On Prime Ideals in Rings Satisfying Certain Functional Identities

Abstract

In this note, we will examine the relationship between a generalized derivation ϝ associated with a derivation ∂ that satisfies specific algebraic identities and the behavior of a factor ring $S/P$, where P is a prime ideal of any ring S. We will also explore several consequences and special cases as applications of our findings. Additionally, various examples have been provided to highlight the necessity of the primeness hypothesis in our theorems.

1. Introduction

Throughout this paper, unless otherwise stated, S will represent an associative ring with center $Z\left(S\right)$. Recall that a ring S is called a prime ring if for any $m, n\in S$, $mSn=0$, then either $m=0$ or $n=0$. S is called semiprime if $mSm=0$ for any $m\in S,$then $m=0$. On other words; A ring S is called prime if $IJ\ne 0$ for any two nonzero ideals $I,J\subseteq S$, and semiprime if it contains no nonzero ideal whose square is zero. A ring S is said to be reduced if it has no non-zero nilpotent elements (i.e., has no non-zero elements with square zero, meaning if $m^2=0$ implies that $m=0$ for any $m\in S$). The term domain is used for a non-zero ring S without non-zero divisors, meaning for all $m, n\in S$, $mn=0$ implies that either $m=0$ or $n=0$. Obviously, every domain is a prime ring, and a prime ring which is reduced is a domain. A commutative domain with an identity is an integral domain, and every integral domain is a reduced ring. The following example shows that prime rings constitute a more general algebraic structure than that of domains.

Example 1.

Let S be the Morita ring as described in ([1] Corollary 2.5). It has been proven that the ring S is semiprime but not reduced. Since a prime ring is semiprime and a domain is a reduced ring, we can conclude that the Morita ring S is prime but not a domain.

In their work [2], Ferreira et al. presented an alternative description of prime rings within the realm of alternative rings. Their theorem states:

Theorem 1

([2] Theorem 1.1). Let S be a 3-torsion free alternative ring. S is a prime ring if and only if $mS.n=0$ (or $m.Sn=0$) implies $m=0$ or $n=0$ for $m,n\in S$.

A proper ideal P of a ring S is considered prime if, for every elements m and n in S, the condition $mSn\subseteq P$ implies that either m belongs to P or n belongs to P. Moreover, an ideal P in the ring S (with unity) is prime if and only if the factor ring $S/P$ is an integral domain. In particular, a commutative ring (with unity) is an integral domain if and only if $\left\{0\right\}$ is a prime ideal.

An additive function $\partial :S⟶S$ is called a derivation if it satisfies the rule $\partial \left(mn\right)=\partial \left(m\right)n+m\partial \left(n\right)$ for all $m, n\in S$. In [3], Posner investigated the relationship between the commutativity of a ring S and a derivation ∂. He proved that a prime ring S admitting a nonzero centralizing derivation is commutative. In $1978,$ Herstein [4] assumed that S is a prime ring that admits a non-zero derivation ∂ satisfying the condition $\partial \left(m\right)\partial \left(n\right)=\partial \left(n\right)\partial \left(m\right)$ for all $m, n\in S$. He proved the following: $\left(i\right)$ If $char\left(S\right)\ne 2,$ then S is a commutative integral domain, $\left(ii\right)$ If $char\left(S\right)=2$, then either S is commutative or an order in a simple algebra which is 4-dimensional over its center. An additive mapping $\mathsf{ϝ}$: $S\to S$ is called a generalized derivation associated with a derivation ∂ if $\mathsf{ϝ}\left(mn\right)=\mathsf{ϝ}\left(m\right)n+m\partial \left(n\right)$ holds for all $m, n\in S$. Familiar examples of generalized derivations are derivations and generalized inner derivations. The latter includes left multipliers when satisfying $\mathsf{ϝ}\left(mn\right)=\mathsf{ϝ}\left(m\right)n$ for all $m, n\in S$. These concepts are also discussed in other algebraic structures such as $BCI$-algebras and d-algebras (for example see [5,6,7]. In light of previous results, recent literature has presented numerous findings on commutativity in prime and semiprime rings with appropriately constrained automorphisms, derivations, generalized derivations, and generalized reverse derivations that centralize or commute on the ring S or on its appropriate subsets (for examples see [8,9,10]).

Building on previous findings, several authors have enhanced the study by adopting a different approach. They have utilized various algebraic identities involving prime ideals without imposing additional assumptions on the ring S. Subsequently, they have examined the behavior of a factor ring $S/P$, where P is a prime ideal of S (for examples, see [11,12,13,14,15]).

Based on the studies mentioned above, this article aims to further investigate the behavior of a factor ring $S/P$, where P is a prime ideal of any ring S that admits a generalized derivation $\mathsf{ϝ}$ that satisfies any of the following functional identities:

(i)

$\mathsf{ϝ}\left(m\right)\circ n-\mathsf{ϝ}\left(mn\right)\in P$, for all $m,n\in S,$

(ii)

$\left[\mathsf{ϝ}\right(m),n]\pm \mathsf{ϝ}\left(mn\right)\in P,$ for all $m,n\in S,$

(iii)

$\mathsf{ϝ}\left(m^2\right)-\partial \left(m\right)m\in P,$ for all $m\in S,$

(iv)

$\mathsf{ϝ}\left(m^2\right)+\mathsf{ϝ}\left(m\right)m\in P,$ for all $m\in S,$

(v)

$\mathsf{ϝ}\left(m^2\right)+m\mathsf{ϝ}\left(m\right)\in P$ for all $m\in S.$

Furthermore, in Proposition 2.1, we address the question posed in ([8] Remark 2.3) by defining a generalized derivation $\mathsf{ϝ}$ associated with a derivation ∂ on the ring $K_{2^n}$ that satisfies the identities (i) $\mathsf{ϝ}\left(m\right)\circ y-\mathsf{ϝ}\left(mn\right)=0$ for all $m, n\in S$ and (ii) $\left[\mathsf{ϝ}\right(m),n]\pm \mathsf{ϝ}\left(mn\right)=0$ for all $m, n\in S$. Moreover, we will include several relevant corollaries. Additionally, we will provide various examples to illustrate that the assumptions in our theorems are not redundant.

