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.
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.
Here's something that might or might not make sense to you. You know that every ideal $I$ of a commutative ring $R$ gives rise to an $R$-module $R/I$; these are precisely the $R$-modules on one generator. The $R$-modules of the form $R/m$ where $m$ is maximal are special among these. In this language, the elements of the Jacobson radical are precisely the ones that act trivially on all $R$-modules of the form $R/m$. It follows, for example, that if $I$ is in the Jacobson radical, then we cannot have $IM = M$ for any module of the form $R/m^k$.
Nakayama's lemma asserts that a similar statement is true for all finitely generated $R$-modules. This should be reasonable if you are aware of, for example, the classification of finitely generated $R$-modules when $R$ is a PID. The Wikipedia describes a geometric interpretation of this, but I'm not familiar enough with it to say more.
This is probably not the answer you're looking for, but here's another proof:
Suppose $M \neq 0$ and let $u_1,\ldots,u_n$ be a minimal set of generators for $M$. Then $u_n \in IM$ so we may write $u_n=a_1u_1+\ldots+a_nu_n$ with $a_i \in I$. Hence $(1-a_n)u_n=a_1u_1+\ldots+a_{n-1}u_{n-1}$. Since $I$ was contained in the Jacobson radical, it follows that $1-a_n$ is a unit. Hence $u_n$ belongs to the submodule generated by the $u_1,\ldots,u_{n-1}$, contradiction.
There is a more primitive and natural version of Nakayama's lemma:
Let $R$ be a commutative ring, and $M$ be a finitely generated $R$-module and $I\subset R$ be an ideal. If $IM=M$ then $\exists f\in I,\forall m\in M,f\cdot m=m$.
The proof is a bit harder than the above version, but this pretty much says "If $IM=M$ then one of the elements of $I$ stablizes all elements of $M$." And I think this is much more natural than the above version. Assuming this version, then we can see that the condition of "Jacobson ideal" is only used to show that "$1-f$ is a unit" which leads to "$M=0$".
For completion, I give a proof of the above version here:
Proof: Let $m_1,...,m_n$ be a set of generators of $M$. Since $IM=M$, we have a matrix $A'=(a_{ij})\in \textrm{M}_{n\times n}(R)$ such that $A'\underline{m}=\underline{m}$ where $\underline{m}=(m_1,...,m_n)^t$ and $a_{ij}\in I,\forall i,j\leq n$. Now let $A=\textrm{Id}-A'$ so we have $A\underline{m}=\underline{0}$.
Note that $\det A\in R$ is well-defined and there exists a matrix $B\in \textrm{M}_{n\times n}(R)$ s.t. $BA=AB=\det A\cdot \textrm{Id}$, the proof is in Corollary 9.161 of Advanced Modern Algebra by Joseph J. Rotman. Also $\det$ commutes with the quotient map $R\rightarrow R/I$, with $A\equiv \textrm{Id} \ (\textrm{mod }I)$ we can deduce that $\det A\in 1+I$, so $\exists f\in I$, s.t. $\det A=1-f$.
To show $f$ stablizes $M$ it suffices to show that $f\cdot m_i=m_i,\forall i$. $$\underline{0}=BA\underline{m}=\det A\cdot \underline{m}=(1-f)\cdot \underline{m}$$ So $$f\cdot \underline{m}=\underline{m}$$ The result follows. Q.E.D
I am quite fond of the following proof of Nakayama's lemma, which is sketched (very briefly) on the first page of Serre's Local Algebra. It uses a couple of standard facts about simple modules and cyclic modules which are useful in their own right. What I like about the proof is that it singles out what makes the zero module special, namely that is the only finitely generated module which doesn't have a maximal submodule.
Fix a commutative ring $A$. We recall that an $A$-module $M$ is cyclic if it is generated by a single element; and $M$ is simple if it has exactly two submodules, namely $0$ and $M$. Thus, the zero module is not simple, in analogy with the convention that $1$ is not prime.
