# The Construction of Fraction Gamma Ring Using Noncommutative Gamma Ring

• Zohre Tabatabaee
• Tahereh Roodbarylor
Open Access
Research Paper

## Abstract

One of the first constructions of algebra is the quotient field of a commutative integral domain constructed as a set of fractions. The noncommutative case is restrictive. In this article, the researchers constructed a fraction gamma ring of noncommutative gamma ring. There should be an appropriate set X of elements in a gamma ring R to be used as denominators.

## Keywords

Γ-ring Fraction Γ-ring Noncommutative Γ-ring

## 1 Introduction

The notation of gamma ring was first introduced by Nobusawa (1964) as a generalization of a classical ring and afterward Barnes (1966) improved the concepts of Nobusawa’s Γ-ring. The researchers have constructed fraction gamma ring of commutative integral domain, but this is very complicated in noncommutative gamma ring. Two Γ-ring of fractions are hypothesized, one with right-hand denominators and one with left-hand denominators (Goodearl and Warfield 2004).

Let R and $$\varGamma$$ be two additive Abelian groups and there exists a mapping $$( x,\gamma,y ) \longmapsto \ \ x \gamma y\ \ of \ \ R \times \varGamma \times R\longrightarrow R$$, which satisfies the conditions:
\begin{aligned} (i)\quad (x + y)\gamma z & = x\gamma z + y\gamma z, \hfill \\ \;\;\;\;\;x(\gamma _{1} + \gamma _{2} ) & = x\gamma _{1} y + x\gamma _{2} y, \hfill \\ \;\;\;\;\;\;\;x\gamma (y + z) & = x\gamma y + x\gamma z, \hfill \\ (ii)\;\;(x\gamma _{1} y)\gamma _{2} z & = x\gamma _{1} (y\gamma _{2} z), \hfill \\ \end{aligned}
for all $$x,y,z \in R$$ and $$\gamma,\gamma _{1},\gamma _{2} \in \varGamma$$. Then, R is called a gamma ring.

If there exists $$\ 1_{R} \in R$$ and $$\gamma _{0} \in \varGamma$$ such that for all $$r\in R$$, $$1_{R} \gamma _{0} r=r \gamma _{0} 1_{R} =r$$, then $$1=1_{R}$$ is called identity element (Paul and Uddin 2010).

Let R be a $$\varGamma {\text{-ring}}$$ with 1. An element $$a \in R$$ is called invertible if there exists $$b \in R$$ such that $$a \gamma _{0} b=b \gamma _{0} a =1$$, also b is unique and called the multiplicative inverse of a and is denoted by $$a^{-1}$$.

An element $$a \in R$$ is said to be zero-divisor if there exists $$b\ne 0$$ such that $$a \gamma _{0} b=b \gamma _{0} a=0$$.

For non-zero element x in R, the right annihilator of x and the left annihilator of x are defined to be
\begin{aligned}&ann_{r}(x)=\{a\in R|\ x{ \gamma }_{0}a=0\}, \\&ann_{l}(x)=\{a\in R|\ a{ \gamma }_{0}x=0\} . \end{aligned}
A regular element in a $$\varGamma {\text{-ring}}$$ R is any nonzero-divisor, i.e., any element $$x\in R$$ such that $$ann_{r}(x)=\{0\} \ {\text{and}} \ ann_{l}(x)=\{0\}$$.

A multiplicatively closed subset of $$\varGamma$$-ring R is a subset X of R such that $$1\in X$$ and $$x_{1} \varGamma x_{2} \subseteq X$$, for all $$x_{1}, x_{2} \in X.$$

A $$\varGamma {\text{-ring}}$$ homomorphism (Ullah and Chaudhry 2012) is a mapping f of $$\varGamma {\text{-ring}} \ \ R$$ to $$\varGamma {\text{-ring}} \ \ R^{'}$$ such that
\begin{aligned}&(i)\ \ f(x+y)= f(x)+f(y),\\&(ii)\ \ f( x \gamma y) =f(x) \gamma f(y), \end{aligned}
for all $$x,y\in R$$ and $$\gamma \in \varGamma$$.
We consider the following condition:
\begin{aligned} (*) \ \ x \alpha y \beta z=x \beta y \alpha z, \end{aligned}
for all $$x, y,z\in R$$ and $$\alpha,\beta \in \varGamma$$ (Dey and Paul 2014).

Right P-condition: Let X be a multiplicatively closed set in $$\varGamma$$-ring R. Then, X satisfies the right P-condition, if for $$x\in X$$ and $$r\in R$$, there exist $$y\in X$$ and $$s\in R$$, such that $$r{ \gamma }_{0}y=x{ \gamma }_{0}s$$, then X is called the right P-set. The left P-condition and the left P-set are defined in the same way.

Let R be a $$\varGamma$$-ring and $$X\subseteq R$$ be a multiplicative closed set of regular elements in R. A right $$\varGamma$$-ring of fractions for R with respect to X is any $$\varGamma$$-ring S ($$R\subseteq S$$) such that
1. (a)

Every element of X is invertible in S.

2. (b)

Every element of S can be expressed in the form $$a{ \gamma }_{0}x^{-1}$$, for some $$a\in R$$ and $$x\in X$$.

The left $$\varGamma$$-rings of fractions are defined similarly.

## 2 Fraction of Noncommutative Gamma Ring

Throughout this section, the word gamma ring means noncommutative gamma ring with 1.

### Lemma 1

Let X be a rightP-set in a$$\varGamma$$-ring R. Then, for all elements$$x_{1},\ldots, x_{n} \in X$$, there exist$$s_{1},\ldots,s_{n} \in R$$ such that$$x_{1}{ \gamma }_{0}s_{1}=\cdots =x_{n}{ \gamma }_{0}s_{n}$$and$$x_{1}{ \gamma }_{0}s_{1} \in X$$, that is,$$x_{1}{ \gamma }_{0}R\cap \cdots \cap x_{n}{ \gamma }_{0}R\cap X\ne \emptyset$$ .

### Proof

By induction. We prove for the cases $$n=2$$ and $$n=3$$. Since X is a right P -set, thus for every $$x_{1} \in X$$ and $$x_{2} \in R$$, there exist $$y\in X$$ and $$s\in R$$ such that $$x_{1}{ \gamma }_{0}y=x_{2}{ \gamma }_{0}s$$, and $$x_{1}{ \gamma }_{0}y\in X$$, because X is a multiplicatively closed set.

But we have $$x_{2} \gamma _{0} s \in X$$ and $$x_{3} \in R$$, thus there exist some $$y^{'} \in X$$ and $$s^{'} \in R$$ such that $$x_{2}{ \gamma }_{0}s \gamma _{0} y^{'}=x_{3}{ \gamma }_{0}s^{'}$$ or $$x_{1} \gamma _{0} y \gamma _{0} y^{'}$$, if we consider $$y^{''}=y{ \gamma }_{0} y^{'}$$, then $$x_{1}{ \gamma }_{0}y^{''}=x_{3}{ \gamma }_{0}s^{'}$$ or $$x_{1}{ \gamma }_{0}y^{''}=x_{2}{ \gamma }_{0}y^{'}=x_{3}{ \gamma }_{0}s^{'}$$. $$\square$$

### Lemma 2

Let R be a$$\varGamma$$-ring that satisfies the condition ($$*$$) and X be a multiplicatively closed set of regular elements in R, and assume that there exists a right$$\varGamma$$-ring of fractions, say S, for R with respect to X. Then
1. (i)

X is a right P-set in R.

