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What are the (monic) divisors of the polynomial $x^p-1$ in the ring $(\mathbb{Z}/p^n\mathbb{Z})[x]$?

For $n = 1$, the ring $(\mathbb{Z}/p\mathbb{Z})[x]$ is a UFD, and we have $x^p - 1 = (x-1)^p$.

For $n > 1$, however, factorizations are no longer unique. Given any $\lambda \in \mathbb{Z}/p^n\mathbb{Z}$ such that $\lambda \equiv 1 \mod{p^{n-1}}$, we have $\lambda^p = 1$, so $x-\lambda$ divides $x^p - 1$ over $\mathbb{Z}/p^n\mathbb{Z}$. Writing $x^p - 1 = f(x)(x-\lambda)$, we find another factor $f(x)$, but $f(x)$ does not have any roots, and I suspect it does not have any monic factors at all. Aside from these, are there any other factors?

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I have figured out the answer: The only monic factors of $x^p-1$ over $\mathbb{Z}/p^n\mathbb{Z}$ are those given above (i.e. either linear or degree $p-1$).

Let $n > 1$, and suppose $x^p - 1 = f(x) g(x)$ in $(\mathbb{Z}/p^n\mathbb{Z})[x]$ with $f$ and $g$ monic. We must show that either $f$ or $g$ is linear, so suppose for the sake of contradiction that neither is. In that case, we can write $f(x) \equiv (x-1)^a \pmod p$ and $g(x) \equiv (x-1)^b \pmod p$ with $a, b > 1$, which means $f(1)$, $f'(1)$, $g(1)$, and $g'(1)$ are all divisible by $p$. This is a contradiction because $$p = \frac{d}{dx}(x^p-1)\big|_{x=1} = f(1)g'(1) + f'(1)g(1).$$

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  • $\begingroup$ This is a nice argument. If you can extend it and show that the factors of $x^{p^n}-1$ can only have the predictable degrees, do post it here also! $\endgroup$ – Jyrki Lahtonen May 11 at 5:36

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