Nakayama's lemma

Equivalently, if $N\subseteq M$ is a submodule and

then $M=N$.

Local-ring form. If $(R,\mathfrak m)$ is a local ring and $M$ is finitely generated, then $J(R)=\mathfrak m$, so the condition becomes:

A common (and very useful) reformulation uses the residue field $k=R/\mathfrak m$: if $m_1,\dots,m_n\in M$ have images that generate the $k$-vector space $M/\mathfrak m M$, then $m_1,\dots,m_n$ already generate $M$ as an $R$-module. In particular, the minimal number of generators of $M$ equals $\dim_k(M/\mathfrak m M)$.

Examples

1. "$t$-divisible" finitely generated modules over a DVR vanish. Let $R=k[[t]]$, a local ring with maximal ideal $\mathfrak m=(t)$. If $M$ is finitely generated and satisfies $M=tM$, then Nakayama gives $M=0$. (Intuitively: a nonzero finitely generated module cannot be infinitely divisible by $t$.)

1. The maximal ideal in a 2-variable local ring is not principal. Let $R=k[x,y]_{(x,y)}$, the localization at the prime $(x,y)$. Its maximal ideal is $\mathfrak m=(x,y)$. Consider the quotient $\mathfrak m/\mathfrak m^2$, a vector space over the residue field $k$. The classes of $x$ and $y$ span $\mathfrak m/\mathfrak m^2$, so $\dim_k(\mathfrak m/\mathfrak m^2)\ge 2$. Nakayama implies $\mathfrak m$ needs at least two generators; in particular, $\mathfrak m$ is not generated by a single element.

1. Lifting a generator from the residue field quotient. Let $R=\mathbb Z_{(p)}$, the localization of $\mathbb Z$ at $(p)$, with maximal ideal $\mathfrak m=(p)$. Take $M=R/p^2R$. The image of $1\in M$ generates $M/\mathfrak m M \cong R/pR$ as a $k$-vector space. By Nakayama, $1$ generates $M$, so $M$ is a cyclic module.