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.
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2$\begingroup$ Consider the automorphisms $\sigma_k: \omega \to \omega^k$ for $k=1,7,11$. They form a group and fix $\alpha$ (easy computation). Thus, using the fundamental theorem of the Galois theory, $\alpha$ generates a subfield of degree at most 6. $\endgroup$– ZardoJun 10, 2015 at 14:20
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2$\begingroup$ Let $U=\{\sigma_k \vert k=1,7,11\}$. Then $U$ is a subgroup of the Galois group $Gal(\mathbb{Q}(\omega):\mathbb{Q})$ of order $3$. By the fundamental theorem $U$ fixes an intermediate field $L$ (which containes $\alpha$ and therefore $\mathbb{Q}(\alpha)$) with $[\mathbb{Q}(\omega):L]=3$. Then $[L:\mathbb{Q}] = 18/3= 6$ and $[\mathbb{Q}(\alpha):\mathbb{Q}] \leq 6$. $\endgroup$– ZardoJun 10, 2015 at 14:36
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1$\begingroup$ Yes, if you give reasoning why the Galois group is not isomorphic to $S_3$. The fundamental theorem helps here as well. $\endgroup$– ZardoJun 10, 2015 at 14:44
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1$\begingroup$ You need to look at every polynomial in $\alpha$, not only its powers, show that $1,\alpha, \alpha^2, \alpha^3$ are linearly indepentent over $\mathbb{Q}$ using a basis consisting of powers of $\omega$. $\endgroup$– ZardoJun 10, 2015 at 17:00
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1$\begingroup$ By using that $G=\mathbb{Z}/19\mathbb{Z}^\times$ is abelian and the Galois group of $K=\mathbb{Q}(\alpha)$ is a quotient of $G$, thus also abelian. You also need that $K$ is normal over $\mathbb{Q}$ which follows from $G$ being abelian. $\endgroup$– ZardoJun 10, 2015 at 21:09
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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 $$[\Bbb{Q}(\omega) : \Bbb{Q}(\alpha)] = 3$$
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$\begingroup$ So in that logic, Gal$(\mathbb{Q}(\alpha) / \mathbb{Q})$ is not $(\mathbb{Z}/19\mathbb{Z})^{\times}$? $\endgroup$– RileyJun 10, 2015 at 14:21
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1$\begingroup$ Exactly. I think that it is the cyclic group of order $6$. To check if this is true, one should compute all conjugates of $\alpha$ (there are 18 of them!). $\endgroup$– CrostulJun 10, 2015 at 14:21
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$\begingroup$ How did you calculate so quickly that $(\omega + \omega^7 + \omega^{11})^{14} = \omega^2 + \omega^3 + \omega^{14}$? $\endgroup$– RileyJun 10, 2015 at 14:46
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1$\begingroup$ I calculated $\sigma_k(\alpha) = \omega^{k} + \omega^{7k} + \omega^{11k}$. $\endgroup$– CrostulJun 10, 2015 at 14:51
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$\begingroup$ Hm, ok, I've calculated the conjugates and found 6 triples of $\sigma$ that are equal. But I fail to see why this implies that $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 6$... $\endgroup$– RileyJun 10, 2015 at 15:26