Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.