# Integral basis of $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$

how i can find a integral basis $\mathbb{Z}_K$, if $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$?

-
Dear P.M.O., Have you tried imitating the usual computation for the integral basis of $\mathbb Q(\sqrt{D})$, namely: write an integer of $K$ in the form $x + y \sqrt{3},$ where $x,y \in \mathbb Q(\sqrt{2})$, and then determine conditions on $x$ and $y$. Regards, –  Matt E Oct 15 '12 at 3:00