In Matsumura's 'Commutative Ring Theory', the following version of NAK is shown:
If $M$ is a finitely generated $A$-module, $I\subseteq A$ an ideal s.t. $IM=M$, then there exists an $a\in A$ with $a\equiv 1\pmod{I}$ s.t. $aM=0$. In particular, if $I\subseteq\operatorname{rad}(A)$, $M=0$.
The proof uses the generalized Cayley-Hamilton theorem. But after NAK, Matsumura mentions that the result can easily be proven by using induction. I wonder, how does that proof work? If I try to do the induction step myself, I can't seem to use the induction hypothesis in the right way. I know the proof of the 'in particular' part by induction, but there it is crucial that $I$ is contained in the Jacobson radical.
I thought an inductive proof of the above result should somehow be related to the proof of the 'in particular' part. So let $M=\langle x_1,...,x_r\rangle_A$. Let $M'=\langle x_1,...,x_{r-1}\rangle_A$. If I'm not wrong, $IM=M$ should imply that $IM'=M'$, and I can use the induction hypothesis to get an $a'\in A$ with $a'\equiv1\pmod{I}$ s.t. $a'M'=0$. If we write $x_r=\sum_{\nu=1}^r i_\nu x_\nu$ with $i_\nu\in I$, take $a$ to be $a'(1-i_r)$ (thanks to Dylan Moreland's comment).
EDIT: But I have had a supposedly wrong implication there regarding $IM'=M'$, this maybe does not hold in general. But how to "easily prove this result by induction"? Does anyone have an idea? Because I can't seem to see how to (canonically) get a module $N\subseteq M$ with fewer generators and $IN=N$, but maybe it's possible.
Thanks for your help in advance!