Re: Amateur takes on Wiles's work
From: Andrzej Kolowski (akolowski_at_hotmail.com)
Date: 08/30/04
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Date: 29 Aug 2004 19:15:45 -0700
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0408290713.3a80d92c@posting.google.com>...
>Now I'm not a professional mathematician. I do post about math on
>Usenet, but that's not an indication of expertise!
>
>I'm not beholding to any mathematical interests though, so I feel no
>compulsion to protect a favored Golden Calf of the modern math world,
>which is an argument that supposedly proves something by one Andrew
>Wiles, which I fear doesn't, and I'll say exactly why I say it
>doesn't.
>
>It'll be up to others to answer the charge, dismiss it, or consider
>that I might be right.
>
>First off despite the assertions of great complexity to the area what
>mathematicians initially noticed isn't that complicated:
>
>They had these things they called modular forms, and these things they
>called elliptic curves, which didn't seem at ALL related.
>
>But there are these 4 numbers that you can get from elliptic curves,
>and find modular forms with the same 4 numbers. Those numbers are
>kind of like a description.
>
>So there's some way that modular forms and elliptic curves could have
>the same description!
>
>Mathematicians would check various elliptic curves and find they could
>always find some modular form to associate with it.
>
>Taniyama and Shimura conjectured that there was a pattern here that
>held, as in fact modular forms and elliptic curves WERE related in
>some deep way, and that what mathematicians were noticing wasn't just
>one of those intriguing coincidences.
>
>But you have the setup for a logical fallacy called Cum Hoc, Ergo
>Propter Hoc, where people see what looks like a pattern, and leap to a
>conclusion, though at this point mathematicians were ok, as it was
>only a conjecture.
>
>It took Andrew Wiles coming in, with an attempt at proof by
>association for the logical fallacy to fully take hold.
>
>The problem for many of you with such a charge is that it can seem
>esoteric. I've had two posters on sci.math where I've discussed this
>for a while actually come back to claim that Cum Hoc, Ergo Propter Hoc
>is about time, so it can't appy to mathematics!!!
>
>But notice, it's actually about false implication, where you see a
>pattern, and your mind plays a trick on you and tells you that the
>pattern is proof of itself!!!
>
>To date, while mathematicians now apparently mostly believe the
>Taniyama-Shimura Conjecture, they can't give you a reason why, or can
>they?
>
>It turns out that if the charge of Cum Hoc, Ergo Propter Hoc is itself
>challenged, the next proper step is to ask for a null test.
>
>What is a null test?
>
>A null test is to go through the argument under challenge with the
>assumption that its conclusion is false, and find a contradiction with
>that assumption!
>
>You see, math proofs begin with a truth and proceed by logical steps
>to a conclusion which then MUST BE TRUE.
>
>But the conclusion follows from the previous steps in the proof, so
>any challenge to the conclusion must contradict a previous logical
>step, or the truth with which the proof begins.
>
>Math proofs are perfectly logical.
>
>There is no way for a math proof to fail a null test.
>
>It is just not logically possible.
>
>Therefore, any math proof can be challenged by assuming the opposite
>of its conclusion, and tracing through it until you reach the logical
>step where you end up with a contradiction.
>
>The resolution to the contradiction, if you have a proof, is that your
>assumption is false and the conclusion IS true!
>
>It's neat. It's beautiful. It's just cool.
>
>Notice also that the null test, which can be requested whenever, and
>not just when you have a case of Cum Hoc, Ergo Propter Hoc, is a great
>way for someone who is not an expert in a particular feel to find a
>limited area to check.
>
>For instance, with my challenge to Wiles's work, someone should find a
>single logical step where the assumption of a non-modular elliptic
>curve will cause a contradiction, and be able to give the exact
>section in his work where it occurs!!!
>
>Then they can explain why it occurs and despite the entire work being
>hundreds of pages you have the ability to look at the crucial link
>without going through the entire thing.
>
>You could call that logical step the keystone.
>
>I'm asking for someone to produce the keystone in Wiles's work, which
>will ring out loud and clear if you assume the existence of a
>non-modular elliptic curve.
>
>Let the full challenge--with witnesses now from alt.math.recreational
>and others throughout the world through the Internet--begin.
>
>
>James Harris
I like this idea of a "null test". Here is how I understand it.
You take the conclusion of a given proof. You state its negation.
You then go through the proof line by line to see if anything
contradicts the negation.
If there is *no* step that contradicts the negation, then the
proof must be wrong.
If there *is* a step that contradicts the negation, then it
is still possible that the proof is wrong. You just have
slight and nonconclusive evidence that it isn't.
So the "null test" is somewhat of a one-sided test, but
still useful.
Applying it to Wiles' 100+ page long and very difficult proof is
going to be pretty hard. Besides, you may get all the way
to the end and then find that the concluding statement is
the only one that contradicts the negation, and that would
not tell you much of anything.
So I would like to try it with a shorter proof.
Below is a copy of the paper "Advanced Polynomial Factorization"
by James Harris, exactly as it appeared in the Southwest
Journal of Pure and Applied Mathematics:
=================================================================
[begin paper]
Electronic Journal: Southwest Journal of Pure and Applied Mathematics
Internet: http://rattler.cameron.edu/swjpam.html
ISSN 1083
Issue 2, December 2003, pp. 6--8.
Submitted: July 25, 2003. Published: December 31 2003.
ADVANCED POLYNOMIAL FACTORIZATION
James Harris
Email Address: jstevh@msn.com
ABSTRACT. Algebraic method for determining distribution of factors
within a polynomial factorization, which breaks through what was seen
as a barrier from overinterpretations of Galois Theory.
