﻿ The Commutativity of a *-Ring with Generalized Left *-α-Derivation

Vol.08 No.02(2018), Article ID:82610,10 pages
10.4236/apm.2018.82009

The Commutativity of a *-Ring with Generalized Left *-α-Derivation

Ahmet Oğuz Balcı1, Neşet Aydin1, Selin Türkmen2

1Department of Mathematics, Faculty of Arts and Sciences, Çanakkale Onsekiz Mart University, Çanakkale, Turkey

2Lapseki Vocational School, Çanakkale Onsekiz Mart University, Çanakkale, Turkey

Received: December 21, 2017; Accepted: February 23, 2018; Published: February 26, 2018

ABSTRACT

In this paper, it is defined that left *-α-derivation, generalized left *-α-derivation and *-α-derivation, generalized *-α-derivation of a *-ring where α is a homomorphism. The results which proved for generalized left *-derivation of R in [1] are extended by using generalized left *-α-derivation. The commutativity of a *-ring with generalized left *-α-derivation is investigated and some results are given for generalized *-α-derivation.

Keywords:

*-Ring, Prime *-Ring, Generalized Left *-α-Derivation, Generalized *-α-Derivation

1. Introduction

Let R be an associative ring with center $Z\left(R\right)$ . $xy+yx$ where $x,y\in R$ is denoted by $\left(x,y\right)$ and $xy-yx$ where $x,y\in R$ is denoted by $\left[x,y\right]$ which holds some properties: $\left[xy,z\right]=x\left[y,z\right]+\left[x,z\right]y$ and $\left[x,yz\right]=\left[x,y\right]z+y\left[x,z\right]$ . An additive mapping α which holds $\alpha \left(xy\right)=\alpha \left(x\right)\alpha \left(y\right)$ for all $x,y\in R$ is called a homomorphism of R. An additive mapping β which holds $\beta \left(xy\right)=\beta \left(y\right)\beta \left(x\right)$ for all $x,y\in R$ is called an anti-homomorphism of R. A homomorphism of R is called an epimorphism if it is surjective. A ring R is called a prime if $aRb=\left(0\right)$ implies that either $a=0$ or $b=0$ for fixed $a,b\in R$ . In private, if $b=a$ , it implies that R is a semiprime ring. An additive mapping $\ast :R\to R$ which holds ${\left(xy\right)}^{\ast }={y}^{\ast }{x}^{\ast }$ and ${\left({x}^{\ast }\right)}^{\ast }=x$ for all $x,y\in R$ is called an involution of R. A ring R which is equipped with an involution * is called a *-ring. A *-ring R is called a prime *-ring (resp. semiprime *-ring) if R is prime (resp. semiprime). A ring R is called a *-prime ring if $aRb=aR{b}^{\ast }=\left(0\right)$ implies that either $a=0$ or $b=0$ for fixed $a,b\in R$ .

Notations of left *-derivation and generalized left *-derivation were given in $abu$ : Let R be a *-ring. An additive mapping $d:R\to R$ is called a left *-derivation if $d\left(xy\right)={x}^{\ast }d\left(y\right)+yd\left(x\right)$ holds for all $x,y\in R$ . An additive mapping $F:R\to R$ is called a generalized left *-derivation if there exists a left *-derivation d such that $F\left(xy\right)={x}^{\ast }F\left(y\right)+yd\left(x\right)$ holds for all $x,y\in R$ . An additive mapping $T:R\to R$ is called a right *-centralizer if $T\left(xy\right)={x}^{\ast }T\left(y\right)$ for all $x,y\in R$ . It is clear that a generalized left *-derivation associated with zero mapping is a right *-centralizer on a *-ring.

A *-derivation on a *-ring was defined by Bresar and Vukman in [2] as follows: An additive mapping $d:R\to R$ is said to be a *-derivation if $d\left(xy\right)=d\left(x\right){y}^{\ast }+xd\left(y\right)$ for all $x,y\in R$ .

A generalized *-derivation on a *-ring was defined by Shakir Ali in Shakir: An additive mapping $F:R\to R$ is said to be a generalized *-derivation if there exists a *-derivation $d:R\to R$ such that $F\left(xy\right)=F\left(x\right){y}^{\ast }+xd\left(y\right)$ for all $x,y\in R$ .

