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The Construction of Fraction Gamma Ring Using Noncommutative Gamma Ring

  • Zohre Tabatabaee
  • Tahereh RoodbarylorEmail author
Open Access
Research Paper
  • 400 Downloads

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.

     

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Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  1. 1.Department of Mathematics, Kerman BranchIslamic Azad UniversityKermanIran

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