# Any automorphism of $\mathbb Q(w)$ for a primitive $n$-th root of unity $w$ maps $w$ to another primitive $n$-th root of unity. [closed]

If $w$ is a primitive $n$-th root of unity, prove that any automorphism of $\mathbb Q(w)$ takes $w$ to another $n$-th root of unity. ($\mathbb Q$ denotes rational numbers.)

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## closed as too localized by cardinal, The Chaz 2.0, Benjamin Lim, Asaf Karagila, t.b.Apr 20 '12 at 21:32

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What have you tried? If this is homework, tag it as such. –  lhf Apr 19 '12 at 22:47
What is $\mathbb Q$? Rational numbers may be given a lot of structures. –  user23211 Apr 19 '12 at 22:57
Please read meta.math.stackexchange.com/questions/1803/… which gives suggestions on asking questions here (even if they are not homework questions). –  Gerry Myerson Apr 19 '12 at 23:42
Related meta thread: meta.math.stackexchange.com/questions/4001/… –  cardinal Apr 19 '12 at 23:55

Let $\phi$ be an automorphism of the field $\mathbb{Q}(\omega)$. Since $\phi(ab)=\phi(a)\phi(b)$, we have $$\phi(\omega^n)=(\phi(\omega))^n.$$ But $\phi(\omega^n)=\phi(1)=1$. It follows that $(\phi(\omega))^n=1$ and therefore $\phi(\omega)$ is an $n$-th root of unity.
Note that nowhere did we use the fact that $\omega$ is a primitive $n$-th root of unity. Now that I have done the easier part, it's your turn. From the title, we want to prove that $\phi(\omega)$ is a primitive $n$-th root of unity. We have done the $n$-th root part, now we turn to the primitive part.
As a start, suppose that $(\phi(\omega))^m=1$, where $0 \lt m \lt n$. What does this say about $\omega^m$?