Showing that $R[X]/(Xf-1) \cong R[1/f]$ 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?
 A: Let $S=R[1/f]$ and let $i:R\to S$ be the obvious map. You can easily check that the pair $(S,i)$ has the following property:

whenever $A$ is a ring and $q:R\to A$ is a morphism such that $q(f)$ is invertible in $A$ there is a unique morphism $\bar q:S\to A$ such that $\bar q\circ i=q$.

On the other hand, let $T=R[X]/(fX-1)$ and let $j:R\to T$ be the obvious map. Show that the pair $(T,j)$ has exactly the same property, that is:

whenever $A$ is a ring and $q:R\to A$ is a morphism such that $q(f)$ is invertible in $A$ there is a unique morphism $\bar q:T\to A$ such that $\bar q\circ j=q$.

Finally, show that if $(U_1,k_1)$ and $(U_2,k_2)$ are two pairs which have the above propeerty, then $U_1\cong U_2$.
A: To complete the argument, you can observe the following : if $P\in R[X]$ verifies $P(\frac{1}{f}) = 0$, writing $P = \sum_{k=0}^d a_k X^k$, you have that $\sum_{k=0}^d a_k f^{d-k} = 0$ which shows that $f$ is root of the polynomial $Q = \sum_{k=0}^d a_{d-k} X^k$. Note that $Q(X) = X^d P(\frac{1}{X})$ and $P(X) = X^d Q(\frac{1}{X})$. Now, as $f$ is root of $Q(X)$, you can write that $Q(X) = (X-f)R(X)$ for some $R(X) \in K[X]$. Then, you can check by hand (writing $R(X) = \sum_{k=0}^{d-1} \beta_k X^k$ and injecting that in $Q(X) = (X-f)R(X)$ and then expanding and regrouping same powers of $X$) that $R \in R[X]$. Now this implies that $X^d Q(\frac{1}{X}) = X^d(\frac{1}{X}-f)R(\frac{1}{X}) = (1-fX) X^{d-1} R(\frac{1}{X})$. The element $X^{d-1} R(\frac{1}{X})$ is a polynomial $S(X)$ in $R[X]$ and, as $P(X) = X^d Q(\frac{1}{X})$, we have $P(X) = (1-fX)S(X)$.
For your second question : you can show that $R[X]/(fX-1)$ verifies the universal property of the localization of $R$ wrt powers of $f$. This will straightforwardly imply the isomorphism, as $R$ being a domain implies that $R[1/f]$ is the localisation of $R$ wrt powers of $f$. To verify the universal property, you have to show that the obvious morphism $R \to R[X]/(fX-1)$ sends powers of $f$ in $(R[X]/(fX-1))^{\times}$, and that every ring morphism $\varphi : B \to R[X]/(fX-1)$ sending powers of $f$ in $(R[X]/(fX-1))^{\times}$ factor through the morphism $R \to R[X]/(fX-1)$.
