Maximal ideal of a local ring

Let $R$ be a commutative ring . Write $R^\times$ for the group of units of $R$, and let

be the set of nonunits.

Theorem (nonunits form the unique maximal ideal)

The following are equivalent:

1. $R$ is a local ring .
2. The set $\mathfrak m$ of nonunits is an ideal of $R$.

When these conditions hold, $\mathfrak m$ is the unique maximal ideal of $R$. Equivalently, an element $x\in R$ is a unit if and only if $x\notin\mathfrak m$.

In the important special case $R_{\mathfrak p}$ from localization at a prime , the unique maximal ideal is $\mathfrak pR_{\mathfrak p}$, and the associated residue field is $R_{\mathfrak p}/\mathfrak pR_{\mathfrak p}$.

This “units vs. maximal ideal” dichotomy is exactly what makes local arguments work; for instance, it is the setup needed in Nakayama's lemma .

Examples

1. $\mathbb Z_{(p)}$. In $R=\mathbb Z_{(p)}$, a fraction $\frac{a}{b}$ (with $p\nmid b$) is a unit iff $p\nmid a$. Thus the nonunits are precisely those with numerator divisible by $p$, i.e. the maximal ideal is $p\mathbb Z_{(p)}$.

2. $k[x]_{(x)}$. In $R=k[x]_{(x)}$, the units are exactly fractions $\frac{f}{g}$ with $f(0)\neq 0$ (and $g(0)\neq 0$ by definition of the localization). Hence the nonunits are those with $f(0)=0$, i.e. the maximal ideal is generated by $x$.

3. A field. In a field $k$, the only nonunit is $0$, so $\mathfrak m=(0)$ is the unique maximal ideal.