Is the integral closure of a valuation ring in a finite separable extension of its fraction field étale?

Question

Let $K$ be a field endowed with a rank (height) one valuation with completion $\hat{K}$, which is not discrete. Let $R$ be the valuation ring of $K$.

Let $L \subset \hat{K}$ be a separable finite field extension of $K$. Let $V$ be the valuation ring of $L$.

Is the inclusion $R \rightarrow V$ étale? Or even weakly étale?

What I know: $R$ is a valuation ring (hence a Prüfer domain), and $V$ is a torsion-free module over $R,$ so it is flat. These rings have same residue fields, and the maximal ideal of $R$ generates the maximal ideal of $V.$ In order to be able to conclude, the only thing that is missing is the condition of finitely presented, which I can't show to be true in general.

Answer 1

EDIT I may have overlooked the assumption that $L$ is contained in $\hat{K}$. In general, if $v$ is a valuation of $K$ and $L/K$ is finite separable then $L \otimes_K \hat{K}$ is reduced and thus isomorphic to $\prod_i \hat{L}_i$ where the $\hat{L}_i$ are the completions of $L$ for the various valuations $v_i$ extending $v$. In your situation one of the $\hat{L}_i$ is equal to $\hat{K}$. Since the dimension of $L \otimes_K \hat{K}$ over $\hat{K}$ is equal to $[L:K]$, this means that there must exist some other completions, hence the extension of $v$ is not unique in your setting and the following examples don't apply directly. I'm leaving them however because they might be useful to answer the question. END OF EDIT

The following example, due to Ostrowski and mentioned in The theory of classical valuations by Ribenboim (Sect. 6.3), gives a situation where $V$ is not finitely presented over $R$, and thus not étale over $R$.

Let $K=\mathbb{Q}_2(2^{1/2^\infty})$ be the extension of $\mathbb{Q}_2$ obtained by adjoining the iterate square roots of $2$, endowed with the $2$-adic valuation $v$, with value group $\mathbb{Z}[\frac12]$, so $v$ has rank 1 and is not discrete. Consider the finite separable extension $L=K(\sqrt{-1})$. Then $v$ has a unique extension $v'$ to $L$ with ramification index and residual degree both equal to 1. Let $R$ be the valuation ring of $v$. Since $v$ has a unique extension to $L$, the integral closure $V$ of $R$ in $L$ is the valuation ring of $v'$. We are in a situation where $\sum_i e_i f_i < [L:K]$ using standard notations from ramification theory, and this implies that $V$ is not a finite $R$-module (see Endler, Valuation theory, Theorem 18.6).

Just after Ostrowski's example, Ribenboim gives another interesting example of a complete valued field $(K,v)$ of characteristic $p$ with $v$ non-discrete, and a separable extension $L/K$ of degree $p$, such that $v$ has a unique extension $v'$ in $L$, and the ramification index and the residual degree of $v'/v$ are both equal to 1. This gives another example where $V$ is not étale over $R$.

Since in your situation finite presentation is the only obstruction, one may wonder whether pro-étale extensions are more appropriate, in particular whether $V$ is always pro-étale over $R$.