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##  eISSN 0454-8124
pISSN 1225-6951

### Articles

Kyungpook Mathematical Journal 2019; 59(1): 1-11

Published online March 23, 2019

### Study of Generalized Derivations in Rings with Involution

Department of Mathematics, Aligarh Muslim University, Aligarh, India, e-mail: muzibamu81@gmail.com and adnan.abbasi001@gmail.com

Govt. HSS, Kaprin, Shopian, Jammu and Kashmir, India, e-mail: ndmdarlajurah@gmail.com

Received: March 20, 2018; Revised: August 8, 2018; Accepted: August 13, 2018

### Abstract

Let R be a prime ring with involution of the second kind and centre Z(R). Suppose R admits a generalized derivation F: RR associated with a derivation d: RR. The purpose of this paper is to study the commutativity of a prime ring R satisfying any one of the following identities: (i) F(x) ? x*Z(R) (ii) F([x, x*]) ± x ? x*Z(R) (iii) F(x?x*) ± [x, x*] ∈ Z(R) (iv) F(x)?d(x*) ± x?x*Z(R) (v) [F(x), d(x*)] ± x?x*Z(R) (vi) F(x) ± x ? x*Z(R) (vii) F(x) ± [x, x*] ∈ Z(R) (viii) [F(x), x*] ? F(x) ? x*Z(R) (ix) F(x ? x*) ∈ Z(R) for all xR.

Keywords: prime ring, generalized derivation, derivation, involution.

### 1. Introduction

Throughout this paper R will represent a prime ring with center Z(R). An additive mapping *: RR is called an involution if * is an anti-automorphism of order 2; that is, (x*)* = x for all xR. An element x in a ring with involution is said to be hermitian if x* = x and skew-hermitian if x* = −x. The sets of hermitian and skew-hermitian elements of R will be denoted by H(R) and S(R), respectively. A ring equipped with an involution * is known as ring with involution or *-ring. If char(R) ≠ 2, involution is said to be of the first kind if Z(R) ⊆ H(R), otherwise it is said to be of the second kind. In the later case, S(R) ∩ Z(R) ≠ (0). A ring R is said to be normal if xx* = x*x for all xR. An example is the ring of quaternions. A description of such rings can be found in , where further references can be found.

A derivation on R is an additive mapping d: RR such that d(xy) = d(x)y + xd(y) for all x, yR. Following Bresar , 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. Basic examples are derivations and generalized inner derivations i.e., maps of type xax + xb for some a, bR. A map f: RR is said to be centralizing on R if [f(x), x] ∈ Z(R) for all xR. In a special case, when [f(x), x] = 0 holds for all xR, a map f is said to be commuting on R. The study of centralizing and commuting mappings on prime rings was initiated by the result of Posner , which states that the existence of a nonzero centralizing derivation on a prime ring implies that the ring has to be commutative. Through the years, a lot of work has been done in this context by a number of authors (see, for example, [1, 5, 7, 8, 13, 17] and references therein).

Very recently in many papers the additive mappings like derivations, generalized derivations have been studied in the setting of rings with involution and in fact it was seen that there is a close connection between these mappings and the commutativity of the ring R. For instance in , it is proved that let R be a prime ring with involution * such that char(R) ≠ 2. Let d be a nonzero derivation of R such that [d(x), x*] ∈ Z(R) for all xR and d(S(R) ∩ Z(R)) ≠ (0). Then R is commutative. Many other results in this direction can be found in [3, 4, 10, 15]. The goal of the present paper is to continue this line of investigation by considering certain identities involving Jordan product in the setting of generalized derivations.

### 2. Main Results

We begin our investigation with the following lemmas, which are essential to prove our main results.

### Lemma 2.1.()

Let R be a prime ring with involution ‘*’ of the second kind. Then [x, x*] ∈ Z(R) for all xR if and only if R is commutative.

### Lemma 2.2.()

Let R be a prime ring with involution ‘*’ of the second kind. Then xx*Z(R) for all xR if and only if R is commutative.