In this paper, motivated by definition of a left *-derivation and a generalized left *-derivation in [1] , it is defined that a left *-α-derivation and a generalized left *-α-derivation are as follows respectively: Let R be a *-ring and α be a homomorphism of R. An additive mapping $d:R\to R$ such that $d\left(xy\right)={x}^{\ast }d\left(y\right)+\alpha \left(y\right)d\left(x\right)$ for all $x,y\in R$ is called a left *-α-derivation of R. An additive mapping f is called a generalized left *-α-derivation if there exists a left *-α-derivation d such that $f\left(xy\right)={x}^{\ast }f\left(y\right)+\alpha \left(y\right)d\left(x\right)$ for all $x,y\in R$ . Similarly, motivated by definition of a *-derivation in [2] and a generalized *-derivation in [3] , it is defined that a *-α-derivation and a generalized *-α-derivation are as follows respectively: Let R be a *-ring and α be a homomorphism of R. An additive mapping t which holds $t\left(xy\right)=t\left(x\right){y}^{\ast }+\alpha \left(x\right)t\left(y\right)$ for all $x,y\in R$ is called a *-α-derivation of R. An additive mapping g is called a generalized *-α-derivation if there exists a *-α-derivation t such that $g\left(xy\right)=g\left(x\right){y}^{\ast }+\alpha \left(x\right)t\left(y\right)$ holds for all $x,y\in R$ .

In [4] , Bell and Kappe proved that if $d:R\to R$ is a derivation holds as a homomorphism or an anti-homomorphism on a nonzero right ideal of R which is a prime ring, then $d=0$ . In [5] , Rehman proved that if $F:R\to R$ is a nonzero generalized derivation with a nonzero derivation $d:R\to R$ where R is a 2-torsion free prime ring holds as a homomorphism or an anti homomorphism on a nonzero ideal of R, then R is commutative. In [6] , Dhara proved some results when a generalized derivation acting as a homomorphism or an anti-homomorphism of a semiprime ring. In [7] , Shakir Ali showed that if $G:R\to R$ is a generalized left derivation associated with a Jordan left derivation $\delta :R\to R$ where R is 2-torsion free prime ring and G holds as a homomorphism or an anti-homomorphism on a nonzero ideal of R, then either R is commutative or $G\left(x\right)=xq$ for all $x\in R$ and $q\in {Q}_{l}\left({R}_{C}\right)$ . In [1] , it is proved that if $F:R\to R$ is a generalized left *-derivation associated with a left *-derivation on R where R is a prime *-ring holds as a homomorphism or an anti-homomorphism on R, then R is commutative or F is a right *-centralizer on R.

The aim of this paper is to extend the results which proved for generalized left *-derivation of R in [1] and prove the commutativity of a *-ring with generalized left *-α-derivation. Some results are given for generalized *-α-derivation.

The material in this work is a part of first author’s Master’s Thesis which is supervised by Prof. Dr. Neşet Aydin.

2. Main Results

From now on, R is a prime *-ring where $\ast :R\to R$ is an involution, α is an epimorphism on R and $f:R\to R$ is a generalized left *-α-derivation associated with a left *-α-derivation d on R.

Theorem 1

1) If f is a homomorphism on R, then either R is commutative or f is a right *-centralizer on R.

2) If f is an anti-homomorphism on R, then either R is commutative or f is a right *-centralizer on R.

Proof. 1) Since f is both a homomorphism and a generalized left *-α-derivation associated with a left *-α-derivation d on R, it holds that for all $x,y,z\in R$

$\begin{array}{c}f\left(xyz\right)=f\left(x\left(yz\right)\right)={x}^{\ast }f\left(yz\right)+\alpha \left(yz\right)d\left(x\right)\\ ={x}^{\ast }f\left(y\right)f\left(z\right)+\alpha \left(y\right)\alpha \left(z\right)d\left(x\right).\end{array}$

That is, it holds for all $x,y,z\in R$

$f\left(xyz\right)={x}^{\ast }f\left(y\right)f\left(z\right)+\alpha \left(y\right)\alpha \left(z\right)d\left(x\right).$ (1)

On the other hand, it holds that for all $x,y,z\in R$

$f\left(xyz\right)=f\left(\left(xy\right)z\right)=f\left(xy\right)f\left(z\right)={x}^{\ast }f\left(y\right)f\left(z\right)+\alpha \left(y\right)d\left(x\right)f\left(z\right).$

