Corollary 5

Let $A$ be a commutative Noetherian ring, $E, F$ twofinitely generated $A$-modules and $u : E \to F$ an $A$-homomorphism. For every maximal ideal $m$ of $A$, let $A(m)$ (resp. $E(m), F(m)$) denote the Hausdorff completion of $A$ (resp. $E, F$) with respect to the $m$-adic topology and $u(m) : E(m) \to F(m)$ the corresponding homomorphism to $u$. For $u$ to be injective (resp. surjective, bijective, zero), it is necessary and sufficient that $u(m)$ be so for every maximal ideal $m \notin A$.

We know that for $u$ to be injective (resp. surjective, bijective, zero), it is necessary and sufficient that $u_m : E_m \to F$, be so for every maximal ideal $m$ of $A$ (Chapter II, § 3, no. 3, Theorem 1). We now note that $A$ is a Noetherian local ring (Chapter II, § 2, no. 4, Corollary 2 to Proposition 10) and hence a Zariski ring and there is a canonical $A$-algebra isomorphism $\hat{A}_m \to A(m)$ (§ 2, no. 13, Proposition 18). On the other hand (beginning of no. 4), there is a commutative diagram

where the horizontal arrows on the left arise from the associativity of the tensor product and the isomorphisms $E_m \to E \otimes_A A_m, F_m \to F \otimes_A A_m$; as $E$ and $F$ are finitely generated $A$-modules, it follows from no. 4, Theorem 3 that the rows of this diagram consist of isomorphisms; thus we are reduced to proving that $u_m$ being injective (resp. surjective, bijective, zero) is equivalent to $u_m \otimes 1$ being so. But this follows from the fact that $\hat{A}_m$ (and hence also $A(m)$) is a faithfully flat $A_m$-module by Proposition 9 (Chapter I, § 3, no. 1, Propositions 1 and 2).