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 ProofOfNakayamasLemma 2013-03-22 13:07:46 2013-03-22 13:07:46 mclase (549) mclase (549) 5 mclase (549) Proof msc 13C99