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- #1

- Mar 10, 2012

- 834

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 $q(x)\in \mathbb Q[x]$.

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 book Problem $13$ and $14$ of section $5.6$ show the irreducibility of $\Phi_m$ using the fact that $\theta_k$ is an automorphism for each $k$ with $\gcd(k,m)=1$.

Note that it is clear that the set of all $\mathbb Q$ automorphisms of $E$ is a subset of $\{\theta_k:\gcd(k,m)=1\}$.