Re: Amateur takes on Wiles's work

From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 09/11/04 

Date: Sat, 11 Sep 2004 22:26:58 +0000 (UTC)

In article <uop0d.18589$H23.10895@newssvr29.news.prodigy.com>,
W. Dale Hall <mailtowd-hall@pacbell.net> wrote:

 [.quoting me.]

>The contradiction? Here's from Arturo Magidin:
>
>
> From: magidin@math.berkeley.edu (Arturo Magidin)
> Newsgroups: sci.math
> Subject: Re: JSH: Assocation does not prove
> Date: Mon, 30 Aug 2004 22:39:16 +0000 (UTC)
>
>... a buncha stuff deleted ...
>
> Again, I am using the PDF copy at
> www.eleves.ens.fr/home/rrichard/wiles.pdf

   [.snip.]

> On pp. 542, line 17 through pp. 544 line 3 is the proof that all
> semistable elliptic curves are modular.
>
> Assume that the given curve E is semistable, and is non-modular.
> Since the argument proceeds by cases, we must change this proof
> a bit. That is, currently, the argument is
>
> (a) Either X or not(X) happens;
> (a.1) If X, then E is modular;
> (a.2) if not(X), then E is modular.
>
> If we assume that E is non-modular, then we need to change the
> first part of the proof to
>
> (a) Either X or not (X) happens;
> (a.1) If X, then E is modular,
> (a'.1) Therefore, not (X).
>
> (a.2) If not(X), then E is modular.
> (a.3) not(X) (from (a'.1).
> (a.4) Contradiction. Therefore, E is modular.
>
> "X" here is "the representation constructed from E on E[3] is
> irreducible".

Perhaps a better summary is as follows: The original proof of this
theorem is by cases.

  Either the representation we get from E on E[3] is irreducible,
  or not irreducible.

     (a.1) If it is irreducible, then the work done previously in the
           paper shows that E is modular.

     (a.2) If it is not irreducible, then:

           (b.1) If the representation we get on E[5] is irreducible,
                 then again, by the work done before, E is modular.

     (a.3) The representation on E[5] will be irreducible, by work
            done previously, unless there exists an elliptic curve F
            which has certain properties, X, Y, and Z.

     (a.4) Again, by work done previously, one can prove that any
           elliptic curve F which has properties X and Y will
           necessarily fail to have property Z.

     (a.5) Since there are no elliptic curves with properties X, Y,
           and Z, it follows that (b.1) holds. Thus, E is modular
           wehther the representation is irreducible or not.

If we attempt to turn this proof into a proof by contradiction, as
noted this will only "reveal itself" in (a.5), since step (a.4) does
not depend in any way on the original curve E. We are forced to assert
that the putative elliptic curve F exists in (a.3), but we conclude it
does not in (a.4). Hardly illuminating. 

-- 
======================================================================
"It's not denial. I'm just very selective about
 what I accept as reality."
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Arturo Magidin
magidin@math.berkeley.edu
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