Lemma 38.25.10. Let $A$ be a local domain which is not a field. Let $S$ be a set of finitely generated ideals of $A$. Assume that $S$ is closed under products and such that $\bigcup _{I \in S} V(I)$ is the complement of the generic point of $\mathop{\mathrm{Spec}}(A)$. Then $\bigcap _{I \in S} I = (0)$.

Proof. Since $\mathfrak m_ A \subset A$ is not the generic point of $\mathop{\mathrm{Spec}}(A)$ we see that $I \subset \mathfrak m_ A$ for at least one $I \in S$. Hence $\bigcap _{I \in S} I \subset \mathfrak m_ A$. Let $f \in \mathfrak m_ A$ be nonzero. Then $V(f) \subset \bigcup _{I \in S} V(I)$. Since the constructible topology on $V(f)$ is quasi-compact (Topology, Lemma 5.23.2 and Algebra, Lemma 10.26.2) we find that $V(f) \subset V(I_1) \cup \ldots \cup V(I_ n)$ for some $I_ j \in S$. Because $I_1 \ldots I_ n \in S$ we see that $V(f) \subset V(I)$ for some $I$. As $I$ is finitely generated this implies that $I^ m \subset (f)$ for some $m$ and since $S$ is closed under products we see that $I \subset (f^2)$ for some $I \in S$. Then it is not possible to have $f \in I$. $\square$

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