Nakayama's lemma

Last updated January 27, 2022

Corollary. (Nakayama’s lemma). Let $I$ be an ideal contained in the Jacobson radical of a ring $R$, and let $M$ be a finitely generated $R$- module. Then

1. If $IM=M$, then $M=0$.
2. If $m_1,\dots,m_n\in M$ have images in $M/IM$ that generate it as an $R$-module, then $m_1,\dots,m_n$ generate $M$ as an $R$-module.

Warning. With regards to the second statement, it is in general incorrect to conclude $M$ is finitely generated by finding finitely many generators of $M/IM$. We need those generators to be the image of (finitely many) things in $M$.

Proof. Apply cor. 1 from ( here) to obtain an $r\in I$ such that $(1-r)M=0$. But notice that $r$ is in $I$ which is in the Jacobson radical, hence $r$ is in every maximal ideal. That means $1-r$ is not in any maximal ideal, i.e. $1-r$ is a unit. But then $(1-r)M=M=0$, so $M=0$.

Corollaries

1. If $(R,P)$ is a local ring and $M$ is a finitely-generated $R$-module such that $M/PM=0$ (i.e. $M=PM$), then $M=0$.
   1. If $N\subset M$ is a submodule such that $PM+N=M$ then $N=M$.