Localization Inverts a Multiplicative Set

Statement

Let $R$ be a commutative ring and $S\subseteq R$ a multiplicative set . Let $S^{-1}R$ be the localization , with canonical map

1. (Elements of $S$ become units.)
   For every $s\in S$, $\varphi(s)$ is a unit in $S^{-1}R$, with inverse $\frac{1}{s}$.

2. (Universal property of inverting $S$.)
   If $T$ is a ring and $f:R\to T$ a ring homomorphism such that $f(s)\in T^\times$ for all $s\in S$, then there exists a unique homomorphism

   $$\widetilde f: S^{-1}R \to T$$

   satisfying $\widetilde f\circ \varphi = f$.

In short: $S^{-1}R$ is the universal ring obtained from $R$ by forcing every element of $S$ to be invertible.

Cross-links

• Construction: localization of a ring , multiplicative set
• Units: unit
• Important special case: localization at a prime and local rings

Examples

1. Inverting a prime power set in $\mathbb Z$.
   Let $R=\mathbb Z$ and $S=\{1,2,2^2,2^3,\dots\}$. Then

   $$S^{-1}\mathbb Z \cong \mathbb Z\!\left[\frac12\right],$$

   and $2$ becomes invertible.

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

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

   so $x$ is a unit with inverse $x^{-1}$.

3. Localizing at a prime ideal.
   If $\mathfrak p$ is a prime ideal in $R$, take $S=R\setminus \mathfrak p$. Then $S^{-1}R = R_{\mathfrak p}$ is a local ring (localization at \u03c0 ), and every element outside $\mathfrak p$ becomes a unit.