Let $F = \mathbb{Q}(\sqrt{13})$. Then its ring of integers is $\mathcal{O}_F = \mathbb{Z}\left [\frac{1 + \sqrt{13}}{2}\right ]$.
I am trying to find the class number of $F$.
$F$ has signature $(r,s) = (2,0)$, discriminant $\Delta_F = 13$, and degree $[F: \mathbb{Q}] = 2$.
Hence, by Minkowski's bound for ideal classes, each ideal class $C \in Cl_F$ contains an ideal $I \subseteq \mathcal{O}_F$ with absolute norm satisfying: $$ N(I) \leq \left ( \frac{4}{\pi} \right )^s \frac{n!}{n^n} \sqrt{| \Delta_F |}$$ Hence, with our example, we get $$N(I) < 2$$ and there is no prime less than $2$, so how do I find the generators for the class group now? The generators of the class group are of the form $[P]$, where $P$ is a prime ideal lying above $p$, where $p$ is a prime satisfying $p < 2$. But there is no such prime? So what can I deduce now?
I assume that this shows that the class group is trivial, but I'm not sure why.