Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Do I remember a remark in "Sketch of a program" or "Letter to Faltings" correctly, that acc. to Grothendieck anabelian geometry should not only enable finiteness proofs, but a proof of FLT too? If yes, how?

Edit: In this transcript, Illusie makes a remark that Grothendieck looked for a connection between "FLT" and "higher stacks". BTW, here a note on (acc. to Illusie) Grothendieck's favored landscape.

share|improve this question
    
The "Sketch of a program" is available at grothendieckcircle.org. On page 40 of the pdf-file there is certainly a reference to FLT. –  Philipp Lampe Oct 18 '09 at 18:43
    
Thanks. Has anyone an idea what he had in mind about that? –  Thomas Riepe Oct 19 '09 at 9:42

2 Answers 2

up vote 4 down vote accepted

See the papers of Minhyong Kim. For example, begin by looking at the MR review 2181717 of his paper Invent. Math. 161 (2005), no. 3, 629--656.

share|improve this answer
    
Thanks! I'd be happy about further explanations on arithmetics and anabelian geometry. –  Thomas Riepe Oct 23 '09 at 9:46
1  
See also ucl.ac.uk/~ucahmki/cambridgews.pdf –  Seamus Sep 20 '10 at 15:48

As Minhyong Kim points out in one of his papers on unipotent fundamental group- it is not quite clear how the section conjecture would imply Faltings theorem. The nature of implication (i.e. section conjecture implies Faltings or FLT) may be known to some experts but I don't know if it is explicitly written in literature.

share|improve this answer
    
I believe that it is not yet known that the section conjecture implies Mordell's conjecture (Faltings' theorem). –  Emerton Mar 10 '10 at 14:21
2  
At some point Deligne thought he had a proof that the section conjecture implied Mordell, but the proof doesn't work. This is all explained in an appendix by Deligne to a paper of Stix: arxiv.org/abs/0910.5009 –  Felipe Voloch Mar 10 '10 at 16:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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