I recently came across this question on MathOverflow asking if there are any polynomials of two variables with rational coefficients, such that the map 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 which are injective.

I believe that it should be possible to also rule out the existence of bijective polynomials 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 . Then for any polynomial of one variable, the surface

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

- (a) The rational points in are Zariski dense in .
- (b) The rational points in are not Zariski dense in .

Consider case (b) first. By definition, this case asserts that the rational points in 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 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 all can have an infinity of rational points, but very few of them should have “enough” rational points that their projection to the third coordinate is “large”. In particular, I believe

- (i) If is birational to an elliptic curve, then the number of elements of of height at most should grow at most polylogarithmically in (i.e., be of order .
- (ii) If is birational to a line but not of the form for some rational , then then the number of elements of of height at most should grow slower than (in fact I think it can only grow like ).

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 of with nonlinear). Assuming these assertions, this would mean that there is a curve of the form that captures a “positive fraction” of the rational points of , as measured by restricting the height of the third coordinate to lie below a large threshold , computing density, and sending to infinity (taking a limit superior). I believe this forces an identity of the form

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

for some rational functions with rational coefficients (in which case we would have and ). But such an identity would contradict the hypothesis that is bijective, since one can take a rational point outside of the curve , and set , in which case we have violating the injective nature of . 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 .

This leaves the scenario in which case (a) holds for a “positive fraction” of . Assuming the Bombieri-Lang conjecture, this implies that for such , any resolution of singularities of fails to be of general type. I would imagine that this places some very strong constraints on , since I would expect the equation to describe a surface of general type for “generic” choices of (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 as a fibre product of the simpler surface and the curve over the line . In any event, I believe the way to handle (a) is to show that the failure of general type of implies some strong algebraic constraint between and (something in the spirit of (1), perhaps), and then use this constraint to rule out the bijectivity of by some further *ad hoc* method.

## 23 comments

Comments feed for this article

8 June, 2019 at 8:43 am

Kevin O'BryantI saw a nice talk by Hector Pasten on this problem in January in Chile. If I recall correctly, which is a dubious hypothesis, he had constructed bijections in topologically similar situations, which put some constraint on what kind of argument could work in the QxQ -> Q problem.

8 June, 2019 at 8:45 am

Kevin O'BryantInjection, not bijection. Found the abstract:

Bivariate polynomial injections of rational points

H´ector PASTEN (P. Universidad Cat´olica de Santiago) ´

Abstract I will prove that there is an affine curve C over Q with a dense set

of rational points, and a polynomial function f on C × C defined over Q with

the property that f induces an injective function C(Q)×C(Q) → Q on rational

points.

12 June, 2019 at 4:06 am

PastenIf you want to see the details: https://arxiv.org/abs/1811.02416

8 June, 2019 at 8:45 am

Will SawinAfter getting the finitely many curves, the next step is to apply Hilbert’s irreducibility theorem, which shows that for any curve with a map to (in this case the coordinate map), either the curve has a section (which would be your identity (1)) or else a density zero subset of points lie in its image.

For ruling out the case of a large system of polynomial identities (1), this seems very likely to be possible if the degrees of the polynomial identities can be bounded. Then constructibility arguments show that there must be a universal polynomial identity that works for a generic polynomial. Possibly the degree-bounding follows from arguments relating to function field versions of Mordell’s conjecture.

8 June, 2019 at 9:00 am

Will SawinFor getting the surface to be general type by base change, a necessary condition is that the curve for general have genus at least two – a surface with a rational fibration has Kodaira dimension at most , and a surface with an elliptic fibration has Kodaira dimension at most .

For genus zero fibrations there is a special argument – any genus zero curve with one rational point has infinitely many, so there cannot be a bijection. For genus one fibrations, I think there is a more complicated arguments. Any genus one fibration is a torsor for an elliptic fibration. Because every algebraic genus one fibration has a multisection, it must be a torsor for the -torsion of an elliptic curve. By Hilbert irreducibility, as an -torsor, it remains nontrivial for most fibers. But as all fibers have a rational point it must be nontrivial as en element of the Tate-Shafarevich group. But the kernel map from to the Tate-Shaferevich group is , so that must be nontrivial, so the fiber must have more than one point.

