Injective polynomials
From a paper of MIT's Bjorn Poonen, I learned of an amazingly simple open problem. I'll just quote the paper (here Q denotes the rational numbers):
"Harvey Friedman asked whether there exists a polynomial f(x, y) in Q(x, y) such that the induced map Q x Q --> 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 be an example. But it seems very difficult to prove that any polynomial works. Both Friedman's question and Zagier's speculation are at least a decade old... but it seems that there has been essentially no progress on the question so far."
Poonen shows that a certain other, more widely-studied hypothesis implies that such a polynomial exists. Of course, such a polynomial does not exist if we replace Q by R, the reals. In fact any injection R x R --> R must be (very) discontinuous.
Suppose an injective polynomial f could be identified, answering Friedman's question; it might then be interesting to look at `recovery procedures' to produce x, y given f(x, y). We can't hope for x, y to be determined as polynomials in f(x, y), but maybe an explicit, fast-converging power series or some similar recipe could be found.
"Harvey Friedman asked whether there exists a polynomial f(x, y) in Q(x, y) such that the induced map Q x Q --> 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 be an example. But it seems very difficult to prove that any polynomial works. Both Friedman's question and Zagier's speculation are at least a decade old... but it seems that there has been essentially no progress on the question so far."
Poonen shows that a certain other, more widely-studied hypothesis implies that such a polynomial exists. Of course, such a polynomial does not exist if we replace Q by R, the reals. In fact any injection R x R --> R must be (very) discontinuous.
Suppose an injective polynomial f could be identified, answering Friedman's question; it might then be interesting to look at `recovery procedures' to produce x, y given f(x, y). We can't hope for x, y to be determined as polynomials in f(x, y), but maybe an explicit, fast-converging power series or some similar recipe could be found.
Finally, all of this should be compared with the study of injective polynomials from the Euclidean plane to itself; this is the subject of the famous Jacobian Conjecture. See Dick Lipton's excellent post for more information.
Labels: general math

13 Comments:
Yeah, I also learnt of this problem recently! Btw, I found some related discussion on mathoverflow about whether one could rule out a polynomial bijection.
By
arnab, at 10:03 PM
Cool, thanks. Yes, a bijection would be even better.
Also, certainly there exist surjections Q^2 --> Q. But not all surjections R^2 --> R are surjections Q^2 --> Q; for example f(x, y) = x^3 misses the value 2. Could one give a test for surjectivity of such maps? Or even surjectivity of polynomial maps from Q --> Q?
By
Andy D, at 10:26 AM
Why would an injective map over the reals have to be discontinuous? Is it obvious?
By
Anonymous, at 9:42 PM
anonymous: suppose f is a continuous map from R^2 --> R.
Now consider the image under f of the unit circle. The inescapable conclusion: f is not injective.
By
Andy D, at 10:25 PM
What a great question. Thanks for sharing it.
By
rjl, at 11:20 PM
Never put off till tomorrow what may be done today..................................................................
By
文王廷, at 9:25 AM
這是一篇來自遙遠的78星雲的留言!要開心哦!..................................................................
By
孫邦柔, at 11:10 PM
Judge not a book by its cover.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
By
江趙雲虹趙雲虹仁昆, at 5:57 AM
i like algerba
By
ali0482, at 8:55 AM
你的分享很不錯.. 謝謝 ..................................................
By
瑰潼, at 7:05 AM
謝謝大大的分享 我會學會反省與寬容 感恩 ∩△∩............................................................
By
江趙雲虹趙雲虹仁昆, at 6:48 PM
Why would an injective map over the reals have to be discontinuous? Is it obvious?
By
ali0482, at 10:16 AM
ali--see my second comment above. The idea is that the image of a circle under a continuous map onto the reals fails to be injective, since as we trace the circle, we have to return to our original image-point. Hope this makes sense.
By
Andy D, at 12:13 PM
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