# Finding Galois group of $x^6 - 3x^3 + 2$

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!

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 $$S_3$$.

• Thank you for your answer, all is clear now! Dec 23, 2014 at 9:01
• $1+\omega+\omega^2=0$ so I don't think this is a basis.. Nov 20, 2021 at 2:52
• @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. 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 $X^3-2$.

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