In computing the etale cohomology of curves, one of the key facts one needs is the torsion in $Pic(X) = H^1(X_{et}, \mathcal{O}_X^*)$ for a smooth projective curve $X$. Namely, one shows that the $n$-torsion (for $n$ prime to the characteristic, at least) is given by $(\mathbb{Z}/n\mathbb{Z})^{2g}$ where $g$ is the genus by invoking the theory of the Jacobian. However, I don't really know anything about the Jacobian (and pretty much all that I do know is in characteristic zero, in which case all you need is Abel's theorem for the aforementioned fact), so I am curious: can this fact, describing the torsion in $Pic(X)$, be proved directly in an elementary fashion?
1 Answer
As far as I know, the standard computation of the torsion in $Pic(X$) proceeds as follows: one considers the exact sequence $0 \to Pic^0(X) \to Pic(X) \to \mathbb Z \to 0,$ the map to $\mathbb Z$ being the degree map. This shows that the torsion subgroups of $Pic(X)$ and $Pic^0(X)$ coincide. One then shows that $Pic^0(X)$ is an abelian variety of dimension $g$ (this is the algebraic theory of the Jacobian), and one applies the general theory of abelian varieties to compute its torsion.
An (apparently) alternative approach is as follows: one uses an $n$-torsion line bundle on $X$ to construct a degree $n$ cyclic cover of $X$, and thus classifies the $n$-torsion of the Jacobian in terms of the maps from the etale $\pi_1(X)$ to $\mathbb Z/n\mathbb Z$. But I don't know how (in positive characteristics, where one can't resort to analytic arguments) to concretely compute the group of such maps without going back to the computation in terms of $Pic^0(X)$ and the theory of abelian varieties.
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$\begingroup$ Thanks! BCnrd commented (over email) that the computation $H^1(X,\mu_n)=\hom(\pi_1(X),\mu_n)$ could be used for $H^1$ if one accepts that etale $\pi_1$ is the same in char. 0 and char. p (apparently in SGA 1). But that still fails to compute the cokernel and to get $H^2$ of $\mu_n$. I guess I should learn about Jacobians and abelian varieties. $\endgroup$ May 5, 2011 at 22:10
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$\begingroup$ Whoops, actually I misunderstood: the statement about $\pi_1$ is true up to prime to $p$ quotients. $\endgroup$ May 5, 2011 at 22:13
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$\begingroup$ Dear Akhil, Yes, I should have mentioned that there are arguments involving etale $\pi_1$ that require lifting to char. $0$, but there are surely no easier than the arguments with abelian varieties, which are in any case important to learn. (Not that etale $\pi_1$ isn't.) Best wishes, Matthew $\endgroup$– EmertonMay 5, 2011 at 23:38