Let $A$ be a local ring with nilpotent maximal ideal $\mathfrak{m}$ (i.e., some power of $\mathfrak{m}$ vanishes), and $M$ an $A$-module (not necessarily finitely generated). Let $\bar{S}\subset M/\mathfrak{m}M$ be a set of generators and $S$ a set of representatives of $\bar{S}$ in $M$. Then is it true that $S$ is a set of generators of $M$? This is a common form of Nakayama's lemma with the assumption of finite generation of $M$ replacing the nilpotence of $\mathfrak{m}$. A passage in Matsumura's book "Commutative Ring Theory" (see Theorem 7.10) seems to imply this result, and I can't figure out why.



Non-finite version of Nakayama's lemma?
Dear Kwan,


Let $N$ be the submodule of $M$ generated by $S$. Then by assumption 
$M = N +\mathfrak m M.$ Iterating this, we find that
$$M = N + \mathfrak m (N + \mathfrak m M) = N + \mathfrak m^2 M = \cdots
= N + \mathfrak m^n M$$
for any $n > 0.$ If we take $n$ large enough then $\mathfrak m^n = 0$ (by hypothesis).
Thus $M = N,$ as desired. 


P.S. I've found this to be quite a useful fact!


