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Jun 6, 2021 at 11:39 comment added D.-C. Cisinski @YuanYang The formula will hold for all schemes if you consider the discrete topology on $\mathbb{Z}_l$. But then the $H^1$ will not be the good one: it will be the one with coefficients in the discrete ring of $l$-adic integers and that is far from classical l-adic cohomology.
Jun 6, 2021 at 9:02 comment added Yuan Yang @Denis-CharlesCisinski Dear Denis, what about the constant sheaf with $\mathbb{Z}_l$ coefficients? Does this formula ($H^1_{et}(X,\underline{Z_l})=Hom_{cont}(\pi_1,Z_l)$)still holds for, let’s say, a normal variety?
Nov 22, 2015 at 18:41 comment added Heer $H^1_{et}(X,Z)=0$ for X a geometrically unibranch (in particular normal implies geometrically unibranch) and irreducible scheme and Z torsion-free group, see SGA7 Exp. VIII. Prop. 5.1
Mar 22, 2013 at 5:26 comment added Benjamin Antieau @Emerton: thanks! Definitely looks useful.
Mar 21, 2013 at 12:17 comment added Emerton Dear Benjamin, If you are still interested in this kind of thing, this discussion in comments at the Secret Blogging Seminar is highly relevant: sbseminar.wordpress.com/2009/04/20/… Regards,
Sep 30, 2012 at 2:45 answer added Thomas Geisser timeline score: 13
Dec 29, 2011 at 22:33 vote accept Benjamin Antieau
Dec 29, 2011 at 22:33 comment added Benjamin Antieau Thanks for the clarification about the etale fundamental group.
Dec 28, 2011 at 10:48 comment added S. Carnahan As Denis-Charles Cisinski mentions, you have to be careful with your definition of $\pi_1^{et}$ - the version defined in SGA1 only uses finite etale covers, while the "gros" version in SGA3 uses all etale covers. One has examples of gros etale fundamental groups that are infinite and discrete, coming from maximally degenerate curves of positive arithmetic genus.
Dec 27, 2011 at 23:49 comment added D.-C. Cisinski We always have the formula $H^1(X,\mathbb{Z})=Hom_{cont}(\pi^{et}_1(X),\mathbb{Z}$ (where the progroup $\pi^{et}_1(X)$ is defined as the fundamental group of the small etale topos of $X$, say). But $\pi_1(X)$ is known to be profinite only for $X$ noetherian and normal, so that this does not contradicts the counter example of Vistoli below.
Dec 27, 2011 at 22:23 answer added Angelo timeline score: 26
Dec 27, 2011 at 21:24 history asked Benjamin Antieau CC BY-SA 3.0