Galois group of $P(X)=X^6−4X^3+2$

Question

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 ?

Answer 1

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.

Answer 2

I will be using Dedekind's theorem relating cycle structures in the Galois group to factorization modulo a prime $p$, where $p$ is not a factor of the discriminant. See Qiaochu's post for an explanation of this theorem.

This time $p=7$ is most illuminating, because modulo $7$ we have the factorization $$ p(x)=(x+1)(x+2)(x+4)(x^3+2). $$ A way to see this is to observe that modulo $7$ the residue class of $3$ can serve as $\sqrt2$. Therefore $2+\sqrt2\equiv5$ is a non-cube modulo $7$, but $2-\sqrt2\equiv-1$ has three modular cubic roots ($-1,-2,-4$).

Let us number the six zeros as $a_1=\alpha$, $a_2=\omega\alpha$, $a_3=\omega^2\alpha$, $a_4=\beta$, $a_5=\omega\beta$, $a_6=\omega^2\beta$. We can then identify $G=Gal(E/\Bbb{Q})$ as a subgroup of $S_6$. Applying Dedekind's theorem to the above modulo $7$ data reveals that $\tau=(123)\in G$.

Because $\sqrt2\in E$, there are automorphisms in $G$ that exchange $\pm\sqrt2$. Let $\sigma$ be one such. Then $\sigma\tau\sigma^{-1}$ must keep the cubic roots of $2+\sqrt2$ fixed, and permute the cubic roots of $2-\sqrt2$ as a 3-cycle. Replacing $\tau$ by $\tau^2$ if necessary, we can then conclude that $\delta=(456)\in G$ also.

Because $\sqrt2,\omega\in E$, we know that $4\mid [E:\Bbb{Q}$. As the automorphisms $\delta$ and $\tau$ generate a subgroup of order $9$, we can conclude that $4\cdot9=36\mid [E:\Bbb{Q}]$.

On the other hand $\alpha$ and $\beta$ are at most cubic over $\Bbb{Q}(\sqrt2,\omega)$, so $[E:\Bbb{Q}]\le36$. Therefore $[E:\Bbb{Q}]=|G|=36$.

The complex conjugation gives the automorphism $\gamma=(23)(56)\in G$.

An element $\phi\in G$ is fully determined, if we know $\phi(\alpha),\phi(\beta)$ and $\phi(\omega)$. There are six alternatives for $\phi(\alpha)$ and two alternatives for $\phi(\omega)$. The choice of $\phi(\alpha)$ uniquely determines $\phi(\alpha^3)=\phi(2+\sqrt2)=2\pm\sqrt2$, in other words $\phi(\alpha)$ determines $\phi(\sqrt2)$ uniquely. Because $(\alpha\beta)^3=2$ knowing $\phi(\alpha^3)$ determines $\phi(\beta^3)$ uniquely. Keeping this in mind we have three choices for $\phi(\beta)$ to each choice of $\phi(\alpha)$. As $|G|=36$ all the thusly allowed $6\cdot2\cdot3=36$ combinations of $\phi(\alpha),\phi(\omega)$ and $\phi(\beta)$ occur (independently from each other). Therefore there exists an automorphism $\xi\in G$ such that $\xi$ keeps $\omega$ fixed, and interchanges $\alpha$ and $\beta$. It follows that as a permutation $$ \xi=(14)(25)(36). $$ We see that $\langle\xi,\gamma\rangle$ is a Sylow $2$-subgroup of $G$, and $\langle\tau,\delta\rangle$ is a Sylow $3$-subgroup of $G$. Together those two subgroups must generate all of $G$.