So, it means that for all $x,y,z\in R$

$f\left(xyz\right)={x}^{\ast }f\left(y\right)f\left(z\right)+\alpha \left(y\right)d\left(x\right)f\left(z\right).$ (2)

Combining Equation (1) and (2), it is obtained that for all $x,y,z\in R$

${x}^{\ast }f\left(y\right)f\left(z\right)+\alpha \left(y\right)\alpha \left(z\right)d\left(x\right)={x}^{\ast }f\left(y\right)f\left(z\right)+\alpha \left(y\right)d\left(x\right)f\left(z\right).$

This yields that for all $x,y,z\in R$

$\alpha \left(y\right)\left(\alpha \left(z\right)d\left(x\right)-d\left(x\right)f\left(z\right)\right)=0.$

Replacing y by yr where $r\in R$ in the last equation, it implies that

$\alpha \left(y\right)\alpha \left(R\right)\left(\alpha \left(z\right)d\left(x\right)-d\left(x\right)f\left(z\right)\right)=\left( 0 \right)$

for all $x,y,z\in R$ . Since α is surjective and R is prime, it follows that for all $x,z\in R$

$\alpha \left(z\right)d\left(x\right)=d\left(x\right)f\left(z\right).$ (3)

Replacing x by xy where $y\in R$ in the last equation, it holds that for all $x,y,z\in R$

$\alpha \left(z\right){x}^{\ast }d\left(y\right)+\alpha \left(z\right)\alpha \left(y\right)d\left(x\right)={x}^{\ast }d\left(y\right)f\left(z\right)+\alpha \left(y\right)d\left(x\right)f\left(z\right).$

Using Equation (3) in the last equation, it implies that for all $x,y,z\in R$

$\left[\alpha \left(z\right),{x}^{\ast }\right]d\left(y\right)+\left[\alpha \left(z\right),\alpha \left(y\right)\right]d\left(x\right)=0.$

Since α is surjective, it holds that for all $x,y,z\in R$

$\left[z,{x}^{\ast }\right]d\left(y\right)+\left[z,\alpha \left(y\right)\right]d\left(x\right)=0.$

Replacing z by ${x}^{\ast }$ in the last equation, it follows that for all $x,y\in R$

$\left[{x}^{\ast },\alpha \left(y\right)\right]d\left(x\right)=0.$

Since α is a surjective, it holds that $\left[{x}^{\ast },y\right]d\left(x\right)=0$ for all $x,y\in R$ . Replacing y by yz where $z\in R$ in the last equation, it gets $\left[{x}^{\ast },y\right]zd\left(x\right)=0$ for all $x,y,z\in R$ . So, it implies that for all $x,y\in R$

$\left[{x}^{\ast },y\right]Rd\left(x\right)=\left(0\right).$

Since R is prime, it follows that $\left[{x}^{\ast },y\right]=0$ or $d\left(x\right)=0$ for all $x,y\in R$ . Let $A=\left\{x\in R|\left[{x}^{\ast },y\right]=0,\forall y\in R\right\}$ and $B=\left\{x\in R|d\left(x\right)=0\right\}$ . Both A and B are

additive subgroups of R and R is the union of A and B. But a group can not be set union of its two proper subgroups. Hence, R equals either A or B.

Assume that $A=R$ . This means that $\left[{x}^{\ast },y\right]=0$ for all $x,y\in R$ . Replacing x by ${x}^{\ast }$ in the last equation, it gets that $\left[x,y\right]=0$ for all $x,y\in R$ . Therefore, R is commutative.

Assume that $B=R$ . This means that $d\left(x\right)=0$ for all $x\in R$ . Since f is a generalized left *-α-derivation associated with d, it follows that f is a right *-centralizer on R.

