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 $\struct {A_S, \iota}$ of $A$ at $S$.

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


 * $\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$

Lemma 1
Let $A_S$ denote the quotient set $\paren {A \times S} / \sim$.

Let $\dfrac a s$ denote the equivalence class of $\tuple {a, s}$ in $\paren {A \times S} / \sim$.

For $\dfrac a s, \dfrac b t \in A_S$, let the following operations 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
Now define $\iota: A \to A_S$ by:
 * $\map \iota a := \dfrac a 1$

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

That is, to be shown is the $(2)$ of.

Let $B$ be a ring.

Let $g: A \to B$ be a mapping such that:
 * $g \sqbrk S \subseteq B^\times$

where $B^\times$ denotes the set of units of $B$.

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

Then we must have:
 * $\map h {\dfrac a 1} = \map g a$

and:
 * $\map h {\dfrac 1 s} \cdot \map h {\dfrac s 1} = 1$

Therefore:
 * $\map h {\dfrac 1 s} \map g s = 1$

so:
 * $\map h {\dfrac 1 s} = \map g s^{-1}$

Therefore:
 * $\map h {\dfrac a s} = \map h {\dfrac a 1} \cdot \map h {\dfrac 1 s} = \map g a \map g s^{-1}$.

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

Therefore, to conclude the proof we pull out:

Lemma 3
Hence the result.

Also see

 * If $A$ is an integral domain and $S = A \setminus \set 0$ then the localization of $A$ at $S$ is precisely the field of quotients of $A$.