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If we have $f=x^3-2 \in \mathbb{Q}[x]$ and it has splitting field $L=\mathbb{Q}(\alpha , \omega )$ where $\alpha =\sqrt[3]{2} \ ,\ \omega =\exp{(2\pi i/3)}$.

I know that since $f$ is irreducible and it's roots are $\alpha , \alpha \omega, \alpha \omega^2 $, there is an element $\sigma \in \text{Aut}(L/\mathbb{Q})$ such that $\sigma (\alpha )=\alpha \omega .$

How do I determine the action of $\sigma $ on the other roots?

Anonmath101
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  • You can't determine it since various things can happen. $\sigma(\alpha\omega)$ is one of $\alpha$ and $\alpha\omega^2$ and both cases occur. – ancient mathematician May 31 '21 at 16:58
  • The proposed duplicate addresses what the title asks, and if you drill down you will see that the body of your Question is fully resolved in accordance with the Comments left above. – hardmath May 31 '21 at 23:44

2 Answers2

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The splitting field is $\mathbb{Q}(\alpha, \sqrt{-3})$ with Galois group $G$ generated by $\sigma : \alpha \to \omega \alpha$ and $\tau : \sqrt{-3} \to -\sqrt{-3}$. Note that $\tau$ acts on $\omega$ because $\omega = (-1 + \sqrt{-3}/2)$.

$\sigma$ is a homomorphism, although its action on the other roots of the polynomial can't be uniquely fixed. All three roots of the polynomial are algebraically 'the same' so there isn't a canonical choice.

Tom
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Since $f$ is irreducible and has 3 roots, the Galois group is a subgroup of $S_3$ which is a group of order 6, depending on which permutations of the roots are algebraically allowed. But the real root $\sqrt[3]{2}$ already generates an extension of rank 3 of $\mathbb{Q}$ (it's isomorphic to $\mathbb{Q}[x]/(f)$). Since you still don't have the two non-real roots, you need another extension to split which must now exhaust the rank of $6$ which you have to play with. So the Galois group must be all $S_3$ and all permutations of the roots allowed.