I know the statement of Nakayama's Lemma
Let $R$ be a local ring with maximal ideal $M$ and $I$ be a finite generated ideal of $I$ such that $MI=I$ then $I=(0)$
In this case I know the demonstration, my question is what happens if we remove the hypothesis that $I$ is finitely generated? the Lemma is still fulfilled and in this case how would the demonstration be? I have been trying to emulate the one of the case $I$ f.g but I have not been successful, any comment or suggestion I would be very grateful.
The ring $k[x^{1/2},x^{1/4},x^{1/8},...]/(x)$ is a local ring that has a nil, idempotent maximal ideal generated by the powers $x^{1/2^n}$. Obviously its maximal ideal isn't zero, so we have an example here of $I=M$ not finitely generated, and $M^2=M\neq \{0\}$.