Is it possible for a finite integral closure of a DVR to not be a PID? Suppose that we have a point $A$(local ring, DVR) of an abstract curve over $k=\bar{k}$ given by a field $k(X)$. Let $k(Y)$ be a finite extension of
$k(X)$ and denote by $B$ the integral closure of $A$ in $K(Y)$.
Is it possible for $B$ to not be a PID? 
 A: I think the following more general statement is true.
Proposition. Let $A$ be a one-dimensional noetherian semi-local domain and let $K$ be its field of fractions. Consider a finite algebraic field extension $K \subseteq L$, and let $B$ be the integral closure of $A$ in $L$. Then, $B$ is a semi-local Dedekind domain, and is therefore a principal ideal domain.
Proof. First, by a consequence of the Krull–Akizuki theorem [Matsumura, Cor. to Thm. 11.7], we see that $B$ is a Dedekind domain with finitely many prime ideals $\mathfrak{n}_1,\mathfrak{n}_2,\ldots,\mathfrak{n}_r$ lying over each maximal ideal $\mathfrak{m}$ in $A$. We note that these $\mathfrak{n}_i$ are maximal ideals in $B$ since $\operatorname{height}\mathfrak{n}_i \ge 1$ (they strictly contain the prime ideal $0 \subseteq B$) and since $\operatorname{height}\mathfrak{n}_i \le 1$ (the ring $B$ has dimension one). Moreover, every maximal ideal of $B$ must lie over a maximal ideal in $A$, since if a prime ideal $\mathfrak{q}$ in $B$ contracts to a non-maximal ideal in $A$, then it must contract to the zero ideal in $A$, and going-up would then imply that $\mathfrak{q}$ is not maximal. Thus, $B$ is a semi-local Dedekind domain.
We now want to show that a semi-local Dedekind domain $B$ is a PID. Let $\mathfrak{n}_1,\mathfrak{n}_2,\ldots,\mathfrak{n}_s$ be the maximal ideals in $B$; since every ideal can be written as a product of powers of the $\mathfrak{n}_i$, it suffices to show that each $\mathfrak{n}_i$ is principal. We fix an $i$ in the argument below.
First, we claim that $\mathfrak{n}_i \smallsetminus \mathfrak{n}_i^2 \ne \emptyset$. If not, then $\mathfrak{n}_iB_{\mathfrak{n}_i} = \mathfrak{n}_i^2B_{\mathfrak{n}_i}$, in which case Nakayama's lemma implies $\mathfrak{n}_iB_{\mathfrak{n}_i} = 0$, contradicting the fact that $B_{\mathfrak{n}_i}$ is a DVR.
Next, we consider the surjection
$$B \longrightarrow \frac{B}{\mathfrak{n}_1\cdots\mathfrak{n}_{i-1}\mathfrak{n}_i^2\mathfrak{n}_{i+1}\cdots\mathfrak{n}_s} \simeq \frac{B}{\mathfrak{n}_1} \times \cdots \frac{B}{\mathfrak{n}_{i-1}} \times \frac{B}{\mathfrak{n}_i^2} \times \frac{B}{\mathfrak{n}_{i+1}}\times\cdots\times\frac{B}{\mathfrak{n}_s},$$
where the isomorphism holds by the Chinese remainder theorem since maximal ideals are coprime. Now choose $x \in \mathfrak{n}_i \smallsetminus \mathfrak{n}_i^2$, which exists by the previous paragraph, and let $b$ be an element in the preimage of $(1,\ldots,1,x,1,\ldots,1)$ in the surjection above. We then see that $b \in \mathfrak{n}_i \smallsetminus \mathfrak{n}_i^2$ but that $b \notin \mathfrak{n}_j$ for all $j \ne i$. Writing $(b)$ as a product of powers of the ideals $\mathfrak{n}_j$ in $B$, we therefore see that $(b) = \mathfrak{n}_i$, hence $\mathfrak{n}_i$ is principal. $\blacksquare$
