Nakayama's Lemma/Proof 2

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:

$M = 0$

Proof

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:

$M = 0$

$\blacksquare$