On 2/19/2013 10:54 AM, Henry Baker wrote:
Q about the proof of FLT:
I realize that Fermat himself wouldn't recognize Wiles's proof, but would Fermat recognize the assumptions?
I.e., does the Wiles proof require any "exotic" assumptions -- e.g., axiom of choice, generalized Riemann hypothesis, etc. ?
It may surprise you to find out that the question of what assumptions Wiles' proof requires is still a subject of debate. To take care of the uncontested parts first, Wiles definitely does not rely upon RH, ERH, GRH or any other unproved hypothesis of number theory. His result is "unconditional". He does (implicitly) use AC, if only because it is embedded deep into the tissue of modern algebra (e.g., starting on page 3 of Atiyah and Macdonald's "Introduction to Commutative Algebra"). The controversy arises from Wiles' use of Grothendieck's development of algebraic geometry, and in particular of so-called "cohomological number theory". As a matter of technical convenience, Grothendieck assumes the existence of a "universe", which is, essentially, a set that can model ZFC. The existence of such a set implies the consistency of ZFC, so (by Godel Incompleteness) it is an assumption strictly stronger than ZFC. The question is whether this technical device (equivalent to assuming the existence of an uncountable strongly inaccessible cardinal) can be removed. One common opinion is that it can in all proofs of interest, including Wiles', but that it may be quite some time before anyone actually succeeds in doing so. I learned most of what I know about this issue from a paper by Colin McLarty published in 2010 in the Bulletin of Symbolic Logic (online here: http://www.cwru.edu/artsci/phil/Proving_FLT.pdf). It is surprisingly readable, but of course it only presents his point of view. In particular, he believes that the usage of universes is removable, but doubts that Wiles' proof can be reduced all the way to Peano arithmetic. -- Fred W. Helenius fredh@ix.netcom.com