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Dec 10, 2021 at 7:47 comment added YCor Also I find it confusing, especially in questions as you're asking, to denote $X_k$ both for "$X$ viewed as $k$-variety" and "the set of $k$-points of $X$". For instance, $X_k$ (set of points) can be empty but $X$ is not.
Dec 10, 2021 at 7:46 comment added YCor Do you assume $X$ is projective? Otherwise you can remove as many points as you like.
Dec 10, 2021 at 5:20 answer added Jérémy Blanc timeline score: 4
Dec 10, 2021 at 5:06 history became hot network question
Dec 9, 2021 at 22:11 answer added Ben C timeline score: 8
Dec 9, 2021 at 20:20 comment added Karl Schwede If you are allowing rational maps, no, I could just let $X$ be the blowup of $P^n$ at some points (including infinitely near points), and there would be no bound I think. On the other hand even for a birational regular map (defined everywhere), one can glue all the closed points of P^n together, and then the target would have one closed point ($X$ would not be normal though). The existence of a map should be fine though, at least for all but finitely many $p>0$.
Dec 9, 2021 at 20:09 history asked user114666 CC BY-SA 4.0