Galois group of the polynomial $x^6-5$ over $\mathbb{Q}$ and $\mathbb{R}$

Question

I know it is posted before: Galois group of the splitting field of $ x^6 - 5$.

I tried the "over $\mathbb{Q}$" case first. Let $\alpha$ be a root of $f(x) = x^6-5$ and $\omega$ the primitive 6th root of $1$. Then the splitting field of $f(x)$ is $\mathbb{Q}[\alpha, \omega]$ which has degree $12$ over $\mathbb{Q}$.

I don't know how to proceed from here, and how to find the Galois group of the polynomial over $\mathbb{R}$? Thank you.

Answer 1

There are only two possibilities for the Galois group of any polynomial over $\Bbb R$: Either the trivial group or $\Bbb Z_2$. And since $x^6-5$ has non-real solutions, that means the answer has to be $\Bbb Z_2$.

Answer 2

Preliminaries

I'll just collect together the results already stated in the question. Let $\omega_6 = e^{\frac{2\pi i}{6}}$ and $\omega_3 = e^{\frac{2\pi i}{3}}$ be primitive sixth and third roots of unity respectively. Then $\omega_3 = -\omega_6^{-1}$ so $\mathbb{Q}(\omega_6) = \mathbb{Q}(\omega_3)$.

As noted in the question, the splitting field of $f(x) = x^6 - 5$ over $\mathbb Q$ is $L = \mathbb{Q}(\alpha, \omega_6) = \mathbb{Q}(\alpha, \omega_3)$, where $\alpha = \sqrt[6]{5}$.

Since $f(x)$ is irreducible (by Eisenstein's Criterion), we have $[\mathbb{Q}(\alpha):\mathbb{Q}] = \deg f = 6$, and since $\mathbb{Q}(\alpha) \subseteq \mathbb{R}$, we have by the Tower Law that $[L :\mathbb{Q}] = 12$.

Actual Answer

Note that $L = \mathbb{Q}(\omega_3, \alpha)$ is the splitting field of $x^6 - 5$ over $\mathbb{Q}(\omega_3)$, and $[L:\mathbb{Q}(\omega_3)]=6$, which means that $x^6 - 5$ is irreducible over $\mathbb{Q}(\omega_3)$, hence the Galois group $\Gamma(L/\mathbb{Q}(\omega_3))$ acts transitively on its roots, so there exists $\sigma \in \Gamma(L/\mathbb{Q}(\omega_3))$ with $\sigma(\alpha) = \omega_6\alpha$. It is clear that for each integer $i$, we have $\sigma(\omega_6^i\alpha) = \omega_6^{i+1}\alpha$, so $\sigma$ acts on the roots of $f$ as a rotation by angle $\pi/3$.

Let $\tau$ be the restriction to $L$ of complex conjugation. Clearly $\tau \in \Gamma(L/\mathbb{Q})$, and $\tau$ acts on the roots by reflection in the real axis.

Since $\sigma$ acts as a rotation by angle $\pi/3$, and $\tau$ acts as a reflection, it is clear that $\sigma, \tau$ generate a subgroup of $\Gamma(L/\mathbb{Q})$ isomorphic to the dihedral group with twelve elements (denoted variously by $D_{12}$ and $D_6$). Since in fact the Galois group has order $12$, we have in fact $$ \Gamma(L/\mathbb Q) = \langle \sigma, \tau \rangle \cong D_{12} $$

Remark

I opted here for a geometrical argument, since it seems to me the quickest way to see the structure of the group. However, we could just as well have checked that the composition $\sigma\tau\sigma\tau$ maps $\omega_6$ and $\alpha$ to themselves, and hence is the identity map on $L$. This means that $\sigma, \tau$ satisfy the defining relations of the presentation $\langle a, b\mid a^6, b^2, abab\rangle$ for $D_{12}$, and the result follows by considering the order of the groups.