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Kyungpook Mathematical Journal 2022; 62(1): 43-55

Published online March 31, 2022

On n-skew Lie Products on Prime Rings with Involution

Shakir Ali and Muzibur Rahman Mozumder*

epartment of Mathematics, Aligarh Muslim University, Aligarh-202002, U. P., India
e-mail : shakir.ali.mm@amu.ac.in and muzibamu81@gmail.com

Received: July 7, 2021; Revised: October 5, 2021; Accepted: October 7, 2021

Let R be a *-ring and n≥ 1 be an integer. The objective of this paper is to introduce the notion of n-skew centralizing maps on *-rings, and investigate the impact of these maps. In particular, we describe the structure of prime rings with involution '*' such that $*[x,d(x)]n∈Z(R)$ for all x∈ R (for n=1, 2), where $d:R→R$ is a nonzero derivation of R. Among other related results, we also provide two examples to prove that the assumed restrictions on our main results are not superfluous.

Keywords: Prime ring, derivation, involution, centralizing mappings, 2-skew Lie product, 2-skew centralizing mappings, n-skew commuting mappings, n-skew centralizing mapping

1. Introduction

This research is motivated by the recent work's of Ali-Dar [1], Qi-Zhang [5] and Hou-Wang [3]. However, our approach is different from that of the authors of [5] and [3]. A ring R with an involution '*' is called a *-ring or ring with involution '*'. Throughout, we let R be a ring with involution '*' and Z(R), the center of the ring R. Moreover, the sets of all hermitian and skew-hermitian elements of R will be denoted by H(R) and S(R), respectively. The involution is called the first kind if $Z(R)⊆H(R)$, otherwise $S(R)∩Z(R)≠(0)$(see [2] for details). A ring R is said to be 2-torsion free if 2x=0 (where x ∈ R) implies x=0. A ring R is called prime if aRb=(0) (where a,b ∈ R) implies a=0 or b=0. A derivation on R is an additive mapping $d:R→R$ such that d(xy)=d(x)y+xd(y) for all x,y ∈ R.

For any x,y ∈ R, the symbol [x,y] will denote the Lie product xy-yx and the symbol $*[x,y]$ will denote the skew Lie product xy-yx*, where '*' is an involution on R. In a recent paper, Hou and Wang [3] extended the concept of skew Lie product as follows: for an integer n ≥ 1, the n-skew Lie product of any two elements x and y is defined by $*[x,y]n=*[x,*[x,y]n−1]$, where $*[x,y]0=y$, $*[x,y]=xy−yx*$ and $*[x,y]2=x2y−2xyx*+y(x*)2$. Obviously, for n=1, the skew Lie product and n-skew Lie product coincides. Note that, for n=2, we call it 2-skew Lie product. In [3], Hou and Wang studied the strong 2-skew commutativity preserving maps in prime rings with involution. In fact, they described the form of strong 2-skew commutativity preserving maps on a unital prime ring with involution that contains a non-trivial symmetric idempotent. In [5], Qi and Zhang studied the properties of n-skew Lie product on prime rings with involution and as an application, they characterized n-skew commuting additive maps, i.e.; an additive mapping f on R into itself such that *[x,f(x)]n=0 for all x∈ R. In definition of n-skew commuting mapping (defined in [5]), if we consider that f is any map (not necessarily additive) then it is more reasonable to call f a n-skew commuting. To give its precise definition, we make a slight modification in Qi and Zhang's definition for n-skew commuting mapping. For an integer n ≥ 1, a map f of a *-ring R into itself is called n-skew commuting mapping on R if $*[x,f(x)]n=0$ for every x∈ R. For an integer n ≥ 1, a map f of a *-ring R into itself is called n-skew centralizing mapping on R if $*[x,f(x)]n∈Z(R)$ for every x ∈ R. In particular, for n = 1, 2, we call them 1-skew commuting (resp. 1-skew centralizing) and 2-skew commuting (resp. 2-skew centralizing) mapping.

