Proposition 4

Let $A$ and $B$ be two commutative Noetherian rings, $h : A \to B$ a ring homomorphism, $3$ an ideal of $A$ and $2$ an ideal of $B$ containing $\mathfrak{J}B$ and contained in the Jacobson radical of $B$. Let $\hat{A}$ be the Hausdorff completion of $A$ with respect to the $3$-adic topology and $\hat{B}$ the Hausdorff completion of $B$ with respect to the $2$-adic topology; $h$ is continuous with these topologies and $h : \hat{A} \to \hat{B}$ therefore makes $\mathbf{8}$ into an $\hat{A}$-algebra. Let $M$ be a finitely generated $B$-module and $\hat{M}$ its Hausdorff completion with respect to the $2$-adic topology; the following properties are equivalent: (a) $M$ is a flat $A$-module. (b) $\hat{M}$ is a flat $A$-module. (c) $\hat{M}$ is a flat $\hat{A}$-module.

As $B$ with the $2$-adic topology is a Zariski ring, $\hat{B}$ is a faithfully flat $B$-module ($§ 3$, no. 5, Proposition 9) and $M$ is canonically isomorphic to $M \otimes_B \hat{B}$ ($§ 3$, no. 4, Theorem 3); it is immediately verified that this canonical isomorphism is an isomorphism of the $A$-module structure on $\hat{M}$ onto the $A$-module structure on $M \otimes_B \hat{B}$ derived from that on $M$. Applying Proposition 4 of Chapter I, $§ 3$, no. 2 with $R$ replaced by $B$, $S$ by $A$, $E$ by $\hat{B}$, $F$ by $M$, we see that for $M$ to be a flat $A$-module, it is necessary and sufficient that $\hat{M}$ be a flat $A$-module. Moreover, $\hat{M}$ is a finitely generated $\hat{B}$-module and $\mathfrak{J} \hat{B}$ is contained in $\mathfrak{Q} = \mathfrak{Q} \hat{B}$ and hence in the Jacobson radical of $\hat{B}$ ($§ 3$, no. 4, Proposition 8); therefore $\hat{M}$ is an ideally Hausdorff $A$-module with respect to $\mathfrak{J} \hat{A}$ (Proposition 2). Conditions (b) and (c) are therefore equivalent by the Corollary to Proposition 3.