Show that a ring of fractions and a quotient ring are isomorphic For a commutative ring $R$ with $1\neq 0$ and a nonzerodivisor $r \in R$, let $S$ be the set 
$S=\{r^n\mid n\in \mathbb{Z}, n\geq 0\}$ and denote $S^{-1}R=R\left[\frac{1}{r}\right]$.
 Prove that there is a ring isomorphism $$R\left[\frac{1}{r}\right]\cong \frac{R[x]}{(rx -1)}.$$
I'm thinking maybe I can find a homomorphism from $R[x]$ to $R\left[\frac{1}{r}\right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?
 A: Tricky is proving the kernel $K = (rx\!-\!1).\,$ A simple way: if $f\in K$ then by nonmonic division
$$ r^n f(x) = (rx\!-\!1)\,q(x) + r',\ \ {\rm for} \ \ r'\in R,\ n\in \Bbb N$$
Evaluating at $\, x = 1/r\,$ shows $\,r'\! = 0\,$ so $\,rx\!-\!1\mid r^n f\,\Rightarrow\,rx\!-\!1\mid f,\,$ by $\,(rx\!-\!1,r) = (1);\,$ more explicitly $\,rx\!-\!1\mid rg\,\Rightarrow\, rx\!-\!1\mid g = x(rg)-(rx\!-\!1)g$.
Remark $\ $ See this answer for another proof and further discussion. If you already know basic (universal) properties of localizations then see the linked dupe for ways to employ them.
A: You should note that $\frac{1}{r}$ has the property $\frac{1}{r}\cdot r=1$. That is, $\frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $s\in R$ to itself for all $s$. Send $x\mapsto \frac{1}{r}$. This is clearly a surjective homomorphism 
$$ R[x]\xrightarrow{\phi} S^{-1}R.$$
Calculate $\ker\phi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.
