Let $f \in \mathbb{Z}[x]$ be a non-constant polynomial where $p$ is a prime that is not a divisor of the leading coefficient of $f$.
Assume that $f = g_{1}g_{2},\ldots, g_{k}$ is a factorization of $f$ into a product of irreducible polynomials over $\mathbb{Z}_p$ (i.e., all $g_i \in Z_p[x]$ and $f$ is viewed as a polynomial over $\mathbb{Z}_p$) and let $h$ be any divisor of $f$ over $\mathbb{Z}$.
I need to prove that some $g_i$ is a divisor of $h$ over $\mathbb{Z}_p$.
This is my attempt so far:
If $h$ is a divisor of $f$ over $\mathbb{Z}$, then $\exists k \in \mathbb{Z}[x]$ such that $f = h \cdot k$.
Now, correct me if I'm wrong, but I'm assuming that $f$ can be considered the same polynomial over both $\mathbb{Z}$ and $\mathbb{Z}_{p}$ here, unless when we say that "$h$ is a divisor of $f$ over $\mathbb{Z}$", we mean that it is the divisor of the image of $f$ (call it $\tilde{f}$) under a map that takes $f$ from $\mathbb{Z}_p \to \mathbb{Z}$, which we know is not a function. In this case, then we have that $\exists k \in \mathbb{Z}[x]$ such that $\tilde{f}= h \cdot k$.
Then, reducing $\tilde{f} \mod p$ we obtain that $f = \overline{h} \cdot \overline{k}$, where $\overline{h}$ is $\,h\mod p\,$ and $\overline{k}$ is $k$ reduced modulo $p$.
Now, we know that $f = \overline{h} \cdot \overline{k} = g_1 g_2 \cdots g_k$. But, all this tells me is that some $g_i$ is a divisor of the product of $h$ over $\mathbb{Z}_p$ and $k$ over $\mathbb{Z}_p$.
At this point, I get completely stuck.
It's entirely possible that this is not the correct way to approach this problem. If someone could help me figure out the right way to do this, I would be most appreciative!
Thank you for your time and petience!
Also, could be related to this question.
I'm not sure if you are talking about cyclotomic fields. Those are obtained by joining a primitive $n$th root of unity to $Q$ the field of rational numbers. The $n$th cyclotomic field is referred to as $Q_n$. It is referred to $Z_n$ in your question. The answer is simple, all polynomials $P(x)$ that define the same field as $Q_x$, prime (powers) $p^k$ dividing $P(x)$ must have the form $1$ $\pmod n$. I don't know if this was what you were looking for.