Nakayama's lemma states that given a finitely generated $A$-module $M$, and $J(A)$ the Jacobson radical of $A$, with $I\subseteq J(A)$ some ideal, then if $IM=M$, we have $M=0$.
I've read the proof, and while being relatively simple, it doesn't give much insight on why this lemma should be true, for example – is there some way to see how the fact that $J(A)$ is the intersection of all maximal ideals related to the result?
Any intuition on the conditions and the result would be of great help.
Best Answer
Suppose your module is of finite length. Then you can consider on it the so called radical filtration, which organises the module into an onion-like thing, with elements of the maximal ideal pushing elements of the module farther in from their starting layer to one right below and, moreover, each layer obtained from the one above it in this way.
Now, the condition $\mathfrak m M=M$ tells you that the outermost layer of the module is actually empty: obviously, then, there is not much in the whole thing and $M=0$. We have just discovered Nakayama's lemma!
If your module is arbitrary, exactly the same happens.