\documentclass[11pt]{amsart} %%The following pakages are available. \usepackage[mathscr]{eucal} \usepackage{amscd} \usepackage{amsfonts} \usepackage{amsmath, amsthm, amssymb} \usepackage{latexsym} \usepackage[dvips]{graphics} %%%%%%%% Theorem style %%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{lipsum} % Package to generate dummy text throughout this template \usepackage{cite} \usepackage[sc]{mathpazo} % Use the Palatino font \usepackage[T1]{fontenc} % Use 8-bit encoding that has 256 glyphs \linespread{1.000} % Line spacing - Palatino needs more space between lines \usepackage{microtype} % Slightly tweak font spacing for aesthetics \usepackage[hmarginratio=1:1,top=30mm]{geometry} % Document margins \usepackage{hyperref} % For hyperlinks in the PDF \usepackage{abstract} % Allows abstract customization \renewcommand{\abstractnamefont}{\normalfont\bfseries} % Set the "Abstract" text to bold \renewcommand{\abstracttextfont}{\normalfont\small\itshape} % Set the abstract itself to small italic text \usepackage{titlesec} % Allows customization of titles \renewcommand\thesection{\Roman{section}} % Roman numerals for the sections \titleformat{\section}[block]{\large\scshape\centering}{\thesection.}{1em}{} % Change the look of the section titles \renewcommand{\thesection}{\arabic{section}} \newtheorem{defi}{Definition}[section] \newtheorem{thm}{Theorem}[section] \newtheorem{lem}{Lemma}[section] \numberwithin{equation}{section} \newtheorem{rem}{Remark}[section] \newtheorem{prop}{Proposition}[section] \newtheorem{cor}{Corollary}[section] \newtheorem{ex}{Example}[section] \newtheorem{fa}{Fact}[section] \newtheorem{con}{Conjecture} \theoremstyle{plain} %\newtheorem{thm}{Theorem}[section] %\newtheorem{cor}{Corollary}[section] %\newtheorem{lem}{Lemma}[section] %\newtheorem{prop}{Proposition}[section] \newcommand{\sq}{\hfill\rule{2mm}{2mm}} %\newcommand{\pp}{\vline} %\newcommand{\zz}{\mbox{$Z\hspace{-2.2mm}Z$}} %\newcommand{\nn}{\mbox{$ I \hspace{-2.5mm} N$}} %\newcommand{\rr}{\mbox{$ I \hspace{-1.5mm} R$}} %\theoremstyle{definition} %\newtheorem{defn}[thm]{Definition} %\newtheorem{ex}{Example}[section] %\theoremstyle{remark} %\newtheorem{rem}{Remark}[section] %%%%%%%%%%%%%%%%%% Equation style %%%%%%%%%%%%%%%%%% %\numberwithin{equation}{section} %\renewcommand{\labelenumi}{(\arabic{enumi})} %\renewcommand{\labelenumii}{(\alph{enumii})} %%%%%%%%%%%%% Add the following for editors %%%%%%%%%%% \pagestyle{plain} %\renewcommand{\thepage}{{\fontsize{14pt}{14pt} %\selectfont {--\arabic{page}--}}} \thispagestyle{empty} \setcounter{page}{1} %%%%%%%%% PRINT SIZE %%%%%%%%%%%%%%%%%%% \setlength{\textwidth}{16.2cm} \setlength{\textheight}{22cm} \setlength{\oddsidemargin}{0cm} \setlength{\evensidemargin}{0cm} \setlength{\topmargin}{0cm} \setlength{\headheight}{0cm} \setlength{\headsep}{0cm} \begin{document} %\maketitle \thispagestyle{empty} %%%%%%% Please send the following part %%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%% HEADER %%%%%%%%%%%%%%%% \title{A note on an Engel condition with generalized derivations in rings\\ %title first line %OF PAPERS %title second line } %\medskip \author{Mohd Arif Raza, Nadeem ur Rehman and Tarannum Bano} \address{Mohd Arif Raza, Department of Mathematics, Aligarh