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I am reading Hartshorne algebraic geometry. On page 71, the paragraph before Proposition 2.2, it is said that "if $V(\mathfrak{a})$ is a closed set" and $D(f) \cap V(\mathfrak{a}) = \emptyset$. But it seems that $V(\mathfrak{a})$ is always closed. In this paragraph, it is proved that for each $\mathfrak{p} \not\in V(\mathfrak{a})$, there is an $f$ such that $\mathfrak{p} \in D(f)$. Could we conclude that $D(f) \cap V(\mathfrak{a}) = \emptyset$? Thank you very much.

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Hartshorne wants to prove there that every open set is a union of basic open subsets (i.e. those of the form $D(f)$). In other words, every closed subset is an intersection of "basic closed" subsets, i.e. those of the form $V(f)$. But this immediately follows from 2.1. (b). I don't know why Hartshorne repeats the argument. Anyway, here is what he does: If $E$ is a closed subset, we can write $E=V(\mathfrak{a})$ for some ideal $\mathfrak{a}$ (this is what is meant by "Let $V(\mathfrak{a})$ be a closed set"). If $\mathfrak{p} \in V(\mathfrak{a})^c$, i.e. $\mathfrak{a} \not\subseteq \mathfrak{p}$, there is some $f \in \mathfrak{a} \setminus \mathfrak{p}$. Then $\mathfrak{p} \in D(f) \subseteq V(\mathfrak{a})^c$ since $V(\mathfrak{a}) \subseteq V(f)$. This shows $E^c = V(\mathfrak{a})^c = \cup_{f \in \mathfrak{a}} D(f)$, so that in fact every open subset is a union of basic open subsets.

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