Xjyuk

Denote by $k$ an algebraically closed field. Can one produce a DVR $A$ over $k$ such that

• the fraction field of $A$ has an automomorphism not preserving $A$
  


• no non-trivial field extension of $k$ maps, as a $k$-algebra, to $A$?

This question is in part inspired by this post (though I guess the connection is not entirely clear, I will try to clarify if this gets responses).

If $k((x))$ has an automorphism not preserving $k[[x]]$, that would mean a positive answer to this question. I do not know if such automorphism exists.

If you do not insist on completeness, an easier example seems to be $A=k[x]_(x)subseteq k(x)$. Then the substitution map $f(x)/g(x) mapsto f(x^-1)/g(x^-1)$ defines an automophism of $k(x)$ sending $A$ to its evil twin $k[x^-1]_(x^-1),$ hence not preserving $A$, and no proper overfield of $k$ can map into $A$ since the residue field of $A$ is $k$.

Every automorphism of $k((x))$ preserves $k[[x]]$. This argument is adapted from an answer of Will Sawin. Let $V$ be the set of valuations $v : k((x))^times to mathbbZ$ which are $0$ on $k^times$. As usual, we put $v(0) = infty$ for any valuation $v$. I claim that $f in k[[x]]$ if and only if $v(f) geq 0$ for all $v in V$.

Clearly, if $f notin k[[x]]$, then $v(f)<0$ for the standard valuation $v$.

In the other direction, let $v in V$. Choose $n$ relatively prime to the characteristic of $k$. Let $f$ be of the form $1+sum_geq 1 a_j x^j$, then $f$ has an $n^j$-th root in $k((x))$ for all $j>0$. So $n^j | v(f)$ and we deduce that $v(f)=0$ for such an $f$. Any $g in k[[x]]$ is the sum of such an $f$ and an element of $k$, so any such $g$ has $v(g) geq 0$.