2. (ii)

For every $$s_{1},\ldots,s_{n} \in S$$ , there exist $$a_{1},\ldots,a_{n} \in R$$ and $$x\in X$$ such that $$s_{i}=a_{i} { \gamma }_{0}x^{-1}$$ , for every $$1\le i \le n$$ .

3. (iii)

Let $$a,b\in R$$ and $$x,y\in X$$ . Then, $$a{ \gamma }_{0}x^{-1}=b{ \gamma }_{0}y^{-1}$$ in S if and only if there exist $$c,d\in R$$ such that $$a{ \gamma }_{0}c=b{ \gamma }_{0}d$$ and $$x{ \gamma }_{0}c=y{ \gamma }_{0}d\in X$$ .

### Proof

1. (i)
Suppose S be a right $$\varGamma$$-ring of fractions for R with respect to X. For $$x,y\in X$$ and $$a,b\in R$$, using ($$*$$), we consider the product of fractions $$a{ \gamma }_{0}x^{-1}$$ and $$b{ \gamma }_{0}y^{-1}$$ in S
\begin{aligned} (a{ \gamma }_{0}x^{-1})\gamma (b{ \gamma }_{0}y^{-1})=a\gamma (x^{-1}{ \gamma }_{0}b){ \gamma }_{0}y^{-1} \end{aligned}
Since every element of S has the form of a fraction with right-hand denominator, the left-hand fraction $$x^{-1}{ \gamma }_{0}b$$ must equal $$c{ \gamma }_{0}z^{-1}$$ for some $$c\in R$$ and $$z\in X$$, hence we have
\begin{aligned} (a{ \gamma }_{0}x^{-1})\gamma (b{ \gamma }_{0}y^{-1})&=a\gamma (x^{-1}{ \gamma }_{0}b){ \gamma }_{0}y^{-1} \\&=a\gamma (c{ \gamma }_{0}z^{-1}){ \gamma }_{0}y^{-1} \\&=a\gamma c{ \gamma }_{0}(z^{-1}{ \gamma }_{0}y^{-1})\\&=a\gamma c{ \gamma }_{0}(y{ \gamma }_{0}z)^{-1}. \end{aligned}
Note that we have a necessary condition: for $$b\in R$$ and $$x\in X$$, there must exist $$c\in R$$ and $$z\in X$$ such that $$x^{-1}{ \gamma }_{0}b=c{ \gamma }_{0}z^{-1}$$, that is, $$b{ \gamma }_{0}z=x{ \gamma }_{0}c$$. This is precisely the right P -condition.

2. (ii)

Since S is a right $$\varGamma$$-ring of fractions, for each $$s_{i}\in S$$ ($$1\le i \le n$$), there exist $$b_{i} \in R$$ and $$x_{i} \in X$$ such that $$s_{i}=b_{i}{ \gamma }_{0}x_{i}^{-1}$$. Using Lemma 1, there exist $$x\in X$$ and $$c_{1},\ldots,c_{n} \in R$$ such that $$x=x_{i}{ \gamma }_{0}c_{i}$$, for all i. Since x and $$x_{i}$$ are both invertible in S, so is $$c_{i}$$ and $$x^{-1}=c_{i}^{-1}{ \gamma }_{0}x_{i}^{-1}$$. Thus, $$x_{i}^{-1}=c_{i}{ \gamma }_{0}x^{-1}$$ and we have $$s_{i}=b_{i}{ \gamma }_{0}x_{i}^{-1}=b_{i}{ \gamma }_{0}c_{i}{ \gamma }_{0}x^{-1}$$. If we consider $$a_{i}=b_{i}{ \gamma }_{0}c_{i}$$, then $$s_{i}=a_{i}{ \gamma }_{0}x^{-1}$$.

3. (iii)
$$\Longleftarrow$$ Suppose that, there exist $$c,d\in R$$ such that $$a{\gamma}_{0}c=b{ \gamma }_{0}d$$ and $$x{\gamma }_{0}c=y{\gamma }_{0}d\in X$$. Then
\begin{aligned} a{\gamma }_{0}x^{-1}&=a{\gamma }_{0}1{\gamma }_{0}x^{-1} \\&=a{\gamma }_{0}(c{\gamma }_{0}c^{-1}){\gamma }_{0}x^{-1} \\&=(a{\gamma }_{0}c){\gamma }_{0}(x{\gamma }_{0}c)^{-1} \\&=(b{\gamma }_{0}d){\gamma }_{0}(y{\gamma }_{0}d)^{-1} \\&=b{\gamma }_{0}(d{\gamma }_{0}d^{-1}){\gamma }_{0}y^{-1} \\&=b{\gamma }_{0}y^{-1}. \end{aligned}
$$\Longrightarrow$$ Suppose that, $$a{\gamma }_{0}x^{-1}=b{\gamma }_{0}y^{-1}$$. Using Lemma 1, there exist $$c,d\in R$$ such that $$x{\gamma }_{0}c=y{\gamma }_{0}d\in X$$. Hence
\begin{aligned} (a{\gamma }_{0}c){\gamma }_{0}(x{\gamma }_{0}c)^{-1}&=a{\gamma }_{0}(c{\gamma }_{0}c^{-1}){\gamma }_{0}x^{-1} \\&=a{\gamma }_{0}x^{-1} \\&=b{\gamma }_{0}y^{-1} \\&=b{\gamma }_{0}(d{\gamma }_{0}d^{-1}){\gamma }_{0}y^{-1} \\&=(b{\gamma }_{0}d){\gamma }_{0}(y{\gamma }_{0}d)^{-1} \\&=(b{\gamma }_{0}d){\gamma }_{0}(x{\gamma }_{0}c)^{-1}, \end{aligned}
and therefore, $$a{\gamma }_{0}c=b{\gamma }_{0}d$$.

$$\square$$

### Proposition 1

Let X be a right P-set of regular elements in a $$\varGamma$$ -ring R. We define a relation $$\sim$$ on $$R\times X$$ as follows:
\begin{aligned} (a,x)\sim (b,y) \Longleftrightarrow \exists c,d\in R,\ \ a{\gamma }_{0}c=b{\gamma }_{0}d,\ \ x{\gamma }_{0}c=y{\gamma }_{0}d. \end{aligned}
Then, $$\sim$$ is an equivalence relation.

### Proof

We show that the relation $$\sim$$ is reflexive, symmetric and transitive.

For every $$a\in R$$ and $$x\in X$$, let $$c=d=1\in R$$. Then, $$a{\gamma }_{0}1=a{\gamma }_{0}1$$ and $$x{\gamma }_{0}1=x{\gamma }_{0}1$$, and therefore $$(a,x)\sim (a,x)$$.