2. Preliminaries

In the following, we will use the symbol $(\mathsf{ϝ},\partial )$ to denote the generalized derivation $\mathsf{ϝ}$ associated with a derivation ∂. For any $m, n\in S$, the symbol $[m, n]$ will denote the commutator $mn-nm$, and the symbol $m\circ n$ will stand for the skew-commutator $mn+nm$. Given an integer $\mathsf{\ell }>1,$ a ring S is said to be $\mathsf{\ell }$-torsion free if for $m\in S$, $\mathsf{\ell }m=0$ implies $m=0$.

To begin, let us provide the following lemma and remark, which will be referred to frequently in the proofs of our theorems.

Lemma 1

([16] Lemma 1.2). Let S be a ring and let P be a prime ideal of S. If $[m,n]\in P$ for all $m,n\in S$, then $S/P$ is a commutative integral domain.

Remark 1.

(Brauer’s trick) Let $(G,+)$ be a group, and let Γ and Λ be proper subgroups of it. Then G cannot be equal to $\mathrm{\Gamma }\cup \mathrm{\Lambda }.$

3. Main Results

It was shown in ([8] Theorem 2.1) that if the prime ring S admits a derivation ∂ such that $\partial \left(m\right)\circ n=\partial \left(mn\right)$ for all $m, n\in S$, then the derivation ∂ is proven to be zero. In the next theorem, we will discuss more general situations by replacing the derivation ∂ with a generalized derivation $(\mathsf{ϝ},\partial )$ and dropping the assumption that S is prime to explore the implications.

Theorem 2.

Let P be a prime ideal of a ring S that admits a generalized derivation $(\mathsf{ϝ}, \partial )$. Then the following statements are true for all $m,n\in S$:

(i) 

$\mathsf{ϝ}\left(m\right)\circ n-\mathsf{ϝ}\left(mn\right)\in P$ if and only if $\partial \left(S\right)\subseteq P$ and $\mathsf{ϝ}\left(S\right)\subseteq P.$

(ii) 

$\left[\mathsf{ϝ}\right(m),n]\pm \mathsf{ϝ}\left(mn\right)\in P$ if and only if $\partial \left(S\right)\subseteq P$ and $\mathsf{ϝ}\left(S\right)\subseteq P.$

Proof. 

(i) It is obvious that if $\partial \left(S\right)\subseteq P$ and $\mathsf{ϝ}\left(S\right)\subseteq P$, then S satisfies the identity $\mathsf{ϝ}\left(m\right)\circ n-\mathsf{ϝ}\left(mn\right)\in P$ for all $m,n\in S$. Therefore, we need to prove the non-trivial direction. Based on our assumption, we have

$$\begin{array}{c}\hfill \mathsf{ϝ}\left(m\right)\circ n-\mathsf{ϝ}\left(mn\right)\in P\mathrm{for}\mathrm{all}m,n\in S.\end{array}$$

This can be expressed as

$$\begin{array}{c}\hfill n\mathsf{ϝ}\left(m\right)-m\partial \left(n\right)\in P\mathrm{for}\mathrm{all}m,n\in S.\end{array}$$

(1)

Replacing n by $nh$ in (1), we obtain

$$nh\mathsf{ϝ}\left(m\right)-m\partial \left(n\right)h-mn\partial \left(h\right)\in P\mathrm{for}\mathrm{all}m,n,h\in S.$$

(2)

Right Multiplying (1) by h and combining it with (2), we get

$$n\mathsf{ϝ}\left(m\right)h-nh\mathsf{ϝ}\left(m\right)+mn\partial \left(h\right)\in P\mathrm{for}\mathrm{all}m,n,h\in S.$$

(3)

Replacing n by $sn$ in (3), we obtain

$$sn\mathsf{ϝ}\left(m\right)h-snh\mathsf{ϝ}\left(m\right)+msn\partial \left(h\right)\in P\mathrm{for}\mathrm{all}m,n,h,s\in S.$$

By left multiplying (3) by s and combining it with the last equation, we can conclude that $[m,s]S\partial \left(h\right)\subseteq P$ for all $m,s,h\in S$. Since P is prime, we deduce that either $\partial \left(h\right)\in P$ for all $h\in S$ or $[m,s]\in P\mathrm{for}\mathrm{all}m,s\in S$. Let us define $\mathrm{\Gamma }=\{h\in S:\partial (h)\in P\}$ and $\mathrm{\Lambda }=\{m\in S:[m,s]\in P\mathrm{for}\mathrm{all}s\in S\}$. It can be easily verified that both $\mathrm{\Gamma }$ and $\mathrm{\Lambda }$ are additive subgroups of S, their union equals S, and $\mathrm{\Gamma }\cap \mathrm{\Lambda }=⌀$. By applying Remark 1, we get either $\mathrm{\Gamma }=S$ or $\mathrm{\Lambda }=S$. Let us first assume that $\mathrm{\Gamma }=S$. Then, $\partial \left(h\right)\in P$ for all $h\in S$, so (2) becomes $nS\mathsf{ϝ}\left(m\right)\subseteq P$ for all $m,n\in S$. Since P is prime, we can conclude that $\mathsf{ϝ}\left(S\right)\subseteq P$. Secondly, if $\mathrm{\Lambda }=S$, that means $[m,s]\in P\mathrm{for}\mathrm{all}m,s\in S$, then by Lemma 1, $S/P$ is a commutative integral domain. So, from (3), we can derive $mS\partial \left(h\right)\subseteq P\mathrm{for}\mathrm{all}m,h\in S$. Once again, since P is prime, we can deduce that $\partial \left(S\right)\subseteq P$. Therefore, as discussed above, we can conclude that $\mathsf{ϝ}\left(S\right)\subseteq P$.

