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If we take a commutative unital ring $R$ and a multiplicative subset $S$, we localise at $S$ using the equivalence relation on $R \times S$ given by $(a,s) \sim (b,t)$ if and only if there is $u \in S$, such that $u(ta-sb)=0$. As far as I can tell, this condition says that we want things to be equivalent if their difference has a zero divisor in $S$, but I don't understand why we would necessarily want that. Can anyone shed some light please?

Orlly
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  • It is useful to think about the universal property of localisation here. If you have some ring $A$ and a homomorphism $\phi : R \to A$ such that the image of $S$ in $A$ is invertible in $A$, then you will find that any $(r, s) \in R \times S$ such that $r s = 0$ must also satisfy $\phi (r) = 0$. – Zhen Lin Apr 21 '21 at 10:54
  • formally, you can write elements in $R\times S$ as $\frac{a}{s}$, where $a\in R,s\in S$, then it's clear that two fractions are viewed as the same iff $ta-sb$ is a zero divisor – LuOH3 Apr 21 '21 at 10:56
  • Look at the case $S={ u^n,n\ge 0}$ then the localization of $R$ at $S$ is $S^{-1}R=R[x]/(ux-1)$. In this quotient ring $u$ has an inverse $x$ and the kernel of $R\to S^{-1}R$ is ${r\in R,\exists n, r u^n=0}$ and $(a,u^n)\in R\times S$ in your construction represents the element $x^n a\in R[x]/(ux-1)$. – reuns Apr 21 '21 at 11:17

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