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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!

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If $IM' = M'$ is indeed true, wouldn't you just (borrowing notation from later) use $a'(1 - i_r)$? It's $1 \bmod I$, the $a'$ kills off the first $r - 1$ generators, and $(1 - i_r)$ knocks $x_r$ into $M'$ where it can be annihilated by $a'$. – Dylan Moreland May 6 '12 at 22:46
Hello @Dylan! Indeed I suspected this element to do what I wanted in my notes, but I was too stupid to see it was $1$ in $A/I$... but of course it is, thank you! – InvisiblePanda May 7 '12 at 7:22

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