Proposition 3

Let $A$ be a commutative ring, $B$ a commutative $A$-algebra, $\mathfrak{g}$ an ideal of $A$ and $M$ a $B$-module. Suppose that $B$ is a Noetherian ring and a flat $A$-module and that $M$ is ideally Hausdorff with respect to $\mathfrak{g} B$. The following conditions are equivalent: (a) $M$ is a flat $B$-module. (b) $M$ is a flat $A$-module and $M/\mathfrak{g} M = M/(\mathfrak{g} B) M$ is a flat $(B/\mathfrak{g} B)$-module. If further the canonical homomorphism $A/\mathfrak{g} \to B/\mathfrak{g} B$ is bijective, conditions (a) and (b) are also equivalent to: (c) $M$ is a flat $A$-module.

Condition (a) implies (b) by Chapter I, §2, no. 7, Corollaries 2 and 3 to Proposition 8 and the fact that $M/\mathfrak{g} M$ is isomorphic to $M \otimes_B (B/\mathfrak{g} B)$. Suppose condition (b) holds; to show that $M$ is a flat $B$-module, we shall apply Theorem 1 of no. 2 with $A$ replaced by $B$ and $\mathfrak{g}$ by $\mathfrak{g} B$. It will therefore be sufficient to show that the canonical mapping $f : \mathfrak{J}B \otimes_B M \to \mathfrak{J}M$ is injective. Let $f_1$ be the canonical mapping $\mathfrak{J} \otimes_A B \to \mathfrak{J}B$ and $f_2$ the canonical isomorphism $\mathfrak{J} \otimes_A M \to (\mathfrak{J} \otimes_A B) \otimes_B M ; f \circ (f_1 \circ 1_M) \circ f_2$ is the canonical mapping $f' : \mathfrak{J} \otimes_A M \to \mathfrak{J}M$, as is easily verified. Now $f'$ is an isomorphism since $M$ is a flat $A$-module, whilst $f_1$ is an isomorphism because $B$ is flat over $A ; f$ is then an isomorphism.

Let $p : A/\mathfrak{J} \to B/\mathfrak{J}B$ be the canonical homomorphism; the $(A/\mathfrak{J})$-module structure on $M/\mathfrak{J}M$ derived by means of $p$ from its $(B/\mathfrak{J}B)$-module structure is isomorphic to that on $M \otimes_A (A/\mathfrak{J})$. Then it follows that, if $M$ is a flat $A$-module, $M/\mathfrak{J}M$ is a flat $(A/\mathfrak{J})$-module and hence also a flat $(B/\mathfrak{J}B)$-module if $p$ is an isomorphism; we have thus proved that (c) $\Rightarrow$ (b) in that case.