Corollary 2

Let $A$ be a Zariski ring and $E$ a finitely generated $A$-module. If $\hat{E}$ is a free $A$-module, $E$ is a free $A$-module.

Let $m$ be a defining ideal of $A$, which is therefore contained in the Jacobson radical of $A$. We apply the criterion of Chapter II, § 3, no. 5, Proposition 5: the canonical mapping $j_E : E \to \hat{E}$ defines a bijection $i_E : E/mE \to \hat{E}/(mE)^{\wedge}$; similarly the canonical mapping $j_A : A \to \hat{A}$ defines a bijection $i_A : A/m \to \hat{A}/\hat{m}$, which is a ring isomorphism. Then $(mE)^{\wedge} = \hat{A}.mE = \hat{m}\hat{E}$ (no. 4, Theorem 3), so that $\hat{E}/(mE)^{\wedge}$ is given an $(\hat{A}/\hat{m})$-module structure and hence (by means of $i_A$) an $(A/m)$-module structure. It is immediate that $i_E$ is $(A/m)$-linear, so that it is an $(A/m)$-module isomorphism. As $\hat{E}/\hat{m}\hat{E}$ is a free $(\hat{A}/\hat{m})$-module, $E/mE$ is a free $(A/m)$-module.

On the other hand, let $\nu : m \otimes_A E \to E$ be the canonical homomorphism; as $(m \otimes_A E) \otimes_A \hat{A}$ is canonically identified with $m \otimes_A \hat{E}$ and $E \otimes_A \hat{A}$ with $\hat{E}$ (no. 4, Theorem 3), the hypothesis that $\hat{E}$ is a free $\hat{A}$-module implies that the homomorphism $\nu \otimes 1 : \hat{m} \otimes_{\hat{A}} \hat{E} \to \hat{E}$ is injective. As $\hat{A}$ is a faithfully flat $A$-module (Proposition 9), we conclude that $\nu$ is injective (Chapter I, § 3, no. 1, Proposition 2) and the conditions for applying the above mentioned criterion are indeed fulfilled.