Nakayama's Lemma/Proof 1

Proof
We induct on the number of generators of $M$.

Base Case:

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

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 = \mathfrak aM = \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:


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

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:
 * $\mathfrak a M = M \implies M = 0$