The problem is from this paper (click for pdf) by Mark van Hoeij.
Let $f \in \mathbb{Z}[X]$ be monic and squarefree.
Let $B$ be a Landau-Mignotte bound for $f$, i.e. for any rational factor $\phi$ of $f$, $||\phi||_\infty < B$, and let $a$ be an integer such that $p^a > 2 B$.
Furthermore, let $f = \prod_{i\in I} f_i$ with $f_i \in \mathbb{Z}_p[X]$ monic and irreducible for all $i$, where $\mathbb{Z}_p[X]$ denotes the $p$-adic integers.
Let $C^a(g)$ denote the polynomial obtained from $g$ by, for each coefficient, picking the representative in $\mathbb{Z}_p/(p^a\mathbb{Z}_p) \cong \mathbb{Z}/(p^a\mathbb{Z})$ with the smallest (Euclidean) absolute value - i.e. the representative in the interval $\left(-p^a/2,\,p^a/2\right]$.
Then, if $S\subset I$ and $g = \prod_{i\in S}f_i$, the following conditions are equivalent:
1) $g \in \mathbb{Z}[X]$
2) $g = C^a(g)$
3) $C^a(g)$ divides $f$ in $\mathbb{Z}[X]$.
How do I show either of the implications
3) $\implies$ 1)
or
3) $\implies$ 2) ?
The other implications I already have, but I need one of the above to complete the missing link.
I tried writing $g = C^a(g) + p^a w$ for some $p$-adic polynomial $w$ (which due to all factors being monic has degree $< \deg(C^a(g)$, writing $g$ like that is based on series representations of $p$-adic integers) and then writing, for some $b\in \mathbb{Z}[X]$ and a $p$-adic polynomial $q$, $$ b C^a(g) = f = g q = q C^a(g) + p^a q w $$
My aim was to show $w=0$.
Some approaches where based on the fact that $C^a(g)$ is monic, so one can try to view both sides of the equation as the result of polynomial long division, but I'm not sure this is of help, because for that, the degree of $qw$ has to be strictly smaller than that of $C^a(g)$...
Writing out all polynomials involved with $p$-adic number series representations also didn't get me very far.
Maybe this is simpler than I thought? What am I missing?
For $u\in \Bbb{Z}_p[X]$ monic you take a polynomial $C_a(u)\in\Bbb{Z}[X]$ with the smallest coefficients such that $C_a(u)\equiv u \bmod p^a$.
Your bound $A_f$ is such that $u | f$ in $\Bbb{Z}[X]$ implies $u=C_a(u)$ for $a\ge A_f$
You factor $f$ in $f=\prod_i f_i = gh\in \Bbb{Z}_p[X]$
Assume $a\ge A_f$ is large enough so that the $f_i$ are all distinct $\bmod p^a$
Start from $g\in \Bbb{Z}[X]$ iff $g=C_a(g)$ iff $f=gh\in \Bbb{Z}[X]$
Then $C_a(g) | f\in \Bbb{Z}[X]$ implies $f=C_a(g)w\bmod p^a$ and this factorization lifts uniquely to $\Bbb{Z}_p$, thus $C_a(g)= g$ and you get your result.