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Localization of a ring $R$ at a multiplicative subset $S$ gives a (smallest) ring $S^{-1}R$ and a map $R\to S^{-1}R$ such that each $S$ has an inverse in $S^{-1}R$.

For simplicity just consider the monoid structure of $R$; recall that this corresponds to a category $BR$ which has a single object and whose unique hom-set is $R$. Then $S$ being a multiplicative subset means that it's a subcategory of $BR$. The localization $S^{-1}R$ is the smallest (in a sense) category $R$ with a functor $\gamma:BR\to S^{-1}R$ and such that each $s\in S$ is an isomorphism in $S^{-1}R$.

On the other hand, size issues aside the localization of a category $C$ at a subset $W\subset \text{Mor}(C)$ always exists (it doesn't have to be a subcategory).

So I wonder if this means that the localization of a ring exists for any subset of it, rather than only for multiplicative subsets, and if so if there is an elementary construction for it.

Thanks in advance.

Daniel Teixeira
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    hi Daniel; even if $S$ is not multiplicatively closed, you can still localize $R$ at $S$ eg by taking $S^{-1}R$ to be the quotient ring $R[t_s:s\in S]\big/\langle t_s s-1:s\in S\rangle$, and this will have the same universal property as when $S$ is multiplicatively closed. do you have in mind something different than this? – Atticus Stonestrom Nov 26 '21 at 20:28
  • thanks, that is exactly what I was looking for, but I didn't find any reference for it. if you wish you can post an answer so I can accept it – Daniel Teixeira Nov 26 '21 at 20:38
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    That is discussed here in many places, e.g. here and here – Bill Dubuque Nov 26 '21 at 20:52
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    Submonoids are allowed to contain $0$, but multiplicatively closed sets are not (by the standard definition). If you allow $0$, then the map $R \to S^{-1}R$ will not be an injection (because $S^{-1}R$ will be ${0}$ if $0 \in S$). – Rob Arthan Nov 26 '21 at 23:23
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    @RobArthan: Multiplicatively closed sets are absolutely allowed to contain $0$. And the map $R\to S^{-1}R$ may not be injective even if $0\not\in S$. – Eric Wofsey Nov 27 '21 at 15:04
  • @EricWofsey: apologies! I misremembered the definition. I didn't mean to suggest that the map $R \to S^{-1}R$ is an injection if $0 \not\in S$ (my comment doesn't actually say that). – Rob Arthan Nov 28 '21 at 01:10

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