Galois group of $x^6+3$ over $\mathbb Q$

I'm having some difficulties finding the Galois group of the polynomial $g(x)=x^6+3$ over $\mathbb Q$.

Here's what I did :
I observed that the roots of the given polynomial are $\sqrt[6]3 \xi_{12}^{k}$ where $\xi_{12}$ is a primitive 12-th root of the unity and $k=1,3,5,7,9,11$. Called $\mathbb{K}$ the splitting field of $g(x)$ over $\mathbb{Q}$, is obvious that $\mathbb{Q}(\sqrt[6]3,\xi_{12})=\mathbb Q(\sqrt[6]{3},i)\supseteq\mathbb{K}$ so $6|[\mathbb K:\mathbb Q]\le12$. But from this point I'm not able to continue rigorously.
Seems to me that $[\mathbb K:\mathbb Q]=6$ but I'm not sure on how to proof that. Can anyone please help me? Thanks in advance!

Solution 1:

Let $\alpha = \sqrt[6]{-3}$ be a root of $g$. I claim that the splitting field $\mathbb{K}$ of $g$ is $\mathbb{Q}(\alpha)$. The roots of $g$ are $\alpha, \zeta \alpha, \cdots, \zeta^5 \alpha$ where $\zeta$ is a primitive $6^\text{th}$ root of unity, so to prove that $\mathbb{Q}(\alpha)$ is the splitting field, it suffices to show that $\mathbb{Q}(\alpha)$ contains a primitive $6^\text{th}$ root of unity.

Note that $\alpha^3$ is a root of $x^2 + 3$, so $\alpha^3 = \pm \sqrt{-3}$. Then $\zeta = \frac{1 + \alpha^3}{2} = \frac{1 \pm \sqrt{-3}}{2}$ is a primitive $6^\text{th}$ root of unity and $\zeta \in \mathbb{Q}(\alpha)$, as desired. Thus $\mathbb{K} = \mathbb{Q}(\alpha)$ is the splitting field of $g$, hence $[\mathbb{K} : \mathbb{Q}] = 6$.

(For computing the Galois group of $g$, see this post.)

Solution 2:

See this post. Once you add $\zeta_{12}\sqrt[6]{3}$ to $\mathbb{Q}$, you have added all of the roots: $ (\zeta_{12}\sqrt[6]{3})^6=3\zeta_{12}^6$, so that $\zeta_{12}^6=i\in K=\mathbb{Q}(\zeta_{12}\sqrt[6]{3})$. Hence,

• $(\zeta_{12}\sqrt[6]{3})i=\zeta_{12}^7\sqrt[6]{3}$, which implies $\zeta_8\in K$
• $(\zeta_{12}\sqrt[6]{3})\zeta_8=\zeta_{12}^9\sqrt[6]{3}$, which implies that $\zeta_{10}\in K$
• $(\zeta_{12}\sqrt[6]{3})\zeta_{10}=\zeta_{12}^{11}\sqrt[6]{3}$...

You get the point. Thus $\zeta_{12}\sqrt[6]{3}, \zeta_{12}^3\sqrt[6]{3},\zeta_{12}^5\sqrt[6]{3},\zeta_{12}^7\sqrt[6]{3},\zeta_{12}^9\sqrt[6]{3},\zeta_{12}^{11}\sqrt[6]{3}\in K$ and $$ [K:\mathbb{Q}]=6, $$ contrary to the other posts.

Solution 3:

Let $\zeta=e^{\frac{2\pi i}{6}}$. Then the roots of $X^6+3$ are $\zeta^i\sqrt[6]{-3},\ 1\leq i\leq 6$ so the splitting field over $\mathbb{Q}$ is $F=\mathbb{Q}(\zeta,\sqrt[6]{-3})$. It is $$\zeta=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\in\mathbb{Q}(\sqrt[6]{-3})\Rightarrow F=\mathbb{Q}(\sqrt[6]{-3})$$

$X^6+3$ is irreducible (by Eisenstein) so $|F:\mathbb{Q}|=|\mathbb{Q}(\sqrt[6]{-3}):\mathbb{Q}|=6$ so for the Galois group it is $|G|=6 \Rightarrow G\cong S_3$ or $\mathbb{Z}_6$. Note that $$(\sqrt[6]{-3})^2=-\sqrt[3]{3}\Rightarrow \sqrt[3]{3}\in F$$ If we consider $\mathbb{Q}(\sqrt[3]{3})/\mathbb{Q}$ we see that it is not a Galois extension hence $G$ is not abelian $\Rightarrow G\cong S_3$