The objective of this paper is to introduce the notion of n-skew centralizing mappings on *-rings. Further, we investigate the impact of these mappings and describe the nature of prime *-rings which satisfy certain *-differential identities. In particular, for an integer n ≥ 1 we prove that if a 2-torsion free prime ring R with involution '*' of the second kind which admits a nonzero derivation d such that $*[x,d(x)]n∈Z(R)$ for all , then R is commutative. Moreover, some more related results are obtained. Further more, examples prove that, the assumed curtailment can not be relaxed as given.

2. Main Results

In order to study the effect of n-skew centralizing mappings, we need the following two lemmas for developing the proofs of our main results. We begin our discussion with the following lemmas:

Lemma 2.1. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If $x2x*∈Z(R)$ for all x∈ R, then R is commutative.

Proof. Linearization of $x2x*∈Z(R)$ gives that $x2y*+xyx*+xyy*+yxx*+yxy*+y2x*∈Z(R)$ for all $x,y∈R.$ Taking x=-x in the last expression and combine it with the above relation, we get

Substituting ky for y, where $k∈S(R)∩Z(R)$, we obtain $(−x2y*+xyx*+yxx*)k∈Z(R)$ for all $x,y∈R.$ Invoking the primeness of R and using the fact that $S(R)∩Z(R)≠(0)$, we get

Combining (2.1) and (2.2), we conclude that $x2y*∈Z(R).$ Replacing y by y*, we get $x2y∈Z(R)$ for all $x,y∈R.$ This can be further written as $[x2y,w]=0$ for all x,y,w∈ R. Replacing y by ry, we get $x2r[y,w]=0$ for all $x,y,w,r∈R$. Hence by the primeness of the ring R, we are force to conclude that R is commutative.

Lemma 2.2. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If $*[x,x*]2∈Z(R)$ for all x ∈ R, then R is commutative.

Proof. By the hypothesis, we have

Replacing x by kx in (2.3) where $k∈S(R)∩Z(R)$, we get

By (2.3) and (2.4), we conclude that $−4x(x*)2k3∈Z(R)$ for all x∈ R. Since R is 2-torsion free prime ring and $S(R)∩Z(R)≠(0)$, we obtain $x(x*)2∈Z(R)$ for all x∈ R. On linearizing we get

Taking x=-x in (2.5) and using (2.5), we obtain

Substitute ky for y, where $k∈S(R)∩Z(R)$ in (2.6), we get $2y(x*)2k∈Z(R)$ for all $x,y∈R.$ This implies that $yx2∈Z(R)$ for all $x,y∈R.$ Henceforth, using the same arguments as we have used in Lemma 2.1, we conclude that R is commutative. This proves the lemma.

Theorem 2.3. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admits a nonzero derivation d such that $*[x,d(x)]n∈Z(R)$ for all $x∈R(n=1,2)$, then R is commutative.

proof. Case (i) First we discuss the case, "when n=1" and "i.e.",

Linearizing the above expression, we get

That is,

This further implies that

Hence

Replacing y by hy in (2.7), where $h∈H(R)∩Z(R)$ and using it, we have

Using the primeness of R, we obtain either d(h)=0 or

First we consider the situation

Substituting kx for x in (2.8), where $k∈S(R)∩Z(R)$ and combining it with (2.8), we get

Since R is 2-torsion free prime ring, we deduce that

Replacing x by z, where z ∈ Z(R), we get [y,r]z=0 for all y,r ∈ R. Henceforth, we conclude that R is commutative. Now consider the case d(h)=0 for all $h∈H(R)∩Z(R)$. This implies that d(k)=0 for all $k∈S(R)∩Z(R).$ Replacing y by ky in (2.7), where $k∈S(R)∩Z(R)$ with d(k)=0 and adding with (2.7), we get

$2(x[d(y),r]+[x,r]d(y)−d(y)[x*,r]−[d(y),r]x*+y[d(x),r]+[y,r]d(x))k=0$

for all x,y,r ∈ R. Since R is 2-torsion free ring and $S(R)∩Z(R)≠(0),$ the above relation implies that

$x,y∈R.$ Taking y=h, where $h∈H(R)∩Z(R)$ and using the fact that d(h)=0, we get [d(x),r]h=0 for all x,r ∈ R and $h∈H(R)∩Z(R)$. This yields that [d(x),r]=0 for all x,r ∈ R. Hence in view of Posner's [4] first theorem, R is commutative.

