Nakayama's Lemma

Lemma
Let $A$ be a commutative ring with unity.

Let $M$ be a finitely generated $A$-module.

Let $\operatorname{Jac} \left({A}\right)$ be the Jacobson radical of $A$.

Let $\operatorname{Jac} \left({A}\right) M = M$.

Then:
 * $M = 0$

Corollary 1
Let $A$ be a commutative ring with unity.

Let $M$ be a finitely generated $A$-module.

Let there exist a submodule $N \subseteq M$ such that:


 * $M = N + \operatorname{Jac} \left({A}\right) M$

Then $M = N$.

Corollary 2
Let $A$ be a commutative ring with unity.

Let $M$ be a finitely generated $A$-module.

Let:
 * $m_1 + \operatorname{Jac} \left({A}\right) M, \dotsc, m_n + \operatorname{Jac} \left({A}\right) M$

generate $M / \operatorname{Jac} \left({A}\right) M$ over $A / \operatorname{Jac} \left({A}\right)$.

Then $m_1,\ldots, m_n$ generate $M$ over $A$.

Proof of Nakayama's Lemma
We induct on the number of generators of $M$.

Base Case:

Let $M$ have a single generator $m_1 \in M$.

Then $\operatorname{Jac} \left({A}\right) m_1 = M$.

So:
 * $m_1 \in \operatorname{Jac} \left({A}\right) m_1$

That is:
 * $m_1 = a m_1$

for some $a \in \operatorname{Jac} \left({A}\right)$.

By Characterisation of Jacobson Radical, $1 - a$ is a unit in $A$.

So:
 * $\left({1 - a}\right)^{-1} \left({1 - a}\right) m = 0$

thus $m = 0$.

Inductive Step:

Suppose that $M$ is generated by $n$ elements:


 * $M = A m_1 + \dotsb + A m_n$

for some $m_1, \dotsc, m_n \in M$.

Then we have:


 * $M = \operatorname{Jac} \left({A}\right) M = \operatorname{Jac} \left({A}\right) m_1 + \dotsb + \operatorname{Jac} \left({A}\right) m_n $

Thus for some $a_1, \dotsc, a_n \in \operatorname{Jac} \left({A}\right)$:


 * $ m_1 = a_1 m_1 + \dotsb + a_n m_n $

Then:


 * $\left({1 - a_1}\right) m_1 = a_2 m_2 + \dotsb + a_n m_n$

By Characterisation of Jacobson Radical, $1 - a$ is a unit in $A$.

Multiplying both sides by $\left({1 - a_1}\right)^{-1}$ gives:


 * $ m_1 = \left({1 - a_1}\right)^{-1} a_2 m_2 + \dotsb + \left({1 - a_1}\right)^{-1} a_n m_n$

so we have $m_1 \in A m_2 + \cdots + A m_n$.

Therefore $M$ has $n - 1$ generators $m_2, \dotsc, m_n$.

By the induction hypothesis:
 * $\operatorname{Jac} \left({A}\right) M = M \implies M = 0$

Proof of Corollary 1
If $M = N + \operatorname{Jac} \left({A}\right) M$ then:


 * $\operatorname{Jac} \left({A}\right) \left({M / N}\right) = M/N$

so by Nakayama's Lemma:
 * $M/N = 0$

and so:
 * $M = N$

Proof of Corollary 2
Let $N$ be the submodule of $M$ generated by $m_1, \dotsc, m_n$.

Then:
 * $M = N + \operatorname{Jac} \left({A}\right) M$

Hence by Corollary 1:
 * $M = N$

Proof using Cayley-Hamilton
Take $\phi$ to be the identity in Cayley-Hamilton Theorem for Finitely Generated Modules, and $a = a_0 + \cdots + a_{n-1}$.

Then $a \in \mathfrak a$ and $1 + a \in \operatorname{Ann}_A \left({M}\right)$ as required.

For the second part, notice that if $\mathfrak a \subseteq \operatorname{Jac} \left({A}\right)$, then by Characterisation of Jacobson Radical $1 + a$ is a unit in $A$.

Thus a unit $\left({1 + a}\right) \in A$ annihilates $M$ and necessarily $M = 0$.

Also see

 * Cayley-Hamilton Theorem