Relation between $S$-ideal class group and usual ideal class group For $\mathcal{O}_{K}$, the integer ring of a global field, we denote $S$ to be any set of primes of a global field $K.$ Let
$$\mathcal{O}_{K,S}:=\{x\in K\mid v_{\mathfrak{p}}\geq 0\text{ for }\mathfrak{p}\notin S\}$$ be the ring of $S$-integers of $K$ (see Neukirch, Schmidt, Wingberg Cohomology of Number Fields, Ch. VIII, § 3).
Ideal class group of $\mathcal{O}_{K,S}$ is called $S$-ideal class group.
　Neukirch, Schmidt, Wingberg Cohomology of Number Fields, Ch. VIII, § 3,$p452$
states that

$S-$ideal class group is the quotient of the usual ideal class group
$Cl_K$ of $K$ by the subgroup generated by the classes of all prime
ideals in $S$.

without no more explanation.
How can I prove this statement?
My try and thought :
　Let $X＝\operatorname{Spec}(\mathcal{O}_{K})$, $X_S＝\operatorname{Spec}(\mathcal{O}_{K,S})$.
Natural map $X_S→X$ induces natural surjective map $f:\operatorname{Pic}(X)→\operatorname{Pic}(X_S)$.Thus, $\operatorname{Pic}X/\ker f$ is isomorphic to $\operatorname{Pic}(X_S)$(Hartshone's proposition $Ⅱ6.5$).
So, I need to prove $\ker f$ is generated by the classes of all prime ideals in $S$.
This is algebraic geometrical point of view, I want to accomplish this kind of proof, but another algebraic number theoretical approach is also appreciated.
 A: This is analogous to the Picard group argument you laid out, but anyway:

*

*Let
$$A_S = \{a \in O_K: v(a) > 0 \text{ for some } v \in S\}.$$
Then $O_{K,S} = A_S^{-1}O_K$ and is still a Dedekind domain.

*Let $Cl(K) = Frac(O_K) / P(O_K)$, the fractional ideals mod principal ideals. We want to see how fractional ideals/principal ideals change from $O_K$ to $O_{K,S}$.

*

*Primes in $O_{K,S}$ correspond to primes in $O_K$ that doesn't intersect $A_S$ (via pullback).

*$Frac(O_K) = \oplus_v \mathbb{Z}v$ for primes $v \in K$. By last bullet, $Frac(O_{K,S}) = \oplus_{v \not\in S} \mathbb{Z}v$ for primes $v \in K$.

*The natural map $Frac(O_K) \to Frac(O_{K,S})$ is the projection map dropping the $v \in S$ components. This is surjective, and clearly descends to $Cl(K) \to Cl_S(K)$.



*Finally, we need to figure out the kernel $Cl(K) \to Cl_S(K)$.

*

*Let $I \in \oplus_v \mathbb{Z}v$. If it becomes principal $ = div(a) \in Cl_S(K)$ for some $a \in K$, this exactly means factorization of $(a)^{-1} I$ only involves $v \in S$, i.e.
$$(a)^{-1}I \in \oplus_{v \in S} \mathbb{Z}v,$$
i.e. $I$ lies in the subgroup of $Cl(K)$ generated by $v \in S$.



This means the kernel is exactly subgroup of $Cl(K)$ generated by $v \in S$.
