Re: Attacking my algebraic integer work
From: Nora Baron (norabaron_at_hotmail.com)
Date: 03/27/05
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Date: 26 Mar 2005 16:49:56 -0800
jstevh@msn.com wrote:
> Nora Baron wrote:
>
> <deleted>
>
> > I got no problem with that! In the most recent available version
> of
> > "Advanced Polynomial Factorization", Harris said:
>
> The paper Advanced Polynomial Factorization has been retired.
>
> Rather than keep with it, mainly as it was too timid in many ways, I
> wrote a new paper and sent that off to the Annals.
>
> >
> > "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."
> >
> > There is no ambiguity there. He clearly states that he is doing
> > his arithmetic in the ring of algebraic integers. One does not
> > need to look at his proof to see that the statement is false. In
> > fact, the statement was disproved in at least 3 different ways
> > before he even submitted the paper to SWJPAM. Since he claims to
> > have proved a false statement, the details of his proof are
> > irrelevant. But we have shown how they are wrong anyway.
>
> I'm not interested in going over work that has been retired, so I'm
> definitely not interested in arguing over what old work may or may
not
> say.
>
So why don't you post the revised paper? Presumably what you are
going to do is simply tell us that the new paper is perfect in all
respects, without even letting us see what it says. If so, it is
exceedingly clear that, far from subjecting yourself to honest
criticism, what you now intend to do is just completely weasel out
of it.
> What I will say is that it is definitely NOT true that given
>
> 65 x^3 - 12 x + 1 = (a1 x + 1)*(a2 x +1)*(a3 x + 1)
>
> in the ring of algebraic integers, one of the a's is coprime to 5, in
> the ring of algebraic integers.
>
> If you look carefully, you will see the statement you give is not
> exactly the same, as it talks about a factorization in the ring of
> algebraic integers, and then simply talks about coprimeness, where it
> seems to me that you might reasonably assume that the declaration is
> still in the ring of algebraic integers.
>
Not only might one reasonably assume it: It is almost certain
that that is what you originally intended. I think you changed your
story later but not the text. I think you submitted the paper
to a journal knowing full well, *as you say above*, that the
editors, reviewers, and readers
"... might reasonably assume that the declaration is still in
the ring of algebraic integers".
Thus the paper as submitted was either (1) false, or (2) deliberately
misleading.
Which was it?
I think what you may now want to say is that, given a
factorization
65 x^3 - 12 x + 1 = (a1 x + 1)*(a2 x +1)*(a3 x + 1)
where a1, a2 and a3 are algebraic integers, two of them are
not coprime to 5 in some yet-to-be-well-defined ring of objects,
while the third IS coprime to 5 in that ring. Can you confirm
this, or is that sticking your neck out too far?
If this is correct, it is difficult to see the point. Your
original goal was to show that one of these was coprime to 5
*in the algebraic integers*, not in some other ring. You must
remember how this all relates back to the first part of your
"proof" of Fermat's Last Theorem. For that crucial step, you
needed that two of a1, a2, and a3 had "factors of" f, while
the third was coprime to f *IN THE RING OF ALGEBRAIC INTEGERS*.
The fact is, the factorization that you originally wanted
can take place entirely in the ring of algebraic integers. There
is no *need* to go to some other ring. The algebraic integers,
for your purposes, are neither incomplete nor deficient. They
may not have a property that you want, but that property is
one that you invented in order to make your argument work. You
have now got the cart before the horse: you have not constructed
an argument to justify a statement; you have constructed a
statement to justify an argument. Essentially you are saying:
"I cannot prove what I originally needed, but if I redefine
what I needed, I can prove that." But it leaves you nowhere
with respect to your original goal. And why is factorization
within the "object ring" even worth mentioning? It is an
artificial and unnecessary construct at best.
Finally: there is a central core to your previous paper. It
involves the generalization from m = 0 (the singular, degenerate,
reducible case) to m > 0 (the irreducible case). You have justified
it by an argument based on the (correct) statement that "constant
terms are constant". I would bet that that is still the central
ingredient in your revised paper, and that it has not changed in
any essential way. Below I have isolated the core of this argument
as a lemma. I want to know if you agree or disagree with the
statement or the proof. If it is wrong, in the spirit of your
current stance regarding criticism, I want to know exactly where
and why, in detail.
------------------------------------------------------------------------------
Lemma. Assume f(m) is a function from the natural integers to
the algebraic integers. Assume that f(m) is divisible by p^2
for all m, but also f(m)/p^2 is coprime to p for all m (where
p is a prime). Assume that there exist two other functions
g(m) and h(m), also from the natural integers to the algebraic
integers, such that
f(m) = g(m) * h(m),
and g(0) = c * p^2, where c is an algebraic integer which is
coprime to p, and h(0) is coprime to p.
Then h(m) is coprime to p for for all values of m, and
g(m) is divisible by p^2 for all m.
Proof:
The *constant terms* of f(m), g(m), and h(m) by definition are
f(0), g(0), and h(0) respectively. Note that
g(m) = g*(m) + g(0), where g*(m) = g(m) - g(0). Similarly
h(m) = h*(m) + h(0), where h*(m) = h(m) - h(0).
We can write
f(m) = (g*(m) + g(0)) * (h*(m) + h(0))
= (g*(m) + c*p^2) * (h*(m) + h(0)).
Note that f(0) = g(0) * h(0) = c*p^2 * h(0).
Therefore
f(m)/p^2 = (g*(m)/p^2 + c) * (h*(m) + h(0)).
Again, equating the constant terms gives
f(0)/p^2 = c * h(0),
which by hypothesis is coprime to p.
Now, f(m)/p^2 and c are algebraic integers, as is
(h*(m) + h(0)).
>>From the fact that f(m)/p^2 is coprime to p for all
m, one must conclude that h*(m) + h(0) is coprime to
p for all m also. But h*(m) + h(0) = h(m). Therefore
h(m) is coprime to p for all m, and g*(m) is divisible
by p for all m. Since g(m) = g*(m) + c * p^2, g(m)
must be divisible by p for all m.
QED.
--------------------------------------------------------------------------
Nora B.
> So let me make this point so that there is no ambiguity, no
> misunderstanding, no room for doubt on this position:
>
> NONE the a's is coprime to 5 in the ring of algebraic integers.
>
> > I think what is going to happen here is the following. Harris
> > will soon return with a post saying, "I tried every possible
> > way to attack my own work, and sure enough, I found that it is
> > without any flaws. It is perfect. I hereby re-declare myself
> > to be the GMOAT (Greatest Mathematician of All Time). I am
> > Ozymandias, King of Kings. Look upon me, ye mighty [i.e., ye
> > Powers That Be], and despair!"
> >
>
> That would be non-productive.
>
>
> James Harris
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