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

Proof
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 $$M=\sum_{i=1}^nAm_i$$ for some $$m_1,\ldots,m_n\in M$$.

Let $$M'=M/(Am_n)$$.

Still $$\Jac A M'=M'$$, and $$M'$$ has $$n-1$$ generators.

So $$M'=0$$ and $$M=Am_n$$.

We are done.