### Theorem 2.3

Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: RR associated with a derivation d: RR, such that F(x) ∘ x*Z(R) for all xR. Then R is commutative.

Proof

By the given hypothesis, we have

$F(x)∘x*∈Z(R) for all x∈R.$

A Linearization of (2.1) yields that

$F(x)∘y*+F(y)∘x*∈Z(R) for all x,y∈R.$

This can be further written as

$[F(x)∘y*,r]+[F(y)∘x*,r]=0 for all x,y,r∈R.$

Replacing y by hy where hH(R) ∩ Z(R), we get

$[F(x)∘(hy)*,r]+[F(hy)∘x*,r]=0 for all x,y,r∈R.$

On solving, we obtain

$[F(x)∘y*,r]h+[F(y)∘x*,r]h+[y∘x*,r]d(h)=0 for all x,y,r∈R$

and hH(R) ∩ Z(R). Using (2.2), we get [yx*, r]d(h) = 0 for all x, y, rR and hH(R) ∩ Z(R). Using the primeness of R we have either [yx*, r] = 0 for all x, y, rR or d(h) = 0 for all hH(R)∩Z(R). First if we consider [yx*, r] = 0 for all x, y, rR. Replacing y by x we get [xx*, r] = 0 for all x, rR. Thus in view of Lemma 2.2, we get R is commutative. Now consider the second case d(h) = 0 for all hH(R) ∩ Z(R). This intern implies d(k) = 0 for all kS(R) ∩ Z(R) and hence d(z) = 0 for all zZ(R). Replacing y by ky where kS(R) ∩ Z(R) in (2.2) and using d(z) = 0 for all zZ(R), We obtain

$-[F(x)∘y*,r]k+[F(y)∘x*,r]=0 for all x,y,r∈R and k∈S(R)∩Z(R).$

Using (2.2) in the previous equation, we get 2[F(y)∘x*, r]k = 0 for all x, y, rR and kS(R)∩Z(R). Since char(R) ≠ 2, we have [F(y)∘x*, r]k = 0for all x, y, rR and kS(R)∩Z(R). Now using the primeness and the fact that S(R)∩Z(R) ≠ (0), we have [F(y)∘x*, r] = 0 for all x, y, rR. That is, [F(y) ∘ x, r] = 0 for all x, y, rR. Taking x = z where zZ(R), we get [2F(x)z, r] = 0 for all x, rR and zZ(R). Since char(R) ≠ 2, we get [F(x)z, r] = 0 for all x, rR and zZ(R). Further since S(R) ∩ Z(R) ≠ (0) and using the primeness of R, we have [F(y), r] = 0 for all y, rR. Replace y by r, we get [F(r), r] = 0 for all rR. Thus in view of [18, Theorem 3.1], R is commutative.

### Theorem 2.4

Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: RR associated with a derivation d: RR, such that F([x, x*]) ± xx*Z(R) for all xR. Then R is commutative.

Proof

We first consider the case

$F([x,x*])+x∘x*∈Z(R) for all x∈R.$

If F is zero, then we get xx*Z(R). Then by Lemma 2.2, we get R is commutative. Now consider F is non zero. Linearizing (2.3), we get

