[FOM] FLT and ZFC, again

Timothy Y. Chow tchow at alum.mit.edu
Thu Dec 6 11:23:22 EST 2007


I recently had an email exchange with a number theorist (who prefers to 
remain anonymous, but who gave permission for his anonymized remarks to 
be forwarded to FOM) regarding the possibility that the current proof of 
Fermat's Last Theorem might require axioms beyond ZFC.  His response:

> Oh, come on! :-) Just because a proof uses a cohomology theory doesn't
> mean that it depends on Grothendieck universes. The issue is the following.
> If you are Grothendieck in the mid-60s then you want to write SGA4 and
> you know what is important now (etale cohomology and a serious
> attempt at a proof of the Weil conjectures), but you don't know what
> will be important in the future, so you make some executive decision
> to set up the entire formalism of cohomology theory using universes
> because at some point you want to take a limit over a collection
> which, in some vast generality, may not be a set [the class of covers
> of a space, where "cover" is an abstract idea satisfying some axioms].
>
> There were hence two issues which Grothendieck had to make decisions 
> about: (1) should he restrict to situations where the limit is over a 
> set? and (2) if not, then how to formulate things? **For better or 
> worse**, Grothendieck decided not to restrict to the case where the 
> limit is a set, and to resolve the ensuing set-theoretic issues by using 
> the language of universes. It is just downright daft to go from this 
> fact to the vague idea (that some people seem still to cling onto, 
> although I can see you're not actually one of them) that there are 
> "set-theoretic issues with Wiles' proof".
>
> Had Grothendieck just stuck to cases where the limit was over a set then 
> the underlying formalism of things would be a lot lumpier; there would 
> have been weird underlying assumptions about his categories, but in 
> practice **nothing would change** regarding etale and fppf cohomology, 
> because these are both situations where the limit is over a set. The 
> problem is that to compute Cech cohomology in some huge generality you 
> need to take limits over covers. Now if X is a top space then the set of 
> its covers is, well, a set! Similarly in the etale and fppf site. What I 
> am saying is not only is it *clear* that there are no set-theoretic 
> issues, but *also* that in a parallel universe where Grothendieck had 
> decided to put in weird extra axioms in his categories instead, that 
> look ugly and weird but which are satisfied in almost all cases 
> *including all cases that Wiles uses*, it would not have even *occurred* 
> to people to suggest that Wiles' proof depended on set-theoretic issues!

When I pointed out that nobody has proved any metatheorem that one can 
quote to eliminate universes "automatically," the response was:

> *But* there exist sufficiently many mathematical pedants that for any 
> result that is important enough for people to use in practice, if it 
> really *does* depend on taking limits over a proper class, there will be 
> someone somewhere who knows this and doesn't stop reminding people of 
> how this "isn't really quite ZFC".

Finally, here's an addendum from a followup email:

> I'm only saying things that are well-known to number theorists (who have 
> spent a moment worrying about these things, at least), whether or not 
> they are well-known to logicians. And I was slightly disingenuous: Mazur 
> has at times used the fpqc topology on the category of schemes in his 
> work, and this topology does *not* have the property that the collection 
> of all [isomorphism classes of] covers of a scheme forms a set: the 
> problem is that an arbitrary extension of fields is faithfully flat and 
> quasi-compact. On the other hand I believe that I've never seen an 
> argument in arithmetic geometry which is both "much used" and "relies 
> essentially on the fpqc topology". I know people who know full well that 
> arguments with the fpqc topology are "arguments with a question mark 
> over them" and treat them as such.

Tim


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