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##  eISSN 0454-8124
pISSN 1225-6951

### Articles

Kyungpook Mathematical Journal 2018; 58(3): 481-488

Published online September 30, 2018

### Generalized Derivations on *-prime Rings

Department of Mathematics, Aligarh Muslim University, Aligarh-202002, India
e-mail : mashraf80@hotmail.com

Department of Mathematics, Integral University, Lucknow-226026, India
e-mail : rashidmaths@gmail.com

Received: December 30, 2015; Revised: August 17, 2018; Accepted: August 20, 2018

Let I be a *-ideal on a 2-torsion free *-prime ring and F : RR a generalized derivation with an associated derivation d : RR. The aim of this paper is to explore the condition under which generalized derivation F becomes a left centralizer i.e., associated derivation d becomes a trivial map (i.e., zero map) on R.

Keywords: derivation, generalized derivation, *-prime ring, centralizer.

### 1. Introduction

Let R be an associative ring. A mapping * : RR is said to be an involution on R if

• * * (x) = x,

• * (x + y) = * (x) + * (y) and

• * (xy) = * (y) * (x)

hold for all x, yR. A ring R equipped with an involution ‘*’ is said to be ring with involution. A ring R with involution ‘*’ is said to be *-prime if aRb = aR*(b) = {0} implies that either a = 0 or b = 0. Note that every prime ring with involution ‘*’ is *- prime but the converse is not true in general. It can be easily seen that a *-prime ring is semiprime. To show this, let aRa = {0}. Therefore, we have ar * (a)sar * (a) = 0 for all r, sR. Also ar * (a)s * (ar * (a)) = ar * (a)sa * (r) * (a) = 0 for all r, sR. Hence by *-primeness, we get ar * (a) = 0. Also, we have assumed that ara = 0. Therefore, by *-primeness, we find that a = 0, as required. An additive mapping d : RR is called a derivation if d(xy) = d(x)y + xd(y) for all x, yR. An additive mapping F : RR is called a generalized derivation if there exists a derivation d : RR such that F(xy) = F(x)y + xd(y) for all x, yR. An additive mapping F : RR is called a left centralizer if F(xy) = F(x)y for all x, yR. The situation when generalized derivation becomes a left centralizer arises when associated derivation is trivial map i.e., zero map. In this paper, our aim is to discuss the situation when derivation associated with the generalized derivation on an appropriate subset of a *-prime ring R becomes zero i.e., generalized derivation becomes a left centralizer.

### 2. Identities Related to Generalized Derivations on *-ideals of *-prime Rings

Over the last three decades, several authors have explored various identities involving automorphisms or derivations on an appropriate subset of a prime or semiprime ring (see [1, 2, 3, 4, 5], where further references can be found). The purpose of this section is to prove some results which are of independent interest and related to generalized derivations on *-prime rings. We begin this section with the following well known result which are needed for developing the proof of the results presented in this section.

### Lemma 2.1

Let R be a *-prime ring and I a nonzero *-ideal of R. If aIb = aI * (b) = {0} or aIb = *(a)Ib = {0} for all a, bI, then either a = 0 or b = 0.

Proof

Given that aIb = aI *(b) = {0} for any a, bR. Hence, axrb = axr *(b) = 0 for all xI and rR. Since R is *-prime, we find that either ax = 0 for all xI or b = 0. But, since I is a *-ideal, ax = 0 implies that arx = ar * (x) = 0 for all xI and rR. Now, *-primeness of R yields that either I = {0} (a contradiction) or a = 0. Similarly, we can prove that if aIb = *(a)Ib = {0} for all a, bI, then either a = 0 or b = 0.

### Theorem 2.2

Let R be a noncommutative *-prime ring and I a nonzero *-ideal of R. Suppose F1, F2 : RR are two generalized derivations with associated derivations d1, d2 : RR respectively such that d1F1 ≠ 0 on I and * commutes with d2, F1. If F1(x)∘F2(y) = 0 for all x, yI or [F1(x), F2(y)] = 0 for all x, yI, then d2 = 0.

