TITLE: Does there exist a prime number $p$ such that $p\mathcal{O}_{K}$ in $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a prime ideal?
QUESTION [9 upvotes]: Problem: Prove or disprove: there exists a prime number $p$ such that $p\mathcal{O}_{K}$ in $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a prime ideal, where $\mathcal{O}_K$ denotes the ring of algebraic integers in $K$.

The following is my idea:
Suppose $p\mathcal{O}_K=p$ is a prime ideal. Then we know that $g=1$, $e=1$, and $f=4$ from $4=efg$. Now, denote by $D$ and $I$ the decomposition group and the inertia group for $p$. Then, $|D|=ef=4$ and $|I|=e=1$.
On the other hand, $K/\mathbb{Q}$ is Galois with $G=\mathbb{Z}_2 \times \mathbb{Z}_2$, and I can find all intermediate fields:
$$K_1=\mathbb{Q}(\sqrt{2}),\qquad K_2=\mathbb Q(\sqrt 3),\qquad K_3=\mathbb Q(\sqrt 6).$$
For a given $p$, I try to determine whether or not $p$ is inert in $K_i$  ($i=1,2,3$). But I guess this cannot work.
That's all I try. Please help me! Thank you very much!

REPLY [3 votes]: If $p$ is totally inert then the reduction map is an isomorphism $D_p \rightarrow$ Gal($\mathbb{F}_{p^4}/\mathbb{F}_p$) which is known to be cyclic of order 4. (Here $\mathbb{F}_{p^4}$ arises as the residue field of $\mathcal{O}_K/p$ and its degree over $\mathbb{F}_p$ is just the residue degree of $p$.)
This means that $D \cong \mathbb{Z}/4\mathbb{Z}$ which is obvious nonsense so no such $p$ can exist.