4 ms·

Gerd Faltings, who proved the Mordell conjecture, wins the Abel Prize

58 points by digital55 8mo ago

RonSFriedman86 8mo ago
[dead]
nhatcher 8mo ago
Oh wow! I wouldn't have expected this so many years later. Mordel's conjecture implies asva special case that for all n>=4 there are only a finite number of solutions to Fermat's equations with relative prime numbers. Brings me back!
- delhanty 8mo ago
  > Mordel's conjecture implies as a special case that for all n>=4 there are only a finite number of solutions to Fermat's equations with relative prime numbers I just learnt that fact from Wikipedia's article on Mordel's conjecture (now Faltings' theorem), was curious whether the theorem could be strengthened to obtain a full proof of Fermat’s Last Theorem (FLT) that is genuinely different from the Taylor–Wiles proof (or its later variants) and so asked an AI (in this case Grok via Twitter). Grok correctly told me "no it's not possible", but then surfaced (as an aside) a nice expository article on the Taylor–Wiles proof by Faltings from AMS notices in July 1995, which I thought I'd share here: https://www.ams.org/notices/199507/faltings.pdf https://www.ams.org/notices/199507/faltings.pdf
011101101 8mo ago
A point is that which has no breadth. The line is a breadthless legth. Mordell conjecture is that only circles or figure contain infinite points, whereas curves with exponents over 3 are finite accumulations.
ljsprague 8mo ago
"He proved that if a curve’s equation has a variable raised to a power higher than 3, then it must have a finite number of [rational] points."
- OgsyedIE 8mo ago
  This must be an incorrect description of what has actually been proved, since x^4 is a counterexample.
  - raphlinus 8mo ago
    My understanding, which is to be taken with a grain of salt, is that there's an additional constraint, not stated in the Scientific American article, that the plane curve be irreducible. The example of x^4 is reducible, it's x^2 * x^2 among other thing. The actual conjecture is expressed in terms of genus, but this follows from the genus-degree formula.
    - deleted 8mo ago
      [deleted]
    - thornhill 8mo ago
      The curve they mean y = x^4 is irreducible but the genus is 0 since it’s isomorphic to the affine line.
      - jlev1 8mo ago
        The correct description is “a smooth curve of genus at least 2”. https://en.wikipedia.org/wiki/Faltings%27_theorem https://en.wikipedia.org/wiki/Faltings%27_theorem The reason for the confusion is that a smooth, projective plane curve of degree d has genus (d-1)(d-2)/2, which is 2 or greater starting at d=4. Hence the phrasing in the article, which is missing the “smooth, projective” hypothesis. The equation y = x^4 doesn’t define a smooth curve when extended to the projective plane, because it has a singularity at infinity.
        thornhill 8mo ago
        I think the theorem applies to any curve, if you take geometric genus.
deleted 8mo ago
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58 points by digital55 8mo ago

[dead]

Oh wow! I wouldn't have expected this so many years later. Mordel's conjecture implies asva special case that for all n>=4 there are only a finite number of solutions to Fermat's equations with relative prime numbers. Brings me back!

delhanty 8mo ago
> Mordel's conjecture implies as a special case that for all n>=4 there are only a finite number of solutions to Fermat's equations with relative prime numbers I just learnt that fact from Wikipedia's article on Mordel's conjecture (now Faltings' theorem), was curious whether the theorem could be strengthened to obtain a full proof of Fermat’s Last Theorem (FLT) that is genuinely different from the Taylor–Wiles proof (or its later variants) and so asked an AI (in this case Grok via Twitter). Grok correctly told me "no it's not possible", but then surfaced (as an aside) a nice expository article on the Taylor–Wiles proof by Faltings from AMS notices in July 1995, which I thought I'd share here: https://www.ams.org/notices/199507/faltings.pdf https://www.ams.org/notices/199507/faltings.pdf

011101101 8mo ago

A point is that which has no breadth. The line is a breadthless legth. Mordell conjecture is that only circles or figure contain infinite points, whereas curves with exponents over 3 are finite accumulations.

ljsprague 8mo ago

"He proved that if a curve’s equation has a variable raised to a power higher than 3, then it must have a finite number of [rational] points."

OgsyedIE 8mo ago
This must be an incorrect description of what has actually been proved, since x^4 is a counterexample.
- raphlinus 8mo ago
  My understanding, which is to be taken with a grain of salt, is that there's an additional constraint, not stated in the Scientific American article, that the plane curve be irreducible. The example of x^4 is reducible, it's x^2 * x^2 among other thing. The actual conjecture is expressed in terms of genus, but this follows from the genus-degree formula.
  - deleted 8mo ago
    [deleted]
  - thornhill 8mo ago
    The curve they mean y = x^4 is irreducible but the genus is 0 since it’s isomorphic to the affine line.
    - jlev1 8mo ago
      The correct description is “a smooth curve of genus at least 2”. https://en.wikipedia.org/wiki/Faltings%27_theorem https://en.wikipedia.org/wiki/Faltings%27_theorem The reason for the confusion is that a smooth, projective plane curve of degree d has genus (d-1)(d-2)/2, which is 2 or greater starting at d=4. Hence the phrasing in the article, which is missing the “smooth, projective” hypothesis. The equation y = x^4 doesn’t define a smooth curve when extended to the projective plane, because it has a singularity at infinity.
      - thornhill 8mo ago
        I think the theorem applies to any curve, if you take geometric genus.

deleted 8mo ago

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