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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?… --- very similar, posted just hours ago. Maybe you two could get together and work out your problems? – Gerry Myerson Nov 14 '12 at 3:46
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. – jonaprieto Nov 14 '12 at 3:50

A direct construction seems easiest.

Let $\zeta_n$ denote a primitive $n$-th root of unity. Then $\mathbb{Q}(\zeta_n)$ consists of elements of the form $ a_0 + a_1 \zeta_n + a_2 \zeta^{c_2}_n + \cdots + a_{\phi(n)} \zeta_n^{c_{\phi(n)}}$ where $\phi$ is Euler's totient function, $c_k$ is the $k$-th smallest number coprime to $n$ and $a_i$ are rational coefficients. Similarly, elements of $\mathbb{Q}(\zeta_m)$ are of the form $ b_0 + b_1 \zeta_m + b_2 \zeta^{d_2}_m + \cdots + b_{\phi(m)} \zeta_m^{d_{\phi(m)}}$ and since $(m,n)=1$ none of the non-constant terms in one corresponds to a non-constant term in the other. So the intersection can only contain the rational numbers.

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