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Dec 22, 2021 at 23:33 vote accept stupid_question_bot
Dec 9, 2021 at 13:19 comment added Will Sawin @stupid_question_bot I'm not sure if it's closed, but it certainly is constructible, because it's a statement in the first-order language of fields, and it doesn't contain the generic point, so it's contained in a proper closed subset.
Dec 9, 2021 at 5:54 comment added stupid_question_bot To fix notation, suppose $k$ is an uncountable field, $S$ a $k$-scheme, and $X = \mathbb{A}^n_S$. Given $d\ge 1$, as you say you can find $n_d$ sections $\sigma_1,\ldots,\sigma_{n_d}$ which don't satisfy any nontrivial degree $d$ equations over the generic point of $S$. Why must the set $\{s\in S \;|\;\;(\sigma_1,\ldots,\sigma_{n_d})\cap X_s \text{ satisfies a nontrivial degree $d$ equation in $X_s$}\}$ be closed?
Dec 9, 2021 at 4:59 history edited Will Sawin CC BY-SA 4.0
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Dec 9, 2021 at 4:59 comment added Will Sawin @stupid_question_bot I'm fixing an embedding to either $\mathbb A^n$ or $\mathbb P^n$. And yes.
Dec 9, 2021 at 4:43 comment added stupid_question_bot Thanks for your answer! In your final paragraph, what do you mean by a "degree $d$ equation"? Are you still considering $\mathbb{A}^1$ over a curve? Also for the final sentence, presumably you wanted to say that the union of countably many proper closed subsets can't be the whole thing?
Dec 9, 2021 at 3:43 history answered Will Sawin CC BY-SA 4.0