$F([x,y*])+F([y,x*])+x∘y*+y∘x*∈Z(R) for all x,y∈R.$

This can be further written as

$[F([x,y*]),r]+[F([y,x*]),r]+[x∘y*,r]+[y∘x*,r]=0 for all x,y,r∈R.$

Replacing y by hy where hH(R) ∩ Z(R), we have

$[F([x,y*]h),r]+[F([y,x*]h),r]+[x∘y*,r]h+[y∘x*,r]h=0$

for all x, y, rR and hH(R) ∩ Z(R). This further implies that

$[F([x,y*]),r]h+[[x,y*],r]d(h)+[F([y,x*]),r]h+[[y,x*],r]d(h)+[x∘y*,r]h$

+[yx*, r]h = 0 for all x, y, rR and hH(R) ∩ Z(R). Using (2.4) in (2.5), we get ([[x, y*], r] + [[y, x*], r])d(h) = 0 for all x, y, rR and hH(R) ∩ Z(R). Using the primeness of R we have either ([[x, y*], r] + [[y, x*], r]) = 0 for all x, y, rR or d(h) = 0 for all hH(R) ∩ Z(R). If we consider ([[x, y*], r] + [[y, x*], r]) = 0 for all x, y, rR. This implies that [x, y*] + [y, x*] ∈ Z(R) for all x, yR. Replacing y by x we get [x, x*] ∈ Z(R) for all xR. Thus in view of Lemma 2.1, we get R is commutative. Now suppose that d(h) = 0 for all hH(R) ∩ Z(R). This further implies that d(z) = 0 for all zZ(R). Replacing y by ky in (2.4) where kS(R) ∩ Z(R), we get

$-F([x,y*]k)+F([y,x*]k)-(x∘y*)k+(y∘x*)k∈Z(R) for all x,y∈R$

and kS(R) ∩ Z(R). This further implies that

$-F([x,y*])k-[x,y*]d(k)+F([y,x*])k+[y,x*]d(k)-(x∘y*)k+(y∘x*)k∈Z(R)$

for all x, yR and kS(R) ∩ Z(R). Using (2.4) and the fact that d(z) = 0 for all zZ(R), we get (F([y, x*]) + yx*)kZ(R) for all x, yR and kS(R) ∩ Z(R). Now using the primeness and the fact that S(R) ∩ Z(R) ≠ (0), we obtain (F([y, x*]) + yx*) ∈ Z(R) for all x, yR. Replace x by x*, we get (F([y, x]) + yx) ∈ Z(R) for all x, yR. Taking y = x, we get x2Z(R) for all xR. Replacing x by x + y and using x2Z(R) for all xR, we get xy + yxZ(R) for all x, yR. This can be further written as [xy + yx, r] = 0 for all x, y, rR. Replace y by z where zZ(R), we get 2[x, r]z = 0 for all x, rR and zZ(R). Finally using the facts that char(R) ≠ 2, S(R) ∩ Z(R) ≠ (0) and the primeness of R, we obtain R is commutative.

The second case can be proved in a similar manner with necessary variations.

### Theorem 2.5

Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: RR associated with a derivation d: RR, such that F(xx*) ± [x, x*] ∈ Z(R) for all xR. Then R is commutative.

Proof

We first consider the positive sign case. If F is zero then we have [x, x*] ∈ Z(R) for all xR. Thus in view of Lemma 2.1, we get R is commutative. If F is nonzero then we have

$F(x∘x*)+[x,x*]∈Z(R) for all x∈R.$

Linearizing (2.6), we get

$F(x∘y*)+F(y∘x*)+[x,y*]+[y,x*]∈Z(R) for all x,y∈R.$

Replacing y by hy where hH(R) ∩ Z(R) and using (2.7), we get

$(x∘y*+y∘x*)d(h)∈Z(R) for all x,y∈Z(R) and h∈H(R)∩Z(R).$

Now using the primeness of R, we have (xy* + yx*) ∈ Z(R) for all x, yR or d(h) = 0 for all hH(R) ∩ Z(R). Replace y by x, we get xx*Z(R) for all xR. Thus by Lemma 2.2, we get R is commutative. Now consider d(h) = 0 for all hH(R) ∩ Z(R). This further implies that d(z) = 0 for all zZ(R). Replacing y by ky in (2.7) where kS(R) ∩ Z(R) and using (2.7), we have 2(F(yx*) + [y, x*])kZ(R) for all x, yR and kS(R) ∩ Z(R). This further implies that F(yx*) + [y, x*] ∈ Z(R) for all x, yR. Replacing x by x*, we get F(yx) + [y, x] ∈ Z(R) for all x, yR. Taking y = x, we get F(x2) ∈ Z(R) for all xR. Linearizing this and using F(x2) ∈ Z(R) for all xR, we get F(xy) ∈ Z(R) for all x, yR. Replacing y by z, where zZ(R) we get 2F(xz) ∈ Z(R) for all xR and zZ(R). This can be further written as 2[F(xz), r] = 0 for all x, rR and zZ(R). Since char(R) ≠ 2 this implies that [F(xz), r] = 0 for all x, rR and zZ(R). Since d(z) = 0 for all zZ(R), we finally arrive at [F(x), r]z = 0 for all x, rR and zZ(R). Now using the primeness and the fact that S(R)∩Z(R) ≠ (0) we have [F(x), r] = 0 for all x, rR. Replace r by x we get [F(x), x] = 0 for all xR. Then by [18, Theorem 3.1], we get R is commutative. Following the same steps, we get R is commutative in the second case as well.

