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!
Best Answer
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.)