I try to find the way to proof the next easy fact to two extension.
Problem: Let $n,\ m$ positive integers such that $(n,m)=1$. If you have $\alpha$ like a $m^{\text{th}}$ primitive root of unity and $\beta$ be $n^{\text{th}}$ primitive root of unity , then see that $ \mathbb{Q(\alpha)} \cap \mathbb{Q(\beta)} = \mathbb{Q}$
My plan to proof.
$\mathbb{Q}\subset \mathbb{Q(\alpha)} \cap \mathbb{Q(\beta)} :$ Because, $ \mathbb{Q(\alpha)}$ and $\mathbb{Q(\beta)}$ both are the extension field $\mathbb{Q}$. So It's trivial by definition.
$ \mathbb{Q(\alpha)} \cap \mathbb{Q(\beta)} \subset \mathbb{Q}$ : Because $(n,m)=1$, with the respective cyclotomic polynomial to $m,\ n$. The only common factor will be $(x-1)$ and this polynomial generate $\mathbb{Q}$
$x^n - 1 = \prod_{d|n} \Phi_d(x) =\Phi_1(x) \prod_{d|n, d\neq 1} \Phi_d(x) = (x-1) \prod_{d|n, d\neq 1} \Phi_d(x) $ $x^m - 1 = \prod_{d|m} \Phi_d(x) =\Phi_1(x) \prod_{d|m, d\neq 1} \Phi_d(x) = (x-1) \prod_{d|m, d\neq 1} \Phi_d(x) $ So, $[\mathbb{Q(\alpha)} \cap \mathbb{Q(\beta)} : \mathbb{Q}] = \partial(x-1) = 1$.
I'm not sure about that. How can you proof that. Thanks guys.