Muslim University, Aligarh-202002 India} \email{\href{mailto:arifraza03@gmail.com}{arifraza03@gmail.com}} \address{Nadeem ur Rehman, Department of Mathematics, Aligarh Muslim University, Aligarh-202002 India} \email{\href{mailto:nu.rehman.mm@amu.ac.in}{nu.rehman.mm@amu.ac.in}} \address{Tarannum Bano, Department of Mathematics, Aligarh Muslim University, Aligarh-202002 India} \email{\href{mailto:tbano.rs@amu.ac.in}{tbano.rs@amu.ac.in}} \date {} \maketitle %\bigskip %%%%%%%%TEXT %%%%%%%%%%%%% \begin{abstract} {Let $R$ be a prime ring with characteristic different from two, $I$ a nonzero ideal of $R$ and $F$ a generalized derivation associated with a nonzero derivation $d$ of $R$. In the present paper we investigate the commutativity of $R$ satisfying the relation $F([x,y]_k)^n=([x,y]_k)^l$ for all $x,y\in I$, where $l,n, k$ are fixed positive integers. Moreover, let $R$ be a semiprime ring, $A=O(R)$ be an orthogonal completion of $R$ and $B=B(C)$ the Boolean ring of $C$. Suppose $F([x,y]_k)^n=([x,y]_k)^l$ for all $x,y\in R$, then there exists a central idempotent element $e$ of $B$ such that $d$ vanishes identically on $eA$ and the ring $(1-e)A$ is commutative.} \end{abstract} \maketitle \vspace{.25cm}\noindent \emph{Key words:} {\small Prime and semiprime rings; Generalized derivation; Generalized polynomial identity (GPI); Ideal.} \vskip 10pt \noindent \emph{2010 Mathematics Subject Classification:} 16N60; 16U80; 16W25. \section{Introduction} Let $R$ be an associative ring with center $Z(R)$. For each $x,y\in R$, define $[x,y]_k$ inductively by $[x,y]_1=xy-yx$ and $[x,y]_k=[[x,y]_{k-1},y]$ for $k > 1$. The ring $R$ is said to satisfy an Engel condition if there exists a positive integer $k$ such that $[x,y]_k=0$ for all $x,y\in R$. Note that an Engel condition is a polynomial $[x,y]_k=\sum_{m=0}^{k}(-1)^{m}{k \choose m} y^mxy^{k-m}$ in non-commutative indeterminates $x,y$ and $[x+z,y]_k=[x,y]_k+[z,y]_k$. Recall that a ring $R$ is prime if $xRy=\{0\}$ implies either $x=0$ or $y=0$, and $R$ is semiprime if $xRx=\{0\}$ implies $x=0$. An additive mapping $d:R\rightarrow R$ is called a derivation if $d(xy)=d(x)y+yd(x)$ holds, for all $x,y\in R$. In particular $d$ is an inner derivation induced by an element $q\in R$, if $d(x)=[q,x]$ holds, for all $x\in R$. An additive mapping $F:R\rightarrow R$ is called generalized derivation associated with a derivation $d$ if $F(xy)=F(x)y+xd(y)$ holds, for all $x,y\in R$. \vspace{.25cm} The Engel type identity with derivation first appeared in the well-known paper of Posner \cite{P} which states that a prime ring admitting a nonzero derivation $d$ must be commutative if $[d(x), x]\in Z(R)$ holds, for all $x\in R$. Since then several authors have studied this kind of Engel type identities with derivations acting on an appropriate subset of prime and semiprime rings (see \cite{HC, FH}, for a partial bibliography). In 1992, Daif and Bell \cite[Theorem 3]{DB} proved that if in a semiprime ring $R$ there exists a nonzero ideal $I$ of $R$ and a derivation $d$ of $R$ such that $d([x,y]) =[x,y]$ for all $x,y\in I$, then $I\subseteq Z(R)$. In addition, if $R$ is a prime ring, then $R$ is commutative. In 2003, Quadri et al. \cite{QN} extended the result of Daif and Bell