Picard groups of non-projective varieties - MathOverflow

Picard groups of non-projective varieties

2010-07-29T17:33:13Z

As far as I know, the main representability result for the relative Picard functor $Pic_{X/k}$, for a noeth. sep. scheme of finite type over a field $k$ is:

If $X$ is proper then $Pic_{X/k}$ is representable by a $k$-scheme loc. of finite type. (This is attributed to Murre and Oort in Bosch-Lüttkebohmert-Raynaud)

I am interested in what can be said once the requirement of properness is dropped, e.g. what can be said for quasi-projective varieties?

Representability is probably to much to ask for (even as an algebraic space), but do you have references or know of examples where the Picard functor of a non-projective quasi-projective variety is representable?

Is there a weaker sense of representability in which sense the "open" Picard functor is representable?

Is the group somehow controlled by (the group of $k$-points of) representable objects. (I have the naive impression that if $X$ is my quasi-projective variety, then a proper hypercovering of $X$ should be able to compute $H^1(X,\mathcal{O}_X^*)$, and that then one might be able to use representability theorems for proper/projective maps, but I know nearly nothing about the involved technical requirements.)

Edit: I should have added that I do not want to assume resolution of singularities.

Answer by Simon Pepin Lehalleur for Picard groups of non-projective varieties

2010-08-05T16:30:27Z

The first thing to consider is the case of affine curves : let $k$ be an algebraically closed field, $C/k$ a smooth affine curve, $\bar{C}/k$ its smooth projective compactification, $\bar{C}=C\cup{p_0,p_1,...,p_n}$, $J=J(\bar{C})$ the jacobian, $\theta:C\rightarrow J$ the map induced by the choice of the base point $p_0$. Then $Pic^0(C)$ is identified with the quotient $J/\langle\theta(p_1),\ldots,\theta(p_n))\rangle$. This is always divisible but depends somehow on what this subgroup of the groups of rational points of an abelian variety look like (does it land in the torsion, etc.).

Let's think about it over $\mathbb{C}$ : there you have the quotient of a complex torus by a finitely generated subgroup : when this subgroup is not discrete the quotient does look like it is not representable as the $\mathbb{C}-$points of a scheme.

*Edit : * As Emerton pointed out in the comments, in this case the correct "geometric" object is the 1-motive associated to C. But there is a general construction of Picard 1-motives associated to varieties over a field of characteristic 0 due to Barbieri-Viale and Srinivas, which encode the $Pic^0$ geometrically :

Albanese and Picard 1-motives Luca Barbieri-Viale - Vasudevan Srinivas Mémoires de la SMF 87 (2001), vi+104 pages

http://arxiv.org/abs/math/9906165

Answer by Fede for Picard groups of non-projective varieties

2010-09-20T20:34:14Z

Message deleted because false.