Weak right η-centralizers of prime rings

Abstract

This work encompasses the study of certain weak maps termed as weak right η-centralizers defined on $\mathfrak{D}$ a noncommutative prime ring to ${\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$. The first section provides an overview of the motivation behind our work. In the second section, our main focus is to establish the characterization of weak right η-centralizers while addressing the theory of functional identities, particularly in the challenging case of low dimensions. Furthermore, in the third section, we derive the characterization of weak Jordan σ-derivations based on the aforementioned characterization.

Keywords:

• Prime ring
• derivation
• maximal left ring of quotients
• polynomial identity (PI)
• generalized polynomial identity (GPI)
• functional identity (FI)

2020 AMS Subject Classifications:

• 16R60
• 16N60
• 16W25
• 16W20

1. Introduction

Until the end of this paper, $\mathfrak{D}$ depicts a prime ring. In the second section of this paper, we always depict $\mathfrak{D}$ with an epimorphism σ. A ring in which for every $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}$, $\mathfrak{g}\mathfrak{D}\mathfrak{h}=0$ yields $\mathfrak{g}=0$ or $\mathfrak{h}=0$ is called prime ring. Let's establish a standardized abbreviation for the maximal left (and similarly, symmetric and right) ring of quotients of $\mathfrak{D}$ as ${\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$, ${\mathfrak{Q}}_{\mathrm{ms}}(\mathfrak{D})$ and ${\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$. Popularly, ${\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ ${\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ and ${\mathfrak{Q}}_{\mathrm{ms}}(\mathfrak{D})$ are also prime rings and all have C as centre regularly named as centroid of $\mathfrak{D}$. See [1], for an in-depth knowledge. Further, for $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}$, $[\mathfrak{g},\mathfrak{h}]$ means the commutator given by $[\mathfrak{g},\mathfrak{h}]=\mathfrak{g}\mathfrak{h}-\mathfrak{h}\mathfrak{g}$ and $\mathfrak{g}o\mathfrak{h}$ means the skew-commutator provided by $\mathfrak{g}o\mathfrak{h}=\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g}$. We abbreviate σ as an isomorphism, which is called inner if there exists a unit element $q\in {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ such that ${\mathfrak{g}}^{\sigma }=q\mathfrak{g}{q}^{-1}$ for all $\mathfrak{g}\in \mathfrak{D}$. If σ is not inner, it is called as outer. We consider throughout σ is non-identity. Since the isomorphism $\sigma$ on $\mathfrak{D}$ can have unique extension as isomorphism of ${\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$, thus $\sigma$ is an isomorphism of C also. The isomorphism $\sigma$ is first kind if it give rise to an identity mapping on C otherwise second kind. Consider a mapping $\mathfrak{E}$ from $\mathfrak{D}$ to ${\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$. $\mathfrak{E}$ the additive map on $\mathfrak{D}$ is referred to as a derivation if it satisfies the equation $\mathfrak{E}(\mathfrak{g}\mathfrak{h})=\mathfrak{g}\mathfrak{E}(\mathfrak{h})+\mathfrak{E}(\mathfrak{g})\mathfrak{h}$ for all $\mathfrak{g}$ and $\mathfrak{h}$ in $\mathfrak{D}$. Further, $\mathfrak{E}$ is called a Jordan derivation if it satisfies the equation $\mathfrak{E}({\mathfrak{g}}^2)=\mathfrak{g}\mathfrak{E}(\mathfrak{g})+\mathfrak{E}(\mathfrak{g})\mathfrak{g}$ for all $\mathfrak{g}$ in $\mathfrak{D}$. Now, introducing a new map $\mathfrak{E}$ from $\mathfrak{D}$ to $Q_{\mathrm{ml}}(\mathfrak{D})$ that fulfils the condition $\mathfrak{E}({\mathfrak{g}}^2)-\mathfrak{g}\mathfrak{E}(\mathfrak{g})-\mathfrak{E}(\mathfrak{g})\mathfrak{g}\in C$ for all $\mathfrak{g}$ in $\mathfrak{D}$, we define it as a weak Jordan derivation.

Let's also introduce another mapping $\mathfrak{G}$ from $\mathfrak{D}$ to $Q_{\mathrm{ml}}(\mathfrak{D})$, following the same additive property. $\mathfrak{G}$ is known as Jordan $\sigma$-derivation (resp. Jordan $\ast$-derivation) if $\mathfrak{G}({\mathfrak{g}}^2)=\mathfrak{G}(\mathfrak{g})\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathfrak{G}(\mathfrak{g})$ (resp. $\mathfrak{G}({\mathfrak{g}}^2)=\mathfrak{G}(\mathfrak{g}){\mathfrak{g}}^{\ast }+\mathfrak{g}\mathfrak{G}(\mathfrak{g})$) for all $\mathfrak{g}\in \mathfrak{D}$. For certain $k\in {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ if $\mathfrak{G}(\mathfrak{g})={\mathfrak{g}}^{\sigma }k-k\mathfrak{g}$ for all $\mathfrak{g}\in \mathfrak{D}$, then Jordan $\sigma$-derivation $\mathfrak{G}$ of $\mathfrak{D}$ is said to be X-inner. If this is not the case, then $\mathfrak{G}$ is called X-outer [2–4].

One can easily support the fact that every derivation is Jordan derivation. Are we free to talk about converse with that ease? To apprehend this Herstein [5, Theorem 2.1] studied and came up with the proof that if we restrict char$(\mathfrak{D})\ne 2$ then a Jordan derivation is merely a derivation on prime ring. In particular, every Jordan derivation of $M_2(\mathrm{GF}(2))$ is a derivation [6, Theorem 3.11]. Lee and Lin [2] studied the case of char$(\mathfrak{D})=2$ and ensured Jordan derivation is a sum of a derivation and a central valued additive map which vanishes on square type elements.

The operator algebras and Lie homomorphisms are intimately related with each other and have been studied by many authors [7,8]. This intrigued researchers to study Lie derivations and its advanced versions in [7,9]. A new concept was introduced which is a generalization of Lie derivation (analogously Jordan derivation) named as weak Lie derivations (resp. weak Jordan derivation). An additive map $L:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ is called a weak Lie derivation (resp. weak Jordan derivation) if $L([x,y])-[L(x),y]-[x,L(y)]\in \mathfrak{C}$ (analogously, $L(x^2)-L(x)x-\mathrm{xL}(x)\in \mathfrak{C}$). This motivated Lin [4] to develop the novel format of weak Jordan derivation on prime rings. Lee et al.  [3,10] shaped the complete form of Jordan $\ast$-derivation. This notable characterization of Jordan $\ast$-derivation helped Siddeeque et al. to work out the characterization of weak Jordan $\ast$-derivation. One may see [11]. Taking these results as backbone to our work, we have worked out the characterization of certain special types of additive maps like weak right η-centralizers, weak Jordan $\sigma$-derivation , etc. See Definitions 1.1 and 1.2.

An additive map $P:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ is a right centralizer if $P(\mathrm{xy})=\mathrm{xP}(y)$ for every $x,y\in \mathfrak{D}.$ This is just equivalent to the statement that there exists certain $\theta \in {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ owing to which $P(x)=\mathrm{x\theta }$ where $\mathfrak{D}$ is a prime ring. For proof visit [12, Lemma 2.1]. Also, if $P(x^2)=\mathrm{xP}(x)$ for every $x\in \mathfrak{D}$ then P is called as Jordan right centralizer. A strong and beneficial work was established by Zalar [13, Proposition 1.4] for any 2-torsion free semiprime ring. Further, this work was carried out by Lee [14, Theorem 1.2] for any arbitrary semiprime ring and he established the same result thereby opening gates to solution of many functional identities.

The novel concepts of weak Lie derivations, weak Lie homomorphisms and usage of Jordan right centralizers in solving non-trivial functional identities are realized by all and grabbing this opportunity, an astounding characterization is given by Lin [4] for weak Jordan derivation. Motivated by this, we introduced a new notion called as weak Jordan right η-centralizer or for brevity let's call weak right η-centralizer defined as below:

Definition 1.1

Consider $\mathfrak{J}:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D}),$ an additive map, such that $\mathfrak{J}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{J}(\mathfrak{g})\in \mathfrak{C}$ for all $\mathfrak{g}\in \mathfrak{D}$, where $\eta =\pm 1$. Then such a map $\mathfrak{J}$ is coined as a weak right η-centralizer. For $\eta =-1$, the operator is simply called as weak right centralizer.

So, we struggled hard to give the characterization of weak right η-centralizers. In Section 2, there is abundant discussion on weak right η-centralizers.

Further, Lee gives the complete characterization of Jordan σ-derivation by using [15, Theorem 2.1] and [3, Theorem 2.3], making extensive use of results on GFI with (anti) automorphisms and derivations [16]. Indeed, [15, Theorem 2.1] deals with a novel FI whereas in [3, Theorem 2.3], a special FI is discussed under the crucial case of not PI-ring. This motivated us to give the following definitions:

Definition 1.2

Consider $\mathfrak{G}:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ an additive map and $\sigma$ an epimorphism of $\mathfrak{D}$. If $\mathfrak{G}({\mathfrak{g}}^2)-\mathfrak{G}(\mathfrak{g})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}$ for all $\mathfrak{g}\in \mathfrak{D}$, then we call $\mathfrak{G}$ as weak Jordan $\sigma$-derivation. Note that the definition can be given while keeping σ as an endomorphism.

The purpose of Section 3 is to provide under what conditions any weak Jordan $\sigma$-derivation is X-inner.

2. Weak right η-centralizers on prime rings

Let $\mathfrak{G}$ be a weak right η-centralizer. We put (1) $\begin{array}{r}\mu (\mathfrak{g})=\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(1) for all $\mathfrak{g}\in \mathfrak{D}$. On linearizing one can establish the following easily: (2) $\begin{array}{r}\lambda (\mathfrak{g},\mathfrak{h})=\mathfrak{G}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{h})+\eta \mathfrak{h}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(2) for all $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}$.

Lemma 2.1

Let $\mathfrak{D}$ be a prime ring with deg$(\mathfrak{D})>2$. Then, every weak right η-centralizer $\mathfrak{G}$ is $\mathcal{Z}(\mathfrak{D})$-linear.

Proof.

Let $f(\mathfrak{g})=\mathfrak{G}(\beta \mathfrak{g})-\beta \mathfrak{G}(\mathfrak{g})$, for $\beta \in \mathcal{Z}(\mathfrak{D})$ and $\mathfrak{g}\in \mathfrak{D}$. Then our aim is to show f is zero. For $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}$, we have $\begin{array}{r}\mathfrak{G}((\beta \mathfrak{g})\mathfrak{h}+\mathfrak{h}(\beta \mathfrak{g}))+\mathrm{\eta \beta }\mathfrak{g}\mathfrak{G}(\mathfrak{h})+\eta \mathfrak{h}\mathfrak{G}(\beta \mathfrak{g})\in \mathfrak{C}.\end{array}$On the other hand, $\begin{array}{r}\mathfrak{G}((\beta \mathfrak{h})\mathfrak{g}+\mathfrak{g}(\beta \mathfrak{h}))+\mathrm{\eta \beta }\mathfrak{h}\mathfrak{G}(\mathfrak{g})+\eta \mathfrak{g}\mathfrak{G}(\beta \mathfrak{h})\\ & \in \mathfrak{C}.\end{array}$On subtracting the above two, we get $\begin{array}{r}\mathfrak{g}(\mathfrak{G}(\beta \mathfrak{h})-\beta \mathfrak{G}(\mathfrak{h}))+\mathfrak{h}(\mathfrak{G}(\beta \mathfrak{g})-\beta \mathfrak{G}(\mathfrak{g}))\in \mathfrak{C}.\end{array}$This implies $\mathfrak{g}f(\mathfrak{h})+\mathfrak{h}f(\mathfrak{g})\in \mathfrak{C}.$ Using [17, Theorem 2.4], we have f=0 for deg$(\mathfrak{D})>2.$

We now discuss the matrix algebra or PI case in the following lemma.

