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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
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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
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Related meta thread: meta.math.stackexchange.com/questions/4001/… –  cardinal Apr 19 '12 at 23:55
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1 Answer 1

up vote 4 down vote accepted

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 $\omega^m$?

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Great answer - just beat me to it. I am especially fond how your choice in how much to give away. –  mixedmath Apr 19 '12 at 23:02
    
@mixedmath: The OP probably would benefit from more than one answer, even if the answers are along similar lines. Even simple variation in notation can contribute to learning. –  André Nicolas Apr 19 '12 at 23:08
    
André, I am experiencing some rendering errors in your answer. Are you also? –  The Chaz 2.0 Apr 20 '12 at 0:24
    
@TheChaz: There seems to be no difficulty, but I am now at a PC that is usually problem-free. Can't say the same of the Mac I usually use, which is ordinarily well-behaved but has fairly frequent troubles with MSE. –  André Nicolas Apr 20 '12 at 0:29
    
I see "As a start, suppose that (ϕ(ω))m=1, where [[0&ltm&ltn]]. What does this say about ωm?"... everything is fine except for the bracketed part. It's no big deal, just thought I'd see if it was a problem on my end (Firefox 3.6.8 MAC) –  The Chaz 2.0 Apr 20 '12 at 0:34
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