2) Since f is both an anti-homomorphism and a generalized left *-α-derivation associated with a left *-α-derivation d on R, it holds that

$f\left(xy\right)=f\left(y\right)f\left(x\right)={x}^{\ast }f\left(y\right)+\alpha \left(y\right)d\left( x \right)$

for all $x,y\in R$ . It means that for all $x,y\in R$

$f\left(y\right)f\left(x\right)={x}^{\ast }f\left(y\right)+\alpha \left(y\right)d\left(x\right).$

Replacing y by xy in the last equation and using that f is an anti-homomorphism, it follows that for all $x,y\in R$

${x}^{\ast }f\left(y\right)f\left(x\right)+\alpha \left(y\right)d\left(x\right)f\left(x\right)={x}^{\ast }f\left(y\right)f\left(x\right)+\alpha \left(x\right)\alpha \left(y\right)d\left( x \right)$

which implies that for all $x,y\in R$

$\alpha \left(y\right)d\left(x\right)f\left(x\right)=\alpha \left(x\right)\alpha \left(y\right)d\left(x\right).$ (4)

Replacing y by zy where $z\in R$ in the last equation, it holds that for all $x,y,z\in R$

$\alpha \left(z\right)\alpha \left(y\right)d\left(x\right)f\left(x\right)=\alpha \left(x\right)\alpha \left(z\right)\alpha \left(y\right)d\left(x\right).$

Using Equation (4) in the above equation, it gets $\left[\alpha \left(z\right),\alpha \left(x\right)\right]\alpha \left(y\right)d\left(x\right)=0$ for all $x,y,z\in R$ . Since $\alpha$ is surjective, it holds that $\left[z,\alpha \left(x\right)\right]yd\left(x\right)=0$ for all $x,y,z\in R$ . That is, for all $x,z\in R$

$\left[z,\alpha \left(x\right)\right]Rd\left(x\right)=\left(0\right).$

Since R is prime, it implies that $\left[z,\alpha \left(x\right)\right]=0$ or $d\left(x\right)=0$ for all $x,z\in R$ . Let $K=\left\{x\in R|\left[z,\alpha \left(x\right)\right]=0,\forall z\in R\right\}$ and $L=\left\{x\in R|d\left(x\right)=0\right\}$ . Both K and L are additive subgroups of R and R is the union of K and L. But a group cannot be set union of its two proper subgroups. Hence, R equals either K or L.

Assume that $K=R$ . This means that $\left[z,\alpha \left(x\right)\right]=0$ for all $x,z\in R$ . Since α is surjective, it holds that $\left[z,x\right]=0$ for all $x,z\in R$ . It follows that R is commutative.

Assume that $L=R$ . Now, required result is obtained by applying similar techniques as used in the last paragraph of the proof of 1).

Lemma 2 If f is a nonzero homomorphism (or an anti-homomorphism) and $f\left(R\right)\subset Z\left(R\right)$ then R is commutative.

Proof. Let f be either a nonzero homomorphism or an anti-homomorphism of R. From Theorem 1, it implies that either R is commutative or f is a right *-centralizer on R. Assume that R is noncommutative. In this case, f is a right *-centralizer on R. Since $f\left(R\right)$ is in the center of R, it holds that $\left[f\left({x}^{\ast }y\right),r\right]=0$ for all $x,y,r\in R$ . Using that f is a right *-centralizer and $f\left(R\right)\subset Z\left(R\right)$ , it yields that for all $x,y,r\in R$

$0=\left[f\left({x}^{\ast }y\right),r\right]=\left[xf\left(y\right),r\right]=\left[x,r\right]f\left( y \right)$

which follows that for all $x,y,r\in R$

$\left[x,r\right]f\left(y\right)=0.$

Since $f\left(R\right)$ is in the center of R, it is obtained that for all $x,y,r\in R$

$\left[x,r\right]Rf\left(y\right)=\left(0\right).$

Using primeness of R, it is implied that either $\left[x,r\right]=0$ or $f\left(y\right)=0$ for all $x,y,r\in R$ . Since f is nonzero, it means that R is commutative. This is a contradiction which completes the proof.

Theorem 3 If f is a nonzero homomorphism (or an anti-homomorphism) and $f\left(\left[x,y\right]\right)=0$ for all $x,y\in R$ then R is commutative.

