1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.