Case (ii) Now, we prove the result for n=2 i.e.,

On expansion we acquire

Replacing x by xh in (2.9), where $h∈H(R)∩Z(R)$, we obtain

Then by the primeness of R we are force to conclude that either $d(h)h2=0$ or $x3−2x2x*+x(x*)2∈Z(R)$ for all x∈ R. First we consider the case

Substituting kx for x in (2.10), where $k∈S(R)∩Z(R)$, we get This further implies that

Subtracting (2.10) from (2.11) and using 2-torsion freeness of R, we obtain Therefore, by Lemma 2.1, R is commutative. Now consider the second case $d(h)h2=0$ for all $h∈H(R)∩Z(R)$. This implies that d(h)=0 for all $h∈H(R)∩Z(R)$. Therefore, d(k)=0 for all $k∈S(R)∩Z(R)$. Replacing x by kx in (2.9), where $k∈S(R)∩Z(R)$ and using the fact that d(k)=0, we obtain

Application of (2.9) yields $4xd(x)x*k3∈Z(R)$ for all x ∈ R. Since R is 2-torsion free ring and $S(R)∩Z(R)≠(0)$, we get

Putting x+h in place of x, where $h∈H(R)∩Z(R)$, we arrive at

Taking x=-x in (2.14) and then combining it with the obtained relation, we get $2d(x)h2∈Z(R)$. This implies that $d(x)∈Z(R)$ for all x∈ R, since the involution '*' is of the second kind. Hence, by Posner's [4] first theorem, R is commutative.

Theorem 2.4. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admits a nonzero derivation d such that $d(*[x,x*]n)∈Z(R)$ for all x ∈ R (n=1,2), then R is commutative.

proof. Case (i) Firstly we are focus to discuss the case when n=1 i.e.,

Linearizing this, we get

This implies that

Replacing y by hy, where $h∈H(R)∩Z(R)$ in (2.15), we obtain

Using the primeness of R, we have either d(h)=0 or

$x[y*,r]+[x,r]y*−y*[x*,r]−[y*,r]x*+y[x*,r]+[y,r]x*−x*[y*,r]−[x*,r]y*=0$

for all x,y,r ∈ R. We first consider the relation (2.16). Replacing y by ky, where $k∈S(R)∩Z(R)$ in (2.16), we get $2(y[x*,r]+[y,r]x*)k=0$ for all $x,y,r∈R$. Since R is 2-torsion free ring and $S(R)∩Z(R)≠(0)$, we obtain $y[x*,r]+[y,r]x*=0$ for all $x,y,r∈R.$ Taking x=k, where $k∈S(R)∩Z(R)$, we get -[y,r]k=0 for all y,r∈ R. Thus -[y,r]=0 for all y,r∈ R. That is, R is commutative. Now consider d(h)=0 for all $h∈H(R)∩Z(R)$. This implies that d(k)=0 for all $k∈S(R)∩Z(R).$ Replacing y by ky, where $k∈S(R)∩Z(R)$ in (2.15) and making use of (2.15), we get

This implies that

Taking x=h where $h∈H(R)∩Z(R)$ and using d(h)=0, we arrive at h[d(y),r]=0 for all y,r ∈ R. Then by the primeness of R and the fact that $S(R)∩Z(R)≠(0)$, we obtain [d(y),r]=0 for all y,r∈ R. Hence by Posner's [4] first theorem, R is commutative.

