Picard groups of non-projective varieties - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:01:11Z http://mathoverflow.net/feeds/question/33814 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33814/picard-groups-of-non-projective-varieties Picard groups of non-projective varieties Lars 2010-07-29T17:33:13Z 2010-09-23T18:42:01Z <p>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:</p> <p>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)</p> <blockquote> 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?</blockquote> <p>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?</p> <p>Is there a weaker sense of representability in which sense the "open" Picard functor is representable?</p> <p>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.)</p> <p><strong>Edit</strong>: I should have added that I do not want to assume resolution of singularities.</p> http://mathoverflow.net/questions/33814/picard-groups-of-non-projective-varieties/34652#34652 Answer by Simon Pepin Lehalleur for Picard groups of non-projective varieties Simon Pepin Lehalleur 2010-08-05T16:30:27Z 2010-09-23T18:42:01Z <p>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.). </p> <p>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.</p> <p>*<em>Edit : *</em> 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 :</p> <p>Albanese and Picard 1-motives Luca Barbieri-Viale - Vasudevan Srinivas Mémoires de la SMF 87 (2001), vi+104 pages </p> <p><a href="http://arxiv.org/abs/math/9906165" rel="nofollow">http://arxiv.org/abs/math/9906165</a></p> http://mathoverflow.net/questions/33814/picard-groups-of-non-projective-varieties/39432#39432 Answer by Fede for Picard groups of non-projective varieties Fede 2010-09-20T20:34:14Z 2010-09-23T07:48:57Z <p>Message deleted because false.</p>