Knowlpedia

Localization inverts a multiplicative set

In the localization S^{-1}R, every element of S becomes a unit, and S^{-1}R is universal with that property.

Let RR be a and let S⊆RS\subseteq R be a . The S−1RS^{-1}R comes with a canonical ring homomorphism

ι:R→S−1R,r↦r1.\iota:R\to S^{-1}R,\qquad r\mapsto \frac{r}{1}.

Theorem (elements of SS become units). For every s∈Ss\in S, the element ι(s)=s/1\iota(s)=s/1 is a unit in S−1RS^{-1}R, with inverse 1/s1/s. In particular, every fraction can be rewritten as

rs=ι(r) ι(s)−1.\frac{r}{s}=\iota(r)\,\iota(s)^{-1}.

Universal property (often used as the definition). If TT is any commutative ring and φ:R→T\varphi:R\to T is a homomorphism such that φ(s)\varphi(s) is a unit for every s∈Ss\in S, then there exists a unique homomorphism ψ:S−1R→T\psi:S^{-1}R\to T with ψ∘ι=φ\psi\circ\iota=\varphi. It is given by

ψ(r/s)=φ(r)φ(s)−1.\psi(r/s)=\varphi(r)\varphi(s)^{-1}.

This perspective explains why produces a : inverting all elements outside a prime ideal forces exactly those elements to become units.

Examples
  1. Inverting a single integer. Take R=ZR=\mathbb Z and S={1,2,22,23,… }S=\{1,2,2^2,2^3,\dots\}. Then S−1R≅Z[1/2]S^{-1}R\cong \mathbb Z[1/2], and 22 becomes a unit with inverse 1/21/2.
  1. Laurent polynomials by inverting a variable. Take R=k[x]R=k[x] and S={1,x,x2,… }S=\{1,x,x^2,\dots\}. Then S−1R≅k[x,x−1]S^{-1}R\cong k[x,x^{-1}], and xx becomes a unit. Its elements are Laurent polynomials because the denominators are powers of xx.
  1. Localizing at a prime ideal. If p\mathfrak p is a prime ideal of RR, set S=R∖pS=R\setminus\mathfrak p. Then S−1RS^{-1}R is the , in which every element outside p\mathfrak p becomes invertible.