Proof. Let f be a homomorphism of R. It holds that R is commutative or f is a right *-centralizer on R from Theorem 1. Assume that R is noncommutative. In this case, f is a right *-centralizer on R. From the hypothesis, it gets that $f\left(\left[x,y\right]\right)=0$ for all $x,y\in R$ . Since f is a homomorphism, it holds that for all $x,y\in R$

$0=f\left(\left[x,y\right]\right)=f\left(xy-yx\right)=f\left(x\right)f\left(y\right)-f\left(y\right)f\left(x\right)=\left[f\left(x\right),f\left(y\right)\right]$

i.e., for all $x,y\in R$

$\left[f\left(x\right),f\left(y\right)\right]=0.$

Replacing x by ${x}^{\ast }z$ in the last equation, using that f is a right *-centralizer on R and using the last equation, it holds that $0=\left[f\left({x}^{\ast }z\right),f\left(y\right)\right]=\left[xf\left(z\right),f\left(y\right)\right]=\left[x,f\left(y\right)\right]f\left(z\right)$ for $x,y,z\in R$ . So, it follows that for all $x,y,z\in R$

$\left[x,f\left(y\right)\right]f\left(z\right)=0.$

Replacing x by xr where $r\in R$ and using the last equation, it holds that $\left[x,f\left(y\right)\right]rf\left(z\right)=0$ for all $x,y,z,r\in R$ . This implies that for all $x,y,z\in R$

$\left[x,f\left(y\right)\right]Rf\left(z\right)=\left(0\right).$

Using the primeness of R, it is obtained that either $\left[x,f\left(y\right)\right]=0$ or $f\left(z\right)=0$ for all $x,y,z\in R$ . Since f is nonzero, it follows that $f\left(R\right)\subset Z\left(R\right)$ . Using Lemma 2, it is obtained that R is commutative. This is a contradiction which completes the proof.

Let f be an anti-homomorphism of R. This holds that R is commutative or f is a right *-centralizer on R from Theorem 1. Assume that R is noncommutative. In this case, f is a right *-centralizer on R. From the hypothesis, it gets that $f\left(\left[x,y\right]\right)=0$ for all $x,y\in R$ . Since f is an anti-homomorphism, it holds that for all $x,y\in R$

$0=f\left(\left[x,y\right]\right)=f\left(xy-yx\right)=f\left(y\right)f\left(x\right)-f\left(x\right)f\left(y\right)=-\left[f\left(x\right),f\left(y\right)\right]$

i.e., for all $x,y\in R$

$\left[f\left(x\right),f\left(y\right)\right]=0.$

After here, the proof is done by the similarly way in the first case and same result is obtained.

Theorem 4 If f is a nonzero homomorphism (or an anti-homomorphism), $a\in R$ and $\left[f\left(x\right),a\right]=0$ for all $x\in R$ then $a\in Z\left(R\right)$ or R is commutative.

Proof. Let f be either a homomorphism or an anti-homomorphism of R. It holds that R is commutative or f is a right *-centralizer on R from Theorem 1. Assume that R is noncommutative. In this case, f is a right *-centralizer on R. From the hypothesis, it yields that for all $x,y\in R$

$0=\left[f\left({x}^{\ast }y\right),a\right]=\left[xf\left(y\right),a\right]=x\left[f\left(y\right),a\right]+\left[x,a\right]f\left(y\right)=\left[x,a\right]f\left( y \right)$

i.e., for all $x,y\in R$

$\left[x,a\right]f\left(y\right)=0.$

Replacing x by xr where $r\in R$ , it holds that $\left[x,a\right]rf\left(y\right)=0$ for all $x,y,r\in R$ . This implies that $\left[x,a\right]Rf\left(y\right)=\left(0\right)$ for all $x,y\in R$ . Using the primeness of R, it implies that $\left[x,a\right]=0$ or $f\left(y\right)=0$ for all $x,y\in R$ . Since f is nonzero, it follows that $a\in Z\left(R\right)$ . That is, it is obtained that either $a\in Z\left(R\right)$ or R is commutative.

Theorem 5 If f is a nonzero homomorphism (or an anti-homomorphism) and $f\left(\left[x,y\right]\right)\in Z\left(R\right)$ for all $x,y\in R$ then R is commutative.

