Timeline for Hasse-Weil Zeta Functions & Fermats Last Theorem
Current License: CC BY-SA 3.0
7 events
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Feb 20, 2017 at 14:40 | comment | added | David Loeffler | There is an invariant -- the genus -- which will tell you that the Fermat curve for a given $n$ has finitely rational points (because of Faltings' proof of the Mordell conjecture). But for FLT we need to show that there are no non-trivial solutions at all, for any value of $n > 2$, which is a very different problem indeed. | |
Feb 20, 2017 at 14:38 | comment | added | Daniel Loughran | @EIns Nell: The Hasse-Weil zeta function of a variety does not tell you whether there is a rational point. The zeta function of any plane conic over $\mathbb{Q}$ is $\zeta(s)\zeta(s-1)$, but of course some conics have rational points while others do not. | |
Feb 20, 2017 at 13:04 | comment | added | Eins Null | I guess what I'm looking for is something similar to the analytic class number formula, where you can read off an invariant ( the numbers of real and complex embeddings) which, by Dirichlets unit theorem, tells you if the Norm of that field equals 1, as a polynomial, has infinitely many solutions or not. Maybe this just is not the case. Thanks for your answer. | |
Feb 20, 2017 at 12:03 | comment | added | David Loeffler | @DanielLoughran You're right. Oops. | |
Feb 20, 2017 at 11:59 | history | edited | David Loeffler | CC BY-SA 3.0 |
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Feb 20, 2017 at 11:20 | comment | added | Daniel Loughran | I would have thought that the Hasse-Weil zeta function of Fermat curves should be much easier to understand than general curves due to the presence of automorphisms. I means the curve $x^3 + y^3 = z^3$ is an elliptic curve with complex multiplication. I imagine modularity in this case is much easier than the general case due to the description of the $L$-function in terms of Hecke characters. Though I agree that the question is very misguided. | |
Feb 20, 2017 at 9:28 | history | answered | David Loeffler | CC BY-SA 3.0 |