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I encountered the following statement without a reference many times. For a smooth variety $X$ over a perfect field $k$.

$Hom(H^1_{et}(X, \mathbb{Z}/n), \mathbb{Z}/n) \cong \pi^{ab}_1(X)/n$

Is there any reference for this? Why this is true?

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Dear Grilo, As for "why is this true", it is the analogue in the etale world of the analogous statement for topological spaces and usual cohomology, namely that $Hom(H^1(X,\mathbb Z/n),\mathbb Z/n)=\pi_1(X)^{ab}/n$, which follows from Hurewicz's theorem relating $\pi_1$ and $H_1$. Regards, Matthew –  Emerton Dec 20 '12 at 15:33
@Emerton: but saying that "it is the analogue" doesn't answer "why"! –  John Pardon Dec 20 '12 at 18:50
@unknown For characteristic zero the result actually follows from Hurewicz (and the Lefschetz principle and comparison between étale and simplicial cohomology) –  Felipe Voloch Dec 20 '12 at 21:28

1 Answer 1

up vote 4 down vote accepted

See Milne's online course notes on Étale Cohomology, Example 11.3, or Lei Fu's Étale Cohomology Theory, Proposition 5.7.20. (By passing to the direct limit over all $n$, you can even prove it for $\mathbf{Q}/\mathbf{Z}$.)

You only need $X$ to be connected Noetherian.

I am interested in alternative proofs not using torsors and Cech cohomology.

BTW, for $X$ normal, it is also true for $\mathbf{Z}$- or $\mathbf{Q}$-coefficients---both sides are $0$ in this case.

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It may be worth noting that the main content lies in the implicit use of Proposition 10.6 in those notes (valid for every abelian sheaf, such as constant coefficients in any abelian group) and that this is all valid for any connected scheme, whereas the identification of ${\rm{H}}^1(X,A)$ with the group of continuous homomorphisms from $\pi_1(X)$ into $A$ fails if we do not assume $A$ is finite (e.g., for the nodal cubic $X$ and $A = \mathbf{Z}$, there are nontrivial etale $\mathbf{Z}$-torsors over $X$), the "problem" being due to finiteness aspects built into the definition of $\pi_1(X)$. –  user29720 Dec 20 '12 at 15:49

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