Minkowski bound

Remarks

The proof embeds $\mathcal{O}_K$ and its ideals as lattices in $\R^n$. Minkowski's theorem on convex bodies then gives, in each nonzero ideal $\mathfrak{b}$, a nonzero element $\alpha$ with $\abs{N_{K/\Q}(\alpha)} \le M_K N(\mathfrak{b})$. Applied to an ideal $\mathfrak{b}$ in the inverse class, the ideal $\alpha \mathfrak{b}^{-1}$ is integral, lies in the given class and has norm at most $M_K$.

Consequence: the class group is generated by the classes of the prime ideals of norm at most $M_K$. Since there are only finitely many ideals of bounded norm, the class group is finite.

Example: for $K = \Q(\sqrt{-5})$ one has $n = 2$, $r_2 = 1$, $d_K = -20$, so $M_K = \sqrt{20} \cdot \frac{4}{\pi} \cdot \frac{2}{4} \approx 2.85$. Only the prime ideal $\mathfrak{p} = (2, 1 + \sqrt{-5})$ of norm $2$ needs checking. It is not principal, since $a^2 + 5b^2 = 2$ has no integer solution, and $\mathfrak{p}^2 = (2)$; so the class number is $2$. For $K = \Q(i)$, $M_K = 2 \cdot \frac{4}{\pi} \cdot \frac{2}{4} \approx 1.27 < 2$, so every class contains the ideal $(1)$ and $\Z[i]$ is a principal ideal domain.

Going deeper

For a student

The bound is the standard tool for computing class numbers by hand for fields of small degree. Since every class contains an integral ideal of norm at least $1$, it also gives $M_K \ge 1$, which yields a lower bound for $\abs{d_K}$; this is the Hermite–Minkowski theorem. The same geometry of numbers proves Dirichlet's unit theorem.