Localization of a ring

Let $R$ be a commutative ring (with $1$) and let $S \subseteq R$ be a multiplicative set .

Definition (construction by fractions)

Define an equivalence relation on $R\times S$ by

The localization of $R$ at $S$ is the set of equivalence classes

and the class of $(r,s)$ is written $\frac{r}{s}$.

Addition and multiplication are defined by

This makes $S^{-1}R$ into a commutative ring, and the map

is a ring homomorphism.

Universal property

In $S^{-1}R$, every $\iota(s)$ (with $s\in S$) is a unit . Moreover, $(S^{-1}R,\iota)$ is universal with this property:

For any commutative ring $T$ and any ring homomorphism $f:R\to T$ such that $f(s)$ is a unit in $T$ for all $s\in S$, there exists a unique ring homomorphism $\tilde f:S^{-1}R\to T$ with

$$> \tilde f\!\left(\frac{r}{s}\right)=f(r)\,f(s)^{-1} > \quad\text{and}\quad > \tilde f\circ\iota=f. >$$

(See also localization inverts a multiplicative set .)

Examples

1. Invert a prime (or any element) in $\mathbb Z$.
   Let $R=\mathbb Z$ and $S=\{1,2,2^2,2^3,\dots\}$. Then

   $$S^{-1}\mathbb Z \cong \mathbb Z\left[\tfrac12\right].$$

2. Invert a polynomial.
   Let $R=k[x]$ and $S=\{1,x,x^2,\dots\}$. Then

   $$S^{-1}k[x] \cong k[x,x^{-1}],$$

   the Laurent polynomial ring in one variable.

3. Localization at a prime ideal.
   If $\mathfrak p$ is a prime ideal of $R$ and $S=R\setminus\mathfrak p$, then

   $$S^{-1}R = R_{\mathfrak p},$$

   the localization at a prime .