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Oct 25, 2017 at 16:48 vote accept João Dias
Oct 23, 2017 at 11:01 history edited Johannes Hahn CC BY-SA 3.0
Fixed typo; $\tau$ isn't defined on the whole algebraic closure.
Oct 23, 2017 at 11:00 comment added Johannes Hahn No, of course not. At the very least because $\tau(1)=1$. But I see what you mean: $\tau$ is not defined on $\overline{\mathbb{F}_p}$ (you can't normalise for extensions of degree divisible by $p$), but on the other field I mentioned. I will correct that.
Oct 23, 2017 at 1:04 comment added João Dias But if the normalized galois trace is $F-$invariant does not mean that is zero?
Oct 23, 2017 at 0:04 history answered Johannes Hahn CC BY-SA 3.0