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

$\displaystyle \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 19.2.1   There is a natural bijection between $ F_K$ and the group of nonzero fractional ideals of $ \O_K$. The correspondence is induced by

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

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

$\displaystyle \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

$\displaystyle \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 class group $ C_K$ of the number field $ K$ is by definition the quotient of $ F_K$ by the image of $ K^*$.

Theorem 19.2.2   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 19.1.12), hence compact. But a compact discrete group is finite. $ \qedsymbol$



Subsections
William Stein 2012-09-24