Let $\Sigma$ be a splitting field in $\mathbb C$ of the polynomial $f(x)=x^4-49$ over $\mathbb Q$.
Construct $\Sigma$ and find the degree of the field extension $\Sigma:\mathbb Q$.
Determine the elements and the structure of the Galois group $\Gamma(\Sigma:\mathbb Q).$
Find all the subgroups of $\Gamma(\Sigma : \mathbb Q)$ and their fixed fields (justify answer).
Find all the subfields of $\Sigma$ (justify answer).
I think I've got the first two figured out with $\Sigma = \mathbb Q(i,\sqrt 7)$ and $[\mathbb Q(i,\sqrt7)]=4$.
Every element $\gamma\in\Gamma(\Sigma:\mathbb Q)$ is completely and uniquely determined by the images $\gamma(\sqrt7)$ and $\gamma(i)$ which must be roots of $x^2-7$ and $x^2+1$ respectively. Thus we have $\gamma(\sqrt 7 )=\pm\sqrt 7$ and $\gamma(i)=\pm i$. Hence the Galois group has the structure of the Klein 4 group where its elements are the automorphisms:
$Id: \begin{cases} \sqrt7 \mapsto \sqrt7 \\ i \mapsto i \end{cases}$
$\phi: \begin{cases} \sqrt7 \mapsto \sqrt7 \\ i \mapsto -i \end{cases}$
$\psi: \begin{cases} \sqrt7 \mapsto -\sqrt7 \\ i \mapsto i \end{cases}$
$\psi\phi: \begin{cases} \sqrt7 \mapsto -\sqrt7 \\ i \mapsto -i \end{cases}$
- I'm struggling a little with part 3 and 4. If I had to guess I would say that the fixed fields were $\mathbb Q(i)$ and $\mathbb Q(\sqrt 7)$ and $\mathbb Q(i\sqrt 7)$ but I'm unsure how to completely prove it.
Could anyone offer any help for the last two parts?