### Theorem 2.6

Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: RR associated with a derivation d: RR, such that F(x) ∘ d(x*) ± xx*Z(R) for all xR. Then R is commutative.

Proof

We first consider the case

$F(x)∘d(x*)+x∘x*∈Z(R) for all x∈Z(R).$

If either F or d or both are zero then we get xx*Z(R) for all xZ(R). Then by Lemma 2.2, we get R is commutative. Now consider the case in which both F and d are nonzero. Linearizing (2.8), we get

$F(x)∘d(y*)+F(y)∘d(x*)+x∘y*+y∘x*∈Z(R) for all x,y∈R.$

Replacing y by hy where hH(R) ∩ Z(R) and using (2.9), we get

$(F(x)∘y*+y∘d(x*))d(h)∈Z(R) for all x,y∈R and h∈H(R)∩Z(R).$

Using the primeness condition, we get either F(x) ∘ y* + yd(x*) ∈ Z(R) for all x, yR or d(h) = 0 for all hH(R) ∩ Z(R). First consider F(x) ∘ y* +yd(x*) ∈ Z(R) for all x, yR. This can be further written as

$[F(x)∘y*,r]+[y∘d(x*),r]=0 for all x,y,r∈R.$

Replacing y by ky where kS(R)∩Z(R) and using (2.10), we get 2[yd(x*), r]k = 0 for all x, y, rR and kS(R) ∩ Z(R). Using the primeness and the facts that char(R) ≠ 2 and S(R) ∩ Z(R) ≠ (0), we get [yd(x*), r] = 0 for all x, y, rR. Replacing x by x* we have [yd(x), r] = 0 for all x, y, rR. Replacing y by z where zZ(R), we get 2[d(x), r]z = 0 for all x, rR and zZ(R). Since char(R) ≠ 2 and S(R) ∩ Z(R) ≠ (0), this implies that [d(x), r] = 0 for all x, rR. Then by posner’s result , we get R is commutative. Now we consider the case d(h) = 0 for all hH(R) ∩ Z(R). This implies that d(z) = 0 for all zZ(R). Replacing y by ky in (2.9) where kS(R) ∩ Z(R) and using (2.9), we obtain

$2(F(y)∘d(x*)+y∘x*)k∈Z(R) for all x,y∈R and k∈S(R)∩Z(R).$

This further implies that F(y) ∘ d(x*) + yx*Z(R) for all x, yR. That is, F(y) ∘ d(x) + yxZ(R) for all x, yR. Taking x = z where zZ(R) and using d(z) = 0 for all zZ(R), we get 2yzZ(R) for all yR and zZ(R). Since char(R) ≠ 2, we get that yzZ(R) for all yR and zZ(R). This can be written as [yz, r] = 0 for all y, rR and zZ(R). This further implies that [y, r] = 0 for all y, rR. That is, R is commutative.

Similarly we can prove the second part.

### Theorem 2.7

Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: RR associated with a derivation d: RR, such that [F(x), d(x*)] ± xx*Z(R) for all xR. Then R is commutative.

Proof

The proof is on the similar lines as in the above theorem.

### Theorem 2.8

Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: RR associated with a derivation d: RR, such that F(x) ± xx*Z(R) for all xR. Then R is commutative.

