proof of Nakayama’s lemma

Let $X=\{x_{1},x_{2},\dots,x_{n}\}$ be a minimal set of generators for $M$ , in the sense that $M$ is not generated by any proper subset of $X$ .

Elements of $\mathfrak{a}M$ can be written as linear combinations $\sum a_{i}x_{i}$ , where $a_{i}\in\mathfrak{a}$ .

Suppose that $|X|>0$ . Since $M=\mathfrak{a}M$ , 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}ba_{i}x_{i}.

But this means that $x_{1}$ is redundant as a generator of $M$ , and so $M$ is generated by the subset $\{x_{2},x_{3},\dots,x_{n}\}$ . This contradicts the minimality of $X$ .

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

Title	proof of Nakayama’s lemma
Canonical name	ProofOfNakayamasLemma
Date of creation	2013-03-22 13:07:46
Last modified on	2013-03-22 13:07:46
Owner	mclase (549)
Last modified by	mclase (549)
Numerical id	5
Author	mclase (549)
Entry type	Proof
Classification	msc 13C99