A Pair of Derivations on Prime Rings with Antiautomorphism

Abstract

This article examines the commutativity of rings with antiautomorphisms, specifically when they are equipped with derivations that satisfy certain algebraic identities. Moreover, we present examples to demonstrate the necessity of the various restrictions imposed in the hypotheses of our theorems.

1. Introduction

In this article, we will use the symbols R to denote a ring and Z to denote a center of $R.$ The anticommutator of elements a and b, given by $ab+ba,$ will be denoted as $a\circ b,$ while the commutator, $ab-ba,$ will be represented by $[a,b],$ where a and b belong to a ring $R.$ A ring R is deemed prime if, for every $a,b$ in $R,$ the equality $aRb=\left(0\right)$ implies that either a or b is equal to zero. A derivation on R is defined as an additive mapping $\mu :R\to R$ that satisfies the property $\mu \left(ab\right)=\mu \left(a\right)b+a\mu \left(b\right)$ for every a and b in $R.$ Let $a_0$ be a fixed element in $R,$ and let $\mu :R\to R$ be a mapping that satisfies the equality $\mu \left(a\right)=[a_0,a]$ for every a in $R.$ In this case, $\mu$ is said to be an inner derivation of R induced by $a_0;$ otherwise, it is an outer derivation of $R.$ An antiautomorphism of R is an additive bijective map $\Lambda :R\to R$ that adheres to the condition $\Lambda \left(ab\right)=\Lambda \left(b\right)\Lambda \left(a\right)$ for every a and b in $R.$ If ${\Lambda }^2\left(a\right)=a$ for every a in $R,$ then $\Lambda$ is known as an involution, denoted by $\ast .$ Within a ring with the involution $\ast ,$ an element a is classified as hermitian if $a^{*}=a$ or skew-hermitian if $a^{*}=-a.$ The involution is categorized as the second-kind if the intersection of Z with the set of all skew-hermitian elements is non-zero; otherwise, it is referred to as the first kind. For more comprehensive information about the above concepts, refer to the books [1] (Chapters I & VII) and [2,3].

Example 1. 

(1)

Any involution ∗ can be regarded as an antiautomorphism Λ satisfying ${\Lambda }^2=I_{id},$ implying that Λ is the involution itself. However, it should be noted that not all antiautomorphisms are involutions, as demonstrated in $\left(2\right).$

(2)

Consider $\mathbb{H},$ a real quaternion ring, and a mapping Λ from $\mathbb{H}$ to itself defined as $\Lambda \left(\chi \right)=(1+i)\overline{\chi }{(1+i)}^{(-1),}$ where $\overline{\chi }$ represents the conjugate of $\chi .$ In other words, if $\chi =u_1+u_{2}i+u_{3}j+u_{4}k,$ where $u_t\in \mathbb{R}$ for $t=1,2,3,4,$ then χ can be expressed as $\overline{\chi }=u_1-u_{2}i-u_{3}j-u_{4}k.$ However, it should be noted that Λ does not possess an order of one or two, indicating that it is not an involution.

A classical problem in ring theory revolves around investigating and establishing conditions that lead to the commutativity of a ring $R.$ Over time, researchers have discovered that derivations of rings and their modules are among the most effective tools for addressing this problem. Extensive literature suggest a close connection between the overall structure of a ring R and the behavior of additive maps defined on R (see [4]). Consequently, considerable interest has been devoted to exploring the relationship between the commutativity of a prime ring R and the behavior of certain special mappings on $R.$ Many authors have examined the commutativity of semiprime and prime rings, studying constrained maps like automorphisms, derivations, and centralizers that operate on appropriate subsets of rings (e.g., Lie ideals, ideals, one sided ideals, etc.). These discussions and investigations can be found in works such as [5,6]. Moreover, the study of ∗-prime rings has further extended several well established results concerning prime rings, as demonstrated in [7,8]. These works also provide additional references on the subject. For more results related to derivations, one may refer to sources such as [9,10,11,12].

In [13], Ashraf and Rehman established a significant result pertaining to prime rings. According to this result, if a prime ring R possesses a derivation, denoted as ‘$\mu$’, which fulfills either of the following conditions: $\mu \left(ab\right)-ab\in Z,$$\mu \left(ab\right)+ab\in Z,$$\mu \left(ab\right)-ba\in Z,$$\mu \left(ab\right)+ba\in Z,$$\mu \left(a\right)\mu \left(b\right)-ab\in Z$ for every $a,b\in J,$ where J represents a non-zero two-sided ideal of $R,$ then it necessarily follows that R must be a commutative ring. In [14] Ali et al. demonstrated that if a prime ring R of char(R)$\ne 2$ with involution of the second-kind allows the existence of a non-zero derivation $\mu$ satisfying $\mu \left(a\right)\mu \left(a^{*}\right)\pm a^{*}a=0$ or $\mu \left(a\right)\mu \left(a^{*}\right)\pm a{a}^{*}=0$ for all a in $R,$ then R must be a commutative ring. The aforementioned results can be viewed as specific cases derived from our more general result stated in Theorem 4 (i) and (ii), where the order of $\Lambda$ is restricted to two and ${\Xi }_1$ is equal to ${\Xi }_2$. Several related generalizations of these results exist in the literature (e.g., [15]).

In the case of a nonempty subset B of a ring $R,$ a map $\mu :B\to R$ is referred to as a centralizing (resp. commuting) map on B if $\left[\mu \right(t),t]\in Z$ (resp. $\left[\mu \right(t),t]=0$) holds for every $t\in B.$ The investigation of centralizing and commuting maps traces its roots back to 1955, when Divinsky [16] demonstrated that a simple Artinian ring becomes commutative if it possesses a commuting automorphism that is distinct from the identity mapping. Shortly thereafter, in 1957, Posner [17] established that a prime ring must also be commutative if it accommodates a nonzero centralizing derivation. In 1970, Luh [18] extended Divinsky’s findings to prime rings. Subsequently, Mayne [19] derived a corresponding result to Posner’s for centralizing automorphisms that are not the identity. The culmination of these investigations is reflected in the comprehensive works of [20,21,22,23]. This research article aims to examine the aforementioned results within the context of prime rings that possess antiautomorphisms.

Consider a ring R with involution ∗ and a nonempty subset B of $R.$ A map $\lambda$ from R to itself is termed ∗-centralizing on B if the expression $\lambda \left(t\right)t^{*}+n{t}^{*}\lambda \left(t\right)$ belongs to the center of R for every $t\in B$ and $n=-1.$ In particular, when $\lambda \left(t\right)t^{*}+n{t}^{*}\lambda \left(t\right)$ equals zero for every $t\in B$ and $n=-1,$ the map $\lambda$ is referred to as ∗-commuting on $B.$ In case $n=1,$ it said to be skew ∗-centralizing and skew ∗-commuting on $B,$ respectively. In 2022, Rehman and Alnoghashi [24] generalized the previous concepts to (skew) $\Lambda$-centralizing and (skew) $\Lambda$-commuting, where $\Lambda$ is an antiautomorphism and $\lambda$ is a generalized derivation on $R,$ that is, $\lambda \left(t\right)\Lambda \left(t\right)\pm \Lambda \left(t\right)\lambda \left(t\right)$ belongs to the center of R or equals zero, for every $t\in R,$ respectively. In [25,26], Ali and Dar, embarked on the investigation of these maps and demonstrated that the presence of a nonzero ∗-centralizing derivation in a prime ring of char(R)$\ne 2$ with a second-kind involution leads to the ring being commutative. They also demonstrated the ∗-version of Posner’s second theorem and its associated issues in addition to characterising these maps in semiprime and prime rings with involution. For more details about Posner’s second theorem see [17]. In 2017, Nejjar et al. [27] (Theorem 3.7) also achieved congruent outcomes, further corroborating the aforementioned results.

