# Localization of Ring Exists/Lemma 3

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## Lemma for Localization of Ring Exists

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

## Proof

Let $a / s = b / t \in A_S$.

Then there is $u \in S$ such that $a t u = b s u$.

Therefore, by the homomorphism property, and the fact that $g \left[{S}\right] \subseteq B^\times$:

- $g \left({a}\right) g \left({t}\right) = g \left({b}\right) g \left({s}\right) \to g \left({a}\right) g \left({s}\right)^{-1} = g \left({b}\right) g \left({t}\right)^{-1}$

We have that $g$ is a ring homomorphism.

As $B$ is a commutative ring it follows that $h$ is also a ring homomorphism.

$\blacksquare$