I have a question about Galois group. Actually, we are considering : $P(X) = X^6-5$. The roots on a algebraic closure are : $X = \{5^{\frac{1}{6}}, z_65^{\frac{1}{6}}, ..., z_{6}^55^{\frac{1}{6}} \}$, where $z_6$ is a primitive root of $X^6-1$. Let $L = \mathbb{Q}(5^{\frac{1}{6}}, z_65^{\frac{1}{6}}, ..., z_{6}^55^{\frac{1}{6}}) = \mathbb{Q(5^{\frac{1}{6}}, i\sqrt3)}$. Now, we want to find the Galois group $G=Gal(L/\mathbb{Q})$.
I've take $\sigma \in G$. So $\sigma_{|\mathbb{Q(5^{\frac{1}{6}})}}$ is a morphism, which is detemined by $\sigma(5^{\frac{1}{6}})$, so we have 6 possibilities $\sigma(5^{\frac{1}{6}}) \in X$. And then, I would like to extend, in a certain way, the restriction of $\sigma$ on $\mathbb{Q(5^{\frac{1}{6}}})$, to the real $\sigma$. We can notice that $\mathbb{Q(5^{\frac{1}{6}}}) \cap \mathbb{Q}(i\sqrt3) = \mathbb{Q}$. So, I would like to extend the little morphism to the big one, saying that :
$\sigma : (5^{\frac{1}{6}}, i\sqrt3) \rightarrow (5^{\frac{1}{6}}z_6^k, +-i\sqrt3)$
is well defined, and go all over the elements of $G$.
But actually, I can't see why this morphism would be well defined. Actually, the condition $\mathbb{Q(5^{\frac{1}{6}}}) \cap \mathbb{Q}(i\sqrt3) = \mathbb{Q}$ seems to be not enough to conclude, as we can see it through the example of $\mathbb{Q}(j, 2^{\frac{1}{3}})$.
So, how to conclude ? And if this way to proceed is not the good one, how to find the Galois group ?
**Edit ** Or maybe, I can conclude by saying that we know that $[L:\mathbb{Q}] = 12$ cause $[L:\mathbb{Q}] = [L : \mathbb{Q}(5^{\frac{1}{6}})] [\mathbb{Q}(5^{\frac{1}{6}}) : \mathbb{Q}]$, and $[\mathbb{Q}(5^{\frac{1}{6}}) : \mathbb{Q}] = 6$ and the first one is $\geq 2$, and as $|G| \leq 12$, we can conclude ?
