# 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]

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 $$(a,s) \sim (b,s') \iff s'a-sb=0$$?

• Otherwise there is no reason to believe that this is an equivalence relation, as $S$ could contain zero divisors. Rational numbers don't have this problem, because $\mathbb{Z}$ is an integral domain. Commented Aug 14, 2019 at 2:23

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 $$at-cs=0$$.