If $$(a,x)\sim (b,y)$$, then there exist $$c,d\in R$$ such that $$a{\gamma }_{0}c=b{\gamma }_{0}d$$ and $$x{\gamma }_{0}c=y{ \gamma }_{0}d\in X$$ and so $$b{\gamma }_{0}d=a{\gamma }_{0}c$$ and $$y{\gamma }_{0}d=x{ \gamma }_{0}c$$ . Thus, $$(b,y)\sim (a,x)$$.

If $$(a,x)\sim (b,y)$$ and $$(b,y)\sim (c,z)$$, then we have
\begin{aligned} (a,x)\sim (b,y) \ \ \Rightarrow \ \ \exists \ d,e\in R \quad {\rm s.t} {\left\{ \begin{array}{ll} \ a{ \gamma }_{0}d=b{ \gamma }_{0}e \\ \ x{ \gamma }_{0}d=y{ \gamma }_{0}e\in X \end{array}\right. } \end{aligned}
(1)
\begin{aligned} (b,y)\sim (c,z) \ \ \Rightarrow \ \ \exists \ f,g\in R \quad {\rm s.t} {\left\{ \begin{array}{ll} \ b{ \gamma }_{0}f=c{ \gamma }_{0}g \\ \ y{ \gamma }_{0}f=z{ \gamma }_{0}g\in X \end{array}\right. } \end{aligned}
(2)
Since y and $$z{ \gamma }_{0}g$$ are in X and all elements in X are invertible in S, thus f is invertible and so $$b=c{ \gamma }_{0}g{ \gamma }_{0}f^{-1}$$ and $$y=z{ \gamma }_{0}g{ \gamma }_{0}f^{-1}$$.
Using (1), we obtain
\begin{aligned} {\left\{ \begin{array}{ll} \ a{ \gamma }_{0}d=b{ \gamma }_{0}e=c{ \gamma }_{0}g{ \gamma }_{0}f^{-1}{ \gamma }_{0}e \\ \ x{ \gamma }_{0}d=y{ \gamma }_{0}e=z{ \gamma }_{0}g{ \gamma }_{0}f^{-1}{ \gamma }_{0}e \end{array}\right. } \end{aligned}
We consider $$h=g{ \gamma }_{0}f^{-1}{ \gamma }_{0}e\in R$$, and so there exist $$d,h\in R$$ such that $$a{ \gamma }_{0}d=c{ \gamma }_{0}h$$ and $$x{ \gamma }_{0}d=z{ \gamma }_{0}h$$, that is $$(a,x)\sim (c,z)$$.

Hence, the proof is complete. $$\square$$

### Definition 1

Let R be a $$\varGamma$$-ring that satisfies the condition ($$*$$) and X be a right P-set of regular elements in R, and assume that [ax] denote the equivalence class of (a, x), and $$RX^{-1}$$ denote the set of equivalence classes for all $$a\in R$$ and $$x\in X$$. We define addition and multiplication for [ax] and [by] in $$RX^{-1}$$ as follows:
1. (i)
Since $$x,y\in X$$ and X is a right P-set, thus using Lemma 1, there exist $$c,d\in R$$ such that $$z=x{ \gamma }_{0}c=y{ \gamma }_{0}d\in X$$ and we have
\begin{aligned} a{ \gamma }_{0}x^{-1}+b{ \gamma }_{0}y^{-1}&=a{ \gamma }_{0}c{ \gamma }_{0}z^{-1}+b{ \gamma }_{0}d{ \gamma }_{0}z^{-1} \\&=(a{ \gamma }_{0}c+b{ \gamma }_{0}d){ \gamma }_{0}z^{-1} \end{aligned}
We define $$[a,x]+[b,y]=[a{ \gamma }_{0}c+b{ \gamma }_{0}d,x{ \gamma }_{0}c]$$.

2. (ii)
Since $$b\in R$$, $$x\in X$$ and X is a right P-set, thus there exist $$c\in R$$ and $$z\in X$$ such that $$b{ \gamma }_{0}z=x{ \gamma }_{0}c$$ or $$x^{-1}{ \gamma }_{0}b=c{ \gamma }_{0}z^{-1}$$. Using ($$*$$), we have
\begin{aligned} (a{ \gamma }_{0}x^{-1})\alpha (b{ \gamma }_{0}y^{-1})&=a\alpha x^{-1}{ \gamma }_{0}b{ \gamma }_{0}y^{-1} \\&=a\alpha (x^{-1}{ \gamma }_{0}b){ \gamma }_{0}y^{-1} \\&=a\alpha (c{ \gamma }_{0}z^{-1}){ \gamma }_{0}y^{-1} \\&=(a\alpha c){ \gamma }_{0}(y{ \gamma }_{0}z)^{-1}. \end{aligned}
We define $$[a,x]\alpha [b,y]=[a\alpha c,y{ \gamma }_{0}z]$$, for all $$\alpha \in \varGamma$$ .

These definitions are well defined, because let $$[a,x]=[a^{'},x^{'}]$$. Then, there exist $$c_{1},d_{1} \in R$$ such that
\begin{aligned}&a{ \gamma }_{0}c_{1}=a^{'}{ \gamma }_{0}d_{1} \ \Rightarrow \ a=a^{'}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{1}^{-1}, \end{aligned}
(3)
\begin{aligned}&x{ \gamma }_{0}c_{1}=x^{'}{ \gamma }_{0}d_{1} \in X \ \Rightarrow \ x=x^{'}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{1}^{-1}. \end{aligned}
(4)
And let $$[b,y]=[b^{'},y^{'}]$$. Then, there exist $$c_{2},d_{2} \in R$$ such that
\begin{aligned}&b{ \gamma }_{0}c_{2}=b^{'}{ \gamma }_{0}d_{2} \ \Rightarrow \ b=b^{'}{ \gamma }_{0}d_{2}{ \gamma }_{0}c_{2}^{-1}, \end{aligned}
(5)
\begin{aligned}&y{ \gamma }_{0}c_{2}=y^{'}{ \gamma }_{0}d_{2} \in X \ \Rightarrow \ y=y^{'}{ \gamma }_{0}d_{2}{ \gamma }_{0}c_{2}^{-1}. \end{aligned}
(6)
1. (i)
Using (3) and (4), we have
\begin{aligned} [a,x] + [b,y] & = [a\gamma _{0} c + b\gamma _{0} d,x\gamma _{0} c] \\& = [a^{{\text{'}}} \gamma _{0} d_{1} \gamma _{0} c_{1}^{{ - 1}} \gamma _{0} c + b^{{\text{'}}} \gamma _{0} d_{2} \gamma _{0} c_{2}^{{ - 1}} \gamma _{0} d_{2} ,x^{{\text{'}}} \gamma _{0} d_{1} \gamma _{0} c_{1}^{{ - 1}} \gamma _{0} c]. \\ \end{aligned}
If we consider $$c_{3}=d_{1}{ \gamma }_{0}c_{1}^{-1}{ \gamma }_{0}c\in R\ {\rm and} \ d_{3}=d_{2}{ \gamma }_{0}c_{2}^{-1}{ \gamma }_{0}d\in R$$, then
\begin{aligned} \,[a,x]+[b,y]&=[a^{'}{ \gamma }_{0}c_{3}+b^{'}{ \gamma }_{0}d_{3},x^{'}{ \gamma }_{0}c_{3}]\\&=[a^{'},x^{'}]+[b^{'},y^{'}]. \end{aligned}
It is enough to prove that $$x^{'}{ \gamma }_{0}c_{3}=y^{'}{ \gamma }_{0}d_{3}$$. Using (4), (6) and definition (i), we have
\begin{aligned} y^{'}{ \gamma }_{0}d_{3}&=y^{'}{ \gamma }_{0}d_{2}{ \gamma }_{0}c_{2}^{-1}{ \gamma }_{0}d\\&=y{ \gamma }_{0}d\\&=x{ \gamma }_{0}c\\&=x^{'}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{1}^{-1}{ \gamma }_{0}c\\&=x^{'}{ \gamma }_{0}c_{3}. \end{aligned}
Thus, the addition is well defined.

