Nakayama's Lemma

Statement

Let $M$ be a finitely generated $R$-module and $I\subset R$ an ideal which is contained in the Jacobson radical. Then $M=0$ if and only if $IM=M$.

Proof

If $M=0$ then $IM=M=0$ trivially.
Let $IM=M$ satisfying the conditions above. Since $M$ is finitely generated, let $\{m_1,\dots ,m_n\}$ be a minimal set of generators. Because $IM=M$ then

$$m_1=\sum _{j=1}^n{i}_j{m}_j{i}_j\in I.$$

Rearranging terms gives

$$\begin{array}{rl}{m}_1& =i_1{m}_1+\sum _{j=2}^n{i}_j{m}_j\\ (1-i_1)m_1& =\sum _{j=2}^n{i}_j{m}_j\end{array}$$

Since $I$ is contained in the Jacobson radical, $(1-i_1)$ is a unit so

$$m_1=(1-i_m)\sum _{j=2}^n{i}_j{m}_j$$

which contradicts the assumption that $\{m_1,\dots ,m_n\}$ is a minimal generating set. Hence, $M=0$.