Some time ago I used some result saying that if $M$ is a finitely generated $R$ module, where $R$ is local noetherian (I don't know if that's necessary), then for $N$ a submodule of $M$ and $I$ a nilpotent ideal, then : $I.M+N=M\implies M=N$.
I wrote that is supposed to be a corollary of Nakayama, $m.M=M\implies M=0$
however I have trouble showing it.
My attempts for proving it didn't work, trying to use the nilpotency of $I$ and trying to get $m(M-N)=M-N$ Can I get some advice for this?
Apply Nakayama's lemma to the quotient module $M/N$. A general element in this module has the form $m+N$. By assumption we can write $m=\sum\limits_{i=1}^r a_im_i+n$ where $m_i\in M, a_i\in I, n\in N$. Then:
$m+N=\sum\limits_{i=1}^r a_im_i+n+N=\sum\limits_{i=1}^r a_im_i+N=\sum\limits_{i=1}^r a_i(m_i+N)\in I\cdot(M/N)$
And so $I\cdot(M/N)=M/N$. Since $I$ is nilpotent, it is contained in the Jaconson radical of $R$, and so by Nakayama $M/N=0$, i.e $M=N$.