Finiteness of the class number

Remarks

The standard proof uses the Minkowski bound. Let $n = [K : \Q]$, let $2r_2$ be the number of non-real embeddings of $K$ into $\C$, and let $d_K$ be the discriminant of $K$. By Minkowski's theorem on lattice points, every ideal class contains an integral ideal $\mathfrak{a}$ with norm $$N(\mathfrak{a}) \le \frac{n!}{n^n} \left(\frac{4}{\pi}\right)^{r_2} \sqrt{\abs{d_K}}.$$ There are only finitely many ideals of $\mathcal{O}_K$ of bounded norm, so there are finitely many classes. Example: for $K = \Q(\sqrt{-5})$, $n = 2$, $r_2 = 1$ and $d_K = -20$, so the bound is $\tfrac12 \cdot \tfrac4\pi \cdot \sqrt{20} \approx 2.85$; the ideals of norm at most $2$ are $\mathcal{O}_K$ and $(2, 1 + \sqrt{-5})$, which gives $h_K \le 2$. The hypothesis that $K$ is a number field is needed: for a general Dedekind domain the class group can be infinite.

Going deeper

For a student

The theorem is one of the two basic finiteness results of algebraic number theory, the other being Dirichlet's unit theorem. Both come from the geometry of numbers: $\mathcal{O}_K$ embeds as a lattice in $\R^n$, and Minkowski's convex body theorem produces small elements. The explicit bound makes class groups computable; for example it shows at once that $\Q(i)$, $\Q(\sqrt{-2})$ and $\Q(\sqrt{-3})$ have class number $1$. The size of $h_K$ is described by the class number formula, which links it to the residue of the Dedekind zeta function at $s = 1$.