Find the minimal polynomial of $\alpha = e^{2\pi \imath/19}$, the Galois group of $\Bbb Q(\alpha) /\Bbb Q$, and the subfields of the extension

Question

Given $\alpha = e^{2 \pi \imath / 19}$, find:

1. the minimal polynomial of $\alpha$ over $\mathbb{Q}$,
2. the Galois group, and
3. subfields of the extension.

My tries:

1. Since $e^{\frac{2\pi \imath x}{19}} = \cos(\frac{2 \pi x}{19}) + \imath \sin(\frac{2 \pi x}{19}) \in \mathbb{Q} \iff \sin(\frac{2 \pi x}{19}) = 0$, we get that the least positive integer satisfying our condition is $19$. So $x^{19} - 1$ is a polynomial with $\alpha$ as a root. We can easily see, that its root is also $1$, so we can devide it by $x - 1$ and we get $x^{18} + x^{17} + \ldots + x + 1$ (each degree $\leq$ 18 has a coefficient $1$). Its roots are $19$th complex roots of $1$ other than $1$. I think this should be a minimal polynomial of $\alpha$, but I'm not sure how to show it is irreducible (now I checked wolfram and it says it is its minimal polynomial, but I still need a proof). Maybe considering it in $\mathbb{Z}_2$ could work, but it would take a really long time, so I guess there is a better way.
2. Let $j = \phi(\alpha)$. Then $j^{19} = \phi(\alpha)^{19} = \phi(\alpha^{19}) = \phi(1) = 1$. So $j$ must be $19$th not real (because it wouldn't be automorphism then) root of $1$. Since (by part 3 if some subpart of it is right) the degrees of extensions generated by other powers of $\alpha$ are all the same, we could just take $\alpha$ to any power of it and it would be $\mathbb{Q}$-automorphism. So we have $18$ distinct automorphisms. Is this right?
3. I'm not sure if "subfields of an extension" must contain that field which is being extended or no. If yes, then I guess all its subfields (as an extensions of $\mathbb{Q}$) must be of degrees $1, 2, 3, 6, 9$ or $18$. But I think that a minimal polynomial for $\alpha$ is also a minimal polynomial for $\alpha^2, \alpha^3, \ldots, \alpha^{17}$, so I guess it does not contain any other subfields than $\mathbb{Q}, \mathbb{Q}(\alpha)$. Is this right?

Answer 1

(1) Hint There's a standard trick (that works for any primitive $p$th root of unity for $p$ prime): Denote $\Phi_{19}(x) = x^{18} + \cdots + x + 1$, and consider the polynomial $\Phi(y + 1)$---can you show that the latter is irreducible?

(2) This is basically correct, but you may want to identify the isomorphism type of the Galois group. One can show that the group is abelian, leaving $\Bbb Z_2 \times \Bbb Z_3 \times \Bbb Z_3$ and $\Bbb Z_2 \times \Bbb Z_9$ as the only possibilities. In general, the Galois group of a cyclotomic polynomial $\Phi_n$ is $(\Bbb Z / n \Bbb Z)^{\times}$.

(3) This is not true: It is possible to have an intermediate field $\Bbb Q(\beta)$, $\Bbb Q \subsetneq \Bbb Q(\beta) \subsetneq \Bbb Q(\alpha)$. For example, $\beta := \alpha + \frac{1}{\alpha} = \alpha + \bar{\alpha}$ is real, so $\Bbb Q(\beta) \subset \Bbb R$ and hence $\alpha \not\in \Bbb Q(\beta)$. On the other hand, rearranging the definition of $\beta$ gives $\alpha^2 - \beta \alpha + 1 = 0$. So the minimal polynomial of $\alpha$ over $\Bbb Q(\beta)$ is $x^2 - \beta x + 1$, and hence $[\Bbb Q(\alpha) : \Bbb Q(\beta)] = 2$; in particular, $\beta \not\in \Bbb Q$. (In fact, this also tells us that the minimal polynomial of $\beta$ over $\Bbb Q$ is $\frac{18}{2} = 9$.)