Let $R$ be a valuation ring containing a field $k$, with residue field $F$ and quotient field $K$. Assume $F/k$ is separable. Is $K/k$ separable?


I have convinced myself that (for a positive answer) it is enough to treat the case of a rank one valuation ("height one" in Bourbaki's terminology), with $F=k$ and $K/k$ finitely generated. (Instaed of the latter condition, we may assume $K$ complete).


The answer is yes for a discrete valuation, because then (if $F=k$) the completion of $R$ is isomorphic to $k[[t]]$.



Valuations and separable extensions
What about this: we have to prove that $K$ and every purely inseparable extension $l/k$ are linearly disjoint over $k$.


Let $x\_1,\ldots ,x\_r\in l$ be $k$-linearly independent elements and assume $0=a\_1x\_1+\ldots +a\_rx\_r$ for some elements $a\_i\in K$. We can divide by the coefficient $a\_j$ with the least value and thus assume $a\_1,\ldots ,a\_r\in R$ with at least one coefficient being a unit of $R$.


There is a unique valuation ring $S$ of $K.l$ dominating $R$. Taking residues modulo the maximal ideal of $S$ yields a linear combination $0=\overline{a\_1}x\_1+\ldots +\overline{a\_r}x\_r$ in the field $F.l$. The separability of $F/k$ yields $\overline{a\_i}=0$ for all $i$, a contradiction.