If $\left[\lambda \right(a),\lambda (b\left)\right]=0$ whenever $[a,b]=0$ for every a and b in $R,$ then a mapping $\lambda :R\to R$ preserves commutativity. The preservation of commutativity has been an active area of research in matrix theory, operator theory, and ring theory (refer to [28] for further details). Let B be a subset of $R,$ a mapping $\lambda$ is referred to as strong commutativity-preserving (SCP) if $\left[\lambda \right(a),\lambda (b\left)\right]=[a,b]$ holds for every a and b in $B.$ Bell and Daif examined the possibility of rings admitting a derivation that is SCP on a nonzero right ideal in [29]. They demonstrated that if a semiprime ring R possesses a derivation $\mu$ satisfying $\left[\mu \right(a),\mu (b\left)\right]=[a,b]$ for any a and b in a right ideal L of $R,$ then $L\subseteq Z.$ Furthermore, if $L=R,$ then R is commutative. Deng and Ashraf later presented a result in [30], stating that a semiprime ring R has a nonzero central ideal if it has a derivation $\mu$ and a mapping $\lambda :L\to R$ defined on a nonzero ideal $L,$ such that $\left[\lambda \right(a),\mu (b\left)\right]=[a,b]$ for any a and b in $L.$ In particular, they established that if $L=R,$ then R is commutative. If a mapping $\lambda :R\to R$ satisfies $\left[\lambda \right(a),\lambda (b\left)\right]+[a,b]=0$ for every a and b in a subset S of $R,$ it is termed skew strong commutativity-preserving (skew SCP). Ali and Huang proposed that a nonzero central ideal is contained in R when R is a 2-torsion-free semiprime ring and $\mu$ is a derivation of R that satisfies the skew SCP on a nonzero ideal L of R [31]. There exist numerous related generalizations of these results within the literature (see for example, [32]).

Inspired by the concept of ∗-SCP derivation, in 2017, Nejjar et al. [27] embarked on investigating a broader and more comprehensive notion by exploring the identity $[\mu \left(t\right),\mu \left(t^{*}\right)]-[t,t^{*}]\in Z$ for every element $t\in R.$ Their objective was to examine the implications of this identity within the context of a prime ring R of char(R)$\ne 2$ with an involution ∗ of the second-kind. In their work [27] (Theorems 3.1, 3.5 and 3.8) they successfully established that if R admits a non-zero derivation, denoted as $\mu ,$ that satisfies the condition $\mu \left(t\right)\mu \left(t^{*}\right)+k\mu \left(t^{*}\right)\mu \left(t\right)+m(t{t}^{*}+n{t}^{*}t)\in Z$ for every element $t\in R$ and $k,m,n\in \{-1,1\},$ then the ring must necessarily be commutative. The previous results are a special case of our result when the order of $\Lambda$ is equal to two and ${\Xi }_1={\Xi }_2$ in Theorem 2.

In their notable work, Mamouni et al. [33] (2021) made a significant contribution to the study of prime rings. They established a compelling result that holds true for prime rings denoted as R. According to their results, if such a prime ring of char(R)$\ne 2$ with an involution ∗ of the second-kind possesses two derivations, denoted as $\mu ,\omega$, that satisfies any of the following conditions: $\mu \left(t\right)t^{*}-t^{*}\omega \left(t\right)\in Z$ ([33] (Theorem 1)), $\mu \left(t^{*}\right)t-t^{*}\omega \left(t\right)\in Z$ ([33] (Theorem 2)), $\mu \left(t\right)\omega \left(t^{*}\right)+k\omega \left(t^{*}\right)\mu \left(t\right)+m(t\circ t^{*})\in Z$ ([33] (Theorem 3)) for every $t\in R$, where $k,m\in \{-1,1\},$ then it can be concluded that the ring R must be a commutative ring. The previous results are a special case of our result when the order of $\Lambda$ is equal to two in Theorems 1 and 2.

Based on the motivations above and observations, the goals and objectives of this article are to establish the following main results:

Theorem 1. 

Let R be a prime ring of char(R)$\ne 2,$ equipped with an antiautomorphism Λ which is Z-nonlinear and let ${\Xi }_1,{\Xi }_2$ be nonzero derivations on $R.$ Then for every $a\in R$ and $k\in \{1,-1\},$ the following conditions are equivalent.

(i) 

${\Xi }_1\left(a\right)\Lambda \left(a\right)+k\Lambda \left(a\right){\Xi }_2\left(a\right)\in Z,$

(ii) 

${\Xi }_1(\Lambda \left(a\right))a+k\Lambda \left(a\right){\Xi }_2\left(a\right)\in Z,$

(iii) 

R is commutative.

Theorem 2. 

Let R be a prime ring of char(R) $\ne 2$, equipped with an antiautomorphism Λ which is Z-nonlinear and let ${\Xi }_1,{\Xi }_2$ be derivations on $R.$ Then for every $a\in R$ and $k,m,n\in \{1,-1\},$ the following conditions are equivalent.

(i) 

${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)+m(a\Lambda \left(a\right)+n\Lambda \left(a\right)a)\in Z,$

(ii) 

R is commutative.

Theorem 3. 

Let R be a prime ring of char(R) $\ne 2$, equipped with an antiautomorphism Λ which is Z-nonlinear and let ${\Xi }_1,{\Xi }_2$ be derivations on $R.$ Then for every $a\in R$ and $k,n\in \{1,-1\},$ the following conditions are equivalent.

(i) 

${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)+na\Lambda \left(a\right)\in Z,$

(ii) 

${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)+n\Lambda \left(a\right)a\in Z,$

(iii) 

R is commutative.

Theorem 4. 

Let R be a prime ring of char(R) $\ne 2$, equipped with an antiautomorphism Λ which is Z-nonlinear and let ${\Xi }_1,{\Xi }_2$ be derivations on $R.$ Then for every $a\in R$ and $k,m,n\in \{1,-1\},$ the following conditions are equivalent.

(i) 

${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+na\Lambda \left(a\right)\in Z,$

(ii) 

${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+n\Lambda \left(a\right)a\in Z,$

(iii) 

${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+m(a\Lambda \left(a\right)+n\Lambda \left(a\right)a)\in Z,$

(iv) 

R is commutative.

2. Preliminaries

This section will present some facts that will assist us in our proofs. Due to their frequent utilization, Facts 1–3 will be implicitly employed in our subsequent proofs without explicit reference.

Fact 1. 

“Ref. [34] (Lemma 4) Let b and $ab$ be in the center of a prime ring $R.$ Then a is in Z or $b=0$".

Fact 2. 

“Ref. [35] (Lemma 42.1) Let R be a prime ring. If $\mu :R\to R$ is a derivation, then for any $z\in Z,$$\mu \left(z\right)\in Z$".

Fact 3. 

Let R be a ring. If $\Lambda :R\to R$ is an antiautomorphism, then for any $z\in Z,$$\Lambda \left(z\right)\in Z$.

Fact 4. 

“Ref. [35] (Lemma 2.5) Let R be a prime ring and I is a non-zero right ideal of $R.$ If $[t,s]\in Z$ ($t\circ s\in Z$) for every $t,s\in I,$ then R is commutative".

Fact 5. 

“Ref. [17] (Theorem 2) Let R be a prime ring. If R admits a derivation μ such that $\left[\mu \right(t),t]\in Z$ for every $t\in R,$ then $\mu =0$ or R is commutative".

Fact 6. 

