Given the polynomial $$p(x) = x^{8}+x^{7}+x^{6}+x^{5}+2x^{4}+x^{3}+x^{2}+x+1$$
The task was to find the splitting field,a primitive element of the extension, the Galois group,including counting all the subgroup and the respective fixed field.
My effort : noticing that $$p(x) =\frac{x^{9}-1}{x-1}+x^{4} =\frac{(x^{5}-1)(x^{4}+1)}{x-1}$$
I was able to affirm that the splitting field of $p(x)$ is $\mathbb{Q}(\zeta_{5},\zeta_{8}) =\mathbb{Q}(\zeta_{40})$.
Besides, knowing that $Q(\zeta_{m}) \cap Q(\zeta_{m}) = Q(\zeta_{gcd(m,n)})$,
I was able to affirm that the Galois group $$Gal(\mathbb{K}/\mathbb{Q}) \cong Gal(\mathbb{Q}(\zeta_{5})/\mathbb{Q}) \times Gal(\mathbb{Q}(\zeta_{8})/\mathbb{Q}) \cong \mathbb{Z}_{4} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$$
With some patience, we discover that there are :
A single subgroup isomorphic to $(\mathbb{Z}_{2})^{3}$, six subgroups isomorphic to $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$, four subgroups isomorphic to $\mathbb{Z}_{4}$, seven subgroup isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{2}$, seven subgroups isomorphic to $\mathbb{Z}_{2}$ and two non-proper subgroups.
Since the group is abelian and I know three generators, I can state that three of those are $\phi,\psi,\sigma$,where : $$ \begin{cases} \phi : \zeta_{5} \longmapsto \zeta_{5}^{2} \\ \psi : \sqrt{2} \longmapsto -\sqrt{2} \\ \sigma : i \longmapsto -i \end{cases}$$
The question concerns the study of the fixed subfield.
I know (thanks to the fundamental theorem of Galois group) that for every subgroup of the Galois group $H$, $H$ is associated to a single fixed subfield,extension of $\mathbb{Q}$,which degree corresponds to the index of the subgroup, $[G:H]$,where $G = Gal(\mathbb{Q}(\zeta_{40})/\mathbb{Q})$.
I'm really struggling to find these fixed fields.
Sometimes product or sums easily work, but when the group is bigger, I don't know a general approach,or what to look for,since it seems difficult to me to find the extensions of $\mathbb{Q}$ and justify their diversity, other than quadratic, which diversity is easily justifiable. For example extensions of degree $8$ or $4$,which correspond to subgroups of order two or four, are the most problematic.
Any help,method or tip would be appreciated.
