1. Introduction

In everything that follows, $R$ denotes an associative ring with center $Z(R).$ Let $Q$ and $Q_{mr}$ stands for the two-sided Martindale quotient ring and right Utumi quotient ring (also known as maximal right ring of quotients) of $R$ respectively. The center of $Q_{mr}$ is called extended centroid of $R$ and is denoted by $C$ (i.e. $C=Z(Q_{mr})$). For the basic idea of these objects we refer the reader to ^[12]. For any $a, b\in R,$ a ring $R$ is called prime ring if $aRb=(0)$ implies $a=0$ or $b=0$ and is called semi-prime ring if $aRa=(0)$ implies $a=0$. An additive mapping $d:R\rightarrow R$ is said to be a derivation of $R$ if $d(xy)=d(x)y+xd(y)$ for all $x, y\in R.$ For some fixed $a\in R,$ the mapping $I_{a}:R\rightarrow R$ such that $x\mapsto [a, x]$ for all $x\in R,$ is a well-known example of a derivation. Specifically, $I_{a}$ is called the inner derivation of $R$ induced by the element $a$. In 1991, Bre$\check{s}$ar ^[13] introduced a generalized notion of a derivation, called generalized derivation. A generalized derivation of a ring $R$ is an additive mapping $F:R\rightarrow R$ uniquely determined by a derivation $d$ of $R$ such that $F(xy)=F(x)y+xd(y)$ for any $x, y\in R.$ Clearly, every derivation is a generalized derivation but the converse is not always true. For any $a, b\in R,$ $F(x)=ax+xb$ and $F(x)=ax$ are the most natural examples of a generalized derivation of $R$ associated with $d=I_{b}$ and $d=0$ respectively. In ^[14], Lee extended the concept of a generalized derivation. Accordingly, let $I$ be a dense right ideal of $R$ and $\delta: I\rightarrow Q_{mr}$ be a derivation. A generalized derivation is an additive mapping $F:I\rightarrow Q_{mr}$ such that $F(xy)=F(x)y+x\delta(y)$ holds for all $x, y\in I$. Further, in this paper Lee also showed that $F$ can be uniquely extended to a generalized derivation of $Q_{mr}$ and defined as $F(x)=ax+\delta(x)$ for some $a\in Q_{mr}$ (see ^[14], Theorem 3.)

Recall that, a nonempty set $J,$ which is an additive subgroup of $R$ is said to be a Jordan ideal of $R$ if $J\circ R\subseteq J.$ The following are some well known facts about Jordan ideals: if $J$ be a nonzero Jordan ideal of a ring and $u\in J,$ then

Ⅰ. $2J[R, R]\subseteq J, 2[R, R]J\subseteq J$ (^[1], Lemma 2.4)

Ⅱ. $4u^{2}R\subseteq J, 4Ru^{2}\subseteq J$ and $2[u^{2}, R]\subseteq J$ (^[15], proof of Lemma 3)

Ⅲ. $4uRu\subseteq J$ (^[15], proof of Theorem 3)

Ⅳ. $4[u, R]u\subseteq J$ and $4u[u, R]\subseteq J$

A classical result of Herstein ^[16] states that if a prime ring $R$ with char($R$) $\neq 2$ admits a derivation $d$ such that $d(x)d(y)=d(y)d(x)$ for all $x, y\in R,$ then $R$ is commutative. Motivated by this situation, Bell and Daif ^[17] without any restriction on the char ($R$), obtained the same conclusion from the identity $d(xy)=d(yx)$ (i.e. $d([x, y])=0$) where $x, y$ varies over a nonzero ideal of $R.$ In an addition to this, recently Oukhtite et al. ^[3] proved the following theorem: Let $R$ be a 2-torsion free prime ring and $J$ be a nonzero Jordan ideal of $R$. If $R$ admits a nonzero derivation $d$ such that $d([x, y])\in Z(R)$ for all $x, y\in J$, then $R$ is commutative. In this paper, we intend to prove this result for generalized derivations.