(ii) Use similar arguments as in Proof $\left(i\right)$, with some necessary modifications to achieve the desired result. □

As a consequence of the previous theorem, the following corollary can be deduced in the case where $\mathsf{ϝ}=\partial$.

Corollary 1.

Let P be a prime ideal of a ring S that admits a derivation ∂. Then the following statements are true for all $m,n\in S$:

(i) 

$\partial \left(m\right)\circ n-\partial \left(mn\right)\in P$ if and only if $\partial \left(S\right)\subseteq P$.

(ii) 

$[\partial (m),n]\pm \partial \left(mn\right)\in P$ if and only if $\partial \left(S\right)\subseteq P$.

When $P=\left\{0\right\}$ in the above theorem, the following corollary can be deduced.

Corollary 2.

Let S be a prime ring that admits a generalized derivation $(\mathsf{ϝ}, \partial )$. Then the following statements are true for all $m,n\in S$:

(i) 

$\mathsf{ϝ}\left(m\right)\circ n-\mathsf{ϝ}\left(mn\right)=0$ if and only if $\partial =0$ and $\mathsf{ϝ}=0.$

(ii) 

$\left[\mathsf{ϝ}\right(m),n]\pm \mathsf{ϝ}\left(mn\right)=0$ if and only if $\partial =0$ and $\mathsf{ϝ}=0.$

If $\mathsf{ϝ}$ is replaced by ∂ in Corollary 2, then there exists no non-trivial derivation that satisfies the following corollary.

Corollary 3.

Let S be a prime ring that admits a derivation ∂. Then the following statements are true for all $m,n\in S$:

(i) 

$\partial \left(m\right)\circ n-\partial \left(mn\right)=0$ if and only if ∂ is a trivial derivation on S.

(ii) 

$[\partial (m),n]\pm \partial \left(mn\right)=0$ if and only if ∂ is a trivial derivation on S.

In ([8] Remark 2.3), the exercise posed was to define a generalized derivation $(\mathsf{ϝ},\partial )$ on $K_{2^{\mathsf{\ell }}}.$ The purpose of the following proposition is to provide an answer to this question and to illustrate the necessity of S being prime in Corollary 2. To achieve this goal, we will define a generalized derivation $\mathsf{ϝ}$ on $K_{2^{\mathsf{\ell }}}$ associated with a non-zero derivation ∂ that satisfies the identities $\mathsf{ϝ}\left(m\right)\circ n-\mathsf{ϝ}\left(mn\right)=0$ and $\left[\mathsf{ϝ}\right(m),n]\pm \mathsf{ϝ}\left(mn\right)=0,$ for all $m, n\in K_{2^{\mathsf{\ell }}}$. Before delving into the proposition, we first need to the following definition (refer to details in [8]).

Definition 1

([8] Definition 2.1). Let us consider the set

$$K_{2^{\mathsf{\ell }}}:=〈m_1,\dots ,m_{\mathsf{\ell }}|m_i{m}_j=m_i,2m_i=0,\forall i,j=1,2,\dots \mathsf{\ell }〉.$$

It is easy to verify that $K_{2^{\mathsf{\ell }}}$ is a non-commutative ring without 1 and with characteristic 2. This ring is neither prime nor a domain, and its center is equal to zero. In fact, $K_{2^{\mathsf{\ell }}}$ is an algebra over ${\mathbb{Z}}_2$ with ℓ-generators.

In particular, when $\mathsf{\ell }=2$, we obtain $K_{2^2}=〈m,n〉=\{0,m,n,h\}$ with the following relations:

$$2m=2n=0,h=m+n,m^2=mn=m,n^2=nm=n.$$

The operations of addition and multiplication defined on $K_{2^2}$ are shown in the Cayley table (the reader can refer back to ([8] Example 2.1)).

Now we are ready to prove the following proposition:

Proposition 1.

Let $\chi =K_{2^{\mathsf{\ell }}}$ be as above where ℓ is an even integer. For $1\le \lambda \le \mathsf{\ell },$ let ${\mathsf{ϝ}}_{\lambda },$ ${\partial }_{\lambda }:\chi ⟶\chi$ be additive maps defined on the generators $m_1,\cdots ,m_{\mathsf{\ell }}\in \chi$ by

$${\mathsf{ϝ}}_{\lambda }\left(m\right)=\left\{\begin{array}{ccc}{\displaystyle \sum _{i=1}^{\mathsf{\ell }}}{m}_i\hfill & ifm=m_1,\cdots ,m_{\lambda -1},m_{\lambda +1},\cdots ,m_{\mathsf{\ell }};\hfill & \\ 0\hfill & ifm=m_{\lambda }.\hfill \end{array}\right.$$

$${\partial }_{\lambda }\left(m\right)=\left\{\begin{array}{ccc}0\hfill & ifm=m_1,\cdots ,m_{\lambda -1},m_{\lambda +1},\cdots ,m_{\mathsf{\ell }};\hfill & \\ {\displaystyle \sum _{i=1}^{\mathsf{\ell }}}{m}_i\hfill & ifm=m_{\lambda }.\hfill \end{array}\right.$$

Then ${\mathsf{ϝ}}_{\lambda }=\mathsf{ϝ}$ is a generalized derivation on χ associated with a non-zero derivation ${\partial }_{\lambda }=\partial$ satisfies the following identities:

(i) 

$\mathsf{ϝ}\left(m\right)\circ n=\mathsf{ϝ}\left(mn\right)$ for all $m,n\in \chi .$

(ii) 

$\left[\mathsf{ϝ}\right(m),n]\pm \mathsf{ϝ}\left(mn\right)=0$ for all $m,n\in \chi .$

Proof. 

