Nakayama's Lemma

Lemma
$\newcommand{\Jac}[1] {\operatorname{Jac} \left({#1}\right)}$ Let $A$ be a commutative ring with unity

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

Let $\Jac A$ be the Jacobson radical of $A$.

If $\Jac A M = M$, then $M = 0$.

Corollary 1
With $A$, $M$ as above, if there exists a submodule $N \subseteq M$ such that:


 * $ M = N + \Jac A M $

Then $M = N$.

Corollary 2
With $A$, $M$ as above, if


 * $ m_1 + \Jac AM,\ldots, m_n + \Jac AM$

generate $M/ \Jac AM$ over $A/ \Jac A$ then $m_1,\ldots, m_n$ generate $M$ over $A$.

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

If $M$ has a single generator $m_1\in M$, then $\Jac A m_1=M$.

So $m_1 \in \Jac A m_1$, i.e. $m_1=am_1$ for some $a \in \Jac A$.

By the characterisation of the Jacobson radical, necessarily $1-a$ is a unit in $A$, so


 * $(1-a)^{-1}(1-a)m=0$

thus $m=0$.

Suppose now that $M=Am_1 + \cdots + A m_n$ for some $m_1,\ldots,m_n\in M$.

Then we have:


 * $M = \Jac A M = \Jac A m_1 + \cdots + \Jac A m_n $

Thus for some $a_1,\ldots,a_n \in \Jac A$:


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

Then:


 * $ (1-a_1)m_1 = a_2 m_2 + \cdots + a_n m_n $

Since $a_1 \in \Jac A$, $1-a_1$ is a unit in $A$ we have $m_1 \in A m_2 + \cdots + A m_n$.

Therefore $M$ has $n-1$ generators $m_2,\ldots,m_n$, and by the induction hypothesis $\Jac A M = M \implies M = 0$.

Proof of Corollary 1
If $M = N + \Jac A M$ then


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

so by Nakayama's Lemma, $M/N = 0$ and $M = N$.

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

Then $M = N + \Jac A M$.

Hence by Corollary 1, $M = N$.

Alternative Proof
Nakayama's lemma can also be proved as a corollary to the Cayley-Hamilton theorem