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!
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 $S$.