# Localization of Ring Exists/Lemma 1

## Lemma for Localization of Ring Exists

Let $A$ be a commutative ring with unity.

Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$.

Let $\sim$ be the relation defined on the Cartesian product $A \times S$ by:

$\left({a, s}\right) \sim \left({b, t}\right) \iff \exists u \in S: a t u = b s u$

The relation $\sim$ is an equivalence relation.

## Proof

Since by definition $1 \in S$ and $a s = a s$ for all $a \in A, s \in S$, it follows that $\sim$ is reflexive.

Also $\sim$ is clearly symmetric because if $a t u = b s u$ then $b s u = a t u$ by symmetry of $=$.

Lastly suppose that:

$\left({a, s}\right) \sim \left({b, t}\right)$

and:

$\left({b, t}\right) \sim \left({c, u}\right)$

Then there are $v, w \in S$ such that:

$a t v = b s v$
$b u w = c t w$

Therefore:

$a u t v w = b u w s v = c t w s v$

and so:

$a u \left({t v w}\right) = c s \left({t v w}\right)$

Since $S$ is multiplicatively closed:

$\left({a, s}\right) \sim \left({c, u}\right)$

Thus $\sim$ is transitive.

$\blacksquare$