Showing isomorphism between fractions and a quotient ring

[Mr Davis 97]

Homework Statement

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 $$R\left[\frac{1}{r}\right]\cong \frac{R[x]}{(rx -1)}.$$

Homework Equations

The Attempt at a Solution

Let $\phi : R[x] \to R\left[ \frac{1}{r} \right]$ such that $\phi(p(x)) = p(\frac{1}{r})$. This is a homomorphism since it is the evaluation homomorphism. Let $a\in R$ and $k\in \mathbb{N}$, then $\phi(ax^k) = \frac{a}{r^k}$, and the latter is a general element of $R\left[ \frac{1}{r} \right]$. Now, we will show that $\ker (\phi) = (rx-1)$. The "$\supseteq$" direction is clear, since if $a\in (rx-1)$, then for some $g(x)\in R[x]$, $a=g(x)(rx-1)$. Then $$\phi(g(x)(rx-1)) = \phi(g(x))\phi(rx-1) = \phi(g(x))(r\cdot \frac{1}{r} - 1) = \phi(g(x))\cdot 0 = 0.$$

The "$\subseteq$" direction is where I get confused. If the underlying ring were a field I could use the division algorithm, but since it is not I am not sure what to do.

 

[fresh_42]

Homework Statement

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 $$R\left[\frac{1}{r}\right]\cong \frac{R[x]}{(rx -1)}.$$

Homework Equations

The Attempt at a Solution

Let $\phi : R[x] \to R\left[ \frac{1}{r} \right]$ such that $\phi(p(x)) = p(\frac{1}{r})$. This is a homomorphism since it is the evaluation homomorphism. Let $a\in R$ and $k\in \mathbb{N}$, then $\phi(ax^k) = \frac{a}{r^k}$, and the latter is a general element of $R\left[ \frac{1}{r} \right]$. Now, we will show that $\ker (\phi) = (rx-1)$. The "$\supseteq$" direction is clear, since if $a\in (rx-1)$, then for some $g(x)\in R[x]$, $a=g(x)(rx-1)$. Then $$\phi(g(x)(rx-1)) = \phi(g(x))\phi(rx-1) = \phi(g(x))(r\cdot \frac{1}{r} - 1) = \phi(g(x))\cdot 0 = 0.$$

The "$\subseteq$" direction is where I get confused. If the underlying ring were a field I could use the division algorithm, but since it is not I am not sure what to do.

I haven't done it (a bit late here), but I assume it is something like:

Let $p(x)=a_n+a_{n-1}x+\ldots +a_1x^{n-1}+a_0x^n \in \operatorname{ker} \phi\,.$ Then $a_nr^n+a_{n-1}r^{n-1}+\ldots +a_1r+a_0 = 0$. So $a_0=r\cdot b_0$ and $0=r\cdot \left(a_nr^{n-1}+\ldots +(a_1+b_0) \right)$ and because $r$ isn't a zerodivisor $0=a_nr^{n-1}+a_{n-1}r^{n-2}+\ldots +a_2r +b_1\,.$
Now it looks as if a minimal $n=\operatorname{deg}(p)$ argument or an "and so on" could apply so we get either $n=1$ or $n=0$.