Number Theory – Relation Between S-Ideal Class Group and Usual Ideal Class Group

algebraic-geometryalgebraic-number-theoryarithmetic-geometryideal-class-groupnumber theory

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, p.$452$
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, 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.

Best Answer

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 $v \in S$.

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