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(edited for clarity)

In a comment on a response to this question, moonface states the following: "Even if you tried to compute H^2 [etale with Z/5Z-coefficients] of a surface fibered in genus 2 curves over a base curve X, then (to compute the cohomology of X with coefficients in the relevant local system) you have to pass to a 125-fold covering of X and compute the Jacobian of that beast."

Would someone be willing to explain this further? Given the importance of etale cohomology and the difficulties in working with it, I would really love to see the details of this spelled out.

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In that question, we were working with mod $5$ coefficients. Also, $125$ should have been $625$. Finally, it is not "equivalent": you need only a piece of the cohomology of the cover. Specifically, if $f : X \rightarrow C$, let $C'$ be the $5$-torsion in the relative Jacobian; its Galois group is a subgroup of $\mathrm{GSp}(\mathbb{F}_5)$, and you need roughly the piece of the cohomology of $C'$ through which $G$ acts by the standard representation. –  moonface Mar 25 '10 at 15:40
    
David- The way you wrote this question is extremely confusing. I think it would have been really helpful to include part of moonface's comment as an actual quote. –  Ben Webster Mar 25 '10 at 15:42
    
Correction to previous comment: The cover $C' \rightarrow C$ defined by $5$-torsion in relative Jacobian isn't Galois; indeed, we are only interested in a "certain piece" of its cohomology, but I'm not sure how to find this "piece" without passing to the "Galois closure"... And the Galois closure of $C' \rightarrow C$ is a cover of monstrous degree indeed. –  moonface Mar 25 '10 at 15:50
    
@Ben: I wasn't sure about the etiquette for quoting other people's answers to previous questions, hence the elliptical wording. I have edited it for clarity. –  David Hansen Mar 25 '10 at 15:58
    
@Moonface: Ahhh, let me guess - by a Leray spectral sequence, computing H^2(X) is reduced to computing H^1(C,H^1(fiber)), and since H^1(fiber) is a local system of $\mathbb{F}_5^{4}$'s you trivialize it by passing to a $5^4$-fold cover. –  David Hansen Mar 25 '10 at 16:02

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