Lemma 2.2

Let $\mathfrak{D}=M_n(\mathcal{F})$ be a matrix ring of order $n\ge 3$ for a field $\mathcal{F}$. Suppose $\mathfrak{G}$ are the additive maps from $\mathfrak{D}$ to itself satisfying $\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathcal{F}$ for all $\mathfrak{g}\in \mathfrak{D}$, where $\eta \ne \pm 1$ a fixed non-zero central element with $\mathfrak{G}$ as $\mathcal{F}$-linear and $\mathfrak{G}(\mathcal{F})\subseteq \mathcal{F}$, having char $\mathcal{F}\ne 2$. Then, $\mathfrak{G}(\mathfrak{g})=0$, for all $\mathfrak{g}\in \mathfrak{D}.$

Proof.

Here we open up with the proof of Lemma by evaluating the required images of unit matrices $e_{\mathrm{ij}}$ owing to map $\mathfrak{G}$.

Step I: On employing 1(1) $\begin{array}{r}\mu (\mathfrak{g})=\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(1) , computing $\mathfrak{G}$ for $\mathfrak{g}=e_{\mathrm{ii}}$, where $i=1,\dots ,n$. For $\mathfrak{g}=e_{11}$, we have $\begin{array}{rl}& \mathfrak{G}(e_{11}^2)+\eta e_{11}\mathfrak{G}(e_{11})\\ & =\left(\begin{array}{cccc}(1+\eta ){\alpha }_{11}^{11}& (1+\eta ){\alpha }_{12}^{11}& \cdots & (1+\eta ){\alpha }_{1n}^{11}\\ \\ {\alpha }_{21}^{11}& {\alpha }_{22}^{11}& \cdots & {\alpha }_{2n}^{11}\\ \\ ⋮& ⋮& \ddots & ⋮\\ \\ {\alpha }_{n1}^{11}& {\alpha }_{n2}^{11}& \cdots & {\alpha }_{\mathrm{nn}}^{11}\end{array}\right)\\ & \in \mathcal{F}.\end{array}$This implies that $\begin{array}{r}(1+\eta ){\alpha }_{11}^{11}={\alpha }_{22}^{11}={\alpha }_{33}^{11}=\cdots ={\alpha }_{\mathrm{nn}}^{11}.\end{array}$So $\begin{array}{rl}\mathfrak{G}(e_{11})& =(\begin{array}{ccccccccccc}{\alpha }_{11}^{11}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{12}^{11}}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{13}^{11}0}& (1+\eta ){\alpha }_{11}^{11}& 00& 0& (1+\eta ){\alpha }_{11}^{11}⋮& ⋮& ⋮0& 0& 0\end{array}\\ & \begin{array}{cc}\cdots & {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{1n}^{11}}\\ \cdots & 0\\ \cdots & 0\\ \ddots & ⋮\\ \cdots & (1+\eta ){\alpha }_{11}^{11}\end{array}).\end{array}$Similarly, we have $\begin{array}{rl}\mathfrak{G}(e_{22})& =(\begin{array}{ccccccccccc}(1+\eta ){\alpha }_{22}^{22}& 0& 0{\displaystyle \frac{(1-\eta )}{2}{\alpha }_{21}^{22}}& {\alpha }_{22}^{22}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{23}^{22}0}& 0& (1+\eta ){\alpha }_{22}^{22}⋮& ⋮& ⋮0& 0& 0\end{array}\\ & \begin{array}{cc}\cdots & 0\\ \cdots & {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{2n}^{22}}\\ \cdots & 0\\ \ddots & ⋮\\ \cdots & (1+\eta ){\alpha }_{22}^{22}\end{array}),\\ \mathfrak{G}(e_{\mathrm{nn}})& =(\begin{array}{ccccccccccccc}(1+\eta ){\alpha }_{\mathrm{nn}}^{\mathrm{nn}}& 0& 00& (1+\eta ){\alpha }_{\mathrm{nn}}^{\mathrm{nn}}& 00& 0& (1+\eta ){\alpha }_{\mathrm{nn}}^{\mathrm{nn}}⋮& ⋮& ⋮0& 0& 0{\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n1}^{\mathrm{nn}}}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n2}^{\mathrm{nn}}}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n3}^{\mathrm{nn}}}\end{array}\\ & \begin{array}{ccc}\cdots & 0& 0\\ \cdots & 0& 0\\ \cdots & 0& 0\\ \ddots & ⋮& \\ \cdots & (1+\eta ){\alpha }_{\mathrm{nn}}^{\mathrm{nn}}& 0\\ \cdots & {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n,n-1}^{\mathrm{nn}}}& {\alpha }_{\mathrm{nn}}^{\mathrm{nn}}\end{array}).\end{array}$Step II: Since $\mathfrak{G}(I_n)=\mathfrak{G}(e_{11})+\mathfrak{G}(e_{22})+\mathfrak{G}(e_{33})+\cdots +\mathfrak{G}(e_{\mathrm{nn}})\in \mathcal{F}$, we have $\begin{array}{rl}& (\begin{array}{ccccc}{\alpha }_{11}^{11}+(1+\eta ){\displaystyle \sum _{i=2}^n{\alpha }_{\mathrm{ii}}^{\mathrm{ii}}}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{12}^{11}{\displaystyle \frac{(1-\eta )}{2}{\alpha }_{21}^{22}}}& {\alpha }_{22}^{22}+(1+\eta ){\displaystyle \sum _{i=1,i\ne 2}^n{\alpha }_{\mathrm{ii}}^{\mathrm{ii}}⋮}& ⋮{\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n1}^{\mathrm{nn}}}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n2}^{\mathrm{nn}}}\end{array}\\ & \begin{array}{cc}\cdots & {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{1n}^{11}}\\ \cdots & {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{2n}^{22}}\\ \ddots & ⋮\\ \cdots & {\alpha }_{\mathrm{nn}}^{\mathrm{nn}}+(1+\eta ){\displaystyle \sum _{i=1,i\ne n}^n{\alpha }_{\mathrm{ii}}^{\mathrm{ii}}}\end{array})\in \mathcal{F}.\end{array}$Step III:

From the above we conclude $\begin{array}{r}{\alpha }_{\mathrm{kk}}^{\mathrm{kk}}+(1+\eta )\sum _{i=1,i\ne k}^n{\alpha }_{\mathrm{ii}}^{\mathrm{ii}}={\alpha }_{k^{\prime }{k}^{\prime }}^{k^{\prime }{k}^{\prime }}+(1+\eta )\sum _{i=1,i\ne k^{\prime }}^n{\alpha }_{\mathrm{ii}}^{\mathrm{ii}},\end{array}$for all $k=1,\dots ,n$. Taking two at a time and solving for all $1\le k,k^{\prime }\le n$, we get $\begin{array}{r}{\alpha }_{11}^{11}={\alpha }_{22}^{22}=\cdots ={\alpha }_{\mathrm{nn}}^{\mathrm{nn}}.\end{array}$So, we have $\begin{array}{rl}\mathfrak{G}(e_{11})& =(\begin{array}{ccccccccccc}{\alpha }_{11}^{11}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{12}^{11}}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{13}^{11}0}& (1+\eta ){\alpha }_{11}^{11}& 00& 0& (1+\eta ){\alpha }_{11}^{11}⋮& ⋮& ⋮0& 0& 0\end{array}\\ & \begin{array}{cc}\cdots & {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{1n}^{11}}\\ \cdots & 0\\ \cdots & 0\\ \ddots & ⋮\\ \cdots & (1+\eta ){\alpha }_{11}^{11}\end{array}).\end{array}$Similarly, we have $\begin{array}{rl}\mathfrak{G}(e_{22})& =(\begin{array}{ccccccccccc}(1+\eta ){\alpha }_{11}^{11}& 0& 0{\displaystyle \frac{(1-\eta )}{2}{\alpha }_{21}^{22}}& {\alpha }_{11}^{11}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{23}^{22}0}& 0& (1+\eta ){\alpha }_{11}^{11}⋮& ⋮& ⋮0& 0& 0\end{array}\\ & \begin{array}{cc}\cdots & 0\\ \cdots & {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{2n}^{22}}\\ \cdots & 0\\ \ddots & ⋮\\ \cdots & (1+\eta ){\alpha }_{11}^{11}\end{array}),\\ \mathfrak{G}(e_{\mathrm{nn}})& =(\begin{array}{ccccccccccccc}(1+\eta ){\alpha }_{11}^{11}& 0& 00& (1+\eta ){\alpha }_{11}^{11}& 00& 0& (1+\eta ){\alpha }_{11}^{11}⋮& ⋮& ⋮0& 0& 0{\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n1}^{\mathrm{nn}}}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n2}^{\mathrm{nn}}}& {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n3}^{\mathrm{nn}}}\end{array}\\ & \begin{array}{ccc}\cdots & 0& 0\\ \cdots & 0& 0\\ \cdots & 0& 0\\ \ddots & ⋮& \\ \cdots & (1+\eta ){\alpha }_{11}^{11}& 0\\ \cdots & {\displaystyle \frac{(1-\eta )}{2}{\alpha }_{n,n-1}^{\mathrm{nn}}}& {\alpha }_{11}^{11}\end{array}).\end{array}$Step IV: By using (1(1) $\begin{array}{r}\mu (\mathfrak{g})=\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(1) ), first calculate $\mathfrak{G}$ for each $\mathfrak{g}=e_{\mathrm{ij}}$, where $i\ne j$ and take help of (2(2) $\begin{array}{r}\lambda (\mathfrak{g},\mathfrak{h})=\mathfrak{G}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{h})+\eta \mathfrak{h}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(2) ) to find $\mathfrak{G}$ for such $\mathfrak{g}$ with $\mathfrak{h}=e_{\mathrm{jj}}$.

For instance, by (1(1) $\begin{array}{r}\mu (\mathfrak{g})=\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(1) ) for $\mathfrak{g}=e_{12}$, we have $\begin{array}{r}-\eta e_{12}\mathfrak{G}(e_{12})=\mathfrak{G}(e_{12}^2)-\mu (e_{12})=-\mu (e_{12})\in \mathcal{F}.\end{array}$That is, $\begin{array}{r}{e}_{12}\mathfrak{G}(e_{12})=\left(\begin{array}{ccccc}{\alpha }_{21}^{12}& {\alpha }_{22}^{12}& {\alpha }_{23}^{12}& \cdots & {\alpha }_{2n}^{12}\\ 0& 0& 0& \cdots & 0\\ 0& 0& 0& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& & 0\end{array}\right)\in \mathcal{F}.\end{array}$Thus $\begin{array}{r}{\alpha }_{21}^{12}={\alpha }_{22}^{12}={\alpha }_{23}^{12}=\cdots ={\alpha }_{2n}^{12}=0,\end{array}$and so $\begin{array}{r}\mathfrak{G}(e_{12})=\left(\begin{array}{ccccc}{\alpha }_{11}^{12}& {\alpha }_{12}^{12}& {\alpha }_{13}^{12}& \cdots & {\alpha }_{1n}^{12}\\ 0& 0& 0& \cdots & 0\\ {\alpha }_{31}^{12}& {\alpha }_{32}^{12}& {\alpha }_{33}^{12}& \cdots & {\alpha }_{3n}^{12}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ {\alpha }_{n1}^{12}& {\alpha }_{n2}^{12}& {\alpha }_{n3}^{12}& \cdots & {\alpha }_{\mathrm{nn}}^{12}\end{array}\right).\end{array}$Let $(\mathfrak{g},v)=(e_{12},e_{11})$. By (2(2) $\begin{array}{r}\lambda (\mathfrak{g},\mathfrak{h})=\mathfrak{G}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{h})+\eta \mathfrak{h}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(2) ), we have $\begin{array}{rl}& \mathfrak{G}(e_{12}{e}_{11}+e_{11}{e}_{12})+\eta e_{11}\mathfrak{G}(e_{12})+\eta e_{12}\mathfrak{G}(e_{11})\\ & =\lambda (e_{12},e_{11})\in \mathcal{F}.\\ & (\begin{array}{ccc}(1+\eta ){\alpha }_{11}^{12}& (1+\eta )({\alpha }_{12}^{12}+{\alpha }_{11}^{11})& (1+\eta ){\alpha }_{13}^{12}\\ 0& 0& 0\\ {\alpha }_{31}^{12}& {\alpha }_{32}^{12}& {\alpha }_{33}^{12}\\ ⋮& ⋮& ⋮\\ {\alpha }_{n1}^{12}& {\alpha }_{n2}^{12}& {\alpha }_{n3}^{12}\end{array}\\ & \begin{array}{cc}\cdots & (1+\eta ){\alpha }_{1n}^{12}\\ \cdots & 0\\ \cdots & {\alpha }_{3n}^{12}\\ \ddots & ⋮\\ \cdots & {\alpha }_{\mathrm{nn}}^{12}\end{array})\in \mathcal{F}.\end{array}$Thus $\mathfrak{G}(e_{12})=-{\alpha }_{11}^{11}{e}_{12}.$ On treading similar path, one can observe that $\begin{array}{r}\mathfrak{G}(e_{\mathrm{ij}})=-{\alpha }_{11}^{11}{e}_{\mathrm{ij}},\mathrm{for}i\ne j.\end{array}$Let $(\mathfrak{g},\mathfrak{h})=(e_{12},e_{31})$, we have $\begin{array}{rl}& \mathfrak{G}(e_{12}{e}_{31}+e_{31}{e}_{12})+\eta e_{31}\mathfrak{G}(e_{12})+\eta e_{12}\mathfrak{G}(e_{31})\\ & =\lambda (e_{12},e_{31})\in \mathcal{F}.\end{array}$This is reduced to the following: $\begin{array}{r}-(1+\eta ){\alpha }_{11}^{11}{e}_{32}\in \mathcal{F},\mathrm{i}.\mathrm{e}.{\alpha }_{11}^{11}=0.\end{array}$Therefore, $\mathfrak{G}(e_{\mathrm{ij}})=0,\text{for all}i\ne j.$

