Group of units of localization

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).

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

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 $$S$$.
• 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? Jul 15, 2022 at 18:17
• This answer assumes $R$ is an integral domain, even though the OP asked for any commutative unital ring $R$. Oct 25, 2023 at 18:03
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.