Ruling out polynomial bijections over the rationals via Bombieri-Lang?

What's new 2019-08-20

I recently came across this question on MathOverflow asking if there are any polynomials ${P}$ of two variables with rational coefficients, such that the map ${P: {\bf Q} \times {\bf Q} \rightarrow {\bf Q}}$ is a bijection. The answer to this question is almost surely “no”, but it is remarkable how hard this problem resists any attempt at rigorous proof. (MathOverflow users with enough privileges to see deleted answers will find that there are no fewer than seventeen deleted attempts at a proof in response to this question!)

On the other hand, the one surviving response to the question does point out this paper of Poonen which shows that assuming a powerful conjecture in Diophantine geometry known as the Bombieri-Lang conjecture (discussed in this previous post), it is at least possible to exhibit polynomials ${P: {\bf Q} \times {\bf Q} \rightarrow {\bf Q}}$ which are injective.

I believe that it should be possible to also rule out the existence of bijective polynomials ${P: {\bf Q} \times {\bf Q} \rightarrow {\bf Q}}$ if one assumes the Bombieri-Lang conjecture, and have sketched out a strategy to do so, but filling in the gaps requires a fair bit more algebraic geometry than I am capable of. So as a sort of experiment, I would like to see if a rigorous implication of this form (similarly to the rigorous implication of the Erdos-Ulam conjecture from the Bombieri-Lang conjecture in my previous post) can be crowdsourced, in the spirit of the polymath projects (though I feel that this particular problem should be significantly quicker to resolve than a typical such project).

Here is how I imagine a Bombieri-Lang-powered resolution of this question should proceed (modulo a large number of unjustified and somewhat vague steps that I believe to be true but have not established rigorously). Suppose for contradiction that we have a bijective polynomial ${P: {\bf Q} \times {\bf Q} \rightarrow {\bf Q}}$ . Then for any polynomial ${Q: {\bf Q} \rightarrow {\bf Q}}$ of one variable, the surface

$\displaystyle S_Q := \{ (x,y,z) \in \mathbb{A}^3: P(x,y) = Q(z) \}$

has infinitely many rational points; indeed, every rational ${z \in {\bf Q}}$ lifts to exactly one rational point in ${S_Q}$ . I believe that for “typical” ${Q}$ this surface ${S_Q}$ should be irreducible. One can now split into two cases:

(a) The rational points in ${S_Q}$ are Zariski dense in ${S_Q}$ .
(b) The rational points in ${S_Q}$ are not Zariski dense in ${S_Q}$ .

Consider case (b) first. By definition, this case asserts that the rational points in ${S_Q}$ are contained in a finite number of algebraic curves. By Faltings’ theorem (a special case of the Bombieri-Lang conjecture), any curve of genus two or higher only contains a finite number of rational points. So all but finitely many of the rational points in ${S_Q}$ are contained in a finite union of genus zero and genus one curves. I think all genus zero curves are birational to a line, and all the genus one curves are birational to an elliptic curve (though I don’t have an immediate reference for this). These curves ${C}$ all can have an infinity of rational points, but very few of them should have “enough” rational points ${C \cap {\bf Q}^3}$ that their projection ${\pi(C \cap {\bf Q}^3) := \{ z \in {\bf Q} : (x,y,z) \in C \hbox{ for some } x,y \in {\bf Q} \}}$ to the third coordinate is “large”. In particular, I believe

(i) If ${C \subset {\mathbb A}^3}$ is birational to an elliptic curve, then the number of elements of ${\pi(C \cap {\bf Q}^3)}$ of height at most ${H}$ should grow at most polylogarithmically in ${H}$ (i.e., be of order ${O( \log^{O(1)} H )}$ .
(ii) If ${C \subset {\mathbb A}^3}$ is birational to a line but not of the form ${\{ (f(z), g(z), z) \}}$ for some rational ${f,g}$ , then then the number of elements of ${\pi(C \cap {\bf Q}^3)}$ of height at most ${H}$ should grow slower than ${H^2}$ (in fact I think it can only grow like ${O(H)}$ ).

I do not have proofs of these results (though I think something similar to (i) can be found in Knapp’s book, and (ii) should basically follow by using a rational parameterisation ${\{(f(t),g(t),h(t))\}}$ of ${C}$ with ${h}$ nonlinear). Assuming these assertions, this would mean that there is a curve of the form ${\{ (f(z),g(z),z)\}}$ that captures a “positive fraction” of the rational points of ${S_Q}$ , as measured by restricting the height of the third coordinate ${z}$ to lie below a large threshold ${H}$ , computing density, and sending ${H}$ to infinity (taking a limit superior). I believe this forces an identity of the form

$\displaystyle P(f(z), g(z)) = Q(z) \ \ \ \ \ (1)$

for all ${z}$ . Such identities are certainly possible for some choices of ${Q}$ (e.g. ${Q(z) = P(F(z), G(z))}$ for arbitrary polynomials ${F,G}$ of one variable) but I believe that the only way that such identities hold for a “positive fraction” of ${Q}$ (as measured using height as before) is if there is in fact a rational identity of the form

$\displaystyle P( f_0(z), g_0(z) ) = z$

for some rational functions ${f_0,g_0}$ with rational coefficients (in which case we would have ${f = f_0 \circ Q}$ and ${g = g_0 \circ Q}$ ). But such an identity would contradict the hypothesis that ${P}$ is bijective, since one can take a rational point ${(x,y)}$ outside of the curve ${\{ (f_0(z), g_0(z)): z \in {\bf Q} \}}$ , and set ${z := P(x,y)}$ , in which case we have ${P(x,y) = P(f_0(z), g_0(z) )}$ violating the injective nature of ${P}$ . Thus, modulo a lot of steps that have not been fully justified, we have ruled out the scenario in which case (b) holds for a “positive fraction” of ${Q}$ .

This leaves the scenario in which case (a) holds for a “positive fraction” of ${Q}$ . Assuming the Bombieri-Lang conjecture, this implies that for such ${Q}$ , any resolution of singularities of ${S_Q}$ fails to be of general type. I would imagine that this places some very strong constraints on ${P,Q}$ , since I would expect the equation ${P(x,y) = Q(z)}$ to describe a surface of general type for “generic” choices of ${P,Q}$ (after resolving singularities). However, I do not have a good set of techniques for detecting whether a given surface is of general type or not. Presumably one should proceed by viewing the surface ${\{ (x,y,z): P(x,y) = Q(z) \}}$ as a fibre product of the simpler surface ${\{ (x,y,w): P(x,y) = w \}}$ and the curve ${\{ (z,w): Q(z) = w \}}$ over the line ${\{w \}}$ . In any event, I believe the way to handle (a) is to show that the failure of general type of ${S_Q}$ implies some strong algebraic constraint between ${P}$ and ${Q}$ (something in the spirit of (1), perhaps), and then use this constraint to rule out the bijectivity of ${P}$ by some further ad hoc method.