# Relative identity component for group algebraic spaces

Let $S$ be a locally noetherian scheme and let $G$ be a separated and smooth $S$-group algebraic space of finite presentation. Does there exist an open sub-(group algebraic space) $G^0 \subset G$ characterized by the requirement that $G_s^0$ be the identity component of $G_s$ for every $s \in S$? (Note that the $S$-fibers of $G$ are automatically schemes.)

In the scheme case, the positive answer (with less restrictive assumptions) follows from the results of EGA IV, section 15.6 (or of http://stacks.math.columbia.edu/tag/055K). It seems that the positive answer in the algebraic space case is used implicitly in the proof of Thm. 1 of section 6.6 of "Neron models" of Bosch, Lutkebohmert, and Raynaud (seemingly it is used to justify the applicability of Thm. 2 in the first paragraph of the said proof on p. 163). Thus I wonder if a positive answer to my question already appears somewhere in the literature?

• 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! – user74230 Dec 10 '14 at 12:26
• 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.. – Question Mark Dec 10 '14 at 16:53
• 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). – user74230 Dec 10 '14 at 17:04

• As noted there, the EGA proof applies almost verbatim for stacks, up to IV$_3$ Lemma 15.5.6 (any $X$ flat and lfp over a dvr $R$ with reduced connected special fiber also has connected generic fiber), whose proof in EGA is scheme-specific (set of generizations of a point is a local Spec). Romagny's Lemma B.2 gives a simpler proof of this ("reduced fiber trick") that works beyond schemes (omitting the easy reduction to the quasi-compact case for a statement about idempotents, q-c needed to pass localization through H$^0$); the same trick is in an earlier paper of Kisin, and perhaps elsewhere. – user74230 Dec 10 '14 at 12:09
• Thanks to both of you! @user74230: in your first parenthetical, you need to also assume that $X$ is connected. On a different note, I wonder if these type of results are purely topological? Is it possible that a similar result could be proved in the generality of locally spectral spaces? After all, both flatness and finite presentation assumptions have a strong topological aspect to them (generizing + Chevalley's theorem); perhaps the assumption of reduced geometric fibers is not so topological though... – Question Mark Dec 10 '14 at 17:16