# Automorphism that maps primitive roots of unity.

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.

• 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 $w_j$?) Apr 17, 2014 at 7:09
• 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... Apr 17, 2014 at 7:14
• @GregMartin Sorry, forgot to @ you. Apr 17, 2014 at 7:23
• 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 $\gcd(j,n)=1$. Apr 17, 2014 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? Apr 17, 2014 at 19:43

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.