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2. Main Results

Lemma 2.1. Let $R$ be a ring and $J$ be a Jordan ideal of $R$. Then $[[J, J], R]\subseteq J.$

Proof. For any $r\in R$ and $x\in J,$ we have $x\circ r\in J.$ That is, $xr+rx\in J.$ For any $y\in J,$ we have $xyr+yrx+yxr-yxr=[x, y]r+y(x\circ r)\in J.$ Again we have $ryx+xry-rxy+rxy=-r[x, y]+(x\circ r)y\in J.$ On combining these two expressions, we obtain $[[x, y], r]+y\circ(x\circ r)\in J$ for any $x, y\in J$ and $r\in R.$ Clearly, $y\circ(x\circ r)\in J.$ Therefore, we have $[[x, y], r]\in J$ for all $x, y\in J$ and $r\in R.$

Lemma 2.2. Let $R$ be a 2-torsion free semi-prime ring and $J\not\subseteq Z(R)$ be a Jordan ideal of $R$. Then $J$ contains a nonzero ideal of $R.$

Proof. By Lemma 2.1, we have $[[x, y], r]\in J$ for any $x, y\in J$ and $r\in R.$ For some $z\in J,$ we find $[x, y]zr-zr[x, y]=[x, y]zr-z[x, y]r+z[x, y]r-zr[x, y]= [[x, y], z]r+z[[x, y], r]=[[x, y], z]r+z[xy, r]-z[yx, r]\in J.$ By Lemma 2.4 in ^[1], we have $2z[xy, r]\in J$ and $2z[yx, r]\in J$ for all $x, y, z\in J$ and $r\in R.$ On combining these expressions, we obtain $2[[x, y], z]r\in J$ for all $x, y, z\in J$ and $r\in R.$ Again, it gives $2[[x, y], z]rs+2s[[x, y], z]r\in J,$ where $x, y, z\in J$ and $r, s\in R.$ It implies that $2R[[J, J], J]R\subseteq J.$ Further, if $2R[[J, J], J]R=(0)$ i.e. $R[[J, J], J]R=(0)$ it forces that $(R[[J, J], J])^{2}=(0),$ which contradicts the semi-primeness of $R$. Hence, $J$ contains a nonzero ideal of $R.$

The following lemma is may be of independent interest.

Lemma 2.3. Let $R$ be a ring and $J$ be a Jordan ideal of $R.$ Then $2R[J^{2}, J]R\subseteq J.$

Proof. It is well known that $2[x^{2}, r]\in J$ for any $x\in J$ and $r\in R.$ For some $y\in J,$ we replace $r$ by $yr$ and get $2(x^{2}yr-yrx^{2})\in J.$ That means, $2(x^{2}y-yx^{2})r+2y(x^{2}r-rx^{2})\in J,$ where $x, y\in J$ and $r\in R.$ Since $2y[x^{2}, r]\in J,$ we must have $2(x^{2}y-yx^{2})r\in J.$ Therefore, $2((x^{2}y-yx^{2})r)s+2s(x^{2}y-yx^{2})r\in J$ for any $x, y\in J$ and $r, s\in R.$ Hence, we obtain $R[2J^{2}, J]R\subseteq J.$

Lemma 2.4. Let $R$ be a 2-torsion prime ring. Let $J\not\subseteq Z(R)$ be a Jordan ideal of $R$ and $d:R\rightarrow R$ be a derivation of $R$. If $x^{2}d(x^{2}) =0$ for all $x\in J,$ then $d=0.$

Proof. By hypothesis, we have $x^{2}(d(x)\circ x)=0$ for any $x\in J.$ By Lemma 2.2, $J$ contains a nonzero ideal $I$ of $R$ i.e. $I\subseteq J,$ where $2R[[J, J], J]R=I.$ That gives, $x^{2}(d(x)\circ x)=0$ for all $x\in I.$ By Kharchenko's theory ^[6] of differential identities, we have the following two cases:

Case 1: If $d$ is a $Q$-outer derivation, then $I$ satisfies the polynomial identity

$x^{2}(y\circ x)=0,$

for all $x, y\in I.$ On replacing $y$ by $2x,$ we have $(2x^{2})^{2}=0$ for all $x\in I.$ Which is a contradiction by Xu ^[11].

Case 2: Suppose $d$ is a $Q$-inner derivation induced by some $q\in Q$ i.e $d(r)=[q, r]$ for all $r\in R.$ For any $x\in I,$ we have

$x^{2}([q, x]\circ x)=0.$

In view of Theorem 1 in ^[7], $Q$ and $I$ satisfy same GPIs. Therefore, we have

$u^{2}([q, u]\circ u)=0,$

for all $u\in Q.$ By Theorem 2.5 and 3.5 in ^[8], $Q$ and $Q\bigotimes_{C}\overline{C}$ both are prime and centrally closed. So, we may replace $R$ by $Q$ or $Q\bigotimes_{C}\overline{C}$ according as $C$ is finite or infinite. In case, $Q$ has infinite center $C,$ we have $u^{2}([q, u]\circ u)=0$ for any $u\in Q\bigotimes_{C}\overline{C},$ where $\overline{C}$ stands for algebraic closure of extended centroid $C.$ Thus, we may assume that $R$ is centrally closed over $C$ (i.e. $RC=R$) which is either finite or algebraically closed and

