Galois Group of $x^{6}-2x^{3}-1$

Question

I was trying to compute the normal closure of $\mathbb{Q}[\alpha]$, where $\alpha = \sqrt[3]{1+\sqrt{2}}$.

I had a reallyyyy hard time proving that $x^{6}-2x^{3}-1$ is irreducible. I proved that it has no roots on $\mathbb{Q}$. Then I didn't know exactly what to do.

I used this lemma: $f(X^p)$ irreducible or $p$th power if $f$ irreducible With the lemma, I proved that it cannot have factors of order 3, as it's polynomial is irreducible or a third power of irreducible on $\mathbb{F}_{3}[x]$ as $x^2-2x-1$ is irreducible here as it has no roots.

Then it can only has factors of order 4 and 2. If it has no factors of order 2, then it is irreducible, as if it has a factor of order 4 it has to have a factor of order 2.

To prove that it is irreducible, or it has a factor of degree $2$. Then, I used the factorization on the splitting field to see that any product of two of those linear factors is not a rational polynomial.

If there is any easier way to prove this, I would like to know. I failed horribly trying to show that it's projections on finite fields are irreducible, or shifting and using Eisenstein.

Well, then as that field contains $\mathbb{Q}[\sqrt{2}, \sqrt{-3}]$ which is a normal extension, I know it has a normal $3$-Sylow subgroup. I know that's not abelian because it has non normal extensions (so it's group has non normal subgroups, so it's not abelian).

I had $D_{12},Q_{12}$ and $A_{4}$. I know that $A_{4}$ has more than one subgroup of order $3$ (it has four $3$-cycles). So it isn't.

But I don't know how to know if it's $D_{12}$ or $Q_{12}$. I know that if $H$ is the group that fixes $\mathbb{Q}[\sqrt{2}, \sqrt{-3}]$, then $ G / H \cong V_{4}$ where $V_{4}$ is the Klein Group.

I feel like I need to extract more information from that field, but it's hard to find good generators or the image of the automorphisms with such $\alpha$. Computing the group of $x^{6}-3$ is way easier because $\sqrt[6]{3}$ and $i$ are more tame than this monster. Any ideas? I feel like I should be more methodic on this, but I have read Dummit section about computing Galois groups and it feels like it's very ad hoc.

Answer 1

Let $\alpha_i$ be the roots of $x^{6}-2x^{3}-1$.

Let $\omega$ be a root of $x^2 - x + 1$ so that $\omega^3 = -1$.

• $\alpha_1 = \sqrt[3]{1+\sqrt{2}}$
• $\alpha_2 = \omega \sqrt[3]{1+\sqrt{2}} = \omega \alpha_1$
• $\alpha_3 = \omega^2 \sqrt[3]{1+\sqrt{2}} = \omega^2 \alpha_1$
• $\alpha_4 = \sqrt[3]{1-\sqrt{2}} = -\omega^2/\alpha_1$
• $\alpha_5 = \omega \sqrt[3]{1-\sqrt{2}} = 1/\alpha_1$
• $\alpha_6 = \omega^2 \sqrt[3]{1-\sqrt{2}} = \omega/\alpha_1$

Thus $\mathbb Q(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha_{5},\alpha_{6}) = \mathbb Q(\omega,\alpha_1)$. And $[\mathbb Q(\omega):\mathbb Q] = 2$, $[\mathbb Q(\alpha_1):\mathbb Q] = 6$ so the composite has degree 6 or 12.

We can produce 12 distinct automorphisms based on mapping $\alpha_1$ to any other $\alpha$ and $\omega$ to $\omega$ or $\omega^2$.

This forms a 6-cycle with an inversion map. This is the group $D_{12}$