Proof

we first consider the case

$F(x)+x∘x*∈Z(R) for all x∈R.$

If F is zero then we have xx*Z(R) for all xR. Then by Lemma 2.2, we get R is commutative. Consider F is nonzero. Linearizing (2.12), we get

$x∘y*+y∘x*∈Z(R) for all x,y∈R.$

Replacing y by ky where kS(R) ∩ Z(R) and using (2.13), we get (yx*)kZ(R) for all x, yR and kS(R) ∩ Z(R). Using primeness and the fact that S(R) ∩ Z(R) ≠ (0), we have yx*Z(R) for all x, yR. Taking y = x, we get xx*Z(R) for all xR. Thus in view of Lemma 2.2, we get R is commutative.

Similarly we can prove the other case.

On similar lines we can prove the following result.

### Theorem 2.9

Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: RR associated with a derivation d: RR, such that F(x) ± [x, x*] ∈ Z(R) for all xR. Then R is commutative.

### Theorem 2.10

Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: RR associated with a derivation d: RR, such that [F(x), x*] ∓ F(x) ∘ x*Z(R) for all xR. Then R is commutative.

Proof

We first consider the case

$[F(x),x*]-F(x)∘x*∈Z(R) for all x∈R.$

Linearizing (2.14), we get

$[F(x),y*]+[F(y),x*]-F(x)∘y*-F(y)∘x*∈Z(R) for all x,y∈Z(R).$

Replacing y by hy where hH(R) ∩ Z(R) and using (2.15), we get

$([y,x*]-y∘x*)d(h)∈Z(R) for all x,y∈R and h∈H(R)∩Z(R).$

Using the primeness condition we have [y, x*] − yx*Z(R) for all x, yR or d(h) = 0 for all hH(R) ∩ Z(R). First suppose [y, x*] − yx*Z(R) for all x, yR. This further implies that [y, x] − yxZ(R) for all x, yR. Replacing y by x we get x2Z(R) for all xR. Replacing x by x+y, we obtain xyZ(R) for all x, yR. That is, [xy, r] = 0 for all x, y, rR. Taking y = z, where zZ(R), we get 2[x, r]z = 0 for all x, rR and zZ(R). Since char(R) ≠ 2, S(R)∩Z(R) ≠ (0) and using the primeness of R, we have [x, r] = 0 for all x, rR. This implies that R is commutative. Now we consider the case d(h) = 0 for all hH(R) ∩ Z(R). This intern implies that d(z) = 0 for all zZ(R). Replacing y by ky where kS(R) ∩ Z(R) in (2.15) and using (2.15), we get

$2([F(y),x*]-F(y)∘x*)k∈Z(R) for all x,y∈R and k∈S(R)∩Z(R).$

Since char(R) ≠ 2 and S(R) ∩ Z(R) ≠ (0), we obtain

$[F(y),x*]-F(y)∘x*∈Z(R) for all x,y∈R.$

Replacing x by x*, we get

$[F(y),x]-F(y)∘x∈Z(R) for all x,y∈R.$

This can be further written as F(y)xxF(y)−F(y)xxF(y) ∈ Z(R) for x, yR. This implies that −2xF(y) ∈ Z(R) for all x, yR. That is, xF(y) ∈ Z(R) for all x, yR, since char(R) ≠ 2. That is, [xF(y), r] = 0 for all x, y, rR. Replacing r by x, we get

$x[F(y),x]=0 for all x,y∈R.$

Replacing y by yx we get x[F(y), x]x+xy[d(x), x]+x[y, x]d(x) = 0 for all x, yR. using (2.18), we get xy[d(x), x] + x[y, x]d(x) = 0 for all x, yR. Taking y = x, we have

$x2[d(x),x]=0 for all x∈R.$

Replacing x by x + z where zZ(R) and using d(z) = 0 for all zZ(R), we obtain (z2 +2xz)[d(x), x] = 0 for all xR and zZ(R). Left multiplying by x we get z2x[d(x), x] + x2[d(x), x]2z = 0 for all xR and zZ(R). Using (2.19), we get z2x[d(x), x] = 0 for all xR and zZ(R). Now using the primeness and the fact that S(R) ∩ Z(R) ≠ (0), we get

$x[d(x),x]=0 for all x∈R.$

Replacing x by x + z and using d(z) = 0 for all zZ(R), we get z[d(x), x] = 0 for all xR and zZ(R). Using primeness and the fact that S(R) ∩ Z(R) ≠ (0), we finally arrive at [d(x), x] = 0 for all xR. Thus by the result of Posner , we get R is commutative.

