# 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 $\U0001d51eM$ can be written as linear combinations^{} $\sum {a}_{i}{x}_{i}$, where ${a}_{i}\in \U0001d51e$.

Suppose that $|X|>0$. Since $M=\U0001d51eM$, 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 $M=0$.

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 |