Proof

Replace y by yr in F1(x) ∘ F2(y) = 0 and also use this condition, to get

$F2(y)[r,F1(x)]+y(F1(x)∘d2(r))+[F1(x),y]d2(r)=0 for all x,y∈I and r∈R.$

Replacing r by F1(x) in the above relation, we find that

$y(F1(x)∘d2(F1(x)))+[F1(x),y]d2(F1(x))=0 for all x,y∈I.$

Now, replacing y by yz, we get

$[F1(x),y]zd2(F1(x))=0 for all x,y,z∈I.$

Also, replacing x by x + t in (2.2) and using (2.2), we get

$[F1(x),y]zd2(F1(t))+[F1(t),y]zd2(F1(x))=0 for all x,y,z,t∈I.$

Using (2.2) and (2.3), we find that

$[F1(x),y]zd2(F1(t))R[F1(x),y]zd2(F1(t)) =-[F1(t),y]zd2(F1(x))R[F1(x),y]zd2(F1(t)) ={0} for all x,y,z,t∈I.$

Since *-primeness of R implies semiprimeness of R, we obtain that

$[F1(x),y]zd2(F1(t))=0 for all x,y,z,t∈I.$

Since I is *-ideal and * commutes with F1, we get

$*([F1(x),y]zd2(F1(t))=0 for all x,y,z,t∈I.$

Also, R is *-prime, we get either [F1(x), y] = 0 for all x, yI or d2(F1(t)) = 0 for all tI. Take [F1(x), y] = 0 for all x, yI. Replace y by yr to get y[F1(x), r] = 0 for all x, yI and rR and hence yz[F1(x), r] = 0 for all x, y, zI and rR. Since I is a *-ideal, we get *(y)z[F1(x), r] = 0 for all x, y, zI and rR. By Lemma 2.1, we get [F1(x), r] = 0 for all xI and rR. Further, replacing x by xr, we find that

$[x,r]d1(x)+x[d1(r),r)=0 for all x∈I and r∈R.$

Replacing x by xz in (2.5) and using (2.5), we get [x, r]zd1(r) = 0 for all x, zI and rR. Substituting r by r + r1, we find that [x, r]zd1(r1) + [x, r1]zd1(r) = 0 for all x, zI and r, r1R. Therefore [x, r]zd1(r1)R[x, r]zd1(r1) = − [x, r1]zd1(r)R[x, r]zd1(r1) = {0}. Hence, we get [x, r]zd1(r1) = 0 for all x, zI and r, r1R. Since I is a *-ideal, by Lemma 2.1, we have either [x, r] = 0 or d1(r1) = 0. But [x, r] = 0 implies that x[s, r] = 0 for all xI, r, sR. Since I is nonzero *-ideal, we find that xt[s, r] = *(x)t[s, r] = 0 for all xI, r, s, tR. Since R is *-prime and I is nonzero, we find that R is commutative which leads to a contradiction. Also, d1(r1) = 0 implies d1 = 0 which is again a contradiction to the assumption that d1F1 ≠ 0 .

Now, take d2(F1(x)) = 0 for all xI. Replace x by xr, we get

$F1(x)d2(r)+d2(x)d1(r)+xd2(d1(r))=0 for all x∈I and r∈R.$

Replacing r by F1(y), we get

$d2(x)d1(F1(y))+xd2(d1(F1(y)))=0 for all x,y∈I.$

Replace x by xz in (2.6) and using (2.6), we find that d2(x)zd1(F1(y)) = 0 for all x, y, zI. Since I is a *-ideal and * commutes with d2, we get *(d2(x))zd1(F1(y)) = 0 for all x, y, zI. Using *-primeness of R and Lemma 2.1, we find that either d2 = 0 or d1F1 = 0 on I, a contradiction. But d2 = 0 on I implies that d2 = 0 on R.

Replace y by yr in [F1(x), F2(y)] = 0 to get

$F2(y)[F1(x),r]+[F1(x),y]d2(r)+y[F1(x),d2(r)]=0 for all x,y∈I and r∈R.$

Now, replacing r by F1(x), we get

$[F1(x),y]d2(F1(x))+y[F1(x),d2(F1(x))]=0 for all x,y∈I.$

Replace y by yz in (2.7) and use (2.7), to find that [F1(x), y]zd2(F1(x)) = 0 for all x, y, zI. Using similar arguments as used after (2.2), we get that d2 = 0.

