Let $L_0$ be the splitting field of $f=x^4 + 2x + 2$ over $K=\mathbb{Q}_2$.
We take two distinct roots $\alpha_1,\alpha_2 \in L_0$ and set $C = (\alpha_2-\alpha_1)^4$, and set $L = L_0(\sqrt[3]{C})$.
Question: What is the Galois group of the extension $L/K$?
Observations which I found and that may help:
- The Galois group of $L_0/K$ is $S_4$ since it can be shown that the resolvent cubic of $f$ is irreducible over $\mathbb{Q}_2$ and the discriminant of $f$ is no square over $\mathbb{Q}_2$.
- The extension $L/L_0$ is Galois since $L_0$ contains some (and therefore all) primitive third roots of unity. Therefore, $L/L_0$ is cyclic of degree $3$.
- Therefore, the Galois group $G=\operatorname{Gal}(L/K)$ must have $\operatorname{Gal}(L/L_0) = C_3$ as a subgroup and $\operatorname{Gal}(L_0/K) = S_4$ is a quotient of $G$.
Can someone give me a hint on how to continue from here?