Maybe this is a bit naive, but consider an equation of the form in Fermats Last Theorem $f_n(x,y,z) = x^n +y^n - z^n$.

Would it be possible to reprove Fermats Last Theorem by considering the Hasse-Weil zeta function $L(X, s)$ of the projective variety $X$ cut out by $f_n$ and analyzing special values on the left side of the right half plane for which $L(f_n,s)$ is defined?

Are there any conjectures/beliefs out there of such a proof?


closed as unclear what you're asking by Felipe Voloch, RP_, Stefan Kohl, Jan-Christoph Schlage-Puchta, abx Feb 24 '17 at 9:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This is stated, of course, in the whole-hearted belief that such a function would have Analytic/Meromorphic Continuation to all of the complex numbers. $\endgroup$ – Eins Null Feb 20 '17 at 4:24

The only method we know of to prove analytic / meromorphic continuation of zeta-functions of alg. varieties over number fields is to go via some kind of modularity, or potential modularity, statement. So even making sense of the statement of your program already requires the key piece of technology developed to prove FLT (and modularity of these high-degree Fermat curves is likely to be vastly harder than elliptic curves).

That is, your "whole-hearted belief" makes this proposed program circular.

EDIT: Dan Loughran's comment alerted me to the fact that this is misleading. Fermat curves have lots of automorphisms, and there are enough of these to force the L-function to factor as a product of L-functions of Groessencharacters of cyclotomic fields, for which analytic continuation + functional equation of the L-function are known; see Aoki (1991). This is, of course, an automorphy statement of a kind, but one that's far easier than automorphy of general higher-genus curves would be.

There are several remaining serious obstacles. For instance, once you go beyond genus 1, the link between the special values of the L-function of a curve and the existence or otherwise of rational points on the curve is very indirect. Rather than points on $X$, the $L$-series gives you information about points on $Jac(X)$. Sometimes, with lots of extra work, you can translate this into information about points on $X$ itself (the Chabauty--Coleman method), but there are lots of cases where this does not apply.

  • $\begingroup$ 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. $\endgroup$ – Daniel Loughran Feb 20 '17 at 11:20
  • $\begingroup$ @DanielLoughran You're right. Oops. $\endgroup$ – David Loeffler Feb 20 '17 at 12:03
  • $\begingroup$ 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. $\endgroup$ – Eins Null Feb 20 '17 at 13:04
  • 2
    $\begingroup$ @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. $\endgroup$ – Daniel Loughran Feb 20 '17 at 14:38
  • $\begingroup$ 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. $\endgroup$ – David Loeffler Feb 20 '17 at 14:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.