# Factorization of $x^p-1$ modulo $p^n$

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?

## 1 Answer

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 $$p = \frac{d}{dx}(x^p-1)\big|_{x=1} = f(1)g'(1) + f'(1)g(1).$$

• 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! – Jyrki Lahtonen May 11 at 5:36