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There exists information on the Picard (and Brauer) group of a reductive algebraic group over a number field k. For example, Sansuc shows (in his big Crelle paper of 1980) that if G is connected and semisimple over a number field k, then Pic G is the group of k-rational points of the character group of the fundamental group of G. In particular, if G is semisimple and simply connected, then Pic G=0. My question is: do there exist results of this type over more general bases? For example, a natural generalization of the equality "Pic G=0 if G semisimple and simply connected over a number field" would be "Pic G=Pic U if G is a semisimple and simply connected group scheme over a Dedekind scheme U". Is the latter true?

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Cristian, have you looked in Raynaud's thesis? One natural idea would be to see to what extent Pic(G) is entirely accounted for by central extensions by $\mathbf{G}_m$, which is what underlies the direct proof over a field (using derived group to split the extension in the simply connected case). I assume when you say $G$ is semisimple over a Dedekind base, you mean it is smooth and affine with all fibers connected semisimple (not just generic fiber). – BCnrd Apr 26 '10 at 20:45
Hi, Cristian! Welcome to Math Overflow! – Pete L. Clark Apr 26 '10 at 21:12

Assuming you know that Pic of the generic fibre is trivial, this seems to follow immediately from the localisation sequence: since G is a group there is a section, so the map Pic U to Pic G is an injection. On the other hand (since G is smooth so Pic = Cl) there is an exact sequence:

$\ \ \ \oplus_x \mathbb{Z}_x \to Pic \ G \to Pic \ G_K \to 0 $

where X runs over the closed points of U and $G_K$ denotes the generic fibre. Since $Pic\ G_K$ is trivial it follows from this that the map Pic U to Pic G is surjective.

(Note that the only property of the closed fibres that is used is that they are reduced and irreducible.)

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OK, the proof seems correct. Thanks! -I should also add that Brian's suggestion that I look into Raynaud's thesis was right on the mark, as it also leads to an answer to my question (and to another one that I had been struggling with). Many thanks to the anonymous poster and to Brian Conrad for their help! – Cristian D. Gonzalez-Aviles Apr 27 '10 at 14:34

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