Let $u$ and $v$ be relatively prime integers, and let $R'$ be the ring obtained from $\mathbb{Z}$ by adjoining an element $\alpha$ with the relation $v\alpha=u$. Prove that $R'$ is isomorphic to $\mathbb{Z}[\frac{u}{v}]$ and also to $\mathbb{Z}[\frac{1}{v}]$.
This is from Artin's chapter on Factoring.
Edit: My professor changed the problem slightly to assume that $R'=\mathbb{Z}[\frac{u}{v}]$, so that we only need to show that it is isomorphic to $\mathbb{Z}[\frac{1}{v}]$. Supposed to make the problem more do-able, but still seems daunting...
So I know by definition that $$\mathbb{Z}[\frac{u}{v}]=\{a_0+a_1(\frac{u}{v})+a_2(\frac{u}{v})^2+\cdots+a_n(\frac{u}{v})^n|a_i\in\mathbb{Z}\}$$
So since $u,v$ are relatively prime, I know there exist integers $r,s$ so that $ur+vs=1$ (not sure if this is helpful information). I also know then that $\frac{u}{v}\notin\mathbb{Z}$ so we must have $\frac{u}{v}\in\mathbb{Q}$.
Also, if you consider a mapping $\phi:\mathbb{Z}[x]\to\mathbb{Z}[\frac{u}{v}]$ which sends $x\to\frac{u}{v}$ note that $vx-u\in ker(\phi)$. Somehow I am supposed to deduce that $R'=\frac{\mathbb{Z}[x]}{(vx-u)}$ (why?). Not sure what to do with this information.
She also said to look at $$\frac{1}{p}=a_0+a_1(\frac{u}{v})+a_2(\frac{u}{v})^2+\cdots+a_n(\frac{u}{v})^n$$ (Why do we set this equal to $\frac{1}{p}$?) And see that $$\frac{v^n}{p}=a_0v^n+a_1uv^{n-1}+a_2u^2v^{n-2}+\cdots+a_nu^n\in\mathbb{Z}$$ is true if $\mathbb{Z}[\frac{u}{v}]$ is a field, but it isn't (why?) but supposedly it is an integral domain...
If you can, I would really appreciate it if you could clarify any of my above questions. And I'm not seeing how this is getting me closer to showing there is an isomorphism from $R'\to\mathbb{Z}[\frac{1}{v}]$...
Not only are $\Bbb Z[\frac{u}{v}]$ and $\Bbb Z[\frac{1}{v}]$ isomorphic, they are equal subrings of $\Bbb Q$ (a stronger condition).
If $\alpha,\beta$ are elements of some common ring extension of $R$ then $\,R[\alpha]\subseteq R[\beta]~\Leftrightarrow~\alpha\in R[\beta]$.
Consider $\frac{u}{v},\frac{1}{v}\in\Bbb Q$ where $R=\Bbb Z$: $\phantom{}$ as $\frac{u}{v}\in\Bbb Z[\frac{1}{v}]$ is obvious, you must show $\frac{\color{Blue}{1}}{v}\in\Bbb Z[\frac{\color{Green}{u}}{v}]$.
We want to write $\color{Blue}1$ in terms of $\color{Green}{u}$ (mod $v$). Bezout says $\frac{1}{v}=\frac{mu+nv}{v}=m\frac{u}{v}+n$ for some $m,n$.