Localization of Ring Exists/Lemma 1
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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:
- $\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$
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:
- $\tuple {a, s} \sim \tuple {b, t}$
and:
- $\tuple {b, t} \sim \tuple {c, u}$
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:
- $a u \paren {t v w} = c s \paren {t v w}$
Since $S$ is multiplicatively closed:
- $\tuple {a, s} \sim \tuple {c, u}$
Thus $\sim$ is transitive.
$\blacksquare$