Chain of intermediate fields $Q ⊂ Q(α_ 8 ) ⊂ Q(α_ 4 ) ⊂ Q(α_ 2 ) ⊂ Q(α)$ Let α be the real positive 16th root of 3 and consider the field F = Q(α) generated by α over
the field of rational numbers. Observe that there is a chain of intermediate fields
$Q ⊂ Q(α_
8
) ⊂ Q(α_
4
) ⊂ Q(α_
2
) ⊂ Q(α) = F$.
(a) Compute the degrees of these intermediate field extensions and conclude they are all distinct.
(b) Show that every intermediate field K between Q and F is one of the above (hint: consider
the constant term of the minimal polynomial of α over K)
I think the degrees are all easy to find by looking at the minimum polynomials of the adjoined elements in each case:
$α_8$ is $x^2 - 3$ --> degree 2
$α_4$ is $x^4 - 3$ --> degree 4
$α_2$ is $x^8 - 3$ --> degree 8
$α$ is $x^{16} - 3$ --> degree 16
As for these being the only intermediate fields, does it have to do with 3 being congruent to 3 mod 4 implying $Z[\sqrt{3}]$ is integrally closed in its field of fractions. I'm not sure thats just a guess.
 A: Let $K$ be the splitting field of the polynomial $x^{16}-3$ then $K$ is generated by $\alpha = \sqrt[16]{3}$ and $\omega = e^{\frac{2\pi i}{16}}$ so a basis for $K$ is $1, \alpha, \alpha^2, \ldots, \alpha^{15}, \omega, \omega \alpha, \omega \alpha^2, \ldots, \omega \alpha ^{15}, \ldots, \omega^i, \omega^i \alpha, \omega^i \alpha^2, \ldots, \omega^i \alpha^{15}, \ldots$ where $i \in \{0, 1,\ldots,7\}$. The dimension of $K$ is then $16*8 = 128$ (in fact $K$ can be considered as the tensor product of $\mathbb{Q}(\alpha)$ with the cyclotomic field of degree $16$). An automorphism of $K$ is thus completely known if its values on $\alpha$ and $\omega$ are known. Since $\alpha$ is a root of $x^{16}-3$ its value must also be a root so there are automorphisms of the form $\phi_k(\alpha) = \omega^k \alpha, \phi_k(\omega) = \omega$ with $k \in \{0,1,\ldots,15\}$. Another type of automorphism is of the form $\psi_i(\alpha) = \alpha, \psi_i(\omega) = \omega^i$ where $i \in \{1,3,5,7,9,11,13,15\}$ (the latter because $\omega$ is a primitive root of unity so its image must also be a primitive root of unity). Every other automorphism of $K$ is the composition of any of these two types $\phi_k$ and $\psi_i$ of automorphisms. In fact the $\phi_k$ form the cyclic group $C_{16}$ of order $16$ (in fact we have $\phi_k = \phi_1^k$), and the $\psi_i$ form the Galois group $H$ of the cyclotomic field of degree $16$ i.e. $C_4 \times C_2$. Inspection shows that $\psi_i^{-1} \phi_k \psi_i$ maps $C_{16}$ to $C_{16}$ so that the Galois group $G$ of $K$ is the semi-direct product $G = C_{16} : (C_4 \times C_2)$. 
The fields in the question don't contain any expression in $\omega$ so it is clear that the corresponding Galois groups all have to contain the $\psi_i$. The smallest such subgroup is $H = C_4 \times C_2$ itself and it corresponds to the field $F$. The next step is to take a group that contains $H$ and intersects $C_{16}$ in a subgroup of order $2$, this is obtained by adding the element $\phi_8$ to $H$ giving a group that fixes $\alpha^2$. This procedure can be continued by adding $\phi_4$ and $\phi_2$ giving subgroups that fixes $\alpha^4$ and $\alpha^8$ respectively.
