Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have already asked this at math.stackexchange, but since no one answered there after my edit, I decided to try here, although it might be a non-research level question.

The following version of Nakayama's lemma is from Matsumura's Commutative Ring Theory:

Let $M$ be a finitely generated $A$-module, $I\subseteq A$ an ideal s.t. $IM=M$. Then there exists an $a\in A$ with $a\equiv 1\pmod{I}$ and such that $aM=0$. In particular, if $I\subseteq\operatorname{rad}(A)$ we have $M=0$.

After the proof of this via a generalized Cayley-Hamilton, he mentions that the result 'can easily be proved [..] by induction on the number of generators of $M$.' I wonder: how? I tried doing it similarly to the inductive proof of the 'in particular' part, but it didn't work out for me (see MSE for more information on what I think I was doing wrong).

Wouldn't I need to be able to find an $N\subseteq M$, $IN=N$, with fewer generators than $M$ in a somewhat obvious way to use the induction hypothesis? How could I do this?

Thanks in advance!

share|improve this question
You need to do it the other way around. Let $N$ be generated by the first generator of $M$, then $M/N$ needs fewer generators than $M$ and satisfies $I.(M/N)=M/N$, and you can take it from there. –  Neil Strickland May 21 '12 at 8:20
@Neil: What should we do with some $a \equiv 1$ mod $I$ satisfying $a M \subseteq N$? @Rand: I think that Matsumura only refers to the particular case that $I \subseteq jac(R)$, this can be done by induction. The general form is equivalent to Cayley-Hamilton and no reasonable induction is possible. –  Martin Brandenburg May 21 '12 at 8:37
I believe Martin is right and said so here: mathoverflow.net/questions/41836/… –  Phillip Williams May 21 '12 at 22:09
Hey InvisiblePanda, I saw your name in chat.SE and couldn't resist contacting you! I think we have a book series in common :-) The name of this room is pretty funny. –  rand al'thor Dec 29 '14 at 21:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.