Let $(\mathfrak{g},\mathfrak{h})=(e_{\mathrm{ij}},e_{\mathrm{ji}})$ for $i\ne j$. By (2(2) $\begin{array}{r}\lambda (\mathfrak{g},\mathfrak{h})=\mathfrak{G}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{h})+\eta \mathfrak{h}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(2) ), we have $\begin{array}{rl}& \mathfrak{G}(e_{\mathrm{ij}}{e}_{\mathrm{ji}}+e_{\mathrm{ji}}{e}_{\mathrm{ij}})+\eta e_{\mathrm{ij}}\mathfrak{G}(e_{\mathrm{ji}})+\eta e_{\mathrm{ji}}\mathfrak{G}(e_{\mathrm{ij}})\\ & =\lambda (e_{\mathrm{ij}},e_{\mathrm{ji}})\in \mathcal{F}.\end{array}$This provides $\begin{array}{r}\mathfrak{G}(e_{\mathrm{ii}})+\mathfrak{G}(e_{\mathrm{jj}})\in \mathcal{F}\end{array}$Likewise, we have $\begin{array}{r}\mathfrak{G}(e_{\mathrm{ii}})+\mathfrak{G}(e_{\mathrm{kk}})\in \mathcal{F}\text{and}\mathfrak{G}(e_{\mathrm{jj}})+\mathfrak{G}(e_{\mathrm{kk}})\in \mathcal{F}\end{array}$for distinct i,j,k and hence $\mathfrak{G}(e_{\mathrm{ii}})\in \mathcal{F}$.

This gives ${\alpha }_{21}^{22}=0={\alpha }_{23}^{22}=\dots ={\alpha }_{2n}^{22}$ as $\mathfrak{G}(e_{11})\in \mathcal{F}$.

Thus proceeding on the similar lines we have $\mathfrak{G}(e_{\mathrm{ij}})=0,$ for all $1\le i,j\le n.$ Thus in all $\mathfrak{G}(\mathfrak{g})=0$ for all $\mathfrak{g}\in \mathfrak{D}.$

Theorem 2.1

Let $\mathfrak{D}$ be a prime PI-ring with deg $\mathfrak{D}>2$ and let $\mathfrak{G}:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ be an additive map satisfying $\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}$ for all $\mathfrak{g}\in \mathfrak{D}$, where $\eta \ne \pm 1$ a fixed non-zero central element. Then $\mathfrak{G}(\mathfrak{g})=0$ for all $\mathfrak{g}\in \mathfrak{D}$ if $\mathfrak{G}(\mathcal{Z}\mathcal{(}\mathfrak{D}\mathcal{)})\subseteq C$, where char$(\mathfrak{D})\ne 2$.

Proof.

In a nutshell, if deg$\mathfrak{D}>2$, Lemma 2.1 yields a $\mathfrak{Z}(\mathfrak{D})$-linear map $\mathfrak{G}$. We regard for $Z(\mathfrak{D})$, the quotient field is $\mathfrak{C}$. Furthermore, $\mathfrak{G}$ has unique extension to a map $\overline{\mathfrak{G}}:\mathfrak{D}\mathfrak{C}\to \mathfrak{D}\mathfrak{C}$, where dim${}_C\mathfrak{D}\mathfrak{C}\ge 9$, as defined here: $\overline{\mathfrak{G}}(\frac{\mathfrak{g}}{\beta })=\frac{\mathfrak{G}(\mathfrak{g})}{\beta }$ for $\mathfrak{g}\in \mathfrak{D}$ and $\beta \in \mathfrak{Z}(\mathfrak{D})-\{0\}$. We can easily justify that $\overline{\mathfrak{G}}$ is a weak right η-centralizer map and a $\mathfrak{C}$-linear as well because $\mathfrak{G}$ is $\mathfrak{Z}(\mathfrak{D})$-linear. Let $\mathcal{F}$ depicts the algebraic closure of C. Besides, $\overline{\mathfrak{G}}$ has the extension as a weak right η-centralizer on $\mathfrak{D}\mathfrak{C}{\otimes }_{\mathfrak{C}}\mathcal{F}$ say $\tilde{\mathfrak{G}}$ and with the following definition: $\begin{array}{r}\tilde{\mathfrak{G}}(\sum _i{\mathfrak{g}}_i\otimes {\alpha }_i)=\sum _i\overline{\mathfrak{G}}({\mathfrak{g}}_i)\otimes {\alpha }_i,\end{array}$where ${\mathfrak{g}}_i\in \mathfrak{D}\mathfrak{C}$ also ${\alpha }_i$ belongs to $\mathcal{F}$, where $\mathfrak{D}\mathfrak{C}{\otimes }_{\mathfrak{C}}\mathcal{F}\cong M_k(\mathcal{F})$ and $k=\sqrt{{\text{dim}}_{\mathfrak{C}}\mathfrak{D}\mathfrak{C}}>2$. Thus $\tilde{\mathfrak{G}}$ is an additive map on $M_k(\mathcal{F})$ to itself, where k>2. Also, we have $\begin{array}{rl}& \tilde{\mathfrak{G}}({\mathfrak{g}}^2)+\eta \mathfrak{g}\tilde{\mathfrak{G}}(\mathfrak{g})\in \mathcal{F},\tilde{\mathfrak{G}}(\alpha \mathfrak{g})=\alpha \tilde{\mathfrak{G}}(\mathfrak{g}),\\ & \text{and}\tilde{\mathfrak{G}}(\mathcal{F})\subseteq \mathcal{F},\end{array}$for all $\mathfrak{g}\in M_k(\mathcal{F})$ and $\alpha \in \mathcal{F}$. As we are reduced to the matrix algebra case so now following Lemma 2.2, we have $\tilde{\mathfrak{G}}(\mathfrak{g})=0$ which enables $\mathfrak{G}(\mathfrak{g})=0$ for all $\mathfrak{g}\in M_k(\mathcal{F})$.

We are now ready to state the first pivotal result of this section.

Theorem 2.2

Let $\mathfrak{D}$ be a prime ring and let $\mathfrak{G}$ be a weak right η-centralizer. Then the following results prevail:

1. For $\eta =-2$, if $\mathfrak{D}$ is PI-ring, then $\mathfrak{G}(\mathfrak{g})=0$ for all $\mathfrak{g}\in \mathfrak{D}$ otherwise there exists some $q\in {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ such that $\mathfrak{G}(\mathfrak{g})=\mathfrak{g}q$, for all $\mathfrak{g}\in \mathfrak{D}$.

2. For $\eta =-1$, there exists some $q\in {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ such that $\mathfrak{G}(\mathfrak{g})=\mathfrak{g}q$, for all $\mathfrak{g}\in \mathfrak{D}$.

3. For $\eta \ne -1,-2$,

   1. $\mathfrak{G}(\mathfrak{g})=\mathfrak{g}q$, for all $\mathfrak{g}\in \mathfrak{D}$ if $\mathfrak{D}$ is not a PI-ring;

   2. in case of PI-ring char$(\mathfrak{D})=2$, we have $\mathfrak{G}=0$ unless $\mathrm{deg}(\mathfrak{D})=2;$

   3. in case of PI-ring char$(\mathfrak{D})\ne 2$, there exists some $q\in {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ such that $\mathfrak{G}(\mathfrak{g})=\mathfrak{g}q$, for all $\mathfrak{g}\in \mathfrak{D}$ unless $\mathrm{deg}(\mathfrak{D})=2.$

Proof.

We bifurcate our proof in two cases:

1. When $\mathfrak{D}$ is not a PI-ring.

   Let $F:\mathfrak{D}\times \mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ be a bi-additive map defined by $F(\mathfrak{g},\mathfrak{h})=\mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})=\mathcal{H}(\mathfrak{g}\mathfrak{h})+\mathcal{H}(\mathfrak{h}\mathfrak{g})$ for $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}.$ [18, Lemma 2.3], for all $w,\mathfrak{g},\mathfrak{h},z\in \mathfrak{D}$, we obtain $\begin{array}{r}F(\mathfrak{g}w,\mathfrak{h}z)-F(\mathfrak{g},w\mathfrak{h}z)=F(z\mathfrak{g}w,\mathfrak{h})-F(z\mathfrak{g},w\mathfrak{h}).\end{array}$From (2(2) $\begin{array}{r}\lambda (\mathfrak{g},\mathfrak{h})=\mathfrak{G}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{h})+\eta \mathfrak{h}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(2) ), it follows that $\begin{array}{rl}& \eta (\mathfrak{g}w\mathfrak{G}(\mathfrak{h}z)+\mathfrak{h}z\mathfrak{G}(\mathfrak{g}w))-\eta (\mathfrak{g}\mathfrak{G}(w\mathfrak{h}z)+w\mathfrak{h}z\mathfrak{G}(\mathfrak{g}))\\ & -\eta (z\mathfrak{g}w\mathfrak{G}(\mathfrak{h})+\mathfrak{h}\mathfrak{G}(z\mathfrak{g}w))+\eta (z\mathfrak{g}\mathfrak{G}(w\mathfrak{h})\\ & +w\mathfrak{h}\mathfrak{G}(z\mathfrak{g}))\in \mathfrak{C}.\end{array}$ This implies that $\begin{array}{rl}& \mathfrak{g}[w(\mathfrak{G}(\mathfrak{h}z)-\mathfrak{G}(w\mathfrak{h}z))]+\mathfrak{h}[z\mathfrak{G}(\mathfrak{g}w)-\mathfrak{G}(z\mathfrak{g}w)]\\ & +z[\mathfrak{g}(\mathfrak{G}(w\mathfrak{h})-w\mathfrak{G}(\mathfrak{h}))]\\ & +w[\mathfrak{h}\mathfrak{G}(z\mathfrak{g})-\mathfrak{h}z\mathfrak{G}(\mathfrak{g})]\in \mathfrak{C}.\end{array}$ Using [17, Theorem 2.4], we have $\mathfrak{h}\mathfrak{G}(z\mathfrak{g})-\mathfrak{h}z\mathfrak{G}(\mathfrak{g})=0$ for all $\mathfrak{g},\mathfrak{h},z\in \mathfrak{D}.$ Again using [17, Theorem 2.5], we have $\mathfrak{G}(z\mathfrak{g})=z\mathfrak{G}(\mathfrak{g})$ for all $\mathfrak{g},\mathfrak{h},z\in \mathfrak{D}$. Clearly, we have $q\in {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ (see [12, Lemma 2.1]). Hence, $\mathfrak{G}$ is settled as asserted. This result holds well on deg$(\mathfrak{D})>4.$

