Suppose you are working with $K=\mathbb{Q}[\sqrt{m},\sqrt{n}]$ and $K_1=\mathbb{Q}[\sqrt{m}],K_2=\mathbb{Q}[\sqrt{n}], K_3=\mathbb{Q}[\sqrt{k}]$ where $m,n$ are squarefree integers and $k = mn/gcd(m,n)^2$.
I want to prove that a prime $p \in \mathbb{Z}$ is never inert in each of the $K_i$
Proof: By contradiction. If $p$ were inert in each of the subfields, then the inertia field $K_E$ must contain each of the subfields since it is the largest subfield such that the ramification index is $1$ and clearly, in $K_i$s the ramification indices are all equal $1$. Then $K_E = K$. Moreover, the decomposition field $K_D$ is the smallest subfield where a prime $Q$ in $K$ lying over $p$ is the only prime in $K$ lying over $p$. However, $K_i$s have this property (I'm not pretty sure about this point, maybe it's better using the fact that $K_D$ is the largest subfield where ramification index and inertia degree equal $1$? What do you think?). Then $K_D = \mathbb{Q}$.
Using Galois correspondence, $D=G(K/\mathbb{Q})$ and $E=\{ 1 \}$. Finally $D/E \cong G(K/\mathbb{Q})$. The latter is the Knlein group which is not cyclic, while $D/E$ is cyclic. Contraddiction
(I used notation as in Marcus'bokk "Number Theory", moreover I used Theorem 29 and Corollary 1 of theorem 28 ).
What do you think? Any suggestions?
