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Let $X$ be a connected proper smooth curve over a finite field (so the generic point of $X$ is the spectrum of a global field $K$), and let $G \rightarrow X$ be an affine $X$-group scheme of finite type. Is $G(X)$ finite?

I will accept any answer that assumes additional hypotheses on $G$, as long as the class of $G$ under consideration includes all reductive group schemes (i.e., smooth affine $G \rightarrow X$ whose geometric fibers are (connected) reductive groups). Note that separatedness of $G \rightarrow X$ ensures that after replacing $G$ by the schematic image of its generic fiber one may without loss of generality assume that $G$ is flat.

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  • $\begingroup$ See math.stackexchange.com/questions/311654 for $G$ abelian (but various other hypotheses removed). $\endgroup$ – David E Speyer Aug 9 '14 at 0:42
  • $\begingroup$ I don't understand something. Since $G$ is affine over $X$, assuming it's of finite type, we can embed $G$ into $\mathbb{A}^n\times X$ for some big $n$. This reduces the question to $G = \mathbb{A}^n\times X$, in which case the answer is trivially yes as $\Gamma(X, \mathcal{O}_X)^n$ is a finite group... $\endgroup$ – Piotr Achinger Aug 9 '14 at 0:50
  • $\begingroup$ Thanks, I meant to assume that $G \rightarrow X$ is of finite type. Why can you embed $G$ into $\mathbb{A}^n_X$ as you claim? I agree that this can be done locally on $X$. $\endgroup$ – Question Mark Aug 9 '14 at 0:57
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    $\begingroup$ @QuestionMark: A slight correction of Piotr's comment works. If $f:Y \rightarrow X$ is an affine morphism of finite type to a proper scheme $X$ over a field $k$, then by expressing the quasi-coherent finite type $O_X$-algebra $f_{\ast}(O_Y)$ as a direct limit of its coherent $O_X$-submodules we can find a coherent $F \subset f_{\ast}(O_Y)$ generating it as an $O_X$-algebra. For Dedekind $X$ and $X$-flat $Y$, $F$ is a vector bundle. Thus, $Y$ is closed in a vector bundle $V$ over $X$, so $Y(X) \subset V(X)$, which is finite if $k$ is. $\endgroup$ – user27920 Aug 9 '14 at 1:53
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    $\begingroup$ Sure, but relax "smooth" to "reduced". By Chow's Lemma there is a surjective map $X' \rightarrow X$ from a projective reduced $k$-scheme $X'$, so $Y(X) \subset Y'(X')$ for $Y' := Y_{X'}$, thereby reducing to the case when $X$ is projective. Then $F$ as above is a quotient of a vector bundle on $X$, so once again $Y$ is closed in a vector bundle over $X$, etc. $\endgroup$ – user27920 Aug 10 '14 at 2:35

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