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

The relation $\sim$ is an equivalence relation.

$\Box$

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

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

$\Box$

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 Definition of Localization of Ring.

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.

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Therefore, to conclude the proof we pull out:

Lemma 3

The map $\map h {\dfrac a s} = \map g a \map g s^{-1}$ is a well defined ring homomorphism $A_S \to B$.

$\Box$

Hence the result.

$\blacksquare$

Notes

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The proof shows additionally that:

$\forall x \in A_S: \exists s \in S: s x \in \iota \sqbrk A$

$\map \ker \iota = \set {a \in A: \exists s \in S: a s = 0}$

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 $A$.