Is it possible to factor a polynomial over a ring which is not an integral domain? I assumed not, but I'm wondering about the following scenario.
Let $f \in \mathbb{Z}[x]$ be monic and irreducible but factor completely over primes $p$ and $q$. Given the roots of $f$ mod $p$ and $q$ we can find roots in $\mathbb{Z}/pq\mathbb{Z}$ by CRT. (It may not be apparent which roots to combine via CRT to get a meaningful result, but it seems possible nonetheless).
Can we find these roots by factoring $f$ over $\mathbb{Z}/pq\mathbb{Z}$ directly? If so, is there an algorithm or reference?
Example:
Take $p = 2$, $q = 3$ and $f = x^2 - x \in \Bbb Z / 6\Bbb Z[x]$.
We then have $f = x(x - 1) = (x - 3)(x - 4)$. Which one should be "the" factorization of $f$?
As you can see, the problem is that $f$ is a polynomial of degree $2$, yet has $4$ different roots in $\Bbb Z / 6 \Bbb Z$. Thus elements of the ring $\Bbb Z/6\Bbb Z[x]$ cannot be factored uniquely.
This also means that you cannot find all the roots by "factoring" the polynomial over $\Bbb Z/6\Bbb Z$.