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This question already has an answer here:

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?

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marked as duplicate by André 3000, darij grinberg, user10354138, Brahadeesh, KReiser Dec 7 '18 at 8:22

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    $\begingroup$ Yes, evaluation homomorphism can do the trick $\phi(p(x))=p\left(\frac{1}{r}\right)$. $\endgroup$ – Anurag A Dec 6 '18 at 22:07
  • $\begingroup$ Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few. $\endgroup$ – André 3000 Dec 6 '18 at 23:31
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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 properties of localizations then see also the linked dupe for ways to exploit these proeprties.

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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.

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  • $\begingroup$ Why is the map clearly a surjection? $\endgroup$ – Wesley Dec 6 '18 at 22:18
  • $\begingroup$ Let $a\in R$ and $r^k\in S$. Is it surjective because $\phi(ax^k) = \frac{a}{r^k}$, and the latter term is a general term of the ring of fractions? $\endgroup$ – Wesley Dec 6 '18 at 22:26
  • $\begingroup$ Yes that is one way to see it. $\endgroup$ – Antonios-Alexandros Robotis Dec 6 '18 at 22:50
  • $\begingroup$ It's clear that $(rx-1) \subseteq \ker(\phi)$, but how would I show that $\ker(\phi) \subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm. $\endgroup$ – Wesley Dec 6 '18 at 23:14
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    $\begingroup$ @Wesley One simple way is to use the nonmonic division algorithm - see my answer. $\endgroup$ – Bill Dubuque Dec 7 '18 at 0:02

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