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

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 '10 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 '10 at 19:47
Shouldn't Jonas repost his comment as an answer? – David Corwin Jul 27 '10 at 22:19
Does the question become any easier with Q replaced by Z, as in $f:\mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}$? – Mikhail Katz Apr 14 '13 at 7:38
Now 16 answers, all deleted. – Gerry Myerson Oct 7 '14 at 5:09

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

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This was posted (by Jonas Meyer) as a comment on 11 April 2010, and got 125 upvotes (so far!) as a comment. Why post it now as an answer? – Gerry Myerson Jan 13 at 5:04
@GerryMyerson Probably on suggestion from David Corwin, and/or on the Stack Exchange principle that important relevant information shouldn't be relegated to comments. I guess Boaz signals, by making it CW, that this isn't for reputation gain, but as a public service. – Todd Trimble Jan 13 at 5:16
@Todd, fair enough. Thanks. – Gerry Myerson Jan 13 at 5:20
@GerryMyerson is right. The issue is that the problem is presented as unsolved, drawing unnecessary attention and time, just to find in the comments that it is as answered as it could be. I believe, MO is not expected to provide answers like complete solutions of P=NP. Rather, answers that are within knowledge, or easy reach, of experts. I can stop doing this service if this is against the policies, but then an alternative solution to the issue I raise here better be found. – Boaz Tsaban Jan 13 at 12:31

## protected by Andrés E. CaicedoDec 5 '13 at 14:23

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