Consider a monic polynomial $f\in\mathbb Z[X]$ (see Note). Assume that we can factor $f\bmod p=gh$ into two monic irreducible polynomials $g\neq h\in\mathbb F_p[X]$ of same degree (or more generally, into $n$ monic irreducible polynomials of same degree). This induces a ring isomorphism $$\frac{\mathbb F_p[X]}{(f\bmod p)}\cong\frac{\mathbb F_p[X]}{(g)}\times\frac{\mathbb F_p[X]}{(h)}.$$
Using Hensel lifting, we find two monic polynomials $\tilde g,\tilde h\in\mathbb Z_{p^r}[X]$ $\big($with $\mathbb Z_{p^r}:=\frac{\mathbb Z}{(p^r)}$ $\big)$ for which
$$\tilde g\bmod p=g,\quad \tilde h\bmod p=h,\quad f\bmod p^r=\tilde g\tilde h.$$
I wonder if this also induce a ring isomorphism $$\frac{\mathbb Z_{p^r}[X]}{(f\bmod p^r)}\cong \frac{\mathbb Z_{p^r}[X]}{(\tilde g)}\times \frac{\mathbb Z_{p^r}[X]}{(\tilde h)}.$$
My problem here is that $\mathbb Z_{p^r}[X]$ is not an integral domain, and it thus does not even make sense to ask whether or not we can make $\tilde g$ and $\tilde h$ irreducible. Otherwise, we could just proceed as usual: start with the map $\mathbb Z[X]\to\frac{\mathbb Z_{p^r}[X]}{(\tilde g)}\times \frac{\mathbb Z_{p^r}[X]}{(\tilde h)}$, notice that the kernel is precisely given by $(f\bmod p^r)$, then conclude.
The authors of this document talk of such an isomorphism in section 3.3, but I am not convinced whether this is actually viable.
Note: In the context that I am interested in, $f$ is equal to $\Phi_m$, the $m$th cyclotomic polynomial, where $m\geq 1$ is not a multiple of $p$.
Yes.
There are $u,v,w\in \Bbb{Z}[x]$ such that $gu+hv = 1+pw$ so
$(gu+hv)^{p^r} = 1 +p^r W$. This can be rewritten as $$gU+hV = 1+p^r W$$ and this is what you need to get an isomorphism $$\Bbb{Z}/(p^r)[x]/(gh) \to \Bbb{Z}/(p^r)[x]/(g)\times \Bbb{Z}/(p^r)[x]/(h)$$