Knowlpedia

Prime correspondence under localization

Prime ideals in a localization S^{-1}R correspond to primes of R disjoint from S via extension and contraction.

Let RR be a and let S⊆RS\subseteq R be a . Write S−1RS^{-1}R for the .

The map

p⟼S−1p\mathfrak p \longmapsto S^{-1}\mathfrak p

induces an inclusion-preserving bijection between:

  • prime ideals p⊆R\mathfrak p\subseteq R with p∩S=∅\mathfrak p\cap S=\varnothing, and
  • prime ideals q⊆S−1R\mathfrak q\subseteq S^{-1}R.

The inverse bijection is contraction:

q⟼qc:={ r∈R:r/1∈q }.\mathfrak q \longmapsto \mathfrak q^c := \{\, r\in R : r/1 \in \mathfrak q \,\}.

In other words, “primes survive localization exactly when they do not meet the set of denominators.” This refines the fact that localization .

Geometric form

On the , this correspondence identifies Spec⁡(S−1R)\operatorname{Spec}(S^{-1}R) with the subspace

{ p∈Spec⁡(R):p∩S=∅ }\{\,\mathfrak p\in \operatorname{Spec}(R) : \mathfrak p\cap S=\varnothing\,\}

of Spec⁡(R)\operatorname{Spec}(R). For S={1,f,f2,… }S=\{1,f,f^2,\dots\}, this subspace is the basic open set D(f)D(f); for a general multiplicative set it need not be open.

Localization at a prime

If S=R∖pS=R\setminus \mathfrak p for a prime ideal p\mathfrak p, then S−1RS^{-1}R is the ring . The primes of RpR_\mathfrak p correspond exactly to primes q⊆R\mathfrak q\subseteq R with q⊆p\mathfrak q\subseteq \mathfrak p.

Examples
  1. Inverting a prime in Z\mathbb{Z}. Take R=ZR=\mathbb{Z} and S={1,p,p2,… }S=\{1,p,p^2,\dots\}. Then S−1R=Z[1/p]S^{-1}R=\mathbb{Z}[1/p]. A prime ideal of Z\mathbb{Z} meets SS exactly when it contains pp, so the primes of Z[1/p]\mathbb{Z}[1/p] correspond to (0)(0) and (ℓ)(\ell) for primes ℓ≠p\ell\neq p.
  1. Localizing away from a hypersurface. Let R=k[x,y]R=k[x,y] and S={1,x,x2,… }S=\{1,x,x^2,\dots\}, so S−1R=RxS^{-1}R=R_x. Primes of RxR_x correspond to primes of k[x,y]k[x,y] that do not contain xx. For instance, (y)(y) survives, while (x,y)(x,y) does not.
  1. Localization at a maximal ideal. In R=k[x,y]R=k[x,y], localizing at m=(x,y)\mathfrak m=(x,y) gives . Its prime ideals correspond to all prime ideals of k[x,y]k[x,y] contained in (x,y)(x,y), including (0)(0), (x)(x), (y)(y), (x−y)(x-y), and (x,y)(x,y).