### Theorem 2.3

Let R be a noncommutative *-prime ring and I be a nonzero *- ideal in R. Suppose F1, F2 : RR are two generalized derivations with associated derivations d1, d2 : RR respectively such that d1F1 ≠ 0 on I and * commutes with d2, F1. If F1(x) ∘ F2(y) = xy for all x, yI or [F1(x), F2(y)] = [x, y] for all x, yI, then d2 = 0.

Proof

It is given that F1(x) ∘ F2(y) = xy holds for all x, yI. Replace y by yr for yI and rR, to get

$(F1(x)∘F2(y))r+F2(y)[r,F1(x)]+[F1(x),y]d2(r)+y(F1(x)∘d2(r))=(x∘y)r+y[r,x].$

Application of given condition yields that

$F2(y)[r,F1(x)]+[F1(x),y]d2(r)+y(F1(x)∘d2(r))=y[r,x] for all x,y∈I and r∈R.$

Replacing r by F1(x), we get

$[F1(x),y]d2(F1(x))+y(F1(x)∘d2(F1(x)))=y[F1(x),x] for all x,y∈I.$

Replacing y by yz in (2.9) and using (2.9), we have

$[F1(x),y]zd2(F1(x))=0 for all x,y,z∈I.$

Arguing with similar lines as used after (2.2), we get the required result.

Take [F1(x), F2(y)] = [x, y] for all x, yI. Replace y by yr for yI, rR and use similar arguments as used above with necessary variations, we get the required result.

### Theorem 2.4

Let R be a noncommutative *-prime ring and I a nonzero *- ideal in R. Suppose F1, F2 : RR are two generalized derivations with associated derivations d1, d2 : RR respectively and * commutes with d2, F1. If F1(x)F2(y) ± xyZ(R) for all x, yI, then either d1 = 0 or d2 = 0.

Proof

Replace y by yz in F1(x)F2(y) ± xyZ(R) for y, zI, we find that

$(F1(x)F2(y)±xy)z+F1(x)yd2(z)∈Z(R) for all x,y,z∈I.$

On commuting with z, we get

$[F1(x)F2(y)±xy,z]z+[F1(x),z]yd2(z)+F1(x)[yd2(z),z]=0 for all x,y,z∈I.$

Using the given hypothesis we obtain

$[F1(x),z]yd2(z)+F1(x)[yd2(z),z]=0 for all x,y,z∈I.$

Replacing y by F1(x)y in (2.10), we find that

$[F1(x),z]F1(x)yd2(z)+F1(x)(F1(x)[yd2(z),z]+[F1(x),z]yd2(z))=0 for all x,y,z∈I.$

Application of (2.10) in (2.11) yields that

$[F1(x),z]F1(x)yd2(z)=0 for all x,y,z∈I.$

Replacing z by z + t in (2.12) and using (2.12), we find that

$[F1(x),z]F1(x)yd2(t)+[F1(x),t]F1(x)yd2(z)=0 for all x,y,z,t∈I.$

Application of (2.12) and (2.13) gives that (2.14)

$[F1(x),z]F1(x)yd2(t)R[F1(x),z]F1(x)yd2(t) =-[F1(x),t]F1(x)yd2(t)R[F1(x),z]F1(x)yd2(z) ={0} for all x,y,z,t∈R.$

As a *-prime ring is also semiprime, (2.14) implies that [F1(x), z]F1(x)yd2(t) = 0 for all x, y, z, tI. Since * commutes with d2 and I is a *-ideal, we have

