TITLE: Is $3$ prime in the ring of integers of the field $\mathbb{Q}(\sqrt{2\sqrt{2}-1})$?
QUESTION [7 upvotes]: I am trying to determine if the number $3$ stays prime in the ring of integers of the quartic field $K=\mathbb{Q}(\sqrt{2\sqrt{2}-1})$, or rather adjoin a real root of $X^4+2X^2-7$. 

I do know that $3$ stays prime in the ring of integers for the quadratic subfield $\mathbb{Q}(\sqrt{2})$ and that $3$ is not ramified in $O_K$.
I am a topologist working on my thesis, and any algebraic ideas on how to think about this question are welcome.

REPLY [4 votes]: To use theorem 27 from Marcus's Number Fields, if we let $\alpha = \sqrt{2\sqrt{2} - 1}$ be a real root of $f(x) = x^4 + 2x^2 - 7$, you just need to check that 3 does not divide $|\mathcal{O}_K/\mathbb{Z}[\alpha]|$, which can be accomplished by noting that $$\mathrm{disc}(1,...,\alpha^3) = |\mathcal{O}_K/\mathbb{Z}[\alpha]|^2 \mathrm{disc}(\mathcal{O}_K). $$ So it suffices to check that 3 does not divide the left hand side, which is equal to $$N^K_\mathbb{Q}(f'(\alpha)) = N^K_\mathbb{Q}(4\alpha(\alpha^2 + 1)),$$ which we can evaluate using multiplicativity and transitivity of the norm. It turns out this norm is just a power of 2 times 7, so we can use theorem 27. But $f$ has no roots $\bmod 3$, and one can check that it does not factor into quadratics $\bmod 3$. So 3 remains prime.