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Is there any polynomial $f(x,y)\in{\mathbb Q}[x,y]{}$ such that $f\colon\mathbb{Q}\times\mathbb{Q} \rightarrow\mathbb{Q}$ is a bijection?

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Z.H.
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    Vaguely related (not the question perhaps, but some of the answers): http://mathoverflow.net/questions/9731/polynomial-representing-all-nonnegative-integers – Kevin Buzzard Apr 11 '10 at 12:40
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    Is it known (or obvious) that there is an injective f? – Tom Leinster Apr 11 '10 at 17:56
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    Quote from http://arxiv.org/abs/0902.3961, Bjorn Poonen, Feb. 2009: "Harvey Friedman asked whether there exists a polynomial $f(x,y)\in Q[x,y]$ such that the induced map $Q × Q\to Q$ is injective. Heuristics suggest that most sufficiently complicated polynomials should do the trick. Don Zagier has speculated that a polynomial as simple as $x^7+3y^7$ might already be an example. But it seems very difficult to prove that any polynomial works. Our theorem gives a positive answer conditional on a small part of a well-known conjecture." – Jonas Meyer Apr 11 '10 at 19:47
  • Notice the surjectivity of f's like x^2-y^2 or x^2y^3... Is there a "non-trivial" example of surjective f? – Yaakov Baruch Apr 11 '10 at 21:46
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    @Kevin: tentatively, perhaps "trivial" should mean that f can be made linear in a variable by either rational linear variable substitutions (y=z-x), or by plugging a value into one variable (x=1/y, x=0), or a combination of both. – Yaakov Baruch Apr 11 '10 at 23:44
  • Up to linear transformations, an irreducible quadratic f has the same values as x^2 - b y^2 for some b; this cannot be surjective because of the existence of inert primes in Q(sqrt(b))/Q. (I deleted a part of this comment about density considerations for Image(f), which I think to have been misguided.) – Yaakov Baruch Apr 17 '10 at 19:29
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    Shouldn't Jonas repost his comment as an answer? – David Corwin Jul 27 '10 at 22:19
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    I highly doubt that this is the case, but I have no rigorous proof. I want to indicate nevertheless the heuristic I am following. If this would happen, we would get a polynomial $p$, which bijects $p:\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$. Since $p$ is injective and hence its derivative is a nonzero vector, I would expect that its inverse exists and is differentiable everywhere. The inverese function would now indeed produce a continuous space filling curve as pointed out by Daniel Miller. I do not know of any differentiable space filling curve, which is injective. – Marc Palm Nov 19 '10 at 06:47
  • Moreover I do not know of any space filling curve, which is differentiable and surjective. The examples I know come mostly from the universal properties of L functions and there the produced curves have dense image. – Marc Palm Nov 19 '10 at 06:50
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    @pm: p will most definitely NOT be injective on $\mathbb{R}$. To see why even consider the simple function $X^3-2X$ which is injective on $\mathbb{Q}$ but not on $\mathbb{R}$ (this follows easily from knowledge of the primes in the ring of Eisenstein integers). – Yaakov Baruch Feb 22 '11 at 14:31
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    If there exists such an $f$, then there does so in any number $n$ of variables, by a simple induction. So does there exist, for some $n \geq 2$, a polynomial $p(x_1,...,x_n)$ in $n$ variables over $\mathbb{Q}$ such that $p : \mathbb{Q}^n \rightarrow \mathbb{Q}$ is bijective ? Replace bijective everywhere by injective if you like. I don't know if this is any easier to answer, but sometimes you can say a lot more about Diophantine equations in many variables. – Peter Hegarty Sep 22 '11 at 13:52
  • @Peter: as far as an injective $p$ in $n$ variables, it seems to be almost certainly possible - see the 3rd comment from top. – Yaakov Baruch Oct 31 '11 at 18:24
  • A few remarks that might help: 1) one can choose its coefficient to be in $\mathbb N$. 2) this condition is expressible in first order logic, so one may replace $\mathbb Q$ by any field having the same theory as $\mathbb Q$. 3) if there is such a map, it extends to $\overline{\mathbb{Q}}$. What can be said about the fibres of this map in $\mathbb{Q}$ ? Are there finite or having at least one finite projection on one of two coordinates? 4) Something true in $\overline {\mathbb{Q}}$ that is first order expressible is also true in $\overline{\mathbb{C}}$. – Drike Nov 04 '11 at 11:22
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    @Drike I can't speak to the first three (1 is reasonable, 2 is clear, and 3 is a question) but 4 is absolutely false (consider $\forall x x^2+1\not=0$) – Richard Rast Nov 14 '11 at 13:12
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    @Drike, because $n \mathbb Q = \mathbb Q$ for any non-zero rational $n$ then a bijection $f$ can be assumed to have all its coefficients in $\mathbb Z$. But I don't see how one can conclude they can be assumed to be in $\mathbb N$. For example, what if there are two terms of even degree in $x$ and $y$ with coefficients of opposit sign? – John R Ramsden Feb 26 '12 at 08:41
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    Does the question become any easier with Q replaced by Z, as in $f:\mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}$? – Mikhail Katz Apr 14 '13 at 07:38
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    Now 16 answers, all deleted. – Gerry Myerson Oct 07 '14 at 05:09
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    I would like to point out this closely related question I asked on M.SE just in case someone wanted to offer a more "divulgative" perspective on this problem to undergraduate students. Best regards. –  Dec 10 '14 at 18:25
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    @GerryMyerson 17 now (deleted) :) – Yemon Choi Jun 07 '19 at 21:01
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    @MarcPalm There are no injective and no differentiable space filling curves. – Alex Ravsky Mar 22 '20 at 10:09
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    What do we know about this in other fields? It seems the case of finite characteristic is already discussed in Terry's post and in Poonen's paper. What can we say in general or particular about the existence of polynomials of the form $f(x)\in k[x],$ for $x=(x_{1}, \dots, x_{n})$ and $k$ a ring, such that $f\colon k^{n}\to k$ is a bijection or an injection. I am particularly interested in what can we already say about extensions of $\mathbb{Q}$: mainly, what do we know about the situation in the algebraic numbers and the real algebraic numbers? And in $\mathbb{R}$ or $\mathbb{C}$? – Hvjurthuk Jul 21 '20 at 19:35
  • @Yaakov Baruch "most definitely NOT" means that it could still be possible? Have you got a proof that rules totally out the possibility suggested (without justification) by @MarcPalm? Or is it just highly possible heuristics? – Hvjurthuk Jul 21 '20 at 19:53
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    @Hvjurthuk. if $p: \mathbb{R}^2\rightarrow\mathbb{R}$ is a bijective polynomial, then for each $y_0$ the univariate polynomial $p(,y_0)$ must surject to $\mathbb{R}$ (or else it would have even degree and not be injective) . Therefore for $y_0\ne y_1$ BOTH $p(,y_0)$ and $p(,y_1)$ surject onto $\mathbb{R}$ and so the target $\mathbb{R}$ is already covered twice, negating the injectivity of $p$. Is this what you were asking? – Yaakov Baruch Jul 22 '20 at 13:12

