# Is every finite separable extension of a strictly henselian DVR totally ramified?

Let $R$ be a discrete valuation ring with field of fraction $K$ and residue field $k$ and let $K'$ be a finite and separable extension of $K$. If $R$ is henselian ("Hensel's lemma holds", e.g. if it is complete) and $k$ is algebraically closed then the extension $K'|K$ is totally ramified (from Hensel's lemma if follows that there's just one prime lying above the prime of $R$ and since $k$ is algebraically closed there's no residual extension).

What happens if we assume only that $R$ is strictly henselian (i.e. that $k$ is only separably closed)?

Equivalently: does it exist a finite separable extension $K'$ of $K$ such that the residual extension is a nontrivial inseparable extension? I believe that such an extension should exist. Moreover, can we find a Galois extension with this property? Can we do this both in characteristics $(0,p)$ and $(p,p)$?

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Let $k$ a separably closed imperfect field of char. $p$. Fix $t \in k$ for which $X^p - t \in k[X]$ has no root in $k$. Set $K = k((s))$, and form the extension $L = K(y) \supset K$ obtained by adjoining to $K$ a root $y$ of the poly. $f(Y) = Y^p + sY - t \in K[Y]$.

Write $A = k[[s]]$ for the integers of $K$. Since the image of $f(Y)$ in $k[Y] = (A/sA)[Y]$ is irreducible, $f(Y)$ is irreducible in $A[Y]$ and hence in $K[Y]$. Since $f'(Y) = s \ne 0$, it follows that $L/K$ is separable.

Now write $B$ for the integral closure of $A$ in $L$; thus $y \in B$. Denoting by $\pi$ a uniformizer for the DVR $B$, note that $s \in \pi B$. Thus the image of $y$ in $\ell = B/\pi B$ is a $p$-th root of $t$, so $\ell$ is a proper (purely) inseparable extension of $k$.

EDIT: In hindsight, it seems that with minor tweaking the same argument gives an answer in the "mixed characteristic" case.

Indeed, let $t \in k$ as before, and now let $A$ a complete (or Henselian) DVR with residue field $k$ and with fractions $K$ of char. 0. Now choose an element $\tau \in A$ whose image in $k$ is $t$. Let $\sigma \in A$ be a uniformizer, and set $f(Y) = Y^p + \sigma Y - \tau \in A[Y]$. Then as before the extension $L = K(y)$ of $K$ is separable, where $y$ is a root of $f(Y)$. And the residue extension is again inseparable.

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Thanks, that's nice. Do you see any example in mixed characteristic? – user1728 Jan 28 '11 at 9:06
Thanks for the nudge (see Edit:). – George McNinch Jan 28 '11 at 12:15
Thank you again! – user1728 Jan 28 '11 at 12:28
Oh, I just realized that moreover if $K$ has characteristic 0 then you can take $\sigma=0$ in your argument, and if $K$ contains the $p$-th roots of unity then the extension $L|K$ is also Galois, wow. (But I have no idea whether this can be easily achieved also in the same char case) – user1728 Jan 28 '11 at 12:32