“Ref. [24] (Lemmas 1.3 and 1.4) Let R be a prime ring with antiautomorphism Λ which is Z-nonlinear. Then R is commutative if any one of the following is satisfied:

(i) 

$[\Lambda (a),a]\in Z$ for every $a\in R,$

(ii) 

$\Lambda \left(a\right)\circ a\in Z$ for every $a\in R,$

(iii) 

$\Lambda \left(a\right)a\in Z$ for every $a\in R,$

(iv) 

$a\Lambda \left(a\right)\in Z$ for every $a\in R$".

In this context, we will introduce the concept of a generalized polynomial identity as defined in [36]. Consider $R,$ a prime ring, $R{C}_C〈X〉,$ a free product of $RC$ over $C,$ and $C〈X〉,$ a free algebra generated by a set of indeterminates $X.$ An additive subgroup R of $RC$ is deemed a generalized polynomial identity over C (referred to as R being GPI over C) if there exists a nonzero element $\Omega (t_1,t_2,\dots ,t_n)$ in $R{C}_C〈X〉$ such that $\Omega (s_1,s_2,\dots ,s_n)=0$ holds for every element $s_j\in R.$

Fact 7. 

“Ref. [37] (Kharchenko’s theorem) If $\Omega (a_i,\mu \left(a_i\right))$ is a generalized polynomial identity for $R,$ where R is a prime ring and μ an outer derivation of $R,$ then R also satisfies the generalized polynomial identity $\Omega (a_i,b_i),$ where $a_i$ and $b_i$ are distinct indeterminates".

3. The Main Results

To prove our results, we require a set of auxiliary lemmas. Let us commence with the following:

Lemma 1. 

If ${\Xi }_1\left(a\right)\Lambda \left(a\right)+n\Lambda \left(a\right){\Xi }_2\left(a\right)\in Z$ for every $a\in R$ and $n\in \{-1,1\},$ then R is commutative.

Proof. 

Assume that

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right)\Lambda \left(a\right)+n\Lambda \left(a\right){\Xi }_2\left(a\right)\in Z\end{array}$$

(1)

for every $a\in R$ and $n\in \{-1,1\}.$ By linearizing (1), we obtain

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right)\Lambda \left(b\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right)+n\Lambda \left(a\right){\Xi }_2\left(b\right)+n\Lambda \left(b\right){\Xi }_2\left(a\right)\in Z\end{array}$$

(2)

for every $a,b\in R$ and $n\in \{-1,1\}.$ Let $0\ne z\in Z.$ Replacing b by $bz$ in (2), we have

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right)\Lambda \left(b\right)\Lambda \left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right)z+b\Lambda \left(a\right){\Xi }_1\left(z\right)+n\Lambda \left(a\right){\Xi }_2\left(b\right)z\hfill \\ \hfill +& n\Lambda \left(a\right)b{\Xi }_2\left(z\right)+n\Lambda \left(b\right){\Xi }_2\left(a\right)\Lambda \left(z\right)\in Z.\hfill \end{array}$$

(3)

Again, replacing a by $az$ in (3), we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right)\Lambda \left(b\right)\Lambda \left(z\right)z+a\Lambda \left(b\right)\Lambda \left(z\right){\Xi }_1\left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right)z\Lambda \left(z\right)+b\Lambda \left(a\right){\Xi }_1\left(z\right)\Lambda \left(z\right)\hfill \\ \hfill +& n\Lambda \left(a\right){\Xi }_2\left(b\right)z\Lambda \left(z\right)+n\Lambda \left(a\right)b{\Xi }_2\left(z\right)\Lambda \left(z\right)+n\Lambda \left(b\right){\Xi }_2\left(a\right)\Lambda \left(z\right)z\hfill \\ \hfill +& n\Lambda \left(b\right)a\Lambda \left(z\right){\Xi }_2\left(z\right)\in Z.\hfill \end{array}$$

That is,

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right)\Lambda \left(b\right)z+a\Lambda \left(b\right){\Xi }_1\left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right)z+b\Lambda \left(a\right){\Xi }_1\left(z\right)+n\Lambda \left(a\right){\Xi }_2\left(b\right)z\hfill \\ \hfill +& n\Lambda \left(a\right)b{\Xi }_2\left(z\right)+n\Lambda \left(b\right){\Xi }_2\left(a\right)z+n\Lambda \left(b\right)a{\Xi }_2\left(z\right)\in Z.\hfill \end{array}$$

Multiplying (2) by z and then using it in the previous expression, we see that

$$\begin{array}{c}\hfill a\Lambda \left(b\right){\Xi }_1\left(z\right)+b\Lambda \left(a\right){\Xi }_1\left(z\right)+n\Lambda \left(a\right)b{\Xi }_2\left(z\right)+n\Lambda \left(b\right)a{\Xi }_2\left(z\right)\in Z.\end{array}$$

Taking a by $az$ in the last relation and using it, we find that

$$(b\Lambda \left(a\right){\Xi }_1\left(z\right)+n\Lambda \left(a\right)b{\Xi }_2\left(z\right))(\Lambda \left(z\right)-z)\in Z.$$

Since there is $0\ne z\in Z$ such that $\Lambda \left(z\right)\ne z,$ we arrive at $b\Lambda \left(a\right){\Xi }_1\left(z\right)+n\Lambda \left(a\right)b{\Xi }_2\left(z\right)\in Z.$ Putting a by ${\Lambda }^{-1}\left(a\right)$ in the last expression, we conclude that

$$\begin{array}{c}\hfill ba{\Xi }_1\left(z\right)+nab{\Xi }_2\left(z\right)\in Z.\end{array}$$

(4)

Replacing b by z in (4), we infer that $a({\Xi }_1\left(z\right)+n{\Xi }_2\left(z\right))\in Z.$ That is, $a\in Z$ for every $a\in R$ or ${\Xi }_1\left(z\right)+n{\Xi }_2\left(z\right)=0.$ In case $a\in Z$ for every $a\in R,$ we have R is commutative. If ${\Xi }_1\left(z\right)+n{\Xi }_2\left(z\right)=0,$ then

$$\begin{array}{c}\hfill {\Xi }_1\left(z\right)=-n{\Xi }_2\left(z\right).\end{array}$$

(5)

Using (5) in (4), we obtain $[b,a]{\Xi }_1\left(z\right)\in Z.$ Hence, $[b,a]\in Z$ or ${\Xi }_1\left(z\right)=0.$ Suppose that $[b,a]\in Z$ for every $a,b\in R,$ and by Fact 4, we find that R is commutative. If ${\Xi }_1\left(z\right)=0,$ then ${\Xi }_2\left(z\right)=0,$ by (5). Using fact that ${\Xi }_1\left(z\right)=0={\Xi }_2\left(z\right)$ in (3), see that

$${\Xi }_1\left(a\right)\Lambda \left(b\right)\Lambda \left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right)z+n\Lambda \left(a\right){\Xi }_2\left(b\right)z+n\Lambda \left(b\right){\Xi }_2\left(a\right)\Lambda \left(z\right)\in Z.$$

Multiplying (2) by z and then using it in the previous expression, we find that

$$({\Xi }_1\left(a\right)\Lambda \left(b\right)+n\Lambda \left(b\right){\Xi }_2\left(a\right))(\Lambda \left(z\right)-z)\in Z.$$

Hence, ${\Xi }_1\left(a\right)\Lambda \left(b\right)+n\Lambda \left(b\right){\Xi }_2\left(a\right)\in Z.$ That is,

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right)b+nb{\Xi }_2\left(a\right)\in Z.\end{array}$$

(6)

Putting $b=z$ in (6), we infer that ${\Xi }_1\left(a\right)+n{\Xi }_2\left(a\right)\in Z,$ and so $D\left(a\right)\in Z$ is a derivation of $R,$ where $D={\Xi }_1+n{\Xi }_2,$ and hence $\left[D\right(a),a]=0$ and by Fact 5, we conclude that R is commutative or $D=0.$ In case $D=0,$ we have

$$\begin{array}{c}\hfill {\Xi }_1=-n{\Xi }_2.\end{array}$$

(7)

By using (7) in (6), we obtain $[{\Xi }_1\left(a\right),b]\in Z.$ In particular, $[{\Xi }_1\left(a\right),a]\in Z$ and by Fact 5, we have ${\Xi }_1=0$ or R is commutative. In case ${\Xi }_1=0,$ and from (7), we see that ${\Xi }_2=0.$ □

Lemma 2. 

