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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?

54321user
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    To give an intuitive explanation of why $r/s = r'/s'$ should hold if there is such a $u$: then we would expect $\frac{r}{s} = \frac{rs'u}{ss'u} = \frac{r'su}{ss'u} = \frac{r'}{s'}$. – Daniel Schepler May 31 '17 at 23:05

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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.

Alex Provost
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  • Do you know of an example of a localization and elements which gives you a nilpotent in the $(rs' - r's)$ part? – 54321user May 31 '17 at 22:08
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    If $R = \mathbb{Q}[x,y] / (xy)$ and $S$ is the powers of $x$, then $y/1 = 0/1$ in $S^{-1} R$ even though $y\cdot 1 - 0\cdot 1 \ne 0$ in $R$. (More generally, the definition is about taking into account zero divisors by elements of $S$, not just nilpotents.) – Daniel Schepler May 31 '17 at 22:18