Is the ring of integers of a DVR also a DVR?

Question

$R$ is a DVR and $L$ is a finite separable extension of $Q(R)$. Can we say that the ring of integers is a DVR?

I know it's a Dedekind domain. So showing it's local is enough.

If it's not a DVR, what is a counterexample?

Answer 1

Here's one way to construct a counter example. Consider the integral extension $\mathbb C[X]\subset \mathbb C[X,Y]/\langle Y^2-X\rangle $ given by $X\mapsto \bar X$. This corresponds to the finite morphism of affine varieties $$ V=V(Y^2-X)\subset \mathbb A^2_C\rightarrow\mathbb A^1_\mathbb C$$ $$(z_1,z_2) \mapsto z_1$$ Now $1$ is a smooth point in $\mathbb A^1_\mathbb C$. So the local ring at $1$, $R$ is a DVR. Also $C[X]\subset \Gamma(V)$ is integral so $\Gamma(V)_m:=\left (\mathbb C[X]\backslash m \right )^{-1}\Gamma(V)$ is integral over $R=\mathbb C[X]_m$ where $m=(X-1)\mathbb C[X]$. So the ring of integers of $\mathbb C[X]$ in $\operatorname{Frac}(\Gamma(V))$ is an integral extension of $\Gamma(V)_m$. But $\Gamma(V)_m$ already contains two distinct maximal ideals namely $\langle X-1,Y\pm 1\rangle$. So the integral closure cannot be local either and hence ain't a DVR.