How do I show that:
Let $f$ be an (irreducible) separable quartic over $K$ and $u$ a root of $f$. There is no field properly between $K$ and $K(u)$ if and only if the Galois group of $f$ is either $A_4$ or $S_4$.
I know that I have to consider the resolvent cubic of $f$, and consider $m=\left[K(a,b,c):K\right]$, where $a,b,c$ are the roots of the resolvent cubic. Then I have to show that $m=3$ or $m=6$. How can this be done?
And how do I work with the other direction?
I believe this should work : $[K(u):K] = 4$, so $K(u)$ corresponds to a subgroup $H$ of the Galois group $G$ such that $[G:H] = 4$. A proper sub-field $K < L < K(u)$ would correspond to a proper subgroup $H < M < G$. In this case, $[G:M] = [M:H] = 2$.
Thus, your problem can be restated as follows : The Galois group $G$ does not have subroups $M$ and $H$ as above iff $G = A_4$ or $S_4$.
So, the question is, which Galois groups admit such subgroups.
Now your options for $G$ in this case are :
$m=1$, and $G = \mathbb{Z}_2\times \mathbb{Z}_2$ : Here, $H = \{e\}$, and $M = \mathbb{Z}_2\times \{0\}$ works.
$m=2$, and $G = \mathbb{Z}_4$ or $D_4$: Here, either $H= \{e\}$ or $H \cong \mathbb{Z}_2$. In either case, the Sylow theorems guarantee a subgroup $M$ as above.
$m=3$, and $G = A_4$ : We know $A_4$ does not have a subgroup of index 2.
$m=6$ and $G = S_4$ : $S_4$ has a unique subgroup of index 2, so $M = A_4$. But then $A_4$ does not have a subgroup of index 2.
This completes the analysis in both directions.