Case (ii) Next, for n=2 we have

On expansion we get

Linearization of (2.17) yields

Substituting -x for x in (2.18) and combining the obtained relation with (2.18), we obtain

Since R is 2-torsion free, the last relation gives

Replacing y by hy, where $h∈H(R)∩Z(R)$ in (2.19) and intermix it with (2.19), we come to

By the primeness of the ring R, we get either d(h)=0 or

$x2y*+xyx*+yxx*−2xx*y*−2xy*x*−2y(x*)2+(x*)2y*+x*y*x*+y*(x*)2∈Z(R)$

for all x,y ∈ R. Replacing y by ky, where $k∈S(R)∩Z(R)$ in (2.20), we arrive at

This implies that

Substituting kx for x in the last relation, we conclude that $y(x*)2∈Z(R)$ for all x,y∈ R. Now proceed as we have already done in Lemma 2.1, we conclude that R is commutative. "Considering the second case in which we have d(h)=0 for all $h∈H(R)∩Z(R).$" This implies that d(k)=0 for all $k∈S(R)∩Z(R).$ Now replacing x by kx in (2.19) and using "d(k)=0, we get"

Since $S(R)∩Z(R)≠(0)$, the last expression implies that

Combining this with (2.19) and using the fact that R is 2-torsion free ring, we arrive at

Replacing x by kx, where $k∈S(R)∩Z(R)$ and combining it with previous expression, we obtain $2d(y(x*)2)∈Z(R)$ for all x,y ∈ R. Replacing x by h, where $h∈H(R)∩Z(R)$ we come to $d(y)h2∈Z(R)$ for all y ∈ R. This implies that $d(y)∈Z(R)$ for all y ∈ R. Hence, R is commutative by Posner's [4] first theorem.

Theorem 2.5. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admits a derivation d such that $*[x,d(x)]2+*[x,x*]2∈Z(R)$ for all x ∈ R, then R is commutative.

Proof. By the hypothesis we assume that

If we take d=0. Then, application of Lemma 2.2, yields the required result. Now consider the case d ≠ 0 and on expansion of (2.21), we get

Replacing x by xh in (2.21), where $h∈H(R)∩Z(R)$, we obtain $(x3−x2x*+x(x*)2)d(h)h2∈Z(R)$ for all x ∈ R. Now by the primeness of R we get either $x3−2x2x*+x(x*)2∈Z(R)$ for all x ∈ R or $d(h)h2=0$. Now, we suppose that

This is same as the relation (2.10) in Theorem 2.3 and hence we conclude that R is commutative. Now we consider the case $d(h)h2=0$ for all $h∈H(R)∩Z(R)$. Since R is prime ring, so we get d(h)=0. This also implies that d(k)=0 for all $k∈S(R)∩Z(R).$ Replacing x by xk in (2.22) and combining with (2.22) we arrive at $(2xd(x)x*−x2x*−(x*)3)k3∈Z(R)$ for all x∈ R. Since $S(R)∩Z(R)≠(0)$, so by the primeness of R, we get

Linearization of (2.24) gives

$2xd(x)y*+2xd(y)x*+2xd(y)y*+2yd(x)x*+2yd(x)y*+2yd(y)x*−x2y*−xyx*−xyy*−yxx*−yxy*−y2x*−(x*)2y*−x*y*x*−x*(y*)2−y*(x*)2−y*x*y*−(y*)2x*$

$∈Z(R)$ for all x,y ∈ R.Replacing x by -x in (2.25) and combining the obtained relation with (2.25), we obtain

$2xd(x)y*+2xd(y)x*+2yd(x)x*−x2y*−xyx*−yxx*−(x*)2y*−x*y*x*−y*(x*)2$

$∈Z(R)$ for all x,y ∈ R. Taking x=h, where $h∈H(R)∩Z(R)$, we get

The primeness of R yields that $d(y)−2y*−y∈Z(R)$ for all y ∈ R. Replacing y by ky and on solving, we get y ∈ Z(R) for all y ∈ R. Hence, this conclude that R is commutative.

Theorem 2.6. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admit two distinct derivations d1 and d2 such that $*[x,d1(x)]2−*[x,d2(x)]2∈Z(R)$ for all x ∈ R, then R is commutative.

