# Jacobian of a curve

Let $C$ be a curve and $J$ be its Jacobian.

What is the relation between $H^1(C,\mathcal{O}_C)$ and $H^1(J,\mathcal{O}_J)$ ?

Can someone point me to an easy reference for this subject?

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They are the same in characteristic 0, because the Abel-Jacobi map induces an isomorphism $$H^0(C, \Omega) \to H^0(J, \Omega)$$ and (the notation below means singular cohomology with complex coefficients of the associated complex manifold) it also induces an isomorphism $$H^1(C) \to H^1(J)$$ but we have (canonically, but not naturally, split) exact sequences (for any $X$) $$0 \to H^0(X, \Omega) \to H^1(X) \to H^1(X, \mathcal{O}) \to 0.$$
I believe the result is true for basically the same reason using an appropriate Weil-type cohomology theory in characteristic $p$ or a lifting argument. Perhaps there is an easier argument. I don't know a reference in characteristic $p$. You might try Milne's notes on his website.
Are you saying it also holds in characteristic $p$ (but you don't know where to find the proof)? – averageman Apr 25 '12 at 16:26
I believe it holds in characteristic $p$ because of the proof I wrote, since the exact sequence I wrote is still true if the $H^1$ in the middle means etale cohomology with coefficients in $\mathbb{Z}_\ell$ for any $\ell \neq p$. – user29743 Apr 25 '12 at 16:31
Yes, please do - I am definitely a little squeamish about characteristic $p$ here. – user29743 Apr 25 '12 at 16:34
My argument is dead wrong in characteristic $p$ - the fields of coefficients of the cohomology groups don't agree! However, I believe the claim is still true using a lifting argument to characteristic 0, but you will need more details from someone else. Editing response. – user29743 Apr 25 '12 at 19:59