2. (ii)
Let $$\alpha \in \varGamma$$. Then using (3), (6) and ($$*$$), we have
\begin{aligned} \,[a,x]\alpha [b,y]&=[a\alpha b,y{ \gamma }_{0}z] \\&=[(a^{'}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{1}^{-1})\alpha c,y^{'}{ \gamma }_{0}d_{2}{ \gamma }_{0}c_{2}^{-1}{ \gamma }_{0}z]\\&=[a^{'} \alpha (d_{1}{ \gamma }_{0}c_{1}^{-1}{ \gamma }_{0}c),y^{'}{ \gamma }_{0}d_{2}{ \gamma }_{0}c_{2}^{-1}{ \gamma }_{0}z]. \end{aligned}
If we consider $$c^{'}=d_{1}{ \gamma }_{0}c_{1}^{-1}{ \gamma }_{0}c\in R$$ and $$z^{'}=d_{2}{ \gamma }_{0}c_{2}^{-1}{ \gamma }_{0}z\in X$$, then
\begin{aligned} \,[a,x]\alpha [b,y]&=[a^{'} \alpha c^{'},y^{'}{ \gamma }_{0}z^{'}]\\&=[a^{'},x^{'}]\alpha [b^{'},y^{'}]. \end{aligned}
It is enough to prove that $$b^{'}{ \gamma }_{0}z^{'}=x^{'}{ \gamma }_{0}c^{'}$$. Using (5) and (4), we have
\begin{aligned} b^{'}{ \gamma }_{0}z^{'}&=b^{'}{ \gamma }_{0}d_{2}{ \gamma }_{0}c_{2}^{-1}{ \gamma }_{0}z\\&=b{ \gamma }_{0}z\\&=x{ \gamma }_{0}c\\&=x^{'}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{1}^{-1}{ \gamma }_{0}c\\&=x^{'}{ \gamma }_{0}c^{'}. \end{aligned}
Thus, multiplication is well defined.

### Lemma 3

If R is a $$\varGamma$$ -ring that satisfies the condition (*) and X be a right P-set of regular elements in R, then $$RX^{-1}$$ is an additive group.

### Proof

It is clear that $$RX^{-1}$$ is closed under addition.

For $$[a_{1},x_{1}],[a_{2},x_{2}],[a_{3},x_{3}]\in RX^{-1}$$, there exist $$c_{1},d_{1} \in R$$ such that $$z_{1}=x_{1}{ \gamma }_{0}c_{1}=x_{2}{ \gamma }_{0}d_{2} \in X$$ and there exist $$c_{2},d_{2} \in R$$ such that $$z_{2}= x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}=x_{3}{ \gamma }_{0}d_{2} \in X$$. We have
\begin{aligned} &([a_{1},x_{1}]+[a_{2},x_{2}])+[a_{3},x_{3}] \\ &\quad= [a_{1}{ \gamma }_{0}c_{1}+a_{2}{ \gamma }_{0}d_{1},x_{1}{ \gamma }_{0}c_{1}]+[a_{3},x_{3}] \\ &\quad = [(a_{1}{ \gamma }_{0}c_{1}+a_{2}{ \gamma }_{0}d_{1}){ \gamma }_{0}c_{2}+a_{3}{ \gamma }_{0}d_{2},x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}].\end{aligned}
(7)
Since $$z_{1}=x_{1}{ \gamma }_{0}c_{1}=x_{2}{ \gamma }_{0}d_{1}$$ and $$z_{2}=x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}=x_{3}{ \gamma }_{0}d_{2}$$, we obtain
\begin{aligned}&z_{1}^{-1}=c_{1}^{-1}{ \gamma }_{0}x_{1}^{-1}, \end{aligned}
(8)
\begin{aligned}&x_{1}^{-1}=c_{1}{ \gamma }_{0}z_{1}^{-1}, \end{aligned}
(9)
\begin{aligned}&x_{2}^{-1}=d_{1}{ \gamma }_{0}z_{2}^{-1}, \end{aligned}
(10)
\begin{aligned}&c_{1}^{-1}{ \gamma }_{0}x_{1}^{-1}=c_{2}{ \gamma }_{0}z_{2}^{-1}, \end{aligned}
(11)
\begin{aligned}&x_{3}^{-1}=d_{2}{ \gamma }_{0}z_{2}^{-1}. \end{aligned}
(12)
And
\begin{aligned} z_{1}&=x_{1}{ \gamma }_{0}c_{1} \Longrightarrow z_{1}{ \gamma }_{0}c_{2}= x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}=z_{2} \\&\quad \Longrightarrow \ \ z_{1}^{-1}=c_{2}{ \gamma }_{0}z_{2}^{-1}. \end{aligned}
(13)
Using (8), (9), (11) and (12), we have
\begin{aligned} & a_{2}{ \gamma }_{0}x_{2}^{-1}+a_{3}{ \gamma }_{0}x_{3}^{-1} \\ \quad &=a_{2}{ \gamma }_{0}d_{1}{ \gamma }_{0}z_{1}^{-1}+a_{3}{ \gamma }_{0}d_{2}{ \gamma }_{0}z_{2}^{-1} \\&=a_{2}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{1}^{-1}{ \gamma }_{0}x_{1}^{-1}+a_{3}{ \gamma }_{0}d_{2}{ \gamma }_{0}z_{2}^{-1} \\ \quad &=a_{2}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{2}{ \gamma }_{0}z_{2}^{-1}+a_{3}{ \gamma }_{0}d_{2}{ \gamma }_{0}z_{2}^{-1}. \end{aligned}
Then
\begin{aligned} & [a_{1},x_{1}] +([a_{2},x_{2}]+[a_{3},x_{3}]) \\ &= [a_{1},x_{1}]+[a_{2}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{2}+a_{3}{ \gamma }_{0}d_{2},x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}]. \end{aligned}
But using (9) and (13), we have
\begin{aligned} &a_{1}{ \gamma }_{0}x_{1}^{-1}+(a_{2}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{2}+a_{3}{ \gamma }_{0}d_{2}){ \gamma }_{0}(x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2})^{-1} \\ &\quad=a_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}z_{1}^{-1} +(a_{2}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{2}+a_{3}{ \gamma }_{0}d_{2}){ \gamma }_{0}z_{2}^{-1} \\ &\quad=a_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}{ \gamma }_{0}z_{2}^{-1}+(a_{2}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{2}+a_{3}{ \gamma }_{0}d_{2}){ \gamma }_{0}z_{2}^{-1}.\end{aligned}
Thus
\begin{aligned} &[a_{1},x_{1}]+[a_{2}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{2}+a_{3}{ \gamma }_{0}d_{2},x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}] \\ &\quad =[a_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}+a_{2}{ \gamma }_{0}d_{1}{ \gamma }_{0}c_{2}+a_{3}{ \gamma }_{0}d_{2},x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}]\\ &\quad =[(a_{1}{ \gamma }_{0}c_{1}+a_{2}{ \gamma }_{0}d_{1}){ \gamma }_{0}c_{2}+a_{3}{ \gamma }_{0}d_{2},x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}]. \end{aligned}
(14)
Using (7) and (14), ($$RX^{-1},+$$) is associative. Now, we prove $$[0,1]\in RX^{-1}$$ is identity element. For every $$a\in R$$, $$x\in X$$ and $$[a,x]\in RX^{-1}$$, if we consider $$c=1\in R$$ and $$d=x\in X$$, then $$x{ \gamma }_{0}c=1{ \gamma }_{0}d$$ and so
\begin{aligned} [a,x]+[0,1]&=[a{ \gamma }_{0}c+0{ \gamma }_{0}d,x{ \gamma }_{0}c]\\ &=[a,x]. \end{aligned}
And also, we consider $$c^{'}=x\in R$$ and $$d^{'}=1\in R$$, then we have
\begin{aligned}{}[0,1]+[a,x]&=[0{ \gamma }_{0}c+a{ \gamma }_{0}d^{'},1{ \gamma }_{0}c^{'}] \\ &=[0+a{ \gamma }_{0}1,x]\\&=[a,x]. \end{aligned}
For every $$[a,x]\in RX^{-1}$$, we know $$[-a,x]\in RX^{-1}$$. If we consider $$c=d=x^{-1} \in R$$, then $$x{ \gamma }_{0}c=x{ \gamma }_{0}d=1$$ and so
\begin{aligned}[a,x]+[-a,x]&=[a{ \gamma }_{0}c-a{ \gamma }_{0}d,x{ \gamma }_{0}c]\\ &=[0,1]. \end{aligned}
Therefore, $$[-a,x]$$ is inverse of [a, x].
Now, we show that ($$RX^{-1},+$$) is Abelian group.
\begin{aligned} [a,x]+[b,y]&=[a{ \gamma }_{0}c+b{ \gamma }_{0}d,x{ \gamma }_{0}c]\ \ \ (z=x{ \gamma }_{0}c=y{ \gamma }_{0}d\in X)\\&=[b{ \gamma }_{0}d+a{ \gamma }_{0}c,y{ \gamma }_{0}d]\ \ \ (R\ is\ Abelian)\\ &=[b,y]+[a,x]. \end{aligned}
Thus, the proof is complete. $$\square$$

