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Suppose $X$ is an integral noetherian scheme and that $Y$ is a closed integral subscheme with generic point $\eta$. If $U\cap Y\neq\emptyset$ for any affine open $U$, then $\eta\in U$. This follows from the fact that $U\cap Y$ is an affine open subset of $Y$ and $\{\eta\}$ is dense in $Y$.

My question is: Suppose $U$ is an open subset of $X$ (but not necessarily affine) and that $U\cap Y\neq\emptyset$. Let $\operatorname{Spec}A\subset U$ be an open affine. Is it true that $\eta\in \operatorname{Spec} Y$?

ponchan
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  • As shown at the linked duplicate, the generic point of a subscheme must be in every nonempty open subset of that scheme. – KReiser Jun 05 '22 at 17:49
  • @KReiser Where specifically does the linked question answer my question? I only see this shown when $U$ is affine. – ponchan Jun 05 '22 at 17:55
  • Literally in the first line of the answer: "1) A point $z$ dense in $Z$ must belong to any non-empty open subset of $Z$, so in particular to $Z\cap U$. " – KReiser Jun 05 '22 at 17:56
  • You can also see this directly from the second linked duplicate: there it is proven that a subset $A\subset S$ is dense iff for all nonempty open $U\subset S$, $A\cap U\neq\emptyset$. Taking $A={\eta}$ and $Y=S$, this says that $\eta$ belongs to every nonempty open subset of $Y$. – KReiser Jun 05 '22 at 21:06

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