Why does one want to have the standard definition of localization?

Question

The standard way of defining the localization of a commutative ring is as follows: given a multiplicatively closed subset $S\subset R$ the localization is defined first by considering the set $R\times S/\sim$ where $$ (r,s) \sim (r',s') \text{ if there exists a } u \in S \text{ such that } u(rs' - r's) =0 $$ the rest is just equipping this set with a ring structure, but my question lies here: why do we want the $u$? Don't we not want nilpotents in our denominator?

Answer 1

If you don't require the existence of such an $u$, $\sim$ won't be transitive in general (and hence not an equivalence relation). This becomes quite apparent when you attempt to prove that transitivity holds.