Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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
add comment

2 Answers 2

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).

share|improve this answer
add comment

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 $S_3$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.