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I came across the following in my commutative algebra notes

If $R$ is a domain with quotient field $K$ and $S$ is a multiplicative set, then we have the inclusions $R \subseteq R_S \subseteq K.$

The set $R_S = \left\{\frac{r}{s} \, | \, r \in R, \, s \in S\right\}$ is the localization of $R$ at $S$.

What do the $\subseteq$'s mean in this context? To me, it can't mean set inclusion in the traditional sense because $R_S$ is a collection of equivalence classes $\frac{r}{s}$. Does $\subseteq$ mean that there is a monomorphism between the rings?

Thanks!

Edit To basically ask the same question with different terminology, does $R \subseteq R_S$, for example, mean that we can embed R into $R_S$?

J. Dunivin
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  • as far as I know the symbol $\subseteq$ just have a meaning in mathematics – Masacroso Nov 21 '21 at 04:46
  • @Masacroso I thought that too. But if that's true, how can an element $r \in R$ also be in $R_S$ if $R_S$ is the set of equivalence classes? – J. Dunivin Nov 21 '21 at 04:48
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    Every fraction $r/s$ is an element of $K$ isn't it? And yes, $R$ is a subset of $S^{-1}R$ (in this definition): every element $r\in R$ is expressible as $r/1\in S^{-1}R$. When you generalize this definition to contexts where zero divisors may appear, that's when you need equivalence classes. – anon Nov 21 '21 at 05:02
  • @runway44 That makes sense! If you make this comment into an answer, I will gladly accept it =) – J. Dunivin Nov 21 '21 at 05:08
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    @runway44 Not unless you perform set-theoretic surgery to splice the domain into the quotient field. It all depends on how the quotient field is constructed set theoretically (which we usually ignore in algebra since one key point of algebra is to abstract away from any internal structure of elements). Here what is important is the universal property of the localization - which yields the sought unique embedding. – Bill Dubuque Nov 21 '21 at 06:34

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