Lemma 1

Let A be a Noetherian ring, $p$ a prime ideal of A, E an A-module whose annihilator contains a power of $p$ and such that $\operatorname{Ass}(E) = \{p\}$ and F a B-module such that $\rho_*(F)$ is a flat A-module. The condition $\mathfrak{P} \in \operatorname{Ass}_B(E \otimes_A F)$ then implies $\rho^{-1}(\mathfrak{P}) = p$.

If n is such that $p^nE = 0$, then $p^nB \subset \operatorname{Ann}(E \otimes_A F)$, whence $p^nB \subset \mathfrak{P}$, which implies $p^n \subset \rho_1(\mathfrak{P})$ and therefore $p \subset \rho_1(\mathfrak{P})$ since $\rho_1(\mathfrak{P})$ is prime. Moreover, if $a \in A - p$, the homothety h with ratio a on E is injective ($§ 1$, no. 1, Corollary 2 to Proposition 2); as $h \otimes 1_F$ is the homothety $h'$ with ratio $\rho(a)$ on $E \otimes_A F$ and $\rho_*(F)$ is flat, $h'$ is injective (Chapter I, § 2, no. 2, Definition 1); this proves that $\rho(a) \notin \mathfrak{P}$, whence $\rho^{-1}(\mathfrak{P}) = p$.