Proof. We assume that

If either d1 or d2 is zero, then we get the required result by Theorem 2.3 above. Now consider both d1, d2 are non-zero. Expansion of (2.26) yields that

$x2d1(x)−2xd1(x)x*+d1(x)(x*)2−x2d2(x)+2xd2(x)x*−d2(x)(x*)2∈Z(R)$

for all x ∈ R. Replacing x by xh in (2.27), where $h∈H(R)∩Z(R)$ and on simplifying with the help of (2.27), we get

This implies either $x3−2x2x*+x(x*)2∈Z(R)$ for all x ∈ R or $(d1(h)−d2(h))h2=0$. If $x3−2x2x*+x(x*)2∈Z(R)$ for all x∈ R, then by using the same steps as we have used after (2.10), we arrive at $x2x*∈Z(R)$ for all x ∈ R. Thus R is commutative, by Lemma 2.1. On the other hand, if $(d1(h)−d2(h))h2=0$ for all $h∈H(R)∩Z(R).$ Then we are force to conclude that $d1(h)=d2(h)$ and hence $d1(k)=d2(k)$ for all $k∈S(R)∩Z(R).$ Replacing x by kx in (2.27), and combining with (2.27) by using the fact that $d1(k)=d2(k)$, we get

Since R is 2-torsion free and $S(R)∩Z(R)≠(0),$ the last relation gives

Linearizing (2.28), we obtain

Replacing x by -x in (2.29) and combining the obtained result with (2.29), we get

$2(xd1(x)y*+yd1(x)x*+xd1(y)x*−xd2(x)y*−xd2(y)x*−yd2(x)x*)∈Z(R)$

for all x,y ∈ R. Since R is 2-torsion free ring, the above expression yields

$xd1(x)y*+yd1(x)x*+xd1(y)x*−xd2(x)y*−xd2(y)x*−yd2(x)x*∈Z(R)$

for all x,y∈ R. Replacing x by kx in (2.30) and on solving with the help (2.30) and using the fact that $d1(k)=d2(k)$, we get $(xd1(x)−xd2(x))y*∈Z(R)$ for all x,y ∈ R. Replacing y by h, where $h∈H(R)∩Z(R)$. Then by the primeness of R and $S(R)∩Z(R)≠(0)$ condition force that $xd1(x)−xd2(x)∈Z(R)$ for all x ∈ R. Linearizing this we get $xd1(y)+yd1(x)−xd2(y)−yd2(x)∈Z(R)$ for all x,y ∈ R. Taking y by h where $h∈H(R)∩Z(R)$ and using $d1(h)=d2(h),$ we obtain $d1(x)−d2(x)∈Z(R)$ for all x ∈ R. This can be further written as

Replacing x by xr in (2.31), we get $[x,r](d1(r)−d2(r))=0$ for all x,r ∈ R. Substitute xu for x in the last relation, we obtain $[x,r]u(d1(r)−d2(r))=0$ for all x,r,u ∈ R. Then by the primeness of R, for each fixed r ∈ R, we get either [x,r]=0 for all x ∈ R or $d1(r)−d2(r)=0$. Define and . Clearly, A and B are additive subgroups of R whose union is R. Hence by Brauer's trick, either A=R or B=R. If A=R, then [x,r]=0 for all x,r∈ R. This implies that R is commutative. If B=R, then $d1(r)=d2(r)$ for all r∈ R, which is a contradiction to our assumption. Hence, we conclude that R is commutative.

Theorem 2.7. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admit derivations d1, d2 such that at least one of them is nonzero and satisfies $d1(*[x,x*]2)+*[x,d2(x*)]2∈Z(R)$ for all x ∈ R, then R is commutative.

