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

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

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Jonas Meyer's answer:

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

3This 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 – 2016-01-13T05:04:17.113

20@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 – 2016-01-13T05:16:58.847

1@Todd, fair enough. Thanks. – Gerry Myerson – 2016-01-13T05:20:46.740

16@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 – 2016-01-13T12:31:23.803

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 – 2010-11-19T06:47:32.693

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 – 2010-11-19T06:50:04.253

4@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 – 2011-02-22T14:31:39.410

5If 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 – 2011-09-22T13:52:33.190

@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 – 2011-10-31T18:24:55.477

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 – 2011-11-04T11:22:02.223

@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 – 2011-11-14T13:12:50.263

1@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 – 2012-02-26T08:41:47.793

6Does the question become any easier with Q replaced by Z, as in $f:\mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}$? – Mikhail Katz – 2013-04-14T07:38:09.140

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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 – 2010-04-11T12:40:22.29348Is it known (or obvious) that there is an

injectivef? – Tom Leinster – 2010-04-11T17:56:56.010142

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 – 2010-04-11T19:47:33.837Notice 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 – 2010-04-11T21:46:02.057

@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 – 2010-04-11T23:44:08.070

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 – 2010-04-17T19:29:44.133

24Now 16 answers, all deleted. – Gerry Myerson – 2014-10-07T05:09:04.320

2

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.

– Dal – 2014-12-10T18:25:10.27013Shouldn't Jonas repost his comment as an answer? – David Corwin – 2010-07-27T22:19:27.590