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Dec 31, 2014 at 1:31 history edited Question Mark
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Dec 10, 2014 at 17:04 comment added user74230 Good point, I had forgotten that in the setup of interest a fiberwise-dense open subscheme is given to us. In general I would guess that $G$ might not be a scheme even though $G^0$ is, but finding such $G$ that is both separated and smooth isn't quite as obvious as I was hoping (e.g., relative Pic is rarely both separated and smooth).
Dec 10, 2014 at 17:00 vote accept Question Mark
Dec 10, 2014 at 16:53 comment added Question Mark The "further issue" does not arise: apply Thm. 2 with $G = \overline{X}^0$, $X = \overline{X}$ (which equals $G$ in your notation), and $Y = X$ to get that $\overline{X}$ is a scheme as desired. I agree that the crucial reliance on Artin's theorem from SGA 3 is slightly unpleasant because (i) Artin's theorem was written in non-algebraic space language (because algebraic spaces hadn't been invented yet!); (ii) (At a first glance) it uses different definition of a strict birational group law than the one in BLR. I am sure none of these are real issues, but still..
Dec 10, 2014 at 12:26 comment added user74230 Given your motivation, it is worth noting that (i) section 6.6 is not used elsewhere in that book and (ii) even after settling the above mild "gap" in the exposition (a simpler matter than the proof of the deep result of Artin invoked there as a black box), there is a further issue you have to deal with: given that $G^0$ is a scheme (by Theorem 2 in loc. cit.), to infer that $G$ is a scheme you have to first make an etale surjective base change on $S$ to get enough sections to do translations, but then to return to the original $S$ you face an effective etale descent problem for schemes!
Dec 10, 2014 at 8:19 answer added Laurent Moret-Bailly timeline score: 6
Dec 10, 2014 at 7:07 history asked Question Mark CC BY-SA 3.0