Polynomial analogue of "prime independence"

Question

In number theory a well-known fact is that congruence modulo distinct primes are 'independent'. That is, to know that $n \equiv a \pmod{p}$ does not change the probability as to what $n \equiv x \pmod{q}$ will be, if $p,q$ are distinct primes. Is there an analogue of this result for polynomials? That is, if we are given co-prime polynomials $p(x), q(x) \in \mathbb{Z}[x]$ say, does it follow that $f(x) \pmod{p(x)}$ and $f(x) \pmod{q(x)}$ are independent?

Answer 1

The closest analogue is to consider polynomials over finite fields $\mathbb{F}_q[x]$. Here the counting is straightforward: there are $q^n$ monic polynomials of degree $n$ over $\mathbb{F}_q$. If $p$ is a monic polynomial then the probability that it divides a monic polynomial of degree $n \ge d$ is exactly $\frac{1}{q^{\deg p}}$ for fixed $n$, and this implies the desired independence result on the nose for sufficiently large $n$, not just asymptotically.

In fact one can say much more than this: as $q, n \to \infty$ the numbers $X_r$ of irreducible factors of degree $r$ approach jointly independent Poisson random variables with mean $\frac{1}{r}$. See, for example, this blog post.

Over $\mathbb{Z}[x]$ things are less independent. For example, $f(x) \bmod (x - p)$ and $f(x) \bmod (x - q)$ are just $f(p)$ and $f(q)$ respectively, and these are related since $f(p) \equiv f(q) \bmod (p-q)$.