If ${\Xi }_1(\Lambda \left(a\right))a+n\Lambda \left(a\right){\Xi }_2\left(a\right)\in Z$ for every $a\in R$ and $n\in \{-1,1\},$ then R is commutative.

Proof. 

Assume that

$$\begin{array}{c}\hfill {\Xi }_1(\Lambda \left(a\right))a+n\Lambda \left(a\right){\Xi }_2\left(a\right)\in Z\end{array}$$

(8)

for every $a\in R$ and $n\in \{-1,1\}.$ By linearizing (8), we have

$$\begin{array}{c}\hfill {\Xi }_1(\Lambda \left(a\right))b+{\Xi }_1(\Lambda \left(b\right))a+n\Lambda \left(a\right){\Xi }_2\left(b\right)+n\Lambda \left(b\right){\Xi }_2\left(a\right)\in Z\end{array}$$

(9)

for every $a,b\in R$ and $n\in \{-1,1\}.$ Let $0\ne z\in Z.$ Replacing b by $bz$ in (9), we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1(\Lambda \left(a\right))bz+{\Xi }_1(\Lambda \left(b\right))a\Lambda \left(z\right)+\Lambda \left(b\right)a{\Xi }_1(\Lambda \left(z\right))+n\Lambda \left(a\right){\Xi }_2\left(b\right)z\hfill \\ \hfill +& n\Lambda \left(a\right)b{\Xi }_2\left(z\right)+n\Lambda \left(b\right){\Xi }_2\left(a\right)\Lambda \left(z\right)\in Z.\hfill \end{array}$$

(10)

Again, replacing a by $az$ in (10), we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1(\Lambda \left(a\right))bz\Lambda \left(z\right)+\Lambda \left(a\right)bz{\Xi }_1(\Lambda \left(z\right))+{\Xi }_1(\Lambda \left(b\right))a\Lambda \left(z\right)z+\Lambda \left(b\right)a{\Xi }_1(\Lambda \left(z\right))z\hfill \\ \hfill +& n\Lambda \left(a\right){\Xi }_2\left(b\right)z\Lambda \left(z\right)+n\Lambda \left(a\right)b{\Xi }_2\left(z\right)\Lambda \left(z\right)+n\Lambda \left(b\right){\Xi }_2\left(a\right)\Lambda \left(z\right)z\hfill \\ \hfill +& n\Lambda \left(b\right)a\Lambda \left(z\right){\Xi }_2\left(z\right)\in Z.\hfill \end{array}$$

Multiplying (9) by $\Lambda \left(z\right)z$ and then using it in the previous expression, we see that

$$\begin{array}{c}\hfill \Lambda \left(a\right)bz{\Xi }_1(\Lambda \left(z\right))+\Lambda \left(b\right)a{\Xi }_1(\Lambda \left(z\right))z+n\Lambda \left(a\right)b{\Xi }_2\left(z\right)\Lambda \left(z\right)+n\Lambda \left(b\right)a\Lambda \left(z\right){\Xi }_2\left(z\right)\in Z.\end{array}$$

(11)

Taking a by $az$ in (11) and then multiplying it by z and then subtracting them, we obtain

$$\Lambda \left(a\right)b(z{\Xi }_1(\Lambda \left(z\right))+n{\Xi }_2\left(z\right)\Lambda \left(z\right))(\Lambda \left(z\right)-z)\in Z.$$

Hence, $ab(z{\Xi }_1(\Lambda \left(z\right))+n{\Xi }_2\left(z\right)\Lambda \left(z\right))\in Z.$ Putting $b=z$ in the last relation, we conclude that

$$\begin{array}{c}\hfill a(z{\Xi }_1(\Lambda \left(z\right))+n{\Xi }_2\left(z\right)\Lambda \left(z\right))\in Z.\end{array}$$

(12)

That is, $a\in Z$ for every $a\in R$ or $z{\Xi }_1(\Lambda \left(z\right))+n{\Xi }_2\left(z\right)\Lambda \left(z\right)=0.$ If $a\in Z$ for every $a\in R,$ then R is commutative. If $z{\Xi }_1(\Lambda \left(z\right))+n{\Xi }_2\left(z\right)\Lambda \left(z\right)=0,$ then

$$\begin{array}{c}\hfill z{\Xi }_1(\Lambda \left(z\right))=-n{\Xi }_2\left(z\right)\Lambda \left(z\right).\end{array}$$

(13)

Multiplying (9) by z and then using it in (10), we have

$$\begin{array}{c}\hfill ({\Xi }_1(\Lambda \left(b\right))a+n\Lambda \left(b\right){\Xi }_2\left(a\right))(\Lambda \left(z\right)-z)+\Lambda \left(b\right)a{\Xi }_1(\Lambda \left(z\right))+n\Lambda \left(a\right)b{\Xi }_2\left(z\right)\in Z.\end{array}$$

(14)

Replacing a by $az$ in the last relation and then multiplying it by z and then subtracting them, we obtain

$$(n\Lambda \left(b\right)a{\Xi }_2\left(z\right)+n\Lambda \left(a\right)b{\Xi }_2\left(z\right))(\Lambda \left(z\right)-z)\in Z.$$

That is, $(\Lambda \left(b\right)a+\Lambda \left(a\right)b){\Xi }_2\left(z\right)\in Z.$ Hence, $\Lambda \left(b\right)a+\Lambda \left(a\right)b\in Z$ or ${\Xi }_2\left(z\right)=0.$

If $\Lambda \left(b\right)a+\Lambda \left(a\right)b\in Z,$ then $\Lambda \left(a\right)a\in Z,$ and by Fact 6, we conclude that R is commutative. Now, in case ${\Xi }_2\left(z\right)=0,$ using the last relation in (13), we see that

$$\begin{array}{c}\hfill {\Xi }_1(\Lambda \left(z\right))=0={\Xi }_2\left(z\right).\end{array}$$

(15)

By using (15) in (14), we find that ${\Xi }_1(\Lambda \left(b\right))a+n\Lambda \left(b\right){\Xi }_2\left(a\right)\in Z.$ That is,

$$\begin{array}{c}\hfill {\Xi }_1\left(b\right)a+nb{\Xi }_2\left(a\right)\in Z.\end{array}$$

(16)

Putting $b=z$ in (16) and using (15), we conclude that ${\Xi }_2\left(a\right)\in Z.$ That is, $[{\Xi }_2\left(a\right),a]=0$ and by Fact 5, we infer that ${\Xi }_2=0$ or R is commutative. In case ${\Xi }_2=0,$ using the last relation in (16), we have ${\Xi }_1\left(b\right)\in Z,$ and so ${\Xi }_1=0$ or R is commutative. □

Based on Lemmas 1 and 2, the proof of Theorem 1 follows.

Lemma 3. 

If ${\Xi }_1\left(a\right){\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right){\Xi }_1\left(a\right)+m(ab+nba)\in Z$ for every $a,b\in R$ and $k,m,n\in \{-1,1\},$ then R is commutative.

Proof. 

