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!
 A: 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.)
A: 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.
A: 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 $\Rightarrow G\cong S_3$
A: First, it can't be true that $ [ \mathbb{K} : \mathbb{Q} ] = 6$ since $[ \mathbb{Q}(\sqrt[6]{3}) : \mathbb{Q} ] = 6$ ($x^6 + 3$ is irreducible by Eisenstein criterion) and $\xi_{12} = \exp (i \pi / 6) = \frac{\sqrt{3}}{2} + \frac{i}{2} \in \mathbb{C}$ and hence $\xi_{12} \notin \mathbb{Q}(\sqrt[6]{3}) \subset \mathbb{R}$. You need to find $[ \mathbb{Q}(\sqrt[6]{3},\xi_{12}) : \mathbb{Q}(\sqrt[6]{3})]$. Note that $$ x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)$$ is reducible. Moreover, $\sqrt{3} = (\sqrt[6]{3})^3 \in \mathbb{Q}(\sqrt[6]{3})$ and $i = \sqrt{-1}$ is algebric of degree 2 over this field, thus $\mathbb{K} = \mathbb{Q}(\sqrt[6]{3},i)$ is of degree 12 over $\mathbb{Q}$ since $[ \mathbb{Q}(\sqrt[6]{3},i) : \mathbb{Q}(\sqrt[6]{3}) ] = 2$ ($x^2 + 1$ is still irreducible there).