Proof. We are given that d1 and d2 are derivations of R such that

If d2 is zero then by Theorem 2.4, we get R is commutative. If d1 is zero then we have $*[x,d2(x*)]2∈Z(R)$ for all x∈ R. Expansion of last relation gives

Replacing x by hx, where $h∈H(R)∩Z(R)$ in (2.33) and combining the obtained expression, we get $*[x,x*]2d2(h)h2∈Z(R)$ for all x ∈ R. Now applying the primeness of the ring R, we get either $*[x,x*]2∈R$ or $d(h)h2=0$. If $*[x,x*]2∈Z(R)$ for all x∈ R, then by Lemma 2.2, we get R is commutative. Now consider the second case in which we have $d2(h)h2=0$ for all $h∈H(R)∩Z(R).$ This implies that $d2(h)=0,$ from here we get $d2(k)=0$ for all $k∈S(R)∩Z(R).$ Replacing x by kx in (2.30) and using the fact that $d2(k)=0$, we get $4xd2(x*)x*k3∈Z(R)$ for all x∈ R. This implies that $xd2(x*)x*∈Z(R)$ for all x ∈ R. Arguing as above after (2.13), we conclude that R is commutative.

Now consider the second case in which both d1 and d2 are nonzero. On expansion of (2.32), we have

$d1(x2x*)−2d1(x(x*)2)+d1((x*)3)+x2d2(x*)−2xd2(x*)x*+d2(x*)(x*)2∈Z(R)$

for all x ∈ R. Replacing x by hx, where $h∈H(R)∩Z(R)$ in (2.34) and solving with the help of (2.34), we get

By the primeness of the ring R, we get either $*[x,x*]2∈Z(R)$ for all x∈ R or $(3d1(h)+d2(h))h2=0$ for all $h∈H(R)∩Z(R).$ If $*[x,x*]2∈Z(R)$ for all x ∈ R, then by Lemma 2.2, we get R is commutative. Now consider the case $(3d1(h)+d2(h))h2=0$. This implies that $d2(h)=−3d1(h)$ and hence $d2(k)=−3d1(k)$ for all $k∈S(R)∩Z(R).$ Now substituting kx for x, where $k∈S(R)∩Z(R)$ in (2.34) and combining the obtained result with (2.34), we get $4(d1(x(x*)2)+xd2(x*)x*)k3∈Z(R)$ for all x ∈ R. Since R is 2-torsion free ring and $S(R)∩Z(R)≠(0)$, then invoking the primeness of R we obtain $d1(x(x*)2)+xd2(x*)x*∈Z(R)$ for all x∈ Z(R). Linearization to the last expression gives

Replacing x by -x in (2.35), we get

$2(d1(xx*y*)+d1(xy*x*)+d1(y(x*)2)+xd2(x*)y*+xd2(y*)x*+yd2(x*)x*)∈Z(R)$

for all x,y ∈ R. Since R is 2-torsion free ring, we get

$d1(xx*y*)+d1(xy*x*)+d1(y(x*)2)+xd2(x*)y*+xd2(y*)x*+yd2(x*)x*∈Z(R)$

for all x,y ∈ R. Substituting ky for y, where $k∈S(R)∩Z(R)$ in (2.36) and combining with (2.36) with use of $d2(k)=−3d1(k)$, we arrive at

$2d1(y(x*)2)k+2yd2(x*)x*k−xx*y*d1(k)+y(x*)2d1(k)+2xy*x*d1(k)∈Z(R)$

for all x,y ∈ R and $k∈S(R)∩Z(R)$. Substitute ky for y in (2.37) yields

Subtracting (2.38) form (2.37), we get $(−2xx*y*+4xy*x*−2y(x*)2)d1(k)k∈Z(R)$ for all $x,y∈R$. Since R is 2-torsion free prime ring and $S(R)∩Z(R)≠(0)$, the last expression forces that either $xx*y*−2xy*x*+y(x*)2∈Z(R)$ for all x,y ∈ R or $d1(k)k=0$. Suppose