Assume that

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right){\Xi }_1\left(a\right)+m(ab+nba)\in Z\end{array}$$

(17)

for every $a\in R$ and $k,m,n\in \{-1,1\}.$ Taking a by $ar$ in (17), where $r\in R,$ we obtain

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right)r{\Xi }_2\left(b\right)+a{\Xi }_1\left(r\right){\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right){\Xi }_1\left(a\right)r+k{\Xi }_2\left(b\right)a{\Xi }_1\left(r\right)+m(arb+nbar)\in Z,\end{array}$$

and so

$$\begin{array}{cc}\hfill & ({\Xi }_1\left(a\right){\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right){\Xi }_1\left(a\right)+m(ab+nba))r+{\Xi }_1\left(a\right)[r,{\Xi }_2\left(b\right)]\hfill \\ \hfill +& a{\Xi }_1\left(r\right){\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right)a{\Xi }_1\left(r\right)+ma[r,b]\in Z.\hfill \end{array}$$

Hence,

$$\begin{array}{c}\hfill [{\Xi }_1\left(a\right)[r,{\Xi }_2\left(b\right)]+a{\Xi }_1\left(r\right){\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right)a{\Xi }_1\left(r\right)+ma[r,b],r]=0.\end{array}$$

(18)

If ${\Xi }_1$ or ${\Xi }_2$ is an outer derivation, then from (17) and by Fact 7, we obtain $c{\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right)c+m(ab+nba)\in Z$ (in case ${\Xi }_1$ is an outer derivation) for every $a,b,c\in R.$ Putting $c=0$ in the last relation, we obtain $ab+nba\in Z,$ and by Fact 4, R is commutative. Now, if ${\Xi }_1$ and ${\Xi }_2$ are inner derivations, then from (18), we obtain

$$\begin{array}{c}\hfill [[a,i_1][r,[b,i_2]]+a[r,i_1][b,i_2]+k[b,i_2]a[r,i_1]+ma[r,b],r]=0\end{array}$$

for every $a,b,r\in R$ and some $i_1,i_2\in R.$ Putting $b=i_2$ in the previous expression, we see that $[a[r,i_2],r]=0.$ Again, putting a by $sa$ in the last expression and using it, where $s\in R,$ we find that $[s,r]a[r,i_2]=0,$ and so $[s,r]=0$ or $[r,i_2]=0.$ If $[s,r]=0$ for every $r,s\in R,$ then R is commutative. If $[r,i_2]=0$ for every $r\in R,$ then $i_2\in Z,$ and so ${\Xi }_2=0.$ Using the previous expression in (17), we infer that $ab+nba\in Z,$ and by Fact 4, we find that R is commutative. □

Lemma 4. 

If ${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)+m(a\Lambda \left(a\right)+n\Lambda \left(a\right)a)\in Z$ for every $a\in R$ and $k,m,n\in \{-1,1\},$ then R is commutative.

Proof. 

Assume that

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)+m(a\Lambda \left(a\right)+n\Lambda \left(a\right)a)\in Z\end{array}$$

(19)

for every $a\in R$ and $k,m,n\in \{-1,1\}.$ By linearizing (19), we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(b\right)+k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)\hfill \\ \hfill +& m(a\Lambda (b)+b\Lambda (a)+n\Lambda (a)b+n\Lambda (b\left)a\right)\in Z\hfill \end{array}$$

(20)

for every $a,b\in R.$ Let $0\ne z\in Z.$ Replacing b by $bz$ in (20), we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))\Lambda \left(z\right)+{\Xi }_1\left(a\right)\Lambda \left(b\right){\Xi }_2(\Lambda \left(z\right))+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))z\hfill \\ \hfill +& b{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(b\right)z+k{\Xi }_2(\Lambda \left(a\right))b{\Xi }_1\left(z\right)\hfill \\ \hfill +& k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)\Lambda \left(z\right)+k\Lambda \left(b\right){\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(z\right))\hfill \\ \hfill +& m(a\Lambda (b)\Lambda (z)+b\Lambda (a)z+n\Lambda (a)bz\hfill \\ \hfill +& n\Lambda \left(b\right)a\Lambda \left(z\right))\in Z.\hfill \end{array}$$

(21)

Putting $b=a$ in (21) and using (19), we find that

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))+a{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)+k{\Xi }_2(\Lambda \left(a\right))a{\Xi }_1\left(z\right)+k\Lambda \left(a\right){\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(z\right))\in Z.\end{array}$$

(22)

Linearizing (22), we arrive at

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right)\Lambda \left(b\right){\Xi }_2(\Lambda \left(z\right))+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))+a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)+b{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)\hfill \\ \hfill +& k{\Xi }_2(\Lambda \left(a\right))b{\Xi }_1\left(z\right)+k{\Xi }_2(\Lambda \left(b\right))a{\Xi }_1\left(z\right)+k\Lambda \left(a\right){\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(z\right))\hfill \\ \hfill +& k\Lambda \left(b\right){\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(z\right))\in Z\hfill \end{array}$$

(23)

for every $a,b\in R.$ Taking a by $az$ in (23), we conclude that

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right)\Lambda \left(b\right){\Xi }_2(\Lambda \left(z\right))z+a\Lambda \left(b\right){\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))\Lambda \left(z\right)\hfill \\ \hfill +& a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)z+b{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)\Lambda \left(z\right)+b\Lambda \left(a\right){\Xi }_1\left(z\right){\Xi }_2(\Lambda \left(z\right))\hfill \\ \hfill +& k{\Xi }_2(\Lambda \left(a\right))b{\Xi }_1\left(z\right)\Lambda \left(z\right)+k\Lambda \left(a\right)b{\Xi }_1\left(z\right){\Xi }_2(\Lambda \left(z\right))+k{\Xi }_2(\Lambda \left(b\right))a{\Xi }_1\left(z\right)z\hfill \\ \hfill +& k\Lambda \left(a\right){\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(z\right))\Lambda \left(z\right)+k\Lambda \left(b\right){\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(z\right))z\hfill \\ \hfill +& k\Lambda \left(b\right)a{\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)\in Z.\hfill \end{array}$$

(24)

Multiplying (23) by z and the using it in (24), we have

$$\begin{array}{cc}\hfill & a\Lambda \left(b\right){\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)+b\Lambda \left(a\right){\Xi }_1\left(z\right){\Xi }_2(\Lambda \left(z\right))+k\Lambda \left(a\right)b{\Xi }_1\left(z\right){\Xi }_2(\Lambda \left(z\right))\hfill \\ \hfill +& k\Lambda \left(b\right)a{\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)+({\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))+b{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)\hfill \\ \hfill +& k{\Xi }_2(\Lambda \left(a\right))b{\Xi }_1\left(z\right)+k\Lambda \left(a\right){\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(z\right)))(\Lambda \left(z\right)-z)\in Z.\hfill \end{array}$$

(25)

Replacing b by $bz$ in (25), we find that

$$\begin{array}{cc}\hfill & a\Lambda \left(b\right){\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)\Lambda \left(z\right)+b\Lambda \left(a\right){\Xi }_1\left(z\right){\Xi }_2(\Lambda \left(z\right))z+k\Lambda \left(a\right)b{\Xi }_1\left(z\right){\Xi }_2(\Lambda \left(z\right))z\hfill \\ \hfill +& k\Lambda \left(b\right)a{\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)\Lambda \left(z\right)+({\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))z+b\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)\hfill \\ \hfill +& b{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)z+k{\Xi }_2(\Lambda \left(a\right))b{\Xi }_1\left(z\right)z+k\Lambda \left(a\right){\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(z\right))z\hfill \\ \hfill +& k\Lambda \left(a\right)b{\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right))(\Lambda \left(z\right)-z)\in Z.\hfill \end{array}$$

Multiplying (25) by z and the using it in the last relation, we obtain

$$\begin{array}{cc}\hfill & (a\Lambda \left(b\right){\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)+k\Lambda \left(b\right)a{\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)+b\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)\hfill \\ \hfill +& k\Lambda \left(a\right)b{\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right))(\Lambda \left(z\right)-z)\in Z.\hfill \end{array}$$

