# 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!

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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 $\zeta_{12}\sqrt[6]{3}$ 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.

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Hey you are really wrong, $3^\frac{1}{6}$ it isn't a root of $x^6+3$. Actually the extension $K/Q$ it is galois with galois group $S_3$.

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It's $\sqrt[6]{3}\cdot \zeta_{12}^k$, with a primitive $12$-th root of unity. –  Daniel Fischer Oct 6 '13 at 19:22

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).

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