and proved that if $R$ is a prime ring, $I$ a nonzero ideal of $R$ and $F$ a generalized derivation associated with a nonzero derivation $d$ such that $F([x,y])=[x,y]$ for all $x,y\in I$, then $R$ is commutative. Very recently, Huang and Davvaz \cite{HD} generalized the result of Quadri et al. and proved that if $R$ is a prime ring and $F$ is a generalized derivation associated with a nonzero derivation $d$ of $R$ such that $F([x,y])^m=[x ,y]^{n}$ for all $x,y \in R$, where $m,n$ are fixed positive integers, then $R$ is commutative. \vspace{.25cm} On the other hand, in 1994 Giambruno et al. \cite{GGM} established that a ring must be commutative if it satisfies $([x,y]_k)^n=[x,y]_k$. Inspired by the above mention results it is natural to investigate what we can say about the commutativity of ring satisfying the relation $F([x,y]_{k})^{n}=([x,y]_k)^l$, where $F$ is a generalized derivation associated with a nonzero derivation $d$ of $R$ and $l,n,k$ are fixed positive integers. \vspace{.25cm} If we take $k=1$, then we obtain the following: \begin{cor}(\cite[Theorem A]{HD}) Let $R$ is a prime ring and $n,l$ are fixed positive integers. If $R$ admits a generalized derivation $F$ associated with a nonzero derivation $d$ such that $F([x,y])^{n}=([x,y])^l$ for all $x,y\in R$, then $R$ is commutative. \end{cor} \section{Generalized derivation in prime ring} Throughout this section, we take $R$ is a prime ring, $I$ is a nonzero ideal, $U$ is the Utumi quotient ring, $C$ is the extended centroid and $Q$ is the symmetric Martindale quotient ring. For a complete and detailed description of the theory of generalized polynomial identities involving derivations, we refer to \cite{BM}. \vspace{.25cm} We denote by $Der(U)$ the set of all derivations on $U$. By a derivation word we mean an additive map $\Delta$ of the form $\Delta = d_1d_2 \hdots d_m$ with each $d_i\in Der(U)$. Then a differential polynomial is a generalized polynomial with coefficients in $U$ of the form $\Phi(\Delta_jx_i)$ involving non-commutative indeterminates $x_i$ on which the derivation words $\Delta_j$ act as unary operations. The differential polynomial $\Phi(\Delta_jx_i)$ is said to be a differential identity on a subset $T$ of $U$ if it vanishes for any assignment of values from $T$ to its indeterminates $x_i$. Let $D_{int}$ be the $C$-subspace of $Der(U)$ consisting of all inner derivations on $U$ and $d$ be a nonzero derivation on $R$. By \cite[Theorem 2]{K}, we have the following result (see also \cite[Theorem 1]{TK}). \vspace{.25cm} If $\Phi(x_1,\hdots,x_n, {^dx_1},\hdots,{^dx_n})$ is a differential identity on $R$, then one of the following assertions holds: \begin{enumerate} \item[(i)] either $d\in D_{int}$; \item [(ii)]or, $R$ satisfies the generalized polynomial identity $\Phi(x_1,\cdots, x_n, y_1,\cdots, y_n)$. \end{enumerate} Before starting our result, we state the following theorem which is very crucial for developing the proof of our main result. \begin{thm}(\cite[ Theorem 3]{TK1})\label{t} Every generalized derivation $F$ on a dense right ideal of $R$ can be uniquely extended to a generalized derivation of $U$ and assumes of the form $F(x)=ax+d(x)$, for some $a\in U$ and a derivation $d$ on $U$. \end{thm} \begin{lem}\label{l0} Let $R$ be a prime ring with characteristics different from two, $n,k$ are fixed positive integers and $b\in Q$ with $b\notin C$ such that $([b,x]_{k+1})^n=0$ for all $x\in R$. Then $R$ satisfies a nonzero generalized polynomial identity (GPI). \end{lem} \begin{proof} By both \cite[Theorem 6.4.1]{BM} and \cite[Theorem 2]{C}, we have $([b,x]_{k+1})^n=0$ for all $x\in Q$. That is, the element $([b,X]_{k+1})^n$ in the free product $T=Q\ast_CC\{X\}$ is a generalized polynomial identity on $R$. As $b\notin C$, we can easily see that the term $(bX^{k+1})^n$ appears nontrivially in the expansion of $([b,X]_{k+1})^n$. So $([b,X]_{k+1})^n$ is a nonzero element in $T=Q\ast_CC\{X\}$. Therefore, $R$ satisfies a nonzero generalized polynomial identity. \end{proof} Now, we prove our main result of this section. \begin{thm}\label{t1} Let $R$ be a prime ring with characteristics different from two and $I$ be a nonzero ideal of $R$. If $R$ admits a nonzero generalized derivation $F$ associated with a nonzero derivation $d$ such that $F([x,y]_k)^n=([x,y]_k)^l$ for all $x,y\in I$, where $l,n,k$ are fixed positive integers, then $R$ is commutative. \end{thm} \begin{proof} Since $R$ is a prime ring and $F([x,y]_k)^n=([x,y]_k)^l$ for all $x,y\in I$. By Theorem \ref{t}, for some $a\in U$ and a derivation $d$ on $U$ such that $I$ satisfies the differential identity $$(a[x,y]_k+d([x,y]_k))^n=([x,y]_k)^l$$ which can be written as \begin{eqnarray}\label{e1} &&\bigg(a\nonumber\big(\sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} y^mxy^{k-m}\big)+\sum\limits_{m=0}^{k}(-1)^{m}{k \choose m}(\sum\limits_{i+j=m-1}y^{i}d(y)y^j)xy^{k-m}\\\nonumber &&+\sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} y^md(x)y^{k-m}+\sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} {y^{m}x} (\sum\limits_{r+s=k-m-1} y^rd(y)y^s)\bigg)^n\\ &&-(\sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} y^mxy^{k-m})^l=0. \end{eqnarray} \vspace{.25cm}\noindent Firstly we assume that $d$ is an outer derivation on $Q$. By Kharchencko's theorem \cite{K}, $I$ satisfies the generalized polynomial identity \[\begin{split} \bigg(a&\big( \sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} y^mxy^{k-m}\big)+\sum\limits_{m=0}^{k}(-1)^{m}{k \choose m}(\sum\limits_{i+j=m-1}y^{i}zy^j)xy^{k-m}\\ &+ \sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} y^mwy^{k-m}+ \sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} {y^{m}x} (\sum\limits_{r+s=k-m-1} y^rzy^s)\bigg)^n\\ &- (\sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} y^mxy^{k-m})^l=0. \end{split}\] In particular $x=z=0$, we have $(\sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} y^mwy^{k-m})^n=0~\text{ for all}~ y,w\in I.$ By Chuang \cite[Theorem 2]{C}, this polynomial identity is also satisfied by $Q$ and hence $R$ as well, i.e., $(\sum\limits_{m=0}^{k}(-1)^{m}{k \choose m} y^mwy^{k-m})^n=0$ for all $y,w\in R$. Substituting $y$ with $[b,w]$, where $b$ is a noncentral element of $R$ in the above identity, we have $([b,w]_{k+1})^n=0$ for all $w\in R$. It follows from both \cite[Theorem 6.4.1]{BM} and \cite[Theorem 2]{C} that $([b,w]_{k+1})^n=0$ for all $w\in Q$. \vspace{.25cm} In case $C$ is infinite, we have $([b,w]_{k+1})^n=0$ for all $w\in Q\otimes_C \overline C$, where $\overline{C}$ is the algebraic closure of $C$. Since both $C$ and $Q\otimes_C \overline C$ are centrally closed \cite[Theorem 2.5 and Theorem 3.5]{EM}, we may replace $R$ by $Q$ or $Q\otimes_C \overline C$ according as $C$ is finite or infinite. Thus we may assume that $R$ is centrally closed over $C$ which is either finite or algebraically closed and $([b,w]_{k+1})^n=0$ for all $w\in R$. By Lemma \ref{l0}, $R$ is a nontrivial generalized polynomial identity (GPI). By Martindale's theorem \cite{MR}, $R$ is a primitive ring and so is isomorphic to a dense subring of linear transformations on a vector space $\mathcal{V}$ over $C$. \vspace{.25cm} Suppose that $\mathcal{V}$ is infinite dimensional over $C$. For any $v\in \mathcal{V}$, we claim that $v$ and $vb$ are $C$-dependent. On contrary suppose that $v$ and $vb$ are $C$-independent. We choose $ v_1,v_2,\cdots,v_k$ such that $v,vb,v_1,\cdots,v_k$ are $C$-dependent. By the density of $R$ on $\mathcal{V}$, there exists $x_0\in R$ such that $$vx_0=0,~vbx_0=v_1,~ v_ix_0=v_{i+1},~ v_kx_0=v, ~\text{where}~i=1,2,\cdots,{k-1}.$$ We see that $$v[b,x_0]_{k+1}=vbx_0^{k+1}=v_1x_0^k=v_2x_0^{k-1}=\cdots=v_kx_0=v,$$ and so $$0=v([b,x_0]_{k+1})^n=v\neq 0, ~~\text{ a contradiction}.$$ Our next goal is to show that there exists $\alpha \in C$ such that $bv=v\alpha$, for any $v\in \mathcal{V}.$ Now choose $v,w\in \mathcal{V}$ such that they are linearly $C$-independent. By the previous argument there exist $\alpha_v, \alpha_w, \alpha_{v+w}\in C$ such that $bv=v\alpha_v,~ bw=w\alpha_w, ~b(v+w)=(v+w)\alpha_{v+w}$. Moreover $v\alpha_v+w\alpha=(v+w)\alpha_{v+w}.$ Hence $v(\alpha_{v}-\alpha_{v+w})+w(\alpha_{w}-\alpha_{v+w})=0,$ and because $v,w$ are linearly $C$-independent, we have $\alpha_v=\alpha_w=\alpha_{v+w}$, that is, $\alpha$ does not depend on the choice of $v$. Now for $r\in R,v\in \mathcal{V}$, we have $bv=v\alpha$, $r(bv)=r(v\alpha)$ and also $b(rv)=(rv)\alpha$. Thus $0=[b,r]v$, for any $v\in \mathcal{V}$, that is $[b,r]\mathcal{V} =0$. Since $\mathcal{V}$ is a left faithful irreducible $R$-module, hence $[b,r]=0$, for all $r\in R$, i.e., $b \in C$, a contradiction. \vspace{.25cm}\noindent So $\mathcal{V}$ must be of finite dimensional, i.e., $R\cong M_t(\mathbb{F})$ for some $t>1$. Now we assume that $t=2$, i.e., $M_2(\mathbb{F})$ satisfies $([b,w]_{k+1})^n=0$. Let $e_{ij}$ is the usual unit matrix with $1$ in $(i,j)$-entry and zero elsewhere. Take $b=\sum_{i,j=1}^{2} b_{ij}e_{ij}$ with $b_{ij}\in \mathbb{F}$ and by choosing $w=e_{11}$, we see that $[b,e_{11}]_{k+1}=(-1)^{k+1}b_{12}e_{12}+b_{21}e_{21}$. Thus we have $0=([b,e_{11}]_{k+1})^{2n}=(-1)^{({k+1})n}(b_{12}b_{21})^ne_{11}+(-1)^{({k+1})n}(b_{12}b_{21})^ne_{22}$ which gives $b_{12}b_{21}=0$ and so either $b_{12}=0$ or $b_{21}=0$. Now we assume that $b_{21}=0$. Let $\chi$ be any automorphism of $R$ such that $\chi(x)=(1+e_{21})x(1-e_{21})$. Therefore $\chi(b)=(b_{11}-b_{12})e_{11}+b_{12}e_{12}+(b_{11}-b_{12}-b_{22})e_{21}+(b_{12}+b_{22})e_{22}$. Since $([\chi(b),w]_{k+1})^n=0$ for all $x\in R$, then it can be easily seen that $b_{12}(b_{11}-b_{12}-b_{22})=0$. Hence either $b_{12}=0$ or $(b_{11}-b_{12}-b_{22})=0$. Suppose that $(b_{11}-b_{12}-b_{22})=0$. If $k$ is even, by