Ideals and Divisors

Suppose that $K$ is a finite extension of $\mathbf{Q}$. Let $F_K$ be the the free abelian group on a set of symbols in bijection with the non-archimedean valuation $v$ of $K$. Thus an element of $F_K$ is a formal linear combination

$$\sum_{v\text{ non arch.}} n_v \cdot v$$

where $n_v\in\mathbf{Z}$ and all but finitely many $n_v$ are 0.

Lemma 21.2.1   There is a natural bijection between $F_K$ and the group of nonzero fractional ideals of $\O _K$. The correspondence is induced by

$$v\mapsto \wp_v = \{x \in \O _K : v(x)<1\},$$

where $v$ is a non-archimedean valuation.

Endow $F_K$ with the discrete topology. Then there is a natural continuous map $\pi:\mathbb{I}_K \to F_K$ given by

$$\mathbf{x}= \{x_v\}_v \mapsto \sum_v \ord _v(x_v)\cdot v.$$

This map is continuous since the inverse image of a valuation $v$ (a point) is the product

$$\pi^{-1}(v) = \pi \O _v^* \quad \times \prod_{w \text{ archimedean}} K_w^*\quad \times \prod_{w\neq v \text{ non-arch.}} \O _w^*,$$

which is an open set in the restricted product topology on $\mathbb{I}_K$. Moreover, the image of $K^*$ in $F_K$ is the group of nonzero principal fractional ideals.

Recall that the $C_K$ of the number field $K$ is by definition the quotient of $F_K$ by the image of $K^*$.

Theorem 21.2.2   The class group $C_K$ of a number field $K$ is finite.

Proof. We first prove that the map $\mathbb{I}_K^1\to F_K$ is surjective. Let $\infty$ be an archimedean valuation on $K$. If $v$ is a non-archimedean valuation, let $\mathbf{x}\in \mathbb{I}_K^1$ be a $1$-idele such that $x_w=1$ at ever valuation $w$ except $v$ and $\infty$. At $v$, choose $x_v = \pi$ to be a generator for the maximal ideal of $\O _v$, and choose $x_\infty$ to be such that $\left\vert x_\infty\right\vert _\infty = 1/\left\vert x_v\right\vert _v$. Then $\mathbf{x}\in\mathbb{I}_K$ and $\prod_{w}\left\vert x_w\right\vert _w = 1$, so $\mathbf{x}\in \mathbb{I}_K^1$. Also $\mathbf{x}$ maps to $v \in F_K$.

Thus the group of ideal classes is the continuous image of the compact group $\mathbb{I}_K^1/K^*$ (see Theorem 21.1.12), hence compact. But a compact discrete group is finite. $\qedsymbol$