Let $A$ be a ring and $S$ a multiplicative closed set. Then the localization of $A$ with respect to $S$ is defined as the set $S^{-1}A$ consisting of equivalence classes of pairs $(a, s)$ where to such pairs $(a,s), (b,t)$ are said to be equivalent if there exists some $u$ in $S$ such that
$$u(at-bs)=0$$
Now, in the Wikipedia article about the localization of a ring, it says that the existence of that $u\in S$ is crucial in order to guarantee the transitive property of the equivalence relation.
I've seen the proof that the equivalence relation defined above is indeed an equivalence relation, but I fail to see how crucial the existence of $u$ is. For example, why doesn't it work if we simply say that two pairs $(a,s),(b,t)$ are equivalent iff $at – bs = 0$? I tried to come up with a counterexample for such case, but failed in the attempt.
Best Answer
You have to think in the case that $A$ is not an integral domain. For example, consider $\mathbb{Z}_8$. Make $a_1=a_3=\overline{1},a_2=s_1=\overline{2}, s_2=\overline{4}$ and $ s_3=\overline{6}$. We have $$a_1s_2=\overline{1}.\overline{4}=\overline{2}.\overline{2}=a_2s_1$$ $$a_2s_3=\overline{2}.\overline{6}=\overline{4}=\overline{1}.\overline{4}=a_3s_2$$ and $$a_1s_3=\overline{1}.\overline{6}=\overline{6}\neq\overline{2}=\overline{1}.\overline{2}=a_3s_1.$$