Theorem 2

Let $\rho : A \to B$ be a ring homomorphism, E an A-module and F a B-module such that $\rho_*(F)$ is a flat A-module. Then

When A is Noetherian, the two sides of (5) are equal.

Let $p \in \operatorname{Ass}_A(E)$; by definition there exists an exact sequence

Since F is a flat A-module we derive an exact sequence

whence $\operatorname{Ass}_B(F/pF) \subset \operatorname{Ass}_B(E \otimes_A F)$, which proves the inclusion (5).

Suppose now that A is Noetherian and let us prove the opposite inclusion.

We proceed in stages:

(i) Suppose first that E is a finitely generated A-module and that Ass_A(E) is reduced to a single element $p$. By § 1, no. 4, Theorem 1 there exists a composition series $(E_i)_{0 \leq i \leq n}$ of E such that $E_i/E_{i+1}$ is isomorphic to $A/p_i$, where $p_i$ is a prime ideal of A; moreover ($§ 1$, no. 4, Theorem 2 and no. 3, Proposition 7) all the $p_i$ contain $p$. As F is a flat A-module, the $E_i \otimes_A F$ form a composition series of $E \otimes_A F$ and $(E_i \otimes_A F)/(E_{i+1} \otimes_A F)$ is identified with

Then by virtue of $§ 1$, no. 1, Proposition 3 We know that E is annihilated by a power of $p$ (no. 1, Remark); Lemma 1 then shows that, for all $\mathfrak{P} \in \operatorname{Ass}_B(E \otimes_A F)$, $\rho^{-1}(\mathfrak{P}) = p$. As $F/p_i F$ is isomorphic to $(A/p_i) \otimes_A F$, $\rho^{-1}(\mathfrak{P}') = p_i$ for all $\mathfrak{P}' \in \operatorname{Ass}_B(F/p_i F)$ by Lemma 1, whence $\operatorname{Ass}_B(E \otimes_A F) \cap \operatorname{Ass}(F/p_i F) = \varnothing$ if $p_i \neq p$, which proves the theorem in the case considered.

(ii) Suppose only that E is a finitely generated A-module. Let $p_i$ ($1 \leq i \leq n$) be the elements of $\operatorname{Ass}_A(E)$ and let $\{0\} = \bigcap^n Q_i$ be a corresponding reduced primary decomposition (no. 3); E is then isomorphic to a submodule of the direct sum of the $E_i = E/Q_i$ and, as F is a flat A-module, $E \otimes_A F$ is isomorphic to a submodule of the direct sum of B-modules $E_i \otimes_A F$. We deduce ($§ 1$, no. 1, Proposition 3 and Corollary 1 to Proposition 3) that

But $E_i$ is a finitely generated A-module such that $\operatorname{Ass}_A(E_i)$ is reduced to a single element $p_i$ (no. 1, Definition 1). By (i), $\operatorname{Ass}_B(E_i \otimes_A F) = \operatorname{Ass}_B(F/p_i F)$, whence the theorem in this case.

(iii) General case. The B-module $E \otimes_A F$ is the union of the submodules $E' \otimes_A F$, where $E'$ runs through the set of finitely generated submodules of the A-module E. If $\mathfrak{P}$ belongs to $\operatorname{Ass}_B(E \otimes_A F)$, then there exists a finitely generated submodule $E'$ of E such that $\mathfrak{P} \in \operatorname{Ass}_B(E' \otimes_A F)$. By (ii), there exists $p \in \operatorname{Ass}_A(E')$ such that $\mathfrak{P} \in \operatorname{Ass}_B(F/pF)$; as $\operatorname{Ass}_A(E') \subset \operatorname{Ass}_A(E)$, this completes the proof of Theorem 2.