Let G be a reductive group over a local field F. Let O be the ring of integers of F.


The following are equivalent (and groups satisfying these conditions are called unramified):


(a) G is quasisplit and split after passing to an unramified extension of F.


(b) G is the generic fibre of a reductive group scheme over O.


This is stated, as far as I can tell without a proof or reference to one, in Tits' Corvallis article. My question is - how does one prove this equivalence?



On unramified p-adic groups
Let us start with (b) => (a). We only need to show that if $\mathcal{G}$ is a reductive $O$-group scheme, then $\mathcal{G}\_F$ is quasi-split and split after an unramified extension. 


Let us first show that $\mathcal{G}\_F$ is quasi-split. Let $\mathrm{Bor}$ be the $O$-scheme parametrizing the Borel subgroups of $\mathcal{G}$, so that $\mathrm{Bor}$ is $O$-smooth (smoothness of the scheme of Borels is true for a reductive group scheme over any base). Over finite fields, reductive groups are quasi-split (ultimately by an application of (an extension of) Lang's theorem), so choose a Borel subgroup of the special fiber of $\mathcal{G}$. Hensel's lemma (or [EGA IV, 18.5.17]) then lifts the chosen residual point of $\mathrm{Bor}$ to an $O$-point. The resulting Borel subgroup of $\mathcal{G}$ shows that $\mathcal{G}$ is quasi-split, and hence so is $\mathcal{G}\_F$.


Now let us show that $\mathcal{G}\_F$ is split after an unramified extension. Reductive group schemes are split etale locally on the base, so $\mathcal{G}$ splits over an etale cover of $\mathrm{Spec } O$. Since $O$ is Henselian, any such cover can be refined by a finite etale cover (look at what covers the closed point), i.e., by the spectrum of the ring of integers of a finite unramified extension $K/F$. Passing to the generic fiber we conclude that $\mathcal{G}\_F$ splits over $K$.


Now let us prove that (a) => (b). Choose a finite unramified extension $K/F$ over which $G$ splits. Let $G\_0$ be the split reductive $O$-group with the same (geometric) root datum $R$ as $G$. Then $G$ is a form of $(G\_0)\_F$, so $G$ corresponds to an element $x \in H^1(K/F, \mathrm{Aut}\_{G\_0})$. Recall that $\mathrm{Aut}\_{G\_0}$ is an extension
$$
1 \rightarrow G\_0/Z\_{G\_0} \rightarrow \mathrm{Aut}\_{G\_0} \rightarrow \mathrm{Aut}(R, \Delta) \rightarrow 1,
$$
where $\Delta$ is a choice of a base of positive roots, and that the extension splits as a semi-direct product after a choice of pinnings (we fix a choice of the latter and of $\Delta$). Let $y \in H^1(K/F, \mathrm{Aut}(R, \Delta))$ be the image of $x$. The $O$-group scheme $\mathrm{Aut}(R, \Delta)$ is constant, so $y$ comes from a unique $z \in H^1(O\_K/O, \mathrm{Aut}(R, \Delta))$ (where $O\_K$ is the ring of integers of $K$). Let $w \in H^1(O, \mathrm{Aut}\_{G\_0})$ be the image of $z$ via the chosen semi-direct product decomposition, and let $\mathcal{G}$ be a resulting form of $G\_0$. Then $\mathcal{G}\_F$ and $G$ are inner forms of each other because their classes in $H^1(F, \mathrm{Aut}\_{G\_0})$ have the same image $y$ in $H^1(F, \mathrm{Aut}(R, \Delta))$. However, both $\mathcal{G}\_F$ and $G$ are quasi-split (the former by (b) => (a)), so they are isomorphic due to the following general fact: over a semi-local base scheme, every reductive group scheme has a *unique* quasi-split inner form.


