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For p a prime and q a prime power (q not a power of p), I'm trying to find the polynomial divisors of $x^p -1$ in $Z_{q}[x]$. In particular, I'm hoping to present a general idea of the minimum degree of a monic divisor of $x^p -1$ in $Z_{q}[x]$.

I was hoping to approach this using cyclotomic polynomials of deg n, where n is ranges through the divisors of p, since $x^N -1$ is equal to the product of the cyclotomic polynomials with degrees ranging through the divisors of N.

However, I am unsure how working in $Z_{q}[x]$ actually comes into this. Is there a problem with the fact that I am attempting to factor $x^p -1$ outside of $Z_{p}[x]$? If so, are there any more effective ways to approach this?

user613048
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    What do you mean the least degree divisor ? do you mean $\min_{f\in \mathbb{Z}_q[x],f\mid x^p-1}{\text{Degree}(f)}$ ? If so, then the answer seems 1. – Zongxiang Yi Apr 07 '19 at 05:05
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    Least degree divisor is trivially $x-1$. What is the question you really wanted to ask? – Jyrki Lahtonen Apr 07 '19 at 05:06
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    If $q=\ell^n$, $\ell$ a prime, $\ell\neq p$, then we can find a full factorization of $x^p-1$ as follows. First factor $x^p-1$ in $\Bbb{Z}\ell[x]$. An old answer of mine gives the degree of the factors (with a bit more we can actually find them). Note that the assumption $\ell\neq p$ is crucial. And it also implies that the factors in $\Bbb{Z}{\ell}[x]$ are distinct. The latter is crucial when applying Hensel's lemma, "lifting" the factorization modulo $\ell$ recursively to factorizations modulo $\ell^k$ for all $k$. – Jyrki Lahtonen Apr 07 '19 at 05:14
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    Here I give an example of Hensel lifting a factorization of $x^7-1$ in $\Bbb{Z}_2[x]$ to a factorization in $\Bbb{Z}_4[x]$. The technique there is quite general. Here I use a trick specific to $\ell=2$ to do it faster (and finding a factorization modulo $8$). There's a fair amount of grunt work to do. – Jyrki Lahtonen Apr 07 '19 at 05:20
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    And here I work towards a criterion of checking which cyclotomic polynomials remain irreducible when reduced modulo a prime. You see, I have touched this general theme from many angles on our site. Anything I could post as an answer would be a reproduction of old stuff, so I'm very reluctant to do that. Hopefully the links help. – Jyrki Lahtonen Apr 07 '19 at 05:23
  • yes that helps quite a lot, I think I am to assume q is not a power of p. – user613048 Apr 07 '19 at 05:23
  • Factorization modulo a prime power is a bit tricky in the sense that it is not unique in general: $x^2=(x-2)(x+2)$ modulo $4$. In the case where we insist that the zeros of the factors should be roots of unity (of order coprime to $q$), Hensel lifting gives a unique outcome, but that result is a bit kludgy to describe. – Jyrki Lahtonen Apr 07 '19 at 05:30

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