Localization of Ring is Unique

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Theorem

Let $A$ be a commutative ring with unity.

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

Let $\left({A_S, \iota}\right)$ and $\left({\tilde A_S, \tilde \iota}\right)$ both satisfy the definition of the localization of $A$ at $S$.


Then there is a canonical isomorphism $\phi: A_S \to \tilde A_S$.


Proof

By the definition of localization, there exist unique homomorphisms:

$g : A_S \to \tilde A_S$
$h : \tilde A_S \to A_S$

such that:

$h \circ \iota = \tilde \iota$

and:

$g \circ \tilde \iota = \iota$

Therefore:

$\tilde \iota = h \circ \iota = h \circ \left({g \circ \tilde \iota}\right) = \left({h \circ g}\right) \circ \tilde \iota$

The identity mapping also satisfies this equality (that is, $\tilde \iota = \left({h \circ g}\right) \circ \tilde \iota$).

Therefore by the uniqueness of $h$ and $g$ we have:

$h \circ g = I_{\tilde A_S}$

where $I_{\tilde A_S}$ denotes the identity mapping on $\tilde A_S$.

Similarly we have that:

$g \circ h = I_{A_S}$

where $I_{A_S}$ denotes the identity mapping on $A_S$.

Therefore by Bijection iff Left and Right Inverse $g = \phi$ is a bijective homomorphisms, that is, an isomorphism.

$\blacksquare$