For genus two and up fibrations, I imagine it must always be possible to obtain general type by base change by polynomials, but I don’t know a reference. If a sufficiently strong result of this form could be proven, one could try to do it by base changing by a polynomial such that which fails to be injective infinitely often, and use that, plus the polynomial identity from earlier, to contradict the injectivity of the earlier map.

14 June, 2019 at 10:04 am

PastenMaybe the non-degeneracy condition on has to do with the homogeneous part of highest degree (to so say, the possible singularities of at infinity, that force the genus to decrease). For instance, assume that has no repeated linear factor. If then I think that the double plane construction (cf. [BPvdV]) gives that for all but finitely constants b, the surface is birational to a smooth projective surface of general type.

14 June, 2019 at 10:37 am

PastenHere are the details: replace by so that the affine zero set of is smooth. Let be the projective closure of in the projective plane, and let if is even, if is odd (=line at infinity in the projective plane). The hypothesis on makes a normal crossings divisor. There is a double cover of the projective plane branched exactly along (since is even). The canonical sheaf of the canonical resolution of is the pull-back of (since is n.c.), which is big when , i.e. . is birational to our surface.

17 June, 2019 at 7:16 am

Terence TaoThat’s great! This construction seems to knock out a lot of the subcases of Case (a) (assuming Bombieri-Lang of course) leaving mostly just the cases of low degree which possibly can be done by hand. The obvious thing to try next is to replace by for larger , but I guess this would tend to make the degree requirement on worse as ?

9 June, 2019 at 7:50 am

Terence TaoAh, this looks like a much simpler route to getting a section than invoking Faltings theorem and then analysing the genus one and genus zero curves by hand.

It occurs to me that one can take to be a rational function rather than a polynomial. The large system of identities (1) is then basically saying that if is bijective, then is more or less bijective also (certainly it is injective, and it looks like it will be surjective on the range in which Case (b) applies). I suppose one could then try to apply Bombieri-Lang for (which maybe follows from three-dimensional Bombieri-Lang over ?) to try to get things like the uniform degree bound? This would be somewhat in the spirit of the work of Caporaso, Harris and Mazur deriving uniform bounds on Faltings’ theorem from Bombieri-Lang for surfaces (see also this followup paper of Abramovich).

10 June, 2019 at 5:13 am

Daniel LoughranHere is my attempt at making clear some of the geometry of the problem (following Will Sawin’s comments).

Let be the affine surface given by

This is equipped with the natural projection to the -coordinate, which is assumed to be a bijection on rational points. Note that is a rational variety, in particular its rational points are Zariski dense.

We can probably assume that the generic fibre is a curve of genus at least , as the small genus cases can be treated separably, as Will suggests.

We then consider the base-change of via the map , for some polynomial . Explicitly, is given by

equipped with the associated map given by the base change of , which is again just projection onto the -coordinate. Note that is surjective on rational points as it arises as the base change of map with the same property.

Now, providing that is separable of sufficiently large degree, the surface should be a surface of general type. Then the Bomberi-Lang conjecture predicts that its rational points are not Zariski dense, so they lie in the union of finitely many curves . We then consider the map induced by ; if each such map is not of degree one then the image is a so-called thin set. But then this contradicts the fact that is surjective on rational points, by Hilbert’s irreducibility theorem. Hence there is some such that is of degree . This therefore gives a section of .

Note that the exact same arguments also applies to more general finite covers , rather than the special covers considered above. This extra flexibility might come in useful.

These arguments all seem reasonable so far. However, the next claim is that this implies that itself must have a section. This seems reasonable, but I don’t see how to prove this. We have the freedom to vary and its degree, so these arguments seem to show that the generic fibre , viewed as a curve over the function field , has a rational point over field extensions of coprime degree, hence has a -cycle of degree one. But in general this need not imply the existence of a rational point.

