# Relatively prime cyclotomic extensions

I'm having some trouble approaching the following problem:

Let $m, n \in \mathbb{Z}^+$ be such that $(m,n)=1$. Let $\alpha$ be a primitive $m$-th root of unity and $\beta$ be a primitive $n$-th root of unity. Show that $\mathbb{Q}(\alpha)\cap\mathbb{Q}(\beta)=\mathbb{Q}$.

I've tried computing the irreducible polynomial of $\alpha$ over $\mathbb{Q}(\beta)$ to show that it has degree $\varphi(m)$. Does this way lead to the desired result?

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You know that the degree of the $n$-cyclotomic field over $\mathbb Q$ is $\varphi(n)$, it seems. And you know the multiplicative property of the Euler function, that when $\gcd(m,n)=1$ you get $\varphi(mn)=\varphi(m)\varphi(n)$? It all ought to fall out from that. – Lubin Nov 14 '12 at 0:14
I am aware of these properties. Let me elaborate a little bit more. Let $p=irr(\alpha,\mathbb{Q}(\beta))$. I know that $\alpha\beta$ is a primitive $mn$-th root of unity. If I could show that $\mathbb{Q}(\alpha\beta)=\mathbb{Q}(\alpha,\beta)$, then it would be immediate that $\varphi(m)\varphi(n)=\varphi(mn)=[\mathbb{Q}(\alpha\beta):\mathbb{Q}]=[\mathbb{‌​Q}(\alpha,\beta):\mathbb{Q}(\beta)][\mathbb{Q}(\beta):\mathbb{Q}]$ and so $\varphi(m)=[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\beta)]=$ degree of $p$. – Klaus Müller Nov 14 '12 at 1:32
Well, I think you have it. $\alpha\in\mathbb Q(\alpha\beta)$ because $(\alpha\beta)^n$ is a primitive $m$-th root of $1$. By symmetry, $\mathbb Q(\alpha,\beta)\subset\mathbb Q(\alpha\beta)$. And $\alpha\beta\in \mathbb Q(\alpha,\beta)$ implies that $\mathbb Q(\alpha\beta)\subset\mathbb Q(\alpha,\beta)$ – Lubin Nov 14 '12 at 1:52
@Lubin, even if you prove ${\bf Q}(\alpha,\beta)={\bf Q}(\alpha\beta)$, you still have to make a connection somehow to ${\bf Q}(\alpha)\cap{\bf Q}(\beta)$. – Gerry Myerson Nov 14 '12 at 2:54
Well, @Gerry, I’ve felt a little hobbled by my uncertainty of how much OP knows. Seems to me that knowing that the degree of the $n$-cycl over $\mathbb Q$ is $\varphi(n)$ should give you everything, as long as you have the right background. The Theorem on Natural Irrationalities gives it to you right away, and somehow I thought that OP saw how that would apply. – Lubin Nov 14 '12 at 3:10