### Theorem 1

LetR be a$$\varGamma$$-ring that satisfies the condition (*) and X be a right P-set of regular elements in R. Then, ($$RX^{-1},+,\cdot$$) is a $$\varGamma$$-ring with an identity element [1, 1].

### Proof

Using Lemma 3, ($$RX^{-1},+$$) is an Abelian group. For every $$[a_{1},x_{1}]$$,$$[a_{2},x_{2}]$$ and $$[a_{3},x_{3}]$$ in $$RX^{-1}$$ and $$\alpha, \beta \in \varGamma$$, we have
\begin{aligned} & ([a_{1},x_{1}]+[a_{2},x_{2}])\alpha [a_{3},x_{3}] \\ &\quad =[a_{1}{ \gamma }_{0}c_{1}+a_{2}{ \gamma }_{0}d_{1},x_{1}{ \gamma }_{0}c_{1}]\alpha [a_{3},x_{3}]\\ &\quad =[(a_{1}{ \gamma }_{0}c_{1}+a_{2}{ \gamma }_{0}d_{1})\alpha c_{2},x_{3}{ \gamma }_{0}z_{1}]. \end{aligned}
(15)
There exist $$c_{1},d_{1} \in R$$ such that
\begin{aligned} x_{1}{ \gamma }_{0}c_{1}=x_{2}{ \gamma }_{0}d_{1}. \end{aligned}
(16)
There exist $$c_{2} \in R$$ and $$z_{1} \in X$$ such that
\begin{aligned} a_{3}{ \gamma }_{0}z_{1}=x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}. \end{aligned}
(17)
Using (16) and (17), we obtain
\begin{aligned} x_{1}{ \gamma }_{0}c_{1} =x_{2}{ \gamma }_{0}d_{1},\ \ x_{2} \in X \ \ \Longrightarrow {\left\{ \begin{array}{l} x_{2}^{-1}{ \gamma }_{0}x_{1}{ \gamma }_{0}c_{1}=d_{1},\\ x_{2}^{-1}{ \gamma }_{0}x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2}=d_{1}{ \gamma }_{0}c_{2}, \end{array}\right. } \end{aligned}
(18)
\begin{aligned} a_{3}{ \gamma }_{0}z_{1}=x_{1}{ \gamma }_{0}c_{1}{ \gamma }_{0}c_{2} \ \ \Longrightarrow \ \ x_{1}^{-1}{ \gamma }_{0}a_{3}=c_{1}{ \gamma }_{0}c_{2}{ \gamma }_{0}z_{1}^{-1}. \end{aligned}
(19)
Using (17) and (18), we have
\begin{aligned} x_{2}^{-1}{ \gamma }_{0}a_{3}{ \gamma }_{0}z_{1}=d_{1}{ \gamma }_{0}c_{2} \ \ \Longrightarrow \ \ x_{2}^{-1}{ \gamma }_{0}a_{3}=d_{1}{ \gamma }_{0}c_{2}{ \gamma }_{0}z_{1}^{-1}. \end{aligned}
(20)
For $$[a_{1},x_{1}]\alpha [a_{3},x_{3}]+[a_{2},x_{2}]\alpha [a_{3},x_{3}]$$, by ($$*$$) and (19), we get
\begin{aligned} (a_{1}{ \gamma }_{0}x_{1}^{-1})\alpha (a_{3}{ \gamma }_{0}x_{3}^{-1})&=a_{1} \alpha x_{1}^{-1}{ \gamma }_{0}a_{3}{ \gamma }_{0}x_{3}^{-1} \\ &=a_{1} \alpha c_{1}{ \gamma }_{0}c_{2}{ \gamma }_{0}z_{1}^{-1}{ \gamma }_{0}x_{3}^{-1} \\ &=(a_{1} \alpha c_{1}{ \gamma }_{0}c_{2}){ \gamma }_{0}(x_{3}{ \gamma }_{0}z_{1})^{-1}. \end{aligned}
(21)
Using ($$*$$) and (20), we have
\begin{aligned} (a_{2}{ \gamma }_{0}x_{2}^{-1})\alpha (a_{3}{ \gamma }_{0}x_{3}^{-1}) &=a_{2} \alpha x_{2}^{-1}{ \gamma }_{0}a_{3}{ \gamma }_{0}x_{3}^{-1} \\ &=a_{2} \alpha d_{1}{ \gamma }_{0}c_{2}{ \gamma }_{0}z_{1}^{-1}{ \gamma }_{0}x_{3}^{-1} \\ &=(a_{2}{ \gamma }_{0}d_{1} \alpha c_{2}){ \gamma }_{0}(x_{3}{ \gamma }_{0}z_{1})^{-1}. \end{aligned}
(22)
Using (21) and (22), we obtain
\begin{aligned} & [a_{1},x_{1}]\alpha [a_{3},x_{3}]+[a_{2},x_{2}]\alpha [a_{3},x_{3}] \\ &\quad=[(a_{1}{ \gamma }_{0}c_{1})\alpha c_{2},x_{3}{ \gamma }_{0}z_{1}]+[(a_{2}{ \gamma }_{0}d_{1})\alpha c_{2},x_{3}{ \gamma }_{0}z_{1}]. \end{aligned}
If we consider $$c=d=1\in R$$, then $$x_{3} \alpha z_{1}{ \gamma }_{0}c=x_{3} \alpha z_{1} { \gamma }_{0}d$$ and so
\begin{aligned} &[(a_{1}{ \gamma }_{0}c_{1})\alpha c_{2},x_{3}{ \gamma }_{0}z_{1}]+[(a_{2}{ \gamma }_{0}d_{1})\alpha c_{2},x_{3}{ \gamma }_{0}z_{1}]\\ & \quad=[(a_{1}{ \gamma }_{0}c_{1})\alpha c_{2}{ \gamma }_{0}1+(a_{2}{ \gamma }_{0}d_{1})\alpha c_{2}{ \gamma }_{0}1, x_{3}{ \gamma }_{0}z_{1}{ \gamma }_{0}1]\\& \quad =[(a_{1}{ \gamma }_{0}c_{1}+a_{2}{ \gamma }_{0}d_{1})\alpha c_{2},x_{3}{ \gamma }_{0}z_{1}]. \end{aligned}
(23)
Using (15) and (22), it is concluded that
\begin{aligned} &([a_{1},x_{1}]+[a_{2},x_{2}])\alpha [a_{3},x_{3}] \\ &\quad= [a_{1},x_{1}]\alpha [a_{3},x_{3}]+[a_{2},x_{2}]\alpha [a_{3},x_{3}].\end{aligned}