We begin with a couple of lemmas which characterise cyclic modules as the quotients of $A$, and simple modules as quotients of $A$ by maximal ideals. The reader familiar with these characterisations may wish to skip this section.
Claim. If $M$ is a simple $A$-module and $m$ is a nonzero element of $M$, then $\langle m\rangle=M$. In particular, $M$ is cyclic.
Proof. Since $\langle m\rangle\neq0$, the simplicity of $M$ implies $\langle m\rangle = M$.
Claim. If $I\subseteq A$ is an ideal, the quotient $A/I$ is a cyclic $A$-module.
Proof. The module $A/I$ is generated by $1$.
Claim. If $M$ is a cyclic $A$-module generated by $m$, then $M$ is isomorphic to the quotient of $A$ by the annihilator $I$ of $m$.
Proof. The linear map $A\to M, \ a\mapsto am$ induces the desired isomorphism.
Claim. If $\mathfrak m\subset A$ is a maximal ideal, the quotient $A/\mathfrak m$ is simple.
Proof. Recall that if $N$ is a submodule of an $A$-module $M$, there is a one-to-one correspondence between submodules of $M$ containing $N$, and submodules of $M/N$. The claim follows immediately from this theorem.
Claim. If $M$ is a simple $A$-module, then $M\cong A/I$ for some maximal ideal $I$.
Proof. Since $M$ is simple and therefore cyclic, there is an ideal $I$ such that $M\cong A/I$. But if $I$ is not maximal, say $I\subset J\subset A$ for some ideal $J$, then the image of $J$ in $A/I$ is a proper nonzero submodule of $A/I$.
With the above theorems in hand, we can begin to understand the structure of finitely generated modules. This is the key to Nakayama.
Claim. A finitely generated nonzero module $M$ has a maximal submodule. That is, if $\mathcal A$ is the collection of proper submodules of $M$, ordered by inclusion, then $\mathcal A$ has a maximal element.
Proof. We can adapt the proof using Zorn's lemma that every nonzero ring has a maximal ideal. Suppose $\mathcal B$ is a chain in $\mathcal A$. If $\mathcal B$ is empty, then since $M$ is nonzero, the zero module is an upper bound of $\mathcal B$. Otherwise, $\mathcal B$ is nonempty, and $N=\bigcup_{P\in\mathcal B}P$ is a submodule of $M$ such that $P\subseteq N$ for every $P\in\mathcal B$. To conclude that $N$ is an upper bound of $\mathcal B$, it thus suffices to check that $N$ is a proper submodule of $M$. Suppose for contradiction that $N=M$, and let $m_1,\dots,m_r$ be a family of generators of $M$. For each $i\in \{1,\dots,r\}$, choose a module $P_i\in\mathcal B$ such that $m_i\in P_i$. Since $\{P_1,\dots,P_r\}$ is a finite subset of the chain $\mathcal B$, it has a maximum element $P$. But then $m_1,\dots,m_r \in P$, so $P=M$, contrary to the fact that $P\in\mathcal B$. We conclude that $\mathcal A$ is inductively ordered, and hence by Zorn it has a maximal element.
Claim. If $N$ is a maximal submodule of an $A$-module $M$, then the quotient $M/N$ is simple.
Proof. This is immediate from the correspondence theorem for modules.
Claim. If $M$ is a simple $A$-module, then the Jacoboson radical $J$ of $A$ annihilates $M$.
Proof. We may assume that $M=A/\mathfrak m$ for some maximal ideal $\mathfrak m\subset A$. Then, $\mathfrak mM=0$, hence $JM=0$.
Theorem (Nakayama's lemma). Suppose $M$ is a finitely generated $A$-module, and $J$ is the Jacobson radical of $A$. Then, $M=JM$ implies $M=0$.
Proof. We prove the contrapositive, i.e. that if $M\neq0$ then $M\neq JM$. Let $N$ be a maximal submodule of $M$. The quotient $M/N$ is simple, and hence annihilated by $J$. It follows that $JM\subseteq N\subset M$, and the proof of Nakayama's lemma is complete.