$u^{2}([q, u]\circ u)=0,$

(1)

for all $u\in R.$ By Theorem 3 of Martindale ^[9], $RC$ (and so $R$) is a primitive ring having nonzero socle $\mho$ with associated division ring $D.$ Now, by a result of Jacobson [^[10], pg. 75], $R$ is isomorphic to a dense ring of linear transformations of some vector space $\mathcal{V}$ over $D$ and $\mho$ contains the linear transformation of $R$ with finite rank. If $\mathcal{V}$ is finite dimensional over $D,$ the density of $R$ on $\mathcal{V}$ implies that $R\cong M_{h}(D),$ where $h=dim_{D}(\mathcal{V}).$ Let us suppose that $dim_{D}(\mathcal{V})\geq 2,$ otherwise we are done.

Next, for any $v\in \mathcal{V},$ we claim that $\{v, qv\}$ is a linearly $D-$dependent set. If $qv=0,$ then there is nothing to prove. Let $qv\neq 0.$ If possible, we assume that $v$ and $qv$ are linearly independent over $D.$ By the density of $R,$ we can find some $x\in R$ such that

$xv=0, \quad xqv=qv$

The equation (1) forces that

$0=(u^{2}([q, u]\circ u))v=-qv,$

which is a contradiction. Thus, $\{v, qv\}$ must be linearly dependent over $D$ for all $v\in \mathcal{V}.$ That means, we can find some $\beta\in D$ such that $qv=v\beta.$ Next, we shall show that $\beta$ is independent of the choice of $v.$ Let us choose linearly independent $u$ and $v$ in $\mathcal{V}.$ By above process, we can find $\beta_{u}, \beta_{v}$ and $\beta_{u+v}$ in $D$ such that

$qu=u\beta_{u}, qv=v\beta_{v}\;\mbox{and} \;q(u+v)=\beta_{u+v}.$

Further, $u\beta_{u}+v\beta_{v}=(u+v)\beta_{u+v}.$ It implies $u(\beta_{u}-\beta_{u+v})+v(\beta_{v}-\beta_{u+v})=0.$ Hence, $\beta_{u}=\beta_{v}=\beta_{u+v},$ as $u$ and $v$ are chosen to be linearly independent.

Now, for any $r\in R$ and $v\in\mathcal{V},$ we have $qv=v\beta, r(qv)=r(v\beta)$ and $q(rv)=rv\beta.$ It implies that $[q, r]v=0$ for all $v\in\mathcal{V}.$ But $\mathcal{V}$ is left faithful irreducible $R-$module, hence $[q, r]=0$ i.e. $q\in Z(R)$ i.e. $d=0.$

Theorem 2.5. Let $R$ be a 2-torsion free prime ring. Let $J$ be a nonzero Jordan ideal of $R$ and $F:R\rightarrow R$ be a generalized derivation of $R$ associated with a nonzero derivation $d.$ If $F([J, J])\in Z(R),$ then $R$ is commutative.

Proof. We divide the proof into the following two cases:

Case 1: If $J\subseteq Z(R).$ With the aid of Lemma 3 of ^[2], $R$ is commutative.

Case 2: If $J\not\subseteq Z(R).$ Firstly, we claim that $Z(R)\cap J\neq (0).$ Let us assume that $Z(R)\cap J=(0).$ By our hypothesis, $F([u, v])\in Z(R)$ for all $u, v\in J.$ We replace $u$ and $v$ by $2u^{2}$ and $2vu^{2}$ respectively in order to get $4F([u^{2}, vu^{2}])\in Z(R).$ It is easy to see that $4F([u^{2}, vu^{2}])=4F([u^{2}, v])u^{2}+4[u^{2}, v](d(u)\circ u)\in J.$ Therefore, we find $4F([u^{2}, v]u^{2})=0$ for all $u, v\in J.$ That gives

$F([u^{2}, v])u^{2}+[u^{2}, v]d(u^{2})=0$

(2)

On replacing $v$ by $2vu^{2}$ in (2), we get

$F[u^{2}, v]u^{4}+[u^{2}, v]d(u^{2})u^{2}+[u^{2}, v]u^{2}d(u^{2})=0$

(3)

On combining Eq. (2) and Eq. (3), we get $[u^{2}, v]u^{2}d(u^{2})=0.$ Substitute $v=2[r, s]v,$ we get $[u^{2}, [r, s]]Ju^{2}d(u^{2})=(0).$ Primeness of $J$ implies that either $[u^{2}, [r, s]]=0$ or $u^{2}d(u^{2})=0.$ Let us assume that $u^{2}d(u^{2})=0$ for all $u\in J.$ It leads to a contradiction with the aid of Lemma 2.4. In the latter case, we have $[u^{2}, [r, s]]=0$ for any $u\in J$ and $r, s\in R.$ Putting $r=sr,$ we get $[u^{2}, s][r, s]=0.$ It implies that $[u^{2}, s]R[r, s]=(0).$ It forces that $u^{2}\in Z(R).$ From the proof of Lemma 5 in ^[3], $J\subseteq Z(R)$, again a contradiction.

