Let A be a subring of B, and C the integral closure of A in B. If f, g are monic polynomials in B[x] such that fg is in C[x], then f, g are in C[x].
The first part of the problem allowed the assumption that B is an integral domain, which gave rise to a proof using a field containing B in which f and g both split into linear factors. For the second part, the existence of zero divisors is a possibility, so no such field is guaranteed to exist.
I'm not clear on how to begin this proof. Is there some way to use a total quotient ring similar to using the field for integral domains, or is some completely different strategy needed? Is this a matter of extending the fact that this works for integral domains into non-integral-domains, or coming up with a completely different proof to accommodate the zero divisors?
Okay, it turns out this is indeed a case of reducing it to the integral domain case. It works like this: suppose $f(x) = \sum_{i=1}^n a_i x^i$, $g(x)=\sum_{j=1}^m b_j x^j$ and $f(x)g(x)=\sum_{k=1} c_k x^k$, and consider the ring $B' = \mathbb{Z}[a_i,b_j]$ where $a_i,b_j$ are treated as indeterminates. Then $B'$ is an integral domain, and since the result is already proved for an integral domain, we know that $f'(x), g'(x)$ in $B'[x]$ with those coefficients have the desired property with $\mathbb{Z}[c'_k]$ (where each $c'_k$ is built from the $a_i, b_j$ the same way the $c_k$ are) as $A'$ and $C'$ as the integral closure of $A'$ in $B'$. From there, we make a homomorphism from $B'$ to $B$ that maps each $a_i,b_j$ indeterminate in $B'$ into the corresponding element of $B$, and all it takes to get to the result is to work out the details of this idea.