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 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 11 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. 


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. 


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 spacefilling curve. But in general, spacefilling 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. 


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. 


protected by Andres Caicedo Dec 5 '13 at 14:23
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