A question about the equivalence relation on the localization of a ring.

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.

Solution 1:

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 $$a_1s_3=\overline{1}.\overline{6}=\overline{6}\neq\overline{2}=\overline{1}.\overline{2}=a_3s_1.$$

Solution 2:

For a more geometric example, let $A = \mathbb{R}[x,y] / (xy)$ and $S = \{ 1, x, x^2, x^3, \ldots \}$. (In algebraic geometry, $A$ represents a ring of functions on the union of the two axes in $\mathbb{R}^2$.) Then in $S^{-1} A$, $y/1 = 0/1$ even though $y \cdot 1 - 0 \cdot 1 \ne 0$ in $A$ - but in fact, $x (y \cdot 1 - 0 \cdot 1) = 0$.

Intuitively, the reason $y = 0$ must hold in $S^{-1} A$ is that $xy = 0$ is inherited from $A$, and then you can multiply both sides by $x^{-1}$. (The localization corresponds to the geometric operation of intersecting with the set where $x \ne 0$.)