easy computation we see that $0=([b,e_{11}+e_{21}]_{k+1})^{2n}=(2b^2_{12})^ne_{11}+(2b^2_{12})^ne_{22}$. It implies that $(2b^2_{12})^n=0$ and so $b_{12}=0$. If $k$ is odd, then we have $0=([b,e_{11}+e_{21}]_{k+1})^{2n}=(-2b^2_{12})^ne_{11}+(-2b^2_{12})^ne_{22}$. Which implies that $(-2b^2_{12})^n=0$ and so $b_{12}=0$. Thus in all, $b$ is a diagonal matrix. As above we know that $\chi(b)=\sum_{i=1}^2b_{ii}e_{ii}+(b_{11}-b_{22})e_{21}$ is a diagonal matrix. Therefore, $b_{11}=b_{22}$, and so, $b$ is central in $R$, a contradiction. \vspace{.25cm} Now we consider the case when $t>2$. Let $b=\sum\limits_{i,j=1}^t$ with $b_{ij}\in \mathbb{F}$. Write $b=\left(\begin{matrix} b_{11}& \mathcal{A}\\ \mathcal{B}&\mathcal{C} \end{matrix} \right)$ where $\mathcal{A}=(b_{12},\cdots,b_{1t})$ $\mathcal{B}=(b_{21},\cdots,b_{t1})^T$ and $\mathcal{C}=(b_{ij})$ with $2\leq i,j\leq t$. Note that $[b,e_{11}]_{k+1}=\left(\begin{matrix} 0& (-1)^{k+1}\mathcal{A}\\ \mathcal{B}& 0 \end{matrix}\right)$. By given hypothesis, one can have $$([b,e_{11}]_{k+1})^{2n}=\left(\begin{matrix} (-1)^{n(k+1)}(\mathcal{AB})^n&0 \\ 0& (-1)^{n(k+1)}(\mathcal{BA})^n \end{matrix} \right).$$ In particular $(-1)^{n(k+1)}(\mathcal{AB})^n=0$ and so $\mathcal{AB}=0$. \vspace{.25cm} Let $\chi_{ij}$ be an inner automorphism of $R$ given by $\chi_{ij}(x)=(1+e_{ij})x(1-e_{ij})$ for $x\in R$. Write $1+e_{21}=\left(\begin{matrix}1&0\\ \mathcal{E}_2& \mathcal{I}_{t-1}\end{matrix}\right)$ where $\mathcal{E}_2=(1,0,\cdots,0)^T$ and $\mathcal{I}_{t-1}$ is the $(n-1)$-identity matrix. Thus $\chi_{21}(b)=\left(\begin{matrix}b_{11}-b_{12}&\mathcal{A}\\ b_{11}\mathcal{E}_2-b_{12}\mathcal{E}_2+\mathcal{B}-\mathcal{C}\mathcal{E}_2& \mathcal{E}_2\mathcal{A}+\mathcal{C}\end{matrix}\right)$. By easy calculation, it follows that $b_{11}b_{12}-b_{12}^2-\mathcal{AC}\mathcal{E}_2=0$. Suppose first that $k$ is even. We can easily see that $[b, e_{11}+e_{21}]_{k+1}=\left(\begin{matrix} b_{12}& -\mathcal{A}\\ \mathcal{J}_1 & -\mathcal{E}_2\mathcal{A}\end{matrix}\right)$ where $\mathcal{J}_1=\mathcal{B}+\mathcal{C}\mathcal{E}_2-\mathcal{E}_2 b_{11}$. Therefore $([b, e_{11}+e_{21}]_{k+1})^2=\left(\begin{matrix} b_{12}^2-\mathcal{A}\mathcal{J}_1& 0\\ \ast & -\mathcal{J}_1\mathcal{A}+b_{12}\mathcal{E}_2\mathcal{A}\end{matrix}\right)$. Making use of both $\mathcal{AB}=0$ and $b_{11}b_{12}-b_{12}^2-\mathcal{AC}\mathcal{E}_2=0$, we get $\mathcal{A}\mathcal{J}_1=-b_{12}^2$.\\ Thus $([b, e_{11}+e_{21}]_{k+1})^2=\left(\begin{matrix} 2b_{12}^2& 0\\ \ast & -\mathcal{J}_1\mathcal{A}+b_{12}\mathcal{E}_2\mathcal{A}\end{matrix}\right)$. Therefore by assumption, we have $0=([b, e_{11}+e_{21}]_{k+1})^{2n}=\left(\begin{matrix} (2b_{12}^2)^n& 0\\ \ast & (-\mathcal{J}_1\mathcal{A}+b_{12}\mathcal{E}_2\mathcal{A})^n\end{matrix}\right)$. In particular, $(2b_{12}^2)^n=0$, and so $b_{12}=0$. Next suppose that $k$ is odd. By computation we have $[b, e_{11}+e_{21}]_{k+1}=\left(\begin{matrix} -b_{12}& \mathcal{A}\\ \mathcal{J}_2 & \mathcal{E}_2\mathcal{A}\end{matrix}\right)$ where $\mathcal{J}_2=\mathcal{B}+\mathcal{C}\mathcal{E}_2-(b_{11}+2b_{12})\mathcal{E}_2 b_{11}$. Thus $$([b, e_{11}+e_{21}]_{k+1})^2=\left(\begin{matrix} b_{12}^2+\mathcal{A}\mathcal{J}_2& 0\\ \ast & \mathcal{J}_2\mathcal{A}+b_{12}\mathcal{E}_2\mathcal{A}\end{matrix}\right).