Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

However, that question dealt only with projective curves.

Question

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 $Hom_{cont}(\pi_1^{et}(X),\mathbb{Q}_l)\cong H^1_c(X,\mathbb{Q}_l)$?

share|improve this question
1  
Why $H^1_c$ instead of $H^1?$ –  shenghao Oct 25 '11 at 3:39
add comment

1 Answer 1

up vote 4 down vote accepted

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 $\mathbb{Z}/l\mathbb{Z}$.

share|improve this answer
    
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? –  Ari Oct 24 '11 at 19:28
    
I'm sure we're assuming connected, otherwise this isn't even true in the usual algebraic topological sense. –  Makhalan Duff Oct 24 '11 at 19:29
    
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? –  Ari Oct 24 '11 at 19:35
    
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 $(\pi_1^{et}(X))^{ab}\otimes \mathbb{Z}/l\mathbb{Z}$... –  Makhalan Duff Oct 24 '11 at 19:38
1  
@Ariyan: The statement about $H^1$ should always be true (on sites in general). @Matt: Thanks, and whoops! –  Akhil Mathew Oct 24 '11 at 23:57
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.