Psi }(\alpha )\mathfrak{t}{\mathfrak{t}}^{\ast }-\mathrm{\Phi }(\alpha ){\mathfrak{t}}^{\ast }\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(42) ), we have (43) $\{\mathrm{\Psi }(\alpha )-\mathrm{\Phi }(\alpha )\}\mathfrak{t}\circ {\mathfrak{t}}^{\ast }\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(43) As $\mathrm{\Phi }(\alpha )\ne \mathrm{\Psi }(\alpha )$ and $\mathrm{S}$ is $\ast$-prime, so from (43(43) $\{\mathrm{\Psi }(\alpha )-\mathrm{\Phi }(\alpha )\}\mathfrak{t}\circ {\mathfrak{t}}^{\ast }\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathrm{K}.$(43) ) we deduce that $\mathfrak{t}\circ {\mathfrak{t}}^{\ast }\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{K}$. Therefore by Lemma 4.1, $\mathrm{S}$ is commutative.

Now suppose that $\mathrm{\Phi }(\mathfrak{t}){\mathfrak{t}}^{\ast }-{\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$. Then by above, $\mathrm{S}$ is commutative. Hence we have ${\mathfrak{t}}^{\ast }\chi (\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$ where $\chi =\mathrm{\Phi }-\mathrm{\Psi }$ is also a derivation on $\mathrm{S}$. Replacing $\mathfrak{t}$ by $\mathfrak{t}\mathfrak{s}+\mathfrak{t}$ in the last equation, we arrive at $\mathfrak{t}{\mathfrak{t}}^{\ast }\chi (\mathfrak{s})=0$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Now it is trivial to observe that $\chi =0$ i.e. $\mathrm{\Phi }=\mathrm{\Psi }$. This finishes the proof.

Corollary 4.2

[10], Theorem 1

Let $\mathrm{S}$ be a noncommutative prime ring equipped with a second kind involution “$\ast$”. If char$(\mathrm{S})\ne 2$ and $\mathrm{\Phi },\mathrm{\Psi }:\mathrm{S}\to \mathrm{S}$ are derivations such that $\mathrm{\Phi }(\mathfrak{t}){\mathfrak{t}}^{\ast }-{\mathfrak{t}}^{\ast }\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{S}$, then $\mathrm{\Phi }=\mathrm{\Psi }=0$.

The following example shows that the condition of ∇-primeness in Theorem 4.3 is essential.

Example 4.3

Let $\mathbb{H}$ and $\mathbb{F}[x]$ denote the ring of real quaternions and the polynomial ring over the field of complex numbers $\mathbb{F}$ respectively. Consider the ring $\mathbb{S}=\mathbb{H}\times \mathbb{F}[x]$. Define the maps $\ast ,\mathrm{\Phi }=\mathrm{\Psi }:\mathbb{S}\to \mathbb{S}$ by $(\mathfrak{p},\varphi (x){)}^{\ast }=(\overline{\mathfrak{p}},\overline{\varphi (x)})$ and $\mathrm{\Phi }(\mathfrak{p},\varphi (x))=(0,{\varphi }^{\mathrm{\prime }}(x))$, where $\overline{\varphi (x)}$ is the polynomial obtained by replacing the coefficients of each term of $\varphi (x)$ by its complex conjugate and ${\varphi }^{\mathrm{\prime }}(x)$ denotes the usual derivative of $\varphi (x)$. Then it is easy to verify that Φ is a derivation and ∇ is an anti-automorphism. Take $\mathrm{K}=\mathbb{S}$. Then except ∇-primeness, all the hypotheses in Theorem 4.3 are satisfied here but $\mathbb{S}$ is noncommutative.

The following example demonstrates that the hypothesis “$\mathrm{K}$ is a ∇-closed ideal” in Theorems 4.1–4.3 is crucial.

