I has recently met with the problem: Show that the class number of $K= \mathbb{Q}[\sqrt{2},\sqrt{-3}]$ is 1.
I has already proved that $\left\{1,\sqrt{2},\frac{1+\sqrt{-3}}{2},\frac{\sqrt{2}+\sqrt{-6}}{2} \right\}$ is the integral basis of $\mathcal{O}_K$ with $\mathcal{O}_K$ is the ring of algebraic integers of $K$. I'm trying to use the Minkowski bound to this problem and the Minkowski bound of $K$ is approximately $3,6$; so that we need to factorize $2 \mathcal{O}_K, 3 \mathcal{O}_K$ and prove that these factors are principal, then we finish.
I has known that $2 \mathcal{O}_K = \left(\sqrt{2} \mathcal{O}_K \right)^2$, $\sqrt{2} \mathcal{O}_K$ is prime (by proving that ${\raise0.5ex\hbox{$\scriptstyle \mathcal{O}_K$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle \sqrt{2}\mathcal{O}_K$}} $ is a field with 4 elements) and obviously $\sqrt{2}\mathcal{O}_K$ is principal. Can we do the similar work to $3 \mathcal{O}_K$ or do we have the other way to find the solution for this problem?
Thank you for all your help.