2. When $\mathfrak{D}$ is a PI-ring.

   1. subcase I:First we consider char($\mathfrak{D})=2$:Then the relation (1(1) $\begin{array}{r}\mu (\mathfrak{g})=\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(1) ) reduces to (3) $\begin{array}{r}\mu (\mathfrak{g})=\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}.\end{array}$(3) By [19, Theorem 2], we see that there exists a non zero $\alpha \in \mathcal{Z}(\mathfrak{D})$, such that $\begin{array}{r}0=\mathfrak{G}([\alpha ,\mathfrak{g}])=\mathrm{\eta \alpha }\mathfrak{G}(\mathfrak{g})+\eta \mathfrak{g}\mathfrak{G}(\alpha )\in \mathfrak{C}.\end{array}$This implies $\mathfrak{G}(\mathfrak{g})=\mathfrak{g}q+\zeta (\mathfrak{g}),$ where $q={\alpha }^{-1}\mathfrak{G}(\alpha )$ and $\zeta (\mathfrak{g})\in \mathfrak{C}$, for all $\mathfrak{g}\in \mathfrak{D}$.Now, substitute $\mathfrak{G}(\mathfrak{g})=\mathfrak{g}q+\zeta (\mathfrak{g})$ in (3(3) $\begin{array}{r}\mu (\mathfrak{g})=\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}.\end{array}$(3) ), we get $\begin{array}{r}\mathfrak{g}(\mathfrak{g}q(\eta +1)+\mathrm{\eta \zeta }(\mathfrak{g}))\in \mathfrak{C}.\end{array}$From the above, we have q=0 or $\eta =-1$ unless deg$(\mathfrak{D})=2.$ If $q=0,$ then this implies $\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}$, which forces $\mathfrak{G}(\mathfrak{g})=0$ for all $\mathfrak{g}\in \mathfrak{D}$. For $\eta =-1$, we have $\zeta (\mathfrak{g})=0$ thus, it asserted $\mathfrak{G}$ to be of the form $\mathfrak{G}(\mathfrak{g})=\mathfrak{g}q$ for all $\mathfrak{g}\in \mathfrak{D}$.

   2. subcase II: Now, we consider char$(\mathfrak{D})\ne 2$:Owing to (2(2) $\begin{array}{r}\lambda (\mathfrak{g},\mathfrak{h})=\mathfrak{G}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{h})+\eta \mathfrak{h}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(2) ), we obtain $\begin{array}{rl}& \lambda (\beta \mathfrak{g},\mathfrak{g})=2\mathfrak{G}(\beta {\mathfrak{g}}^2)+\mathrm{\eta \beta }\mathfrak{g}\mathfrak{G}(\mathfrak{g})\\ & +\eta \mathfrak{g}\mathfrak{G}(\beta \mathfrak{g})\in \mathfrak{C}.\\ & \lambda (\beta ,{\mathfrak{g}}^2)=2\mathfrak{G}(\beta {\mathfrak{g}}^2)+\mathrm{\eta \beta }\mathfrak{G}({\mathfrak{g}}^2)\\ & +\eta {\mathfrak{g}}^2\mathfrak{G}(\beta )\in \mathfrak{C}.\end{array}$ Comparing last two equations, we obtain $\begin{array}{rl}& \beta (\mathfrak{G}({\mathfrak{g}}^2)-\mathfrak{g}\mathfrak{G}(\mathfrak{g}))\\ & +\mathfrak{g}(-\mathfrak{G}(\beta \mathfrak{g})+\mathfrak{g}\mathfrak{G}(\beta ))\in \mathfrak{C}.\end{array}$ The above relation reduces to the following by employing (1(1) $\begin{array}{r}\mu (\mathfrak{g})=\mathfrak{G}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{G}(\mathfrak{g})\in \mathfrak{C}\end{array}$(1) ) and $\mathfrak{Z}(\mathfrak{D})$-linearity of $\mathfrak{G}$, we have $\begin{array}{r}\mathfrak{g}(-\beta (\eta +2)\mathfrak{G}(\mathfrak{g})+\mathfrak{g}\mathfrak{G}(\beta ))\in \mathfrak{C}.\end{array}$Thus we obtain for $\eta \ne -2,$ $\begin{array}{rl}& -\beta (\eta +2)\mathfrak{G}(\mathfrak{g})+\mathfrak{g}\mathfrak{G}(\beta )=0,\\ & \text{unless}\mathrm{deg}(\mathfrak{D})=2.\end{array}$ This gives the following form of $\mathfrak{G}$: $\begin{array}{r}\mathfrak{G}(\mathfrak{g})=\mathfrak{g}q,\text{unless}\mathrm{deg}(\mathfrak{D})=2.\end{array}$For $\eta =-2,$ $\mathfrak{G}(\beta )=0,$ this invokes the Theorem 2.1 to obtain $\mathfrak{G}(\mathfrak{g})=0.$

Some numerical applications of Theorems 2.1 and  2.2

Suppose $\mathfrak{F}$ be a derivation on $\mathfrak{D}$ satisfying $\mathfrak{F}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{F}(\mathfrak{g})\in \mathfrak{C}.$ By employing Theorem 2.2 (which obviously involves Theorem 2.1), we have various conclusions which we will deal one by one.

1. For $\eta =-2$, if $\mathfrak{D}$ is PI-ring, then $\mathfrak{F}(\mathfrak{g})=0$ for all $\mathfrak{g}\in \mathfrak{D}$ otherwise there exists some $q\in {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ such that $\mathfrak{F}(\mathfrak{g})=\mathfrak{g}q$, for all $\mathfrak{g}\in \mathfrak{D}$. Now we put $\mathfrak{F}(\mathfrak{g})=\mathfrak{g}q$ in $\mathfrak{F}({\mathfrak{g}}^2)+\eta \mathfrak{g}\mathfrak{F}(\mathfrak{g})\in \mathfrak{C}.$ Thus we have ${\mathfrak{g}}^{2}q-2{\mathfrak{g}}^{2}q\in \mathfrak{C},\text{i.e.}q=0$. Finally, we have $\mathfrak{F}=0.$

2. For $\eta =-1$, there exists some $q\in {\mathfrak{Q}}_{\mathrm{ml}}(\mathfrak{D})$ such that $\mathfrak{F}(\mathfrak{g})=\mathfrak{g}q$, for all $\mathfrak{g}\in \mathfrak{D}$. Since derivation maps centre to centre, we have $\mathfrak{F}(\mathfrak{g})=\mathfrak{g}q$, for all $\mathfrak{g}\in \mathfrak{D}$ and $q\in \mathfrak{C}.$

3. For $\eta \ne -1,-2$,

   1. $\mathfrak{F}(\mathfrak{g})=\mathfrak{g}q$, for all $\mathfrak{g}\in \mathfrak{D}$ and $q\in \mathfrak{C}$, if $\mathfrak{D}$ is not a PI-ring;

   2. in case of PI-ring char$(\mathfrak{D})=2$, we have $\mathfrak{F}=0$ unless $\mathrm{deg}(\mathfrak{D})=2;$

   3. in case of PI-ring char$(\mathfrak{D})\ne 2$, there exists some $q\in \mathfrak{C}$ such that $\mathfrak{F}(\mathfrak{g})=\mathfrak{g}q$, for all $\mathfrak{g}\in \mathfrak{D}$ unless $\mathrm{deg}(\mathfrak{D})=2.$

3. Weak Jordan σ-derivations on prime rings

Suppose from $\mathfrak{D}$ to ${\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$, $\mathcal{H}$ be a weak Jordan σ-derivation.

For every $\mathfrak{g}\in \mathfrak{D}$, we have (4) $\begin{array}{r}\mu (\mathfrak{g})=\mathcal{H}({\mathfrak{g}}^2)-\mathcal{H}(\mathfrak{g})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}.\end{array}$(4) The linearization yields (5) $\begin{array}{rl}\lambda (\mathfrak{g},\mathfrak{h})& =\mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})-\mathcal{H}(\mathfrak{g})\mathfrak{h}-\mathcal{H}(\mathfrak{h})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})\\ & -{\mathfrak{h}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}\end{array}$(5) for all $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}.$

To strengthen the term weak Jordan σ-derivations, we have developed the following example.

Example 3.1

Consider $\mathcal{R}=\mathbf{Z}[i][x]$ be the polynomial ring over the ring of Gaussian integers $\mathbf{Z}[i]$. Then ${\mathfrak{Q}}_{\mathrm{mr}}(\mathcal{R})$ is $\mathbf{Q}[i](x)$, which is the function field of $\mathbf{Q}[i][x]$, where $\mathbf{Q}$ is the field of rationals. Here, the extended centroid C is ${\mathfrak{Q}}_{\mathrm{mr}}(\mathcal{R})$. Let the homomorphism on the ring $\mathcal{R}$ be denoted by σ and defined as the conjugate map. Then every additive map $\mathfrak{G}$ on $\mathcal{R}$ is a weak Jordan σ-derivation.

Lemma 3.1

Consider $q_1,q_2,\dots ,q_n\in {\mathfrak{Q}}_{\mathrm{mr}}$ be $\mathfrak{C}$-linearly independent and assume $q_1^{\prime },q_2^{\prime },\dots ,q_n^{\prime }\in {\mathfrak{Q}}_{\mathrm{mr}}$ with $q_1^{\prime }\ne 0$. If $Y=\{\sum _{i=1}^m{q}_i^{\prime }\mathfrak{g}{q}_i|\mathfrak{g}\in {\mathfrak{Q}}_{\mathrm{mr}}\}$ is finite dimensional linear space over $\mathfrak{C}$, then $Y\ne 0.$

Lemma 3.2

[20, Remark 2.1]

Let $a,b\in {\mathfrak{Q}}_{\mathrm{mr}}$ and $\mathcal{H}$ be a generalized derivation of $\mathfrak{D}$. If I is a non-zero ideal such that $a(\mathcal{H}(\mathfrak{g})b{)}^n=0$ for all $\mathfrak{g}\in I$, where n be a fixed positive integer, then

1. a=0 or b=0;

2. there exists $a_1,b_1\in {\mathfrak{Q}}_{\mathrm{mr}}$ so that $\mathcal{H}(x)=a_{1}x+x{b}_1,$ where either $b_{1}b=0$ and $b{a}_1=0$ or $a{a}_1=0$.

Lemma 3.3

[15, Lemma 4.3]

For finitely many $p_i,q_i\in {\mathfrak{Q}}_{\mathrm{mr}}$, the following are equivalent:

1. $\sum _{i=1}^m{p}_i[\mathfrak{g},\mathfrak{h}]q_i=0$, for all $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}$;

2. $\sum _{i=1}^m{q}_i\mathfrak{g}{p}_i\in \mathfrak{C}$ for all $\mathfrak{g}\in \mathfrak{D}$.

Lemma 3.4

Consider the non-commutative prime ring $\mathfrak{D}$. If $q_1,q_2$ and $q_3\in {\mathfrak{Q}}_{\mathrm{mr}}$ and $q_1\mathfrak{g}+q_2\mathfrak{g}{q}_3\in \mathfrak{C}$, for every $\mathfrak{g}\in \mathfrak{D}.$ If $q_1\ne 0$, then $q_3\in \mathfrak{C}$ and $q_1+q_2{q}_3=0$ otherwise either $q_2=0$ or $q_3=0$.

Proof.

Our assumption states that $q_1\mathfrak{g}+q_2\mathfrak{g}{q}_3\in \mathfrak{C}$ for every $\mathfrak{g}\in \mathfrak{D}.$ Take $q_1\ne 0$ and suppose on the contrary that $q_3\notin \mathfrak{C},\text{i.e.}1$ , and $q_3$ are linearly independent over $\mathfrak{C}$.