Therefore, we must have $0\neq w\in Z(R)\cap J.$ By our hypothesis, we have $F([u, v])\in Z(R)$ for all $u, v\in J.$ Replace $v$ by $2v^{2}w$, we get $F([u, 2v^{2}])w+[u, 2v^{2}]d(w)\in Z(R).$ Since $F([u, 2v^{2}])$ and $w$ are in $Z(R),$ so we find $[[u, 2v^{2}], r]d(w)=0$ for all $u, v\in J.$ It implies that $[[u, 2v^{2}], r]Rd(w)=(0).$ Therefore, either $[[u, 2v^{2}], r]=0$ or $d(w)=0.$ Let us consider $[[u, 2v^{2}], r]=0.$ Put $v=v+w$, we get $[[u, 2vw], r]=0,$ since $w\in Z(R).$ It implies that $[[x, y], r]w=0$ $\Rightarrow$ $[[x, y], r]Rw=(0).$ But $w\neq 0,$ so only possibility is $[[x, y], r]=0$, where $x, y\in J$ and $r\in R.$ That is $[J, J]\subseteq Z(R).$ Hence, $J\subseteq Z(R)$ by Lemma 3 of ^[4], which is not possible.

On other side if $d(w)=0.$ For some $r\in R,$ we substitute $2rw$ in the place of $u$ in the equation $F([u, v])\in Z(R),$ we get $F([r, v])w\in Z(R).$ It implies that $F([r, v])\in Z(R).$ Replacing $y$ by $2sw$, by the same reasons we get $F([r, s])\in Z(R)$ for all $r, s\in R.$ Let $\zeta(r, s)=rs-sr,$ a multilinear polynomial in $R.$ Then we have $F(\zeta(r, s))\in Z(R)$ i.e. $[F(\zeta(r, s)), \zeta(r, s)]=0.$ By Theorem 2 in ^[5], either $\zeta(r, s)$ is central valued or $F(x)=\lambda x$ for all $x\in R$ and for some $\lambda\in C.$ In case, $F(x)=\lambda x,$ our hypothesis yields that $\lambda[r, s]\in Z(R).$ Since $F\neq 0$ so $\lambda\neq 0$ and hence $[r, s]\in Z(R).$ It implies that $R$ is commutative.

It is trivial that, if $F$ is a generalized derivation of $R$ associated with a derivation $d,$ then so is $F\pm I,$ where $I$ is the identity map on $R.$

Corollary 2.1. Let $R$ be a 2-torsion free prime ring. Let $J$ be a nonzero Jordan ideal of $R$ and $F:R\rightarrow R$ be a generalized derivation of $R$ associated with a nonzero derivation $d.$ If any one of the following:

1. $F([x, y])+ [x, y]\in Z(R)$

2. $F([x, y])-[x, y]\in Z(R)$

holds on $J,$ then $R$ is commutative.

We conclude with the following remark, which shows that our main result can't be extended to the class of semiprime rings.

Remark 2.6. Let $R^{1}$ be any noncommutative semiprime ring and $S^{1}$ be any commutative integral domain. Evidently, $R=S^{1}\times R^{1}$ is a semiprime ring and $J= S^{1}\times \{0\}$ is a nonzero Jordan ideal of $R.$ Let $F:R^{1}\rightarrow R^{1}$ be a generalized derivation of $R^{1}$ associated with a derivation $d.$ We define a mapping $\mathcal{F}:R\rightarrow R$ as $(s, r)\mapsto (0, F(r))$ and a mapping $\delta:R\rightarrow R$ as $(s, r)\mapsto (0, d(r))$. Note that, $\mathcal{F}$ is a generalized derivation of $R$ associated with derivation $\delta.$ Now, it is easy to check that $\mathcal{F}([J, J])\in Z(R),$ but $R$ is not commutative.

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Acknowledgments

The authors are greatly indebted to the anonymous referee for his/her valuable comments. They are also thankful to Dr. Basudeb Dhara for helpful discussions.

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Conflict of Interest

No potential conflict of interest was reported by the authors.

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