## Polynomial bijection from QxQ to Q?

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 it known (or obvious) that there is an injective f? – Tom Leinster Apr 11 2010 at 17:56
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 Apr 11 2010 at 19:47
Seven incorrect answers posted (counting three deleted ones). Is this a record for most incorrect answers to an MO question? – Gerry Myerson Aug 16 2011 at 23:45
Now up to 11 answers, 7 of them deleted. – Gerry Myerson Dec 4 2011 at 22:07
Now with 14 answers, 10 of them deleted. – David Roberts Apr 29 at 22:37

Edit[ The following is wrong ~ see comments]

I don't think so.

Suppose $f$ to be surjective. Let $x\mapsto 0$ and $y\mapsto 1$. Now consider two distinct paths $\gamma,\eta:[0,1]\to\mathbb Q\times\mathbb Q$ from $x$ to $y$. Since $f$ is continuous it maps these paths surjectively onto $[0,1]$ (more exactly $[0,1]\subset f\gamma([0,1])$ and $[0,1]\subset f\eta([0,1])$). Thus, $f$ cannot be injective.

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How can one map [0,1] to Q x Q? Oh---maybe you mean R x R? But if f isn't injective on R x R it can still be injective on Q x Q, right? Am I missing something or is something missing? – Kevin Buzzard Apr 11 2010 at 12:54
Haha... i was expecting the intermediate value theorem to hold for $\mathbb Q$ – Garlef Wegart Apr 11 2010 at 12:59
I don't see how you can possibly use ideas of continuity etc to prove results about an incomplete field like the rationals. – Kevin Buzzard Apr 11 2010 at 15:43
Related to Kevin Buzzard’s comments, is there any continuous and bijective mapping from ℚ×ℚ to ℚ? – Tsuyoshi Ito Aug 12 2010 at 17:32
Any (nonempty) countable metrizable space without isolated points is homeomorphic to $\mathbb{Q}$. – Gerald Edgar Aug 15 2010 at 16:21
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I would be inclined to say no, for the following reasons. First, note that the function $f^{-1}$ is a bijection $\mathbb{Q}\mapsto\mathbb{Q}\times\mathbb{Q}$, and as such, is something that resembles a space-filling curve. But in general, space-filling curves are highly complex, "messy" objects, not something one would expect from the inverse of a polynomial in two variables. Furthermore, note that the polynomials $p(x)=f(x,y)$ and $q(y)=f(x,y)$, for fixed $y$, satisfy $p^{-1}(x)\in\mathbb{Q}$ and $q^{-1}(x)\in\mathbb{Q}$ whenever $x\in\mathbb{Q}$, i.e., the equation $p(x)-r=0$ has a rational solution for every $r\in\mathbb{Q}$. However, as far as I know, the only polynomials that satisfy this are linear functions, which could not provide the bijection required.

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After you fix $y$, $p(\mathbb Q)$ is a proper subset of $\mathbb Q$ and so $p(x)-r=0$ doesn't have to be solvable. – Gjergji Zaimi Jul 18 2010 at 13:55
Duh, your right. – Daniel Miller Jul 18 2010 at 18:21
Moreover, the inverse map $f: \mathbb{Q} \to \mathbb{Q} \times \mathbb{Q}$ will not be the restriction of a differentiable map $\mathbb{R} \to \mathbb{R} \times \mathbb{R}$. (If that were true, the result werer easy to prove, because there is no $C^1$-surjection $\mathbb{R} \to \mathbb{R} \times \mathbb{R}$; Sards theorem!). – Johannes Ebert Feb 11 2012 at 10:45

Let us consider a simpler case first. Let $Q_1=Q\cap [0,1]$. Now let us assume $f:Q_1\times Q_1\to Q_1$ is a uniformly continuous bijection. Then, according to Rudin (I believe it is an exercise), there is unique continuous extension, $g$, such that:

1) $f=g$ on $Q_1\times Q_1$.

2) $g$ is uniformly continuous on $\overline{Q_1\times Q_1}=[0,1]^2$.

3) Namely, $g(x_0)=\lim_{x\to x_0} f(x)$.

Since the image of a connected, compact set is connected and compact, then $Im(g)=[0,1]$. But, this is impossible because if we consider three distinct rational points $a,b,c$ in $[0,1]^2$, then $g$ restricted to the connected set $[0,1]^2-${$a,b,c$} is still continuous, but the image will not be connected since g is bijective on $Q_1\times Q_1$.

I think the above case now follows. That is if we consider $f:Q\times Q\to Q$ and $f$ is a polynomial, then $f$ restricted to $Q_1\times Q_1$ would could be a uniformly continuous bijection. However, we won't know what the image is. But this doesn't matter, since when we extend to $g$ the continuous image of a connected compact set is connected and compact. But the only connected compact subsets of $R$ are bounded closed intervals. So, the image would be some $[a,b]$ and would get another contradiction.

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Your argument doesn't work because there is no reason why some $(x_1,x_2) \in \mathbb{R}^2$ with $x_i \notin \mathbb{Q}$ cannot map to an element of $\mathbb{Q}$. – ulrich Oct 3 2011 at 10:42

No, there is none. Suppose that there was such an $f$. Then for every $a\in\mathbb{Q}$ the function $f(a,\cdot):\mathbb{Q}\rightarrow \mathbb{Q}$ would be an injection. So there should infinitely many injections from $\mathbb{Q}$ to $\mathbb{Q}$, that have disjoint images. But there are only two possibilities for their limits when $x\rightarrow +\infty$. Contradiction.

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Two functions can have disjoint images and the same limit when $x\to\infty$. – Guillaume Brunerie Dec 4 2011 at 18:10
So you don't believe there is any polynomial injection, much less bijection. This flies in the face of Jonas Meyer's comment below the OP. No less an authority than Don Zagier believes $x^7 + 3 y^7$ could work. – Todd Trimble Dec 4 2011 at 18:19
No, Todd, the argument anyway use surjectivity in a very strong way. – Valerio Capraro Dec 4 2011 at 18:21
Sorry to be dense, Valerio, but I don't follow. Where is surjectivity being used? – Todd Trimble Dec 4 2011 at 18:45
If one could convert it to a proof, when it probably worked for $\mathbb{Z}$. So, is there a polynomial bijection $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$? (The diagonal counting of pairs of $\mathbb{N}$ using infinite matrix gives a polynomial bijection $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$.) – Lev Glebsky Dec 8 at 16:53