Let $(K,v)$ be valued field (not necessarily discrete, moreover the value group needs not be a subgroup of $\mathbb R$) and $char(K)=char(Kv)$ where $Kv$ is the residue field. Let $(F,w)$ be the maximal unramified extension.

  1. Is it true that any automorphism of the separable closure $K^{sep}$ of $K$ over $K$ fixes $F$ setwise?
  2. Suppose that $K_0$, and $F_0$ are liftings of the corresponding residue fields. Is it true that $F=KF_0$ and that any automorphism of the separable closure $K^{sep}$ as above fixes $F_0$?
  • $\begingroup$ Thanks for the comment. I meant $F_0$. It is not just any set of representatives though. It is a lifting. So it’s a subfield of the corresponding valuation ring inside $F$, as well as being a set of representatives. $\endgroup$ – ugur efem Dec 14 '17 at 18:11
  • $\begingroup$ Yes of course. They must be. I’ll add the assumption. Thanks. $\endgroup$ – ugur efem Dec 14 '17 at 18:29

I have an answer now. I more or less had this argument in mind but wasn't sure, but now I am (with some help from a friend).

Without further ado, here is the outline of the argument.

The first question is basically saying the maximal unramified extension is a Galois extension. This is (as brought to my attention) is well known.

For the second part, let $L/K_0$ be a finite subextension of $F_0$. Then $KL$ has the residue field $L$, and it is an unramified extension. Also $L$ is relatively algebraically closed in $KL$. Then any automorphism of $KL$ over $K$ must fix $L$. Since $F_0$ is algebraic over $L_0$, the result follows by taking unions over all such finite extensions.


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