I am trying to follow the answers to this question about an intersection of affine open sets being covered by affine open sets. However, I can't seem to figure out why given $\operatorname{Spec} A \cap \operatorname{Spec} B$, it is possible to find a distinguished open set $D_{A}(f) \subseteq \operatorname{Spec} A \cap \operatorname{Spec} B$. How would one show this?
2026-03-26 08:14:51.1774512891
Finding a distinguished open set in the intersection of two $\operatorname{Spec}$s
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Note that $\operatorname{Spec} A$ is open, as is $\operatorname{Spec} B$, so their intersection $\operatorname{Spec} A \cap \operatorname{Spec} B$ is again open. This means that the complement of $\operatorname{Spec} A \cap \operatorname{Spec} B$ inside either $\operatorname{Spec} A$ or $\operatorname{Spec} B$ is closed in that particular affine patch, and therefore one may pick a function $f$ which is zero on it and take $D_A(f)$.