Localization at a prime ideal

Let $R$ be a commutative ring and let $\mathfrak p\subset R$ be a prime ideal. The complement

is a multiplicative set. The localization of $R$ at $\mathfrak p$ is the ring

i.e. the localization of a ring obtained by inverting exactly the elements not lying in $\mathfrak p$.

Basic properties

• Every element $s\in R\setminus\mathfrak p$ maps to a unit in $R_{\mathfrak p}$.
• The ring $R_{\mathfrak p}$ is a local ring. Its unique maximal ideal is
  $$\mathfrak pR_{\mathfrak p}=\left\{\frac{r}{s}: r\in\mathfrak p,\ s\notin\mathfrak p\right\}.$$
  This description is part of the general phenomenon summarized in prime correspondence under localization.
• The quotient $R_{\mathfrak p}/\mathfrak pR_{\mathfrak p}$ is the residue field at $\mathfrak p$.

Geometrically, points of Spec(R) are prime ideals, and $R_{\mathfrak p}$ is the algebraic “stalk” of $R$ at the point $\mathfrak p$.

Examples

1. Integers localized at $(p)$. For $R=\mathbb Z$ and $\mathfrak p=(p)$ with $p$ prime,
   $$\mathbb Z_{(p)}=\left\{\frac{a}{b}\in\mathbb Q:\ p\nmid b\right\}.$$
   The maximal ideal is $p\mathbb Z_{(p)}$, and the residue field is $\mathbb F_p$.

1. Local ring at the origin on the line. For $R=k[x]$ and $\mathfrak p=(x)$,
   $$k[x]_{(x)}=\left\{\frac{f(x)}{g(x)}:\ g(0)\neq 0\right\}.$$
   Its maximal ideal is generated by $x$, and the residue field is naturally isomorphic to $k$.

1. Local ring at the origin in the plane. For $R=k[x,y]$ and $\mathfrak p=(x,y)$,
   $$k[x,y]_{(x,y)}$$
   consists of fractions $\frac{f}{g}$ with $g(0,0)\neq 0$. The maximal ideal is $(x,y)k[x,y]_{(x,y)}$, and the residue field is $k$.