We will only prove the proposition for the generators $\{m_1,\dots ,m_{\mathsf{\ell }}\}$ of $\chi$. Other elements of the ring $\chi$ will follow naturally. Clearly, $\mathsf{ϝ}\ne \partial \ne 0$ by definition.

(i) We will prove that ${\mathsf{ϝ}}_{\lambda }=\mathsf{ϝ}$ is a generalized derivation associated with ${\partial }_{\lambda }=\partial ,$ and that

$$\mathsf{ϝ}\left(m\right)\circ n=\mathsf{ϝ}\left(mn\right),\mathrm{for}\mathrm{all}m,n\in \{m_1,\dots ,m_{\mathsf{\ell }}\}\subseteq \chi .$$

(4)

We will do this in the following four cases:

Case I: Let $m=m_{\lambda }$ and $n=m_{\lambda }.$ Then

$$\mathsf{ϝ}\left(m_{\lambda }{m}_{\lambda }\right)=\mathsf{ϝ}\left(m_{\lambda }\right)=0.$$

On the other hand

$$\begin{array}{ccc}\hfill \mathsf{ϝ}\left(m_{\lambda }\right)m_{\lambda }+m_{\lambda }\partial \left(m_{\lambda }\right)& =& 0m_{\lambda }+m_{\lambda }\left(\sum _{i=1}^{\mathsf{\ell }}{m}_i\right)=\sum _{i=1}^{\mathsf{\ell }}{m}_{\lambda }{m}_i=n{m}_{\lambda }=0.\hfill \end{array}$$

Therefore, ϝ is a generalized derivation associated with ∂. For the identity (4),

$$\begin{array}{ccc}\hfill \mathsf{ϝ}\left(m_{\lambda }\right)\circ m_{\lambda }& =& 0\circ m_{\lambda }=0\hfill \\ & =& \mathsf{ϝ}\left(m_{\lambda }\right)=\mathsf{ϝ}\left(m_{\lambda }{m}_{\lambda }\right).\hfill \end{array}$$

Hence, the identity also holds.

Case II: Let $m\ne m_{\lambda }$ and $n\ne m_{\lambda }.$ Then

$$\mathsf{ϝ}\left(mn\right)=\mathsf{ϝ}\left(m\right)=\sum _{i=1}^{\mathsf{\ell }}{m}_i.$$

On the other hand

$$\begin{array}{ccc}\hfill \mathsf{ϝ}\left(m\right)n+m\partial \left(n\right)& =& \left(\sum _{i=1}^{\mathsf{\ell }}{m}_{i}n\right)+m0=\left(\sum _{i=1}^{\mathsf{\ell }}{m}_{i}n\right)=\sum _{i=1}^{\mathsf{\ell }}{m}_i.\hfill \end{array}$$

Hence, ϝ is a generalized derivation associated with ∂. For the identity (4),

$$\begin{array}{ccc}\hfill \mathsf{ϝ}\left(m\right)\circ n=\mathsf{ϝ}\left(m\right)n+n\mathsf{ϝ}\left(m\right)& =& \left(\sum _{i=1}^{\mathsf{\ell }}{m}_i\right)n+n\left(\sum _{i=1}^{\mathsf{\ell }}{m}_i\right)\hfill \\ & =& \left(\sum _{i=1}^{\mathsf{\ell }}{m}_{i}n\right)+\left(\sum _{i=1}^{\mathsf{\ell }}n{m}_i\right)\hfill \\ & =& \sum _{i=1}^{\mathsf{\ell }}{m}_i+\sum _{i=1}^{\mathsf{\ell }}n=\sum _{i=1}^{\mathsf{\ell }}{m}_i+\mathsf{\ell }n=\sum _{i=1}^{\mathsf{\ell }}{m}_i.\hfill \end{array}$$

where

$$\sum _{i=1}^{\mathsf{\ell }}n{m}_i=\mathsf{\ell }n=0\mathrm{because}n\mathrm{is}\mathrm{even}.$$

On the other hand

$$\mathsf{ϝ}\left(mn\right)=\mathsf{ϝ}\left(m\right)=\sum _{i=1}^{\mathsf{\ell }}{m}_i.$$

Hence, the identity also holds.

Case III: Let $m=m_{\lambda }$ and $n\ne m_{\lambda }.$ Then

$$\mathsf{ϝ}\left(m_{\lambda }n\right)=\mathsf{ϝ}\left(m_{\lambda }\right)=0.$$

and

$$\mathsf{ϝ}\left(m_{\lambda }\right)n+m_{\lambda }\partial \left(n\right)=0+m_{\lambda }0=0.$$

Therefore, again ϝ is a generalized derivation associated with ∂. For the identity (4),

$$\mathsf{ϝ}\left(m_{\lambda }\right)\circ n=0\circ n=0=\mathsf{ϝ}\left(m_{\lambda }n\right)=\mathsf{ϝ}\left(m_{\lambda }\right)=0.$$

Thus, the identity also holds.

