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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.

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    Is there an echo in here? http://math.stackexchange.com/questions/236832/relatively-prime-cyclotomic-extensions#comment524549_236832 --- very similar, posted just hours ago. Maybe you two could get together and work out your problems?2012-11-14
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    yes, I had not read these post. But, I found the orther way by L.Washington, the proof is so short in Introduction to cyclotomic fields (Lawrence C. Washington), Preposition 2.4, Chapter 2.2012-11-14

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