Let $L/K$ be a finite separable field extension with $O_v$ valuation ring of $K$. Here I will not assume $O_v$ complete and similarly for $K$ as well.(i.e. There is a completion $\hat{O}_v$ of $O_v$ and $\hat{K}=Frac(\hat{O}_v)$ both of which are complete.)
Now consider integral closure of $O_v$ in $L$ which I will denote as ring $S$. Now since $O_v$ is valuation ring, it is clear that $S$ is Dedekind for sure.
$\textbf{Q:}$ Is $S$ DVR in general? Note that if $O_v$ is complete with $K$ complete, then it follows that $S$ is DVR from theorem of general local complete field extensions.
The reason I ask this question is the following. Consider a complex Riemann surface $X$ and $x\in X$. Denote $M_x$ the meromorphic germs at $x$ and set $z$ as the valuation parameter of $M_x$.(This is in particular DVR but not complete as not all cauchy against valuation converging. The completion is formal power series and formal laurent expansions respectively.) Let $L/M_x$ be any finite field extension. Since $char(M_x)=0$, it follows that the extension is separable. Now by Thm on Puisseux series(Forster Lectures on Riemann Surfaces Chpt 1), one deduces $L=M_x[T]/(T^d-z)$. Note this extension is always totally ramified and valuation ring as well.
$\textbf{Q':}$ It seems from Riemann surface persepctive, this is always DVR? This forces the extension always totally ramified.
$\textbf{Q'':}$ It seems that whether the valuation ring is complete is irrelevant here? I do recall that completeness is required to construct the non-trivial norm by inducting on the degree of extensions(Taylor, Frolich, Algebraic Number Theory Chpt 3) and there might be a way to bypass it. So for general valuation ring $O_v$ of field $K$ and $L/K$ finite separable extensions, I can still consider $L\otimes_O O_v$ which will split into product of "local fields"? The story goes along for complete DVR for sure but does it hold for non-complete DVR? What should I watch out for this procedure?