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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.

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    $\begingroup$ A useful fact that I don't see mentioned often is: the equivalence relation is exactly the equivalence relation generated by $(a, s) \sim (a s', s s')$ for $a \in A$, $s, s' \in S$. So, if you have another equivalence relation which equates $(a, s)$ with $(a s', s s')$ for all $a, s, s'$, then that relation contains the equivalence relation. This can be used, for example, to prove the addition operation on $S^{-1} A$ is well-defined with much shorter calculations than in the direct proof. $\endgroup$ May 31, 2017 at 22:34
  • $\begingroup$ In the absence of OP's response to requests for clarification, I went ahead with the anonymously proposed edit. $\endgroup$
    – hardmath
    Jul 26, 2018 at 22:53

2 Answers 2

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

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  • $\begingroup$ Thank you for your answer! Just a silly question. Everywhere I look, $S$ is assumed to not contain $0$, whereas in your example, $s_1s_2 = \overline{2}\ \overline{4} = \overline{0} \in S$ if we want $S$ to be a multiplicative closed set, right? How is that? $\endgroup$
    – user313212
    Apr 30, 2017 at 16:30
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    $\begingroup$ If $0$ belongs to the multiplicative closed set, then the ring of fractions correspondent is the trivial. $\endgroup$
    – Rafael
    Apr 30, 2017 at 16:36
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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 $x \ne 0$.)

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