I'm trying to find the Galois group of $$f(x)= x^6 - 3x^3 + 2$$ over $\mathbb{Q}$.

Now I can factorise this as $$f(x) = (x-1)(x^2 + x + 1)(x^3 - 2)$$

I can see the splitting field must be $\mathbb{Q}(\omega, \sqrt[3]{2})$ (where $\omega$ is a 3rd root of unity) which has degree $6$ and so I know the Galois group must be order $6$. Now how do I distinguish between the cases $S_3$ or $C_6$?

I'm pretty sure it's $S_3$ but I'm having trouble seeing what the automorphisms actually are. I know that an element $\sigma$ in the Galois group must permute the roots of each irreducible factor and I also know that an automorphism is determined by it's action on $\sqrt[3]{2}$ and $\omega$ but I'm having trouble seeing what the actual automorphisms are in this case?

E.g. I can see sending $\omega \to \omega^2$ permutes the roots of $x^2 + x + 1$, but then what do I do to $\sqrt[3]{2}$? Can that be left fixed?

Thanks for any help!


2 Answers 2


A basis for the splitting field of $x^6 - 3x^3 + 2$ over $\mathbb{Q}$ is $\{1, \omega, \sqrt[3]{2}, \omega\sqrt[3]{2}, \sqrt[3]{2}^2, \omega\sqrt[3]{2}^2\}$.

An element $\sigma \in \operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$ is completely determined by $\sigma(\omega)$ and $\sigma(\sqrt[3]{2})$; note however that not every choice results in an element of $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$ (e.g. $\sigma(\omega) = \sigma(\sqrt[3]{2}) = 1)$.

As $\sigma \in \operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$ permutes the roots of minimal polynomials, we see that $\sigma(\omega) \in \{\omega, \omega^2\}$ and $\sigma(\sqrt[3]{2}) \in \{\sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^2\sqrt[3]{2}\}$.

Let $\alpha(\omega) = \omega^2$ and $\alpha(\sqrt[3]{2}) = \sqrt[3]{2}$; $\alpha$ extends to an element of $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$. In particular

\begin{align*} \alpha : 1 &\mapsto 1\\ \omega &\mapsto \omega^2\\ \omega^2 & \mapsto \omega\\ \sqrt[3]{2} &\mapsto \sqrt[3]{2}\\ \omega\sqrt[3]{2} &\mapsto \omega^2\sqrt[3]{2}\\ \omega^2\sqrt[3]{2} &\mapsto \omega\sqrt[3]{2}. \end{align*}

Likewise, let $\beta(\omega) = \omega$ and $\beta(\sqrt[3]{2}) = \omega\sqrt[3]{2}$; $\beta$ extends to an element of $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$. In particular

\begin{align*} \beta: 1 &\mapsto 1\\ \omega &\mapsto \omega\\ \omega^2 & \mapsto \omega^2\\ \sqrt[3]{2} &\mapsto \omega\sqrt[3]{2}\\ \omega\sqrt[3]{2} &\mapsto \omega^2\sqrt[3]{2}\\ \omega^2\sqrt[3]{2} &\mapsto \sqrt[3]{2}. \end{align*}

As $\alpha$ has order two and $\beta$ has order three, they generate $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$. By comparing $\alpha\circ\beta$ and $\beta\circ\alpha$, we can determine whether $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$ is abelian or not, and hence whether it is $C_6$ or $S_3$.

  • $\begingroup$ Thank you for your answer, all is clear now! $\endgroup$
    – Wooster
    Dec 23, 2014 at 9:01
  • $\begingroup$ $1+\omega+\omega^2=0$ so I don't think this is a basis.. $\endgroup$ Nov 20, 2021 at 2:52
  • 1
    $\begingroup$ @JingjieYANG: You are correct, what I initially had was not a basis. I have modified the basis in my answer; the rest of the answer is unaffected. $\endgroup$ Jan 3, 2022 at 1:27

Let $\alpha := \sqrt[3]{2}$. As you said, any automorphism $\sigma \in Aut_{\mathbb{Q}} L $ is complete determined by

$$\sigma (1) \in \{1\}\\\sigma(\omega) \in \{\omega,\omega^2\}\\\sigma(\alpha) \in \{\alpha, \alpha \omega,\alpha \omega^2\}*$$

Now set the table of possibilities, to find all automorphisms.

$(*)$ Notice that $\alpha, \alpha \omega,\alpha \omega^2$ are the roots of $X^3-2$.

  • $\begingroup$ If you have any question feel free to ask. $\endgroup$ Dec 22, 2014 at 18:49

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