Substituting ky for y, where $k∈S(R)∩Z(R)$ in (2.39) and combining with (2.39), we get $2y(x*)2k∈Z(R)$ for all x,y ∈ R. Taking x=k, we obtain $2yk3∈Z(R)$ for all y∈ R. Since R is 2-torsion free prime ring and $S(R)∩Z(R)≠(0)$, we conclude that R is commutative. Now consider the case in which we have d1(k)k=0 for all $k∈S(R)∩Z(R).$ This implies that d1(k)=0 for all $k∈S(R)∩Z(R)$. This further implies that $d2(k)=0$. Substitute k for x in (2.36), to get

Replacing y by ky, where $k∈S(R)∩Z(R)$ in (2.40) and combining the obtained relation with (2.40), finally we get $2d1(y)k3∈Z(R)$ for all y∈ R. Since R is 2-torsion free ring and $S(R)∩Z(R)≠(0)$, we obtain $d1(y)∈Z(R)$ for all y ∈ R. Hence, by Posner's [4] first theorem, R is commutative.

As an immediate consequence of the above theorem, we get the following corollary:

Corollary 2.8. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admits a nonzero derivation d such that $d(*[x,x*]2)+*[x,d(x*)]2∈Z(R)$ for all x ∈ R, then R is commutative.

The following example shows that the second kind involution assumption is essential in Theorem 2.3 and Theorem 2.4.

Example 2.9. Let Of course, R with matrix addition and matrix multiplication is a noncommutative prime ring. Define mappings $*,d1,d2:R→R$ such that $β1β2β3β4*=β4−β2−β3β1$, $d1β1β2β3β4=0−β2β30$ and $d2β1β2β3β4=0β2−β30$

Obviously, Then $x*=x$ for all $x∈Z(R)$, and hence $Z(R)⊆H(R)$, which shows that the involution * is of the first kind. Moreover, d1 and d2 are nonzero derivations of R such that $*[x,d1(x)]2∈Z(R)$ and $*[x,d1(x)]2−*[x,d2(x)]2∈Z(R)$ for all x∈ R. However, R is not commutative. Hence, the hypothesis of second kind involution is crucial in Theorem 2.3 & 2.4. Our next example shows that Theorems 2.3 and 2.4 are not true for semiprime rings.

Example 2.10. Let $S=R×ℂ,$ where R is same as in Example 2.9 with involution '*' and derivations d1 and d2 same as in above example, $ℂ$ is the ring of complex numbers with conjugate involution τ. Hence, S is a 2-torsion free noncommutative semiprime ring. Now define an involution α on S, as $(x,y)α=(x*,yτ)$. Clearly, α is an involution of the second kind. Further, we define the mappings D1 and D2 from S to S such that $D1(x,y)=(d1(x),0)$ and $D2(x,y)=(d2(x),0)$ for all (x,y)∈ S. It can be easily checked that D1, D2 are derivations on S and satisfying $α[X,D1(X)]2∈Z(S)$ and $α[X,D1(X)]2−α[X,D2(X)]2∈Z(S)$ for all X ∈ S, but S is not commutative. Hence, in Theorems 2.3 & 2.4, the hypothesis of primeness is essential.

We conclude the paper with the following Conjectures.

Conjecture 2.11. Let $n>2$ be an integer, R be a prime ring with involution '*' of the second kind and with suitable torsion restrictions on R. Next, let d be a nonzero derivation on R such that $*[x,d(x)]n∈Z(R)$ for all x∈ R. Then what we can say about the structure of R or the form of d?

Conjecture 2.12. Let $n>2$ be an integer, R be a prime ring with involution '*' of the second kind and with suitable torsion restrictions on R. Next, let d be a nonzero derivation on R such that $d(*[x,x*]n)∈Z(R)$ for all x∈ R. Then what we can say about the structure of R or the form of d?

Conjecture 2.13. Let $n>2$ be an integer, R be a prime ring with involution '*' of the second kind and with suitable torsion restrictions on R. Next, let d be a nonzero derivation on R such that $d(*[x,x*]n)+*[x,d(x*)]n∈Z(R)$ for all x∈ R. Then what we can say about the structure of R or the form of d?

Acknowledgments

The authors are greatful to the learned referee for carefully reading the manuscript. The valuable suggestions have simplified and clarified the paper greatly.

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