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Is every abelian scheme $\mathcal{A}/X$ under suitable conditions on $X$ a quotient of a Picard scheme of a curve $\mathcal{C}/X$? I need it for $X/\mathbf{F}_q$ smooth projective.

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    $\begingroup$ Does your definition of abelian schemes include "projective over $X$"? This at least is necessary. (I assume that by "curve" you mean "smooth proper curve"). $\endgroup$
    – Laurent Moret-Bailly
    Commented Nov 15, 2012 at 16:12
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    $\begingroup$ As far as I see from the literature, the question is not settled already if $X$ is the spectrum of a finite field. $\endgroup$
    – Matthieu Romagny
    Commented Nov 15, 2012 at 17:10
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    $\begingroup$ @Matthieu: it is true over finite fields. The proof, as in the classical case, relies on Bertini's theorem over finite fields and it is proved by Gabber and by Poonen's around 2000. $\endgroup$
    – Qing Liu
    Commented Nov 15, 2012 at 21:15
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    $\begingroup$ @Timo: the statement should hold over local (maybe noetherian) rings because Bertini's theorem is true in this case. $\endgroup$
    – Qing Liu
    Commented Nov 15, 2012 at 21:17
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    $\begingroup$ Liu, good point! Of course Poonen's Bertini theorems involve sections with hypersurfaces of large degree. I thought that in the classical proof that abelian varieties are quotients of jacobians, it was important to cut by hyperplanes. Now that I think about it again, I see that it is not. $\endgroup$
    – Matthieu Romagny
    Commented Nov 19, 2012 at 20:20

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Let me sketch an idea why there are not enough Jacobians. At some points I have asked for references/proofs in []. Edit: Didn't work out as hoped.

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    $\begingroup$ The Kodaira-Parshin construction produces lots of non-isotrivial proper and smooth curves over a smooth projective curve. See for example M. Martin-Deschamps, Astérisque vol. 127 (1985). $\endgroup$
    – Laurent Moret-Bailly
    Commented Feb 2, 2013 at 11:14
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    $\begingroup$ also, the complement of a divisor has no reason to be affine -- take $\mathbb{P}^1 \times \mathbb{P}^1 \setminus 0 \times \mathbb{P}^1 = \mathbb{A}^1 \times \mathbb{P}^1$. $\endgroup$
    – Vivek Shende
    Commented Feb 2, 2013 at 16:53
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    $\begingroup$ $M_{g,n}$ is very far from being affine. For every $g\geq 3$, there is a complete curve passing through any point. This follows from the existence of a compactification with boundary of codimension >1 (the Satake compactification). The problem is only the complement of a very ample divisor is guaranteed to be affine. $\endgroup$
    – Simon Pepin Lehalleur
    Commented Feb 8, 2013 at 19:15

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