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?
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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 $p(x)$.
André Nicolas
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Thanks. Is there a way to close down this post? I feel silly. – Rankeya Oct 06 '11 at 05:58
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1@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 06 '11 at 06:00
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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 06 '11 at 06:01
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Don't worry about it, everyone slips up sometimes, I more than most. There is a way of flagging a moderator, and moderators are all-powerful. – André Nicolas Oct 06 '11 at 06:03
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André is correct, if you really want to delete this question, a moderator (such as myself) will be able to do it. But I do encourage you to leave it up, there is nothing wrong with seeking to improve your understanding. – Zev Chonoles Oct 06 '11 at 06:15
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The answer leaves the question open for non-constant polynomials $q$. – Gerry Myerson Oct 06 '11 at 06:21
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Okay. Can I edit this question to one where $q(x)$ is non-constant? Perhaps even that has a straightforward counter-example. – Rankeya Oct 06 '11 at 06:22
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Ahh, Gerry and I had the same thought at almost the same time. – Rankeya Oct 06 '11 at 06:22
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3More substantial editing will be needed. Let $q(x)=2(x^2+1)$ and $p(x)=x(x-1)(x^2+1)$. – André Nicolas Oct 06 '11 at 06:26
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This is nice. Thanks. – Rankeya Oct 06 '11 at 06:27
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Andre one more small edit. Sorry! – Rankeya Oct 06 '11 at 06:33
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2If $R$ 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 06 '11 at 06:52
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@Rankeya: For $\mathbb{Z}$, and close relatives, the positive assertion would be that $q(x)$ divides $cp(x)$ for some non-zero $c$. – André Nicolas Oct 06 '11 at 14:50
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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.
Ted
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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 06 '11 at 17:56
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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 07 '11 at 04:12