Now, we show that
\begin{aligned} [a_{1},x_{1}](\alpha +\beta )[a_{2},x_{2}]=[a_{1},x_{1}]\alpha [a_{2},x_{2}]+[a_{1},x_{1}]\beta [a_{2},x_{2}]. \end{aligned}
Since $$x_{1} \in X$$ and X is a right P-set, so there exist $$c\in R$$ and $$z\in X$$ such that $$a_{2}{ \gamma }_{0}z=x_{1}{ \gamma }_{0}c$$. If we set $$c^{'}=d^{'}=1$$, then we have
\begin{aligned}&[a_{1},x_{1}]\alpha [a_{2},x_{2}]+[a_{1},x_{1}]\beta [a_{2},x_{2}]\\ &\quad =[a_{1} \alpha c,x_{2}{ \gamma }_{0}z]+[a_{1} \beta c,x_{2}{ \gamma }_{0}z]\\ &\quad=[a_{1} \alpha c { \gamma }_{0}1+a_{1} \beta c { \gamma }_{0}1,x_{2}{ \gamma }_{0}z{ \gamma }_{0}1]\\ &\quad=[a_{1}(\alpha +\beta )c,x_{2} { \gamma }_{0} z]\\ &\quad =[a_{1},x_{1}](\alpha +\beta )[a_{2},x_{2}]. \end{aligned}
For $$([a_{1},x_{1}]\alpha [a_{2},x_{2}])\beta [a_{3},x_{3}]=[a_{1},x_{1}]\alpha ( [a_{2},x_{2}]\beta [a_{3},x_{3}])$$, we have
\begin{aligned} ([a_{1},x_{1}]\alpha [a_{2},x_{2}])\beta [a_{3},x_{3}]&=[a_{1} \alpha c_{1},x_{2}{ \gamma }_{0}z_{1}]\beta [a_{3},x_{3}]\\ &\quad =[a_{1} \alpha c_{1} \beta c_{2},x_{3}{ \gamma }_{0}z_{2}]. \end{aligned}
(24)
There exist $$c_{1} \in R$$, $$z_{1} \in X$$ such that
\begin{aligned} a_{2}{ \gamma }_{0}z_{1}=x_{1}{ \gamma }_{0}c_{1}. \end{aligned}
(25)
There exist $$c_{2} \in R$$, $$z_{2} \in X$$ such that
\begin{aligned} a_{3}{ \gamma }_{0}z_{1}=x_{2}{ \gamma }_{0}z_{1}{ \gamma }_{0}c_{2}. \end{aligned}
(26)
Using (25) and (26), we obtain
\begin{aligned} a_{2}{ \gamma }_{0}z_{1}=x_{1}{ \gamma }_{0}c_{1} \ \ \Longrightarrow \ \ x_{1}^{-1}{ \gamma }_{0}a_{2}{ \gamma }_{0}z_{1}=c_{1}, \end{aligned}
(27)
\begin{aligned} a_{3}{ \gamma }_{0}z_{2}=x_{2}{ \gamma }_{0}z_{1}{ \gamma }_{0}c_{2} \ \ \Longrightarrow \ \ x_{2}^{-1}{ \gamma }_{0}a_{3}=z_{1}{ \gamma }_{0}c_{2}{ \gamma }_{0}z_{2}^{-1}. \end{aligned}
(28)
Using (28) and ($$*$$), we have
\begin{aligned} (a_{2}{ \gamma }_{0}x_{2}^{-1})\beta (a_{3} { \gamma }_{0}x_{3}^{-1})&=a_{2} \beta (x_{2}^{-1}{ \gamma }_{0}a_{3}){ \gamma }_{0}x_{3}^{-1} \\&=a_{2} \beta (z_{1} { \gamma }_{0}c_{2}{ \gamma }_{0}z_{2}^{-1}){ \gamma }_{0}x_{3}^{-1} \\&=a_{2} \beta z_{1} { \gamma }_{0}c_{2}{ \gamma }_{0}(x_{3}{ \gamma }_{0}z_{2})^{-1} \\&=(a_{2} { \gamma }_{0} z_{1})\beta c_{2} { \gamma }_{0}(x_{3}{ \gamma }_{0}z_{2})^{-1}. \end{aligned}
Therefore
\begin{aligned} \,[a_{1},x_{1}]\alpha ( [a_{2},x_{2}]\beta [a_{3},x_{3}])=[a_{1},x_{1}]\alpha [(a_{2}{ \gamma }_{0}z_{1})\beta c_{2},x_{3}{ \gamma }_{0}z_{2}]. \end{aligned}
Using (27) and ($$*$$), we obtain
\begin{aligned} &(a_{1}{ \gamma }_{0}x_{1}^{-1})\alpha (a_{2}{ \gamma }_{0}z_{1})\beta c_{2}{ \gamma }_{0}(x_{3}{ \gamma }_{0}z_{2}^{-1})\\ &\quad=a_{1} \alpha x_{1}^{-1}{ \gamma }_{0} a_{2}{ \gamma }_{0}z_{1} \beta c_{2}{ \gamma }_{0}(x_{3}{ \gamma }_{0}z_{2})^{-1} \\ &\quad=a_{1} \alpha c_{1} \beta c_{2}{ \gamma }_{0}(x_{3}{ \gamma }_{0}z_{2})^{-1}. \end{aligned}
And then
\begin{aligned} \,[a_{1},x_{1}]\alpha ( [a_{2},x_{2}]\beta [a_{3},x_{3}])=[a_{1} \alpha c_{1} \beta c_{2},x_{3}{ \gamma }_{0}z_{2}]. \end{aligned}
(29)
Using (24) and (29), it is concluded that
\begin{aligned} ([a_{1},x_{1}]\alpha [a_{2},x_{2}])\beta [a_{3},x_{3}]=[a_{1},x_{1}]\alpha ( [a_{2},x_{2}]\beta [a_{3},x_{3}]). \end{aligned}
For all $$[a,x]\in RX^{-1}$$, we have
\begin{aligned} (a{ \gamma }_{0}x^{-1}){ \gamma }_{0}(1{ \gamma }_{0}1)=(a{ \gamma }_{0}x^{-1}){ \gamma }_{0}1=a{ \gamma }_{0}x^{-1}. \end{aligned}
Thus, $$[a,x]{ \gamma }_{0}[1,1]=[a,x]$$ and similarly $$[1,1]{ \gamma }_{0}[a,x]=[a,x]$$; therefore, $$[1,1]\in RX^{-1}$$ is an identity element and, hence, the proof is complete. $$\square$$

