Nakayama's Lemma

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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.


Also see