Galois group of $x^3 + x^2 - 2x - 1$.

Question

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!

Answer 1

Well I think the point is that $\beta \in \mathbb{Q}(\alpha)$. So since the polynomial is of third degree $E$ will also contain the third root and therefore is the splitting field. It is normal since splitting fields are normal. The degree is three and there is only one group of order three.

Answer 2

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.