### Definition 2

The $$\varGamma$$-ring $$RX^{-1}$$ is called the $$\varGamma$$-ring of right fraction of R with respect to X.

## 3 Homomorphisms of Noncommutative Gamma Rings

### Theorem 2

Let R be a $$\varGamma$$ -ring that satisfies the condition (*) and $$RX^{-1}$$ be a $$\varGamma$$ -ring of right fraction. Then, the mapping $$f:R \ \ \longrightarrow \ \ RX^{-1}$$ such that $$f(r)=[r,1]$$ is a $$\varGamma$$ -ring homomorphism.

### Proof

At first, we prove that f is well defined. If $$r_1 =r_2$$, since $$r_1 =r_1 \gamma _0 1$$ and $$r_2 =r_2 \gamma _0 1$$, then it is enough to set $$c=d=1$$ and so $$r_1 \gamma _0c=r_{2} \gamma _{0}d$$, $$1 \gamma _0c=1\gamma _{0}d \in X$$ and, therefore, $$[r_1,1]=[r_2,1]$$.

Now we prove that $$f(r_1 +r_2 )=f(r_1 )+f(r_2 )$$ and $$f(r_1 \alpha r_2 )=f(r_1 )\alpha f(r_2 )$$, for all $$r_1, r_2 \in R$$ and $$\alpha \in \varGamma$$. We have $$r_{1} \gamma _{0}1^{-1}+r_{2} \gamma _{0} 1^{-1}=(r_{1}+r_{2}) \gamma _{0} 1^{-1}=(r_{1}+r_{2}) \gamma _{0} 1$$ so let $$c=d=1$$. Then
\begin{aligned} f(r_1 )+f(r_2 )&=[r_1,1]+[r_2,1]\\ &=[r_1 \gamma _0 1+r_2 \gamma _0 1,1\gamma _0 1] \\ &=[r_1 +r_2,1] \\ &=f(r_1 +r_2 ). \end{aligned}
Also using ($$*$$), we have $$(r_{1} \gamma _{0} 1^{-1}) \alpha (r_{2} \gamma _{0} 1^{-1})=(r_{1} \alpha r_{2}) \gamma _{0} 1$$ so let $$z=1 \in X$$ and $$c=r_{2} \in R$$. Then, $$r_{2} \gamma _{0}z=1 \gamma _{0} c$$, we get

$$f(r_1 )\alpha f(r_2 )=[r_1,1]\alpha [r_2,1]=[r_1 \alpha c,1\gamma _{0}z]=[r_{1} \alpha r_{2}, 1]=f(r_1 \alpha r_2 )$$.

Hence, the theorem is proved. $$\square$$

### Lemma 4

Let R and $$R^{'}$$ be a $$\varGamma$$ -rings with identity elements and $$f:R\longrightarrow R^{'}$$ be a $$\varGamma$$ -ring epimorphism. Then $$f(1_{R})=1_{R^{'}}$$ .

### Proof

We prove that $$f(1_R )\gamma _0 r^{'}=r^{'} \gamma _0 f(1_R )=r^{'}$$, for all $$r^{'} \in R^{'}$$. Since f is surjective, there exists $$r\in R$$ such that $$f(r)=r^{'}$$. We have
\begin{aligned}&f(1_R )\gamma _0 r^{'}=f(1_R )\gamma _0 f(r)=f(1_R \gamma _0 r)=f(r)=r^{'}, \\&r^{'} \gamma _0 f(1_R )=f(r)\gamma _0 f(1_R )=f(r\gamma _0 1_R )=f(r)=r^{'}. \end{aligned}
Hence $$f(1_R )=1_{R^{'}}$$. $$\square$$

### Lemma 5

If R and $$R^{'}$$ are $$\varGamma$$ -ring with identity elements, without zero-divisor and $$f:R\longrightarrow R^{'}$$ is a $$\varGamma$$ -ring homomorphism, then $$f(1_{R})=1_{R^{'}}$$ .

### Proof

We have
\begin{aligned} f(1_R )&=f(1_R \gamma _{0} 1_R )=f(1_R )\gamma _0 f(1_R )\\&\Rightarrow \ \ f(1_R )-f(1_R )\gamma _0 f(1_R )=0\\&\Rightarrow \ \ f(1_R )\gamma _0 (1_{R^{'}}-f(1_R ))=0\\&\Rightarrow \ \ 1_{R^{'}}-f(1_R )=0. \end{aligned}
Hence $$f(1_R )=1_{R^{'}}$$. $$\square$$

### Lemma 6

Let R and $$R^{'}$$ be $$\varGamma$$ -rings with identity elements, without zero-divisor and $$f:R\longrightarrow R^{'}$$ is a $$\varGamma$$ -ring homomorphism. Then, $$f(a^{-1})=(f(a))^{-1}$$ .

### Proof

Suppose $$a\in R$$ is invertible and $$a^{-1}$$ is inverse of a. Using Lemma 5, we have
\begin{aligned} f(a\gamma _0 a^{-1})&=f(1_R )=f(a^{-1} \gamma _0 a)\\&\Rightarrow \ \ f(a)\gamma _0 f(a^{-1})=1_{R^{'}}=f(a^{-1})\gamma _0 f(a)\\&\Rightarrow \ \ f(a^{-1})={(f(a))}^{-1}. \end{aligned}
$$\square$$

### Theorem 3

Suppose R and $$R^{'}$$ be $$\varGamma$$-rings and $$g:R\longrightarrow R^{'}$$ be a $$\varGamma$$-ring homomorphism such that g(x) is invertible in $$R^{'}$$ for all $$x\in X$$. Then, there exists a unique $$\varGamma$$-ring homomorphism $$h:RX^{-1} \longrightarrow R^{'}$$ such that $$g=hof$$, where $$f:R\longrightarrow RX^{-1}$$ and $$f(a)=[a,1]$$.

