= Class number one for Q square root of fourteen
{title2=$h(\mathbb Q(\sqrt{14}))=1$}
Every <ideal class> of this real quadratic <number field> contains an <integral ideal> of <ideal norm> at most $\sqrt{14}<4$, by the diamond form of the <Minkowski convex body theorem>. The only possible norms are one, two and three. The sole <prime ideal> above two has norm two and is principal, generated by $4+\sqrt{14}$, whose <field norm> is two. The prime three is inert, since $X^2-14$ is irreducible modulo three, so there is no ideal of norm three. Every class therefore has a principal small representative, and the <ideal class group> is trivial.
This illustrates why the small-ideal bound can prove a class number without listing arbitrary ideals: only prime ideals of a few possible norms need to be tested. Real quadratic fields require the absolute <field norm>; the generator here happens to have positive norm.
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