Is there a finite extension with a non-trivial class group of any PID? Let $R$ be a PID with infinitely many prime ideals. Does there always exist a finite extension $R\subset R'$ with $R'$ being a Dedekind domain with a non-trivial class group?
 A: Counterexample. Let $S$ be an infinite set of primes (of $\mathbf Z$) of density $0$. Let $R$ be the localisation of $\mathbf Z$ away from $S$, i.e. the elements $\tfrac{a}{b}$ with $p \nmid b$ for all $p \in S$. Then
$$\operatorname{Spec} R = S \cup \{\eta\}$$
where $\eta$ is the generic point.
Let $R \subseteq R'$ be a finite extension of Dedekind domains, let $K = \operatorname{Frac} R'$, and write $\mathcal O_K$ for the (usual) ring of integers in $K$. Then $R'$ is the localisation of $\mathcal O_K$ away from $S$, the natural map $\operatorname{Cl}(\mathcal O_K) \to \operatorname{Cl}(R')$ is surjective, and the primes of $\mathcal O_K$ lying above $S$ have density $0$.
Let $H$ be the Hilbert class field of $K$. The isomorphism $\operatorname{Cl}(\mathcal O_K) \stackrel\sim\to \operatorname{Gal}(H/K)$ takes prime ideal classes $[\mathfrak p]$ to the Frobenius $\operatorname{Fr}_\mathfrak p$ at $\mathfrak p$, and the Chebotarev density theorem implies that every ideal class $[I] \in \operatorname{Cl}(\mathcal O_K)$ can be represented by a positive density set of primes.
In particular, $[I]$ contains a prime $\mathfrak p$ not above $S$, hence maps to $0$ under $\operatorname{Cl}(\mathcal O_K) \to \operatorname{Cl}(R')$ since $\mathfrak pR' = R'$. Since $[I]$ is arbitrary and $\operatorname{Cl}(\mathcal O_K) \to \operatorname{Cl}(R')$ surjective, we conclude that $\operatorname{Cl}(R') = 0$. $\square$
Remark. It might be possible to make a more elementary argument if you choose the $S$ to be sufficiently sparse. For example you could try to take $S = \{p_1,p_2,\ldots,\}$ with $p_{i+1} > 2^{p_i}$ or some other bound. (You have to produce a lot of relations in $\operatorname{Cl}(\mathcal O_K)$ involving relatively small primes. I tried this but couldn't quite work it out by hand.)
