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The $n$th cyclotomic polynomial remains irreducible when reduced modulo $p$ if and only if $p$ is a generator of $\mathbb{Z}_n^\times$. Suppose that is not the case, and I know that the polynomial can be factored over $\mathbb{F}_p$. What can I say about the degrees of the irreducible factors?

For example, the 13th cyclotomic polynomial is reducible modulo 3, since $3^3 \equiv 1$ modulo 13. A (long, tedious) factorisation attempt reveals that there are four cubic irreducible factors. Should I have known this a priori?

probablystuck
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  • $13$th? Over $\Bbb{F}_3$? Were you looking at cyclic ternary codes by any chance? Or possibly designsíng a system for football pools? – Jyrki Lahtonen Apr 30 '18 at 18:08
  • Unfortunately I'm not doing anything in application. I'm trying to learn some theory and attempting to get some workable examples that survey the more straightforward things that can happen to some extend. – probablystuck Apr 30 '18 at 22:02

1 Answers1

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Let $p\nmid n$. Then the cyclotomic polynomial $\Phi_n$ factors over $\Bbb F_p$ into $\phi(n)/r$ irreducible factors each of degree $r$, where $r$ is the multiplicative order of $p$ modulo $n$.

To see this, consider a primitive $n$-th root of unity $\zeta$ in an extension of $\Bbb F_p$. The number of conjugates of $\zeta$ over $\Bbb F_p$ is the least positive integer $r$ with $\zeta^{p^r}=\zeta$, that is $p^r\equiv1\pmod n$.

Angina Seng
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  • I don't quite follow your "to see this" - could you elaborate? – probablystuck Apr 30 '18 at 17:51
  • @probablystuck The zeros of $\Phi_n$ are the primitive $n$-th roots of unity, so to find the degree of the corresponding factor of $\Phi_n$ all you need to do is count the number of conjugates it has. – Angina Seng Apr 30 '18 at 17:54