I was just wondering if this is a valid proof. I am assuming knowledge that if $\phi$ is an automorphism of a numeric field the $\phi$ fixes $\mathbb{Q}$. Also, if $\phi \in$ Gal($\mathbb{E}$/$\mathbb{F}$) and ${f} \in \mathbb{F}[x]$ has a root $x_{0}$, then $\phi (x_{0})$ is a root of ${f}$ as well.

Theorem: $\phi (\zeta)=\zeta ^{k}$ where $\phi$ is an automorphism of a field $ \mathbb{F}$ containing $\zeta$ and where $\zeta$ and $\zeta ^{k}$ are $n^{th}$ primitive roots of unity.

Proof: Since $\zeta \in \mathbb{F}$ it follows $$ K_{n}=\{\zeta ^{i}: 0 < i \leq n , i \in \mathbb{N}\} \subset \mathbb{F} $$ By a previous theorem, $\phi$ must send a root of unity to a root of unity. Since $\phi$ is an automorphism it follows that $\phi (K_{n}) = K_{n}$. So $\displaystyle \phi (\zeta^{k}) = \zeta ^{m}$. Let $\zeta ^{k}$ be primitive and suppose $\zeta ^{m}$ is not primitive. Since $\zeta^{k}$ is primitive $\displaystyle K_{n}=\{{(\zeta ^{k})}^{i}: 0<i\leq n\}$. Since $\phi$ is an automorphism, $$ \phi \left({(\zeta^{k})}^{i}\right)=\phi {(\zeta ^{k})}^{i}=({\zeta ^{m}})^{i} $$ So, it follows that $$ \displaystyle \phi (K_{n}) = \{ \phi \left( {(\zeta^{k})}^{i}\right): 0 < i \leq n \} = \{{(\zeta ^{m})}^{i}: 0<i\leq n\} \neq K_{n} $$ since $ \zeta ^{m} $ is not primitive. This contradicts with $ \phi (K_{n}) = K_{n} $. Therefore $ \zeta ^{m} $ is primitive.

Thanks in advance!


What you wrote is not necessarily wrong. However, I think there is a slightly different/easier way to look at this.

Let us consider the cyclotomic field $K = \mathbb{Q}[\zeta_n]$ where $\zeta_n$ is a primitive $n$th root of unity. Perhaps this cyclotomic field is embedded in a larger field extension, or perhaps it is not. In either case, the proof will work the same.

Now consider some automorphism $\phi:K \rightarrow K$. If we want, we can add the restriction that $\phi \in Aut(K/\mathbb{Q})$. Even though $\phi$ is a field automorphism, we can also view some of what it does as a group isomorphism (automorphism) of $G = \{\zeta_n^k: 0 \leq k \leq n-1\}$ such that $\phi:G \rightarrow G$.

From here, we simply note any group isomorphism preserves the order of the elements. An $n$-th root of unity is primitive $\iff$ it has order $n$. Therefore, if $\zeta_n^k$ is a primitive root, then $\phi(\zeta_n^k)$ must also have order $n$ and be a primitive root.

  • 1
    $\begingroup$ This is a wonderfully simple and intuitive approach! Thank you! $\endgroup$ – Thomas Hughes Jun 16 '14 at 4:53

Your Answer

