If $\gcd(f(x), g(x))\ne1$, then $F[x]/(fg)$ is not isomorphic to $F[x]/(f)\times F[x]/(g)$ So I thought up this question as an extension to the corresponding one for $Z_m$, $Z_n$, $(m, n)\ne1$.
The problems is I am unable to prove it, or disprove it. I try, but I keep getting tripped up since an isomorphism doesn't necessarily have to be the identity when restricted to $F$. I managed to prove it for all finite fields and fields generated by their identity (my 'proof' is far to long to put here though). Can anyone shine some light on the general case? I would really appreciate it.
The converse is clearly true, but the problem is that there is no clear way to extend the proof since 1 doesn't generate the arbitrary field.
 A: The ideals of $F[x]/(fg)$ are going to correspond to divisors of $fg$, and the ideals of the product ring are going to correspond to pairs of divisors, one divisor of $f$ with one divisor of $g$. When $f$ and $g$ are not coprime, there are going to be more pairs than there will be divisors of $fg$. (This is my intuition, anyway.) 
A: Even though one might genuinely be interested in the "absolute" question posed, there are arguments in favor of asking the "natural" version of the question, namely, whether there would be a "natural" isomorphism except in the case that $f,g$ are coprime. At the very least, answering the "natural" question first constrains the "absolute" version in various useful (and philosophical) ways, and gives a hint about the "absolute" version.
So, the "natural" question, is whether or not for a PID $R$ the natural homomorphism $R/ab \to R/a\oplus R/b$ by $r\mod ab \to (r\mod a)\oplus (r \mod b)$ is an isomorphism. (When $a,b$ are relatively prime, the Sun-Ze isomorphism (pathetically-often known as the "Chinese Remainder Theorem") gives the inverse. For $d=\gcd(a,b)>1$, there is an element in $R/ab$ annihilated by $d^2$ but not by $d$, while in $R/a\oplus R/b$ there is NO such element.
(One could argue that thinking about the "natural" map was a "red herring", methodologically, since it played no formal role at all, but, I argue, the context and sense of it may be helpful in delimiting the issues... and when it turns out not to be critically relevant... well, ... great!)
Thus, perhaps as though by accident, we discover that a device to prove that the natural (and arguably an isomorphism if destiny says there should be...) map cannot be an isomorphism proves more, namely, that no map can be an isomorphism. A methodology point as well as mathematical-fact-ual.
[Of course, other peoples' intuitions prefer other methodological advice...]
