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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