Given $\alpha = e^{2 \pi \imath / 19}$, find:
- the minimal polynomial of $\alpha$ over $\mathbb{Q}$,
- the Galois group, and
- subfields of the extension.
My tries:
- 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.
- 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?
- 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?