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Let $\omega$ be a primitive $12$th root of unity.
(i) What is $[ \mathbb{Q}(\omega) : \mathbb{Q}]?$

(ii) List the distinct conjugates of $\omega + \omega^{-1}$.

(iii) What is $Aut(\mathbb{Q}(\omega + \omega^{-1})/ \mathbb{Q})$? Is $\mathbb{Q}(\omega + \omega^{-1})$ Galois over $\mathbb{Q}$?

(iv) Diagram the lattice of subfields of $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$, giving generators for each.

(i) The given index is $\phi(12) = \phi(3) \phi(4) = 2(2^2 - 2) = 4$.

(ii) $\mathbb{Q}(\omega)$ is the composite of $\mathbb{Q}(\omega^3)$ and $\mathbb{Q}(\omega^4)$, where $\omega^3$ (resp. $\omega^4$) is a primitive $4$th (resp. $3$rd) root of unity. Both of these fields are Galois over $\mathbb{Q}$ and their intersection is $\mathbb{Q}$. Therefore $G = Aut(\mathbb{Q}(\omega)/ \mathbb{Q})$ is the direct sum of $Aut(\mathbb{Q}(\omega^3) = \{ \omega^3 \mapsto \omega^3, \omega^3 \mapsto \omega^9\}$ and $Aut(\mathbb{Q}(\omega^4)/ \mathbb{Q}) = \{ \omega^4 \mapsto \omega^8\}$.

The conjugates of $\omega + \omega^{-1}$ are the elements $\psi(\omega + \omega^{-1}) : \psi \in G $, and the number of distinct conjugates is the degree of $\mathbb{Q}(\omega + \omega^{-1})$ over $\mathbb{Q}$. Since $\omega = \omega^4 \omega^{-3}$, we have the elements $\omega^4 \omega^{-9}, \omega^8 \omega^{-3}, \omega^8 \omega^{-9}$, so the set of conjugates is $$\{ \omega^4 \omega^{-9} + \omega^{-4} \omega^9, \omega^8 \omega^{-3} + \omega^{-8} \omega^3, \omega^8 \omega^{-9} + \omega^{-8} \omega^9, \omega + \omega^{-1} \} $$ $$ = \{ \omega^{-5} + \omega^5, \omega^5 + \omega^{-5}, \omega^{-1} + \omega, \omega + \omega^{-1} \} = \{ \omega + \omega^{-1}, \omega^5 + \omega^{-5} \}$$ We can assume $\omega = e^{\frac{2 \pi i}{12}}$ so that $\omega^5 = e^{\frac{10 \pi i}{12}}$. Thus $\omega + \omega^{-1} = 2Re(\omega) = 2 \cos(\pi/6) = \sqrt 3$, and $\omega^{-5} + \omega^5 = 2Re(\omega^5) = 2 \cos(5 \pi/6) = - \sqrt{3}$. So $\sqrt{3}, - \sqrt{3}$ are the conjugates.

(iii) $\mathbb{Q}(\omega + \omega^{-1})$ is of course Galois over $\mathbb{Q}$, because $G$ is abelian and every subgroup is normal. Since $\omega + \omega^{-1} = \sqrt{3}$, the given field is $\mathbb{Q}(\sqrt 3)$, so the Galois group is $\mathbb{Z}/2 \mathbb{Z}$.

(iv) The subfields are $\mathbb{Q}(\omega^3) = \mathbb{Q}[i]$ and $\mathbb{Q}(\omega^4) = \mathbb{Q}[ \frac{-1 + i \sqrt{3}}{2}]$, with respective degrees $4$ and $3$ over $\mathbb{Q}$, as well as $\mathbb{Q}(\omega + \omega^{-1}) = \mathbb{Q}(\sqrt 3)$, also degree $2$ over $\mathbb{Q}$. I don't understand. The Galois group of $\mathbb{Q}(\omega)$ is $\mathbb{Z}/2 \mathbb{Z} \times \mathbb{Z}/2 \mathbb{Z}$, which has $4$ subgroups, two nontrivial proper. Shouldn't there be only $2$ proper intermediate fields of $\mathbb{Q}(\zeta)$?

D_S
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    Everything looks ok up to the last part. The group $\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}$ has three non-trivial proper subgroups. The Galois group has three non-trivial elements. They are all of order two, and every one of them generates a subgroup of order two. – Jyrki Lahtonen Jun 11 '14 at 16:47
  • You're right, I overlooked ${(1,1), (0,0 }$ as a subgroup. Thank you! – D_S Jun 11 '14 at 16:55
  • Is the Galois group $\mathbb{Z}_4$? – Libertron Dec 27 '14 at 00:18
  • No, $\mathbb{Q}(\omega)$ is a compositum of the third and fourth cyclotomic extensions over $\mathbb{Q}$, both of which are Galois. Since these fields' intersection is $\mathbb{Q}$, the Galois group of $\mathbb{Q}(\omega)$ over $\mathbb{Q}$ is the direct product of these latter two Galois groups, so $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. – D_S Dec 27 '14 at 02:16

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