Localization inverts a multiplicative set

Let $R$ be a commutative ring and let $S\subseteq R$ be a multiplicative set . The localization $S^{-1}R$ comes with a canonical ring homomorphism

Theorem (Elements of $S$ become units).
For every $s\in S$, the element $\iota(s)=s/1$ is a unit in $S^{-1}R$, with inverse $1/s$. In particular, every fraction can be rewritten as

Universal property (often used as the definition).
If $T$ is any commutative ring and $\varphi:R\to T$ is a homomorphism such that every $\varphi(s)$ (for $s\in S$) is a unit of $T$, then there exists a unique homomorphism $\psi:S^{-1}R\to T$ with $\psi\circ \iota=\varphi$. Concretely, $\psi$ is forced to satisfy $\psi(r/s)=\varphi(r)\varphi(s)^{-1}$.

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

Examples

1. Inverting a single integer.
   Take $R=\mathbb Z$ and $S=\{1,2,2^2,2^3,\dots\}$. Then $S^{-1}R\cong \mathbb Z[1/2]$, and $2$ becomes a unit with inverse $1/2$.

2. Laurent polynomials by inverting a variable.
   Take $R=k[x]$ and $S=\{1,x,x^2,\dots\}$. Then $S^{-1}R\cong k[x,x^{-1}]$, and $x$ becomes a unit. Every element looks like a Laurent polynomial because denominators are powers of $x$.

3. Localizing at a prime ideal.
   If $\mathfrak p$ is a prime ideal of $R$, set $S=R\setminus \mathfrak p$. Then the localization $S^{-1}R$ is the localization $R_{\mathfrak p}$ , where every element not in $\mathfrak p$ becomes invertible; this is the basic way to construct a local ring from $R$.