Proof. Let f be a nonzero homomorphism of R. It implies that either R is commutative or f is a right *-centralizer on R from Theorem 1. Assume that R is noncommutative. In this case, f is a right *-centralizer on R. Since f is a homomorphism and $f\left(\left[x,y\right]\right)\in Z\left(R\right)$ for all $x,y\in R$ , it holds that for all $x,y\in R$

$\begin{array}{c}f\left(\left[x,y\right]\right)=f\left(xy-yx\right)=f\left(xy\right)-f\left(yx\right)\\ =f\left(x\right)f\left(y\right)-f\left(y\right)f\left(x\right)=\left[f\left(x\right),f\left(y\right)\right]\end{array}$

i.e., for all $x,y\in R$

$\left[f\left(x\right),f\left(y\right)\right]\in Z\left(R\right).$

It means that $\left[\left[f\left(x\right),f\left(y\right)\right],r\right]=0$ for all $x,y,r\in R$ . Replacing x by ${x}^{\ast }z$ where $z\in R$ in the last equation, it holds that for all $x,y,z,r\in R$

$\begin{array}{c}0=\left[f\left({x}^{\ast }z\right),f\left(y\right)\right],r\right]=\left[\left[xf\left(z\right),f\left(y\right)\right],r\right]\\ =\left[x,r\right]\left[f\left(z\right),f\left(y\right)\right]+\left[\left[x,f\left(y\right)\right],r\right]f\left(z\right)+\left[x,f\left(y\right)\right]\left[f\left(z\right),r\right]\end{array}$

which implies that for all $x,y,z,r\in R$

$\left[x,r\right]\left[f\left(z\right),f\left(y\right)\right]+\left[\left[x,f\left(y\right)\right],r\right]f\left(z\right)+\left[x,f\left(y\right)\right]\left[f\left(z\right),r\right]=0.$

Replacing x by $f\left(y\right)$ and r by $f\left(z\right)$ , it is obtained that for all $x,y,z\in R$

$\left[f\left(y\right),f\left(z\right)\right]\left[f\left(z\right),f\left(y\right)\right]=0.$

The last equation multiplies by r from right and using that $\left[f\left(x\right),f\left(y\right)\right]\in Z\left(R\right)$ for all $x,y\in R$ , it follows that for all $x,y,z,r\in R$

$\left[f\left(y\right),f\left(z\right)\right]r\left[f\left(z\right),f\left(y\right)\right]=0$

i.e., for all $x,y,z,r\in R$ .

$\left[f\left(z\right),f\left(y\right)\right]R\left[f\left(z\right),f\left(y\right)\right]=\left(0\right).$

Using primeness of R, it is implied that for all $y,z\in R$

$\left[f\left(z\right),f\left(y\right)\right]=0.$

From Theorem 4, it holds that either $f\left(y\right)\in Z\left(R\right)$ for all $y\in R$ or R is commutative. By using Lemma 2, it follows that R is commutative. This is a contradiction which completes the proof.

Let f be a nonzero anti-homomorphism of R. It implies that either R is commutative or f is a right *-centralizer on R from Theorem 1. Assume that R is noncommutative. In this case, f is a right *-centralizer on R. From the hypothesis, it gets that $f\left(\left[x,y\right]\right)\in Z\left(R\right)$ for all $x,y\in R$ . Since f is an anti-homomorphism, it is obtained that for all $x,y\in R$

$f\left(\left[x,y\right]\right)=f\left(xy-yx\right)=f\left(y\right)f\left(x\right)-f\left(x\right)f\left(y\right)=-\left[f\left(x\right),f\left(y\right)\right]$

i.e., for all $x,y\in R$

$\left[f\left(x\right),f\left(y\right)\right]\in Z\left(R\right).$

After here, the proof is done by the similar way in the first case and same result is obtained.

Theorem 6 If f is a nonzero homomorphism (or an anti-homomorphism) and $f\left(\left(x,y\right)\right)=0$ for all $x,y\in R$ then R is commutative.

Proof. Let f be a homomorphism of R. It holds that R is commutative or f is a right *-centralizer on R from Theorem 1. Assume that R is noncommutative. In this case, f is a right *-centralizer on R. So, it gets that for all $x,y\in R$

$0=f\left(\left(x,y\right)\right)=f\left(xy+yx\right)=f\left(xy\right)+f\left(yx\right)=f\left(x\right)f\left(y\right)+f\left(y\right)f\left(x\right).$

It means that for all $x,y\in R$

$f\left(x\right)f\left(y\right)+f\left(y\right)f\left(x\right)=0.$

Replacing x by ${x}^{\ast }z$ where $z\in R$ in the above equation and using that f is a right * the last equation, it is obtained that

