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Let $ w_1,...,w_{ \phi(n)}$ be the primitive $n$th roots of unity of $ t^n -1 \in \mathbb Q[t]$. Show that for each $ 1 \le i \le \phi (n)$, there exists an $ \sigma\in Aut \mathbb Q(w_1)$ satisfies $ \sigma ( w_1) = w_i$.

I know the converse is true, that is given any $ \sigma\in Aut \mathbb Q(w_1)$, $ \sigma (w_1) $ always maps to some $ w_i$. But for this direction I have trouble on construct the automorphism. Any help is appreciated.

user112564
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  • Well, for any automorphism $\sigma$ of a field containing $\Bbb Q$, what is $\sigma(r)$ when $r$ is a rational number? And if an automorphism $\sigma$ of $\Bbb Q(w_1)$ satisfies $\sigma(w_1)=w_i$, what are $\sigma(w_2),...,\sigma(\w_{\phi(n)})$? (What is the relationship between $w_1$ and the other $w_j$?) – Greg Martin Apr 17 '14 at 07:09
  • Well, it's clear that $\sigma (r) = r$ when $ r$ is rational. And I think I can get other $w_j$ by rising $w_1$ to some power, but I don't know how... – user112564 Apr 17 '14 at 07:14
  • @GregMartin Sorry, forgot to @ you. – user112564 Apr 17 '14 at 07:23
  • Indeed, the primitive $n$th roots of unity in $\Bbb C$ are $e^{2\pi i k/n}$ for all $1\le k\le n$ such that $\gcd(k,n)=1$. Finding a power $p$ to raise $e^{2\pi i j/n}$ to to get $e^{2\pi i k/n}$ is equivalent to solving $pj\equiv k\pmod n$ for $p$, which is possible since $\gcd(j,n)=1$. – Greg Martin Apr 17 '14 at 19:35
  • @GregMartin Thanks. And this is actually as far as I reach. I have trouble solving that equation. Can you give any hint? – user112564 Apr 17 '14 at 19:43
  • http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm – Greg Martin Apr 17 '14 at 22:48
  • @GregMartin But how to prove that $\sigma(r \omega) = \sigma(r)\sigma(\omega)$ for $r \in \mathbb Q$ and $\omega$ a root of unity? – Twnk Jun 16 '14 at 03:55
  • Any ring homomorphism $\sigma$ satisfies $\sigma(ab)=\sigma(a)\sigma(b)$. Moreover, any automorphism $\sigma$ of an extension of $\Bbb Q$ satisfies $\sigma(r)=r$ for $r\in\Bbb Q$. – Greg Martin Jun 16 '14 at 08:37

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Your $w_i$ are the roots of $\Phi_n$, the $n$-th cyclotomic polynomial. So your statement is then equivalent to the irreducibility of $\Phi_n$ over $\mathbb Q$ (if $w$ is any root of $\Phi_n$, ${\mathbb Q}[w]$ is isomorphic to the quotient ring $\frac{{\mathbb Q}[X]}{\Phi_n(X)}$).

Do you already know that $\Phi_n$ is irreducible ? If not, you might take a look here for example.

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Ewan Delanoy
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