About a root in the topic of automorphism and fixed field. HELP

[julypraise]

About a root in the topic of automorphism and fixed field. HELP!

Homework Statement

Let $m$ be a positive integer.

Let $\xi = \exp (2pi/m)$. Then $\xi$ is a primitive m-th root of unity. (I.e., $\xi$ is a solution of

$\Phi_{m}(X):=(X^m - 1)/(X-1)$.)

If $\phi \in \mbox{G}(\mathbb{Q}(\xi)/\mathbb{Q})$, i.e., $\phi$ is an automorphism of $\mathbb{Q}(\xi)$ fixing the elements in $\mathbb{Q}$, then $\phi (\xi) = \xi^i$ for some $i$ with $\gcd (i,m) = 1$

Homework Equations

$\mathbb{Q}(\xi) = \mathbb{Q}(\xi^i)$ for any $i$ with $\gcd(i,m)=1$ but not with $i$ such that $\gcd(i,m) \neq 1$

The Attempt at a Solution

I know the above relevant equation holds. And also I know that $\phi(\xi) = \xi^i$ for some integer $i$ but I cannot prove $\gcd (i,m) = 1$. I've tried to use Isomorphism Extension Theorem, but not really works.

 

[lippyka]

Can someone help me to prove this statement? Proof: Let \phi \in \mbox{G}(\mathbb{Q}(\xi)/\mathbb{Q}) , i.e., \phi is an automorphism of \mathbb{Q}(\xi) fixing the elements in \mathbb{Q} such that \phi (\xi) = \xi^i for some integer i . Let \gcd (i,m) \neq 1 . We will show that \mathbb{Q}(\xi) \neq \mathbb{Q}(\xi^i) . Since \gcd (i,m) \neq 1 , there exist nonzero integers a, b such that ai-bm = 0 . Since \xi^m = 1 , it follows that \xi^{ai-bm} = 1 .Now, \xi^i = \phi(\xi) \neq \xi^{ai-bm} = 1 , so \mathbb{Q}(\xi) \neq \mathbb{Q}(\xi^i) . This contradiction implies that \gcd (i,m) = 1 , as desired.