Q:

Find the degree of field extension

To find the degree of then extenstion $$\mathbb{Q}(\sqrt[p]{2}) \leq \mathbb{Q}(\sqrt[2]{2}, \omega_p)$$ where $\omega_p=e^{\frac{2 \pi i}{p}}$
we have to find the degree of $$Irr(\omega_p , \mathbb{Q}(\sqrt[p]{2}))$$
right?? But how could I do that??

A:

Okay so here goes a full solution: To prove irreducibility we will use Eisenstein's criterion extensively. 
I assumed $p$ is prime. First of all note that $\sqrt[p]{2}$ is root of $x^{p}-2$ which is clearly irreducible by the above criterion. Therefore $[\mathbb{Q}(\sqrt[p]{2}):\mathbb{Q}]=p$. 
Note that $\omega_p$ is satisfied by $f(x)=x^{p-1}+x^{p-2}+\cdots+x+1$. Observe that
$$f(x+1)=\sum_{i=0}^{p-1} (x+1)^i=\frac{(x+1)^{p}-1}{x}=\sum_{i=0}^{p-1} \binom{p}{i+1}x^{i}$$
Again the criterion, helps us to conclude $f$ is irreducible over $\mathbb{Q}$. Hence $[\mathbb{Q}(\omega_p):\mathbb{Q}]=p-1$
Hence by using this result we have
$$[\mathbb{Q}(\sqrt[p]{2},\omega_p):\mathbb{Q}]=p(p-1)$$
Hence
$$[\mathbb{Q}(\sqrt[p]{2},\omega_p):\mathbb{Q}(\sqrt[p]{2})]=\frac{[\mathbb{Q}(\sqrt[p]{2},\omega_p):\mathbb{Q}]}{[\mathbb{Q}(\sqrt[p]{2}):\mathbb{Q}]}=\frac{p(p-1)}{p}=p-1$$