Then owing to Lemma 3.1, one can pick $0\ne \theta \in \mathfrak{C}$ for certain $\mathfrak{g}={\mathfrak{g}}_0$ so that $\begin{array}{r}\theta =q_1{\mathfrak{g}}_0+q_2{\mathfrak{g}}_0{q}_3\in \mathfrak{C},\text{for every}\mathfrak{g},\mathfrak{h}\in \mathfrak{D}.\end{array}$Also on replacing $\mathfrak{g}\in \mathfrak{D}$ by $\mathfrak{g}\mathfrak{h}\in \mathfrak{D}$, we have $\begin{array}{r}{q}_1\mathfrak{g}\mathfrak{h}+q_2\mathfrak{g}\mathfrak{h}{q}_3\in \mathfrak{C},\text{for every}\mathfrak{g},\mathfrak{h}\in \mathfrak{D}.\end{array}$Using previous two relations, we find $(\theta -q_2{\mathfrak{g}}_0{q}_3)\mathfrak{h}+q_2{\mathfrak{g}}_0\mathfrak{h}{q}_3\in \mathfrak{C}$. This can be rewritten as $\theta \mathfrak{h}+q_2{\mathfrak{g}}_0[\mathfrak{h},q_3]\in \mathfrak{C}$, for every $\mathfrak{h}\in \mathfrak{D}$. Since $\mathfrak{D}$ and ${\mathfrak{Q}}_{\mathrm{mr}}$ satisfy same GPI, one can see $\theta \mathfrak{h}+q_2{\mathfrak{g}}_0[\mathfrak{h},q_3]\in \mathfrak{C}$, for every $\mathfrak{h}\in {\mathfrak{Q}}_{\mathrm{mr}}$. Replace $\mathfrak{h}$ by $q_3$, we have $\theta q_3\in \mathfrak{C}$. Hence, $q_3\in \mathfrak{C}$.

If $q_1=0$, then due to our assumption $q_2\mathfrak{g}{q}_3\in \mathfrak{C}$, for every $\mathfrak{g}\in \mathfrak{D}.$ From Lemma 3.3, we have $q_3[\mathfrak{g},\mathfrak{h}]q_2=0$, for every $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}$. Fix $\mathfrak{h}={\mathfrak{h}}_0\in \mathfrak{D}$ in previous relation so that $q_3[\mathfrak{g},{\mathfrak{h}}_0]q_2=0$, for every $\mathfrak{g}\in \mathfrak{D}$. Now use Lemma 3.2 such that there exists ${\mathfrak{h}}_0\in {\mathfrak{Q}}_{\mathrm{mr}}$ so that either ${\mathfrak{h}}_0{q}_2=0$ and $q_2{\mathfrak{h}}_0=0$ or $q_3{\mathfrak{h}}_0=0$. By choosing ${\mathfrak{h}}_0$ to be arbitrary, we have $\mathfrak{h}{q}_2=0$ or $q_3\mathfrak{h}=0$ for every $\mathfrak{h}\in \mathfrak{D}$. Hence, either $q_2=0$ or $q_3=0.$

Theorem 3.1

Suppose $\mathfrak{D}$ is a prime ring which is noncommutative in structure and σ is an endomorphism of $\mathfrak{D}$. If σ is not injective, then every weak Jordan σ-derivation $\mathcal{H}$ of $\mathfrak{D}$ is X-inner σ-derivation.

Proof.

We start the proof with the abbreviation of the kernel of σ by ker $(\sigma )$. Due to noninjectiveness of σ, ker $(\sigma )$ is a non-zero ideal of $\mathfrak{D}$. Let $\mathfrak{g}\in \ker (\sigma )$. Then from (4(4) $\begin{array}{r}\mu (\mathfrak{g})=\mathcal{H}({\mathfrak{g}}^2)-\mathcal{H}(\mathfrak{g})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}.\end{array}$(4) ), we have $\begin{array}{r}\mathcal{H}({\mathfrak{g}}^2)-\mathcal{H}(\mathfrak{g})\mathfrak{g}\in \mathfrak{C}.\end{array}$It follows from [17, Proposition 2.1.10] that ${\mathfrak{Q}}_{\mathrm{mr}}$(ker $(\sigma ))={\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$. Henceforth, one can view $\mathcal{H}$ as an additive map from ker $(\sigma )$ to ${\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$. By Theorem 2.2, there exists $q\in {\mathfrak{Q}}_{\mathrm{mr}}$ such that $\mathcal{H}(\mathfrak{g})=q\mathfrak{g}$ for all $\mathfrak{g}\in$ ker $(\sigma )$. Let $\mathfrak{g}\in$ ker $(\sigma )$ and $\mathfrak{h}\in \mathfrak{D}$. Then $\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g}\in$ ker $(\sigma )$ and so $\mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})=q(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g}).$

We easily attain the following central identity: $\begin{array}{rl}& \mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})-\mathcal{H}(\mathfrak{g})\mathfrak{h}-\mathcal{H}(\mathfrak{h})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})\\ & -{\mathfrak{h}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}.\end{array}$This implies $\begin{array}{r}q(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})-q\mathfrak{g}\mathfrak{h}-\mathcal{H}(\mathfrak{h})\mathfrak{g}-{\mathfrak{h}}^{\sigma }q\mathfrak{g}\in \mathfrak{C}.\end{array}$From the above, we have $(q\mathfrak{h}-\mathcal{H}(\mathfrak{h})-{\mathfrak{h}}^{\sigma }q)\mathfrak{g}\in \mathfrak{C}.$ Hence, $\mathcal{H}(\mathfrak{h})=q\mathfrak{h}-{\mathfrak{h}}^{\sigma }q$ for all $\mathfrak{h}\in \mathfrak{D}$, due to [21, Lemma 4.6].

As a proof to show non-commutativity is essential in Theorem 3.1, we have constructed the following example.

Example 3.2

Suppose $\mathfrak{F}:{\mathbf{Z}}_2[i][x]\to Q_{\mathrm{mr}}({\mathbf{Z}}_2[i][x])={\mathbf{Z}}_2[i](x)$ be a weak Jordan σ derivation defined as $\mathfrak{F}(\sum _{i=1,i=n}{a}_t{x}^t)=\sum _{i=1,i=n}t{a}_t{x}^{t-1},$ where $\sigma (\sum _{i=1,i=n}{a}_t{x}^t)=\sum _{i=1,i=n}{a}_t^2{x}^{2t}$ (not injective) for all $a_t\in {\mathbb{Z}}_2[i][x].$ For $\mathfrak{F}$ to be X-inner σ derivation $\mathfrak{F}$ should take the form $\mathfrak{F}(\mathfrak{g})=a\mathfrak{g}-{\mathfrak{g}}^{\sigma }a$, for some fixed $a\in {\mathbf{Z}}_2[i](x)$ and for all $\mathfrak{g}\in \mathfrak{D}.$ but this will pose a problem because a will not be defined for $p(x)=1$ (also a is x dependent).

Theorem 3.2

Suppose that $\mathfrak{D}$ is a prime ring with σ as an X-inner automorphism with deg($\mathfrak{D})>2$. Let $\mathcal{H}:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ be additive maps satisfying $\begin{array}{r}\mu (\mathfrak{g})=\mathcal{H}({\mathfrak{g}}^2)-\mathcal{H}(\mathfrak{g})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}\end{array}$for all $\mathfrak{g}\in \mathfrak{D}$. Then $\mathcal{H}$ is derivation on $\mathfrak{Z}(\mathfrak{D})$. Moreover, if char($\mathfrak{D})\ne 2,$ then $\mathcal{H}(\beta \mathfrak{g})-\beta \mathcal{H}(\mathfrak{g})$ is $\mathfrak{Z}(\mathfrak{D})$-linear map for every $\beta \in \mathfrak{Z}(\mathfrak{D})$ and $\mathfrak{g}\in \mathfrak{D}$.

Proof.

