How can I find the degree of the extension? Let $\omega_7=e^{2\pi i/7}$ . How can I find the degree of the extension $\mathbb{Q} \leq \mathbb{Q}(\omega_7+\omega_7^5)$??
Could you give me some hints??
 A: Hint: We have $\Bbb Q \leq \Bbb Q(\omega_7 + \omega_7^5) \leq \Bbb Q(\omega_7)$.
A: If you've studied the cyclotomic polynomial the answer becomes quite simple! Since $\omega_7=e^{2\pi i/7}$ is a root of the seventh cyclotomic polynomial $\Phi_7$, and $\Phi_7$ can be shown to be irreducible, the degree of the extension is $\deg\Phi_7 = \varphi(7) = 6$, because $7$ is prime ($\varphi$ denotes the Euler totient function). In this proof we also use what Arthur has pointed out, namely that we can drop $\omega_7^5 $ (which I still leave to prove).
A: Extended hint:
The field $\Bbb{Q}(\omega_7+\omega_7^5)$ is contained in the 7th cyclotomic field $\Bbb{Q}(\omega_7)$. That field is an abelian extension of $\Bbb{Q}$, so all intermediate fields are themselves Galois extensions of $\Bbb{Q}$. This follows from Galois correspondence as all subgroups of an abelian group are normal. 
To make much progress here you should be familiar with the Galois group of $G=Gal(\Bbb{Q}(\omega_7)/\Bbb{Q})$. That group is cyclic of order six, and an automorphism $\sigma\in G$ is fully determined by $\sigma(\omega_7)=\omega_7^k$, where $k$ can be any integer in the range $1,2,3,4,5,6$. Denote that automorphism by $\sigma_k$.
The degree of any algebraic extension $K(a)/K$ is the degree of the minimal polynomial of $a$ over $K$. When the extension is separable (which is the case here, because ________ ) the zeros of the minimal polynomial are all simple and exactly the conjugates of $a$. This gives us the idea:

Count the number of conjugates of $\omega_7+\omega_7^5$. These are the numbers $\sigma_k(\omega_7+\omega_7^5)$. Write all of them ($k=1,2,\ldots,6$) in terms of low powers of $\omega$, and check how many of them are distinct.

