Lemma 38.13.1. Let R \to S be a ring map of finite presentation. Let N be a finitely presented S-module. Let \mathfrak q \subset S be a prime ideal lying over \mathfrak p \subset R. Set \overline{S} = S \otimes _ R \kappa (\mathfrak p), \overline{\mathfrak q} = \mathfrak q \overline{S}, and \overline{N} = N \otimes _ R \kappa (\mathfrak p). Then we can find a g \in S with g \not\in \mathfrak q such that \overline{g} \in \mathfrak r for all \mathfrak r \in \text{Ass}_{\overline{S}}(\overline{N}) such that \mathfrak r \not\subset \overline{\mathfrak q}.
Proof. Namely, if \text{Ass}_{\overline{S}}(\overline{N}) = \{ \mathfrak r_1, \ldots , \mathfrak r_ n\} (finiteness by Algebra, Lemma 10.63.5), then after renumbering we may assume that
\mathfrak r_1 \subset \overline{\mathfrak q}, \ldots , \mathfrak r_ r \subset \overline{\mathfrak q}, \quad \mathfrak r_{r + 1} \not\subset \overline{\mathfrak q}, \ldots , \mathfrak r_ n \not\subset \overline{\mathfrak q}
Since \overline{\mathfrak q} is a prime ideal we see that the product \mathfrak r_{r + 1} \ldots \mathfrak r_ n is not contained in \overline{\mathfrak q} and hence we can pick an element a of \overline{S} contained in \mathfrak r_{r + 1}, \ldots , \mathfrak r_ n but not in \overline{\mathfrak q}. If there exists g \in S mapping to a, then g works. In general we can find a nonzero element \lambda \in \kappa (\mathfrak p) such that \lambda a is the image of a g \in S. \square
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