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Let $R$ be a commutative ring with $1$. Let $S \subset R$ be a multiplicatively closed set. What are the units of $S^{-1}R$ ?

This question is probably too broad, so let's focus on integral domains $R$, and $0 \not \in S$ (so that the localization is not the zero ring and the natural morphism $i : R \to S^{-1}R$ is injective).

I think I proved that $$A:= \left\{ \dfrac{a}{s} \;\Big\vert\; s \in S, a \in R^{\times} \cup S \right\}$$ is a subgroup of $R^{\times}$. Notice that if $R$ is a domain and $0 \not \in S$, then $a/s$ is a unit iff there is $(a',s') \in R \times S$ such that $aa'=ss'$. I'm not sure that $(S^{-1}R)^{\times} = A$ holds. Anyway, it would be nice to have some explicit description of $(S^{-1}R)^{\times}$. I found quite nothing on that topic (except maybe this or this).

Thank you very much for your comments!

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    $\begingroup$ Isn't this also related? $\endgroup$
    – user26857
    Nov 3, 2016 at 20:57
  • $\begingroup$ @user26857 : Thank you. I wasn't aware of this recent question. I didn't search "invertible elements" but only "group of units"… $\endgroup$
    – Watson
    Nov 3, 2016 at 21:02
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    $\begingroup$ These questions may also be related: (1), (2), (3). $\endgroup$
    – Watson
    Nov 3, 2016 at 21:16

2 Answers 2

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Let us say the localization is $R[S^{-1}]$. Let $r \in R$ and $s \in S$. In $F$, the field of fractions of $R$, the inverse of $r/s$ is $s/r$. But since $s/r$ is an equivalence class and $s/r =(sq)/(rq),\; \forall q \in R - {0}$. Therefore $ s/r \in R[S^{-1}]$ if $(sq)/(rq) \in R[S^{-1}]$ for some choice of $q \in R-{0}$. We conclude that the set of invertible elements of $R[S^{-1}]$ is $ D = \{rs^{-1} \in F $ such that $qr\in S\}$. This set $D$ is known to be the saturation of $S$.

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  • $\begingroup$ Call two multiplicative subsets $S$ and $T$ “equivalent” if $R[S^{-1}]$ and $R[T^{-1}]$ are isomorphic $R$-algebras, and call a multiplicative subset “maximal” if it is maximal within its equivalence class. (For example, the maximal multiplicative subset of $\mathbb Z$ containing the powers of $18$ is the multiplicative submonoid generated by $2$ and $3$.) Is there any problem in the theory of localization if we require rings to be localized at maximal multiplicative subsets? Does the saturation admit a simpler expression in this case? $\endgroup$
    – isekaijin
    Jul 15, 2022 at 18:17
  • $\begingroup$ This answer assumes $R$ is an integral domain, even though the OP asked for any commutative unital ring $R$. $\endgroup$ Oct 25, 2023 at 18:03
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Denote $\ell:R\to S^{-1}R$ to the localization map. An element $a/s\in S^{-1}R$ is a unit if and only if $a\in T=\ell^{-1}((S^{-1}R)^\times)$, so it suffices to characterize $T$.

Recall that a subset $\Lambda\subset R$ is said to be saturated if $xy\in \Lambda$ implies $x,y\in \Lambda$. Define $$ \overline{S}=\{x\in R\mid\exists y\in R,xy\in S\}. $$ Then it is an exercise to show that $\overline{S}$ is the smallest saturated multiplicatively closed subset of $R$ containing $S$ (the set $\overline{S}$ is called the saturation of $S$). We claim $T=\overline{S}$. On the one hand, if $x\in T$, then $\frac{x}{1}\frac{y}{s}=\frac{1}{1}$ in $S^{-1}R$, for some $y\in R$, $s\in S$. Thus, there is $u\in S$ such that $xyu=u$. By definition, $x\in \overline{S}$, i.e., $T\subset\overline{S}$. Conversely, it is easy to see that $T$ is a saturated multiplicatively closed subset of $R$ containing $S$; hence, $\overline{S}\subset T$.

One also has that $A-\overline{S}$ equals the union of the prime ideals of $A$ disjoint from $S$ [ref], i.e., $\overline{S}$ is the intersection of all multiplicative closed subsets of $A$ containing $S$ that are of the form $A\setminus\mathfrak{p}$, where $\mathfrak{p}\subset A$ is a prime.

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