That is,

$$\begin{array}{c}\hfill (a\Lambda \left(b\right)+k\Lambda \left(b\right)a+b\Lambda \left(a\right)+k\Lambda \left(a\right)b){\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)\in Z.\end{array}$$

Hence, $a\Lambda \left(b\right)+k\Lambda \left(b\right)a+b\Lambda \left(a\right)+k\Lambda \left(a\right)b\in Z$ or ${\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)=0.$

In case $a\Lambda \left(b\right)+k\Lambda \left(b\right)a+b\Lambda \left(a\right)+k\Lambda \left(a\right)b\in Z.$ Putting $b=a$ in the previous expression, we see that $a\Lambda \left(a\right)+k\Lambda \left(a\right)a\in Z,$ and by Fact 6, R is commutative. Suppose that ${\Xi }_2(\Lambda \left(z\right)){\Xi }_1\left(z\right)=0.$

Case (I):

$$\begin{array}{c}\hfill {\Xi }_2(\Lambda \left(z\right))=0={\Xi }_1\left(z\right).\end{array}$$

(26)

Multiplying (20) by z and subtracting it from (21) and using (26), we infer that

$$\begin{array}{c}\hfill ({\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))+k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)+m(a\Lambda \left(b\right)+n\Lambda \left(b\right)a))(\Lambda \left(z\right)-z)\in Z.\end{array}$$

That is,

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))+k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)+m(a\Lambda \left(b\right)+n\Lambda \left(b\right)a)\in Z.\end{array}$$

Hence,

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right){\Xi }_1\left(a\right)+m(ab+nba)\in Z.\end{array}$$

Using Lemma 3 in the previous expression, we find that R is commutative.

Case (II):

$$\begin{array}{c}\hfill {\Xi }_2(\Lambda \left(z\right))=0\ne {\Xi }_1\left(z\right).\end{array}$$

(27)

Using (27) in (23), we conclude that

$$\begin{array}{c}\hfill (a{\Xi }_2(\Lambda \left(b\right))+b{\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right))b+k{\Xi }_2(\Lambda \left(b\right))a){\Xi }_1\left(z\right)\in Z.\end{array}$$

Thus,

$$\begin{array}{c}\hfill a{\Xi }_2(\Lambda \left(b\right))+b{\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right))b+k{\Xi }_2(\Lambda \left(b\right))a\in Z.\end{array}$$

Taking b by $bz$ in the previous expression and using (27) and then multiplying it by z and then subtracting them, we have $(a{\Xi }_2(\Lambda \left(b\right))+k{\Xi }_2(\Lambda \left(b\right))a)(\Lambda \left(z\right)-z)\in Z.$ Hence, $a{\Xi }_2(\Lambda \left(b\right))+k{\Xi }_2(\Lambda \left(b\right))a\in Z.$ That is,

$$\begin{array}{c}\hfill a{\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right)a\in Z.\end{array}$$

(28)

Putting $a=z$ in (28), we obtain $(1+k){\Xi }_2\left(b\right)\in Z.$ If $k\ne -1,$ then ${\Xi }_2\left(b\right)\in Z,$ and so $[{\Xi }_2\left(b\right),b]=0,$ and by Fact 5, we obtain ${\Xi }_2=0$ or R is commutative. In case ${\Xi }_2=0,$ using the last relation in (19), we see that $a\Lambda \left(a\right)+n\Lambda \left(a\right)a\in Z,$ and by Fact 6, R is commutative. In case $k=-1,$ we infer that $[a,{\Xi }_2\left(b\right)]\in Z.$ In particular, $[a,{\Xi }_2\left(a\right)]\in Z,$ and by Fact 5, we conclude that ${\Xi }_2=0$ or R is commutative. In case ${\Xi }_2=0,$ use similar arguments as the above.

Case (III): ${\Xi }_2(\Lambda \left(z\right))\ne 0={\Xi }_1\left(z\right).$ Now, applying similar arguments as used in Case (II), we obtain ${\Xi }_1\left(a\right)b+kb{\Xi }_1\left(a\right)\in Z,$ and then using the same technique as above, we find that R is commutative. □

By utilizing Lemma 4, we obtain the proof for Theorem 2.

Lemma 5. 

If ${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)+ma\Lambda \left(a\right)\in Z$ for every $a\in R$ and $k,n\in \{-1,1\},$ then R is commutative.

Proof. 

Assume that

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)+ma\Lambda \left(a\right)\in Z\end{array}$$

(29)

for every $a\in R$ and $k,m\in \{-1,1\}.$ By linearizing (29), we have

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(b\right)+k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)\hfill \\ \hfill +& m(a\Lambda (b)+b\Lambda (a\left)\right)\in Z\hfill \end{array}$$

(30)

for every $a,b\in R$ and $k,m\in \{-1,1\}.$ Let $0\ne z\in Z.$ Replacing b by $bz$ in (30), we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))\Lambda \left(z\right)+{\Xi }_1\left(a\right)\Lambda \left(b\right){\Xi }_2(\Lambda \left(z\right))+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))z+b{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)\hfill \\ \hfill +& k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(b\right)z+k{\Xi }_2(\Lambda \left(a\right))b{\Xi }_1\left(z\right)+k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)\Lambda \left(z\right)\hfill \\ \hfill +& k\Lambda \left(b\right){\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(z\right))+m(a\Lambda \left(b\right)\Lambda \left(z\right)+b\Lambda \left(a\right)z)\in Z.\hfill \end{array}$$

(31)

Putting $b=a$ in (31), we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))\Lambda \left(z\right)+{\Xi }_1\left(a\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))+{\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))z+a{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)\hfill \\ \hfill +& k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)z+k{\Xi }_2(\Lambda \left(a\right))a{\Xi }_1\left(z\right)+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)\Lambda \left(z\right)\hfill \\ \hfill +& k\Lambda \left(a\right){\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(z\right))+m(a\Lambda \left(a\right)\Lambda \left(z\right)+a\Lambda \left(a\right)z)\in Z.\hfill \end{array}$$

Multiplying (30) by z and then using it in the last relation, we obtain

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))+a{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)+k{\Xi }_2(\Lambda \left(a\right))a{\Xi }_1\left(z\right)+k\Lambda \left(a\right){\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(z\right))\in Z.\end{array}$$

(32)

By linearizing (32), we find that

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right)\Lambda \left(b\right){\Xi }_2(\Lambda \left(z\right))+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))+a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)+b{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)\hfill \\ \hfill +& k{\Xi }_2(\Lambda \left(a\right))b{\Xi }_1\left(z\right)+k{\Xi }_2(\Lambda \left(b\right))a{\Xi }_1\left(z\right)+k\Lambda \left(a\right){\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(z\right))\hfill \\ \hfill +& k\Lambda \left(b\right){\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(z\right))\in Z.\hfill \end{array}$$

Taking b by $az$ in the last relation and using (32), we infer that

$$(a\Lambda \left(a\right)+k\Lambda \left(a\right)a){\Xi }_1\left(z\right){\Xi }_2(\Lambda \left(z\right))\in Z.$$

That is, $a\Lambda \left(a\right)+k\Lambda \left(a\right)a\in Z$ or ${\Xi }_1\left(z\right){\Xi }_2(\Lambda \left(z\right))=0.$ In case $a\Lambda \left(a\right)+k\Lambda \left(a\right)a\in Z,$ and by Fact 6, R is commutative. If ${\Xi }_1\left(z\right){\Xi }_2(\Lambda \left(z\right))=0,$ then ${\Xi }_1\left(z\right)=0$ or ${\Xi }_2(\Lambda \left(z\right))=0.$

