Timeline for Polynomial bijection from $\mathbb Q\times\mathbb Q$ to $\mathbb Q$?
Current License: CC BY-SA 4.0
39 events
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| Jan 5, 2021 at 3:02 | answer | added | Jackson Morrow | timeline score: 56 | |
| Jul 22, 2020 at 13:12 | comment | added | Yaakov Baruch | @Hvjurthuk. if $p: \mathbb{R}^2\rightarrow\mathbb{R}$ is a bijective polynomial, then for each $y_0$ the univariate polynomial $p(,y_0)$ must surject to $\mathbb{R}$ (or else it would have even degree and not be injective) . Therefore for $y_0\ne y_1$ BOTH $p(,y_0)$ and $p(,y_1)$ surject onto $\mathbb{R}$ and so the target $\mathbb{R}$ is already covered twice, negating the injectivity of $p$. Is this what you were asking? | |
| Jul 21, 2020 at 19:53 | comment | added | Hvjurthuk | @Yaakov Baruch "most definitely NOT" means that it could still be possible? Have you got a proof that rules totally out the possibility suggested (without justification) by @MarcPalm? Or is it just highly possible heuristics? | |
| Jul 21, 2020 at 19:35 | comment | added | Hvjurthuk | What do we know about this in other fields? It seems the case of finite characteristic is already discussed in Terry's post and in Poonen's paper. What can we say in general or particular about the existence of polynomials of the form $f(x)\in k[x],$ for $x=(x_{1}, \dots, x_{n})$ and $k$ a ring, such that $f\colon k^{n}\to k$ is a bijection or an injection. I am particularly interested in what can we already say about extensions of $\mathbb{Q}$: mainly, what do we know about the situation in the algebraic numbers and the real algebraic numbers? And in $\mathbb{R}$ or $\mathbb{C}$? | |
| S Mar 24, 2020 at 9:51 | history | bounty ended | CommunityBot | ||
| S Mar 24, 2020 at 9:51 | history | notice removed | user153451 | ||
| Mar 22, 2020 at 10:09 | comment | added | Alex Ravsky | @MarcPalm There are no injective and no differentiable space filling curves. | |
| S Mar 20, 2020 at 21:50 | history | bounty started | CommunityBot | ||
| S Mar 20, 2020 at 21:50 | history | notice added | user153451 | Canonical answer required | |
| S Jun 9, 2019 at 22:11 | history | suggested | CommunityBot | CC BY-SA 4.0 |
LaTeX code improved.
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| Jun 9, 2019 at 21:26 | review | Suggested edits | |||
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| Jun 8, 2019 at 16:18 | answer | added | zeraoulia rafik | timeline score: 35 | |
| Jun 7, 2019 at 21:01 | comment | added | Yemon Choi | @GerryMyerson 17 now (deleted) :) | |
| Jun 12, 2016 at 14:42 | review | Suggested edits | |||
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| Jan 13, 2016 at 1:06 | answer | added | Boaz Tsaban | timeline score: 90 | |
| Dec 10, 2014 at 18:25 | comment | added | user60665 | I would like to point out this closely related question I asked on M.SE just in case someone wanted to offer a more "divulgative" perspective on this problem to undergraduate students. Best regards. | |
| Oct 7, 2014 at 5:09 | comment | added | Gerry Myerson | Now 16 answers, all deleted. | |
| Dec 5, 2013 at 14:23 | history | protected | Andrés E. Caicedo | ||
| S Jun 29, 2013 at 14:02 | history | suggested | Stefan Hamcke | CC BY-SA 3.0 |
latex added in the title
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| Jun 29, 2013 at 13:51 | review | Suggested edits | |||
| S Jun 29, 2013 at 14:02 | |||||
| Apr 14, 2013 at 7:38 | comment | added | Mikhail Katz | Does the question become any easier with Q replaced by Z, as in $f:\mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}$? | |
| Feb 26, 2012 at 8:41 | comment | added | John R Ramsden | @Drike, because $n \mathbb Q = \mathbb Q$ for any non-zero rational $n$ then a bijection $f$ can be assumed to have all its coefficients in $\mathbb Z$. But I don't see how one can conclude they can be assumed to be in $\mathbb N$. For example, what if there are two terms of even degree in $x$ and $y$ with coefficients of opposit sign? | |
| Nov 14, 2011 at 13:12 | comment | added | Richard Rast | @Drike I can't speak to the first three (1 is reasonable, 2 is clear, and 3 is a question) but 4 is absolutely false (consider $\forall x x^2+1\not=0$) | |