First we prove $\mathcal{H}(\beta \mathfrak{g})-\beta \mathcal{H}(\mathfrak{g})$ is $\mathfrak{Z}(\mathfrak{D})$-linear map for every $\beta \in \mathfrak{Z}(\mathfrak{D})$ and $\mathfrak{g}\in \mathfrak{D}$, if char($\mathfrak{D})\ne 2$. Let $\mathcal{H}:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ be additive maps satisfying $\begin{array}{r}\mu (\mathfrak{g})=\mathcal{H}({\mathfrak{g}}^2)-\mathcal{H}(\mathfrak{g})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}\end{array}$for all $\mathfrak{g}\in \mathfrak{D}$. Thus there exist a unit $a\in {\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ such that $\mathcal{H}({\mathfrak{g}}^2)-\mathcal{H}(\mathfrak{g})\mathfrak{g}-a\mathfrak{g}{a}^{-1}\mathcal{H}(\mathfrak{g})\in \mathfrak{C}$. Now, from (5(5) $\begin{array}{rl}\lambda (\mathfrak{g},\mathfrak{h})& =\mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})-\mathcal{H}(\mathfrak{g})\mathfrak{h}-\mathcal{H}(\mathfrak{h})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})\\ & -{\mathfrak{h}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}\end{array}$(5) ), we have (6) $\begin{array}{rl}2\mathcal{H}({\beta }^2{\mathfrak{g}}^2)& =2\mathcal{H}({\mathfrak{g}}^2){\beta }^2+\mathcal{H}({\beta }^2){\mathfrak{g}}^2+({\mathfrak{g}}^{\sigma }{)}^2\mathcal{H}({\beta }^2)\\ & +\lambda ({\beta }^2,{\mathfrak{g}}^2).\end{array}$(6) Assume that char$(\mathfrak{D})\ne 2.$ This gives us the following relation: (7) $\begin{array}{rl}2\mathcal{H}({\beta }^2{\mathfrak{g}}^2)& =2{\beta }^2(\mathcal{H}(\mathfrak{g})\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g})+\mu (\mathfrak{g}))\\ & +(2\mathcal{H}(\beta )\beta +\mu (\beta )){\mathfrak{g}}^2\\ & +({\mathfrak{g}}^{\sigma }{)}^2(2\mathcal{H}(\beta )\beta +\mu (\beta ))+\lambda ({\beta }^2,{\mathfrak{g}}^2).\end{array}$(7) Let's rewrite the above relation as (8) $\begin{array}{rl}2\mathcal{H}({\beta }^2{\mathfrak{g}}^2)& =2{\beta }^2(\mathcal{H}(\mathfrak{g})\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g})+\mu (\mathfrak{g}))\\ & +2\beta (\mathcal{H}(\beta ){\mathfrak{g}}^2+({\mathfrak{g}}^{\sigma }{)}^2\mathcal{H}(\beta ))\\ & +\mu (\beta )({\mathfrak{g}}^2+({\mathfrak{g}}^{\sigma }{)}^2)+\lambda ({\beta }^2,{\mathfrak{g}}^2).\end{array}$(8) Also from (5(5) $\begin{array}{rl}\lambda (\mathfrak{g},\mathfrak{h})& =\mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})-\mathcal{H}(\mathfrak{g})\mathfrak{h}-\mathcal{H}(\mathfrak{h})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})\\ & -{\mathfrak{h}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}\end{array}$(5) ), we have $\begin{array}{r}2\mathcal{H}((\beta \mathfrak{g}{)}^2)=2\mathcal{H}(\beta \mathfrak{g})\beta \mathfrak{g}+2\beta {\mathfrak{g}}^{\sigma }\mathcal{H}(\beta \mathfrak{g})+\lambda (\beta \mathfrak{g},\beta \mathfrak{g}).\end{array}$This gives us the following relation: $\begin{array}{rl}2\mathcal{H}({\beta }^2{\mathfrak{g}}^2)& =\beta (\beta \mathcal{H}(\mathfrak{g})+\mathcal{H}(\beta )\mathfrak{g}+\beta \mathcal{H}(\mathfrak{g})\\ & +{\mathfrak{g}}^{\sigma }\mathcal{H}(\beta )\lambda (\beta ,\mathfrak{g}))\mathfrak{g}\\ & +\beta {\mathfrak{g}}^{\sigma }(\beta \mathcal{H}(\mathfrak{g})+\mathcal{H}(\beta )\mathfrak{g}\\ & +\beta \mathcal{H}(\mathfrak{g})+{\mathfrak{g}}^{\sigma }\mathcal{H}(\beta )+\lambda (\beta \mathfrak{g},\beta \mathfrak{g}).\end{array}$Let's rewrite the above relation as (9) $\begin{array}{rl}2\mathcal{H}({\beta }^2{\mathfrak{g}}^2)& =2{\beta }^2(\mathcal{H}(\mathfrak{g})\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g}))\\ & +\beta (\mathcal{H}(\beta ){\mathfrak{g}}^2+({\mathfrak{g}}^{\sigma }{)}^2\mathcal{H}(\beta ))\\ & +\mathrm{\beta \lambda }(\beta ,\mathfrak{g})(\mathfrak{g}+{\mathfrak{g}}^{\sigma })+2\beta {\mathfrak{g}}^{\sigma }\mathcal{H}(\beta )\mathfrak{g}\\ & +\lambda (\beta \mathfrak{g},\beta \mathfrak{g}).\end{array}$(9) On comparing (8(8) $\begin{array}{rl}2\mathcal{H}({\beta }^2{\mathfrak{g}}^2)& =2{\beta }^2(\mathcal{H}(\mathfrak{g})\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g})+\mu (\mathfrak{g}))\\ & +2\beta (\mathcal{H}(\beta ){\mathfrak{g}}^2+({\mathfrak{g}}^{\sigma }{)}^2\mathcal{H}(\beta ))\\ & +\mu (\beta )({\mathfrak{g}}^2+({\mathfrak{g}}^{\sigma }{)}^2)+\lambda ({\beta }^2,{\mathfrak{g}}^2).\end{array}$(8) ) and (9(9) $\begin{array}{rl}2\mathcal{H}({\beta }^2{\mathfrak{g}}^2)& =2{\beta }^2(\mathcal{H}(\mathfrak{g})\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g}))\\ & +\beta (\mathcal{H}(\beta ){\mathfrak{g}}^2+({\mathfrak{g}}^{\sigma }{)}^2\mathcal{H}(\beta ))\\ & +\mathrm{\beta \lambda }(\beta ,\mathfrak{g})(\mathfrak{g}+{\mathfrak{g}}^{\sigma })+2\beta {\mathfrak{g}}^{\sigma }\mathcal{H}(\beta )\mathfrak{g}\\ & +\lambda (\beta \mathfrak{g},\beta \mathfrak{g}).\end{array}$(9) ), we have (10) $\begin{array}{r}{\mathcal{N}}_{\beta }(\mathfrak{g})\mathfrak{g}+{\mathfrak{g}}^{\sigma }{\mathcal{L}}_{\beta }(\mathfrak{g})\in \mathfrak{C},\end{array}$(10) where ${\mathcal{N}}_{\beta }$ and ${\mathcal{L}}_{\beta }$ are additive maps given as follows: (11) $\begin{array}{rl}& {\mathcal{N}}_{\beta }(\mathfrak{g})=(\mathcal{H}({\beta }^2)-\beta \mathcal{H}(\beta ))\mathfrak{g}-\mathrm{\beta \lambda }(\beta ,\mathfrak{g})\text{and}\\ & {\mathcal{L}}_{\beta }(\mathfrak{g})={\mathfrak{g}}^{\sigma }(\mathcal{H}({\beta }^2)-\beta \mathcal{H}(\beta ))-2\beta \mathcal{H}(\beta )\mathfrak{g}\\ & -\mathrm{\beta \lambda }(\beta ,\mathfrak{g}).\end{array}$(11) Linearize relation (10(10) $\begin{array}{r}{\mathcal{N}}_{\beta }(\mathfrak{g})\mathfrak{g}+{\mathfrak{g}}^{\sigma }{\mathcal{L}}_{\beta }(\mathfrak{g})\in \mathfrak{C},\end{array}$(10) ) to obtain the following: $\begin{array}{r}{\mathcal{N}}_{\beta }(\mathfrak{g})\mathfrak{h}+{\mathcal{N}}_{\beta }(\mathfrak{h})\mathfrak{g}+{\mathfrak{g}}^{\sigma }{\mathcal{L}}_{\beta }(\mathfrak{h})+{\mathfrak{h}}^{\sigma }{\mathcal{L}}_{\beta }(\mathfrak{g})\in \mathfrak{C}.\end{array}$Replace $\mathfrak{h}$ by $\eta \mathfrak{h}$, $\begin{array}{r}{\mathcal{N}}_{\beta }(\mathfrak{g})\eta \mathfrak{h}+{\mathcal{N}}_{\beta }(\eta \mathfrak{h})\mathfrak{g}+{\mathfrak{g}}^{\sigma }{\mathcal{L}}_{\beta }(\eta \mathfrak{h})+\eta {\mathfrak{h}}^{\sigma }{\mathcal{L}}_{\beta }(\mathfrak{g})\in \mathfrak{C}.\end{array}$Also $\begin{array}{r}\eta {\mathcal{N}}_{\beta }(\mathfrak{g})\mathfrak{h}+\eta {\mathcal{N}}_{\beta }(\mathfrak{h})\mathfrak{g}+\eta {\mathfrak{g}}^{\sigma }{\mathcal{L}}_{\beta }(\mathfrak{h})+\eta {\mathfrak{h}}^{\sigma }{\mathcal{L}}_{\beta }(\mathfrak{g})\in \mathfrak{C}.\end{array}$Thus on comparing last two relations we arrive at the following: (12) $\begin{array}{r}({\mathcal{N}}_{\beta }(\eta \mathfrak{h})-\eta {\mathcal{N}}_{\beta }(\mathfrak{h}))\mathfrak{g}+{\mathfrak{g}}^{\sigma }(\eta {\mathcal{L}}_{\beta }(\mathfrak{h})-{\mathcal{L}}_{\beta }(\eta \mathfrak{h}))\in \mathfrak{C}.\end{array}$(12) Fix $p={\mathcal{N}}_{\beta }(\eta \mathfrak{h})-\eta {\mathcal{N}}_{\beta }(\mathfrak{h})$ and $q=\eta {\mathcal{L}}_{\beta }(\mathfrak{h})-{\mathcal{L}}_{\beta }(\eta \mathfrak{h})$ such that $p\mathfrak{g}+a\mathfrak{g}{a}^{-1}q\in \mathfrak{C}.$ From Lemma 3.4, if $p\ne 0$, then we have p=−q and $a^{-1}q\in \mathfrak{C}.$ This gives $[p\mathfrak{g}-{\mathfrak{g}}^{\sigma }p,\mathfrak{g}]=0.$ From [22, Theorem 1.1], we have $(p-\theta )\mathfrak{g}-{\mathfrak{g}}^{\sigma }p=0,$ where $\theta \in \mathfrak{C}.$ On lifting to ${\mathfrak{Q}}_{\mathrm{mr}}$ and putting $\mathfrak{g}=1,$ we have $\theta =0.$ Thus $p\mathfrak{g}+a\mathfrak{g}{a}^{-1}q=0,$ for every $\mathfrak{g}\in \mathfrak{D}$. This is a contradiction as owing to Lemma 3.1, one can pick $0\ne \theta \in \mathfrak{C}$ for certain $\mathfrak{g}={\mathfrak{g}}_0$ so that $\theta =q_1{\mathfrak{g}}_0+q_2{\mathfrak{g}}_0{q}_3\in \mathfrak{C},\text{for some}{\mathfrak{g}}_0\in \mathfrak{D}.$ Therefore, p=q=0 follows directly from Lemma 3.4. Thus the maps ${\mathcal{L}}_{\beta }$ and ${\mathcal{N}}_{\beta }$ are $\mathfrak{Z}(\mathfrak{D})$-linear maps. This forces $\mathrm{\beta \lambda }(\beta ,u)$ to be a $\mathfrak{Z}(\mathfrak{D})$-linear map.

In this effect, for any $\alpha \in \mathfrak{Z}(\mathfrak{D})$ we have $\mathrm{\beta \lambda }(\beta ,\alpha \mathfrak{g})=\mathrm{\alpha \beta \lambda }(\beta ,\mathfrak{g}),$ which furnishes $\beta ({\mathcal{T}}_{\beta }(\alpha \mathfrak{g})-\alpha {\mathcal{T}}_{\beta }(\mathfrak{g}))=0,$ where $\begin{array}{r}{\mathcal{T}}_{\beta }(\mathfrak{g})=\mathcal{H}(\beta \mathfrak{g})-\beta \mathcal{H}(\mathfrak{g}).\end{array}$Thus ${\mathcal{T}}_{\beta }$ is $\mathfrak{Z}(\mathfrak{D})$-linear map. If this procedure is done for arbitrary $\beta \in \mathfrak{Z}(\mathfrak{D}),$ then ${\mathcal{T}}_{\beta }$ is $\mathfrak{Z}(\mathfrak{D})$-linear map for every $\beta \in \mathfrak{Z}(\mathfrak{D})$. We are done.

Next we claim $\mathcal{H}$ is derivation on $\mathfrak{Z}(\mathfrak{D})$. Observe that $\begin{array}{rl}2\mathcal{H}((\mathrm{\beta \gamma })\mathfrak{g})& =2\mathrm{\beta \gamma }\mathcal{H}(\mathfrak{g})+\mathcal{H}(\mathrm{\beta \gamma })\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathrm{\beta \gamma })\\ & +\lambda (\mathrm{\beta \gamma },\mathfrak{g})\text{and}\\ 2\mathcal{H}(\beta (\gamma \mathfrak{g}))& =2\beta \mathcal{H}(\gamma \mathfrak{g})+\mathcal{H}(\beta )\gamma \mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}(\beta )\gamma \mathfrak{g}\\ & +\lambda (\beta ,\gamma \mathfrak{g}).\end{array}$Compare the last two relations to obtain $\begin{array}{rl}& (\mathcal{H}(\mathrm{\beta \gamma })-\gamma \mathcal{H}(\beta )-\beta \mathcal{H}(\gamma ))\mathfrak{g}+{\mathfrak{g}}^{\sigma }(\mathcal{H}(\mathrm{\beta \gamma })-\gamma \mathcal{H}(\beta )\\ & -\beta \mathcal{H}(\gamma ))\in \mathfrak{C}.\end{array}$This is similar to relation (12(12) $\begin{array}{r}({\mathcal{N}}_{\beta }(\eta \mathfrak{h})-\eta {\mathcal{N}}_{\beta }(\mathfrak{h}))\mathfrak{g}+{\mathfrak{g}}^{\sigma }(\eta {\mathcal{L}}_{\beta }(\mathfrak{h})-{\mathcal{L}}_{\beta }(\eta \mathfrak{h}))\in \mathfrak{C}.\end{array}$(12) ), thus we have $\mathcal{H}(\mathrm{\beta \gamma })=\gamma \mathcal{H}(\beta )+\beta \mathcal{H}(\gamma ).$ Therefore, ${\mathcal{H}}_{|}{}_{\mathfrak{Z}(\mathfrak{D})}$ is a derivation.

In case char$(\mathfrak{D})=2$, we have $\begin{array}{rl}& \mathcal{H}(\beta \mathfrak{h}+\mathfrak{h}\beta )-\mathcal{H}(\beta )\mathfrak{h}-\mathcal{H}(\mathfrak{h})\beta -\beta \mathcal{H}(\mathfrak{h})\\ & -{\mathfrak{h}}^{\sigma }\mathcal{H}(\beta )\in \mathfrak{C}.\end{array}$This reduces to $\begin{array}{r}\mathcal{H}(\beta )\mathfrak{h}+{\mathfrak{h}}^{\sigma }\mathcal{H}(\beta )\in \mathfrak{C}.\end{array}$Now, we proceed according to the observation made after relation (12(12) $\begin{array}{r}({\mathcal{N}}_{\beta }(\eta \mathfrak{h})-\eta {\mathcal{N}}_{\beta }(\mathfrak{h}))\mathfrak{g}+{\mathfrak{g}}^{\sigma }(\eta {\mathcal{L}}_{\beta }(\mathfrak{h})-{\mathcal{L}}_{\beta }(\eta \mathfrak{h}))\in \mathfrak{C}.\end{array}$(12) ) that is $\mathcal{H}(\beta )=0.$ Therefore, ${\mathcal{H}}_{|}{}_{\mathfrak{Z}(\mathfrak{D})}$ is a derivation.