Case (I): Suppose that

$$\begin{array}{c}\hfill {\Xi }_1\left(z\right)=0={\Xi }_2(\Lambda \left(z\right)).\end{array}$$

(33)

Using (33) in (31), we have

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))\Lambda \left(z\right)+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))z+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(b\right)z+k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)\Lambda \left(z\right)\hfill \\ \hfill +& m(a\Lambda (b)\Lambda (z)+b\Lambda (a\left)z\right)\in Z.\hfill \end{array}$$

Multiplying (30) by z and then using it in the last relation, we get

$$({\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))+k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)+ma\Lambda \left(b\right))(\Lambda \left(z\right)-z)\in Z.$$

This implies that

$$({\Xi }_1\left(a\right){\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right){\Xi }_1\left(a\right)+mab)(\Lambda \left(z\right)-z)\in Z,$$

and so

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2\left(b\right)+k{\Xi }_2\left(b\right){\Xi }_1\left(a\right)+mab\in Z.\end{array}$$

(34)

Suppose that ${\Xi }_1$ or ${\Xi }_2$ is an outer derivation, and by Fact 7, we obtain $mab\in Z,$ and hence R is commutative. Now, we assume that ${\Xi }_1$ and ${\Xi }_2$ are inner derivations, thus (34) becomes $[i_1,a][i_2,b]+k[i_2,b][i_1,a]+mab\in Z$ for some $i_1,i_2\in R.$ Putting $b=z$ in the previous expression, we see that $maz\in Z,$ that is $a\in Z$ for every $a\in R.$ Therefore, R is commutative.

Case (II): Suppose that

$$\begin{array}{c}\hfill {\Xi }_1\left(z\right)\ne 0={\Xi }_2(\Lambda \left(z\right)).\end{array}$$

(35)

Using the last relation in (31), we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))\Lambda \left(z\right)+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right)z)+b{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(b\right)z\hfill \\ \hfill +& k{\Xi }_2(\Lambda \left(a\right))b{\Xi }_1\left(z\right)+k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)\Lambda \left(z\right)+m(a\Lambda \left(b\right)\Lambda \left(z\right)+b\Lambda \left(a\right)z)\in Z.\hfill \end{array}$$

Multiplying (30) by z and using it in the last relation, we obtain

$$\begin{array}{cc}\hfill & ({\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))+k{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(a\right)+ma\Lambda \left(b\right))(\Lambda \left(z\right)-z)\hfill \\ \hfill +& (b{\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right))b){\Xi }_1\left(z\right)\in Z.\hfill \end{array}$$

Putting $b=z$ in the previous expression and using (35), we see that

$$\begin{array}{c}\hfill ma\Lambda \left(z\right)(\Lambda \left(z\right)-z)+z({\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right))){\Xi }_1\left(z\right)\in Z.\end{array}$$

Now, in case $k=-1,$ we have $ma\in Z,$ and hence R is commutative. Suppose that $k=1.$ Then

$$\begin{array}{c}\hfill ma\Lambda \left(z\right)(\Lambda \left(z\right)-z)+2z{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)\in Z.\end{array}$$

(36)

Replacing a by $az$ in the previous expression and using (35), we find that

$$\begin{array}{c}\hfill ma\Lambda \left(z\right)(\Lambda \left(z\right)-z)z+2z{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(z\right)\Lambda \left(z\right)\in Z.\end{array}$$

Multiplying (36) by $\Lambda \left(z\right)$ and using the last relation, we conclude that

$$\begin{array}{c}\hfill ma\Lambda \left(z\right){(\Lambda \left(z\right)-z)}^2\in Z.\end{array}$$

This implies that $ma\in Z.$ Thus, R is commutative.

Case (III): Suppose that ${\Xi }_1\left(z\right)=0\ne {\Xi }_2(\Lambda \left(z\right)).$ Use similar arguments as in Case (II). □

Lemma 6. 

If ${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+k{\Xi }_2(\Lambda \left(a\right)){\Xi }_1\left(a\right)+m\Lambda \left(a\right)a\in Z$ for every $a\in R$ and $k,n\in \{-1,1\},$ then R is commutative.

Proof. 

Using the same arguments as we have used in the proof of Lemma 5, we obtain the required result. □

Lemma 7. 

If ${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+n\Lambda \left(a\right)a\in Z$ for every $a\in R$ and $n\in \{-1,1\},$ then R is commutative.

Proof. 

Assume that

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+n\Lambda \left(a\right)a\in Z\end{array}$$

(37)

for every $a\in R.$ By linearizing (37), we have

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+n\Lambda \left(a\right)b+\Lambda \left(b\right)a\in Z\end{array}$$

(38)

for every $a,b\in R.$ Replacing a by $az$ in (38), where $z\in Z,$ we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))z+a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))\Lambda \left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))\hfill \\ \hfill +& n\Lambda \left(a\right)b\Lambda \left(z\right)+\Lambda \left(b\right)az\in Z\hfill \end{array}$$

(39)

for every $a,b\in R.$ Multiplying (38) by z and using (39), we obtain

$$\begin{array}{c}\hfill ({\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+n\Lambda \left(a\right)b)(\Lambda \left(z\right)-z)+a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))\in Z\end{array}$$

(40)

for every $a,b\in R.$ Taking a by $az$ in (39), we see that

$$\begin{array}{cc}\hfill & ({\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+n\Lambda \left(a\right)b)(\Lambda \left(z\right)-z)\Lambda \left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))(\Lambda \left(z\right)-z)\hfill \\ \hfill & +a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)z+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))\Lambda \left(z\right)\in Z\hfill \end{array}$$

(41)

for every $a,b\in R.$ Multiplying (40) by $\Lambda \left(z\right)$ and using (41), we find that

$$({\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))-a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right))(\Lambda \left(z\right)-z)\in Z.$$

Hence,

$$\begin{array}{c}\hfill {\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))-a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)\in Z\end{array}$$

(42)

for every $a,b\in R.$ Replacing a by $za$ in (42) and using it, we have

$${\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(z\right))(\Lambda \left(z\right)-z)\in Z$$

and so ${\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(z\right))\in Z,$ that is ${\Xi }_2(\Lambda \left(z\right))=0$ or ${\Xi }_1\left(b\right)\in Z.$

Case (I): Suppose that ${\Xi }_2(\Lambda \left(z\right))=0.$ Using the last relation in (42), we obtain

$$a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)\in Z$$

implies that $a{\Xi }_2\left(b\right){\Xi }_1\left(z\right)\in Z,$ that is ${\Xi }_2\left(b\right){\Xi }_1\left(z\right)\in Z,$ and so ${\Xi }_1\left(z\right)=0$ or ${\Xi }_2\left(b\right)\in Z.$

Subcase 1: Suppose that ${\Xi }_2\left(b\right)\in Z.$ This means $[{\Xi }_2\left(b\right),b]=0$ and by Fact 5, we obtain R is commutative or ${\Xi }_2=0.$ In case ${\Xi }_2=0.$ Using the last relation in (37), we see that $n\Lambda \left(a\right)a\in Z,$ it follows that $\Lambda \left(a\right)a\in Z,$ and by Fact 6, we find that R is commutative.

Subcase 2: Suppose that ${\Xi }_1\left(z\right)=0.$ From (40), we have

$$({\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+n\Lambda \left(a\right)b)(\Lambda \left(z\right)-z)\in Z$$

and so ${\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+n\Lambda \left(a\right)b\in Z.$ Putting $b=z$ in the previous expression, we obtain $n\Lambda \left(a\right)z\in Z,$ that is $a\in Z.$ Hence, R is commutative.

Case (II): Suppose that ${\Xi }_1\left(b\right)\in Z.$ Using the same arguments as used in Subcase 1. □

Lemma 8. 

