Corollary

Suppose that $pB$ is prime for all $p \in \operatorname{Ass}_A(E/E')$. Then, if $p_1, \ldots, p_n$ are the minimal elements of $\operatorname{Ass}_A(E/E')$, the $p_iB$ are minimal elements of

It follows from Proposition 11 that in this case $p_iB \neq p_jB$ for $i \# j$.

Examples (1) Let us take $B = S^{-1}A$, where $S$ is a multiplicative subset of $A$; if $A$ is Noetherian, the hypotheses of Proposition 11 are satisfied and we recover a part of Proposition 6 of no. 4. (2) Let $A$ be a Noetherian ring, $m$ an ideal of $A$ and $B$ the Hausdorff completion of $A$ with respect to the $m$-adic topology; then $B$ is a flat $A$-module and Theorem 2 may be applied with $F = B$; but in general the hypotheses of Proposition 11 are not satisfied for the prime ideals of $A$ (Chapter III, § 2, Exercise 15 (b)). (3) Let $A$ be a Noetherian ring and $B$ the polynomial algebra $A[X_1, \ldots, X_n]$; $B$ is Noetherian and is a free $A$-module and therefore flat. Also, if $p$ is a prime ideal of $A$, $B/pB$ is isomorphic to $(A/p)[X_1, \ldots, X_n]$, which is an integral domain, and hence $pB$ is prime; the hypotheses of Proposition 11 are therefore satisfied for every $A$-module $E$ and every submodule $E'$ of $E$. (4) Let $A$ be a finitely generated algebra over a field $k$, $K$ an extension of $k$ and $B = A \otimes_k K$ the algebra over $K$ obtained by extension of scalars; $A$ and $B$ are Noetherian and $B$ is a free $A$-module and hence Theorem 2 may be applied to $F = B$. In certain cases (for example if $k$ is algebraically closed) it can be shown that for every prime ideal $p$ of $A$, $pB$ is prime or equal to $B$; we shall return later to this example.