# every abelian scheme quotient of a Picard scheme?

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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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"). – Laurent Moret-Bailly Nov 15 '12 at 16:12
As far as I see from the literature, the question is not settled already if $X$ is the spectrum of a finite field. – Matthieu Romagny Nov 15 '12 at 17:10
@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. – Qing Liu Nov 15 '12 at 21:15
@Timo: the statement should hold over local (maybe noetherian) rings because Bertini's theorem is true in this case. – Qing Liu Nov 15 '12 at 21:17
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. – Matthieu Romagny Nov 19 '12 at 20:20

also, the complement of a divisor has no reason to be affine -- take $\mathbb{P}^1 \times \mathbb{P}^1 \setminus 0 \times \mathbb{P}^1 = \mathbb{A}^1 \times \mathbb{P}^1$. – Vivek Shende Feb 2 '13 at 16:53
$M_{g,n}$ 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. – Simon Pepin Lehalleur Feb 8 '13 at 19:15