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Jul 13, 2019 at 16:06 comment added David Lampert 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 $k[[x]]$.
Jul 13, 2019 at 6:36 comment added Laurent Moret-Bailly 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 $k$.
Jul 12, 2019 at 19:57 history edited David E Speyer CC BY-SA 4.0
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Jul 12, 2019 at 19:36 history answered David E Speyer CC BY-SA 4.0