Localization of a PID is a PID I would like a verification of a proof for the following statement. Let $S$ be a multiplicatively closed subset of a ring $R$. If $R$ is a PID, then $S^{-1}R$ is a PID.
Let $I = \left<r_1/s_1, r_2/s_2\ldots\right>$ be an ideal of $S^{-1}R$. Since $1/s_i$ is a unit in $S^{-1}R$, we have $\left<r_i/s_i\right> = \left<r_i\right>$. But then $I = \left<r_1/1,r_2/1,\ldots\right>$ and so we may think of $I$ as an ideal in $R$. Since $R$ is a PID, we have $I = \left<x\right>$ for some $x \in R$. So $I = \left<x/1\right>$, thus $S^{-1}R$ is a PID.
Thanks!
 A: Let me give a proof to 

If $A$ is a PID, and $\mathfrak{p}$ a prime ideal in $A$, then the localization $A_{\mathfrak{p}}$ is also a PID. 

Proof. Suppose $\mathfrak{a} \subset A_{\mathfrak{p}}$ be an ideal. We need to show that $\mathfrak{a}$ is principal, i.e. $\mathfrak{a}= (f)$ for some $f\in A_{\mathfrak{p}} $. Recall (see Prop. 3.11 in Atiyah-Macdonald) that every ideal in the localization is an extended ideal. Hence there exists an ideal $\mathfrak{b} =(g)\subset A$ such that $(g)^e = \mathfrak{a}$. Now given the cannonical map $\varphi:A\to A_{\mathfrak{p}}, \ x \mapsto x/1$, 
$$
\mathfrak{a} = \mathfrak{b}^e :=\langle \varphi(g) \rangle = (g/1).  
$$ 
Thus $A_{\mathfrak{p}}$ is a PID.
This is just a remark that above result gives a nice way of describing the Zariski-open sets in the scheme $\mathrm{Spec}A_{\mathfrak{p}}$ if $A$ is a PID and $\mathfrak{p} \subset A$ is a prime ideal. For a PID $A$, one shows that open sets are same as basic open sets $D(f)$. Indeed, if $U\subset \mathrm{Spec}A$ is open, then $U=V(\mathfrak{a})^c=V(f)^c=:D(f)$ for some $f\in A$ that generates the ideal $\mathfrak{a}.$ Because of the above result in the grey box, we know $A_{\mathfrak{p}}$ is a PID, and hence every open set in $\mathrm{Spec}A_{\mathfrak{p}}$ is of the form $D(g)$ for some $g\in A_{\mathfrak{p}}.$ 
A: the idea behind your method is sound, but it is a little confusing, perhaps, to argue merely that we may think of I as an ideal in R. 
the difficulty that this phraseology avoids is that we cannot a priori rule out the presence of a non-finitely generated ideal in $S^{-1}R$. to rule this out we may seek to use the result that the localization of a Dedekind domain at a multiplicative set is again a Dedekind domain. thus any ideal of the localized domain $S^{-1}R$ is a product of prime ideals. this finiteness condition allows us to slide the ideal back up into $R$ through multiplication by a suitably chosen element of $S$.
A: Since every ideal of a localization is an extended ideal, so the result is true.
