Localization of Ring Exists/Lemma 1

Lemma
The relation $\sim$ is an equivalence relation.

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

Also $\sim$ is clearly symmetric because if $atu = bsu$ then $bsu = atu$ by symmetry of $=$.

Lastly suppose that $(a,s) \sim (b,t)$ and $(b,t) \sim (c,u)$.

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


 * $atv = bsv,\quad buw = ctw$

Therefore,


 * $autvw = buwsv = ctwsv$

and $au(tvw) = cs(tvw)$.

Since $S$ is multiplicatively closed, so $(a,s) \sim (c,u)$.

Therefore $\sim$ is transitive.