TITLE: Galois group of $P(X)=X^6−4X^3+2$
QUESTION [9 upvotes]: I would like to find Galois group of $P(X)=X^6−4X^3+2$.
I have that $(2+\sqrt{2})^{1/3}$ is a root of $P$, denote it $\alpha$. By Eisenstein $P(X)$ is irreducible over $\Bbb{Q}$ then $[\Bbb{Q}(\alpha):\Bbb{Q}]=6.$ 
If we let $\beta:=(2-\sqrt{2})^{1/3}$ then the root of $P(X)$ are $\alpha,\omega\alpha,\omega^2\alpha,\beta,\omega\beta,\omega^2\beta$ where $\omega$ is non trivial $3-$root of unity; then $\Bbb{Q(\alpha,\beta,\omega)}$ is the splitting field of $P(X)$, noted $E.$ It seems that $[E:\Bbb{Q}]=36.$
Not sure how can I find the Galois group ?

REPLY [3 votes]: Here is a solution that I feel is slightly different than the one given by Jyrki Lahtonen.
Let's first prove that $[E:\mathbb{Q}]$ is indeed $36$.
Consider the following field extensions:
$$
\mathbb{Q} \subset \mathbb{Q}(\omega) \subset \mathbb{Q}(\omega, \sqrt{2}) \subset \mathbb{Q}(\omega, \alpha) \subset \mathbb{Q}(\omega, \alpha, \beta).
$$
Then

$[\mathbb{Q}(\omega): \mathbb{Q}] = 2$, since the minimal polynomial of $\omega$ is $X^2+X+1$.  This is a Galois extension.
$[\mathbb{Q}(\omega, \sqrt{2}): \mathbb{Q}(\omega)] = 2$, since the minimal polynomial of $\sqrt{2}$ is $X^2-2$. This is a Galois extension.
$[\mathbb{Q}(\omega, \alpha): \mathbb{Q}(\omega, \sqrt{2})]=3$, since the minimal polynomial of $\alpha$ is $X^3 - (2+\sqrt{2})$.  This is a Galois extension.
$[\mathbb{Q}(\omega, \alpha, \beta): \mathbb{Q}(\omega, \alpha)] = 3$.  This is a bit more difficult to prove. Since $X^3 - (2-\sqrt{2})$ has $\beta$ as a root, then $[\mathbb{Q}(\omega, \alpha, \beta): \mathbb{Q}(\omega, \alpha)]$ is either $1$ or $3$.  One way to quickly see that it must be $3$ is to realize that $X^9-12X^6-6X^3-64$ is the minimal polynomial of $\alpha + \beta$ over $\mathbb{Q}$, so $[\mathbb{Q}(\omega, \alpha, \beta): \mathbb{Q}]$ has to be divisible by $9$.  

Thus $[E:\mathbb{Q}] = 36$.
Now, to compute the Galois group.  Note that, by the Galois correspondence, there are two subgroups of order $3$, corresponding to the intermediate extensions $\mathbb{Q}(\omega, \alpha) \subset \mathbb{Q}(\omega, \alpha, \beta)$ and $\mathbb{Q}(\omega, \beta) \subset \mathbb{Q}(\omega, \alpha, \beta)$.  These subgroups are generated by elements $A$ and $B$, respectively, who fix $\omega$ and are defined by 
$$
A(\alpha) = \alpha, \ A(\beta)=\omega\beta, \ B(\alpha)=\omega\alpha, \ B(\beta)=\beta.
$$
Note that $A$ and $B$ commute, so they generate a $3$-Sylow subgroup isomorphic to $(\mathbb{Z}/3)^2$.
Note that complex conjugation, which we will denote by $C$, is also an element of the Galois group.  Moreover, the extension $\mathbb{Q}(\omega, \alpha+\beta)/\mathbb{Q}$ is of degree $18$ by the above.  By the Galois correspondence, it corresponds to a subgroup of order $2$, generated by an element $D$ that fixes $\omega$ and exchanges $\alpha$ and $\beta$.  Note that $D$ and $C$ commute, so they generate a subgroup isomorphic to $(\mathbb{Z}/2)^2$.
This tells us that the Galois group is generated by $A,B,C$ and $D$.  Relations inside the group are, for instance: $A^3=B^3=C^2=D^2=1, AB=BA, CD=DC, AC=CA^2, BC=CB^2, DA=BD$ and $DB=AD$.  I believe these constitute a presentation of the Galois group.