3 Answers3

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Jonas Meyer's comment:

Quote from arxiv.org/abs/0902.3961, Bjorn Poonen, Feb. 2009: "Harvey Friedman asked whether there exists a polynomial $f(x,y)\in Q[x,y]$ such that the induced map $Q × Q\to Q$ is injective. Heuristics suggest that most sufficiently complicated polynomials should do the trick. Don Zagier has speculated that a polynomial as simple as $x^7+3y^7$ might already be an example. But it seems very difficult to prove that any polynomial works. Our theorem gives a positive answer conditional on a small part of a well-known conjecture." – Jonas Meyer

Added June 2019 Poonen's paper is published as:

Bjorn Poonen, Multivariable polynomial injections on rational numbers, Acta Arith. 145 (2010), no. 2, pp 123-127, doi:10.4064/aa145-2-2, arXiv:0902.3961.

David Roberts
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Boaz Tsaban
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    This was posted (by Jonas Meyer) as a comment on 11 April 2010, and got 125 upvotes (so far!) as a comment. Why post it now as an answer? – Gerry Myerson Jan 13 '16 at 05:04
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    @GerryMyerson Probably on suggestion from David Corwin, and/or on the Stack Exchange principle that important relevant information shouldn't be relegated to comments. I guess Boaz signals, by making it CW, that this isn't for reputation gain, but as a public service. – Todd Trimble Jan 13 '16 at 05:16
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    @GerryMyerson is right. The issue is that the problem is presented as unsolved, drawing unnecessary attention and time, just to find in the comments that it is as answered as it could be. I believe, MO is not expected to provide answers like complete solutions of P=NP. Rather, answers that are within knowledge, or easy reach, of experts. I can stop doing this service if this is against the policies, but then an alternative solution to the issue I raise here better be found. – Boaz Tsaban Jan 13 '16 at 12:31
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There is a new manuscript on the arXiv by Giulio Bresciani, A higher dimensional Hilbert irreducibility theorem, arXiv:2101.01090, which shows that assuming the weak Bombieri--Lang conjecture, there cannot be a polynomial bijection from $\mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}$.

The author writes that:

Our strategy is essentially the one followed in a "polymath project" led by T. Tao, see [Tao19], hence this result should be credited to the polymath project as a whole.

[Tao19] https://terrytao.wordpress.com/2019/06/08/ruling-out-polynomial-bijections-over-the-rationals-via-bombieri-lang/

David Roberts
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Jackson Morrow
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This is a link to a new, crowdsourced attempt to resolve this question (at least conditional on the assumption of some strong number-theoretic conjectures) being led by Terry Tao.

Sam Hopkins
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zeraoulia rafik
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    While what is outlined there is a good strategy of attack, Terry does not suggest it is a rigorous proof, and posts it in the hopes that specialists will fill in the holes and reduce the problem to a consequence of one or more conjectures in algebraic geometry. Gerhard "That Means It's Conditionally False" Paseman, 2019.06.08. – Gerhard Paseman Jun 08 '19 at 16:30