Question about polynomials Let $R$ be an infinite domain and $p(x), q(x) \in R[x]$ such that $q(x) \neq 0$. If for all but a finitely many $s \in R, q(s)|p(s)$, then is is true that $q(x)|p(x)$ in $R[x]$? This seems a little silly, but I cannot figure out why it should be true or false. What about the case where $q(x)$ is not a constant and $q(x)$ is monic?
 A: Let $R=\mathbb{Z}$, let $q(x)=2$ (identically), and let $p(x)=x(x-1)$. Then $q(a)$
divides $p(a)$ for every integer $a$, but $q(x)$ does not divide $p(x)$.
A: Since $q(x)$ is monic, we can divide $p(x)$ by $q(x)$ in $R[x]$: $$p(x) = q(x) f(x) + r(x)$$ where $f(x), r(x) \in R[x]$ and either $r(x)=0$ or $\deg r < \deg q$.  If $r(x)=0$, then $q(x)|p(x)$.  Otherwise, since $q(s)|p(s)$ for all but finitely many $s$, then $q(s)|r(s)$ for all but finitely many $s$.  In other words, $(q,r)$ satisfy the same hypotheses as the initial $(q,p)$, except now we have the extra condition that $\deg r < \deg q$. So to prove the original statement, we must show that $\deg r < \deg q$ cannot happen if $q(s)|r(s)$ for all but finitely many $s$.
In the case $R=\mathbb{Z}$, this is easy: If $\deg r < \deg q$ then $|r(s)|<|q(s)|$ for sufficiently large $|s|$, so that we cannot have $q(s)|r(s)$ for such $s$.  
More generally, if $R$ is a subring of the ring of integers of $\mathbb{Q}(\sqrt{d})$ where $d<0$, then $R$ is an unbounded, discrete subset of $\mathbb{C}$, so a similar argument works. 
On the other hand, as in my comment, if $R$ is a field, then the original statement is false because any polynomials $p$ and $q$ satisfy the hypotheses.