Case IV: Finally, let $m\ne m_{\lambda }$ and $n=m_{\lambda }.$ Then, for the generalized derivation

$$\mathsf{ϝ}\left(m{m}_{\lambda }\right)=\mathsf{ϝ}\left(m\right)=\sum _{i=1}^{\mathsf{\ell }}{m}_i.$$

On the other hand,

$$\begin{array}{ccc}\hfill \mathsf{ϝ}\left(m\right)m_{\lambda }+m\partial \left(m_{\lambda }\right)& =& \left(\sum _{i=1}^{\mathsf{\ell }}{m}_i\right)m_{\lambda }+m0=\left(\sum _{i=1}^{\mathsf{\ell }}{m}_i{m}_{\lambda }\right)=\sum _{i=1}^{\mathsf{\ell }}{m}_i.\hfill \end{array}$$

Therefore, ϝ is a generalized derivation associated with ∂. The identity (4),

$$\begin{array}{ccc}\hfill \mathsf{ϝ}\left(m\right)\circ m_{\lambda }=\mathsf{ϝ}\left(m\right)m_{\lambda }+m_{\lambda }\mathsf{ϝ}\left(m\right)& =& \left(\sum _{i=1}^{\mathsf{\ell }}{m}_i\right)m_{\lambda }+m_{\lambda }\left(\sum _{i=1}^{\mathsf{\ell }}{m}_i\right)\hfill \\ & =& \left(\sum _{i=1}^{\mathsf{\ell }}{m}_i{m}_{\lambda }\right)+\left(\sum _{i=1}^{\mathsf{\ell }}{m}_{\lambda }{m}_i\right)\hfill \\ & =& \sum _{i=1}^{\mathsf{\ell }}{m}_i+\sum _{i=1}^{\mathsf{\ell }}{m}_{\lambda }=\sum _{i=1}^{\mathsf{\ell }}{m}_i+\mathsf{\ell }{m}_{\lambda }\hfill \\ & =& \sum _{i=1}^{\mathsf{\ell }}{m}_i.\hfill \end{array}$$

where

$$\mathsf{\ell }{m}_{\lambda }=0\mathrm{because}\mathsf{\ell }\mathrm{is}\mathrm{even}.$$

On the other hand,

$$\mathsf{ϝ}\left(m{m}_{\lambda }\right)=\mathsf{ϝ}\left(m\right)=\sum _{i=1}^{\mathsf{\ell }}{m}_i.$$

Hence, identity (4) holds and the proposition is proved.

(ii) By utilizing arguments and techniques similar to those used in proof $\left(i\right)$, we can verify that ϝ is a generalized derivation of $\chi$ associated with ∂ that satisfies the identity $\left[\mathsf{ϝ}\right(m),n]\pm \mathsf{ϝ}\left(mn\right)=0$ for all $m,n\in \chi .$ □

Remark 2.

The proposition above proves that in all cases, $(\mathsf{ϝ}, \partial )$ is a generalized derivation of χ that satisfies the following conditions: $\left(i\right)$ $\mathsf{ϝ}\left(m\right)\circ n=\mathsf{ϝ}\left(mn\right)$, $\left(ii\right)$ $\left[\mathsf{ϝ}\right(m),n]\pm \mathsf{ϝ}\left(mn\right)=0$ for all $m,n\in \{m_1,m_2,\dots m_{\mathsf{\ell }}\}\subseteq \chi$. However, χ is not prime as stated in Definition 1. Additionally, neither ∂ nor ϝ is equal to zero. This confirms the necessity of our assumption that $\chi =S$ is prime in Corollary 2.

The following example is intended to illustrate the necessity of our hypothesis that S is a prime in part $\left(i\right)$ of Corollary 2.

Example 2.

Let $S=\left\{\left(\begin{array}{cc}m& n\\ 0& h\end{array}\right)\mid m,n,h\in \mathbb{Z}\right\}$, where $\mathbb{Z}$ is the ring of integers. Define $\mathsf{ϝ}=\partial :S\to S$ by $\mathsf{ϝ}\left(\begin{array}{cc}m& n\\ 0& h\end{array}\right)=\left(\begin{array}{cc}0& m+n-h\\ 0& 0\end{array}\right)$. Clearly, S is not prime because $\left(\begin{array}{cc}0\hfill & m\hfill \\ 0\hfill & 0\hfill \end{array}\right)\left(\begin{array}{cc}m\hfill & 0\hfill \\ 0\hfill & 0\hfill \end{array}\right)=\left(\begin{array}{cc}0\hfill & 0\hfill \\ 0\hfill & 0\hfill \end{array}\right)$. It is easy to see that $(\mathsf{ϝ}, \partial )$ is a generalized derivation of S that satisfies the identity $\mathsf{ϝ}\left(\mathfrak{A}\right)\circ \mathfrak{B}=\mathsf{ϝ}\left(\mathfrak{AB}\right)$, for all $\mathfrak{A},\mathfrak{B}\in S$. However, neither ∂ nor ϝ is equal to zero. Therefore, in Corollary 2, the hypothesis of primeness cannot be omitted.

As a generalization of the concept of a derivation, Sandhu et.al. ([15] Definition 1) introduced the concept of $P$-derivation as a mapping $\partial :S⟶S$ that satisfies the following axioms: (i) $\partial (m+n)-\partial \left(m\right)-\partial \left(n\right)\in P$, for all $m, n\in S$; (ii) $\partial \left(mn\right)-\partial \left(m\right)n-m\partial \left(n\right)\in P$, for all $m, n\in S$. They discussed the behavior of a factor ring $S/P$ when S admits P-derivations ∂ and ð that satisfy any one of the identities: (i) $\partial \left(m^2\right)-mð\left(m\right)\in P$, (ii) $\partial \left(m^2\right)-ð\left(m\right)m\in P$ for all $m\in S$. In the following theorems, instead of P-derivations ∂ and ð mentioned above, we will discuss similar situations in the context of a generalized derivation $\mathsf{ϝ}$ associated with a derivation ∂.

Theorem 3.

Let P be a prime ideal of a ring S with $char(S/P)\ne 2$. If S admits a generalized derivation $(\mathsf{ϝ}, \partial )$, then S satisfies $\mathsf{ϝ}\left(m^2\right)-\partial \left(m\right)m\in P$ for all $m\in S$ if and only if $S/P$ is a commutative integral domain, $\partial \left(S\right)\subseteq P$ and $\mathsf{ϝ}\left(S\right)\subseteq P$.

Proof. 

Assume that $S/P$ is a commutative integral domain, $\partial \left(S\right)\subseteq P$ and $\mathsf{ϝ}\left(S\right)\subseteq P$. It can be easily verified that S satisfies $\mathsf{ϝ}\left(m^2\right)-\partial \left(m\right)m\in P$ for all $m\in S$.

