Is there any polynomials of two variables with rational coefficients, such that the map is a bijection? This is a famous 9-years old open question on MathOverflow. Terry Tao initiated a sort of polymath attempt to solve this problem conditioned on some conjectures from arithmetic algebraic geometry. This project is based on an plan by Tao for a solution, similar to a 2009 result by Bjorn Poonen who showed that conditioned on the Bombieri-Lang conjecture, there is a polynomial so that the map is injective. (Poonen’s result answered a question by Harvey Friedman from the late 20th century, and is related also to a question by Don Zagier.)

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The status of the project is recorded here: https://terrytao.wordpress.com/2019/06/08/ruling-out-polynomial-bijections-over-the-rationals-via-bombieri-lang/#comment-518118

Comment by Gil Kalai — July 5, 2019 @ 5:40 am |

Hi, its nice article about media print, we all know media is a great source of information.

Comment by blog — July 19, 2019 @ 1:27 am |

Estimated Gil Kalai. Exists Collatz Conjecture polymath? How can be created? If I am not a professional, can I participate?I am looking a place to discus about the conjecture and share ideas,but I can not find it.may be I am too much crank but exist a 0.01% possibly that a not professional,with help ,could participate on solving the conjecture.Thanks.

Comment by Alberto Ibañez — August 16, 2019 @ 9:33 pm |

This looks like the Jacobian conjecture over C, which states conjecturally that every polynomial map from C^n to C^n with nonvanishing jacobian determinant is an automorphism. Not proved yet. Many attempted to prove it. H.Bass claims to have reduced it to cubic polynomials and others, ( Wang, Moh, …) to other particular cases. No decisive result up to now, as far as I know. Returning to this question, the claim is if we take K=Q instead of C, and n=2, ( this claim is trivially true for K=C and n=1), then the Jacobian conjecture implies it. ( It’s know to be untrue in prime characteristic fields).

If the claim , ( I’m tempted to baptise it Tao’s Rational Jacobian Conjecture, TRJC for short!), is right, a natural idea is to look for a possible connection with the Jacobian conjecture for K=Q and n=2. W need such a P and Q to try to construct a Jacobian pair and prove the Jacobian conjecture in this case.

conversely, if Jacobian conjecture Jac(Q,2) is true, it woudl furnish all p’s answering TRJC.

But i guess proving Jac(Q,2) must be at leasta s hard as Jac(C,2, ( that my thesis supervisor Le Dung Trang claimed with C.Weber to prove in 1992, but all their results are wrong, although many published in French CRAS, Commentrai Helvetici and Kodai Journal 20 years later!!!)

Jac(C,2) is very hard to prove, ( but Jac(R,2) was proved false by Pinchuk!), and I remember having met Deligne in ICM 1994 and asked him if he can prove it. He told me he woudl write me, but never did.

PS: I’m sorry, my 1st comment was wrong! I’ll contact the administartrs to withdraw it. this 2nd comme,nt is more sound, ( I hope!), because I inadvertently did’nt see that P is a 2 variable, not 1, polynomial. Therefore, there is no direct link with Jac(C,1).

Comment by Jalel Bouharb — October 6, 2020 @ 12:13 pm |

“… there is a polynomial so that the map P : ℚ → ℚ × ℚ is injective”

Sorry to be fussy, but shouldn’t this read more like

“… there is an injective polynomial map (P,Q) : ℚ → ℚ × ℚ”

?

Comment by Dan Asimov — November 30, 2020 @ 12:04 am |

Thanks, I hope its ok on Terry Tao’s blog. Dan, are you involved in some topological high dimensional HEX?

Comment by Gil Kalai — January 17, 2021 @ 6:09 pm |

Yes, but only in even dimensions.

Comment by Dan Asimov — January 17, 2021 @ 6:27 pm |

Dan, this is great! Even dimensions is fine. Even just dimension four is extremely interesting. I’d love to know some details. Please please send me (even partial handwritten things).

Comment by Gil Kalai — January 17, 2021 @ 6:30 pm |

I’m South Korea Freelance writer

Somebody help plz

1 ~ 45 lotto

Select 6

Ex) 1, 2, 3, 24, 35, 36 form

.

One digit numbers all different

.

1, 2, 3, 24, 35, 36

2, 3, 4, 25, 36, 31

…

3, 4, 5, 26, 31, 32

4, 5, 6, 23, 31, 32

.

Total number

6C3 x 3C2 x 1C1 = 20 x 3 x 1 = 60

If one lotto = 0.1 $

Total Chase = 6$

.

45C6 = 8145060

10C6 = 10C4 = 210

.

1 / 8145060 —-> 1/210

.

Could a more effective way be possible?

.

Comment by Hyun seok — February 20, 2021 @ 2:46 am |