[proofplan] We first make explicit the convention for $A_S$: it is the subring of $F$ obtained by allowing denominators supported only at primes in $S$. This localization is again a Dedekind domain, and its nonzero prime ideals are exactly the extensions of the nonzero prime ideals of $A$ outside $S$. The $K_0$ and $K_1$ computations then follow from the standard low-degree $K$-theory of Dedekind domains and rings of $S$-integers. Finally, applying the Dedekind-domain localization boundary to $A_S\subset F$ identifies the boundary with the divisor map over the remaining primes, so the primes in $S$ do not appear. [/proofplan]

[step:Identify the prime ideals of the localized Dedekind domain]Let $T_S\subset A$ denote the multiplicative set of elements whose prime ideal factorization involves only prime ideals in $S$. Equivalently, \begin{align*} T_S=\{a\in A\setminus\{0_A\}:v_{\mathfrak q}(a)=0\text{ for every nonzero prime ideal }\mathfrak q\notin S\}. \end{align*} With the convention in the statement, $A_S=T_S^{-1}A$ as a subring of the fraction field $F=\operatorname{Frac}(A)$. Since $A=\mathcal O_F$ is a Dedekind domain and localization of a Dedekind domain at a multiplicative set is again a Dedekind domain, $A_S$ is a Dedekind domain. The nonzero prime ideals of $A_S$ are precisely the ideals \begin{align*} \mathfrak p A_S \end{align*} where $\mathfrak p$ ranges over the nonzero prime ideals of $A$ satisfying $\mathfrak p\notin S$. If $\mathfrak p\in S$, then $\mathfrak p A_S=A_S$, because every element of $\mathfrak p$ becomes invertible after localizing at $T_S$ in the sense that the whole prime is removed from the spectrum. Thus the localization deletes exactly the prime ideals in $S$ and leaves the primes outside $S$.[/step]

[guided]The first point is to remove the ambiguity in the notation $\mathcal O_{F,S}$. In this proof, $S$ is the finite set of primes that are inverted. Define $T_S\subset A$ to be the multiplicative set of nonzero elements of $A$ whose valuation at every prime outside $S$ is zero: \begin{align*} T_S=\{a\in A\setminus\{0_A\}:v_{\mathfrak q}(a)=0\text{ for every nonzero prime ideal }\mathfrak q\notin S\}. \end{align*} Then localizing $A$ at $T_S$ gives exactly \begin{align*} A_S=T_S^{-1}A=\{x\in F:v_{\mathfrak p}(x)\ge 0\text{ for every nonzero prime ideal }\mathfrak p\notin S\}. \end{align*} Indeed, an element of $T_S^{-1}A$ has the form $a/t$ with $a\in A$ and $t\in T_S$. Since $t$ has no valuation contribution outside $S$, the valuation $v_{\mathfrak p}(a/t)$ is nonnegative for every $\mathfrak p\notin S$. Conversely, if $x\in F$ has nonnegative valuation outside $S$, the denominator ideal of $x$ is supported only on primes in $S$, so multiplying by an element of $T_S$ clears the denominator and shows $x\in T_S^{-1}A$. Now we use the standard commutative algebra fact that a localization of a Dedekind domain is again a Dedekind domain. The ring $A=\mathcal O_F$ is Dedekind, so $A_S$ is Dedekind. The correspondence between [prime ideals under localization](/theorems/2848) says that prime ideals of $T_S^{-1}A$ correspond to prime ideals of $A$ disjoint from $T_S$. A nonzero prime ideal $\mathfrak p\subset A$ is disjoint from $T_S$ exactly when $\mathfrak p\notin S$: primes in $S$ meet the multiplicative set and disappear, while primes outside $S$ remain. Therefore the nonzero prime ideals of $A_S$ are precisely \begin{align*} \mathfrak p A_S \end{align*} with $\mathfrak p\notin S$.[/guided]

[step:Compute $K_0(A_S)$ from the Dedekind-domain formula] Because $A_S$ is a Dedekind domain, [citetheorem:8688] applied to the Dedekind domain $A_S$ gives a natural isomorphism \begin{align*} K_0(A_S)\cong \mathbb Z\oplus \operatorname{Pic}(A_S), \end{align*} sending the class of a finitely generated projective $A_S$-module $P$ to \begin{align*} (\operatorname{rank}(P),[\det(P)]). \end{align*} For a Dedekind domain, the Picard group agrees with the ideal class group by [citetheorem:8647]. Applying this to $A_S$ yields \begin{align*} \operatorname{Pic}(A_S)\cong \operatorname{Cl}(A_S). \end{align*} Composing these isomorphisms gives \begin{align*} K_0(A_S)\cong \mathbb Z\oplus \operatorname{Cl}(A_S). \end{align*} [/step]

[step:Compute $K_1(A_S)$ from the arithmetic Dedekind-domain formula] The ring $A_S=\mathcal O_{F,S}$ is a ring of $S$-integers in the number field $F$. Therefore [citetheorem:8689] applies and gives an isomorphism induced by the determinant: \begin{align*} K_1(A_S)\cong A_S^\times. \end{align*} This proves the asserted $K_1$ computation. [/step]

[step:Identify the localization boundary with valuations outside $S$] Apply the Dedekind-domain localization boundary [citetheorem:8691] to the Dedekind domain $A_S$ with fraction field $F$. Its nonzero prime ideals are the primes $\mathfrak pA_S$ for $\mathfrak p\notin S$, and the residue field of $\mathfrak pA_S$ is canonically identified with $A/\mathfrak p$. Thus the boundary map has the form \begin{align*} \partial:K_1(F)\longrightarrow \bigoplus_{\mathfrak p\notin S}K_0(A/\mathfrak p). \end{align*} Under the identifications $K_1(F)\cong F^\times$ and $K_0(A/\mathfrak p)\cong \mathbb Z$, [citetheorem:8691] identifies this boundary with \begin{align*} x\longmapsto (v_{\mathfrak pA_S}(x))_{\mathfrak p\notin S}. \end{align*} For each $\mathfrak p\notin S$, the valuation $v_{\mathfrak pA_S}$ on $F^\times$ equals the original discrete valuation $v_{\mathfrak p}$ of $A$. Hence \begin{align*} \partial(x)=(v_{\mathfrak p}(x))_{\mathfrak p\notin S}. \end{align*} The primes in $S$ are absent because they are no longer nonzero prime ideals of $A_S$. Therefore inverting a prime removes its divisor summand from the boundary map and makes elements with nonzero valuation at that prime eligible to be units in $A_S$. This completes the proof. [/step]