Henceforth, we will now prove the non-trivial direction of the given hypothesis, which states:

$$\mathsf{ϝ}\left(m^2\right)-\partial \left(m\right)m\in P\mathrm{for}\mathrm{all}m\in S.$$

(5)

By linearizing (5) and using it, we can deduce that

$$\mathsf{ϝ}(m\circ n)-\partial \left(m\right)n-\partial \left(n\right)m\in P\mathrm{for}\mathrm{all}m,n\in S.$$

(6)

Substituting n with $nm$ in (6) and combining it with (5), we obtain

$$(m\circ n)\partial \left(m\right)-n\partial \left(m\right)m\in P\mathrm{for}\mathrm{all}m,n\in S.$$

(7)

Replacing n with $hn$ in (7) and utilizing it, we can deduce that $[m,h]S\partial \left(m\right)\subseteq P\mathrm{for}\mathrm{all}m,h\in S$. Since P is prime, we can conclude that either $\partial \left(m\right)\in P$ for all $m\in S$ or $[m,h]\in P\mathrm{for}\mathrm{all}m,h\in S$. Let us define $\mathrm{\Gamma }=\{m\in S:\partial (m)\in P\}$ and $\mathrm{\Lambda }=\{m\in S:[m,h]\in P\mathrm{for}\mathrm{all}h\in S\}$. It is easy to verify that both $\mathrm{\Gamma }$ and $\mathrm{\Lambda }$ are additive subgroups of S, their union equals S, and $\mathrm{\Gamma }\cap \mathrm{\Lambda }=⌀$. By applying Remark 1, we obtain either $\mathrm{\Gamma }=S$ or $\mathrm{\Lambda }=S$. If $\mathrm{\Gamma }=S$, then we have $\partial \left(m\right)\in P$ for all $m\in S$. Therefore, (5) simplifies to $\mathsf{ϝ}\left(m\right)m\in P$ for all $m\in S$. By linearizing this relation and utilizing it, we find $\mathsf{ϝ}\left(m\right)n+\mathsf{ϝ}\left(n\right)m\in P$ for all $m,n\in S$. Substituting m with $mh$ in the previous expression and using it, we obtain $\mathsf{ϝ}\left(m\right)[h,n]\in P\mathrm{for}\mathrm{all}m,n,h\in S$. Replacing h with $hs$ in the previous equation and using it, we arrive at $\mathsf{ϝ}\left(m\right)S[s,n]\subseteq P\mathrm{for}\mathrm{all}m,s,n\in S$. Applying the primeness of P, we can deduce that either $\mathsf{ϝ}\left(m\right)\in P$ for all $m\in S$ or $[s,n]\in P\mathrm{for}\mathrm{all}s,n\in S$. If $\mathsf{ϝ}\left(m\right)\in P$ for all $m\in S$, then $\mathsf{ϝ}\left(S\right)\subseteq P$. Alternatively, if $[s,n]\in P\mathrm{for}\mathrm{all}s,n\in S$, then by Lemma 1 $S/P$ is a commutative integral domain. Hence, (6) becomes $2\mathsf{ϝ}\left(m\right)n\in P$ for all $m,n\in S$. Since $char(S/P)\ne 2$, the previous equation simplifies to $\mathsf{ϝ}\left(m\right)n\in P$ for all $m,n\in S$. Therefore, we can conclude that $\mathsf{ϝ}\left(S\right)\subseteq P$. If $\mathrm{\Lambda }=S$, then $[m,h]\in P\mathrm{for}\mathrm{all}m,h\in S$, and so $S/P$ is a commutative integral domain, as shown by Lemma 1. Thus, (7) becomes $mS\partial \left(m\right)\subseteq P\mathrm{for}\mathrm{all}m,n\in S.$ By utilizing the primeness of P, we can deduce that $\partial \left(S\right)$ is a subset of P. Based on Equation (6) and the previous discussion, we can conclude that $\mathsf{ϝ}\left(S\right)$ is also a subset of P. □

If the ring S in Theorem 3 is prime, the following corollary can be derived:

Corollary 4.

Let S be a prime ring. If S admits a non zero generalized derivation $(\mathsf{ϝ}, \partial )$ such that $\mathsf{ϝ}\left(m^2\right)=\partial \left(m\right)m$ for all $m\in S$, then $\partial =0$ (in this case, ϝ becomes a left multiplier) and S is a commutative ring with $char\left(S\right)=2$.

In Theorem 3, if we let $\mathsf{ϝ}=\partial$, we obtain the following corollary without imposing any restrictions on the characteristic of the ring $S/P$. Specifically, instead of using two P-derivations ∂ and ð on S, as stated in ([15] Proposition 6), we will weaken it by assuming that S is a ring that admits an ordinary derivation ∂ satisfying the identity $\partial \left(m^2\right)-\partial \left(m\right)m\in P$ for all $m\in S$.

Corollary 5.

Let P be a prime ideal of a ring S. If S admits a derivation ∂, then S satisfies $\partial \left(m^2\right)-\partial \left(m\right)m\in P$ for all $m\in S$ if and only if $\partial \left(S\right)\subseteq P$.

Theorem 4.

Let P be a prime ideal of a ring S with $char(S/P)\ne 2$. If S admits a generalized derivation $(\mathsf{ϝ}, \partial )$, then S satisfies $\mathsf{ϝ}\left(m^2\right)+\mathsf{ϝ}\left(m\right)m\in P$ for all $m\in S$ if and only if one of the following conditions holds:

(i) 

$\partial \left(S\right)\subseteq P$ and $\mathsf{ϝ}\left(S\right)\subseteq P$.

(ii) 

$\partial \left(S\right)\subseteq P$ and $S/P$ is a commutative integral domain.

