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Jun 16, 2014 at 19:23 vote accept Wanderer
Sep 5, 2011 at 6:07 history edited Wanderer CC BY-SA 3.0
edited title
Sep 4, 2011 at 23:25 comment added Wanderer OK, I see why the question was confusing. This should be better.
Sep 4, 2011 at 23:24 history edited Wanderer CC BY-SA 3.0
corrected question
Aug 30, 2011 at 17:53 comment added Jason Starr @Moret-Bailly: That was also the counterexample I had in mind.
Aug 30, 2011 at 16:22 answer added Taylor Dupuy timeline score: 1
Aug 30, 2011 at 7:18 comment added Laurent Moret-Bailly Take $k$ algebraically closed, $S=\mathbb{G}_{m,k}$, $G=\mu_2$, and $X=S\to S$ the squaring map.
Aug 30, 2011 at 6:04 comment added Wanderer @Emerton: might be (not sure) but anyway, $\text{Pic}\ \overline{S} = 0$ implies $\text{Pic}^0\ \overline{S} = 0$... @S. Carnahan: yes :) @Jason Starr: you can impose this as en extra condition, however in the specific situation I'm looking at $S$ will not be proper (there $S$ will be the affine line with some points removed and $G$ will be a torus).
Aug 30, 2011 at 2:44 comment added Emerton Maybe you mean $Pic^0 \overline{S}$, rather than $Pic \overline{S}$, in your final sentence?
Aug 30, 2011 at 0:20 comment added S. Carnahan Minor nitpick: you probably want the isomorphism $\bar{X} \to \bar{G}$ to be $\bar{G}$-equivariant.
Aug 29, 2011 at 23:58 comment added Jason Starr Is $S$ proper over $k$? Otherwise there are counterexamples where $S$ equals the multiplicative group scheme $\mathbb{G}_{m,k}$.
Aug 29, 2011 at 23:25 history edited Wanderer CC BY-SA 3.0
edited title
Aug 29, 2011 at 22:59 comment added Michael Thaddeus How is the right-hand side different from $\overline{G}$?
Aug 29, 2011 at 22:49 history asked Wanderer CC BY-SA 3.0