Galois group of $x^6-5$ over $\Bbb{Q}$.

Let $K$ be the splitting field of $x^6-5$ over $\Bbb{Q}$,

(a) Let $\omega_6$ be a primitive sixth root of unity over $\Bbb{Q}$. Compute the Galois group of $K$ over $\Bbb{Q}(\omega_6)$.

(b) Compute the Galois group of $K$ over $\Bbb{Q}$.

(a) Let $\alpha = 5^{1/6}$. We know that the roots of $x^6-5$ are $\alpha$, $\alpha\omega_6$, $\alpha\omega_6^2$, $\alpha\omega_6^3$, $\alpha\omega_6^4$, and $\alpha\omega_6^5$.

Since $x^6-5$ is irreducible over $\Bbb{Q}$ by Eisenstein's criteria, $[\Bbb{Q}(\alpha, \omega_6):\Bbb{Q}] \leq 6$. We also know that

$$[\Bbb{Q}(\alpha, \omega_6):\Bbb{Q}]=[\Bbb{Q}(\alpha,\omega_6):\Bbb{Q}(\omega_6)][\Bbb{Q}(\omega_6):\Bbb{Q})].$$

Since $[\Bbb{Q}(\omega_6):\Bbb{Q}]=\phi(6)=(3-1)(2-1)=2$, $[\Bbb{Q}(\alpha,\omega_6):\Bbb{Q}(\omega_6)]=1$ or $3$. But since $\alpha \not\in \Bbb{Q}(\omega_6)$, it must be of degree 3.

So $[\Bbb{Q}(\alpha,\omega_6):\Bbb{Q}(\omega_6)] = |Gal(\Bbb{Q}(\alpha,\omega_6)/\Bbb{Q}(\omega_6))|=3$. Since there is only one group of order 3 up to isomorphism, $Gal(\Bbb{Q}(\alpha,\omega_6)/\Bbb{Q}(\omega_6)) \cong \Bbb{Z}_3$.

(b) We know that $|Gal(K/\Bbb{Q})|=6$, and there are only two groups of order 6 up to isomorphism, $\Bbb{Z}_6$ and $S_3$.

We know that there is an automorphism $\sigma: \alpha \rightarrow \alpha\omega_6$, which is cyclic of order 6, so $Gal(K/\Bbb{Q}) \cong \Bbb{Z}_6$.

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@tomasz: $\omega_6^2 = -1 + \omega_6$. $[\mathbf{Q}(\omega_6) : \mathbf{Q}] = 2$. –  Brandon Carter Jul 29 '13 at 1:39
@tomasz: And? $- \omega_6^2$ is a primitive sixth root of unity, hence $\mathbf{Q}(\omega_6) = \mathbf{Q}(\omega_6^2)$. –  Brandon Carter Jul 29 '13 at 2:38
@tomasz: Where did you get that I was saying $\omega_6^2 = 2$ from? My point is that $\mathbf{Q}(\omega_6)$ is a quadratic extension of $\mathbf{Q}$, hence $\omega_6^2$ is of the form $a + b \omega_6$ and both of your comments above are incorrect. If you are still unclear, I'd suggest we move the conversation to chat. –  Brandon Carter Jul 29 '13 at 2:42
@BrandonCarter You put two math expressions side by side without a word in between, and the dot looked like multplication, as if you wrote $\omega_6^2=-1+\omega_6\cdot[\mathbf{Q}(\omega_6) : \mathbf{Q}] = 2$. It doesn't make sense, that's part of why I was so confused. You are correct, but you should avoid writing like that. –  tomasz Jul 29 '13 at 2:44
@Artus: Yes. (extra characters) –  Brandon Carter Jul 29 '13 at 21:38

Your first sentence beginning with the word “Since” is not correct. The right conclusion to draw from the irreducibility of $X^6-5$ over $\mathbb Q$ is that $[\mathbb Q(\alpha)\colon\mathbb Q]=6$. Now the question is whether you get a quadratic extension of $\mathbb Q(\alpha)$ by adjoining $\omega_6=\frac12(1+\sqrt{-3}\,)$, the quantity that’s missing to give you the splitting field.
I would attack the problem by following the suggested strategy and looking at $\mathbb Q(\omega_6)=k$ as your base field. I would continue and use information that maybe you don’t have, namely that the ring of integers of $k$, namely $\mathbb Z[\omega_6]$, is a principal ideal domain. And its primes are the ordinary primes congruent to $2$ modulo $3$, plus pairs of primes $\pi$, $\pi'$, one such pair for each $p\equiv1\pmod3$, and satisfying $\pi\pi'=p$. Plus a special prime $\theta$ for which $\theta^2=3$, the last of these statements being inaccurate because I suppressed mention of any unit factor. So of course, $\theta=\sqrt{-3}$ and $\theta^2$ is actually equal to $-3$. The important fact, though, is that $5$ is still prime in $\mathbb Z[\omega_6]$, so that $X^6-5$ is still Eisenstein, giving a degree of six for $K$ over $k$.The rest I think you can do.
Thanks a lot, but I'm not sure if I could understand you answer (especially the second paragraph)...I tried another way of proving this, but I'm not sure if I'm right. I said that $[\Bbb{Q}(\alpha):\Bbb{Q}]=6$. So now I want to compute $[\Bbb{Q}(\omega_6, \alpha): \Bbb{Q}(\alpha)]$. We know that, since $[\Bbb{Q}:(\omega_6):\Bbb{Q}]=2$, there exists a polynomial of degree two over the rationals with $\omega_6$ as a root. Since $\Bbb{Q}(\alpha)$ contains that polynomial, we just need to check if there is a polynomial of a lower degree than 2 in $\Bbb{Q}(\alpha)$ with $\omega_6$ as a root. –  user58289 Jul 29 '13 at 23:53
This looks very good to me, @Artus. My way of looking at it: $\mathbb Q(\alpha)$ has a real embedding but the field gotten by adjoining to it the square root of $-3$ clearly does not. So the two fields are different. –  Lubin Jul 30 '13 at 19:06
@Libertron, no: If that were the case, then $[\Bbb Q(\alpha,\omega_6)\colon\Bbb Q(\omega_6)]=3$ as well, but since $\Bbb Q(\omega_6)=\Bbb Q_(\omega_3)$, this field is quadratic over $\Bbb Q$. But the total degree is $12$, and $2\cdot3\ne12$. So that Galois group is cyclic of order $6$. –  Lubin Jan 4 at 20:24
Isn’t it the dihedral group of order $12$? The automorphisms should all be of the type $\{\omega_6\mapsto\omega_6^{\pm1}, \alpha\mapsto\omega_6^n\alpha\}$, for $0\le n\le5$, no? –  Lubin Jan 4 at 21:32
Just to check that all’s right with the world, you might verify that each of the automorphisms that moves $\omega_6$ will be its own inverse. –  Lubin Jan 4 at 21:44