From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: FLT, JSH and "Elementary" Proofs
Date: 3 Dec 1998 17:04:39 GMT
In article <3665D2F7.EE932077@sympatico.ca>, t wrote:
>I've been following the various attempts of JSH to provide an elementary
>proof and the following came to mind. I would like to ask if anyone has
>an example of a theorem whose proof required very deep methods, but
>later was established by elementary techniques.
[example illustrating that 'elementary' is supposed to mean 'easy' rather
than 'not using elaborate machinery']
Algebraic topology is often a good source for such examples because it
develops tools which manage to say with some precision things which can
be visualized to be 'obvious'.
Here's one example:
Theorem (Nielsen, about 1914 I think). Every subgroup of a free group is free.
I'm told the original algebraic proof is a mess (I've never seen it).
Here'a a complete and elementary proof:
(1) A free group is the fundamental group of a bouquet of circles.
(2) Subgroups of a fundamental group are fundamental groups of covers.
(3) Covers of graphs (1-dimensional CW complexes) are also graphs.
(4) The fundamental group of a graph is a free group.
Done!
I _think_ this example counts because each of the steps is fairly intuitive
(once the terminology is understood). But at some point this becomes
'cheating' because we simply relegate the difficulties to 'technical lemmas'.
Allow me to quote, verbatim, from Joe Silverman's new number-theory text:
"Proof (sketch) of Fermat's Last Theorem.
(1) Let p >= 3 be a prime, and suppose that there is a solution
(A,B,C) to A^p + B^p = C^p with A, B, C nonzero integers
satisfying gcd(A,B,C) = 1.
(2) Let E_{A,B} be the Frey curve y^2 = x(x+A^p)(x-B^p).
(3) Wiles' Theorem tells us that E_{A,B} is modular; that is, its
p-defects a_p follow a Modularity Pattern.
(4) Ribet's Theorem tells us that E_{A,B} is so strange that it cannot
possibly be modular.
(5) The only way out of this seeming contradiction is the conclusion that the
equation A^p + B^p = C^p can have no solutions in nonzero integers."
How 'bout that -- it would even fit into a book margin!
(Moreoever, those pesky 'lemmas' used in steps (3) and (4) are 'intuitively
obvious' -- everyone has known for decades that elliptic curves over the
rationals are modular, and that curves with a discriminant which is a
perfect (2p)-th power are strange.)
dave