### Proof

We have $$a{ \gamma }_{0}x^{-1}=a{ \gamma }_{0}(1^{-1}{ \gamma }_{0}1){ \gamma }_{0}x^{-1}=(a{ \gamma }_{0}1^{-1}){ \gamma }_{0}(1{ \gamma }_{0}x^{-1})$$ and so $$[a,x]=[a,1]{ \gamma }_{0}[1,x]$$. Using Lemma 6, we define
\begin{aligned} h([a,x])&=h([a,1]{ \gamma }_{0}[1,x])\\&=h(f(a){ \gamma }_{0}(f(x))^{-1})\\&=h(f(a)){ \gamma }_{0}h(f(x))^{-1} \\&=g(a){ \gamma }_{0}g(x)^{-1}. \end{aligned}
Therefore, $$h([a,x])=g(a){ \gamma }_{0}g(x)^{-1}$$. We prove that h is a $$\varGamma$$-ring homomorphism.
For all $$[a_{1},x_{1}],[a_{2},x_{2}]\in RX^{-1}$$ and $$\alpha \in \varGamma$$, since X is the right P-set using Lemma 1, there exist $$c,d\in R$$ such that $$z=x_{1}{ \gamma }_{0}c=x_{2}{ \gamma }_{0}d$$, we have
\begin{aligned} h([a_{1},x_{1}]+[a_{2},x_{2}])& =h([a_{1}{ \gamma }_{0}c+a_{2}{ \gamma }_{0}d,x_{1}{ \gamma }_{0}c])\\ &=g(a_{1}{ \gamma }_{0}c+a_{2}{ \gamma }_{0}d){ \gamma }_{0}g(x_{1}{ \gamma }_{0}c)^{-1} \\ &=(g(a_{1}{ \gamma }_{0}c)+g(a_{2}{ \gamma }_{0}d)){ \gamma }_{0}g(x_{1}{ \gamma }_{0}c)^{-1} \\ & =g(a_{1}){ \gamma }_{0}g(c){ \gamma }_{0}g(x_{1}{ \gamma }_{0}c)^{-1}+g(a_{2}{ \gamma }_{0}d){ \gamma }_{0}g(x_{2}{ \gamma }_{0}d)^{-1} \\ &=g(a_{1}){ \gamma }_{0}g(c){ \gamma }_{0}g(c)^{-1}{ \gamma }_{0}g(x_{1})^{-1}\\ &\quad +g(a_{2}){ \gamma }_{0}g(d){ \gamma }_{0}g(d)^{-1}{ \gamma }_{0}g(x_{2})^{-1} \\ &=g(a_{1}){ \gamma }_{0}g(x_{1})^{-1}+g(a_{2}){ \gamma }_{0}g(x_{2})^{-1} \\ &=h([a_{1},x_{1}])+h([a_{2},x_{2}]). \end{aligned}
For all $$[a_{1},x_{1}],[a_{2},x_{2}]\in RX^{-1}$$, since X is the right P-set, there exist $$c\in R$$ and $$z\in X$$ such that $$a_{2}{ \gamma }_{0}z=x_{1}{ \gamma }_{0}c\in X$$ or $$c=x_{1}^{-1}{ \gamma }_{0}a_{2}{ \gamma }_{0}z$$, we have
\begin{aligned} h([a_{1},x_{1}]\alpha [a_{2},x_{2}])&=h([a_{1} \alpha c,x_{2}{ \gamma }_{0}z])\\ &=g(a_{1} \alpha c){ \gamma }_{0}g(x_{2}{ \gamma }_{0}z)^{-1} \\ &=g(a_{1}) \alpha g(c){ \gamma }_{0}g(z)^{-1}{ \gamma }_{0}g(x_{2})^{-1} \\ &=g(a_{1})\alpha g(x_{1}^{-1}{ \gamma }_{0}a_{2}{ \gamma }_{0}z){ \gamma }_{0}g(z)^{-1}{ \gamma }_{0}g(x_{2})^{-1} \\ &=g(a_{1})\alpha g(x_{1})^{-1}{ \gamma }_{0}g(a_{2}){ \gamma }_{0}g(z)\\ &\quad\times{ \gamma }_{0} g(z)^{-1}{ \gamma }_{0}g(x_{2})^{-1} \\ &=g(a_{1}){ \gamma }_{0}g(x_{1})^{-1} \alpha g(a_{2}){ \gamma }_{0}g(x_{2})^{-1} \\ &=h([a_{1},x_{1}])\alpha h([a_{2},x_{2}]). \end{aligned}
$$\square$$

### Corollary 1

If $$g:R\longrightarrow R^{'}$$ is a $$\varGamma$$ -ring homomorphism such that g has following properties:
1. (i)

For all $$x\in X$$ , g (x) in $$R^{'}$$ is invertible.

2. (ii)

If $$g(a)=0$$ , then there exists $$t\in X$$ such that $$a{ \gamma }_{0}t=0$$ .

3. (iii)

Every element of $$R^{'}$$ is the form $$g(a){ \gamma }_{0}g(x)^{-1}$$ , for some $$a\in R$$ and some $$x\in X$$ , then the $$\varGamma$$ -ring homomorphism $$h:RX^{-1} \longrightarrow R^{'}$$ such that $$h([a,x])=g(a){ \gamma }_{0}g(x)^{-1}$$ is isomorphism.

## References

1. Barnes WE (1966) On the $$\varGamma$$-ring of Nobusawa. Pac J Math 18:411–422
2. Dey KK, Paul AC (2014) Commutativity of prime gamma rings with left centralizers. J Sci Res 6:69–77
3. Dey KK, Paul AC (2011) Free action on prime and semiprime gamma rings. Gen Math Notes 5:7–14Google Scholar
4. Dey KK, Paul AC, Rakhimov S (2012) Dependent elements in prime and semiprime gamma rings. Asian J Algebra 5:11–20
5. Dumitru M (2009) Gamma rings; some interpretation used in the study of their radicals. UPB Sci Bull Series A 71:9–22
6. Fazlul MD, Paul AC (2011) On centralizers of semiprime gamma rings. Int Math Forum 6:627–638
7. Goodearl KR, Warfield RB (2004) An introduction to noncommutative noteherian rings. Cambridge University Press, Cambridge
8. Kyuno S (1978) On the prime gamma rings. Pac J Math 75:185–190
9. Nobusawa N (1964) On the generalization of the ring theorey. Osaka J Math 1:81–89
10. Paul AC, Uddin S (2010) Lie structure in simple gamma ring. Int J Pure Appl Sci Tech 4:63–70Google Scholar
11. Selvarage C, Petchimuthu S (2008) On strongly prime gamma ring and morita equvalence of rings. Southeast Asian Bull Math 19:1137–1147Google Scholar
12. Ullah Z, Chaudhry MA (2012) On $$K$$-centralizers of semiprime gamma rings. Int J Algebra 6:1001–1010