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Let $\mathbb{F}$ be some finite field.

Let $Q(x, y)\in\mathbb{F}[X,Y]$ and $P(x)\in\mathbb{F}[X]$ such that $Q(x, P(x)) = 0$ for every $x\in\mathbb{F}$.

Prove that $Q(x, y) = (y - P(x))A(x, y)$ for some polynomial $A(x, y)$, i.e. $y-P|Q$.

I must say I am stuck and I don't know how to prove this fact.

Alejandro Nasif Salum
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Don Fanucci
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  • The same question is here. See the comments there. – Dietrich Burde Nov 28 '17 at 19:47
  • Hmm. Say the field has order $q = p^e$. Let $Q(x,y) = y$ and $P(x) = x^q-x$. Then $Q(x,P(x)) = P(x) = x^q-x$, which is equal to $0$ for every $x \in \mathbb{F}$. And yet $y$ is not divisible by $y-(x^q-x)$. Can I ask how sure are you that this statement is supposed to be true? – Zach Teitler Nov 29 '17 at 03:43
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    Is it possible that the statement is supposed to be the following?: If $Q(x,P(x))$ is identically equal to $0$ (i.e., the $0$ polynomial; all coefficients in $Q(x,P(x))$ are equal to $0$) then $(y-P) \mid Q$. ..? This is easy to prove by the division algorithm: $Q = (y-P)A+r$, where $r=r(x)$ has no $y$'s (can you see why?). Then substituting $P(x)$ for $y$ gives $r(x) = Q(x,P(x))-(P-P)A(x,P(x)) = 0-0=0$ (identically!). – Zach Teitler Nov 29 '17 at 03:48
  • I was thinking that when I read your previous comment: maybe $Q(X,P(X))\equiv 0$ was thought to be equivalent to $Q(x,P(x))=0 \forall x\in \mathbb{F}$ (and is an easy mistake to make.) – Alejandro Nasif Salum Nov 29 '17 at 03:52
  • Now, what I wonder is whether there is such a division algorithm in $\mathbb{F}[X,Y]$ or in $\mathbb{F}[X][Y]$. – Alejandro Nasif Salum Nov 29 '17 at 03:53
  • @AlejandroNasifSalum (1) Yes, certainly I had a time in the past when I thought that $Q=0$ was the same as $Q(x)=0$ for all $x$; I will not confess how recently this time was :-) (2) Note that $y-P$ has leading coefficient equal to $1$, so that it can certainly divide into any terms that have $y$. More generally there is a Division Algorithm for multivariate polynomials in terms of a monomial ordering; see for example the book Ideals, Varieties, and Algorithms by Cox, Little, O'Shea. But really this special case with $1y$ is much simpler. – Zach Teitler Nov 29 '17 at 04:03
  • You're right! The euclidean function depends on the monomial ordering. And no, I won't ask you how recent you found out the distinction if you promise not to ask me either. ;) – Alejandro Nasif Salum Nov 29 '17 at 04:08

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(This was too long to post as a comment). The comments in the related link conclude that this is not true in general, but the counterexample is over $\mathbb{C}$, not over a finite field. I don't actually know if it's true, but if you consider elements of $\mathbb{F}[X,Y]$ as elements of $\mathbb{F}[X][Y]$ (that is, as polynomials in the indeterminate $Y$ with coefficients in $\mathbb{F}[X]$ then the statement is an immediate consequence of the remainder theorem, which can be proven if the set of polynomials is an euclidean domain. Since $\mathbb{F}[X]$ is not a field, it is not necessarily true that $\mathbb{F}[X][Y]$ is euclidean, but maybe the fact that $\mathbb{F}$ is a finite field, gives you that property. Not sure, I will have to think more.

Alejandro Nasif Salum
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    $y-x^2$ is irreducible over any field. – Zach Teitler Nov 29 '17 at 03:01
  • Indeed, but that is not a counterexample since $y-x^2|y-x^2$. – Alejandro Nasif Salum Nov 29 '17 at 03:29
  • Ugh, the linked question has a typo and the "counterexample" given there is not a counterexample. Okay... let me think about this... – Zach Teitler Nov 29 '17 at 03:35
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    $k[X,Y] \cong k[X][Y]$ is definitely not Euclidean (it is not a PID, e.g., the ideal $(X,Y)$ is not principal). There is definitely not a euclidean function, or a division algorithm in the sense of Euclidean domains. However it is still true that for any $Q(x,y)$ and $P(x)$, we can write $Q(x,y) = (y-P(x))A(x,y)+r(x)$. Simply use "long division" as if it were a Euclidean domain. – Zach Teitler Nov 29 '17 at 04:23
  • That's true. Actually I had in mind the fact that it cannot be a PID but was thinking the inclusion $ED\subset PID$ the other way round (my bad!). Anyway, I guess that all that's said and your last remark close the case. – Alejandro Nasif Salum Nov 29 '17 at 04:59