The following Lemma generalizes and improves [14, Lemma 2.6]. As we have abolished the condition char $\mathfrak{D}=2$ from the hypothesis.

Lemma 3.5

Suppose that $\mathfrak{D}$ is not a PI-ring and let σ be an endomorphism of $\mathfrak{D}$. Let $\sigma ,\mathcal{A},\mathcal{B}:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ be additive maps satisfying $\begin{array}{r}\mathcal{H}(\mathfrak{g}\mathfrak{h})-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})=\mathcal{A}(\mathfrak{g})\mathfrak{h}+\mathcal{B}(\mathfrak{h})\mathfrak{g}\end{array}$for all $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}$. Then $\mathcal{A}$ is a σ-derivation. In addition, if σ is an X-outer automorphism, then $\mathcal{B}=0$.

Remark 3.1

Note that a second kind automorphism σ of ring $\mathfrak{D}$ is never inner automorphism. For instance, if σ is inner, then $\sigma (\beta )=\beta$, for every $\beta \in \mathfrak{C}$, which is a contradiction. Thus a second kind automorphism is always X-outer.

Lemma 3.6

Let $\mathfrak{D}$ be a noncommutative prime PI-ring with char $\mathfrak{D}\ne 2$, equipped with X-outer automorphism $\sigma$ and a weak Jordan σ-derivation $\mathcal{H}$. Then $\mathcal{H}$ is the sum of generalized X-inner $\sigma$-derivation and a central valued additive map.

Proof.

Let $\mathcal{H}$ be a weak Jordan $\sigma$-derivation of $\mathfrak{D}$. As $\mathfrak{D}$ is PI ring and if σ is first kind, then employing Kharchenko's result [3, Theorem 3.2], we conclude σ is X-inner which is against our assumption. So σ is second kind, one can consider a nonzero $\beta \in \mathfrak{C}$ such that ${\beta }^{\sigma }\ne \beta .$ Take $({\beta }^2,\mathfrak{g})$ in position of $(\mathfrak{g},\mathfrak{h})$ in (5(5) $\begin{array}{rl}\lambda (\mathfrak{g},\mathfrak{h})& =\mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})-\mathcal{H}(\mathfrak{g})\mathfrak{h}-\mathcal{H}(\mathfrak{h})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})\\ & -{\mathfrak{h}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}\end{array}$(5) ), we have the following relation: $\begin{array}{rl}2\mathcal{H}({\beta }^2\mathfrak{g})& =\mathcal{H}(\mathfrak{g})({\beta }^2+({\beta }^2{)}^{\sigma })+\mathcal{H}({\beta }^2)\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}({\beta }^2)\\ & +\lambda ({\beta }^2,\mathfrak{g})\end{array}$and $\begin{array}{rl}2\mathcal{H}(\beta (\beta \mathfrak{g}))& =\mathcal{H}(\beta \mathfrak{g})(\beta +{\beta }^{\sigma })+\mathcal{H}(\beta )\beta \mathfrak{g}\\ & +(\beta \mathfrak{g}{)}^{\sigma }\mathcal{H}(\beta )+\lambda (\beta ,\beta \mathfrak{g})\\ & =\frac{1}{2}[\mathcal{H}(u)(\beta +{\beta }^{\sigma })+\mathcal{H}(\beta )\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}(\beta )\\ & +\lambda (\beta ,\mathfrak{g})](\beta +{\beta }^{\sigma })\\ & +\mathcal{H}(\beta )\beta \mathfrak{g}+(\beta \mathfrak{g}{)}^{\sigma }\mathcal{H}(\beta )+\lambda (\beta ,\beta \mathfrak{g})\\ & =\frac{1}{2}\mathcal{H}(\mathfrak{g})(\beta +{\beta }^{\sigma }{)}^2\\ & +[\frac{1}{2}\mathcal{H}(\beta )(\beta +{\beta }^{\sigma })+\mathcal{H}(\beta )\beta ]\mathfrak{g}\\ & +{\mathfrak{g}}^{\sigma }[\frac{1}{2}\mathcal{H}(\beta )(\beta +{\beta }^{\sigma })+\mathcal{H}(\beta ){\beta }^{\sigma }]\\ & +\frac{1}{2}(\beta +{\beta }^{\sigma })\lambda (\beta ,\mathfrak{g})+\lambda (\beta ,\beta \mathfrak{g}).\end{array}$Comparing the above two equations, we have $\begin{array}{rl}& \mathcal{H}(\mathfrak{g})[{\beta }^2+({\beta }^2{)}^{\sigma }-\frac{1}{2}(\beta +{\beta }^{\sigma }{)}^2]\\ & +[\mathcal{H}({\beta }^2)-\mathcal{H}(\beta )\beta -\frac{1}{2}\mathcal{H}(\beta )(\beta +{\beta }^{\sigma })]\mathfrak{g}\\ & +{\mathfrak{g}}^{\sigma }[\mathcal{H}({\beta }^2)-\mathcal{H}(\beta ){\beta }^{\sigma }-\frac{1}{2}\mathcal{H}(\beta )(\beta +{\beta }^{\sigma })]\in \mathfrak{C}.\end{array}$This implies $\begin{array}{rl}& \frac{1}{2}({\beta }^{\sigma }-\beta {)}^2\mathcal{H}(\mathfrak{g})+[\mu (\beta )+\frac{1}{2}({\beta }^{\sigma }-\beta )\mathcal{H}(\beta )]\mathfrak{g}\\ & +{\mathfrak{g}}^{\sigma }[\mu (\beta )-\frac{1}{2}({\beta }^{\sigma }-\beta )\mathcal{H}(\beta )]\in \mathfrak{C}.\end{array}$We can write the above equation as follows: $\begin{array}{r}\mathcal{H}(\mathfrak{g})=a\mathfrak{g}+{\mathfrak{g}}^{\sigma }b+\chi (\mathfrak{g}),\text{for all}\mathfrak{g}\in \mathfrak{D}\mathfrak{.}\end{array}$Here, $\begin{array}{rl}& a=-2({\beta }^{\sigma }-\beta {)}^{-2}[\mu (\beta )+\frac{1}{2}({\beta }^{\sigma }-\beta )\mathcal{H}(\beta )],\\ & b=-2({\beta }^{\sigma }-\beta {)}^{-2}[\mu (\beta )-\frac{1}{2}({\beta }^{\sigma }-\beta )\mathcal{H}(\beta )]\text{and}\\ & \chi :\mathfrak{D}\to C\text{is an additive map.}\end{array}$On addition of a and b, we have $\begin{array}{rl}& a+b=-4({\beta }^{\sigma }-\beta {)}^{-2}\mu (\beta )=\theta \in \mathfrak{C}.\\ & \text{Thus, we have}\mathcal{H}(\mathfrak{g})=a\mathfrak{g}-{\mathfrak{g}}^{\sigma }a+\theta {\mathfrak{g}}^{\sigma }+\chi (\mathfrak{g}).\end{array}$

Definition 3.1

An epimorphism map is said to be X-inner if it is of the form ${\mathfrak{g}}^{\sigma }=a\mathfrak{g}{a}^{-1}$ for $a\in {\mathfrak{Q}}_{\mathrm{ms}}(\mathfrak{D})$, otherwise it is X-outer.

Theorem 3.3

Suppose that $\mathfrak{D}$ is a noncommutative prime ring and let σ be an X-outer epimorphism of $\mathfrak{D}$. Let $\mathcal{H}:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ be an additive map satisfying $\begin{array}{r}\mu (\mathfrak{g})=\mathcal{H}({\mathfrak{g}}^2)-\mathcal{H}(\mathfrak{g})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C},\end{array}$for all $\mathfrak{g}\in \mathfrak{D}$. Then

1. if $\mathfrak{D}$ is a prime PI-ring and σ is a second kind automorphism with char$(\mathfrak{D})\ne 2$, then $\mathcal{H}(\mathfrak{g})=a\mathfrak{g}+{\mathfrak{g}}^{\sigma }b+\chi (\mathfrak{g})$ for every $\mathfrak{g}\in \mathfrak{D}$ and χ is a central valued additive map otherwise;

2. $\mathcal{H}$ is a σ-derivation.

Proof.

In view of Theorem 3.1, we can assume σ to be an X-outer automorphism.

Case 1: $\mathfrak{D}$ is not a prime PI-ring.

We claim that there exist two additive maps $\mathcal{A},\mathcal{B}$ from $\mathfrak{D}$ to ${\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ such that (13) $\begin{array}{r}\mathcal{H}(\mathfrak{g}\mathfrak{h})-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})=\mathcal{A}(\mathfrak{g})\mathfrak{h}+\mathcal{B}(\mathfrak{h})\mathfrak{g},\end{array}$(13) holds for all $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}$. Let $F:\mathfrak{D}\times \mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ be a bi-additive map defined by $F(\mathfrak{g},\mathfrak{h})=\mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})=\mathcal{H}(\mathfrak{g}\mathfrak{h})+\mathcal{H}(\mathfrak{h}\mathfrak{g})$ for $\mathfrak{g},\mathfrak{h}\in \mathfrak{D}.$ From [18, Lemma 2.3], it follows that for all $w,\mathfrak{g},\mathfrak{h},z\in \mathfrak{D}$, we have $\begin{array}{r}F(\mathfrak{g}w,\mathfrak{h}z)-F(\mathfrak{g},w\mathfrak{h}z)=F(z\mathfrak{g}w,\mathfrak{h})-F(z\mathfrak{g},w\mathfrak{h}).\end{array}$From (5(5) $\begin{array}{rl}\lambda (\mathfrak{g},\mathfrak{h})& =\mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})-\mathcal{H}(\mathfrak{g})\mathfrak{h}-\mathcal{H}(\mathfrak{h})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})\\ & -{\mathfrak{h}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}\end{array}$(5) ), it follows that $\begin{array}{rl}& w^{\sigma }[{\mathfrak{h}}^{\sigma }(\mathcal{H}(z\mathfrak{g})-z^{\sigma }\mathcal{H}(\mathfrak{g}))]+{\mathfrak{g}}^{\sigma }[w^{\sigma }\mathcal{H}(\mathfrak{h}z)-\mathcal{H}(w\mathfrak{h}z)]\\ & +{\mathfrak{h}}^{\sigma }[z^{\sigma }\mathcal{H}(\mathfrak{g}w)-\mathcal{H}(z\mathfrak{g}w)]\\ & +z^{\sigma }[{\mathfrak{g}}^{\sigma }(\mathcal{H}(w\mathfrak{h})-w^{\sigma }\mathcal{H}(\mathfrak{h}))]\\ & +[(\mathcal{H}(\mathfrak{h}z)-\mathcal{H}(\mathfrak{h})z)\mathfrak{g}]w+[\mathcal{H}(w\mathfrak{h})z-\mathcal{H}(w\mathfrak{h}z)]\mathfrak{g}\\ & +[\mathcal{H}(z\mathfrak{g})w-\mathcal{H}(z\mathfrak{g}w)]\mathfrak{h}\\ & +[(\mathcal{H}(\mathfrak{g}w)-\mathcal{H}(\mathfrak{g})w)\mathfrak{h}]z\in \mathfrak{C}.\end{array}$[3, Theorem 2.3] guarantees the existence of certain bi-additive maps ${\mathcal{P}}_1,{\mathcal{P}}_2,\text{and}{\mathcal{P}}_3:{\mathfrak{D}}^2\to {\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ such that for all $\mathfrak{g},w,\mathfrak{h}\in \mathfrak{D}$, we have $\begin{array}{rl}& {\mathfrak{g}}^{\sigma }(\mathcal{H}(w\mathfrak{h})-w^{\sigma }\mathcal{H}(\mathfrak{h}))=-{\mathcal{P}}_1(w,\mathfrak{g})\mathfrak{h}-{\mathcal{P}}_2(\mathfrak{g},\mathfrak{h})w\\ & -{\mathcal{P}}_3(w,\mathfrak{h})\mathfrak{g}.\end{array}$By [3, Theorem 2.3], there exist additive maps $\mathcal{A},\mathcal{B}:\mathfrak{D}\to {\mathfrak{Q}}_{\mathrm{mr}}(\mathfrak{D})$ such that for all $\mathfrak{g},w\in \mathfrak{D},$ we have $\begin{array}{r}\mathcal{H}(w\mathfrak{h})-w^{\sigma }\mathcal{H}(\mathfrak{h})=\mathcal{A}(w)\mathfrak{h}+\mathcal{B}(\mathfrak{h})w.\end{array}$Hence (13(13) $\begin{array}{r}\mathcal{H}(\mathfrak{g}\mathfrak{h})-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})=\mathcal{A}(\mathfrak{g})\mathfrak{h}+\mathcal{B}(\mathfrak{h})\mathfrak{g},\end{array}$(13) ) holds. Thus we obtain $\mathcal{A}$ is a σ-derivation and $\mathcal{B}=0$, by employing Lemma 3.5. On putting $\mathcal{B}=0$ in (13(13) $\begin{array}{r}\mathcal{H}(\mathfrak{g}\mathfrak{h})-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})=\mathcal{A}(\mathfrak{g})\mathfrak{h}+\mathcal{B}(\mathfrak{h})\mathfrak{g},\end{array}$(13) ), we have $\begin{array}{r}\mathcal{H}(\mathfrak{g}\mathfrak{h})-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})=\mathcal{A}(\mathfrak{g})\mathfrak{h}.\end{array}$This gives us $(\mathcal{H}-\mathcal{A})(w)w\in \mathfrak{C}$. Since $\mathfrak{D}$ is not PI, therefore $\mathfrak{D}$ does not satisfy $S_4$. By using [21, Lemma 4.16], $\mathcal{H}=\mathcal{A}$, that is, $\mathcal{H}$ is a σ-derivation.

