While self-studying algebraic number theory, I came across the following problem:
Prove that $\mathbb{A} \cap \mathbb{Q}(\sqrt{2},\sqrt{-3})$ is a PID.
where $\mathbb{A} \cap \mathbb{Q}(\sqrt{2},\sqrt{-3})$ denotes the ring of algebraic integers in $\mathbb{Q}(\sqrt{2},\sqrt{-3})$. My first intuition was to find the Minkowski bound and hope that it's below $2$, but this turns out to not be the case. Let $R := \mathbb{A} \cap \mathbb{Q}(\sqrt{2},\sqrt{-3})$; then we know that $|\text{disc}(R)| = 16\cdot2\cdot3\cdot6$, so the Minkowski bound is $$ \frac{4!}{4^4}\left(\frac{4}{\pi}\right)^2\sqrt{16\cdot 6 \cdot 3 \cdot 2} \approx 3.64. $$
Thus, I need only show that the primes lying over $2$ and $3$ are principally generated, but I'm having trouble showing that. I don't (in general) know how primes split in this ring, so I tried to use the different. We know that $\| \text{diff}(R) \| = |\text{disc}(R)| = 2^6\cdot 3^2$, so we can conclude $$(3) = (\frak{p}_3\frak{p}_3')^2. $$
I'm not sure how this helps, though. Any help is appreciated.