In the definition of localization of a ring $R$, why do we require the existence of $t$ in the multiplicative subset to define the equiv relation? [duplicate]

Question

In the definition of localization of a ring $R$ where $S$ the multiplicative subset, we say $(a,s) \sim (b,s') \iff \exists t \in S$ such that $t(s'a-sb)=0$.

Why do we require such a $t\in S$?

What if we just say $(a,s) \sim (b,s') \iff s'a-sb=0$?

Answer 1

The idea is because $\sim$ might not be an equivalence relation otherwise.

If you have $(a,s)\sim (b,s')$ and $(b,s')\sim (c,t)$ in your proposed definition, then $as'-bs=bt-cs'=0$, why does that imply $at-cs=0$?

In the case of integral domains, where you don't need the extra element of $S$, you can just multiply $as'-bs$ by $t$ and $cs'-bt$ by $s$, add the two expressions to cancel out the $bst$, and then factor out the $s'$ and$\textbf{ use the nonexistence of zero divisors}$ to conclude that $at-cs=0$.