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Aug 23 at 9:57 comment added Adrien MORIN This is exactly what I am claiming, yes.
Feb 20 at 6:56 comment added Nobody Are you claiming that [1, II.2.13] still holds without the condition "$mF=0$ for some $m$ that is invertible on $U$"?
Sep 8, 2022 at 11:00 comment added Adrien MORIN I agree with you that H^2 must be finite, but then for H^i with i>=3 I am not sure the isomorphism with cohomology at the real places still holds, so I'm not sure the quantity you're studying still deserves the name of Euler characteristic (we're not sure that cohomolgy is periodic in degree >=3 anymore so it doesn't make sense to truncate in degree <=2 imo)
Sep 6, 2022 at 15:50 comment added Nobody @AdrienMORIN The claim that "finiteness also holds for cases with the order of $M$ not an $S$-unit" was written in the paragraph after Prop. 1.2.2, page 5, but it does not appear to be proven in the text. I think the point is that as a subgroup of the etale cohomology $H^2_{et}(\text{Spec} \mathcal{O}_S, M)$ which is finite by Prop. 4.1.2, the group $H^2(G_{K,S},M)$ is always finite.
Sep 6, 2022 at 8:53 comment added Adrien MORIN I did not find your claim in the notes. Could you cite the precise proposition ? Note that proposition 1.2.2 insists that #M is a S-unit. The edge map from Galois cohomology to étale cohomology is also shown to be an isomorphism only when the sheaf is locally constant constructible with stalk of order a S-unit.
Sep 5, 2022 at 10:30 comment added Nobody The galois cohomology groups $H^i(G_{K,S},M)$ are always finite if $S$ and $M$ are finite, see math.stanford.edu/~conrad/BSDseminar/Notes/L4.pdf.
Sep 2, 2022 at 17:39 history edited Adrien MORIN CC BY-SA 4.0
Added a few paragraphs to explain further the link with the original question
Sep 2, 2022 at 16:44 comment added Adrien MORIN We can say that the cohomology groups $H^i(U,F)$ are finite though because the $H^i_c(U,F)$'s are and Galois cohomology of $\mathbb{R}$/ a p-adic field with finite coefficients is finite; this gives a formula for the alternating product of the $[H^i(U,F)]$ for $i=0$ to $3$. I will write this down more explicitely.
Sep 2, 2022 at 16:35 comment added Adrien MORIN This was mostly an attempt to answer the third question. I would not say that the G_K,S cohomology is the wrong thing to study but it certainly seems messier, and actually I don't know if the G_K,S cohomology is finite without the invertibility condition ! My main reference is Milne's book and he always assumes the order is invertible.
Sep 2, 2022 at 14:11 comment added David Loeffler (Or are you claiming that the $G_{K, S}$-cohomology is simply the wrong thing to study when $S$ doesn't contain all primes dividing $\#M$, and we should abandon it and work with these other theories instead?)
Sep 2, 2022 at 13:57 comment added David Loeffler Could you perhaps go into a little more detail about how this helps to answer the original question? When the order of M is not invertible in $O_{K, S}$, is the relation between Galois and (Weil-)etale cohomology strong enough that we can use it to compute the Euler characteristic of $H^*(G_{K, S}, M)$?
Sep 2, 2022 at 10:26 history answered Adrien MORIN CC BY-SA 4.0