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Suppose $R$ is a ring and $S$ a multiplicative set in $R$. Then the localization $S^{-1}R$ satisfies the universal property that every element of $S$ maps to an invertible element in $S^{-1}R$ and if $R\to T$ is a ring homomorphism such that every element of $S$ maps to an invertible element in $T$, then there is a unique ring homomorphism $S^{-1}R \to T$.

I was wondering whether the following property holds. If $W$ is a multiplicative subset of $S$. Then is there a unique homomorphism (satisfying certain mild conditions) $W^{-1}R \to S^{-1}R$? I think there is a natural map which sends $r/w \to r/w$, but is it the only one?

Essentially if we invert more elements than $S$ we get a map from the localization. If we invert less elements than $S$ do we get a map to the localization?

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Every element of W maps to an invertible element under $R\to S^{-1}R$. – Jason Polak Oct 21 '11 at 17:08
up vote 1 down vote accepted

Sure, apply the universal property of $W^{-1}R$ to the the map $R \rightarrow S^{-1}R$ (so I am saying let $T = S^{-1}R$). That is, the map $R \rightarrow S^{-1}R$ factors through $W^{-1}R$ for any multiplicative subset $W \subset S$.

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Ah, nice. Thanks. – Marcus Lode Oct 21 '11 at 17:08

In the context of integral domains, a sometimes-handy construction is the idea of the ${\bf{saturation}}$ of a multiplicative subset $S$ of $R$, denoted by

$\overline{S}:=\{x\in R\,\vert\,\,x\vert s\textrm{ for some } s\in S\}$

(of course, we're keeping with the convention that $0\notin S$).

With this, it's easy to prove that for any two multiplicative subsets $S,W$ of $R$, we have $R_W\subseteq R_S$ if and only if $\overline{W}\subseteq \overline{S}$. It also follows that $R_S=R_{\overline{S}}$.

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