Let $R$ be a Zariski ring with $I$-adic topology, $I \subset J(R)$. Let $M$ be a finitely generated $R$-module. Now I have to show that if the $I$-adic completion $\widehat{M}$ is a free $\widehat{R}$-module, then $M$ is a free $R$-module.
I only understand: Since $\widehat{R} \otimes_{R} M \cong \widehat{M}$ as $\widehat{R}$-module and $M$ is finitely generated $R$-module, $\widehat{M}$ is free of finite rank over $\widehat{R}$, say of rank $n$. I've to use the fact that $\widehat{R}$ is faithfully flat extension of $R$ and from last line came up with $\widehat{R} \otimes_{R} M \cong \widehat{R} \otimes_{R}R^{n}$. What should I think next ? I need some help.
Best Answer
We will assume that $\widehat M$ is a free $\widehat R$-module of rank $n$ with generators $\widehat m_1, \dots, \widehat m_n.$ By this post, there exist generators $m_1, \dots, m_n$ of $M$ such that $m_i + IM$ is the image of $\widehat m_i$ under the canonical surjection $\rho : \widehat M \to M / IM.$ Consequently, there exists a surjective $R$-linear map $\pi : R^n \to M$ that sends $(r_1, \dots, r_n) \mapsto r_1 m_1 + \cdots + r_n m_n.$ We have a short exact sequence $$0 \to \ker \pi \to R^n \to M \to 0$$ of $R$-modules. By hypothesis that $R$ is Zariski, we have that $\widehat R$ is flat as an $R$-module, hence $$0 \to \widehat R \otimes_R \ker \pi \to \widehat R \otimes_R R^n \to \widehat R \otimes_R M \to 0$$ is a short exact sequence of $R$-modules. Of course, we have that $\widehat R \otimes_R R^n \cong \widehat R^n$ and $\widehat R \otimes_R M \cong \widehat M,$ hence the map $\widehat R \otimes_R R^n \to \widehat R \otimes_R M$ can be identified with the map $(\widehat r_1, \dots, \widehat r_n) \mapsto \widehat r_1 \widehat m_1 + \cdots + \widehat r_n \widehat m_n.$ Considering that $\widehat M$ is a free $\widehat R$-module, this map is injective, hence $\widehat R \otimes_R \ker \pi = 0.$ But as $\widehat R$ is faithfully flat over $R,$ we have that $\ker \pi = 0.$