Example 4.4

Consider the ring $\mathrm{S}=\underset{k-\mathrm{times}}{\underbrace{\mathbb{H}\times \mathbb{H}\times \cdots \times \mathbb{H}}}$, where $\mathbb{H}$ is the ring of real quaternions and $k\ge 2$. Then the map $\mathrm{\nabla }:\mathrm{S}\to \mathrm{S}$ given by $\mathrm{\nabla }({\mathfrak{u}}_1,{\mathfrak{u}}_2,{\mathfrak{u}}_3,\dots ,{\mathfrak{u}}_{k-1},{\mathfrak{u}}_k)=({\overline{\mathfrak{u}}}_k,{\overline{\mathfrak{u}}}_1,{\overline{\mathfrak{u}}}_2,{\overline{\mathfrak{u}}}_3,\dots ,{\overline{\mathfrak{u}}}_{k-1})$, where $\overline{\mathfrak{u}}$ denotes the conjugate of $\mathfrak{u}$, is an anti-automorphism of the second kind and $\mathrm{S}$ is ∇-prime. For a fixed noncentral element $\mathfrak{p}$ of $\mathbb{H}$, the map $\mathrm{\Phi }:\mathrm{S}\to \mathrm{S}$ given by $\mathrm{\Phi }({\mathfrak{u}}_1,{\mathfrak{u}}_2,{\mathfrak{u}}_3,\dots ,{\mathfrak{u}}_{k-1},{\mathfrak{u}}_k)=(0,[\mathfrak{p},{\mathfrak{u}}_2],[\mathfrak{p},{\mathfrak{u}}_3],\dots ,[\mathfrak{p},{\mathfrak{u}}_{k-1}],[\mathfrak{p},{\mathfrak{u}}_k])$ is a nonzero derivation satisfying $\mathrm{\Phi }(\mathfrak{r})\mathfrak{r}-\eta \mathfrak{r}\mathrm{\Phi }(\mathfrak{r})\in \mathcal{Z}(\mathrm{S})$ for all $\mathfrak{r}\in \mathbb{H}\times \{0\}\times \{0\}\times \cdots \times \{0\}$ and $\mathrm{\Phi }(\mathfrak{r})\mathrm{\nabla }(\mathfrak{r})-\eta \mathrm{\nabla }(\mathfrak{r})\mathrm{\Phi }(\mathfrak{r})\in \mathcal{Z}(\mathrm{S})$ for all $\mathfrak{r}\in \mathbb{H}\times \{0\}\times \{0\}\times \cdots \times \{0\}$, where $\eta \in \{-1,0,1\}$. Note that $\mathbb{H}\times \{0\}\times \{0\}\times \cdots \times \{0\}$ is a nonzero ideal of $\mathrm{S}$ which is not ∇-closed.

In Ref. [30], Ashraf et al. studied the commutative property of a prime ring $\mathrm{S}$ with derivation Φ satisfying any one of the conditions given as below: (i) $\mathrm{\Phi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S},$ (ii) $\mathrm{\Phi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S},$ (iii) $\mathrm{\Phi }(\mathfrak{t})\mathrm{\Phi }(\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{S}$, where $\eta \in \{1,-1\}$. Below we shall prove analogous results for ∇-prime rings.

Theorem 4.4

Consider a ∇-prime ring $\mathrm{S}$ with $\mathrm{K}$ as its nonzero ∇-closed ideal. Suppose that $\eta \in \{1,-1\}$ and $\mathrm{\Psi }:\mathrm{S}\to \mathrm{S}$ is a derivation such that any one among the following prevails.

1. $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K},$

2. $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K},$

3. $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K},$

4. $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K},$

   then $\mathrm{S}$ is commutative.

Proof.

