Suppose $f\in \mathbb{Z}[x]$ is an irreducible polynomial with Galois Group $S_4$. If $\theta$ is a root of $f$. Say $K=\mathbb{Q}(\theta)$
- Prove that there is no field poperly contained in $\mathbb{Q}(\theta)/\mathbb{Q}$
- Is $\mathbb{Q}(\theta)/\mathbb{Q}$ a Galois Extension?
- Is $\theta$ constructible by ruler and compass?
If $E$ is the splitting field of $f$ over $\mathbb{Q}$. And since we are working over a field of characteristic $0$ and with an irreducible polynomial. So I will use the Fundamental Theorem of Galois Theory and using that I shall decide things about $\mathbb{Q}(\theta)$ I believe that $[\mathbb{Q}(\theta):\mathbb{Q}]=4$ and hence the corresponding Galois Group $H=\operatorname{Gal}(E/K)$ has order $6$
Now the only possibilites for $H$ are $\mathbb{Z}_6$ and $S_3$. Now I feel $H=S_3$ and if there are intermediate fields between $\mathbb{Q}(\theta)$ and $\mathbb{Q}$, then it gives me a subgroup of $S_4$ of order $12$ containing $S_3$. Which is a contradiction.
Am I correct? Can any help me with the solution because I am trying to understand things.