(iii) 

$S/P$ is a commutative integral domain and $char(S/P)=3$.

Proof. 

It is easy to verify that if any of cases $\left(i\right)$, $\left(ii\right)$, or $\left(iii\right)$ hold, then S satisfies the identity $\mathsf{ϝ}\left(m^2\right)+\mathsf{ϝ}\left(m\right)m\in P$ for all $m\in S.$

Conversely, we will assume that

$$\mathsf{ϝ}\left(m^2\right)+\mathsf{ϝ}\left(m\right)m\in P\mathrm{for}\mathrm{all}m\in S.$$

(8)

By linearizing (8) and using it, we find that

$$\mathsf{ϝ}(m\circ n)+\mathsf{ϝ}\left(m\right)n+\mathsf{ϝ}\left(n\right)m\in P\mathrm{for}\mathrm{all}m,n\in S.$$

(9)

By replacing n with $nm$ in (9) and using it with (8), we get

$$(m\circ n)\partial \left(m\right)+n\partial \left(m\right)m\in P\mathrm{for}\mathrm{all}m,n\in S.$$

(10)

Replace n with $hn$ in (10) and use it to arrive at $[m,h]S\partial \left(m\right)\subseteq P\mathrm{for}\mathrm{all}m,h\in S$. Since P is prime, we get either $\partial \left(m\right)\in P$ for all $m\in S$ or $[m,h]\in P\mathrm{for}\mathrm{all}m,h\in S$. Define $\mathrm{\Gamma }=\{m\in S:\partial (m)\in P\}$ and $\mathrm{\Lambda }=\{m\in S:[m,h]\in P\mathrm{for}\mathrm{all}h\in S\}$. It is easy to verify that both $\mathrm{\Gamma }$ and $\mathrm{\Lambda }$ are additive subgroups of S, their union equals S, and $\mathrm{\Gamma }\cap \mathrm{\Lambda }=⌀$. By applying Remark 1, we get either $\mathrm{\Gamma }=S$ or $\mathrm{\Lambda }=S$. If $\mathrm{\Gamma }=S$, then $\partial \left(m\right)\in P$ for all $m\in S$. Therefore, (8) becomes $2\mathsf{ϝ}\left(m\right)m\in P$ for all $m\in S$. Since $char(S/P)\ne 2$, the last relation becomes $\mathsf{ϝ}\left(m\right)m\in P$ for all $m\in S$. Therefore, as discussed in Theorem 3, we obtain either $\mathsf{ϝ}\left(S\right)\subseteq P$ or $S/P$ is a commutative integral domain. On the other hand, if $\mathrm{\Lambda }=S$, then $[m,h]\in P\mathrm{for}\mathrm{all}m,h\in S$, and so by Lemma 1, $S/P$ is a commutative integral domain. Thus, (10) becomes $3mn\partial \left(m\right)\in P\mathrm{for}\mathrm{all}m,n\in S.$ That is, $\left(3m\right)S\partial \left(m\right)\subseteq P\mathrm{for}\mathrm{all}m\in S.$ By using the primeness of P, we can easily conclude that either $char(S/P)=3$ or $\partial \left(m\right)\in P\mathrm{for}\mathrm{all}m\in S.$ In the latter case, we obtain $\partial \left(S\right)\subseteq P$. The proof is complete. □

By setting $\mathsf{ϝ}=\partial$ in Theorem 4, we can derive the following corollary without assuming that $char(S/P)\ne 2$.

Corollary 6.

Let P be a prime ideal of a ring S. If S admits a derivation ∂, then S satisfies $\partial \left(m^2\right)+\partial \left(m\right)m\in P$ for all $m\in S$ if and only if $\partial \left(S\right)\subseteq P$ or S is a commutative integral domain with $char(S/P)=3$.

As an application of Theorem 4, we obtain the following corollary in the context of a prime ring:

Corollary 7.

Let S be a prime ring of characteristic not 2. Suppose that S admits a generalized derivation $(\mathsf{ϝ}, \partial )$ such that $\mathsf{ϝ}\left(m^2\right)+\mathsf{ϝ}\left(m\right)m=0$ for all $m\in S$. If $\partial \ne 0$, then S is a commutative and $char\left(S\right)=3$.

The following example will demonstrate the importance of assuming that S is prime in Corollary 7.

Example 3.

Let χ, ϝ and ∂ be as stated in Proposition 1. By using similar techniques and arguments with necessary modifications, it can be shown that $(\mathsf{ϝ}, \partial )$ is a generalized derivation on the non-prime ring χ that satisfies the identity $\mathsf{ϝ}\left(m^2\right)+\mathsf{ϝ}\left(m\right)m=0$ for all $m\in \{m_1,m_2,\dots m_{\mathsf{\ell }}\}\subseteq \chi$. However, χ is non-commutative and its characteristic is not equal to 3. Therefore, in Corollary 7, the assumption that $S=\chi$ is prime cannot be eliminated.

Theorem 5.

Let P be a prime ideal of a ring S with $char(S/P)\ne 2$. If S admits a generalized derivation $(\mathsf{ϝ}, \partial )$, then S satisfies $\mathsf{ϝ}\left(m^2\right)+m\mathsf{ϝ}\left(m\right)\in P$ for all $m\in S$ if and only if one of the following conditions holds:

(i) 

$\partial \left(S\right)\subseteq P$ and $\mathsf{ϝ}\left(S\right)\subseteq P$.

(ii) 

$\partial \left(S\right)\subseteq P$ and $S/P$ is a commutative integral domain.

(iii) 

$S/P$ is a commutative integral domain and $char(S/P)=3$.

Proof. 

