Let $K$ be a local field, e.g. $\mathbb{Q}\_p$ or $\mathbb{F}\_p((t))$. Let $G$ be a connected reductive group over $K$. Is it true that $G$ is already defined over a global field? More precisely, does there exist a global field $F$, a place $v$ in $F$ with $F\_v\simeq K$ and a connected reductive group $\tilde{G}$ over $F$ such that as groups
$$\tilde{G}\otimes\_F F\_v\simeq G?$$
I am particularly interested in the case where $K$ is of equal characteristic. Thanks in advance!



Is every connected reductive group over a local field already defined over a global field?
Pick a global field $E$ and finite place $w$ with $E\_w=K$. The fraction field $k$ over $E$ of the henselization of the "algebraic" local ring at $w$ is the direct limit of finite separable sub extensions $F/E$ for which the place $v$ on $F$ from the valuation on $k$ satisfies $F\_v=K$. Thus, it suffices to "algebraize" $G$ over $k$. We therefore forget about number field and focus on a henselian valued field $k$ with completion denoted $K$ and aim to descend a connected reductive $K$-group $G$ to a connected reductive $k$-group.


The Galois groups of $k$ and $K$ are naturally isomorphic, so if $R$ denotes the root datum of $G\_{K\_s}$ then ${\rm{H}}^1(k, {\rm{Aut}}(R)) \rightarrow {\rm{H}}^1(K, {\rm{Aut}}(R))$ is bijective. This says that every quasi-split connected reductive $K$-group descends to a quasi-split connected reductive $k$-group which is moreover unique up to isomorphism (since these H$^1$'s classify quasi-split forms with a given geometric root datum). Every connected reductive $K$-group $G$ has a (unique) quasi-split inner form $G\_0$, so $G$ is obtained from $G\_0$ via twisting against a class in ${\rm{H}}^1(K, G\_0^{\rm{ad}})$ for the adjoint semisimple $G\_0^{\rm{ad}} := G\_0/Z\_{G\_0}$. Thus, it suffices to show that ${\rm{H}}^1(k, H) \rightarrow {\rm{H}}^1(K,H\_K)$ is surjective for every smooth connected affine $k$-group $H$ (such as $H$ being the quasi-split $k$-descent of $G\_0^{\rm{ad}}$). Even better:


**Theorem**: For any smooth affine $k$-group $H$, the natural map
$${\rm{H}}^1(k,H) \rightarrow {\rm{H}}^1(K,H)$$
is bijective.


Proof: By Galois-twisting, injectivity reduces to triviality of the kernel. 
In other words, if $E$ is an $H$-torsor over $k$ which has a $K$-point then it has a $k$-point. More generally, if
$X$ is a smooth $k$-scheme then $X(k)$ is dense in $X(K)$ for the valuation topoloy. This is Zariski-local on $X$, so
we can assume there is an etale map $f:X \rightarrow \mathbf{A}^n\_k$. The open image $V=f(X)$ is dense open in $\mathbf{A}^n\_k$, so
$V(k)$ is dense in $V(K)$ due to density of $k$ in $K$. By the Zariski-local structure theorem for etale morphisms and the $K$-analytic inverse
function theorem, for each $x \in X(K)$ and $v=f(x)\in V(K)$, every $v'$ sufficiently near $v$ admits $x'\in f^{-1}(v')$ near $x$ in $X(K)$. By openness of $X(K) \rightarrow V(K)$, for any $x \in X(K)$ and open $\Omega \subset X(K)$ around $x$ we can find an open $U \subset V(K)$ around $v=f(x)$ such that every 
$u \in U$ is the image of a $K$-point in $\Omega$. Consider such $u \in V(k) \cap U$ (as exists by density of $V(k)$ in $V(K)$). The fiber scheme $f^{-1}(u)$ is finite etale over $k$, so the equivalence of Galois theories of $k$ and $K$ shows that every $K$-point in $f^{-1}(u)$ comes from a unique $k$-point of $f^{-1}(u)$. Hence, we can find a $k$-point in $\Omega$. This completes the proof of injectivity.


For surjectivity, choose a closed $k$-subgroup inclusion $j:H \hookrightarrow {\rm{GL}}\_n:=G$ and let $X=G/H$ (a smooth $k$-scheme). Thus, there is a natural surjection
$$G(k)\backslash X(k) \rightarrow {\rm{H}}^1(k,H)$$
and likewise for $K$ (since ${\rm{H}}^1(k, {\rm{GL}}\_n)=1$ and likewise for $K$). It therefore suffices to show that the natural map
$$X(k) \rightarrow G(K) \backslash X(K)$$
is surjective. Since $X(k)$ is dense in $X(K)$ by the above, it suffices to show that all $G(K)$-orbits in $X(K)$ are open. But
each orbit map $G\_K \rightarrow X\_K$ through a $K$-point is a smooth map since $H$ is smooth, so the induced map on $K$-points is open (hence has open image) by
the $K$-analytic implicit function theorem (using the Zariski-local structure of smooth morphisms).


QED