$0=f\left({x}^{\ast }z\right)f\left(y\right)+f\left(y\right)f\left({x}^{\ast }z\right)=xf\left(z\right)f\left(y\right)+f\left(y\right)xf\left(z\right).$

Using that $f\left(x\right)f\left(y\right)=-f\left(y\right)f\left(x\right)$ for all $x,y\in R$ in the last equation

$\begin{array}{c}0=xf\left(z\right)f\left(y\right)+f\left(y\right)xf\left(z\right)=-xf\left(y\right)f\left(z\right)+f\left(y\right)xf\left(z\right)\\ =\left[f\left(y\right),x\right]f\left( z \right)\end{array}$

i.e. for all $x,y,z\in R$

$\left[f\left(y\right),x\right]f\left(z\right)=0.$

Replacing x by xr, it follows that $\left[f\left(y\right),x\right]Rf\left(z\right)=\left(0\right)$ for all $x,y,z\in R$ . Using primeness of R, it holds that either $\left[f\left(y\right),x\right]=0$ or $f\left(z\right)=0$ for all $x,y,z\in R$ . Since f is nonzero, it implies that $f\left(R\right)\subset Z\left(R\right)$ . Using Lemma 2, it yields that R is commutative. This is a contradiction which completes the proof.

Let f be an anti-homomorphism of R. It holds that R is commutative or f is a right *-centralizer on R from Theorem 1. Assume that R is noncommutative. In this case f is a right *-centralizer on R. Using hypothesis, it gets that for all $x,y\in R$

$0=f\left(\left(x,y\right)\right)=f\left(xy+yx\right)=f\left(xy\right)+f\left(yx\right)=f\left(y\right)f\left(x\right)+f\left(x\right)f\left( y \right)$

i.e., for all $x,y\in R$

$f\left(y\right)f\left(x\right)+f\left(x\right)f\left(y\right)=0.$

After here, the proof is done by the similar way in the first case and same result is obtained.

Now, $g:R\to R$ is a generalized *-α-derivation associated with a *-α-derivation t on R.

Theorem 7 Let R be a *-prime ring where * be an involution, α be a homomorphism of R and $g:R\to R$ be a generalized *-α-derivation associated with a *-α-derivation t on R. If g is nonzero then R is commutative.

Proof. Since g is a generalized *-α-derivation associated with a *-α-derivation t on R, it holds that $g\left(xy\right)=g\left(x\right){y}^{\ast }+\alpha \left(x\right)t\left(y\right)$ for all $x,y\in R$ . So it yields that for all $x,y,z\in R$

$\begin{array}{c}g\left(xyz\right)=g\left(\left(xy\right)z\right)=g\left(xy\right){z}^{\ast }+\alpha \left(xy\right)t\left(z\right)\\ =\left(g\left(x\right){y}^{\ast }+\alpha \left(x\right)t\left(y\right)\right){z}^{\ast }+\alpha \left(x\right)\alpha \left(y\right)t\left(z\right)\\ =g\left(x\right){y}^{\ast }{z}^{\ast }+\alpha \left(x\right)t\left(y\right){z}^{\ast }+\alpha \left(x\right)\alpha \left(y\right)t\left( z \right)\end{array}$

that is, it holds that for all $x,y,z\in R$

$g\left(xyz\right)=g\left(x\right){y}^{\ast }{z}^{\ast }+\alpha \left(x\right)t\left(y\right){z}^{\ast }+\alpha \left(x\right)\alpha \left(y\right)t\left(z\right).$ (5)

On the other hand, it implies that for all $x,y,z\in R$

$\begin{array}{c}g\left(xyz\right)=g\left(x\left(yz\right)\right)=g\left(x\right){\left(yz\right)}^{\ast }+\alpha \left(x\right)t\left(yz\right)\\ =g\left(x\right){z}^{\ast }{y}^{\ast }+\alpha \left(x\right)\left(t\left(y\right){z}^{\ast }+\alpha \left(y\right)t\left(z\right)\right)\\ =g\left(x\right){z}^{\ast }{y}^{\ast }+\alpha \left(x\right)t\left(y\right){z}^{\ast }+\alpha \left(x\right)\alpha \left(y\right)t\left( z \right)\end{array}$

so, it gets that for all $x,y,z\in R$

$g\left(xyz\right)=g\left(x\right){z}^{\ast }{y}^{\ast }+\alpha \left(x\right)t\left(y\right){z}^{\ast }+\alpha \left(x\right)\alpha \left(y\right)t\left(z\right).$ (6)

