Find $Gal(\mathbb{Q}(e^{2\pi i/6},e^{2\pi i / 10}) / \mathbb{Q})$

Question

First I need to show that the extension $\mathbb{Q}(e^{2\pi i/6},e^{2\pi i / 10}) / \mathbb{Q}$ is actually Galois extension, and then I need to find the group itself.

Denote $\alpha = e^{2\pi i/6} $, $\beta = e^{2\pi i / 10}$.

I guess the first step is to find the minimal polynomial of $\alpha,\beta$, but I'm not really sure how to do it. I know that the following polynomials are zeroed by $\alpha, \beta$:

$$\sum_{i=0}^{29} x^i,\quad (\sum_{i=0}^5 x^i) \cdot (\sum_{i=0}^9 x^i)$$

Both are reducible, but I don't know how can I reduce them further.

Answer 1

Disclaimer: This answer uses cyclotomic field theory.

Let $K_n$ denote the $n$^th cyclotomic field.

Then your field extension is $K_6 K_{10}$ (compositum).

Then $K_6 K_{10} = K_{\operatorname{lcm}(6,10)} = K_{30}$, so its Galois group is $(\Bbb Z/30\Bbb Z)^\times$.