$[F1(x),z]F1(x)yd2(t)=0=[F1(x),z]F1(x)y*(d2(t)) for all x,y,z,t∈I.$

Since R is *-prime and I is *-ideal, by Lemma 2.1, we find that either [F1(x), z]F1(x) = 0 or d2(t) = 0 for all x, z, tI. But d2 = 0 on I implies that d2 = 0 on R. Now, take [F1(x), z]F1(x) = 0 for all x, zI. On replacing z by yz for z, yI, we find that [F1(x), y]zF1(x) = 0 for all x, y, zI. Replacing x by x + t for x, tI in [F1(x), y]zF1(x) = 0 and using the similar arguments as used after (2.14), we get [F1(x), y]zF1(t) = 0 for all x, y, z, tI. Since I is a *-ideal and * commutes with F1, we find that [F1(x), y]zF1(t) = [F1(x), y]z * (F1(t)) = 0 for all x, y, z, tI. By Lemma 2.1, we have either [F1(x), y] = 0 for all x, yI or F1(t) = 0 for all tI. If F1(t) = 0 for all tI, then F1(ty) = 0 for all t, yI. This yields that td1(y) = 0 for all t, yI. Replace t by xt for x, tI, we find that xtd1(y) = 0. Since I is *-ideal, xtd1(y) = *(x)td1(y) = 0 for all t, x, yI. By Lemma 2.1, we find that either I = {0} or d1 = 0 on I. But I ≠ {0} implies that d1 = 0 on I which yields that d1 = 0 on R. Now, consider [F1(x), z] = 0 for all x, zI. Replace x by xz in [F1(x), z] = 0 to find that

$[x,z]d1(z)+x[d1(z),z]=0 for all x,z∈I.$

Again replacing x by xy in (2.15) and using (2.15), we get [x, z]yd1(z) = 0 for all x, y, zI. Use the similar steps as used after (2.14) with necessary variations, we find that [x, z]yd1(s) = 0 for all x, y, z, sI. But, since I is a *-ideal, we find that [x, z]yd1(s) = 0 = *([x, z])yd1(s) for all x, y, z, sI. By Lemma 2.1, we get that either I is commutative or d1 = 0 on I. But, in a semiprime ring if an ideal is commutative then it is central and since *-prime ring is semiprime, we get I is central and hence R is commutative which is a contradiction.

### Theorem 2.5

Let R be a noncommutative *-prime ring and I a nonzero *-ideal in R. Suppose F1, F2 : RR are two generalized derivations with associated derivations d1, d2 : RR respectively. If F1(xy) ± F2(yx) = ±[x, y] for all x, yI, then d1 = d2 = 0.

Proof

Let F1(xy) + F2(yx) = ±[x, y] for all x, yI. Replace y by yx in F1(xy) + F2(yx) = ±[x, y] for all x, yI, to find that

$(F1(xy)+F2(yx))x+xyd1(x)+yxd2(x)=±[x,y]x for all x,y∈I.$

Using the given hypothesis, we get

$xyd1(x)+yxd2(x)=0 for all x,y∈I.$

Replace y by yz for y, zI to get 0 = xyzd1(x) + yzxd2(x) = [x, y]zd1(x) for all x, y, zI. Using the similar steps as used after (2.14) with necessary variations, we find that [x, z]yd1(s) = 0 for all x, y, z, sI. As I is a *-ideal, we find that [x, y]zd1(s) = *([x, y])zd1(s) for all x, y, z, sI. By Lemma 2.1, we get that either d1 = 0 on I or I is commutative. But, if d1 = 0 on I, then d1 = 0 on R. Also, if I is a commutative ideal of *-prime ring R, then I is central ideal of R. But if I is central, then R is commutative which leads to a contradiction.

If we replace x by xy in F1(xy) + F2(yx) = ±[x, y] and use the similar arguments as used above, then we find that d2 = 0.

Now, take F1(xy) − F2(yx) = ±[x, y] for all x, yI. Using the similar arguments as used above, we get the required result.

### Theorem 2.6

Let R be a noncommutative *-prime ring and I a nonzero *-ideal in R. Suppose F1, F2 : RR are two generalized derivations with associated derivations d1, d2 : RR respectively. If F1(x)y ± yF2(x) = 0 for all x, yI, then d1 = d2 = 0.

