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Let $Q_r(X)$ be the $r^{th}$ cyclotomic polynomial over $F_p$. Polynomial $Q_r(X)$ divides $X^r-1$ and factors into irreducible factors of degree $o_r(p)$. Let $h(X)$ be one such irreducible factor.

In the paper (Prime is in P), the authors say that "since $h(X)$ is a factor of the cyclotomic polynomial $Q_r(X)$, then $X$ is a primitive $r^{th}$ root of unity in $F=F_p[X]/h(X)$".

For me it's easy to see that it is a $r^{th}$ root of unity, but I can't see why it is a primitive $r^{th}$ root of unity. I can't get why the order of $X$ is exactly $r$ in $F$.

Edit: p is prime

Dietrich Burde
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bbarcom
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  • You seem to have another definition of cyclotomic polynomials which is per se irreducible. – Wuestenfux Aug 19 '19 at 11:37
  • I forgot to say that p is prime. With this there is a theorem that states that you can factor $Q_r(X)$ into irreducible factors of degree $o_r(p)$, assuming that p is prime. I'll add that information, thank you. – bbarcom Aug 19 '19 at 11:51
  • I seem to have answered this very question here. You need the assumption $p\nmid r$ for the claim to hold. This is because no extension field of $\Bbb{F}_p$ has primitive roots of unity of order $p$ (we get the inseparable case). There is probably room for improvement in my answer. – Jyrki Lahtonen Aug 19 '19 at 13:05
  • Normally I am reluctant to use my dupehammer privilege, when I have answered the target myself. This time I will make an exception. Also, I happen to have the dupehammer on all the four tags chosen by the asker, so I cannot really dodge the responsibility :-( – Jyrki Lahtonen Aug 19 '19 at 13:07
  • Mind you, I do upvote the question for you located the main problem yourself. The idea in the answer is basically a counting argument. Assumin $\gcd(r,p)=1$ the polynomial $X^r-1$ has exactly $r$ simple zeros in an appropriate algebraic closure. The same applies then to cyclotomic polynomials $\Phi_d(x), d\mid r$ that are the factors of $X^r-1$. Implying that the minimal polynomials of roots of unity of order $d$, $d\mid r$ a proper factor, are those other factors of $X^r-1$. – Jyrki Lahtonen Aug 19 '19 at 13:12

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