proof of Nakayama’s lemma

Let $X=\{x_1,x_2,\mathrm{\dots },x_n\}$ be a minimal set of generators for $M$, in the sense that $M$ is not generated by any proper subset of $X$.

Elements of $𝔞M$ can be written as linear combinations $\sum a_i{x}_i$, where $a_i\in 𝔞$.

Suppose that $|X|>0$. Since $M=𝔞M$, we can express $x_1$ as a such a linear combination:

$$x_1=\sum a_i{x}_i.$$

Moving the term involving $a_1$ to the left, we have

$$(1-a_1)x_1=\sum _{i>1}{a}_i{x}_i.$$

But $a_1\in J(R)$, so $1-a_1$ is invertible, say with inverse $b$.Therefore,

$$x_1=\sum _{i>1}b{a}_i{x}_i.$$

But this means that $x_1$ is redundant as a generator of $M$, and so $M$ is generated by the subset $\{x_2,x_3,\mathrm{\dots },x_n\}$. This contradicts the minimality of $X$.

We conclude that $|X|=0$ and therefore $M=0$.