On computing the Galois group $\mathrm{Aut}\left(\Bbb Q (\alpha,\omega) / \Bbb Q( \omega + \omega^2 + \omega^4)\right)$

Question

Let $\alpha:= \sqrt[7]{2},\omega:= e^{\frac{2\pi i }{7}}\in \Bbb C$. We set $E:=\Bbb Q(\alpha,\omega)$ and $B:=\Bbb Q(\omega+\omega^2+\omega^4)\leq E$.

Thus, we have the Tower of Fields $$\Bbb Q \leq B \leq E.$$ We can prove that $|\mathrm{Aut}(E/\Bbb Q)|=42$ and that $\theta,\sigma\in \mathrm{Aut}(E/\Bbb Q)$, given by $$\theta:\alpha \longmapsto \alpha \omega,\ \omega \longmapsto \omega$$ $$\sigma: \alpha \longmapsto \alpha,\ \omega \longmapsto \omega^3.$$

Question: How can we prove that $\mathrm{Aut}(E/B)=\langle \theta,\sigma^2\rangle$?

A first thought is to compute the Galois group $\mathrm{Aut}(E/B)$, but this seems to be extremely hard, because of $B$. Another might be to use somehow the Fundamental Theorem of Galois Theory, so to exploit the relations $[E:B]=|\mathrm{Aut}(E/B)|$ and then find the order of the subgroup $\langle \theta,\sigma^2 \rangle$.

Note that $\theta^i(\alpha)=\alpha\omega^i$ and $\sigma^j(\omega)=\omega^{3^j}$.

Answer 1

You meant $\alpha=\sqrt[7]{2}$. Let $K=\Bbb{Q}(\omega)$.

We get $E/K/B/\Bbb{Q}$ where $E/\Bbb{Q}$ and $K/B$ are Galois.

We know that $Aut(K/\Bbb{Q})=\Bbb{Z/7Z}^\times=\langle \sigma\rangle,Aut(E/K)=\Bbb{Z/7Z}$. It is quite immediate that both mix together to give $Gal(E/\Bbb{Q})=Aff(\Bbb{Z/7Z})$ (the group of affine transformations $x\to ax+b$ which corresponds to $\omega^x\alpha\to \omega^{ax+b}\alpha$).

$B$ is a subfield of $K$ fixed by $\sigma^2$, and it is not $\Bbb{Q}$, so it is the subfield fixed by $\sigma^2$ and hence $ Aut(K/B)=\langle \sigma^2\rangle$.

$K$ is the subfield of $E$ fixed by $\theta$.

Extend $\sigma\in Aut(K/\Bbb{Q})$ to an element $\sigma'\in Aut(E/\Bbb{Q})$.

Then it is immediate that $B$ is the subfield of $E$ fixed by $(\sigma')^2$ and $\theta$ which means that $E/B$ is Galois with Galois group $Aut(E/B)=\langle(\sigma')^2,\theta\rangle$.

$[E:B]=[E:K][K:B]=21$ and $Aut(E/B)= \{x\to a^2 x+b\}$.

Answer 2

Below please find a few pieces that help you to onoe way of proving the claims.

• The calculation here reveals that $B=\Bbb{Q}(\sqrt{-7})$. Therefore $[B:\Bbb{Q}]=2$, and you know that $[E:B]=21$.
• You can easily verify that $\theta$ and $\sigma^2$ leave $B$ fixed pointwise, so they belong to $Gal(E/B)$.
• The calculation showing that $G=Gal(E/\Bbb{Q})$ is generated by $\theta$ and $\sigma$ more specifically describes the Galois group as a semidirect product $$G\simeq C_7\rtimes C_6,$$ where the normal subgroup $\simeq C_7$ is generated by $\theta$ and the non-normal factor by $\sigma$. This follows also from the relation $$\sigma\theta\sigma^{-1}=\theta^3$$ that you can verify by calculating their effects on $\alpha,\omega$.
• It is the easy to check that the subgroup generated by $\theta$ and $\sigma^2$ has order $21$ (see reuns's answer for a way of making that explicit), so it must be all of $Gal(E/B)$.