Anyway, let me finish the argument. Assuming that has a section, we deduce a contradiction as follows. As has a Zariski dense set of rational points and the image of a section is a closed subset, there must be a rational point not lying in the section; however this contradicts that is bijective on rational points.

11 June, 2019 at 6:41 am

Terence TaoIt seems that perhaps the (strong) abc conjecture for function fields or polynomials (which is actually a theorem) could give a uniform degree bound for and in (1) in terms of and , as long as is “non-degenerate” in some sense. I found a paper on this strong function field abc conjecture at https://arxiv.org/abs/0811.3153 but am still digesting it. (If one is willing to assume the analogous strong abc conjecture for , this presumably also gives a polynomial bound for the heights of in terms of the height of for any rational , again assuming some suitable non-degeneracy condition on , though I am not sure what one would do with such a bound.)

ADDED: presumably Elkies observation that the abc conjecture implies an effective version of Mordell’s conjecture (Faltings’ theorem) can be adapted to the polynomial case and indeed get a uniform degree bound on in terms of the degree of , assuming we are in the case where the level sets of have genus at least two, which it seems we can reduce to by other comments.

8 June, 2019 at 8:52 am

AnonymousThe set in the definition of seems to be undefined.

[This is my notation for affine space – T.]8 June, 2019 at 10:01 am

AnonymousIs it possible to generalize this problem to polynomials over other coefficient fields?

8 June, 2019 at 9:46 pm

TRHow about bijection and for that matter bijection?

12 June, 2019 at 6:22 pm

David SpeyerThe map is a bijection . Proof of injectivity: If , then and, since is not square, we must have . Proof of surjectivity: Every element of can be written as , and we can write , so every element of is .

The analogous trick does . I suspect some trickery could do but I don’t see it. Since no one know what is, I’m not going to address that.

13 June, 2019 at 11:46 am

Terence TaoNice observation! I wonder what part of the strategy outlined in the post fails in positive characteristic. It seems that the surface has a Zariski-dense set of -points for any -rational function , so either the naive version of Bombieri-Lang is false in or these surfaces are not of general type. Presumably it is the latter (I guess the fibres are now some sort of double line? Characteristic 2 geometry is so strange…)

14 June, 2019 at 4:22 am

PastenThis example goes back to Proposition 8 in G. Cornelissen’s article “Stockage diophantien et hypothèse abc généralisée”. Regarding BL in global function fields of characteristic p>0, already for curves (i.e. Mordell’s conjecture –Samuel’s theorem in this case) one needs to allow exceptions in the isotrivial case. Note that becomes isomorphic to a Fermat curve (constant coefficients) after base change, thus, such varieties are isotrivial.

14 June, 2019 at 1:52 am

Daniel LoughranI should say that this example already appears in Remark 1.6 of Poonen’s paper mentioned above. These examples also indeed do not appear to give surfaces of general type, and really use pathological behaviour in positive characteristic.

9 June, 2019 at 7:51 am

MartiniNeglecting the dependency on one argument, there are many polynomials which give a bijection, e.g., the identity or x cubed.

Where in the above arguments is it assumed that P really depends on two variables?

11 June, 2019 at 9:03 pm

Anonymousx cubed is not a bijection since there are many rationals with no rational cube root.

13 June, 2019 at 12:05 pm

OmarA polynomial that only depends on one of the two variables can never be injective: f(x,y0) = f(x,y1), if f(x,y) only depends on x.

9 June, 2019 at 10:09 am

A sort of Polymath on a famous MathOverflow problem | The polymath blog[…] that the map is a bijection? This is a famous 9-years old open question on MathOverflow. Terry Tao initiated a sort of polymath attempt to solve this problem conditioned on some conjectures from arithmetic […]

10 June, 2019 at 11:57 am

AnonymousThe problem formulation is very simple, but its difficulty is not necessarily related to its simple formulation (e.g. the case of FLT)