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$E$ is the splitting field of a $f(x)(\in\mathbb{Q}[x])$ whose degree is $8$. Say $E=\mathbb{Q}(\sqrt{2},\sqrt{3}, \alpha)$ (Here $\alpha = \sqrt{(2+\sqrt{2})(3+\sqrt{3})}$). Find the group isomorphic to $G(E/\mathbb{Q}) $.

Let me show my solution. By tower theorem, $[E : \mathbb{Q}]=8$ (I could easily check the $deg$ $irr(\alpha, E_1) =2$ for $E_1=\mathbb{Q}(\sqrt{2},\sqrt{3})).$ I found the all the element of $G(E/\mathbb{Q})$ like the below considering the roots of $irr(\alpha, E_1)$ are $\pm\alpha$

$\sigma(\in G(E/\mathbb{Q})) : \sqrt2 \to \pm \sqrt2, \sqrt3 \to \pm \sqrt3, \alpha \to \pm\alpha$

Therefore, $\forall \sigma(\in G(E/\mathbb{Q})), \vert \sigma \vert =2 $. Hence $G(E/\mathbb{Q})\simeq \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 $

Is my solution right? Aren't there any solutions except for mine?

Seewoo Lee
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kechang lee
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    Some of your maps $\sigma$ aren't valid field automorphisms. For example, if $\sigma(\sqrt 2) = \sqrt 2$ but $\sigma(\sqrt 3) = -\sqrt 3$, then $\sigma$ needs to send $\alpha$ to one of the two square roots of $(2 + \sqrt 2)(3 - \sqrt 3)$, not to a square root of $(2 + \sqrt 2)(3 + \sqrt 3)$. – Ravi Fernando Sep 06 '22 at 04:57
  • @RaviFernando, indeed! I didn't think that far – kechang lee Sep 06 '22 at 05:08
  • Your solution is incomplete. First show that the extension is Galois, see here. Then consider the Galois group, compare with this post. Be careful, the claim $Q_8$ is not true there. But the answer points this out correctly. – Dietrich Burde Sep 06 '22 at 09:20

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