The Stacks project

Proof. The meaning of the statement is as follows. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q' \subset S$ be a prime $\mathfrak q' \in \text{Supp}(N)$ Then there exists a prime $\mathfrak q \subset \mathfrak q'$, $\mathfrak q \in \text{Supp}(N)$ lying over $\mathfrak p$. As $N$ is flat over $R$ we see that $N_{\mathfrak q'}$ is flat over $R_{\mathfrak p'}$, see Lemma 10.39.18. As $N_{\mathfrak q'}$ is finite over $S_{\mathfrak q'}$ and not zero since $\mathfrak q' \in \text{Supp}(N)$ we see that $N_{\mathfrak q'} \otimes _{S_{\mathfrak q'}} \kappa (\mathfrak q')$ is nonzero by Nakayama's Lemma 10.20.1. Thus $N_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p')$ is also not zero. We conclude from Lemma 10.39.15 that $N_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p)$ is nonzero. Let $J \subset S_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p)$ be the annihilator of the finite nonzero module $N_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p)$. Since $J$ is a proper ideal we can choose a prime $\mathfrak q \subset S$ which corresponds to a prime of $S_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p)/J$. This prime is in the support of $N$, lies over $\mathfrak p$, and is contained in $\mathfrak q'$ as desired. $\square$