Let $(k,v)$ be a henselian field, with $\mathcal{O}$ and $\bar{k}$ being respectively its valuation ring and its residue field. If $K/k$ a finite separable field extension (on which $v$ thus extends uniquely) is it true that the finite field extension $\bar{K}/\bar{k}$ is also separable?
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The answer to your question is no. Start with the field called $\mathscr{E}$ in $p$-adic Hodge theory. It is the set of Laurent series $\sum_r a_i t^r$ such that $a_r\in \mathbf{Q}_p$ satisfy $|a_r|\to 0$. This is a complete DVR with respect to the valuation $v(\sum a_r t^r) = \inf_r v_p(a_r)$. Note that the residue field of $\mathscr{E}$ is $\mathbf{F}_p(\!(t)\!)$.
The extension $\mathscr{E}\to \mathscr{E}$ induced by $t\mapsto t^p$ is finite separable, but the induced extension of residue fields is $t\mapsto t^p\colon \mathbf{F}_p(\!(t)\!)\to \mathbf{F}_p(\!(t)\!)$, which is not separable.
answered Jul 13, 2015 at 17:08
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2$\begingroup$ For any henselian dvr $\mathcal{O}$, any finite extension $\overline{k}'/\overline{k}$ is the residue field of a finite separable $k'/k$. Indeed, using a tower reduces us to when $p:={\rm{char}}(\overline{k})>0$ and $\overline{k}'=\overline{k}(a_0^{1/p})=\overline{k}[t]/(t^p-a_0)$ with $a_0\in\overline{k}-\overline{k}^p$. Then for $a\in\mathcal{O}^{\times}$ lifting $a_0$ and a uniformizer $\pi\in\mathcal{O}$, $f:=T^p-\pi T-a\in k[T]$ is separable irreducible with $k':=k[T]/(f)$ having residue field $\overline{k}'/\overline{k}$. $\endgroup$– grghxyCommented Jul 13, 2015 at 20:19