Let $\Phi_n(x)$ denote the $n^\text{th}$ cyclotomic polynomial. Suppose it has a root $\alpha$ in the finite field $\Bbb{F}_p$ and $p \nmid n$. Does it follow that $\mathrm{ord}_p(\alpha) = n$?

In the case where we're working with $\Bbb{C}$, then this is more or less a trivial result by definition of cyclotomic polynomials. However, it is no longer clear when working with finite fields. We clearly have $\mathrm{ord}_p(\alpha) \mid n$, but must equality hold? Also, what if we generalise it to $\Bbb{F}_{p^n}$?

  • $\begingroup$ @EwanDelanoy thanks. Can you explain why that implies a negative answer to this question? Also, after some brief research it appears that we can only conclude that's the case if $\frac{n}{m}$ is a prime power (see this). $\endgroup$
    – user711891
    Jun 30, 2020 at 9:49
  • $\begingroup$ @yellowello I have added an answer using elementary arguments. $\endgroup$ Jun 30, 2020 at 10:17
  • $\begingroup$ @yellowello I have deleted my comment, as it's made obsolete by Shubhrajit Bhattacharya's answer $\endgroup$ Jun 30, 2020 at 12:40
  • $\begingroup$ Related. Given that I happened to answer that variant I should not use the dupehammer here. I don't want to suggest a merge either, because this question specifies the prime field and the other is a bit more general. Also, I think seeing these answers benefits those who stumble upon the other, so linking the two will do. +1 to y'all. $\endgroup$ Jun 30, 2020 at 18:53

2 Answers 2


The following statements are equivalent:

$(1)$ $p\mid\Phi_n(\alpha)$ for some $\alpha\in\mathbb{Z}$ and for some prime $p$ such that $\gcd(p,n)=1$.

$(2)$ $\mathrm{ord}_p(\alpha)=n$


We proceed by induction on $n$. For $n=1$ it is trivial since $\Phi_1(X)=X-1$ and hence it has a root at $x\equiv1\pmod{p}$. Now suppose the hypothesis is true for all $k<n$. We will prove it for $n$.

Suppose $p\mid\Phi_n(\alpha)$. Since $\Phi_n(X)\mid X^n-1$, therefore $\alpha^n\equiv1\pmod{p}$. Then we have $\mathrm{ord}_p(\alpha)\mid n$. Let, if possible, $\mathrm{ord}_p(\alpha)=r<n$. Since $r\mid n$ we have $\gcd(r,p)=1$. Hence by induction hypothesis we have $\Phi_r(\alpha)\equiv0\pmod{p}$. Now $p\nmid rn$. Then letting $P(X)=X^{rn}-1$, we have $$\gcd(P(X),P'(X))=\gcd(X^{rn}-1,rnX^{rn-1})=1$$ in $\mathbb{F}_p[X]$. Hence $X^{rn}-1$ has no non-constant repeated factor in $\mathbb{F}_p[X]$. Let $\gcd(\Phi_n(X),\Phi_r(X))=m(X)$ in $\mathbb{F}_p[X]$. Then $m(X)^2\mid X^{rn}-1$ in $\mathbb{F}_p[X]$ implies $m(X)=1$. Hence $\gcd(\Phi_n(X),\Phi_r(X))=1$ in $\mathbb{F}_p[X]$. This contradicts the fact that $\Phi_n(\alpha)\equiv\Phi_r(\alpha)\equiv0\pmod{p}$. Therefore $\mathrm{ord}_p(\alpha)=n$

Conversely, let $\mathrm{ord}_p(\alpha)=n$. Then $\alpha^n\equiv1\pmod{p}$. Since $$\alpha^n-1=\prod_{d\mid n}\Phi_d(\alpha)$$ therefore $\Phi_l(\alpha)\equiv0\pmod{p}$ for some $l\mid n$. If $l<n$, then induction hypothesis would imply that $\mathrm{ord}_p(\alpha)=l<n$, which contradicts the assumption. Therefore $l=n$. This completes the inductive step and hence the proof.

  • 1
    $\begingroup$ Brilliant answer, just what I'm looking for! Just one question, why does $m(X)^2 \mid X^{rn} - 1$? $\endgroup$
    – user711891
    Jun 30, 2020 at 10:44
  • $\begingroup$ Since both $\Phi_n(X)$ and $\Phi_r(X)$ divides $X^{rn}-1$ and both $\Phi_r(X)$ and $\Phi_n(X)$ are multiples of $m(X)$. Therefore $m(X)\mid X^{rn}-1$ $\endgroup$ Jun 30, 2020 at 10:46
  • $\begingroup$ That doesn't sound right. $\gcd(X, X) = X$, and obviously $X \mid X$, but $X^2 \nmid X$. $\endgroup$ Jun 30, 2020 at 10:49
  • 3
    $\begingroup$ $$X^{rn}-1=\prod_{d\mid rn}\Phi_d(X)$$ in the right hand side both $\Phi_r(X)$ and $\Phi_n(X)$. $\endgroup$ Jun 30, 2020 at 11:01
  • $\begingroup$ I see, thanks a lot! $\endgroup$
    – user711891
    Jun 30, 2020 at 11:11

For $p \mid n$: $Φ_4 = X^2 + 1$ has root $1$ in $\mathbb F_2$.

Otherwise, let $q$ be a power of $p$ and $ζ$ be a primitive $q-1$-st root of unity in an algebraic closure of $ℚ_p$. By Hensel’s lemma, since $X^{q-1} - 1$ is separable over $\mathbb F_q$, we have an isomorphism of the group of $q-1$-st roots of unity in $ℚ_p(ζ)$ and the multiplicative group in $\mathbb F_q$, so $μ_{q-1,ℚ_p(ζ)} \cong \mathbb F_q^×$.

Then, since $p \not\mid n$, $X^n - 1$ is also separable over $\mathbb F_q$ and so again by Hensel’s lemma, the above isomorphism restricts to $μ_{n,ℚ_p(ζ)} \cong μ_{n,\mathbb F_q}$, with the roots of $Φ_n$ in $ℚ_p(ζ)$ corresponding to its roots in $\mathbb F_q$.

  • $\begingroup$ Thank you. Apologies, but I forgot to include the condition that $p \nmid n$. $\endgroup$
    – user711891
    Jun 30, 2020 at 9:28
  • $\begingroup$ @yellowello I added a solution, which is hopefully right. I’m rusty on $p$-adic theory. $\endgroup$
    – k.stm
    Jun 30, 2020 at 9:51

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