# Galois Group of $x^3-2$ over $\mathbb{Q}$

I have a question regarding the computation of the Galois Group of $x^3-2$ over $\mathbb{Q}$. I am reading the example given in section 14.1 of Dummit and Foote. We have that the splitting field of $x^3-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{2},\rho)$, where $\rho$ is a primitive cube root of unity. Then any automorphism $\sigma$ maps $\sqrt{2}$ to one of $\sqrt{2}, \rho\sqrt{2},\rho^2\sqrt{2}$. Now, the book says that $\sigma$ maps $\rho$ to either itself or $\rho^2$. My question is why can't $\rho$ be mapped to any of the other roots? Doesn't $\sigma$ map roots of $x^3-2$ to other roots of $x^2-2$?

• Remember: the cube roots of unity aren't roots of $x^3-2$. – TPace Feb 17 '18 at 21:14
• Ah, yes. So then since the minimal polynomial of $\rho$ over $\mathbb{Q}$ is the cyclotomic polynomial $x^2+x+1$, $\rho$ would have to be mapped to other roots of $x^2+x+1$, namely either itself or $\rho^2$. – ponchan Feb 17 '18 at 21:17
• The extension $K = \Bbb Q(\sqrt{2}, \rho)$ has a basis as a $\Bbb Q$-vector space and any automorphism of $K$ is uniquely determined by how $G = \operatorname{Gal}(K/\Bbb Q)$ acts on these basis elements. Also, knowing how $G$ acts on these basis elements makes it much easier to determine fixed fields of subgroups of $G$. – Edward Evans Feb 17 '18 at 21:21

$\rho$ itself is not a root of $x^3-2$. It is a cube root of unity, and its minimal polynomial is the cyclotomic polynomial $\Phi(x)=x^2+x+1$. Thus, $\sigma$ must map $\rho$ to another root of $\Phi$, either itself or $\rho^2$.
• You should go on further and compute explicitly the Galois group $G$ of $X^3-2$ . Even better, show that $G \cong$ the dihedral group $D_6 \cong$ the symmetric group $S_3$ . – nguyen quang do Feb 19 '18 at 7:30