(i) Suppose (44) $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(44) If $\mathrm{\Psi }=0$, then ${\mathrm{K}}^2\subseteq \mathcal{Z}(\mathrm{S})$. Note that ${\mathrm{K}}^2$ is a nonzero ∇-closed ideal of $\mathrm{S}$. Therefore by Lemma 3.6, $\mathrm{S}$ is commutative. Henceforth, we assume that $\mathrm{\Psi }\ne 0$. If $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}\ne \{0\}$, then putting α for $\mathfrak{s}$ in (44(44) $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(44) ), we find that $\alpha [\mathrm{\Psi }(\mathfrak{t}),\mathfrak{t}]=0$ for every $\mathfrak{t}\in \mathrm{K}$ and $\alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. Now $\mathrm{\nabla }(\mathcal{Z}(\mathrm{S})\cap \mathrm{K})\subseteq \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$, so by ∇-primeness of $\mathrm{S}$, the last expression yields $[\mathrm{\Psi }(\mathfrak{t}),\mathfrak{t}]=0$ for every $\mathfrak{t}\in \mathrm{K}.$ Therefore by Theorem 4.1, $\mathrm{S}$ is commutative. Next if $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}=\{0\}$, then from (44(44) $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(44) ), we have $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})=\eta \mathfrak{t}\mathfrak{s}$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Therefore $\eta \mathfrak{t}\mathfrak{s}\mathfrak{u}=\mathrm{\Psi }(\mathfrak{t}\mathfrak{s}\mathfrak{u})=\eta \mathfrak{t}\mathfrak{s}\mathfrak{u}+\mathfrak{t}\mathfrak{s}\mathrm{\Psi }(\mathfrak{u})$ for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$. Thus $\mathfrak{t}\mathfrak{s}\mathrm{\Psi }(\mathfrak{u})=0$ which further implies $\mathfrak{t}\mathfrak{s}{\mathfrak{u}}^2=0$ for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$. And, hence for every $\mathfrak{t}\in \mathrm{K}$, we have ${\mathfrak{t}}^4=0$, which is not possible by Herstein [[2]. Lemma 2.2.1].

(ii) Suppose (45) $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(45) If $\mathrm{\Psi }=0$, then ${\mathrm{K}}^2\subseteq \mathcal{Z}(\mathrm{S})$. Therefore by Lemma 3.6, $\mathrm{S}$ is commutative. Hence assume that $\mathrm{\Psi }\ne 0$. If $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}\ne \{0\}$, then replacing $\mathfrak{s}$ by α in (45(45) $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(45) ), we find that $\alpha \mathrm{S}[\mathrm{\Psi }(\mathfrak{t}),\mathfrak{t}]=\{0\}$ for every $\mathfrak{t}\in \mathrm{K}$ and $\alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. Therefore by ∇-primeness of $\mathrm{S}$, we obtain $[\mathrm{\Psi }(\mathfrak{t}),\mathfrak{t}]=0$ for every $\mathfrak{t}\in \mathrm{K}$ whence by Theorem 4.1, $\mathrm{S}$ is commutative. Next if $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}=\{0\}$, then from (45(45) $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(45) ), we have $\mathrm{\Psi }(\mathfrak{t})\mathfrak{s}+\mathfrak{t}\mathrm{\Psi }(\mathfrak{s})=\eta \mathfrak{s}\mathfrak{t}$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Now replacing $\mathfrak{s}$ by $\mathfrak{s}\mathfrak{u}$ in the previous relation, we get $\{\mathrm{\Psi }(\mathfrak{t})\mathfrak{s}+\mathfrak{t}\mathrm{\Psi }(\mathfrak{s})\}\mathfrak{u}+\mathfrak{t}\mathfrak{s}\mathrm{\Psi }(\mathfrak{u})=\eta \mathfrak{s}\mathfrak{u}\mathfrak{t}$ for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$. Therefore $\eta \mathfrak{s}\mathfrak{t}\mathfrak{u}+\mathfrak{t}\mathfrak{s}\mathrm{\Psi }(\mathfrak{u})=\eta \mathfrak{s}\mathfrak{u}\mathfrak{t}$ for every $\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}$. Replacing $\mathfrak{s}$ by $\mathfrak{t}$ and $\mathfrak{u}$ by ${\mathfrak{t}}^2$ in the last relation, we find that ${\mathfrak{t}}^4=0$ for every $\mathfrak{t}\in \mathrm{K}$, which is a contradiction by Herstein [[2], Lemma 2.2.1].

