The Construction of Fraction Gamma Ring Using Noncommutative Gamma Ring

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.

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

• (a) Every element of X is invertible in S.

• (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

• (i) X is a right P-set in R.

• (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 $1\le i \le n$ .

• (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

• (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.

• (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 $s_{i}=a_{i}{ \gamma }_{0}x^{-1}$.

• (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 [a, x] 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 [a, x] and [b, y] in $RX^{-1}$ as follows:

• (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 $[a,x]+[b,y]=[a{ \gamma }_{0}c+b{ \gamma }_{0}d,x{ \gamma }_{0}c]$.

• (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)

• (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.

• (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:

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

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

• (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.