Q:

Finding the intermediate fields of $\Bbb{Q}(\zeta_7)$.

Let $F= \Bbb{Q}(\zeta_7)$ with $\zeta_7 = e^{2\pi/7}$.
a) What is the Galois group of $F$ over $\Bbb{Q}$?
b)Find all intermediate fields between $\Bbb{Q}$ and $F$. (Write each in the form $\Bbb{Q}(\alpha)$ for some specific $\alpha \in F$.)
c) For each intermediate field $E$ above, give the Galois group of $E$ over $\Bbb{Q}$.

a) Since $\Bbb{Q}(\zeta_7)$ is the splitting field for the cyclotomic polynomial $x^6+x^5+...+1$, $Gal(\Bbb{Q}(\zeta_7)/\Bbb{Q}) \cong U_7 \cong \Bbb{Z}^{\times}_7 \cong \Bbb{Z}_6$.
b) $Gal(\Bbb{Q}(\zeta_7)/\Bbb{Q}) \cong \Bbb{Z}_6$
So we have the following diagram for $\Bbb{Z}_6$,

And a corresponding diagram for $\Bbb{Q}(\zeta_7)$ for some $\alpha$ and $\beta$

Since $\Bbb{Q}(\zeta_7)$ has complex numbers, we can look at $\Bbb{Q}(i)$ to see if it has degree 2 or 3 over $\Bbb{Q}$. Since $x^2+1$ is its minimal polynomial, we see that $[\Bbb{Q(i)} : \Bbb{Q}]=2$ and so it must correspond to $\Bbb{Z}_3$, and $\beta=i$.
Sending every number to its conjugate is one possible automorphism. Of course, it has order 2 and so is isomorphic to $\Bbb{Z}_2$. Since it fixes every real number, the corresponding intermediate field must be of the form $\Bbb{Q}(\alpha)$ for some real number $\alpha$, and must have degree 3 over $\Bbb{Q}$.
If we look at $\zeta_7 + \zeta_7^6$, we see that
$\cos(2\pi/7) + i\sin(2\pi/7) + cos(12\pi/7) + i\sin(12\pi/7)$
$= \cos(2\pi/7) + i\sin(2\pi/7) + \cos(-2\pi/7) + i\sin(-2\pi/7)$
$= 2\cos(2\pi/7)$
which is irrational. So $\Bbb{Q}(\zeta_7 + \zeta_7^6)$ is either a proper subfield of $\Bbb{Q}(\alpha)$ or is equal to it. But from the diagram, we know that there is only one other intermediate field, so $\Bbb{Q}(\zeta_7 + \zeta_7^6) = \Bbb{Q}(\alpha)$.
c) The corresponding Galois group of $\Bbb{Q}(i)$ is isomorphic to $\Bbb{Z}_3$ and that of $\Bbb{Q}(\zeta_7 + \zeta_7^6)$ is isomorphic to $\Bbb{Z}_2$.
Do you think this is correct?
Thanks in advance

A:

Here's a trick that often works in this kind of problem. 
Take an element $\alpha$ of $F$ --- an element that generates $F$ over the ground field. 
Take a subgroup $H$ of the Galois group. 
Form $\sum_{\sigma{\rm\ in\ }H}\sigma(\alpha)$. This is guaranteed to be invariant under $H$ and thus to live in some subfield. Unless you're unlucky, it generates the fixed field of $H$. 
In particular, this is what @ccorn has done in the other answer. 

A:

There is no reason to assume $\mathrm{i}\in F$, so try another $\beta$. I would suggest
$$\beta = \zeta_7 + \zeta_7^2 + \zeta_7^4$$
You will easily verify that $\beta+\bar{\beta}=-1$ and $\beta\bar\beta=2$ (thus $\beta$ is irrational) and that $$\mathbb{Q}[X]\ni\operatorname{minpoly}_{\mathbb{Q}}(\beta)=X^2-(\beta+\bar{\beta})X+(\beta\bar{\beta}) = X^2+X+2$$
Thus $[\mathbb{Q}(\beta):\mathbb{Q}] = 2$.
Edit: Fixed the sloppy notation for $\operatorname{minpoly}_{\mathbb{Q}}(\beta)$. See Gerry Myerson's answer for finding a candidate $\beta$ in general.
Edit 2: $\mathrm{i}\not\in F$ because $\mathrm{i}\zeta_7$ is a primitive $28$th root of unity, so $F(i)$ contains $\zeta_{28}$, whereas $[\mathbb{Q}(\zeta_{28}):\mathbb{Q}] = 12 \neq 6 = [F:\mathbb{Q}]$.
Edit 3: Clarified way to the minimal polynomial.

