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Let $E$ be the splitting field of $x^6-2$ over $\mathbb{Q}$. Show that $Gal(E/\mathbb{Q})\cong D_6$, the dihedral group of the regular hexagon.

I've shown that $E=\mathbb{Q}(\zeta_6, \sqrt[6]{2})$, where $\zeta_6$ is a (fixed) primitive sixth root of unity, and thus that $[E:\mathbb{Q}]=12$.

I'm getting a little mixed up working out the automorphisms, though. I know the Galois group is determined by the action on the generators $\zeta_6$ and $\sqrt[n]{2}$. So then the possibilities appear to be: \begin{align*}\sqrt[6]{2}&\mapsto\zeta_6^n\sqrt[6]{2}\;\;\;\mbox{ for } n=0,1,\ldots ,5 \\ \zeta_6&\mapsto \zeta_6^j\;\;\;\;\;\;\;\;\mbox{ for } j=1,5\,.\end{align*}

Does this make sense? Something doesn't quite feel right, but I'm not sure where the issue might be. I know that in some sense the generators are "independent", because I definitely can't get one generator from the other. (For example, it'd be different if we had fourth roots of unity because we could get $\sqrt{2}$ from both generators.)

Any help is appreciated

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What is sufficient to show/justify the claim that $E=\mathbb{Q}(\zeta_6, \sqrt[6]{2})$ ? I certainly believe it to be true, but am having a hard time proving a similar statement! – The Chaz 2.0 May 2 '11 at 12:57
I was looking through my old questions and realized I never provided an answer to this question for you. It's been so long that you may not still be curious, but if you'd still like an answer let me know – Bey Jan 4 '13 at 22:08
Thanks for the update. I should be able to get by without further discussion of this matter! – The Chaz 2.0 Jan 5 '13 at 2:35
up vote 17 down vote accepted

By definition of the splitting field, it is generated by the roots of $x^6 - 2$. These roots form the vertices of a regular hexagon on the complex plane. Any element of the Galois group must permute the roots, and is completely determined by what it does to the roots, since the splitting group is generated by them.

Let $\sigma$ be an element of the Galois group. Now consider what happens to two adjacent vertices, $\zeta_6^i\sqrt[6]{2}$ and $\zeta_6^{i+1}\sqrt[6]{2}$. By considering $$\frac{\zeta_6^{i+1}\sqrt[6]{2}}{\zeta_6^{i}\sqrt[6]{2}} = \zeta_6,$$ the action of $\sigma$ on the two adjacent vertices determines the action on $\zeta_6$. Since $\zeta_6$ satisfies $\Phi_6(x) = x^2-x+1$, the image of $\zeta_6$ must be another root of $x^2-x+1$, which is either $\zeta_6$ or $\zeta_6^5$, as you note. And if you know what happens to $\zeta_6^i\sqrt[6]{2}$ and to $\zeta_6$, then you know what happens to $\sqrt[6]{2}$ as well. And of course, if you know what happens to both of these, then you know what happens to the adjacent vertices $\sqrt[6]{2}$ and $\zeta_6\sqrt[6]{2}$. So knowing the action on at least two adjacent vertices is equivalent to knowing the action on $\sqrt[6]{2}$ and on $\zeta_6$, and vice versa, and this in turn tells you what is happening on all the vertices. This shows that what you are saying is correct.

Now, to finish off, try showing that two adjacent vertices are always mapped to two adjacent vertices, so that any permutation of the hexagon induced by $\sigma\in\mathrm{Gal}(E/\mathbb{Q})$ is actually a rigid motion of the hexagon. This will show the Galois group is contained in the dihedral group, and the computation of the order that you have already done will finish off the problem.

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Thank you so much! – Bey Mar 28 '11 at 3:09

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