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 $\mathbb{Q}(\sqrt[3]{2},\rho)$, where $\rho$ is a primitive cube root of unity. Then any automorphism $\sigma$ maps $\sqrt[3]{2}$ to one of $\sqrt[3]{2}, \rho\sqrt[3]{2},\rho^2\sqrt[3]{2}$. 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 $x^2-2$?
[Math] Galois Group of $x^3-2$ over $\mathbb{Q}$
abstract-algebrafield-theorygalois-theory
Best Answer
$\rho$ 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 $\rho^2$.