Timeline for $X$-points of reductive group schemes, if $X$ is a proper smooth curve over a finite field
Current License: CC BY-SA 3.0
8 events
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| Aug 10, 2014 at 2:35 | comment | added | user27920 | 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. | |
| Aug 9, 2014 at 12:55 | comment | added | Question Mark | Thanks! I wonder: what about if $X$ is higher dimensional (so just proper smooth over a finite field); does the finiteness of $G(X)$ continue to hold (at least for reductive $G$)? | |
| Aug 9, 2014 at 1:53 | comment | added | user27920 | @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. | |
| Aug 9, 2014 at 0:57 | comment | added | Question Mark | 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 $X$. | |
| Aug 9, 2014 at 0:56 | history | edited | Question Mark | CC BY-SA 3.0 |
added 15 characters in body
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| Aug 9, 2014 at 0:50 | comment | added | Piotr Achinger | 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... | |
| Aug 9, 2014 at 0:42 | comment | added | David E Speyer | See math.stackexchange.com/questions/311654 for $G$ abelian (but various other hypotheses removed). | |
| Aug 9, 2014 at 0:35 | history | asked | Question Mark | CC BY-SA 3.0 |