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If $p$ is a prime, $p \nmid n$ and $n \nmid p - 1$, how can I prove that the $n$th cyclotomic polynomial $\Phi_n(x)$ has no roots modulo $p$?

I know that $n \nmid (p - 1)$ implies that there are no elements of order $n$ modulo $p$, and I also know that $x^n - 1 = \prod_{d \mid n}\Phi_d(x)$, which means that if we suppose by contradiction there was a root $a$, it would satisfy $a^n - 1 \equiv 0 \pmod p$ which means it has order $d$ for some divisor $d$ of $n$ and therefore it is also a root of $\Phi_d(x)$. Can I somehow derive a contradiction here?

user26857
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b_pcakes
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    I think you more or less already derived a contradiction. It may be easier to think as follows. If $a$ where a root modulo $p$ of $\Phi_n(x)$, then $a^n-1\equiv0\pmod p$. Therefore the order $\ell$ of $a$ mod $p$ has to be a factor of $n$. Thus $\ell$ is a factor of $\gcd(n,p-1)=1$. So $\ell=1$. So $a=1$. But $p\nmid\Phi_n(1)$ unless $n$ is a power of $p$, which case was excluded. – Jyrki Lahtonen Nov 11 '16 at 09:09
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    @JyrkiLahtonen Thanks, but why can it not be the case that $n$ is a multiple of $p - 1$? In that case we would just get $(n, p-1) = p-1$ which doesn't help much since the order of any element divides $p-1 = \phi(p)$ – b_pcakes Nov 11 '16 at 09:33

2 Answers2

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Trying to fix my incorrect argument in the comments.

We know that we have the factorization $$ P(x):=x^n-1=\prod_{d\mid n}\Phi_d(x).\qquad(*) $$ This factorization holds over both $\Bbb{Q}$ and $\Bbb{Z}_p$, but we are only interested in it over $\Bbb{Z}_p$.

Because $p\nmid n$, the derivative $P'(x)=nx^{n-1}$ has no common factors with $P(x)$ in the ring $\Bbb{Z}_p[x]$. Therefore all the zeros of $P(x)$ (in any extension field of $\Bbb{Z}_p$) are simple.

Assume that $a\in \Bbb{Z}_p$ is a zero of $\Phi_n(x)$. Necessarily $a\neq0$, so $a$ is also a zero of $x^{p-1}-1$. If the order of $a$ is $\ell$, then $\ell\mid p-1$. Furthermore, $a$ is then also a zero of $$x^\ell-1=\prod_{m\mid \ell}\Phi_\ell(x),$$ so $a$ is also a zero of some factor $\Phi_m(x), m\mid\ell$.

So unless $\ell=n=m$ the element $a$ will be a zero of two factors on the right hand side of $(*)$, contradicting the fact that all the zeros of $P(x)$ are simple.

Therefore $\ell=n$. But this implies that $n=\ell\mid p-1$, which was ruled out in the assumptions.

Jyrki Lahtonen
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  • Thank you, I knew it had to be exactly that contradiction but I had no idea how to show it was a contradiction. Can you provide reference for the statement in your third paragraph? I am not familiar with ring theory (even though in this case it appears to be just terminology), but either way it is not immediately clear to me why the fact that $P'(x)$ has no common factors with $P(x)$ implies that the roots are simple. – b_pcakes Nov 11 '16 at 11:11
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    Actually maybe it is clear; If a root has multiplicity greater than $1$, after taking derivative it has multiplicity at least $1$? Can you confirm? – b_pcakes Nov 11 '16 at 11:13
  • If $(x-a)^2\mid P(x)$, then $(x-a)\mid P'(x)$ (think: product rule of derivatives). Therefore $x-a$ would be a common factor of both $P(x)$ and $P'(x)$. – Jyrki Lahtonen Nov 11 '16 at 11:13
  • Right, that is what I was trying to say. Thanks! – b_pcakes Nov 11 '16 at 11:14
  • @JyrkiLahtonen I wonder why $P'(x)$ has no common factors with $P(x)$? Can we conclude it from the form of $P'(x)$ and $P(x)$ directly? – Gang men Nov 20 '22 at 14:40
  • @Gangmen Absolutely. $P'(x)=nx^n$, so its only irreducible factor is $x$. But $x\nmid P(x)$. – Jyrki Lahtonen Nov 20 '22 at 16:37
  • @JyrkiLahtonen Maybe you mean $P'(x)=nx^{n-1}$ and its only reducible factor is x, and I get it now, thank you so much! – Gang men Nov 20 '22 at 18:19
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We consider the following theorem

Theorem. Let $P(x)$ be a polynomial over either $\mathbb{R}[x], \mathbb{Q}[x], \mathbb{Z}[x]$ or $\mathbb{Z}_p[x]$. Then there exists a non constant polynomial $m(x)$ so that $m(x)^2 \mid P(x)$ if $\gcd (P(x),P'(x)) \ne 1$.

Proof. Let $g(x) \mid \gcd (P(x),P'(x))$ with all root of $g(x)$ are zeros with multiplicity $1$, or this means $\gcd (g(x),g'(x))=1$.

Then $P'(x)=g(x)h(x)$, we also have $P(x)=g(x)H(x)$ so $P'(x)= g'(x)H(x)+g(x)H'(x)$. This implies $g(x) \mid g'(x)H(x)$. Since $\gcd (g(x),g'(x))=1$ so $g(x) \mid H(x)$. Hence, $g(x)^2 \mid P(x)$. $\square$

From this lemma, we can prove the following proposition.

Proposition. Let $m,n$ be two positive integers and $p$ be a prime so that $p \nmid nm$. Then $\gcd (\Phi_m(x),\Phi_n(x))=1$ over $\mathbb{Z}_p[x]$.

Proof. By applying the theorem over $\mathbb{Z}_p[x]$ we find $x^{mn}-1$ has no repeated factors. Now suppose $\gcd (\Phi_m(x),\Phi_n(x))=g(x) \ne 1$ then $g(x)^2 \mid x^{mn}-1$, a contradiction. $\square$

From this proposition, we can say that for such $m,n,p$, $\Phi_m(x)$ and $\Phi_n(x)$ cannot be both divisible by $p$ for the same value of $x$.

Now, back to the problem. From what you found, $a$ is root modulo $p$ of $\Phi_n(x)$ and $\Phi_d(x)$, or $\Phi_n(a) \equiv \Phi_d(a) \equiv 0 \pmod{p}$, a contradiction since $p \nmid nd$.

In fact, your statement can be rephrased as

Let $p$ be a prime. For all positive integer $n$ and integer $a$ so that $\gcd (n,p)=1$ we have $$p \mid \Phi_n(a) \iff \text{ord}_p(a)=n.$$

J. W. Tanner
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Tengu
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