Proposition 15

*Let $A, B$ be two rings, $\phi : A \to B$ a homomorphism and $N$ a $B$-module. The following properties are equivalent: (a) $N$ is a flat $A$-module. (b) For every maximal ideal $n$ of $B$, $N_n$ is a flat $A$-module. (c) For every maximal ideal $n$ of $B$, if we write $m = \phi(n)$, $N_n$ is a flat $A_m$-module.

For all $a \notin m$, the homothety of $N_n$ induced by $a$ is bijective, hence $N_n$ is canonically identified with $(N_n)_m$ and the equivalence of (b) and (c) follows from the

Remark following Proposition 13; the fact that (a) implies (b) is a special case of Proposition 14. It remains to prove that (b) implies (a), that is, that, if (b) holds, for every injective A-module homomorphism $u : M \to M'$, the homomorphism $v = 1 @ u : N \otimes_A M \to N @^* M'$ is injective. Now, $v$ is also a B-module homomorphism and, for it to be injective, it is necessary and sufficient that $v_n : (N \otimes_A M)_n \to (N \otimes_A M')_n$ be so for every maximal ideal $n$ of $B$ (no. 3, Theorem 1). As

$v_n$ is just the homomorphism $1 \otimes u : N_n \otimes_A M \to N_n \otimes_A M'$, which is injective since $N_n$ is a flat A-module by hypothesis.