Proposition 11

Suppose that $A$ and $B$ are Noetherian and that $B$ is a flat $A$-module. Let $E$ be an $A$-module and $E'$ a submodule of $E$ such that, for every ideal $p \in \operatorname{Ass}_A(E/E')$, $pB$ is a prime ideal of $B$ or equal to $B$. Let $E' = \bigcap_{p \in \operatorname{Ass}(E/E')} Q(p)$ be a reduced primary decomposition of $E'$ in $E$, $Q(p)$ being $p$-primary for all $p \in \operatorname{Ass}(E/E')$. (i) If $p \in \operatorname{Ass}(E/E')$ and $pB = B$, then $Q(p) \otimes_A B = E \otimes_A B$. (ii) If $p \in \operatorname{Ass}(E/E')$ and $pB$ is prime, $Q(p)$ is $pB$*-primary* in $E \otimes_A B$. (iii) If $\Phi$ is the set $\mathfrak{p} \in \operatorname{Ass}(E/E')$ such that $pB$ is prime, then

and this relation is a reduced primary decomposition of $E' \otimes_A B$ in $E \otimes_A B$.

If $pB = B$, Theorem 2 applied to $E/\mathbf{Q}(p)$ and $B$ shows that

and, as $B$ is Noetherian and is a flat $A$-module, we conclude (§ 1, no. 1, Corollary 1 to Proposition 2) that $\mathbf{Q}(p) \otimes_A B = E \otimes_A B$. Assertion (ii) follows from Corollary 2 to Theorem 2, taking $\mathfrak{P} = pB$. Finally the relation $E' \otimes_A B = \bigcap_{p \in \Phi} (\mathbf{Q}(p) \otimes_A B)$ follows from the fact that $B$ is a flat $A$-module (Chapter I, § 2, no. 6, Proposition 6); as $p = \overline{\rho}^{-1}(pB)$ for $p \in \Phi$ (Lemma 1), $pB \neq p'B$ for two distinct ideals $p, p'$ of the set $\Phi$; on the other hand, by Theorem 2; we conclude from no. 3, Proposition 4 that is a reduced primary decomposition.