Nakayama's Lemma
Theorem
Let $A$ be a commutative ring with unity.
Let $M$ be a finitely generated $A$-module.
Let $\map {\operatorname{Jac} } A$ be the Jacobson radical of $A$.
Let $\mathfrak a \subseteq \map {\operatorname{Jac} } A$ be an ideal of $A$.
Suppose $\mathfrak a M = M$.
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 1
We induct on the number of generators of $M$.
- Basis for the Induction
Let $M$ have a single generator $m_1 \in M$.
Then $\mathfrak a m_1 = M$.
So:
- $m_1 \in \mathfrak a m_1$
That is:
- $m_1 = a m_1$
for some $a \in \mathfrak a$.
By Characterisation of Jacobson Radical, $1 - a$ is a unit in $A$.
So:
- $\paren {1 - a}^{-1} \paren {1 - a} m = 0$
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 = \mathfrak a M = \mathfrak a m_1 + \dotsb + \mathfrak a m_n $
Thus for some $a_1, \dotsc, a_n \in \mathfrak a$:
- $ m_1 = a_1 m_1 + \dotsb + a_n m_n $
Then:
- $\paren {1 - a_1} m_1 = a_2 m_2 + \dotsb + a_n m_n$
By Characterisation of Jacobson Radical, $1 - a$ is a unit in $A$.
Multiplying both sides by $\paren {1 - a_1}^{-1}$ gives:
- $m_1 = \paren {1 - a_1}^{-1} a_2 m_2 + \dotsb + \paren {1 - a_1}^{-1} a_n m_n$
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:
- $\mathfrak a M = M \implies M = 0$
$\blacksquare$
Proof 2
Let $\phi : M \to M$ be the identity mapping on $M$, i.e.:
- $\forall x \in M : \map \phi x = x$
Since $\mathfrak a M = M$ by hypothesis, $\phi$ is an endomorphism of $M$ such that:
- $\map \phi M \subseteq a M$
By Cayley-Hamilton Theorem, there exist $a_0, \ldots , a_{n-1} \in \mathfrak a$ such that:
- $(1):\quad \phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$
Evaluating $(1)$ at $x=1$, especially, we have:
- $(2):\quad 1 + a= 0$
where:
- $a := a_{n - 1} + \cdots + a_1 + a_0$
Since:
- $a \in \mathfrak a \subseteq \map {\operatorname {Jac} } A$
by Characterisation of Jacobson Radical, $1 + a$ is a unit in $A$.
Thus for each $x \in M$, we have:
\(\ds x\) | \(=\) | \(\ds \paren {1 + a}^{-1} \paren {1 + a} x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 + a}^{-1} 0\) | by $(2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
That is:
- $M = 0$
$\blacksquare$
Source of Name
This entry was named for Tadashi Nakayama.