Proof

Replacing y by yz in F1(x)yyF2(x) = 0, we find that [F1(x), y]z = 0 for all x, y, zI. As I is a nonzero *-ideal and R is *-prime ring, by Lemma 2.1, we find that [F1(x), y] = 0 for all x, yI. Replacing x by xy in [F1(x), y] = 0, we get

$[x,y]d1(y)+x[d1(y),y]=0 for all x,y∈I.$

Replacing x by xz in (2.17) and using (2.17), we find that [x, y]zd1(y) = 0 for all x, y, zI. Using the similar steps as used after (2.14) with necessary variations, we find that [x, z]yd1(s) = 0 for all x, y, z, sI. Hence either d1 = 0 on I, and hence on R or I is commutative, hence I is central. But if I is central, then R is commutative which leads to a contradiction. Further, replacing x by xy in F1(x)yyF2(x) = 0, we find that

$xd1(y)y-yxd2(y)=0 for all x,y∈I.$

Replacing x by xz in (2.18) and using (2.18), we find that [x, y]zd2(y) = 0. Again, using the similar steps as used after (2.14) with necessary variations, we find that [x, z]yd2(s) = 0 for all x, y, z, sI. This implies that either d2 = 0 on I and hence on R or I is central i.e., R is commutative, a contradiction. Now, take F1(x)y + yF2(x) = 0 for all x, yI and use the similar arguments as used above with necessary variations, we get the required result.

### Theorem 2.7

Let R be a noncommutative *-prime ring and I a nonzero *-ideal in R. Suppose F1, F2 : RR are two generalized derivations with associated derivations d1, d2 : RR respectively and * commutes with d1. If F1(xy) + F2(yx) = F1(x)y + yF2(x) for all x, yI, then either d1 = 0 or d2 = 0.

Proof

Let F1(xy) + F2(yx) = F1(x)y + yF2(x) for all x, yI. Replace y by yx in F1(xy) + F2(yx) = F1(x)y + yF2(x), to get

$(F1(xy)+F2(yx))x+xyd1(x)+yxd2(x)=F1(x)yx+yxF2(x) for all x,y∈I.$

Using the given hypothesis, we find that

$yF2(x)x+xyd1(x)+yxd2(x)=yxF2(x) for all x,y∈I.$

This implies that

$y[F2(x),x]+xyd1(x)+yxd2(x)=0 for all x,y∈I.$

Replacing y by d2(z)y in (2.19) and using (2.19), we find that [x, d2(z)]yd1(x) = 0 for all x, y, zI. Using the similar steps as used after (2.14) with necessary variations, we find that [x, d2(z)]yd1(s) = 0 for all x, y, z, sI. This implies that either d1 = 0 on I and hence d1 = 0 on R or [x, d2(z)] = 0 for all x, zI. Take [x, d2(z)] = 0 for all x, zI. Replacing z by yx in [x, d2(z)] = 0, we find that [x, y]d2(x) = 0 for all x, yI. This yields that [x, y]zd2(x) = 0 for all x, y, zI and hence either I is central or d2 = 0 on I i.e., d2 = 0 on R. If I is central, then R is commutative, which is a contradiction.

### Theorem 2.8

Let R be a noncommutative *-prime ring and I be a nonzero *- ideal in R. Suppose F1, F2 : RR are two generalized derivations with associated derivations d1, d2 : RR respectively. If F1(xy) ± F2(yx) = 0 for all x, yI, then d1 = d2 = 0.

Proof

It is given that F1(xy) ± F2(yx) = 0 for all x, yI. Replacing y by yx, we find that

$F1(xy)x+xyd1(x)±(F2(yx)x+yxd2(x))=0 for all x,y∈I.$

Using the given condition, we get

$xyd1(x)±yxd2(x)=0 for all x,y∈I.$

Replacing y by yz for y, zI, we find that 0 = xyzd1(xyzxd2(x) = [x, y]zd1(x) for all x, y, zI. Replacing x by x+s for x, sI, we get [x, y]zd1(s)+[s, y]zd1(x) = 0. Hence, we find that [x, y]zd1(s)R[x, y]zd1(s) = − [s, y]zd1(s)R[x, y]zd1(x) = {0} for all x, y, z, sI. Now, since *-prime ring is semiprime, the above yields that [x, y]zd1(s) = 0 for all x, y, z, sI. Since I is *-prime ring, we find that *([x, y])zd1(s) = [x, y]zd1(s) = 0 for all x, y, z, sI. By, Lemma 2.1, either I is commutative, hence central which implies that R is commutative, a contradiction or d1 = 0 on I and hence d1 = 0 on R.

Similarly, we can show that d2 = 0.

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