Existence of a model over S-integers

Question

Let $G$ is a connected reductive group over a number field $F$. Let $S$ be a set of places containing the archimedean places for which $G$ has a model over the ring of $S$-integers in $F$. Is it true that for $v \notin S$, $G \times_F F_v$ is quasisplit and is split over an unramified extension?

Answer 1

(I assume $S$ is finite, the essential case.) The answer is affirmative if and only if $G$ extends to a reductive $O_{F,S}$-group scheme (i.e., a smooth affine $O_{F,S}$-group with connected reductive fibers). This is a genuine restriction, since for any Dedekind domain $A$ and any smooth affine group $H$ over the fraction field $K$ of $A$ (e.g., $A=O_F$ and $G$ as above) there exists a smooth affine $A$-group with generic fiber $H$ (indeed, take the schematic closure of $H$ in ${\rm{GL}}_{n,A}$ relative to a choice of $K$-subgroup inclusion of $H$ into some ${\rm{GL}}_n$ to get a flat affine $A$-group of finite type with generic fiber $H$, and then apply the "group smoothening" process of Neron-Raynaud). So merely having an "$A$-model" in the sense of a smooth affine $A$-group is not a genuine constraint on a given smooth affine $K$-group; it is "spreading out" the reductivity condition that is the non-trivial feature.

In the given situation over a number field, there is some finite $S' \supset S$ so that $G$ is the $F$-fiber of a reductive $O_{F,S'}$-group $\mathscr{G}$. Indeed, this follows from general "denominator-chasing" principles adapted to the context of reductive group schemes (see either [SGA3, XIX, 2.1.6] combined with constructibility for the locus of geometrically connected fibers [EGA IV$_3$, 9.7.7], or Prop. 3.1.9(1) or Corollary 3.1.11 in the article "Reductive Group Schemes" from the Proceedings of the 2011 Luminy summer school on SGA3). It is then automatic from smoothness of the schemes of Borel subgroups and of maximal tori for a reductive group scheme, as well as finiteness of the residue fields for $F$ at its finite places, that for every $v \not\in S'$ the $F_v$-group $G_{F_v}$ is quasi-split and becomes split over a finite unramified extension of $F_v$ (see Corollary 5.2.14 in "Reductive Group Schemes").

What to do about the finitely many places $v \in S'-S$? Note that any reductive $O_{F,S}$-group with generic fiber $G$ agrees with $\mathscr{G}$ over $O_{F,S"}$ for some $S" \supset S'$. Thus, at the cost of possibly increasing $S'$, general principles for making schemes of finite type over a Dedekind base by "gluing" in integral models over the completion at finitely many maximal ideals (see 1.4/1 and 6/2/D.4(b) in the book Neron Models, a vast generalization of Weil's adelic approach to describing vector bundles over a smooth curve) reduce our task to one over the completion $O_{F_v}$ for each non-archimedean place $v$ of $F$: a connected reductive $F_v$-group $H$ (such as $G_{F_v}$) is the generic fiber of a reductive $O_{F_v}$-group if and only if $H$ is quasi-split and splits over an unramified extension of $F_v$. The implication "$\Rightarrow$" has been addressed already, so it remains to show "$\Leftarrow$".

So now consider a non-archimedean local field $K$ (such as $F_v$ above) and a connected reductive $K$-group $H$ that is quasi-split and becomes split over an unramified finite extension $K'/K$. We seek a reductive $O_K$-group with generic fiber $H$. Let $\mathscr{H}_0$ be the Chevalley group over $O_K$ with the same root datum $R$ as $H_{K_s}$, $H_0$ its $K$-fiber, and $\Delta$ a basis of the underlying root system of $R$. Then $H$ is a quasi-split $K$-form of $H_0$, and these are classified up to isomorphism by the pointed set ${\rm{H}}^1(K, {\rm{Aut}}(R, \Delta))$ (trivial Galois action on ${\rm{Aut}}(R,\Delta)$). By functoriality in $K$, the condition that $H$ splits over an unramified finite extension $K'/K$ is the condition that the class $[H] \in {\rm{H}}^1(K, {\rm{Aut}}(R,\Delta))$ has trivial restriction in ${\rm{H}}^1(K', {\rm{Aut}}(R,\Delta))$. By describing these ${\rm{H}}^1$'s in terms of cocycles we see via unramifiedness of $K'/K$ that the kernel ${\rm{H}}^1(K'/K, {\rm{Aut}}(R,\Delta))$ of restriction over $K'$ is exactly the same as the kernel of the restriction map $${\rm{H}}^1(O_K, {\rm{Aut}}(R,\Delta)) \to {\rm{H}}^1(O_{K'}, {\rm{Aut}}(R,\Delta)).$$ But ${\rm{H}}^1(O_K, {\rm{Aut}}(R,\Delta))$ classifies the (necessarily quasi-split!) $O_K$-forms of $\mathscr{H}_0$, so the one whose $K$-fiber is $[H]$ is the desired reductive $O_K$-group with $K$-fiber $H$.