### 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

## 16 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

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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

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By Anonymous, at 5:57 AM

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By ali0482, at 8:55 AM

你的分享很不錯.. 謝謝 ..................................................

By Anonymous, at 7:05 AM

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By Anonymous, 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

if the polynomial f:R-> R an injective polynomial in some domain in D in R then what happens with the injectivity of f' in D

By Gaurav, at 12:35 AM

Gaurav -- if I understand your question, consider e.g. f(x) = x. It's injective everywhere but f'(x) = 1, constant and non-injective on any domain |D| > 1. -Andy

By Anonymous, at 2:39 AM

Anonymous-: If f:C->C injective then can we say f' is also injective?

By Gaurav, at 4:28 AM

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