# Galois group of $\mathbb{Q}(\sqrt[3]{5},\omega)$

Let $L = \mathbb{Q}(\sqrt[3]{5},\omega)$ be an extension of the field of rational numbers $\mathbb{Q}$, where $\omega = \frac{-1+\sqrt{3}}{2}$.

(i) Show that the Galois group $G(L/\mathbb{Q})$ is isomorphic to $S_3$.

(ii) Let $H\subset G$ be the unique Sylow 3-subgroup of $G$. Find an element $\alpha \in L$ such that $LH = \mathbb{Q}(\alpha)$ where $LH$ is the intermediate field $\mathbb{Q} \subset LH \subset L$ fixed by $H$.

For (i), I know that we have $$L = \mathbb{Q}(\sqrt[3]{5},\omega) = \mathbb{Q}(\sqrt[3]{5}, 1+\sqrt{-3}) = \mathbb{Q}(\sqrt[3]{5},\sqrt{-3}) = \mathbb{Q}(\sqrt{-3})(\sqrt[3]{5}).$$ And the minimal polynomials for $\sqrt[3]{5}$ and $\sqrt{-3}$ are $x^3-5$ and $x^2+3$, respectively. Now $[L : \mathbb{Q}]\leq 3\cdot2=6.$ Now To show we actually have equality, we just need to show that $\sqrt[3]{5}\notin\mathbb{Q}(\sqrt{-3}).$ Indeed, let $a,b\in\mathbb{Q}$, and suppose $\sqrt[3]{5} = a+b\sqrt{-3}.$ Then

\begin{align} (a+b\sqrt{-3})^3 &= 5\\ a^3-9ab^2+(3a^2b-3b^3)\sqrt{-3}&=5\\ \frac{5-a^3+9ab^2}{3a^2b-3b^3}&=\sqrt{-3}, \end{align} which implies that $\sqrt{-3}\in\mathbb{Q}$, a contradiction. Hence the extension has degree 6. I'm not sure how I would show the Galois group is nonabelian to conclude that $G\cong S_3.$

Alternatively, I know that $L$ is the splitting field for $(x^3-5)(x^2+3)$. But the Galois group for this polynomial appears to be $D_6$, the dihedral group of order 12. I determined this just by listing out all of the automorphisms of the roots. Where am I going wrong here?

• Well $D_6$ and $S_3$ are isomorphic. Pick two automorphisms and evaluate them at an element to show that they don't commute. Commented Jul 25, 2016 at 17:39
• $D_6$ and $S_3$ have different orders, so they aren't isomorphic. Commented Jul 25, 2016 at 17:48
• I think $D_6$ here refers to the dihedral group of order $6$, not the symmetries of a hexagon. There's always some confusion about whether to use $D_n$ or $D_{2n}$ to refer to the Dihedral group of an $n$-gon. Commented Jul 25, 2016 at 18:00

After you know the degree of the extension is 6, you don't need to compute anything else. $L$ is the splitting field of $x^3-5$ over $\mathbb{Q}$, hence is Galois and the Galois group permutes the root of the polynomial $x^3-5$. So it's an order 6 subgroup of $S_3$, therefore all of it.

Now the desired subgroup is generated by the automorphism which maps $\sqrt[3]{5} \mapsto \omega \sqrt[3]{5}$ and fixes $\omega$. Hence the associated fixed field is $\mathbb{Q}(\omega)$, which one verifies by checking degrees.

As you mention, we have the fields tower

$$\Bbb Q\subset\Bbb Q(\sqrt[3]5)\subset L:=\Bbb Q(\sqrt[3]5, \omega)\;,\;\;\;\omega=e^{2\pi i/3}$$

and as you pointed out, we in fact have that $\;L=\Bbb Q(\sqrt[3]5,\,\sqrt{3}\,i)\;$ (and in fact $\;L=\Bbb Q(\sqrt[3]5+\sqrt3\,i)$)

Now, every automorphism of $\;L/\Bbb Q\;$ maps roots of some irreducible polynomial to roots of this same polynomial, so you have here two automorphisms:

$$\sigma:\begin{cases}\sqrt[3]5\mapsto\sqrt[3]5\\{}\\\sqrt3\,i\mapsto-\sqrt3\,i\end{cases}\;\;,\;\;\;\;\tau:\begin{cases}\sqrt[3]5\mapsto\sqrt[3]5\,\omega\\{}\\\sqrt3\,i\mapsto\sqrt3\,i\end{cases}$$

and now check that $\;\sigma\tau\neq\tau\sigma\;$ , by observing that

$$\sigma\omega=\sigma\left(-\frac12+\frac12\sqrt3\,i\right)=-\frac12-\frac12\sqrt3\,i=\omega^2$$

$$\tau\omega=\tau\left(-\frac12+\frac12\sqrt3\,i\right)=-\frac12+\frac12\sqrt3\,i=\omega$$

and then:

$$\begin{cases}\sigma\tau(\sqrt[3]5\,\omega)=\sigma(\sqrt[3]5\,\omega^2)=\sqrt[3]5\,\omega\\{}\\\tau\sigma(\sqrt[3]5\,\omega)=\tau(\sqrt[3]5\,\omega^2)=\sqrt[3]5\,\omega^3=\sqrt[3]5\end{cases}$$