Now we consider the case

$[F(x),x*]+x∘x*∈Z(R) for all x∈R.$

Using the same steps as we did in the above case, we get

$[F(y),x*]+F(y)∘x*∈Z(R) for all x,y∈R.$

Replacing x by x*, we have

$[F(y),x]+F(y)∘x∈Z(R) for all x,y∈R.$

This can be further written as

$F(y)x-xF(y)+F(y)x+xF(y)∈Z(R) for all x,y∈R.$

That is, 2F(y)xZ(R) for all x, yR. Since char(R) ≠ 2, this implies that F(y)xZ(R) for all x, yR. That is, [F(y)x, r] = 0 for all x, y, rR. This intern implies that F(y)[x, r] + [F(y), r]x = 0 for all x, y, rR. Taking r = x, we get

$[F(y),x]x=0 for all x,y∈R.$

Replacing y by yx, we arrive at

$[F(y),x]x2+y[d(x),x]x+[y,x]d(x)x=0 for all x,y∈R.$

Using (2.22), we get

$y[d(x),x]x+[y,x]d(x)x=0 for all x,y∈R.$

Replacing y by z where zZ(R). We obtain

$z[d(x),x]x=0 for all x∈R and z∈Z(R).$

Using the primeness and the fact that S(R) ∩ Z(R) ≠ (0), we have [d(x), x]x = 0 for all xR. Replacing x by x+z, where zZ(R), using (2.23) and the fact that d(z) = 0 for all zZ(R), we get [d(x), x]z = 0 for all xR and zZ(R). This further implies that [d(x), x] = 0 for all xR. Hence in view of Posner’s result , we get R is commutative.

### Theorem 2.11

Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: RR associated with a derivation d: RR, such that F(xx*) ∈ Z(R) for all xR. Then R is commutative.

Proof

Proceeding on the similar lines as we did in the previous result, we get F(xy*) ∈ Z(R) for all x, yR (where d(Z(R)) = (0)). Replacing y by z* where zZ(R), we get 2F(x)zZ(R) for all xR and zZ(R). Using the conditions that R is prime, char(R) ≠ 2 and S(R) ∩ Z(R) ≠ (0), we finally arrive at F(x) ∈ Z(R) for all xR. This implies that [F(x), r] = 0 for all x, rR. Then by the result , we get R is commutative.

At the end of paper, let us write an example which shows that the restriction of the second kind involution in our results is not superfluous

### Example 2.21

Let $R={(α1α2α3α4)|α1,α2,α3,α4∈Z2}$. Of course R under matrix addition and matrix multiplication is a noncommutative prime ring. Define mappings F, D, *: R −→ R such that

$F (α1α2α3α4)=(0α2α3α4),D (α1α2α3α4)=(0α2α3α4),(α1α2α3α4)*=(α4α2α3α1).$

Obviously,

$Z(R)={(α100α1)|α1∈Z2}.$

Then x* = x for all xZ(R), and hence Z(R) ⊆ H(R), which shows that the involution * is of the first kind. Moreover, F, D are nonzero generalized derivation and derivation such that the following conditions hold

• F(x) ∘ x*Z(R),

• F([x, x*]) ± xx*Z(R),

• F(xx*) ± [x, x*] ∈ Z(R),

• F(x) ∘ D(x*) ± xx*Z(R),

• [F(x), D(x*)] ± xx*Z(R),

• F(x) ± xx*Z(R),

• F(x) ± [x, x*] ∈ Z(R),

• [F(x), x*] ∓ F(x) ∘ x*Z(R),

• F(xx*) ∈ Z(R),

for all xR. However, R is not commutative. Hence, the hypothesis of second kind involution is crucial in our results.

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