Let R R R commutative ring
and let S ⊆ R S\subseteq R S ⊆ R multiplicative set
. The localization of R R R S S S is a ring, denoted S − 1 R S^{-1}R S − 1 R
ι : R ⟶ S − 1 R
\iota:R\longrightarrow S^{-1}R
ι : R ⟶ S − 1 R such that every element of S S S S − 1 R S^{-1}R S − 1 R localization inverts the multiplicative set
).
Construction (fractions) As a set, S − 1 R S^{-1}R S − 1 R ( r , s ) ∈ R × S (r,s)\in R\times S ( r , s ) ∈ R × S
( r , s ) ∼ ( r ′ , s ′ ) ⟺ ∃ t ∈ S such that t ( r s ′ − r ′ s ) = 0 in R .
(r,s)\sim (r',s') \quad \Longleftrightarrow \quad \exists\,t\in S\text{ such that } t(rs'-r's)=0 \text{ in }R.
( r , s ) ∼ ( r ′ , s ′ ) ⟺ ∃ t ∈ S such that t ( r s ′ − r ′ s ) = 0 in R . Write the class of ( r , s ) (r,s) ( r , s ) r s \frac{r}{s} s r
r s + r ′ s ′ = r s ′ + r ′ s s s ′ , r s ⋅ r ′ s ′ = r r ′ s s ′ .
\frac{r}{s}+\frac{r'}{s'}=\frac{rs'+r's}{ss'},\qquad
\frac{r}{s}\cdot\frac{r'}{s'}=\frac{rr'}{ss'}.
s r + s ′ r ′ = s s ′ r s ′ + r ′ s , s r ⋅ s ′ r ′ = s s ′ r r ′ . The canonical map is ι ( r ) = r 1 \iota(r)=\frac{r}{1} ι ( r ) = 1 r
If 0 ∈ S 0\in S 0 ∈ S ι ( 0 ) \iota(0) ι ( 0 ) 1 = 0 1=0 1 = 0 S − 1 R S^{-1}R S − 1 R S − 1 R S^{-1}R S − 1 R
Universal property The localization is characterized by the following universal mapping property:
If A A A φ : R → A \varphi:R\to A φ : R → A φ ( s ) \varphi(s) φ ( s ) A A A s ∈ S s\in S s ∈ S φ ~ : S − 1 R → A \widetilde\varphi:S^{-1}R\to A φ : S − 1 R → A φ ~ ∘ ι = φ \widetilde\varphi\circ\iota=\varphi φ ∘ ι = φ
φ ~ ( r s ) = φ ( r ) φ ( s ) − 1 .
\widetilde\varphi\!\left(\frac{r}{s}\right)=\varphi(r)\,\varphi(s)^{-1}.
φ ( s r ) = φ ( r ) φ ( s ) − 1 . This same “invert S S S localization of a module
, and it can be viewed as a special case of extension of scalars
.
A basic structural fact is that primes of S − 1 R S^{-1}R S − 1 R R R R S S S prime correspondence under localization
.
Examples Inverting a prime number. Take R = Z R=\mathbb Z R = Z S = { 1 , p , p 2 , … } S=\{1,p,p^2,\dots\} S = { 1 , p , p 2 , … }
S − 1 Z ≅ Z [ 1 p ] = { a p n : a ∈ Z , n ≥ 0 } ⊆ Q .
S^{-1}\mathbb Z \cong \mathbb Z\!\left[\frac{1}{p}\right]
=\left\{\frac{a}{p^n}:a\in\mathbb Z,\ n\ge 0\right\}\subseteq\mathbb Q.
S − 1 Z ≅ Z [ p 1 ] = { p n a : a ∈ Z , n ≥ 0 } ⊆ Q . Laurent polynomials. If R = k [ x ] R=k[x] R = k [ x ] S = { 1 , x , x 2 , … } S=\{1,x,x^2,\dots\} S = { 1 , x , x 2 , … }
S − 1 R ≅ k [ x , x − 1 ] ,
S^{-1}R \cong k[x,x^{-1}],
S − 1 R ≅ k [ x , x − 1 ] , since x x x
Localizing at a prime ideal. If p ⊂ R \mathfrak p\subset R p ⊂ R S = R ∖ p S=R\setminus\mathfrak p S = R ∖ p S − 1 R S^{-1}R S − 1 R localization at the prime
R p R_{\mathfrak p} R p local ring
.