(iii) Suppose (46) $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(46) If $\mathrm{\Psi }=0$, then ${\mathrm{K}}^2\subseteq \mathcal{Z}(\mathrm{S})$. Therefore by Lemma 3.6, $\mathrm{S}$ is commutative. So assume that $\mathrm{\Psi }\ne 0$. First suppose that $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}\ne \{0\}$. Then substituting $\alpha \mathfrak{s}$ for $\mathfrak{s}$ in (46(46) $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(46) ), we get $\alpha \mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})+\{\mathrm{\Psi }(\alpha )\mathrm{\Psi }(\mathfrak{t})-\mathrm{\eta \alpha }\mathfrak{t}\}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$ and $\alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. Therefore $\alpha [\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s}),\mathfrak{s}]=0$ which, by using (46(46) $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(46) ), implies that $\alpha [\mathfrak{t},\mathfrak{s}]\mathfrak{s}=0$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$ and $\alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. In view of ∇-primeness of $\mathrm{S}$, it follows from the last relation that $[\mathfrak{t},\mathfrak{s}]\mathfrak{s}=0$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Replacing $\mathfrak{s}$ by $\alpha +\mathfrak{s}$, we have $\alpha [\mathfrak{t},\mathfrak{s}]=0$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$ and $\alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. Therefore $\mathrm{S}$ is commutative. Next suppose $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}=\{0\}$. Putting $\mathfrak{s}\mathfrak{t}$ in place of $\mathfrak{t}$ in (46(46) $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(46) ), we find that $\mathrm{\Psi }(\mathfrak{s})\mathrm{K}\mathrm{\Psi }(\mathfrak{s})=\{0\}$ for every $\mathfrak{t}\in \mathrm{K}$. Therefore by Lemma 3.7, $\mathrm{\Psi }(\mathfrak{s})=0$ for every $\mathfrak{s}\in \mathrm{K}$. Applying Lemma 4.2, it follows that $\mathrm{\Psi }=0$, which is not possible. Hence from the above discussion we infer that $\mathrm{S}$ is commutative.

