# A DVR algebra with weird automorphisms

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 $$v : k((x))^{\times} \to \mathbb{Z}$$ 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 $$f \not\in k[[x]]$$, 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 $$v(g) \geq 0$$.
• This argument seems to prove only that every automorphism preserving $k$ preserves $k[[x]]$. Fortunately, $k$ is assumed algebraically closed, so every discrete valuation on $k((x))$ is trivial on $k$. – Laurent Moret-Bailly Jul 13 '19 at 6:36
• For any field $k$ every discrete valuation on $k((x))$ is trivial on $k$, as noted in the comment by YCor after the answer of Will Sawin in the link above, and thus every automorphism of $k((x))$ preserves $k[[x]]$. – David Lampert Jul 13 '19 at 16:06