$$ Applying both $\mathcal{AB}=0$ and $b_{11}b_{12}-b_{12}^2-\mathcal{AC}\mathcal{E}_2=0$, we get $\mathcal{A}\mathcal{J}_2=-3b_{12}^2$. Thus $$([b, e_{11}+e_{21}]_{k+1})^2=\left(\begin{matrix} -2b_{12}^2& 0\\ \ast & \mathcal{J}_2\mathcal{A}+b_{12}\mathcal{E}_2\mathcal{A}\end{matrix}\right),$$ and so $$0=([b, e_{11}+e_{21}]_{k+1})^{2n}=\left(\begin{matrix} (-2b_{12}^2)^n& 0\\ \ast & (\mathcal{J}_2\mathcal{A}+b_{12}\mathcal{E}_2\mathcal{A})^n\end{matrix}\right).$$ In particular, $(-2b_{12}^2)^n=0$, and so $b_{12}=0$. \vspace{.25cm} Now we claim that $b$ is a diagonal matrix. Since $([\chi_{j2}(b),x]_{k+1})^n=0$ for all $x\in R$, where $j>2$, as what has been shown, we get that $-b_{1j}=\chi_{j1}(b)_{12}=0$. So $b_{1j}=0$ for $j>1$. For $10$ such that $$R_M \models \Phi(\phi_M(\vec{c})) \Rightarrow \Psi_i(\phi_M(\vec{a})).$$ Then there exist a natural number $k>0$ and pairwise orthogonal idempotents $e_1,e_2,\cdots,e_k \in B$ such that $e_1 +e_2 +\cdots+e_k =1$ and $e_iR \models \Psi_i(e_i \vec{a})$ for all $e_i\neq 0$. \end{lem} Now, we prove our main result of this section. \begin{thm}\label{t2} Let $R$ is a $2$-torsion free semiprime ring and $F$ is a nonzero generalized derivation associated with a nonzero derivation $d$ of $R$ such that $F([x,y]_k)^n=([x,y]_k)^l$ for all $x,y\in R$, where $l,n,k$ are fixed positive integers. Further, let $A=O(R)$ is the orthogonal completion of $R$ and $B=B{C}$, where $C$ is the extended centroid of $R$. Then there exists a central idempotent element $e\in B$ such that $d$ vanishes identically on $eA$ and the ring $(1-e)A$ is commutative. \end{thm} \begin{proof} By the given hypothesis, we have $R$ satisfies $$F([x,y]_k)^n=([x,y]_k)^l.$$ By Theorem \ref{t}, the generalized derivation $F$ can be extended uniquely to a generalized derivation on $U$. Since $U$ and $R$ satisfy the same differential identities (see \cite{TK}), we have $(a[x,y]_k+[q,[x,y]_k])^n=([x,y]_k)^l$ for all $x,y\in U$. According to (\cite{BM}, Remark 3.1.16) $d(A)\subseteq A$ and $d(e)=0$ for all $e\in B$. Therefore, $A$ is an orthogonally complete $\Omega$-$\Delta$- ring where $\Omega = \{0,+,−,.,d\}$. Consider the formulas\\ $\Phi=( \forall x)(\forall y)\parallel (a[x,y]_k+[q,[x,y]_k])^n-([x,y]_k)^l=0\parallel,$\\ $\Psi_1 =( \forall x)(\forall y)\parallel xy=yx \parallel,$\\ $\Psi_2 =( \forall x)\parallel d(x)=0 \parallel.$\\ One can easily verify that $\Phi$ is a hereditary first-order formula and $^\neg\Phi$, $\Psi_1$, $\Psi_2$ are Horn formulas. Using Theorem \ref{t1}, we can easily check that all the conditions of Lemma \ref{l2} are fulfilled. Hence there exist two orthogonal idempotent $e_1$ and $e_2$ such that $e_1+e_2 = 1$ and if $e_i\neq 0$, then $e_iA \models \Psi_i$, $i=1 ,2$. This completes the proof. \end{proof} \begin{center} \begin{thebibliography}{99} \bibitem{BM} {Beidar, K. I., Martindale III, W. S., Mikhalev, A. V.}, {Rings with Generalized Identities}, {\emph{\ Pure and Applied Mathematics, Marcel Dekker 196, New York}, 1996.} \bibitem{B} {Bresar, M.}, {\emph{\ On the distance of the composition of two derivations to be the generalized derivations}, Glasgow Math. 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