If ${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+na\Lambda \left(a\right)\in Z$ for every $a\in R$ and $n\in \{-1,1\},$ then R is commutative.

Proof. 

Using the same arguments as we have used in the proof of Lemma 7, we obtain the required result. □

Lemma 9. 

If ${\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+m(a\Lambda \left(a\right)+n\Lambda \left(a\right)a)\in Z$ for every $a\in R$ and $m,n\in \{-1,1\},$ then R is commutative.

Proof. 

Assume that

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(a\right))+m(a\Lambda \left(a\right)+n\Lambda \left(a\right)a)\in Z.\end{array}$$

(43)

Linearizing (43), we have

$$\begin{array}{c}\hfill {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+m(a\Lambda \left(b\right)+b\Lambda \left(a\right)+n\Lambda \left(a\right)b+n\Lambda \left(b\right)a)\in Z\end{array}$$

(44)

Replacing a by $az$ in (44), we obtain

$$\begin{array}{cc}\hfill & {\Xi }_1\left(a\right){\Xi }_2(\Lambda \left(b\right))z+a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)+{\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))\Lambda \left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))\hfill \\ \hfill +& m(a\Lambda (b)z+b\Lambda (a)\Lambda (z)+n\Lambda (a)b\Lambda (z)+n\Lambda (b\left)az\right)\in Z.\hfill \end{array}$$

(45)

Multiplying (44) by z and then using (45), we obtain

$$\begin{array}{cc}\hfill & a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))+({\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))\hfill \\ \hfill +& mb\Lambda \left(a\right)+mn\Lambda \left(a\right)b\left)\right(\Lambda \left(z\right)-z)\in Z.\hfill \end{array}$$

(46)

Taking a by $az$ in (46), we see that

$$\begin{array}{cc}\hfill & a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)z+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))\Lambda \left(z\right)+({\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+mb\Lambda \left(a\right)\hfill \\ \hfill +& mn\Lambda \left(a\right)b)(\Lambda \left(z\right)-z)\Lambda \left(z\right)+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))(\Lambda \left(z\right)-z)\in Z.\hfill \end{array}$$

(47)

Multiplying (46) by $\Lambda \left(z\right)$ and then using (47), we find that

$$\begin{array}{c}\hfill a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)(z-\Lambda \left(z\right))+{\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))(\Lambda \left(z\right)-z)\in Z.\end{array}$$

Hence, ${\Xi }_1\left(b\right)\Lambda \left(a\right){\Xi }_2(\Lambda \left(z\right))-a{\Xi }_2(\Lambda \left(b\right)){\Xi }_1\left(z\right)\in Z.$ Now, application of similar arguments as used in (42), we obtain R is commutative, except Subcase 2, we obtain ${\Xi }_2(\Lambda \left(z\right))=0={\Xi }_1\left(z\right)$ and by using the previous expression in (46), we obtain

$$({\Xi }_1\left(b\right){\Xi }_2(\Lambda \left(a\right))+mb\Lambda \left(a\right)+mn\Lambda \left(a\right)b)(\Lambda \left(z\right)-z)\in Z.$$

That is,

$${\Xi }_1\left(b\right){\Xi }_2\left(a\right)+mba+mnab\in Z.$$

If ${\Xi }_1$ or ${\Xi }_2$ is outer, then we can put ${\Xi }_1\left(b\right)$ or ${\Xi }_2\left(a\right)$ by any element in R, by Fact 7, let any one of them be zero, and so we obtain $mba+mnab\in Z.$ This implies that $ba+nab\in Z,$ and by Fact 4, R is commutative. Now, if ${\Xi }_1$ and ${\Xi }_2$ are inner, then

$$\begin{array}{c}\hfill [b,i_2][a,i_1]+mba+mnab\in Z\end{array}$$

(48)

for some $i_1,i_2\in R.$ Taking $b=z$ in (48), we find that $a+na\in Z.$ Putting $n=1,$ we conclude that $2a\in Z$ and so $a\in Z$ for every $a\in R.$ Hence, R is commutative. Suppose that $n=-1.$ We have

$$\begin{array}{c}\hfill [b,i_2][a,i_1]+m[b,a]\in Z.\end{array}$$

(49)

Replacing a by $a{i}_1$ in (49), we obtain $([b,i_2][a,i_1]+m[b,a])i_1+ma[b,i_1]\in Z.$ Using (49) in the previous expression, we find that $[ma[b,i_1],i_1]=0.$ That is $[a[b,i_1],i_1]=0.$ Taking a by $ra$ in the previous expression and using it, we obtain $[r,i_1]a[b,i_1]=0.$ This implies that $[r,i_1]R[b,i_1]=\left(0\right).$ Thus, $[r,i_1]=0$ or $[b,i_1]=0.$ In both cases it implies that $i_1\in Z,$ and so (49) reduce to $m[b,a]\in Z$ and so $[b,a]\in Z$ and by Fact 4, we find that R is commutative. □

Example 2. 

(i) 

The example demonstrates the significance of the condition “Λ is Z-nonlinear" in Theorems 2–4: Consider $({\Xi }_1,{\Xi }_2)=(0,0)$ and Λ correspond to any element in the prime ring of real quaternions such that it maps to its conjugate.

(ii) 

To illustrate the importance of the hypothesis “the primeness of R" is essential in our results: Let $R=\mathbb{Z}\left[X\right]\times \mathbb{Z}\left[X\right]\times M_2\left(\mathbb{Z}\right),$$\Lambda (P_1\left(X\right),P_2\left(X\right),U)=(P_2\left(X\right),P_1\left(X\right),adj\left(U\right)),$ and $\Xi (P_1\left(X\right),P_2\left(X\right),U)=(P_1{\left(X\right)}^{\prime },P_2{\left(X\right)}^{\prime },0)$ for every $(P_1\left(X\right),P_2\left(X\right),U)\in R,$ where $\Xi ={\Xi }_1={\Xi }_2.$ Then Λ is a Z-nonlinear antiautomorphism, and Ξ is derivation on $R,$ and R is not commutative.

Remark 1. 

$\left(1\right)$

In the case where Λ represents an automorphism, all the results presented in this article remain valid.

$\left(2\right)$

It is important to emphasize that our results in this article hold valid even when the different assumptions are presumed to be true on a non-zero ideal rather than the entire ring $R.$

4. Future Research

For future research, two main directions can be pursued to extend the current results: Firstly, exploring the concept of a semiprime ring R instead of a prime ring R in the theorems can be a fruitful direction. Investigating the behavior of derivations and special mappings within semiprime rings may reveal new insights into their commutativity. Secondly, substituting the concept of derivations ${\Xi }_1,{\Xi }_2$ with two generalized derivations in the theorems could open up new possibilities for understanding the relationships between mappings and the structure of rings.

5. Conclusions

In this article, we have successfully extended the results previously established by Nejjar et al. [27] and Mamouni et al. [33]. Notably, when we set ${\Xi }_1={\Xi }_2$ and $\Lambda =\ast$ (indicating that an antiautomorphism $\Lambda$ is an involution ∗) in our findings, we recover the results presented by Nejjar et al. [27]. Similarly, setting $\Lambda =\ast$ in our results yields the results of Mamouni et al. [33]. Furthermore, we have demonstrated the $\Lambda$-version of Posner’s second theorem [17], as seen in Theorem 1 (i). Additionally, we have explored the concept of $\Lambda$-SCP derivations for two derivations with an antiautomorphism $\Lambda ,$ rather than the traditional ∗-SCP derivation for a single derivation with an involution $\ast ,$ as shown in Theorem 2. The introduction of this new concept has enriched the understanding of the subject. Lastly, we have provided various examples to emphasize the importance of the restrictions imposed in the assumptions of our results. These examples highlight the significance of these constraints in the context of our findings.