Case 2: $\mathfrak{D}$ is a prime PI-ring.

If σ is of the first kind, i.e. σ is the identity map on C, then employing Kharchenko's result [3, Theorem 3.2], we have, σ is X-inner which is diametrically opposite to our assumption. So, we can consider σ is of the second kind. Thus there exists some $0\ne \beta$ such that ${\beta }^{\sigma }\ne \beta$ and let char $\mathfrak{D}=2,$ then from relation (5(5) $\begin{array}{rl}\lambda (\mathfrak{g},\mathfrak{h})& =\mathcal{H}(\mathfrak{g}\mathfrak{h}+\mathfrak{h}\mathfrak{g})-\mathcal{H}(\mathfrak{g})\mathfrak{h}-\mathcal{H}(\mathfrak{h})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{h})\\ & -{\mathfrak{h}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}\end{array}$(5) ), we have $\begin{array}{r}\mathcal{H}(\mathfrak{g})(\beta +{\beta }^{\sigma })+\mathcal{H}(\beta )\mathfrak{g}+{\mathfrak{g}}^{\sigma }\mathcal{H}(\beta )\in \mathfrak{C}.\end{array}$Thus (14) $\begin{array}{r}\mathcal{H}(\mathfrak{g})=a\mathfrak{g}+{\mathfrak{g}}^{\sigma }a+\eta (\mathfrak{g}),\end{array}$(14) where $a=(\beta +{\beta }^{\sigma }{)}^{-1}\mathcal{H}(\beta )$ and η is a central valued map.

On plugging this characterization of $\mathcal{H}$ in (4(4) $\begin{array}{r}\mu (\mathfrak{g})=\mathcal{H}({\mathfrak{g}}^2)-\mathcal{H}(\mathfrak{g})\mathfrak{g}-{\mathfrak{g}}^{\sigma }\mathcal{H}(\mathfrak{g})\in \mathfrak{C}.\end{array}$(4) ), we have $\begin{array}{rl}& a{\mathfrak{g}}^2+({\mathfrak{g}}^{\sigma }{)}^{2}a+\eta ({\mathfrak{g}}^2)-a{\mathfrak{g}}^2-{\mathfrak{g}}^{\sigma }a\mathfrak{g}-\eta (\mathfrak{g})\mathfrak{g}-{\mathfrak{g}}^{\sigma }a\mathfrak{g}\\ & -({\mathfrak{g}}^{\sigma }{)}^{2}a-{\mathfrak{g}}^{\sigma }\eta (\mathfrak{g})\in \mathfrak{C}.\end{array}$This returns the following relation, if char$\mathfrak{D}=2$, (15) $\begin{array}{r}\eta (\mathfrak{g})(\mathfrak{g}+{\mathfrak{g}}^{\sigma })\in \mathfrak{C}.\end{array}$(15) Suppose we have sets $\mathfrak{X}$ and $\mathfrak{Y}$, defined as follows: $\mathfrak{X}$ consists of elements $\mathfrak{g}$ in $\mathfrak{D}$ such that $\eta (\mathfrak{g})$ equals zero and $\mathfrak{Y}$ consists of elements $\mathfrak{h}$ in $\mathfrak{D}$ such that ${\mathfrak{h}}^{\sigma }-\mathfrak{h}$ belongs to $\mathfrak{C}$. According to Equation (15(15) $\begin{array}{r}\eta (\mathfrak{g})(\mathfrak{g}+{\mathfrak{g}}^{\sigma })\in \mathfrak{C}.\end{array}$(15) ), it is evident that $\mathfrak{D}$ is equal to the union of $\mathfrak{X}$ and $\mathfrak{Y}$.

Considering that $\mathfrak{X}$ and $\mathfrak{Y}$ are certain additive subgroups of $\mathfrak{D}$ with whose combination forms $\mathfrak{D}$, it is important to note that $\mathfrak{D}$, being an additive group, cannot be expressed as the union of two proper subgroups. Consequently, it must be either $\mathfrak{X}=\mathfrak{D}$ or $\mathfrak{Y}=\mathfrak{D}$. Let's assume that $\mathfrak{D}=\mathfrak{Y}$.

In this case, we have ${\mathfrak{g}}^{\sigma }=\mathfrak{g}+\kappa (\mathfrak{g})$ for all $\mathfrak{g}$ in D, where $\kappa (\mathfrak{g})$ belongs to $\mathfrak{C}$. Consequently, ${\mathfrak{g}}^{\sigma }{\mathfrak{h}}^{\sigma }=(\mathfrak{g}\mathfrak{h}{)}^{\sigma }=\mathfrak{g}\mathfrak{h}+\kappa (\mathfrak{g}\mathfrak{h})$. This implies that $(\mathfrak{g}+\kappa (\mathfrak{g}))(\mathfrak{h}+\kappa (\mathfrak{h}))=\mathfrak{g}\mathfrak{h}+\kappa (\mathfrak{g}\mathfrak{h})$ for all $\mathfrak{g},\mathfrak{h}$ in D, which can be further simplified as $\mathfrak{h}\kappa (\mathfrak{g})+\mathfrak{g}\kappa (\mathfrak{h})$ belongs to $\mathfrak{C}$.

Now, let's select α in $\mathfrak{Z}(\mathfrak{D})$ such that ${\alpha }^{\sigma }$ is not equal to α. Taking $\mathfrak{h}=\alpha$, we find that $\mathfrak{g}\kappa (\alpha )$ belongs to $\mathfrak{C}$. Since $\kappa (\alpha )={\alpha }^{\sigma }-\alpha \ne 0$, we can deduce that $\mathfrak{D}$ is commutative, leading to a contradiction.

Therefore, we conclude that $\mathfrak{X}=\mathfrak{D}$, which implies that η equals zero. Consequently, we can express $\mathfrak{H}(\mathfrak{g})=a\mathfrak{g}+{\mathfrak{g}}^{\sigma }a$, where a belongs to $\mathfrak{Z}(\mathfrak{D})$.

Now, the only left out case is of char$(\mathfrak{D})\ne 2$ and σ is second kind, then from Lemma 3.6, we arrive at the following relation: (16) $\begin{array}{r}\mathcal{H}(\mathfrak{g})=a\mathfrak{g}+{\mathfrak{g}}^{\sigma }b+\chi (\mathfrak{g}),\text{for all}\mathfrak{g}\in \mathfrak{D}.\end{array}$(16) In the upcoming example, we show that the choice of σ to be X-outer is essential in Theorem 3.3.

Example 3.3

Suppose $\mathfrak{D}=M_2(\mathfrak{C})$ and $\mathfrak{H}$ be a weak Jordan σ-derivation. Let σ be an identity automorphism (i.e. σ is X-inner epimorphism), then from observation [4, Theorem 1.3 (i)], we can find certain form of $\mathfrak{H}$ as the following: $\begin{array}{r}\mathfrak{H}(\mathfrak{g})=\mathfrak{c}(\mathfrak{g})+[\mathfrak{a},\mathfrak{g}]+\mathfrak{L}(\mathfrak{g})+\eta (\mathfrak{g}),\end{array}$where $\mathfrak{a}\in M_2(\mathfrak{C}),\mathfrak{c}$ a derivation, $\mathfrak{L},\eta :M_2(\mathfrak{C})\to M_2(\mathfrak{C})$ are $\mathfrak{C}$-linear maps such that $\mathfrak{L}$ and η are of the following forms: $\begin{array}{rl}& \mathfrak{L}(\mathfrak{x})=\left(\begin{array}{cc}0& {\beta }_1{\mathfrak{x}}_{21}\\ {\beta }_2{\mathfrak{x}}_{12}+{\beta }_3{\mathfrak{x}}_{21}& 0\end{array}\right),\\ & \eta (\mathfrak{x})=({\beta }_4({\mathfrak{x}}_{11}-{\mathfrak{x}}_{22})+{\beta }_5{\mathfrak{x}}_{12}+{\beta }_6{\mathfrak{x}}_{21})\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ & \text{where}\mathfrak{x}=({\mathfrak{x}}_{\mathrm{ij}})\in M_2(\mathfrak{C}),1\le i,j\le 2.\end{array}$At least for some ${\beta }_i\ne 0$, $\mathfrak{H}$ will not be a derivation. Thus we have encountered with certain form of $\mathfrak{H}$ which is not a derivation, once we adopt a X-inner σ-epimorphism.

Further we show that non-commutativity is essential in Theorem 3.3 in the following example.

Example 3.4

Suppose $\mathfrak{F}:\mathbf{Z}[i][x]\to Q_{\mathrm{mr}}(\mathbf{Z}[i][x])=\mathbf{Q}[i](x)$ be a weak Jordan σ derivation defined as $\mathfrak{F}(\sum _{i=1,i=n}{a}_t{x}^t)=\sum _{i=1,i=n}t{a}_t{x}^{t-1},$ where $\sigma (\sum _{i=1,i=n}{a}_t{x}^t)=\sum _{i=1,i=n}\overline{a_t}{x}^t,$ where $\overline{(a+\mathrm{ib})}=a-\mathrm{ib},$ for all $a_t\in \mathbf{Z}[i][x].$ For $\mathfrak{F}$ to be inner or X-inner σ derivation $\mathfrak{F}$ should take the form $\mathfrak{F}(\mathfrak{g})=a\mathfrak{g}-{\mathfrak{g}}^{\sigma }a$, for some fixed $a\in \mathbf{Q}[i](x)$ and for all $\mathfrak{g}\in \mathfrak{D}.$ but this will pose a problem because $a=\frac{\sum _{i=1,i=n}\overline{a_t}{x}^t}{\sum _{i=1,i=n}\overline{2\mathrm{Im}(a_t)}{x}^t}$ will not be defined for $\sum _{i=1,i=n}{a}_t{x}^t$ with purely real coefficients (also a is x dependent). For instance $p(x)=1+x.$

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