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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 it known (or obvious) that there is an injective f? –  Tom Leinster Apr 11 '10 at 17:56
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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 Apr 11 '10 at 19:47
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Seven incorrect answers posted (counting three deleted ones). Is this a record for most incorrect answers to an MO question? –  Gerry Myerson Aug 16 '11 at 23:45
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Now up to 11 answers, 7 of them deleted. –  Gerry Myerson Dec 4 '11 at 22:07
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Now with 14 answers, 10 of them deleted. –  David Roberts Apr 29 '13 at 22:37
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4 Answers

I would be inclined to say no, for the following reason:

Suppose $f$ satisfies the condition. Then by interpolation, the coefficients of $f$ are all rational numbers. By multiplying $f$ with an integer, we may assume without loss of generality that $f$ has integer coefficients.

Let $d=$deg$(f)$. It is obvious that $d$ can not be 1. It cannot be 2 either, for it is well known that if a conic curve on the plane with rational coefficients has a rational point, then it has infinitely many. As a result, $d\ge 3$.

[The following is only an idea, not a rigorous proof.]

For $x\in \mathbb{Q}$, define the height function $ht(x)$ to be the maximum of the denominator and the numerator of $x$. For a point $p=(x,y)$ in $\mathbb{Q}^2$, define $ht(p)=max(ht(x),ht(y))$. Then, for $p\in \mathbb{Q}^2$, we can expect that $ht(f(p))$ is roughly $ht(p)^d$. Since there are roughly $N^2$ points on $\mathbb{Q}^2$ with height no more than $N$, and roughly $N$ such points on $\mathbb{Q}$, we can see that only when $d=2$ can $f$ be a 1-1 correspondence. But we have already proved that $d\ge 3$, which yields a contradiction.

The problem is, I can't prove that $ht(f(p))$ is approximately $ht(p)^d$ in any sense. If this can be established, I guess I could give a rigorous proof.

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3  
A quick comment about the degree $d=2$ case: The curve $x^2+y^2=0$ has exactly one rational solution. –  Morgan Sherman Jul 19 '13 at 16:40
    
To @MorganSherman and those who upvoted his comment. Do you really doubt that $d \geq 3$? How about looking at the possible degrees of $g(t):=f(p(t),q(t))$ for arbitrary injective polynomial curves $(p,q): \mathbb Q \mapsto (\mathbb Q \times \mathbb Q)$ and injective $f$? Because $g$ is injective, the degree can not be $0$ or $2$. Otherwise $g(t)=c$ or $g(t)=at^2+bt+c$, and hence $g(t)=g(-\frac{b}{a}-t)$ contradicts the injectivity of $g$. It also can not be $1$, for otherwise $g(\mathbb Q) = \mathbb Q$ contradicts the injectivity of $f$. –  Thomas Klimpel Jul 21 '13 at 14:00
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@ThomasKlimpel, I just wanted to point out the error. I honestly haven't thought deeply about the problem. –  Morgan Sherman Jul 21 '13 at 22:57
    
The trouble is that $ht(f(x,y))$ can sometimes be much smaller. EG there are infinitely many rational points on $y^2=x^3-2$, so $ht(x^3-y^2)$ can be $2$ for arbitrarily large $(x,y)$. Of course, that example is bad the other way, failing to be injective, but I think it makes the point. –  David Speyer Dec 5 '13 at 21:23
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Also, for any polynomial $g(x)$ in $\mathbb{Q}[x]$, the map $(x,y) \mapsto (x, y+g(x))$ is a bijection from $\mathbb{Q}^2$ to itself. So, if we had a solution, we could compose with a map like this one to get a solution of arbitrarily high degree. –  David Speyer Dec 5 '13 at 21:24
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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 '10 at 12:54
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Haha... i was expecting the intermediate value theorem to hold for $\mathbb Q$ –  Garlef Wegart Apr 11 '10 at 12:59
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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 '10 at 15:43
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Related to Kevin Buzzard’s comments, is there any continuous and bijective mapping from ℚ×ℚ to ℚ? –  Tsuyoshi Ito Aug 12 '10 at 17:32
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Any (nonempty) countable metrizable space without isolated points is homeomorphic to $\mathbb{Q}$. –  Gerald Edgar Aug 15 '10 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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9  
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 '10 at 13:55
    
Duh, your right. –  Daniel Miller Jul 18 '10 at 18:21
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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 '12 at 10:45
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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 '11 at 10:42
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protected by Andres Caicedo Dec 5 '13 at 14:23

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