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.

  • $\begingroup$ In order for $R \to V$ to be smooth, you also need the extension of residue fields to be separable. $\endgroup$ Feb 15, 2019 at 15:57
  • $\begingroup$ Edited the question accordingly. In fact, I consider $L$ to be embedded in a completion of $K.$ Are you saying it is true with this added hypothesis? $\endgroup$
    – userm
    Feb 15, 2019 at 16:06
  • $\begingroup$ Dear userm, you're right that for $R \to V$ to be étale, it should be finitely presented. Since $V$ is integral over $R$, this implies that $V$ should be a finitely generated $R$-module. I don't think this always holds (although I can't think of a counterexample in your situation) -- this condition is related to the notion of Japanese rings (see also Nagata ring). Also $V$ should be flat over $R$. If these two technical conditions are satisfied, then $R \to V$ is étale because the geometric fibers are non-singular. $\endgroup$ Feb 15, 2019 at 16:45
  • $\begingroup$ Thank you for your answer! I think the map is flat: R is a Prüfer domain, so a torsion-free module should be flat. As you say it, all of the conditions for it to be étale are satisfied, except for possibly finitely presented. I edited accordingly. $\endgroup$
    – userm
    Feb 15, 2019 at 17:19
  • $\begingroup$ you're right $V$ is automatically flat over $R$. Still there is something I don't get when you say $V$ is the integral closure of $R$: in general, the integral closure will rather be semi-local. Also, maybe it would be good if you put all your assumptions in your question, since otherwise it is difficult to give a definite answer. $\endgroup$ Feb 15, 2019 at 18:45

1 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 $R$.

  • $\begingroup$ Dear François, thanks for these examples anyway! And yes, it would already be helpful to know whether this map is weakly étale. $\endgroup$
    – userm
    Feb 18, 2019 at 11:39

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