Timeline for Are hensel valuation rings N2?

6 events

when toggle format	what		by	license	comment
Mar 16, 2013 at 9:54	vote	accept	name
Mar 15, 2013 at 13:51	history	edited	Torsten Schoeneberg	CC BY-SA 3.0	corrected subtlety
Mar 15, 2013 at 2:05	comment	added	name		Yes, Exercise V.1.19b must be a mistake. Thanks for the stackexchange link. The henselisation of Exercise VI.8.3b works too and is a discrete valuation ring. I'll put it as an answer because there is not enough room here.
Mar 15, 2013 at 1:37	comment	added	Torsten Schoeneberg		Hmm, very strange. I don't see how to conclude in ex. V.1.19b without $A$ Noetherian; if it is true in general, it contradicts my answer, as it is in char. 0. (But so are ex. VI.8.2 and VI.8.4 when $char(k) = 0$, aren't they?) Very confusing. -- Btw, regarding your original question, I have found math.stackexchange.com/questions/167993
Mar 14, 2013 at 21:36	comment	added	name		Did you see Exercise V.1.19b in Bourbaki's Algèbre Commutative?: If $A$ is integrally closed with fraction field $K$ and $\{a \in K : a^p \in A\}$ is a finite $A$-module ($p$ is the exponential characteristic of $K$), then for any finite extension $E$ of $K$, the integral closure of $A$ in $L$ is a finite $A$-module (reduce to the case $E/K$ is normal). Your example is characteristic zero right? In the Bourbaki exercise there doesn't seem to be any assumption that $A$ is Noetherian, so there is a contradiction somewhere? (I admit, I haven't done the exercise).
Mar 14, 2013 at 14:54	history	answered	Torsten Schoeneberg	CC BY-SA 3.0