Now, combining the Equations (5) and (6), it holds that for all $x,y,z\in R$

$\begin{array}{l}g\left(x\right){y}^{\ast }{z}^{\ast }+\alpha \left(x\right)t\left(y\right){z}^{\ast }+\alpha \left(x\right)\alpha \left(y\right)t\left(z\right)\\ =g\left(x\right){z}^{\ast }{y}^{\ast }+\alpha \left(x\right)t\left(y\right){z}^{\ast }+\alpha \left(x\right)\alpha \left(y\right)t\left( z \right)\end{array}$

which follows that

$g\left(x\right)\left[{y}^{\ast },{z}^{\ast }\right]=0$

for all $x,y,z\in R$ . Replacing y by ${y}^{\ast }$ and z by ${z}^{\ast }$ , it holds that for all $x,y,z\in R$

$g\left(x\right)\left[y,z\right]=0.$

Replacing y by ry where $r\in R$ in the last equation, it yields that for all $x,y,z,r\in R$

$0=g\left(x\right)\left[ry,z\right]=g\left(x\right)r\left[y,z\right]+g\left(x\right)\left[r,z\right]y.$

Using $g\left(x\right)\left[y,z\right]=0$ for all $x,y,z\in R$ in above equation, it is obtained that for all $x,y,z,r\in R$

$g\left(x\right)r\left[y,z\right]=0$ (7)

i.e., for all $x,y,z\in R$

$g\left(x\right)R\left[y,z\right]=\left(0\right).$ (8)

Replacing y by ${y}^{\ast }$ and z by $-{z}^{\ast }$ , it follows that for all $x,y,z\in R$

$g\left(x\right)R{\left(\left[y,z\right]\right)}^{\ast }=\left(0\right).$ (9)

Now, combining the Equations (8) and (9),

$g\left(x\right)R\left[y,z\right]=g\left(x\right)R{\left(\left[y,z\right]\right)}^{\ast }=\left( 0 \right)$

is obtained for all $x,y,z\in R$ . Using *-primeness of R, it follows that $g\left(x\right)=0$ or $\left[y,z\right]=0$ for all $x,y,z\in R$ . Since g is nonzero, R is commutative.

Theorem 8 Let R be a semiprime *-ring where * be an involution, α be an homomorphism of R and $g:R\to R$ be a nonzero generalized *-α-derivation associated with a *-α-derivation t on R then $g\left(R\right)\subset Z\left(R\right)$ .

Proof. Equation (7) multiplies by s from left, it gets that for all $x,y,z,r,s\in R$

$sg\left(x\right)r\left[y,z\right]=0.$ (10)

Replacing r by sr in the Equation (7), it holds that for all $x,y,z,r,s\in R$

$g\left(x\right)sr\left[y,z\right]=0.$ (11)

Now, combining the Equation (10) and (11),

$sg\left(x\right)r\left[y,z\right]=g\left(x\right)sr\left[y,z\right]$

is obtained for all $x,y,z,r,s\in R$ . It follows that for all $x,y,z,r,s\in R$

$\left[s,g\left(x\right)\right]r\left[y,z\right]=0.$

This implies that

$\left[s,g\left(x\right)\right]R\left[y,z\right]=\left( 0 \right)$

for all $x,y,z,s\in R$ . Replacing s by y and z by $g\left(x\right)$ in the last equation, it yields that

$\left[y,g\left(x\right)\right]R\left[y,g\left(x\right)\right]=\left( 0 \right)$

for all $x,y\in R$ . Using semiprimeness of R, it is implied that for all $x,y\in R$

$\left[y,g\left(x\right)\right]=0.$

That is,

$g\left(R\right)\subset Z\left( R \right)$

which completes the proof.

Cite this paper

Balc, A.O., Aydin, N. and Türkmen, S. (2018) The Commutativity of a *-Ring with Generalized Left *-α-Derivation.Advances in Pure Mathematics, 8, 168-177.
https://doi.org/10.4236/apm.2018.82009

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