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I am actually interested on cyclotomic polynomials of the form $x^n+1$ (i.e. $n$ is a power of 2, for large $n$).

Are these polynomials irreducible modulo a prime $q$? I already saw that $X^2+1$ is not irreducible modulo 5 because $Z_5$ contains roots of -1. What about $x^{512}-1$? Is there a general criterion?

Thanks

Tal-Botvinnik
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2 Answers2

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For any $n \in \mathbb{N}$ and $p$ a prime that does not divide $n$,

$$ \Phi_n(x) \text{ is irreducible in } \mathbb{Z}_p \iff p \text{ is a primitive root modulo } n$$

Wikipedia article: Primitive root modulo n.

For powers of $2$, only $2$ and $4$ have primitive roots. Primitive roots modulo $n$ correspond to generators of $\mathbb{Z}_n^*$. For $n=2^k$ and $k\geq 3$, $\mathbb{Z}_n^*$ is not cyclic and thus there are no primitive roots modulo $n$.

Thus for higher powers of $2$, the cyclotomic polynomial will be reducible modulo any odd prime.

Ken Duna
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  • Thanks. How to check this? Writing the whole list? Is there any other criterion? – Tal-Botvinnik Jun 30 '16 at 16:53
  • Primitive roots mod $n$ correspond to generators of $\mathbb{Z}_n^$. For $n=2^k$ for $k\geq 3$, $\mathbb{Z}_n^$ is not cyclic and thus no primitive roots exist. – Ken Duna Jun 30 '16 at 16:58
  • Ok, thank you. Is there a result concerning the cardinality of the set of inversible elements of $Z_q[X]/(X^n+1)$ for $n>4$ power of 2? – Tal-Botvinnik Jul 01 '16 at 08:58
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I find this relatively general answer from the Finite Field book by Rudolf, theorem 2.47, helpful:

The cyclotomic field $K^{(n)}$ is a simple algebraic extension of $K$. If $K=\mathbb{F}_q$ with $\gcd(q,n)=1$, then the cyclotomic polynomial $Q_n$ factors into $\phi(n)/d$ distinct monic irreducible polynomials in $K[x]$ of the same degree $d$. Here $d$ is the least positive integer such that $q^d \equiv 1 \mod n$.

InanimateBeing
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Maomao Xu
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  • Your answer is fine. Don't loose time answering old questions already having a reasonable answer. – reuns Jan 11 '23 at 18:03