Suppose that $F$ is a nonarchimedean local field, and that $G\_1$, $G\_2$ are connected, simply connected algebraic groups over $F$. Suppose moreover $G\_1$ and $G\_2$ are semisimple. Suppose $H$ is a hyperspecial maximal compact subgroup of $G\_1\times G\_2$. Is $H$ necessarily a product $H\_1\times H\_2$ where the $H\_i$ are hyperspecial maximal compact subgroups of the $G\_i$?


Edited: I had originally written that the Lie algebras of the $G\_i$ were semisimple, which is not such a sensible thing to write. I have changed this to say the $G\_i$ themselves semisimple.


Note: Brian Conrad's answer, below, shows we need assume nothing more than connected reductive hypotheses on $G\_1$ and $G\_2$.



Hyperspecial subgroup of a product of semisimple algebraic groups
To allow all characteristics, the semisimplicity requirement on the Lie algebras should be replaced with the requirement that the $G\_i$ are semisimple as $F$-groups. (In positive characteristic the Lie algebras of connected semisimple groups can often fail to be semisimple.) In fact, the simply connected and semisimplicity hypotheses are stronger than necessary; connected reductive is sufficient. Taking into account the definition of "hyperspecial maximal compact subgroup" (of the group of $F$-rational points), the question in this modified form immediately reduces to a more "algebraic" assertion having nothing to do with local fields, as follows. 


Consider a reductive group scheme $\mathcal{G}$ (with connected fibers, following the convention of SGA3) over a normal noetherian scheme $S$, and let $G$ be its fiber over the scheme $\eta$ of generic points of $S$. Assume $G$ decomposes as a direct product $G = G\_1 \times G\_2$ with $G\_i$ necessarily connected reductive over $\eta$. Then does there exist a pair of reductive closed $S$-subgroup schemes (with connected fibers) $\mathcal{G}\_i$ in $G\_i$ such that $\mathcal{G}\_ 1 \times \mathcal{G}\_2 = \mathcal{G}$ compatibly with the given identification on generic fibers? Below is a proof that the answer is ``yes''. 


First observe that such $\mathcal{G}\_i$ are necessarily unique if they exist, being the Zariski closures of the $G\_i$ in $\mathcal{G}$ (since $S$ is reduced). In view of this uniqueness, by descent theory it follows that to prove the existence we may work \'etale-locally on the base. (This step tends to ruin connectedness hypotheses, since $S' \times\_S S'$ is generally cannot be arranged to be connected, even if $S$ is connected and $S' \rightarrow S$ is a connected \'etale cover.) Hence, we now assume that $\mathcal{G}$ is $S$-split in the sense that it admits a (fiberwise) maximal $S$-torus $\mathcal{T}$ that is $S$-split. We may also work separately over each connected component of $S$, so we can now assume $S$ is connected (as we will make no further changes to $S$ in the argument). 


The maximal $\eta$-torus $T := \mathcal{T}\_ {\eta}$ in $G$ uniquely decomposes as $T = T\_1 \times T\_2$ for necessarily $\eta$-split maximal $\eta$-tori $T\_i$ in $G\_i$. By the classification of split pairs (i.e., fiberwise connected reductive group equipped with a split maximal torus) over any base scheme in terms of root data, to give a direct product decomposition of $(\mathcal{G}, \mathcal{T})$ is the same as to decompose its root datum into a direct product (i.e., direct product of the $X$ and $X^{\vee}$ parts, and corresponding disjoint union for the $R$ and $R^{\vee}$ parts). By connectedness, the normal $S$ is irreducible (i.e., $\eta$ is a single point), so the root data of $(\mathcal{G}, \mathcal{T})$ and its generic fiber $(G,T)$ are canonically identified. QED


