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Let $R$ be an integral domain with quotient field $K$. Let $0 \neq f \in R$. I want to prove
Statement: $R[X]/(Xf-1) \cong R[1/f]$.
Argument: Consider the epimorphism $\phi: R[X] \rightarrow R[1/f]$ which sends $X$ to $1/f$. The goal is to show that the kernel of $\phi$ is the ideal generated by $Xf-1$. If $\phi(p[X])=0$ then $p(1/f)=0$. This shows that $1/f$ is a root of $p(X)$. Viewing $p[X]$ as an element of $K[X]$, this implies that there exists some $g(X) \in K[X]$ such that $p(X) = (X-1/f) g(X) = (Xf-1) \left[g(X) /f\right]$. I will be done if i can show that $g(X)/f \in R[X]$.
Question 1: How to complete the argument?
Question 2 (optional): Is there any other proof to the statement?