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Let $X$ be a scheme. Let $Y$ be a locally closed subset of $X$ and $U$ an open subset of $X$ such that $U \cap Y$ is non-empty and closed in $U$. I would like to deduce there is a closed point of $Y \cap U$.

What I have done: Let $z \in U \cap Y$ and take the affine open neighborhood $z \in \operatorname{Spec}A \subset U$. Then $\operatorname{Spec}A \cap Y$ is a closed subset of $\operatorname{Spec}A$, hence an affine scheme and it follows that there exists a closed point $z_0$ with respect to the topology of $\operatorname{Spec}A \cap Y$ . Now I am trying to prove that $\overline{ \{ z_0 \} }^{Y \cap U} = \{z_0\}$, but I seem to be stuck...

Johnny T.
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1 Answers1

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You can't do this without further conditions: let $X=Y=U$ be a scheme with no closed points. The most common assumption used to guarantee existence of a closed point is quasi-compactness, see for instance here for a proof.

KReiser
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  • Thank you... in the context I'm reading, $X$ is a finite type over an algebraically closed field, which I probably shouldn't have omitted in the question. But even with this additional assumption, I'm still not seeing it because it's not clear to me that $Y \cap U$ will be quasi-compact.. – Johnny T. Jun 09 '20 at 05:51
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    I agree, you should not have omitted this. To solve your question in this case, a scheme finite type over a noetherian scheme is again noetherian, and then every subscheme is quasicompact. Alternatively, you can solve this with just the Nullstellensatz, see for instance this question. – KReiser Jun 09 '20 at 06:08
  • I see.. thank you again – Johnny T. Jun 09 '20 at 07:04