Let $K$ be a quadratic extension and let $F/K$ be an abelian Galois extension with Galois group $H=Gal(F/K)$. Assume that $F/\mathbb Q$ is Galois with Galois group given by the semidirect product $Gal(F/\mathbb Q)\simeq Gal(K/\mathbb Q)\ltimes H$, where the generator $\tau$ of the quotient $G/H\simeq Gal(K/\mathbb Q)$ of $G$ acts on the subgroup $H$ of $G$ by $g\mapsto \tilde{\tau}g\tilde{\tau}^{-1}$, and $\tilde{\tau}$ is any lift of $\tau$ to $G$. Show that if a prime $l$ of $\mathbb Q$ is inert in $K$, then it splits completely in $F$.
The sense is that the unique prime ideal $(l)=l\mathcal{O}_K$ of $K$ above $l$ splits completely in $F$.
I have no idea how to work with this problem. Some ideas?
I think this is not true. Consider $K = \mathbb{Q}(\sqrt{-7})$ and $F = \mathbb{Q}[x]/(x^6 - 3x^5 + x^4 + 3x^3 + x^2 - 3x + 1)$. Then, Gal$(F/\mathbb{Q}) = S_3$ and $5$ is ramified in $F$. But in $K$, it is inert.
See : LMFDB