I'm considering a polynomial $f(x)\in\mathbb{Z}[x]$ and looking at it both in $\mathbb{Z}_n[x]$ ($n$ composite) and in $\mathbb{F}_p[x]$, to gather information about possible roots and irreducibility in $\mathbb{Z}[x]$.
I thought I understood the situation well but it is clear I do not. At first I thought that considering roots/irreducibility in $\mathbb{Z}_n[x]$ gave you nothing. But if there's a homomorphism $\phi:\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$, then if $\alpha$ is a root of $f(x)\in\mathbb{Z}[x]$, then we have for $f(x) = f_0 + f_1x + ... + f_tx^t$:
$0=\phi(0)=\phi(f_0 + f_1\alpha + ... + f_t\alpha^t)=[f_0]_n+[f_1]_n[\alpha]_n + ...+[f_t]_n[\alpha]_n^t$
And thus by the contrapositive, if $f(x)$ has no root in $\mathbb{Z}_n[x]$ then it has no root in $\mathbb{Z}[x]$. Or does the fact that $\mathbb{Z}_n[x]$ is not an integral domain somehow make this not follow? I suspect it does, but I'm having trouble seeing exactly how and why.
As a somewhat related question. How does the homomorphism $\phi:\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ differ from the homomorphism $\psi:\mathbb{Z}[x]\rightarrow\mathbb{F}_p[x]$? Since these are both surjective homs I thought it would be in the kernel, but the kernels don't seem to be qualitatively different. The theorem in my professor's notes says that irreducibility in $\mathbb{F}_p[x]$ implies irreducibility in $\mathbb{Z}[x]$, so there must be something special about $p$ prime.
Also what about if you have say a 4th degree polynomial which factors into irreducible quadratics in $\mathbb{F}_p[x]$, then it's not irreducible in $\mathbb{F}_p[x]$ but it doesn't have any roots in there, can we say it then has no roots in $\mathbb{Z}[x]$?
As you can see I'm harboring a number of related confusions about what's going on here, if anyone could help enlighten me it would be much appreciated, thanks.