TITLE: Galois group of $x^3 + x^2 - 2x - 1$.
QUESTION [9 upvotes]: Suppose $\alpha$ is a root of $x^3 + x^2 - 2x - 1$. Let $E = \mathbb{Q}(\alpha)$. We are given that $\beta = \alpha^2 - 2$ is also a root, and I need to find $Gal(E/\mathbb{Q})$ and show that $E$ is normal over $\mathbb{Q}$.
This is what I was thinking: because we have a relation between two roots, then they are both complex or both real. If they are both real, then you have 2 real roots so thus all 3 roots are real. In that case, then the Galois group is $S_3$ or $A_3$..? Or can we say that $\gamma = \beta^2 -2$ is also a root, so then they must all be real? I'm having trouble with figuring our the Galois group when a relation is given like this.
Thanks!

REPLY [3 votes]: Just for the record, this is the minimal polynomial of $z + z^{-1}$, where $z$ is a primitive $7$-th root of unity. This hint also leads to the result that the Galois group is cyclic of order $3$.
This explains the fact that if $$\alpha = z + z^{-1}$$ is a root, then $$\alpha^{2} - 2 = (z^{2} + z^{-2} + 2) - 2 = z^{2} + (z^{2})^{-1}$$ is also a root.