Explicit representation of $R[\frac{x}{y}]$ where $x, y\in R$ for non-Euclidean PIDs $R$?

Question

It's a fact proven by Pendleton, Gilmer, and Ohm (as an obvious corollary of their work, anyways) that PIDs are QR-domains, meaning every overring (ring between the domain and the quotient field) is a ring of quotients. I'm trying to find a counterexample for something else not being a QR-domain, and so understanding how to get an explicit construction would probably help me more. It's fairly easy to find explicit representations in Euclidean domains if I have something like $R[\frac{x}{y}]$ where $x,y\in R$ and show I indeed get $y$ is a unit in $R[\frac{x}{y}]$ by getting an explicit representation of $\frac{1}{y}$ in $R[\frac{x}{y}]$. My question is, how can I do this kind of thing for a PID that is not a Euclidean domain? I know I theoretically can, and I think it's going to be of the form $R[\frac{1}{y}]$, but how can I get an explicit proof that $y$ indeed becomes a unit (or if y doesn't, then some other nonunit in $R$ does) when passing to the overring, via getting an explicit representation of $\frac{1}{y}$? If you have a concrete example, that would be amazing too.

Thanks!

Answer 1

By PID, we have $(x,y)=(z)$, so $z=ax+by$ and $x=cz$, $y=dz$ for some $a,b,c,d$, then

$$\frac{1}{d} = \frac{z}{y} = a \frac{x}{y} + b$$

$$ \frac{x}{y} =c \frac{1}{d}$$

so $R[\frac{x}{y}]=R[\frac{1}{d}]$. It's not $R[\frac{1}{y}]$ in general because some prime ideal could divide both $x$ and $y$ in equal quantities.