Trivially, if any of $\left(i\right)$, $\left(ii\right)$, or $\left(iii\right)$ is true, then S satisfies the identity $\mathsf{ϝ}\left(m^2\right)+m\mathsf{ϝ}\left(m\right)\in P$ for all $m\in S$. To prove the non-trivial case, assume that

$$\mathsf{ϝ}\left(m^2\right)+m\mathsf{ϝ}\left(m\right)\in P\mathrm{for}\mathrm{all}m\in S.$$

(11)

By linearizing (11) and applying it, we find that

$$\mathsf{ϝ}(m\circ n)+m\mathsf{ϝ}\left(n\right)+n\mathsf{ϝ}\left(m\right)\in P\mathrm{for}\mathrm{all}m,n\in S.$$

(12)

Substituting n with $nm$ in (12) and utilizing it, we get

$$(m\circ n)\partial \left(m\right)+mn\partial \left(m\right)+nm\mathsf{ϝ}\left(m\right)-n\mathsf{ϝ}\left(m\right)m\in P\mathrm{for}\mathrm{all}m,n\in S.$$

(13)

By substituting n with $hn$ in the previous equation and applying it, we get

$$[m,h]n\partial \left(m\right)+mhn\partial \left(m\right)-hmn\partial \left(m\right)\in P\mathrm{for}\mathrm{all}m,n,h\in S.$$

That is, $2[m,h]S\partial \left(m\right)\subseteq P\mathrm{for}\mathrm{all}m,h\in S.$ Because $char(S/P)\ne 2$, the previous equation becomes $[m,h]S\partial \left(m\right)\subseteq P\mathrm{for}\mathrm{all}m,h\in S.$ Applying the primeness of P, we get either $\partial \left(m\right)\in P$ for all $m\in S$ or $[m,h]\in P\mathrm{for}\mathrm{all}m,h\in S$. Define $\mathrm{\Gamma }=\{m\in S:\partial (m)\in P\}$ and $\mathrm{\Lambda }=\{m\in S:[m,h]\in P\mathrm{for}\mathrm{all}h\in S\}$. It is easy to check that both $\mathrm{\Gamma }$ and $\mathrm{\Lambda }$ are additive subgroups of S, their union equals S, and $\mathrm{\Gamma }\cap \mathrm{\Lambda }=⌀$. By applying Remark 1, we get either $\mathrm{\Gamma }=S$ or $\mathrm{\Lambda }=S$. Consider that $\mathrm{\Gamma }=S$. Then, $\partial \left(m\right)\in P$ for all $m\in S$. Therefore, (13) reduces to $[m,\mathsf{ϝ}(m\left)\right]\in P\mathrm{for}\mathrm{all}m\in S$. By linearizing the last relation and using it, we find $[m,\mathsf{ϝ}(n\left)\right]+[n,\mathsf{ϝ}(m\left)\right]\in P\mathrm{for}\mathrm{all}m,n\in S.$ By replacing n with $nh$ in the last equation and using it, we obtain $\mathsf{ϝ}\left(n\right)[m,h]+n[h,\mathsf{ϝ}(m\left)\right]\in P\mathrm{for}\mathrm{all}m,n,h\in S.$ By substituting n with $\tau n$ in the last equation and using it, we can deduce that $\mathsf{ϝ}\left(\tau \right)S[m,h]\subseteq P\mathrm{for}\mathrm{all}m,\tau ,h\in S.$ Applying the primeness of P, we can conclude that either $[m,h]\in P\mathrm{for}\mathrm{all}m,h\in S$ or $\mathsf{ϝ}\left(m\right)\in P$ for all $m\in S.$ In the first scenario, $S/P$ is a commutative integral domain, as stated in Lemma 1. The second scenario leads to $\mathsf{ϝ}\left(S\right)\subseteq P$. Alternatively, if $\mathrm{\Lambda }=S$, we have $[m,h]\in P\mathrm{for}\mathrm{all}m,h\in S.$ Therefore, $S/P$ is a commutative integral domain. Hence, (13) becomes $\left(3m\right)S\partial \left(m\right)\subseteq P$ for all $m\in S$. Since P is prime, we can conclude that either $\partial \left(S\right)\subseteq P$ or $char(S/P)=3$. □

By setting $\mathsf{ϝ}=\partial$ in Theorem 5 and removing the constraint that $char(S/P)\ne 2$, we can derive the following corollary:

Corollary 8.

Let P be a prime ideal of a ring S. If S admits a derivation ∂, then S satisfies $\partial \left(m^2\right)+m\partial \left(m\right)\in P$ for all $m\in S$ if and only if $\partial \left(S\right)\subseteq P$ or S is a commutative integral domain with $char(S/P)=3$.

If the ring imposed in Theorem 5 is prime, then we can derive the following corollary:

Corollary 9.

Let S be a prime ring of characteristic not 2. Suppose that S admits a generalized derivation $(\mathsf{ϝ}, \partial )$ such that $\mathsf{ϝ}\left(m^2\right)+m\mathsf{ϝ}\left(m\right)=0$ for all $m\in S$. If $\partial \ne 0$, then S is a commutative and $char\left(S\right)=3$.

4. Conclusions

The classical theorem in ring theory addresses how to make a ring commutative. Many researchers have utilized different methods, such as additive mappings on the ring S that satisfy specific algebraic criteria to achieve this goal. Our study aimed to build upon previous findings in the literature. In order to do so, we removed any restrictions on a ring S and investigated the connection between a factor ring $S/P$ and a generalized derivation (ϝ, ∂) that satisfies certain algebraic identities involving P, a prime ideal of S. Throughout our discussions, we delved into several significant implications of our theorems. Additionally, we have included several non-trivial examples to underscore the importance of the primeness hypothesis of P in our theorems.

5. Open Question

Recently, researchers such as Bruno L.M. Ferreira [2] have made significant contributions to expanding studies on functional identity to broader non-associative settings. Their work has played a key role in connecting associative and non-associative algebraic structures.

In this context, our open question is whether the results obtained in this manuscript still hold true in a broader context, including non-associative structures, specifically alternative rings and algebras.