Nakayama corollary: generators mod m generate

Corollary (Nakayama: lifting generators).
Let $(R,\mathfrak m)$ be a local ring with residue field $k=R/\mathfrak m$ (see residue field ). Let $M$ be a finitely generated $R$-module, and let $m_1,\dots,m_n\in M$.

If the images of $m_1,\dots,m_n$ generate the $k$-vector space $M/\mathfrak m M$, then $m_1,\dots,m_n$ generate $M$ as an $R$-module.

Equivalently: if $N\subseteq M$ is a submodule with

then $M=N$.

A useful consequence is the minimal number of generators formula

where $\mu_R(M)$ is the smallest size of a generating set of $M$.

This follows from Nakayama's lemma .

Related knowls.

• Nakayama's lemma
• Local ring
• Residue field
• Finitely generated module

Examples

1. Maximal ideal needs two generators.
   Let $R = k[x,y]_{(x,y)}$ with maximal ideal $\mathfrak m=(x,y)$.
   Then $\mathfrak m/\mathfrak m^2$ is a $k$-vector space with basis given by the classes of $x$ and $y$.
   Hence $\dim_k(\mathfrak m/\mathfrak m^2)=2$, so $\mathfrak m$ cannot be generated by one element, and $\{x,y\}$ is a minimal generating set.

2. An ideal that is actually principal.
   Let $R=k[x]_{(x)}$ with $\mathfrak m=(x)$, and take $I=(x^2,x^3)\subset R$.
   Then $I=(x^2)$, and indeed $I/\mathfrak m I$ is 1-dimensional over $k$, so one generator suffices.

3. Reading minimal generators from the residue field.
   In the local ring $R=k[x,y]_{(x,y)}$, consider $I=(x^2, y)$.
   One checks that the classes of $x^2$ and $y$ are nonzero in $I/\mathfrak m I$ and span it, so $\dim_k(I/\mathfrak m I)=2$.
   Nakayama implies $I$ needs exactly two generators (and $\{x^2,y\}$ is minimal).