# Why if the maximal ideal is nilpotent then this module is free?

On matsumura is proved the following proposition:

let $(R,m)$ be a local ring and $M$ a flat $R$-module. If $x_1,\ldots,x_n\in M$ are such that their images $\bar{x}_1,\ldots,\bar{x}_n$ in $\bar{M}=M/mM$ are linearly independet over $R/m$ then $x_1,\ldots,x_n$ are linearly independent over $R$.

Then it says if $M$ is finitely generated then this implies $M$ free, and I see this. But It also says if $m$ is nilpotent then $M$ is free, I don't see this, could you explain it to me, please?

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Let $\{ x_i\}_i\subseteq M$ be a lifting to $M$ of a basis of $M/mM$. It is free by the proposition you quote. So it is enough to show it generates $M$. This is in fact a form of Nakayama's Lemma. Let $N$ be the submodule of $M$ generated by the $x_i$'s. Then $$M = N + mM = N + m(N+mM)=N+m^2M.$$ Now start again with $$M=N+m^2(N+m^2M)=N+m^4M.$$ We see by induction that $M=N+m^rM$ for all $r\ge 1$. As $m$ is nilpotent, $m^r=0$ for some $r\ge 1$. So $M=N$.