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Let $S$ be a multiplicative system. Then the localization of $R$ is defined by an equivalence relation on $R \times S$. The relation is $(a,b) \sim (c,d)$ if there is an $s \in S$ such $s(ad-bc)=0$.

Regarding this, I can't show that transitivity works. Could anyone show me how to prove that $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$ then $(a,b) \sim (e,f)$?

Watson
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Keith
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1 Answers1

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Suppose $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$, so that there are $s$ and $t$ in $S$, with $s(ad-bc) = t(cf-de) = 0$. Since $S$ is multiplicative, and $d,s,t \in S$ we have $dst \in S$, and

$$ dst (af-be) = tf(sad) - bs(tde) = tf(sbc) - bs(tcf)=0.$$

Zavosh
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