Remark

Suppose that $A$ and $B$ are Noetherian. Let $\mathfrak{P}$ be a prime ideal of $B$ and let $p = \overline{\rho}^{-1}(\mathfrak{P})$; let us write $S = A - p$ and let $k(p) = S^{-1}(A/p)$ be the field of fractions of $A/p$. Since $\mathfrak{P}$ contains $pB$, $\mathfrak{P}/pB$ is a prime ideal of $B/pB$. If $\rho'$ is the composite homomorphism $A \xrightarrow{\rho} B \to B/pB$, we know that $S^{-1}(B/pB)$ is identified with the ring $(\rho'(S))^{-1}(B/pB)$ and $\mathfrak{P}' = S^{-1}(\mathfrak{P}/pB)$ with an ideal of this ring (Chapter II, § 2, no. 2, Proposition 6); as $\mathfrak{P}/pB$ does not meet $\rho'(S)$, $\mathfrak{P}'$ is a prime ideal of $S^{-1}(B/pB)$ (Chapter 11, § 2, no. 5, Proposition 11); moreover there are canonical isomorphisms between $S^{-1}(B/pB)$, $S^{-1}((A/p) \otimes_A B)$ and $(S^{-1}(A/p)) \otimes_A B = k(p) \otimes_A B$; similarly $S^{-1}(F/pF)$ is canonically identified with $k(p) \otimes_A F$. This being so, under the hypotheses of Theorem 2, in order that $\mathfrak{P} \in \operatorname{Ass}_B(E \otimes_A F)$, it is necessary and sufficient that $p \in \operatorname{Ass}_A(E)$ and

For by Theorem 2 and its Corollary 1, it amounts to verifying that the conditions are equivalent; but, as $B$ is Noetherian, this follows from § 1, no. 2, Corollary to Proposition 5 and the above identifications.