Show that $q(\omega)\mapsto q(\omega^k)$ is an automorphism of $\mathbb Q$ given $\gcd(k,m)=1$. Let $p(x)=x^m-1$ be a polynomial over $\mathbb Q$ and $E$ be the splitting field for $p$ over $\mathbb Q$. We know that $p$ has $\phi(m)$ primitive roots in $E$, where $\phi$ is the Euler's totient function. Let $\omega$ be a primitive root of $p$. 
Define $\theta_k:E\to E$ as $\theta_k(q(\omega))=q(\omega^{k})$ for all $q[x]\in \mathbb Q[x]$.
I want to show that $\theta_k$ is an isomorphism for each $k$ satisfying $\gcd(k,m)=1$.
The only problem here is to show that $\theta_k$ is well defined. This is equivalent to showing that $q(\omega)=0 \Rightarrow q(\theta_k(\omega))=0$, for each $q(x)\in \mathbb Q[x]$ and this is where I am stuck.
I don't want to use that $\Phi_m$, the $m$-th cyclotomic polynomial, is irreducible over $\mathbb Q$. The reason for this is that in Herstein's Topics in Algebra (2nd Edition) Problem $13$ and $14$ of section $5.6$ ask us to show the irreducibility of $\Phi_m$ over $\mathbb Q$ by first proving that $\theta_k$ is an automorphism for each $k$ with $\gcd(k,m)=1$ and then using it. 
The following may be useful in solving the above:
$\mathbb Q(\omega)=\mathbb Q(\omega^k)\iff \gcd(k,m)=1$.
The set of all $\mathbb Q$ automorphisms of $E$ is a subset of $\{\theta_k:\gcd(k,m)=1\}$.
I had posted this on http://mathhelpboards.com/linear-abstract-algebra-14/automorphisms-splitting-field-m-th-cyclotomic-polynomial-6185.html#post28229 but I was not convinced by the argument given by the helper.
Thank you.
 A: One argument I have seen applied here is the following. Hopefully it helps.
Let $h(x)$ be the minimal polynomial of $\omega$ over $\mathbb{Q}$. It is a factor of $p(x)$, so we know that it has coefficients in $\mathbb{Z}$. Furthermore
$p(x)=h(x)k(x)$ for some other polynomial $k(x)\in\mathbb{Z}[x]$.
Let $q<m$, be a prime not dividing $m$. I claim that $\omega^q$ is then also a zero of $h(x)$. Assume contrariwise that this is not the case. As we know that $\omega^q$ is a zero of $p(x)$, it must be the case that $\omega^q$ is a zero of $k(x)$. It follows that $\omega$ is a zero of $k(x^q)$, so $h(x)\mid k(x^q)$ in the
ring $\mathbb{Z}[x]$. Now we pass to the quotient ring $\mathbb{Z}_q[x]$. Let us denote the images of the polynomials by $\overline{h}(x)$ et cetera. This makes sense, because the polynomials have integer coefficients. 
Modulo $q$ the Freshman's dream states that
$$
\overline{k}(x^q)=\overline{k}(x)^q.
$$
But also $\overline{h}(x)\mid \overline{k}(x^q)$, so this implies that $\overline{h}(x)$ and $\overline{k}(x)$ have a non-trivial common factor. Thus the polynomial $\overline{p}(x)$ has a repeated factor. This contradicts the derivative test for repeated factors. Namely, $x$ is the only irreducible factor of $\overline{p}'(x)=mx^{m-1}$, so $\gcd_{\mathbb{Z}_q[x]}(\overline{p},\overline{p}')=1$, implying that $\overline{p}(x)$ cannot have any repeated factors. Here it was crucial that $q\nmid m$ as otherwise $\overline{p}'(x)=0$.
So we know that $h(\omega^q)=0$. This implies easily that your automorphism $\theta_q$ is well defined. Any integer $k$ coprime $n$ is the product of such primes $q$ (possibly with repetitions). The mapping $\theta_k$ is then the composition of those mappings $\theta_q$, and hence it is a well-defined automorphism.
