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!
Best Answer
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 $\textit{saturation }$ of S.