Let $p$ be a prime, $k$ a positive integer. Let $f,h \in \mathbb{Z}[x]$ be polynomials such that
- $h | f \mod p^k$ in $ (\mathbb{Z}/p^k\mathbb{Z})[x]$
- $h \mod p$ is irreducible in $\mathbb{F}_p$
Then $f$ has an irreducible factor $h_0$ in $\mathbb{Z}[x]$ such that $h | h_0 \mod p$. Here, $\sum_i a_ix^i (\mod p)$ means $\sum_i (a_i \mod p)x^i$.
Why is this true? Are there some general results connecting divisibility in $\mathbb{Z}[x], \mathbb{Z}/p^k\mathbb{Z}[x]$ and $\mathbb{F}_p[x]$?
This statement is part of the proposition 2.5 in this article.
Here is a clean way to describe the situation: factorizations refine in residues.
If $a\in A$ and $a=a_1\cdots a_n$ and $\pi:A\to B$ and $B$ is a UFD and $\pi(a)=b_1\cdots b_m$ is its factorization into irreducibles (repetitions allowed) then we can partition the factors $b_i$ in such a way that each multiset of factors is precisely the multiset of irreducible factors of one of the $\pi(a_i)$s up to units.
The proof of this is quite straightforward: use the fact that $\pi(a)=\pi(a_1)\cdots\pi(a_n)$, that each of these $\pi(a_i)$s factors into irreducibles, and that cumulative total of these must be $\{b_i\}$ since $B$ is a unique factorization domain.
In particular, this applies with $A=\Bbb Z[x]$ and $B=\Bbb F_p[x]$. Every irreducible factor of $\overline{f}(x)$, i.e. the mod $p$ residue of some $f(x)\in\Bbb Z[x]$, "comes from" an irreducible factor of the original $f(x)$, i.e. divides the mod $p$ residue of some irreducible factor in $\Bbb Z[x]$ of $f(x)$.
For this conclusion to be drawn from our hypotheses, it is enough that we work mod $p$; working modulo higher powers of $p$ is superfluous.
Notice that when going from $\Bbb Z$ to $\Bbb F_p$, factorizations are allowed to "break down" further. So given a residual factorization, in order to lift it back to $\Bbb Z$ we would have to "clump" atoms together. If we extend our scalars to the $p$-adic integers $\Bbb Z_p$, we don't necessarily have to do any clumping, and can keep the same exact shape of factorization. Given $a(x)=a_1(x)a_2(x)\cdots a_n(x)$ for polynomials in $\Bbb F_p[x]$, under some mild conditions this lifts up to $A(x)=A_1(x)A_2(x)\cdots A_n(x)$ for polynomials in $\Bbb Z_p[x]$, where $A(x)\equiv a(x)$ and $A_i(x)\equiv a_i(x)$ mod $p$ are lifts; if the original factors were in fact irreducible, then we can guarantee the lifts are too. A special case of this is in lifting linear factors, which amounts to lifting roots mod $p$, and this is done with Hensel's lemma (a non-archimedean form of Newton's method for root approximation).