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Sep 1, 2015 at 11:08 comment added Oli Gregory Thinking about it, I guess that a generic quintic threefold satisfies the hypotheses and provides a counter-example. Thanks to Daniel Litt for pointing this out; I definitely should've thought through that case.
Jul 28, 2015 at 7:53 comment added Oli Gregory @WillSawin Yes, I had come to the same conclusion but I was rather hoping that somebody here would tell me that the hypotheses excludes this sort of thing. For example, in the K3 surface case I mentioned, this all works out because $k$ is perfect. Unfortunately I tend to agree with you that this seems false, which is rather a nuisance for the construction I had in mind...
Jul 28, 2015 at 4:10 comment added Will Sawin This seems false, because the Picard group might be $\mathbb Z$, in which case your formula shows that the cohomology group is $\mathbb Q_p/\mathbb Z_p$.
Jul 27, 2015 at 9:25 history edited Oli Gregory CC BY-SA 3.0
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Jul 27, 2015 at 9:16 history asked Oli Gregory CC BY-SA 3.0