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 pair $\left({A_S, \iota}\right)$ satisfying the definition of the localisation 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
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
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 localisation.

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
Hence the result.

Also see

 * If $A$ is an integral domain and $S = A \setminus \left\{{0}\right\}$ then the localisation of $A$ at $S$ is precisely the field of fractions of $A$.