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## 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