If $\widehat{M}$ is a free $\widehat{R}$-module of rank $n$ then $M$ has a generating set of $n$ elements as an $R$-module.

Question

With reference to my last question If $\widehat{M}$ is a free $\widehat{R}$-module, then $M$ is a free $R$-module, $R$ is a Zariski ring. I want to ask the following question.

Let $R$ be a Zariski ring with $I$-adic topology, $I \subset J(R)$. Let $M$ be a finitely generated $R$-module such that the $I$-adic completion $\widehat{M}$ is a free $\widehat{R}$-module of rank $n$. Then how can I show that $M$ has a generating set of $n$ elements as an $R$-module.

I need help.

Answer 1

Consider $n$ generators of $\widehat M$, $x_1,...,x_n$.

Let $y_1,...,y_n$ denote their image in $M/IM$. Then, $y_1,...,y_n$ generate $M/IM$.

Indeed, $\widehat M\to M/IM$ is surjective ($M\to \widehat M\to M/IM$ is surjective), so if $z\in M/IM$, let $w$ be any antecedent, $w= \sum_i \lambda_i x_i$ implies that $z =\sum_i \mu_i y_i$, with $\mu_i$ the image of $\lambda_i$ under $\widehat R\to R/I$.

But now since $I\subset J(R)$, Nakayama's lemma tells you that any antecedents of $y_1,...,y_n$ generate $M$ (here, use the assumption that $M$ is finitely generated)