Let $A$ be a complete $DVR$ with fraction field $K$. Let $L/K$ be a finite separable extension of $K$ and let $B$ be the integral closure of $A$ in $L$. Then $B$ is also a complete $DVR$ and let us denote by $\mathfrak{p} \subset A$ the maximal ideal of $A$ and $\mathfrak{q} \subset B$ the maximal ideal of $B$. We also set $k=A/\mathfrak{p},\,l=B/\mathfrak{q}$. Pick $\alpha \in B$ such that $L=K(\alpha)$ and assume that $\alpha$ is a root of some nonzero $f \in A[t]$ such that its class $\overline{f} \in k[t]$ is separable. Does it follow that the extension $B/A$ is unramified (and $B=A[\alpha]$)?
I think about the following argument: set $n=[L:K]$ and let $g \in A[t]$ be the minimal polynomial of $\alpha$ (monic of degree $n$). Then $g|f$ so $\overline{g}|\overline{f}$ and $\overline{g} \in k[t]$ must be separable (since $\overline{f}$ is). Note also that $\overline{g}$ is irreducible (otherwise $g$ is not irreducible by Hensel lemma, here it is important that $\overline{g}$ is separable so we can indeed apply Hensel lemma). Let $\overline{\alpha} \in l$ be the class of $\alpha$. Consider the extension $l/k$. Note that $\overline{\alpha}$ is a root of the separable irreducible monic polynomial $\overline{g} \in k[t]$ of degree $n$. It follows that $[l:k] \geqslant n=[L:K]$ but on the other hand we always have $[l:k] \leqslant [L:K]$ so we conclude that $[l:k]=[L:K]=n$ so $l=k(\overline{\alpha})$, hence, $l/k$ is separable (use that $\overline{g}$ is separable) of degree $[L:K]$ so the extension $B/A$ is unramified. Then it is easy to conclude that $B=A[\alpha]$ using that $A,\,B$ are $DVR$‘s.