Gerd Faltings, who proved the Mordell conjecture, wins the Abel Prize
58 points by digital55 6 days ago | 8 comments
nhatcher 5 days 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!
reply011101101 16 hours ago
A point is that which has no breadth.
replyThe 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 15 hours 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."
replyOgsyedIE 12 hours ago
This must be an incorrect description of what has actually been proved, since x^4 is a counterexample.
replyraphlinus 11 hours 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.
replythornhill 10 hours ago
The curve they mean y = x^4 is irreducible but the genus is 0 since it’s isomorphic to the affine line.
replyjlev1 8 hours ago
The correct description is “a smooth curve of genus at least 2”.
replyhttps://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.