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

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1. Introduction

Let R be an associative ring with center $Z\left(R\right)$ . $xy+yx$ where $x\mathrm{,}y\in R$ is denoted by $\left(x\mathrm{,}y\right)$ and $xy-yx$ where $x\mathrm{,}y\in R$ is denoted by $\left[x\mathrm{,}y\right]$ 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 $x\mathrm{,}y\in R$ 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 $x\mathrm{,}y\in R$ 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 $a\mathrm{,}b\in R$ . In private, if $b=a$ , it implies that R is a semiprime ring. An additive mapping $\ast \mathrm{:}R\to R$ which holds ${\left(xy\right)}^{\ast}={y}^{\ast}{x}^{\ast}$ and ${\left({x}^{\ast}\right)}^{\ast}=x$ for all $x\mathrm{,}y\in R$ 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 $a\mathrm{,}b\in R$ .

Notations of left *-derivation and generalized left *-derivation were given in $abu$ : Let R be a *-ring. An additive mapping $d\mathrm{:}R\to R$ is called a left *-derivation if $d\left(xy\right)={x}^{\ast}d\left(y\right)+yd\left(x\right)$ holds for all $x\mathrm{,}y\in R$ . An additive mapping $F\mathrm{:}R\to R$ 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 $x\mathrm{,}y\in R$ . An additive mapping $T\mathrm{:}R\to R$ is called a right *-centralizer if $T\left(xy\right)={x}^{\ast}T\left(y\right)$ for all $x\mathrm{,}y\in R$ . 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 $d\mathrm{:}R\to R$ is said to be a *-derivation if $d\left(xy\right)=d\left(x\right){y}^{\ast}+xd\left(y\right)$ for all $x\mathrm{,}y\in R$ .

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

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 $d\mathrm{:}R\to R$ such that $d\left(xy\right)={x}^{\ast}d\left(y\right)+\alpha \left(y\right)d\left(x\right)$ for all $x\mathrm{,}y\in R$ 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 $x\mathrm{,}y\in R$ . 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 $x\mathrm{,}y\in R$ 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 $x\mathrm{,}y\in R$ .

In [4] , Bell and Kappe proved that if $d\mathrm{:}R\to R$ 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 $F\mathrm{:}R\to R$ is a nonzero generalized derivation with a nonzero derivation $d\mathrm{:}R\to R$ 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 $G\mathrm{:}R\to R$ is a generalized left derivation associated with a Jordan left derivation $\delta \mathrm{:}R\to R$ 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 $F\mathrm{:}R\to R$ 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 $\ast \mathrm{:}R\to R$ is an involution, α is an epimorphism on R and $f\mathrm{:}R\to R$ 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 $x\mathrm{,}y\mathrm{,}z\in R$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Since α is a surjective, it holds that $\left[{x}^{\ast},y\right]d\left(x\right)=0$ for all $x\mathrm{,}y\in R$ . 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 $x\mathrm{,}y\mathrm{,}z\in R$ . So, it implies that for all $x\mathrm{,}y\in R$

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

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

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

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

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

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

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

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

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

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

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

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 $x\mathrm{,}y\mathrm{,}z\in R$ . Since $\alpha $ is surjective, it holds that $\left[z,\alpha \left(x\right)\right]yd\left(x\right)=0$ for all $x\mathrm{,}y\mathrm{,}z\in R$ . That is, for all $x\mathrm{,}z\in R$

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

Since R is prime, it implies that $\left[z,\alpha \left(x\right)\right]=0$ or $d\left(x\right)=0$ for all $x\mathrm{,}z\in R$ . Let $K=\left\{x\in R\mathrm{}|\mathrm{}\left[z,\alpha \left(x\right)\right]=0,\forall z\in R\right\}$ and $L=\left\{x\in R\mathrm{}|\mathrm{}d\left(x\right)=0\right\}$ . 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 $x\mathrm{,}z\in R$ . Since α is surjective, it holds that $\left[z,x\right]=0$ for all $x\mathrm{,}z\in R$ . 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 $x\mathrm{,}y\mathrm{,}r\in R$ . Using that f is a right *-centralizer and $f\left(R\right)\subset Z\left(R\right)$ , it yields that for all $x\mathrm{,}y\mathrm{,}r\in R$

