Localization at a prime ideal

The ring R_p obtained by inverting all elements outside a prime ideal p.
Localization at a prime ideal

Let RR be a and let pR\mathfrak p\subset R be a prime ideal. The complement

S=Rp S=R\setminus \mathfrak p

is a . The localization of RR at p\mathfrak p is the ring

Rp:=S1R, R_{\mathfrak p} := S^{-1}R,

i.e. the obtained by inverting exactly the elements not lying in p\mathfrak p.

Basic properties

  • Every element sRps\in R\setminus\mathfrak p maps to a unit in RpR_{\mathfrak p}.
  • The ring RpR_{\mathfrak p} is a . Its unique maximal ideal is pRp={rs:rp, sp}. \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 .
  • The quotient Rp/pRpR_{\mathfrak p}/\mathfrak pR_{\mathfrak p} is the at p\mathfrak p.

Geometrically, points of are prime ideals, and RpR_{\mathfrak p} is the algebraic “stalk” of RR at the point p\mathfrak p.

Examples

  1. Integers localized at (p)(p). For R=ZR=\mathbb Z and p=(p)\mathfrak p=(p) with pp prime,

    Z(p)={abQ: pb}. \mathbb Z_{(p)}=\left\{\frac{a}{b}\in\mathbb Q:\ p\nmid b\right\}.

    The maximal ideal is pZ(p)p\mathbb Z_{(p)}, and the residue field is Fp\mathbb F_p.

  2. Local ring at the origin on the line. For R=k[x]R=k[x] and p=(x)\mathfrak p=(x),

    k[x](x)={f(x)g(x): g(0)0}. k[x]_{(x)}=\left\{\frac{f(x)}{g(x)}:\ g(0)\neq 0\right\}.

    Its maximal ideal is generated by xx, and the residue field is naturally isomorphic to kk.

  3. Local ring at the origin in the plane. For R=k[x,y]R=k[x,y] and p=(x,y)\mathfrak p=(x,y),

    k[x,y](x,y) k[x,y]_{(x,y)}

    consists of fractions fg\frac{f}{g} with g(0,0)0g(0,0)\neq 0. The maximal ideal is (x,y)k[x,y](x,y)(x,y)k[x,y]_{(x,y)}, and the residue field is kk.