Welcome to MathOverflow

Question

3

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!

edited May 21 2012 at 17:49

David White
4,858●1●20●27

asked May 21 2012 at 7:48

Rand al'Thor
115●4

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 2012 at 8:20

1

@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 2012 at 8:37

1

I believe Martin is right and said so here: mathoverflow.net/questions/41836/… – Phillip Williams May 21 2012 at 22:09

Inductive proof of a version of Nakayama’s lemma

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.