A.M.S. (MOS) Subject Classification Codes. 11R04,11R09
KEY WORDS AND PHRASES. Polynomial factorization, Galois theory,
Factorization lemma, Ring of algebraic integers
ADVANCED POLYNOMIAL FACTORIZATION APPROACHED.
Determining the distribution of factors within irrational algebraic
integers has long been considered impossible as it is not possible to
do using Galois Theory. However a simple technique through the
introduction of more variables makes it possible. To highlight the
standard belief consider the algebraic integer roots of x^2 + x - 5.
While you know that the algebraic integer factors are themselves
factors of 5, can either not have non unit factors of 5? How do you
know?
In looking to consider distribution of algebraic integer factors
within a factorization I'll be using a more complicated example than
x^2 + x - 5.
This paper will show, using basic algebraic methods, that given the
factorization, in the ring of algebraic integers,
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
one of the a's is coprime to 5.
First I'll need a simple lemma to generalize beyond factors of a
polynomial that are themselves polynomials.
FACTORIZATION LEMMA:
Given a factor g of a polynomial P(x), further defined as a factor for
all x, which means that the value of g for a value 'a' of x is a factor
of P(a), within the ring of algebraic integers, there exists r and c
such that
g = r + c
where r=0, or varies as x varies, and c is a factor of the constant term
P(0) and is itself constant.
Let x=0, then g must be a factor of P(0), so at that point c = g.
If when x does not equal 0, g=c, r=0. If when x does not equal 0, g!=c
there must exist r which varies with x. That is, r=g. []
As an example consider sqrt(x + 1) which is a non polynomial factor of
x+1, and while there are an infinity of irrational solutions consider
the rational solution at x=35.
Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,
and c=1. But for different values of x, g and r will vary, while c will
not.
PRIMARY ARGUMENT.
Given
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
in the ring of algebraic integers. Let
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f)
Here f is a non unit, non zero algebraic integer coprime to 3 and x,
and u a non unit, non zero algebraic integer coprime to f. Note P(m)
has a factor that is f^2 .
That expression comes from expanding (v^3 + 1) x^3 - 3 vxy^2 + y^3,
using the substitutions v = -1 + m f^2, and y = uf, where additional
variables provide an additional degree of freedom.
Now consider the factorization
P(m) = (a1 x + uf)(a2 x + uf)(a3 x + uf)
where multiplying out shows that
a1 a2 a3 = m^3 f^6 - 3 m^2 f^4 + 3m f^2 = f^2 (m^3 f^4 - 3 m^2 f^2 + 3m)
so
a1 a2 a3 = m f^2 (m^2 f^4 - 3 m f^2 + 3).
Therefore, at least one of the a's cannot be coprime to m, and at
least one of the a's must equal 0 when m=0.
(Note: The a's are roots of a monic polynomial with algebraic integer
coefficients so they are algebraic integers.)
Notice that the constant term P(0) is given by P(0) = f^2 (3x u^2 +
u^3 f) and also that P(0)/f^2 = 3x u^2 + u^3 f, which is coprime to f.
Then I have the factor of P(m), g1, where g1 = a1 x + uf, where here
I also have that a1 is not coprime to m.
From my factorization lemma, I have that, when m=0
g1 = c = uf
meaning f is a factor of the constant term.
Therefore, exactly two of the a's equal 0, when m=0, to get the factor
f^2 in the constant term P(0), while one must not equal 0, or f^3 would
be the factor.
Now as noted before in general P(m) has a factor that is f^2, and
separating that factor off, gives a constant term coprime to f;
therefore, given g1 = a1 x + uf where with m = 0, g1 gives a factor of
f it must have that same factor in general, proving that two of the a's
have a factor that is f .
Therefore, one factor is coprime to f.
Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't
change the a's, I have
(m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3(-1 + mf^2)xu^2 + u^3 = 65x^3 - 12x + 1
which may be more easily seen from using v = -1 + mf^2 = 4, y=1 with
(v^3 + 1) x^3 - 3vxy^2 + y^3 .
Therefore, with the factorization
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
one of the a's is coprime to 5, which shows where some of the algebraic
integer factors distribute despite the factors being irrational.
[end of paper]
----------------------------------------------------------------------
So the conclusion of this paper is that if the following polynomial
is factored as
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
where a1, a2, and a3 are algebraic integers, then one of a1,
a2, or a3 is coprime to 5.
And the *negation* of this statement is that if the polynomial
is factored as above, then EACH of a1, a2, and a3 is NOT
coprime to 5.
So that's your 'null test' statement.
Clearly it contradicts the conclusion of the paper. It
was deliberately constructed to be the negation of the
statement of the paper.
So evidently the paper basically passes this "null test":
the negation of the conclusion is contradicted by a
statement in the paper. I cannot conclude from this
that the proof in the paper is wrong.
There is just one little problem.
The null test statement is *true*.
You can prove it in at least 4 different ways.
Proofs have been given by Hall, Magidin, Winter,
Decker, and Baron.
This means that a statement in the paper (namely,
the conclusion) is contradicted by a *true* statement.
Therefore that statement in the paper must be
false.
Therefore the claimed "proof" in the paper is
wrong!
Thanks for the null test idea! Works great!
Andrzej
- Next message: John Morriss: "Re: Why to call it pseudorandom?"
- Previous message: Daniel W. Johnson: "Re: The mathematics of Triangulation"
- In reply to: James Harris: "Amateur takes on Wiles's work"
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