# 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 $\sqrt3 \xi_{12}^{k}$ 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 $\mathbb{Q}(\sqrt3,\xi_{12})=\mathbb Q(\sqrt{3},i)\supseteq\mathbb{K}$ 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!

• What is $[\mathbf{Q}(i\root 6\of 3):\mathbf{Q}]$? Which roots of unity can you find in that field? – Jyrki Lahtonen Sep 16 '11 at 15:51
• IOW: use $k=3$ as the "first root to adjoin". – Jyrki Lahtonen Sep 16 '11 at 16:24

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

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

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

• $\zeta_{12}^6$ is not $i$, it's $\zeta_2 = -1$: since $(\zeta_{12}^6)^2 = \zeta_{12}^{12} = 1$, then it is a root of $x^2 - 1$. But now I see you probably meant $\zeta_{12}^3 = \zeta_4 = i$. – André 3000 Nov 28 '17 at 5:50

Let $\alpha = \sqrt{-3}$ 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.)

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

• see my answer, which contradicts your post. – Justine Jun 5 '15 at 14:32
• It's $\sqrt{-3}$, not $\sqrt{3}$, that is a root of $x^6 + 3$. – André 3000 Nov 28 '17 at 5:57