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Extras:

• $\Bbb{Q}(\omega)/\Bbb{Q}$ is Galois with an abelian Galois group isomorphic to $\Bbb{Z}/7\Bbb{Z}^*\simeq C_6$ generated the restriction of $\sigma$. Therefore the intermediate field $B$ must also be Galois over $\Bbb{Q}$.
• You see that $$\sigma(\omega+\omega^2+\omega^4)=\omega^3+\omega^6+\omega^5=-1-(\omega+\omega^2+\omega^4)$$ implying that $\sigma$ does not fix $B$ element wise but $\sigma^2$ does. This also implies that $[B:\Bbb{Q}]=2$ without the calculation relating $B$ to $\sqrt{-7}$.

Answer 3

Taking into account Jyrki's answer, I tried to collect the pieces. Please, have a look.

(1) By the first lines of his answer: If $[B:\Bbb{Q}]=2$, then $$[E:\Bbb Q]=[E:B][B:\Bbb Q] \implies 42=[E:B]\cdot 2 \iff [E:B]=21=|\mathrm{Aut}(E/B)|.$$ Since $B=\Bbb Q(i\sqrt 7)=\langle1,i\sqrt 7\rangle$, we can see that $\theta(b)=\sigma^2(b)=b,\ \forall b\in B$. Hence, $$\theta,\sigma^2 \in \mathrm{Aut}(E/B)\implies \langle \theta,\sigma^2 \rangle \subseteq \mathrm{Aut}(E/B).$$ But also we have $|\langle \theta,\sigma^2\rangle|=|\langle \theta\rangle | \cdot |\langle \sigma^2\rangle|=7 \cdot 3 =21=|\mathrm{Aut}(E/B)|$. Therefore, $$\mathrm{Aut}(E/B)=\langle \theta,\sigma^2\rangle.$$

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(2) By the extras: We observe that $$\theta(\omega+\omega^2+\omega^4)=\omega+\omega^2+\omega^4,\\ \sigma^2(\omega+\omega^2+\omega^4)=\dotsb =\omega+\omega^2+\omega^4, $$ so, $\omega+\omega^2+\omega^4 \in \mathrm{Fix}(\langle \theta,\sigma^2 \rangle) $. Then, we have the tower of fields $$\Bbb Q \leq B=\Bbb Q (\omega+\omega^2+\omega^4) \leq \mathrm{Fix}(\langle \theta,\sigma^2 \rangle).$$ But, \begin{align*} |G: \langle \theta,\sigma ^2 \rangle |=[\mathrm{Fix}(\langle \theta,\sigma^2\rangle):\Bbb Q] & \iff \frac{|G|}{|\langle \theta,\sigma^2\rangle|}=[\mathrm{Fix}(\langle \theta,\sigma^2\rangle):\Bbb Q] \\ & \iff \frac{42}{21}=[\mathrm{Fix}(\langle \theta, \sigma^2\rangle):\Bbb Q] \\ & \iff [\mathrm{Fix}(\langle \theta,\sigma^2\rangle):\Bbb Q]=2. \end{align*} So, $B=\Bbb Q$ or $B=\mathrm{Fix}(\langle \theta,\sigma^2\rangle)$. But $\omega+\omega^2+\omega^4 \notin \Bbb Q$. So, $$B=\mathrm{Fix}(\langle \theta,\sigma^2\rangle) \implies [B:\Bbb Q]=2.$$ Therefore, $$[E:\Bbb Q]=[E:B][B:\Bbb Q] \iff [E:B]=21 \iff |\mathrm{Aut}(E/B)|=21.$$ Now $[\underbrace{\Bbb Q(\omega+\omega^2+\omega^4)}_{B}:\Bbb Q]=2 < \infty \iff \omega+\omega^2+\omega^4$ is algebraic over $F$ and therefore it determines uniquely the basis of $B$. So, for showing that $\theta,\sigma^2 \in \mathrm{Aut}(E/B)$, it sufficies to show that $\theta,\sigma^2 $ leave $\omega+\omega^2+\omega^4$ fixed, and indeed that happens. So, $\langle \theta,\sigma^2 \rangle \subseteq \mathrm{Aut}(E/B)$. Also, like (1), $|\langle \theta,\sigma^2 \rangle|=21=|\mathrm{Aut}(E/B)|$ and eventually $$\mathrm{Aut}(E/B)=\langle \theta,\sigma^2 \rangle,$$ as wanted.