4
$\begingroup$

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?

  • 0
    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.2012-11-14

1 Answers 1

0

You can find the calculations here http://www.math.umass.edu/~weston/cn/notes.pdf on page 13, and lemmas on page 114 and 115.