Suppose $\alpha = \omega + \omega^7 + \omega^{11}$, where $\omega$ is a primitive root of unity of order 19. I want to determine the Galois group of $\mathbb{Q}(\alpha) / \mathbb{Q}$. Because $\omega$ is a primitive root of unity of order 19, we have that $[\mathbb{Q}(\omega) : \mathbb{Q}] = 18$ and thus, the corresponding Galois group is $(\mathbb{Z}/ 19 \mathbb{Z})^{\times}$ . If I now prove that $\mathbb{Q}(\alpha) = \mathbb{Q}(\omega)$, then the galois group is $(\mathbb{Z}/ 19 \mathbb{Z})^{\times}$. Clearly, $\mathbb{Q}(\alpha) \subset \mathbb{Q}(\omega)$. But I'm not seeing why the inverse inclusion holds, i.e. why $\omega$ is in $\mathbb{Q}(\alpha)$. I've also considered the product formula $$[\mathbb{Q}(\omega) : \mathbb{Q}] = [\mathbb{Q}(\omega) : \mathbb{Q}(\alpha)]\cdot[\mathbb{Q}(\alpha) : \mathbb{Q}],$$ so if I can prove that $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 18$, we're done, but I can't seem to finish this line of thought. Any ideas on how to progress further?
Any help would be greatly appreciated.
[Math] Galois group of primitive root of unity/ field extensions
abstract-algebraextension-fieldgalois-theory
Best Answer
For $k=1, \dots , 18$ denote by $\sigma_k$ the automorphism of $\Bbb{Q}(\omega)$ acting as $\omega \mapsto \omega^{k}$. Then $$\sigma_2(\alpha)= \omega^2+\omega^3+\omega^{14} = \sigma_3(\alpha) = \sigma_{14}(\alpha)$$
So the polynomial $f_{\alpha}=\prod_k (x- \sigma_k(\alpha))$ has a multiple root. In particular, $f_{\alpha}$ is a power of the minimal polynomial of $\alpha$, so $\Bbb{Q}(\omega) \neq \Bbb{Q}(\alpha)$.
I suggest you to compute some other conjugate elements of $\alpha$, and to see how many of them are equal. In my opinion we should have that $$[\Bbb{Q}(\omega) : \Bbb{Q}(\alpha)] = 3$$