Localizing the ring of integers of a number field to get a PID Let $K$ be a number field with ring of integers $R$. Let $I_1,\ldots, I_h$ be (integral) ideals representing generators for the class group of $K$, and let $S$ be the set of prime ideals (a.k.a. finite places) of $R$ dividing $I_1I_2\cdots I_h$. The set of $S$-integers is
$$R_S = \{ a \in K \mid v(a)\geq 0 \ \forall v \notin S\}.$$
Why is $R_S$ a principal ideal domain?
This is claimed in a book I’m reading and I’m sure it is an easy or standard result so references would be fine.
 A: First the geometric perspective. Let $X=\newcommand\Spec{\operatorname{Spec}}\Spec R$, $X_S=\Spec R_S$. Noting that $X_S$ is an open subset of $X$, we have a surjective map $\newcommand\Pic{\operatorname{Pic}}\Pic X\to \Pic X_S$ in the usual way. However $X_S$ is just the open subscheme of $X$ where we deleted the union of the supports of the generators of $\Pic X$, so trivially this map is the 0 map. Hence $\Pic X_S$ is 0. Thus $R_S$ is a PID, since it has trivial ideal class group.
On the other hand, if we go the classical ANT route, we have the following. (This is the same as above, but phrased in more classical language)
Since $R$ and $R_S$ have the same fraction field, any fractional ideal of $R_S$ is a fractional ideal of $R$. Moreover for any fractional ideal $\Lambda$ of $R$, $\Lambda R_S$ is a fractional ideal of $R_S$. By the first observation, this map $\Lambda\mapsto \Lambda R_S$ is surjective. Note that it is clear that this is a group homomorphism, since $\Lambda_1\Lambda_2R_S=(\Lambda_1R_S)(\Lambda_2R_S)$. Hence we have a surjective map from the group of fractional ideals of $R$ to the group of fractional ideals of $R_S$. Next we note that this map sends principal fractional ideals of $R$ to principal fractional ideals of $R_S$, since if $f\in \operatorname{Frac} R$, then $fR\mapsto fR_S$. Hence this map induces a surjective group homomorphism from the ideal class group of $R$ to the ideal class group of $R_S$.
Now we just consider what happens to $I_j$ for any of the ideals in your question.
First note that since $I_j$ was an ideal of $R$, $I_jR_S$ is therefore an ideal of  $R_S$ (as in not a fractional ideal). We'll show that $I_jR_S$ is not contained in any prime ideals in $R_S$, and therefore $I_jR_S$ is the unit ideal, and hence trivial in the ideal class group of $R_S$. By surjectivity of the homomorphism, this will allow us to conclude that the ideal class group of $R_S$ is in fact trivial, and that $R_S$ is therefore a PID.
Now to see that $I_jR_S$ is the unit ideal, we just observe that it isn't contained in any prime of $R_S$. This follows since the primes of $R_S$ are of the form $\newcommand\pp{\mathfrak{p}}\pp R_S$ where $\pp\not\in S$, and have the property that $\pp R_S\cap R=\pp$, so if $I_jR_S \subseteq \pp R_S$, then $I_j\subseteq I_j R_S \cap R \subseteq \pp R_S \cap R = \pp$, which is a contradiction. Hence $I_jR_S$ is the unit ideal. Thus we are done.
A: Let $R$ be a dedekind domain with finite class number and $\mathfrak{p}$ a prime ideal. Let $k$ be the order of $\mathfrak{p}$ in the class group, $\mathfrak{p}^k = (\alpha)$ and let $S = R[\alpha^{-1}]$. Then $S$ is a Dedekind domain. $\alpha \in \mathfrak{p} S$ so that $\mathfrak{p} S= S$. Whence
$$\mathcal{I}_S = \mathcal{I}_R / \langle \mathfrak{p} \rangle,\qquad \mathcal{P}_S = \mathcal{P}_R / (\mathcal{P}_R \cap \langle \mathfrak{p} \rangle) = \mathcal{P}_R/ \langle (\alpha) \rangle,\qquad \mathcal{C}_S=\mathcal{I}_S/\mathcal{P}_S=\frac{\mathcal{I}_R / \langle \mathfrak{p} \rangle}{\mathcal{P}_R / (\mathcal{P}_R \cap \langle \mathfrak{p} \rangle)}=\mathcal{C}_R / \langle \mathfrak{p}\rangle$$
Doing so with all the generators of $\mathcal{C}_R$ yields a Dedekind domain with trivial class group, a PID.