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

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

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

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

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

Using primeness of R, it is implied that either $\left[x,r\right]=0$ or $f\left(y\right)=0$ for all $x\mathrm{,}y\mathrm{,}r\in R$ . 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 $x\mathrm{,}y\in R$ 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 $x\mathrm{,}y\in R$ . Since f is a homomorphism, it holds that for all $x\mathrm{,}y\in R$

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

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

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

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 $x\mathrm{,}y\mathrm{,}z\in R$ . So, it follows that for all $x\mathrm{,}y\mathrm{,}z\in R$

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

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 $x\mathrm{,}y\mathrm{,}z\mathrm{,}r\in R$ . This implies that for all $x\mathrm{,}y\mathrm{,}z\in R$

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

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 $x\mathrm{,}y\mathrm{,}z\in R$ . 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 $f([x,y])=0$ for all $x\mathrm{,}y\in R$ . Since f is an anti-homomorphism, it holds that for all $x\mathrm{,}y\in R$

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

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

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

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 $x\mathrm{,}y\in R$

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

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

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

Replacing x by xr where $r\in R$ , it holds that $\left[x,a\right]rf\left(y\right)=0$ for all $x\mathrm{,}y\mathrm{,}r\in R$ . This implies that $\left[x,a\right]Rf\left(y\right)=\left(0\right)$ for all $x\mathrm{,}y\in R$ . Using the primeness of R, it implies that $\left[x,a\right]=0$ or $f\left(y\right)=0$ for all $x\mathrm{,}y\in R$ . 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 $f\left(\left[x\mathrm{,}y\right]\right)\in Z\left(R\right)$ for all $x\mathrm{,}y\in R$ 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 $f\left(\left[x\mathrm{,}y\right]\right)\in Z\left(R\right)$ for all $x\mathrm{,}y\in R$ , it holds that for all $x\mathrm{,}y\in R$

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

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

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

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

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

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

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

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

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

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 $x\mathrm{,}y\in R$ , it follows that for all $x\mathrm{,}y\mathrm{,}z\mathrm{,}r\in R$

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

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

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

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

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

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 $x\mathrm{,}y\in R$ . Since f is an anti-homomorphism, it is obtained that for all $x\mathrm{,}y\in R$

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

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

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

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 $f\left(\left(x\mathrm{,}y\right)\right)=0$ for all $x\mathrm{,}y\in R$ 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 $x\mathrm{,}y\in R$

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

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

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

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 $x\mathrm{,}y\in R$ in the last equation

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

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

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

Replacing x by xr, it follows that $\left[f\left(y\right)\mathrm{,}x\right]Rf\left(z\right)=\left(0\right)$ for all $x\mathrm{,}y\mathrm{,}z\in R$ . Using primeness of R, it holds that either $\left[f\left(y\right),x\right]=0$ or $f\left(z\right)=0$ for all $x\mathrm{,}y\mathrm{,}z\in R$ . 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 $x\mathrm{,}y\in R$

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

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

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

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

Now, $g\mathrm{:}R\to R$ 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 $g\mathrm{:}R\to R$ 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 $x\mathrm{,}y\in R$ . So it yields that for all $x\mathrm{,}y\mathrm{,}z\in R$

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

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

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

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

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

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

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

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

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

which follows that

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

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

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

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

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

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

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

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

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

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

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

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

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

is obtained for all $x\mathrm{,}y\mathrm{,}z\in R$ . Using *-primeness of R, it follows that $g\left(x\right)=0$ or $\left[y,z\right]=0$ for all $x\mathrm{,}y\mathrm{,}z\in R$ . 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 $g\mathrm{:}R\to R$ 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 $x\mathrm{,}y\mathrm{,}z\mathrm{,}r\mathrm{,}s\in R$

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

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

$g\left(x\right)sr\left[y,z\right]=0.$ (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 $x\mathrm{,}y\mathrm{,}z\mathrm{,}r\mathrm{,}s\in R$ . It follows that for all $x\mathrm{,}y\mathrm{,}z\mathrm{,}r\mathrm{,}s\in R$

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

This implies that

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

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

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

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

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

That is,

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

which completes the proof.

References

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