| Nov 4, 2011 at 11:22 | comment | added | Drike | A few remarks that might help: 1) one can choose its coefficient to be in $\mathbb N$. 2) this condition is expressible in first order logic, so one may replace $\mathbb Q$ by any field having the same theory as $\mathbb Q$. 3) if there is such a map, it extends to $\overline{\mathbb{Q}}$. What can be said about the fibres of this map in $\mathbb{Q}$ ? Are there finite or having at least one finite projection on one of two coordinates? 4) Something true in $\overline {\mathbb{Q}}$ that is first order expressible is also true in $\overline{\mathbb{C}}$. | |
| Oct 31, 2011 at 18:24 | comment | added | Yaakov Baruch | @Peter: as far as an injective $p$ in $n$ variables, it seems to be almost certainly possible - see the 3rd comment from top. | |
| Sep 22, 2011 at 13:52 | comment | added | Peter Hegarty | If there exists such an $f$, then there does so in any number $n$ of variables, by a simple induction. So does there exist, for some $n \geq 2$, a polynomial $p(x_1,...,x_n)$ in $n$ variables over $\mathbb{Q}$ such that $p : \mathbb{Q}^n \rightarrow \mathbb{Q}$ is bijective ? Replace bijective everywhere by injective if you like. I don't know if this is any easier to answer, but sometimes you can say a lot more about Diophantine equations in many variables. | |
| Feb 22, 2011 at 14:31 | comment | added | Yaakov Baruch | @pm: p will most definitely NOT be injective on $\mathbb{R}$. To see why even consider the simple function $X^3-2X$ which is injective on $\mathbb{Q}$ but not on $\mathbb{R}$ (this follows easily from knowledge of the primes in the ring of Eisenstein integers). | |
| Nov 19, 2010 at 6:50 | comment | added | Marc Palm | Moreover I do not know of any space filling curve, which is differentiable and surjective. The examples I know come mostly from the universal properties of L functions and there the produced curves have dense image. | |
| Nov 19, 2010 at 6:47 | comment | added | Marc Palm | I highly doubt that this is the case, but I have no rigorous proof. I want to indicate nevertheless the heuristic I am following. If this would happen, we would get a polynomial $p$, which bijects $p:\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$. Since $p$ is injective and hence its derivative is a nonzero vector, I would expect that its inverse exists and is differentiable everywhere. The inverese function would now indeed produce a continuous space filling curve as pointed out by Daniel Miller. I do not know of any differentiable space filling curve, which is injective. | |
| Oct 26, 2010 at 22:39 | history | edited | Andrés E. Caicedo | CC BY-SA 2.5 |
added 1 characters in body; added 4 characters in body
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| Jul 27, 2010 at 22:19 | comment | added | David Corwin | Shouldn't Jonas repost his comment as an answer? | |
| Jul 14, 2010 at 20:20 | history | edited | Greg Kuperberg |
edited tags
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| Apr 17, 2010 at 19:29 | comment | added | Yaakov Baruch | Up to linear transformations, an irreducible quadratic f has the same values as x^2 - b y^2 for some b; this cannot be surjective because of the existence of inert primes in Q(sqrt(b))/Q. (I deleted a part of this comment about density considerations for Image(f), which I think to have been misguided.) | |
| Apr 11, 2010 at 23:44 | comment | added | Yaakov Baruch | @Kevin: tentatively, perhaps "trivial" should mean that f can be made linear in a variable by either rational linear variable substitutions (y=z-x), or by plugging a value into one variable (x=1/y, x=0), or a combination of both. | |
| Apr 11, 2010 at 21:46 | comment | added | Yaakov Baruch | Notice the surjectivity of f's like x^2-y^2 or x^2y^3... Is there a "non-trivial" example of surjective f? | |
| Apr 11, 2010 at 19:47 | comment | added | Jonas Meyer | 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." | |
| Apr 11, 2010 at 17:56 | comment | added | Tom Leinster | Is it known (or obvious) that there is an injective f? | |
| Apr 11, 2010 at 12:40 | comment | added | Kevin Buzzard | Vaguely related (not the question perhaps, but some of the answers): mathoverflow.net/questions/9731/… | |
| Apr 11, 2010 at 12:03 | history | asked | Z.H. | CC BY-SA 2.5 |