## Polynomial bijection from $\mathbb Q\times\mathbb Q$ to $\mathbb Q$?

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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?

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 – 2010-04-11T12:40:22.293

48Is it known (or obvious) that there is an injective f? – Tom Leinster – 2010-04-11T17:56:56.010

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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 – 2010-04-11T19:47:33.837

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 – 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

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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.

– Dal – 2014-12-10T18:25:10.270

13Shouldn't Jonas repost his comment as an answer? – David Corwin – 2010-07-27T22:19:27.590

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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