Before me, the following was asked: etale fundamental group and etale cohomology of curves

However, that question dealt only with projective curves.


Let $X$ be any scheme (or if you prefer something more concrete, a variety over some field), and let $l$ be some prime different from the characteristics of the residue fields of $X$ (respectively, the characteristic of the field over which the variety is defined), then is there an isomorphism $Hom_{cont}(\pi_1^{et}(X),\mathbb{Q}_l)\cong H^1_c(X,\mathbb{Q}_l)$?

  • 2
    $\begingroup$ Why $H^1_c$ instead of $H^1?$ $\endgroup$
    – shenghao
    Commented Oct 25, 2011 at 3:39

1 Answer 1


In general, it's always true (for a connected scheme) that $H^1_{et}(X, \mathbb{Z}/l \mathbb{Z}) = \hom(\pi_1^{et}(X), \mathbb{Z}/l\mathbb{Z})$ (not compactly supported). Taking inverse limits over $l$ then gives the claim.

The reason this is true is that $H^1_{et}(X, \mathbb{Z}/l\mathbb{Z}$) can be computed by Cech cocycles, and from this it follows that elements of this group classify torsors over $\mathbb{Z}/l\mathbb{Z}$ (i.e. sheaves with a $\mathbb{Z}/l\mathbb{Z}$-action which are locally the constant $\mathbb{Z}/l \mathbb{Z}$ (in the etale topology)). But these, by descent theory, are the same as Galois covers of $X$ with Galois group $\mathbb{Z}/l\mathbb{Z}$, and (by Galois theory) classified by maps from the etale fundamental group into $\mathbb{Z}/l\mathbb{Z}$.

  • $\begingroup$ So this is true for any scheme? We don't need any connectedness assumptions as was suggested by Jared Weinstein's answer quoted by Makhalan? $\endgroup$ Commented Oct 24, 2011 at 19:28
  • $\begingroup$ I'm sure we're assuming connected, otherwise this isn't even true in the usual algebraic topological sense. $\endgroup$ Commented Oct 24, 2011 at 19:29
  • $\begingroup$ Ok. I think we only use this when applying "Galois theory". I'm pretty sure the first statement Akhil makes is true without the connectedness assumption. Or am I wrong? $\endgroup$ Commented Oct 24, 2011 at 19:35
  • $\begingroup$ What do you mean by the first statement? That $H^1_{et}(X,\mathbb{Z}/l\mathbb{Z})=hom(\pi_1^{et}(X),\mathbb{Z}/l\mathbb{Z})$? This would be false in algebraic topology. Think of $hom(\pi_1^{et}(X),\mathbb{Z}/l\mathbb{Z})$ as really being $(\pi_1^{et}(X))^{ab}\otimes \mathbb{Z}/l\mathbb{Z}$... $\endgroup$ Commented Oct 24, 2011 at 19:38
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    $\begingroup$ @Ariyan: The statement about $H^1$ should always be true (on sites in general). @Matt: Thanks, and whoops! $\endgroup$ Commented Oct 24, 2011 at 23:57

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