Suppose $R$ is $\mathbb{Z}$ graded ring and $0\neq f \in R\_1.$ Then I want to show that $R\_f \cong S[X,X^{-1}],$ where $S=(R\_f)\_0$ and $X$ transcendental over $S.$


I wanted to use the isomorphism $R\_f\cong R[X]/(Xf-1)$ but I was not able to get the result. 



$R$ is $\mathbb{Z}$ graded ring and $0\neq f \in R_1,$ show that $R_f \cong S[X,X^{-1}]$
It suffices to note that $A:=R\_f$ is a $\mathbb{Z}$-graded ring with an invertible element $f\in A\_1$. The $A\_0$-homomorphism $A\_0[X,X^{-1}]\to A$ sending $X$ to $f$ respects $\mathbb{Z}$-gradings and is an isomorphism in degree $0$. So it must be bijective because multiplication by powers of $X$ (resp. $f$) in $A\_0[X,X^{-1}]$ (resp. $A$) induces isomorphisms between homogeneous pieces. 


