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Is there a description of how $x^{p^n}-1$ factors in $\mathbb{Z}/p^n$?

I'm aware of the situation for $n = 1$.

EDIT: I really mean $\mathbb{Z}/p^n$, not $\mathbb{F}_{p^n}$.

user26857
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    In $\mathbb Z[x]$, the prime factorization of $x^{p^n}-1$ is the inductive one: $$(x-1)\prod_{k=0}^{n-1}\frac{x^{p^{k+1}}-1}{x^{p^k}-1}$$ – Thomas Andrews Aug 28 '17 at 20:51
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    Note that factorization is not necessarily unique over a non-field. – Thomas Andrews Aug 28 '17 at 20:58
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    I'm not sure if and how it shows in this question, but do observe Thomas Andrews' point. For example modulo $2^2=4$ we have $(x+2)(x-2)=x^2$ and $(x+1)^2=(x-1)^2$. – Jyrki Lahtonen Aug 28 '17 at 21:34
  • @ThomasAndrews That's a good point. I suppose my question isn't well defined. I'll delete it in a bit. –  Aug 29 '17 at 04:18

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A report of my findings. Anything but an answer.

The difficulties arising from the possibility of non-unique factorization do make an appearance here. Somewhat against my expectations the smallest case $p=n=2$ behaves nicely. At least I couldn't find factorizations other than the trivial. $$x^4-1=(x-1)(x+1)(x^2+1)\pmod4.$$

With $p=3,n=2, f(x)=x^9-1$ we do get complications. We have the usual $$ f(x)=(x-1)(x^2+x+1)(x^6+x^3+1),\tag{1} $$ but given that $f(x+3)\equiv f(x)\pmod 9$, we also have $$ f(x)=(x+2)(x^2+7x+4)(x^6+x^3+1).\tag{2} $$ Observe that $x^6+x^3+1$ is invariant under the substitution $x\mapsto x\pm3$, but the lower degree characteristic zero factors are not. It is easy to see that $(1)$ and $(2)$ are essentially different factorizations. For example $x-1$ is not a factor of any of the factors in $(2)$.

I think that more can be said though. May be a $p$-adic tool can shed more light to this question?

Jyrki Lahtonen
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  • Wish I knew, @reuns. I just wanted to get the ball rolling. It may still turn out that this is a dull problem, but at the moment I don't think so. – Jyrki Lahtonen Aug 30 '17 at 19:42
  • All the factorizations are of the same factorization type ? If so your group of elements such that $x^{p^n}-1 = (x+a)^{p^n}-1 \in \mathbb{Z}/{p^n} \mathbb{Z}[x]$ might tell the answer – reuns Aug 30 '17 at 19:43
  • IIRC the cyclotomic factor $\sum_{k=0}^{p-1}x^{k p^{n-1}}=(x^{p^n}-1)/(x^{p^{n-1}}-1)$ remains irreducible over the $p$-adics. That probably shows somehow. – Jyrki Lahtonen Aug 30 '17 at 19:45
  • If $f \in \mathbb{F}_p[x]$ then $f = 0$ iff $\forall a \in \overline{\mathbb{F}_p}, f(a) = 0$. The equivalent for $f \in \mathbb{Z}/p^n\mathbb{Z}[x]$ would be a set of (commuative) matrix algebras $\mathbb{Z}/p^n\mathbb{Z}[M_i]$. – reuns Aug 30 '17 at 19:54
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    Thanks for these examples! They definitely indicate that factorization is probably not a good concept to think about in non-domains. My original goal was to understand the annihilator of $x-1$ modulo $1+x+x^2+\cdots+x^{p^n-1}$, and for that it is indeed much easier to look at things over $\mathbb{Z}_p$, and then note that $x-1$ is a uniformizer there, so its annihilator in $\mathbb{Z}_p/p^n[x]/(1+x+\cdots+x^{p^n-1})$ is just a power of itself, equal to $n$ times the ramification index minus 1. –  Aug 31 '17 at 18:24
  • Thanks for the extras @rtz. I was wondering about the motivation :-) Still, I wouldn't mind learning more about this. May be some other time :-) – Jyrki Lahtonen Aug 31 '17 at 18:57