Let $p$ a prime number. Find a splitting field $E$ of the polynomial $x^p-2 \in \mathbb{Q}[x]$.
I have done the following:
The solutions of $x^p-2=0$ are : $$\sqrt[p]{2}, \sqrt[p]{2}\omega, \dots, \sqrt[p]{2}\omega^{p-1}, \text{ where } \omega=e^{\frac{2 \pi i}{p}}$$
Therefore, the splitting field is $E=\mathbb{Q(\sqrt[p]{2}, \omega)}$
Is this correct so far??
How can I find the degree $[E:\mathbb{Q}]$ ??
EDIT:
$[E:\mathbb{Q}]=[\mathbb{Q(\sqrt[p]{2}, \omega)} : \mathbb{Q}]=[\mathbb{Q(\sqrt[p]{2}, \omega)} : \mathbb{Q}(\sqrt[p]{2})][\mathbb{Q(\sqrt[p]{2})} : \mathbb{Q}]=[\mathbb{Q(\sqrt[p]{2}, \omega)} : \mathbb{Q}(\sqrt[p]{2})]p$
Since $Irr(\sqrt[p]{2}, \mathbb{Q})=x^p-2 \Rightarrow [\mathbb{Q(\sqrt[p]{2})} : \mathbb{Q}]=p$.
But how can we find $[\mathbb{Q(\sqrt[p]{2}, \omega)} : \mathbb{Q}(\sqrt[p]{2})]$ ??
EDIT: Proof that the degree is the product of p and p-1:
$$[\mathbb{Q}(\omega, \sqrt[p]{2}) :\mathbb{Q}]=[\mathbb{Q}(\omega, \sqrt[p]{2}):\mathbb{Q}(\sqrt[p]{2})][\mathbb{Q}(\sqrt[p]{2}):\mathbb{Q}]=a \cdot p$$
$$[\mathbb{Q}(\omega, \sqrt[p]{2}):\mathbb{Q}]=[\mathbb{Q}(\omega, \sqrt[p]{2}):\mathbb{Q}(\omega)][\mathbb{Q}(\omega):\mathbb{Q}]=b \cdot (p-1)$$
$$a \cdot p = b \cdot (p-1) \Rightarrow p \mid a \cdot p \overset{(p, p-1)=1}{ \Longrightarrow } p \mid b \Rightarrow p \leq b \tag 1 $$ $$\Rightarrow a \cdot p \leq a \cdot b \Rightarrow b \cdot (p-1) \leq a \cdot b \Rightarrow p-1 \leq a \tag 2 $$
$$\mathbb{Q} \leq \mathbb{Q}(\omega) \leq \mathbb{Q}(\omega, \sqrt[p]{2}) \Rightarrow Irr(\sqrt[p]{2}, \mathbb{Q}(\omega)) \mid Irr(\sqrt[p]{2}, \mathbb{Q}) $$ $$\Rightarrow \deg Irr(\sqrt[p]{2}, \mathbb{Q}(\omega)) \leq \deg Irr(\sqrt[p]{2}, \mathbb{Q}) \Rightarrow b \leq p \tag 3$$
$$\mathbb{Q} \leq \mathbb{Q}(\sqrt[p]{2}) \leq \mathbb{Q}(\omega, \sqrt[p]{2}) \Rightarrow Irr(\omega, \mathbb{Q}(\sqrt[p]{2})) \mid Irr(\omega, \mathbb{Q}) $$ $$ \Rightarrow \deg Irr(\omega, \mathbb{Q}(\sqrt[p]{2})) \leq \deg Irr(\omega, \mathbb{Q}) \Rightarrow a \leq p-1 \tag 4$$
From $(1)$ and $(3)$ we have that $b=p$ and from $(2)$ and $(4)$ we have that $a=p-1$
Therefore, $[\mathbb{Q}(\omega, \sqrt[p]{2}) :\mathbb{Q}]=p(p-1)$.
You've done most of the work for yourself, and your proposed splitting field is correct.
We can consider two towers of fields: $$\mathbb{Q} \subset \mathbb{Q}[\sqrt[p]{2}] \subset \mathbb{Q}[\sqrt[p]{2}, \omega]$$ $$\mathbb{Q} \subset \mathbb{Q}[\omega] \subset \mathbb{Q}[\sqrt[p]{2}, \omega]$$ where $\omega$ is a $p^{\text{th}}$ root of unity.
In the first tower, the intermediate field is of degree $p$ over $\mathbb{Q}$ (Why?). In the second tower, the intermediate field is of degree $p-1$ over $\mathbb{Q}$. (again, why?)
Now apply the fact that $[K:F] = [K:E] \cdot [E:F]$ for a tower of fields $F \subset E \subset K$. Given this, both $p$ and $p-1$ divide the degree of the splitting field. Further, the degree of the splitting field is at most $p(p-1)$ since, worst case scenario, the minimal polynomial for $\sqrt[p]{2}$ is irreducible over $\mathbb{Q}[\omega]$.
Therefore...