# Show that a ring of fractions and a quotient ring are isomorphic [duplicate]

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?

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

• Yes, evaluation homomorphism can do the trick $\phi(p(x))=p\left(\frac{1}{r}\right)$. – Anurag A Dec 6 '18 at 22:07
• 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. – André 3000 Dec 6 '18 at 23:31

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.

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.

• Why is the map clearly a surjection? – Wesley Dec 6 '18 at 22:18
• 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? – Wesley Dec 6 '18 at 22:26
• Yes that is one way to see it. – Antonios-Alexandros Robotis Dec 6 '18 at 22:50
• 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. – Wesley Dec 6 '18 at 23:14
• @Wesley One simple way is to use the nonmonic division algorithm - see my answer. – Bill Dubuque Dec 7 '18 at 0:02