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If for $m>1$, we denote the $m$-th cyclotomic polynomial by $\Phi_m(X)\in\mathbb Z[X]$, and we let $\overline{\Phi_m}(X)$ be the projection of $\Phi_m(X)$ into $\mathbb F_q[X]$, where $q>1$ a prime which is not a prime factor of $m$. We then have the following:

$\overline{\Phi_m}(X)$ has $\phi(m)$ (Euler totient) distinct roots in the algebraic closure $\overline{\mathbb F_q}$, and they are all primitive $m$-th roots of unity.

I understand the first part, namely that $\overline{\Phi_m}(X)$ has $\phi(m)$ distinct roots in $\overline{\mathbb F_q}$, and that they are all $m$-th roots of unity. But why are they necessarily primitive?

Zuy
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    I think I covered your question in this old answer? Admittedly that answer could still use a bit of cleaning up. Anyway, if $\zeta$ has order $m$, then so does $\zeta^k$ for all $k$ such that $\gcd(k,m)=1$. This is a basic property of cyclic groups. – Jyrki Lahtonen Mar 06 '23 at 10:14
  • @JyrkiLahtonen Thank you. I think you answer is very clear. – Zuy Mar 06 '23 at 10:37
  • In that case I vote to close this as a duplicate. As I have a dupehammer privilege in two of you tags, it means that my vote takes immediate effect. If there are points needing extra clarification, please comment here. I will be @-pinged, and can react and/or reopen the question. The duplicate closure has no effect on your account (unlike closures for other reasons). You can let the question stay here. In fact, I just upvoted it because you put in a non-trivial amount of effort in describing exactly what you needed to gete explained. – Jyrki Lahtonen Mar 06 '23 at 11:14

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