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Jun 8, 2018 at 9:14 comment added Will Sawin @zzy You are right, I have never once in my life remembered the name "Krasner's lemma".
Jun 7, 2018 at 23:29 comment added Zhiyu Thank you, what you say looks like Krasner's lemma. Before I found something related in the tilting approach to the proof of Fontaine-Winterberger theorem while he didn't give a proof and I come across this problem while searching on mathoverflow.
Jun 6, 2018 at 21:43 comment added Will Sawin @zzy I think always yes. The main thing to check is that every algebraic extension of $K(\hat{A})$ is defined over $K(A)$. To do that, you want to check that if you perturb the coefficients of the defining equation slightly, you get the same Galois extension. This is some famous theorem.
Jun 6, 2018 at 9:35 comment added Zhiyu Do $K(A)$ and $K(\hat A)$ have isomorphic absolute Galois groups?
May 21, 2012 at 3:57 history edited Will Sawin CC BY-SA 3.0
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May 20, 2012 at 6:40 history edited Will Sawin CC BY-SA 3.0
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May 20, 2012 at 6:23 history edited Will Sawin CC BY-SA 3.0
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May 20, 2012 at 5:57 history answered Will Sawin CC BY-SA 3.0