# Localization of Ring Exists

## Theorem

Let $A$ be a commutative ring with unity.

Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$.

Then there exists a localization $\left({A_S, \iota}\right)$ of $A$ at $S$.

## Proof

Define a relation $\sim$ on the Cartesian product $A \times S$ by:

$\left({a, s}\right) \sim \left({b, t}\right) \iff \exists u \in S: a t u = b s u$

### Lemma 1

The relation $\sim$ is an equivalence relation.

$\Box$

Let $A_S$ be used to denote the $\left({A \times S}\right) / \sim$.

Let $a / s$ be the equivalence class of $\left({a, s}\right)$ in $\left({A \times S}\right) / \sim$.

For $\dfrac a s, \dfrac b t \in A_S$, let the following be defined:

$\dfrac a s + \dfrac b t = \dfrac{a t + b s}{s t}$
$\dfrac a s \cdot \dfrac b t = \dfrac{a b}{s t}$

### Lemma 2

The operations $+$ and $\cdot$ are well defined on $A_S$.

$\Box$

Now define $\iota: A \to A_S$ by:

$\iota \left({a}\right) := \dfrac a 1$

It is to be shown that $\left({A_S, \iota}\right)$ satisfy the universal property for localization.

Let $B$ be a ring.

Let $g: A \to B$ be a mapping such that:

$g \left({S}\right) \subseteq B^\times$

Suppose that $h: A_S \to B$ is a ring homomorphism with $h \circ \iota = g$.

Then we must have:

$h \left({\dfrac a 1}\right) = g \left({a}\right)$

and:

$h \left({\dfrac 1 s}\right) \cdot h \left({\dfrac s 1}\right) = 1$

Therefore:

$h \left({\dfrac 1 s}\right) g \left({s}\right) = 1$

so:

$h \left({\dfrac 1 s}\right) = g \left({s}\right)^{-1}$

Therefore:

$h \left({\dfrac a s}\right) = h \left({\dfrac a 1}\right) \cdot h \left({\dfrac 1 s}\right) = g \left({a}\right) g \left({s}\right)^{-1}$.

So if such $h$ exists it must equal $g \left({a}\right) g \left({s}\right)^{-1}$, so is unique.

Therefore, to conclude the proof we pull out:

### Lemma 3

The map $h \left({a / s}\right) = g \left({a}\right) g \left({s}\right)^{-1}$ is a well defined ring homomorphism $A_S \to B$.

$\Box$

Hence the result.

$\blacksquare$