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?

-

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 $p(x)$.

-
Thanks. Is there a way to close down this post? I feel silly. –  Rankeya Oct 6 '11 at 5:58
@Rankeya: This is not a silly question, no need to feel bad! And we should leave this up in case anyone else comes along with the same question. –  Zev Chonoles Oct 6 '11 at 6:00
I was actually trying to show that if $R$ is an integrally closed domain, then so is $R[x]$, and at some point I came across a statement of this nature. Oh well. –  Rankeya Oct 6 '11 at 6:01
More substantial editing will be needed. Let $q(x)=2(x^2+1)$ and $p(x)=x(x-1)(x^2+1)$. –  André Nicolas Oct 6 '11 at 6:26
If $R$ is a field, then $q(s)|p(s)$ for all $s$ such that $q(s) \ne 0$, no matter what $p$ and $q$ are. –  Ted Oct 6 '11 at 6:52

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.

-
I think this is quite a satisfactory answer, considering my question stemmed from a problem I was stuck on in Algebraic Number Theory. It would, however, be an interesting question, if my domain $R$ did not have a natural order. –  Rankeya Oct 6 '11 at 17:56
It's also true for the ring of integers of any number field, since any counterexample would yield a counterexample in $\mathbb{Z}$ by taking the product of all conjugates of $p$ and and all conjugates of $q$. In fact, so far I can't think of any counterexamples other than fields; if $\deg r < \deg q$ and $q(s)|r(s)$ for most $s$, somehow it looks like the ring would have to have "a lot" of units so it's pretty close to being a field. –  Ted Oct 7 '11 at 4:12