I want to compute the class number of $K=\mathbb{Q}(\sqrt{6})$, the result should be $1$.
The discriminant $d_K$ is $24$, so the Minkowski bound is $M_K = \sqrt{6} < 3$. So every ideal class contains an ideal in $\mathcal{O}_K$ with ideal norm less than or equal to $M_K$.
I know that $2$ is a norm of the principal ideal $(2 + \sqrt{6})$, in fact $2 = -N(2 + \sqrt{6})$. This is a principal ideal, so the class number is $1$.
Now I wonder if one can, after computing $M_K$, see that the class number is $1$ by knowing that $2$ is ramified (because $6 \equiv 2 \bmod 4$)? That means $2 \mathcal{O}_K = P^2$ for a prime ideal $P$.