(iv) Suppose (47) $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(47) If $\mathrm{\Psi }=0$, then $\mathrm{S}$ is commutative. Hence assume that $\mathrm{\Psi }\ne 0$. First suppose that $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}\ne \{0\}$. If $\mathrm{\Psi }(\alpha )=0$ for some $0\ne \alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$, then replacing $\mathfrak{s}$ by α in (47(47) $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(47) ), we get $\alpha \mathfrak{t}\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t}\in \mathrm{S}$. Therefore by ∇-primeness of $\mathrm{S}$, we deduce that $\mathrm{K}\subseteq \mathcal{Z}(\mathrm{S})$, whence by Lemma 3.6, it follows that $\mathrm{S}$ is commutative. Next suppose that $\mathrm{\Psi }(\alpha )\ne 0$ for some $\alpha \in \mathcal{Z}(\mathrm{S})\cap \mathrm{K}$. Substituting $\alpha \mathfrak{s}$ for $\mathfrak{s}$ in (47(47) $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(47) ), we get $\mathrm{\Psi }(\alpha )\mathrm{\Psi }(\mathfrak{t})\mathfrak{s}\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Now replacing $\mathfrak{s}$ by $\alpha \mathrm{\Psi }(\mathfrak{s})$ in the previous relation and using (47(47) $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(47) ), we arrive at $\alpha \mathrm{\Psi }(\alpha )\mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})$ for every $\mathfrak{t},\mathfrak{s}\in \mathrm{K}$. Therefore ${\mathrm{K}}^2\subseteq \mathcal{Z}(\mathrm{S})$. And, hence by Lemma 3.6, $\mathrm{S}$ is commutative. Now if $\mathcal{Z}(\mathrm{S})\cap \mathrm{K}=\{0\}$, then from (47(47) $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathrm{S})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$(47) ), we have (48) $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})\mathrm{\Psi }(\mathfrak{u})=\eta \mathfrak{u}\mathfrak{t}\mathfrak{s}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s},\mathfrak{u}\in \mathrm{K}.$(48) Therefore $\eta {\mathfrak{t}}^2\mathfrak{s}\mathfrak{u}=\mathrm{\Psi }(\mathfrak{t}\mathfrak{s}\mathfrak{u})\mathrm{\Psi }(\mathfrak{t})=\mathrm{\Psi }(\mathfrak{t})\mathfrak{s}\mathfrak{u}\mathrm{\Psi }(\mathfrak{t})+\mathfrak{t}\mathrm{\Psi }(\mathfrak{s}\mathfrak{u})\mathrm{\Psi }(\mathfrak{t})=\mathrm{\Psi }(\mathfrak{t})\mathfrak{s}\mathfrak{u}\mathrm{\Psi }(\mathfrak{t})+\eta {\mathfrak{t}}^2\mathfrak{s}\mathfrak{u}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$ Hence $\mathrm{\Psi }(\mathfrak{t}){\mathrm{K}}^2\mathrm{\Psi }(\mathfrak{t})=\{0\}$ for every $\mathfrak{t}\in \mathrm{K}$. Therefore by Lemma 3.7, $\mathrm{\Psi }(\mathfrak{t})=0$ for every $\mathfrak{t}\in \mathrm{K}$. Finally by Lemma 4.2, we see that $\mathrm{\Psi }=0$, a contradiction.

Remark 4.1

From the proof of the above theorem the reader may notice that there exists no derivation $\mathrm{\Psi }:\mathrm{S}\to \mathrm{S}$ such that either $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})=\eta \mathfrak{t}\mathfrak{s}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K},$ or $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})=\eta \mathfrak{s}\mathfrak{t}\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathrm{K}.$

We conclude this article with the example given below which demonstrates that the condition of ∇-primeness in Theorem 4.4 is essential.

Example 4.5

Consider the noncommutative ring $\mathcal{R}=\{\left[\begin{array}{cccc}0& 0& a& 0\\ c& 0& b& d\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\mid a,b,c,d\in \mathcal{A}\},$where $\mathcal{A}$ is any noncommutative ring and the map $\mathrm{\Psi }:\mathcal{R}\to \mathcal{R}$ given by $\mathrm{\Psi }\left(\left[\begin{array}{cccc}0& 0& a& 0\\ c& 0& b& d\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\right)=\left[\begin{array}{cccc}0& 0& a& 0\\ 0& 0& b& d\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right].$Then it can be easily verified that Ψ is a derivation and for $\eta \in \{-1,1\}$

1. $\mathrm{\Psi }(\mathfrak{t})\mathfrak{t}-\mathfrak{t}\mathrm{\Psi }(\mathfrak{t})\in \mathcal{Z}(\mathcal{R})\mathrm{for}\mathrm{every}\mathfrak{t}\in \mathcal{R},$

2. $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathcal{R})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathcal{R},$

3. $\mathrm{\Psi }(\mathfrak{t}\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathcal{R})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathcal{R},$

4. $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{t}\mathfrak{s}\in \mathcal{Z}(\mathcal{R})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathcal{R},$

5. $\mathrm{\Psi }(\mathfrak{t})\mathrm{\Psi }(\mathfrak{s})-\eta \mathfrak{s}\mathfrak{t}\in \mathcal{Z}(\mathcal{R})\mathrm{for}\mathrm{every}\mathfrak{t},\mathfrak{s}\in \mathcal{R}.$

Note that $\mathcal{R}$ is not a semiprime ring and hence not a ∇-prime for any (anti)-automorphism ∇.

Disclosure statement

No potential conflict of interest was reported by the author(s).