Re: My paper, and the cheaters

From: Nora Baron (norabaron_at_hotmail.com)
Date: 09/18/04 

Date: 18 Sep 2004 11:58:59 -0700

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409171443.51d8ff25@posting.google.com>...
> norabaron@hotmail.com (Nora Baron) wrote in message news:<36024859.0409170732.566a5e55@posting.google.com>...
> > jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409161404.365eaa1e@posting.google.com>...
> > > I've made some pointed criticisms of math society but if you actually
> > > pay attention you'll realize that the evidence against math society is
> > > nastier than I go into detail about usually.
> > >
> > > Supposedly a correct math result is what's important, but despite my
> > > having a quite correct paper, which was to be published, sci.math'ers
> > > managed to get it yanked IN ONE DAY with some hostile emails.
> > >
> > > The paper has been by a lot of editors and mathematicians and to date,
> > > no major error has been found, which I say because there was one minor
> > > error that actually got pointed out by a sci.math poster.
> > >
> >
> > Please do not ever again claim to be an honest person.
> > Your "proof" contradicts known mathematics. At least 4
> > counterproofs have been given. Dale Hall's counterproof
> > can be checked with simple arithmetic. The exact spot
> > in your "proof" where you make your major error has been pointed
> > out repeatedly. You know all this and yet you continue to
> > deny it.
>
> That's not true.
>
> In fact, I don't disagree with much of what you say.
>
> And I'd like you to admit that in a post, so that I can move on to
> what I actually *DO* say.
>
> First though, I need you to read what I have below and acnowledge that
> you see the agreement on a key point.
>
>
> >
> > Below is one of the simplest counterproofs. Feel free
> > to point out errors (assuming you are still interested in
> > math as opposed to social studies). This is, incidentally,
> > for your future reference, what real proof is supposed to
> > look like.
> >
> > ================================================================
> >
> > James Harris has written a paper called "Advanced Polynomial
> > Factorization" in which he claims the following: if the
> > polynomial
> >
> > 65*x^3 - 12*x + 1
> >
> > is factored in the form
> >
> > (a1*x + 1)*(a2*x + 1)*(a3*x + 1),
> >
> > where a1, a2 and a3 are algebraic integers, then exactly
> > two of a1, a2, and a3 are divisible in the algebraic integers
> > by sqrt(5), and the third one is coprime to 5.
> >
> > It should be noted that -a1, -a2, and -a3 are roots of
> >
> > x^3 + 12*x^2 - 65.
> >
> > That Harris's claim is false can be seen from the following:
> >
> > ==========================================================================
> >
> > Assume m(x) is an irreducible monic polynomial with integer coefficients
> > and algebraic integers a and b are two roots of m(x).
> >
> > Theorem. If p is a nonzero rational integer and a is divisible by
> > sqrt(p) in the algebraic integers, then b is also divisible
> > by sqrt(p).
>
> That last should say, in the algebraic integers.
>

  Agreed.

> Theorem accepted. Proof not needed here so I deleted it out to focus
> on what's important. It is in fact true that if 'a' is divisible by
> sqrt(p) in the algebraic integers then b is also divisible by sqrt(p)
> ***in the ring of algebraic integers***.
>
> Concede agreement "Nora Baron" and then I'll explain the rest.
>

  Of course I "concede". That's what I proved.

  The question now becomes, what did you claim to be true in
"Advanced Polynomial Factorization."

  Just below is the full text of APF exactly as it was submitted
to the SWJPAM:

==========================================================================
[Begin APF]

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
Eail 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 APF]
----------------------------------------------------------------------

  It is clear from this text that you were working in the
ring of algebraic integers. For example, you say:

  "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."

  Isn't that EXACTLY what you said in the paper that you
submitted, and about which you are now complaining that
mathematicians have lied, etc. ?

  Over and over again, throughout APF, you refer to factorization
in the algebraic integers. There is no hint that you might be
thinking of any other ring. If in fact you were thinking the
conclusion applied to some other ring, would you not have stated
that? The paper makes absolutely no sense on any level without
such a statement. What is the reader to think if suddenly, without
warningat the end, you are supposedly talking about factorization
and coprimeness in some other ring?

  It is thus utterly clear that that is what you *intended*. Now
I think you want to rewrite history and claim that the whole
theorem was not about algebraic integers, but about numbers in
some other ring which is not even defined, not even MENTIONED,
in the paper!

>
> >
> > This polynomial is monic and irreducible with integer coefficients.
> > By the theorem above, if one of the roots is divisible by sqrt(5),
> > then they all are. This is not possible because the product of the
> > roots is 65 = 5 * 13. Therefore Harris's claim that ANY of the
> > roots are divisible by sqrt(5) is false.
>
>
> The poster has shifted from the true claim--that the result applies to
> algebraic integers--to a far vaguer and more inclusive claim,
> basically that it applies in general.
>

  No, absolutely not. When I said "divisible" above, I meant
divisible in the ring of algebraic integers. Just as you did
in APF when you said one of the a's is coprime to 5. Never in
that paper did you suggest you might be doing arithmetic in
any other ring.

> Repeatedly, posters keep making claims true in the ring of algebraic
> integers, I accept those claims in that ring, and they then act like I
> don't!!!
>

  READ YOUR OWN PAPER. Again, I quote:

  "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."

  Now, when you say "... factorization, in the ring of algebraic
  integers", any sensible person would assume that you mean a1,
  a2, and a3 are algebraic integers. When you say "one of the a's
  is coprime to 5", any sensible person would assume you meant
  "coprime to 5 in the algebraic integers". That is CLEARLY what
  your paper is about. Any reader of the journal in which it was
  published (for a few hours) would assume that. The paper makes
  no sense otherwise.

> I repeat, the result "Nora Baron" gave is correct *in the ring of
> algebraic integers* and I now challenge that poster and others in that
> camp to finally and for once concede that I am not arguing against
> that point.
>

  Certainly you appear to be accepting what I proved. But then you
must logically also accept that it contradicts what you claim to
have proved in APF.

> Then I'd like, with the permission of "Nora Baron" to explain exactly
> what my proof actually says.
>

  Go for it.

> But first I want the poster "Nora Baron" to accept that I am not
> challenging on this specific point.
>

  From what you say, we are in agreement on this point.

> That's the first step.
>
> It's up to that poster to respond.
>

  Consider it done. Your turn.

  Nora B.

Postscript:

  I suspect you now want to invoke the ring of "objects".
In the past you have said that the ring of algebraic
integers is somehow "incomplete" and that that is what
leads to the problem with APF. You have said that your
ring of objects is bigger than the ring of algebraic
integers. That is, the ring of objects *contains* the
ring of algebraic integers.

  Given this, note that:

  (1) Any factorization of a polynomial in the ring
      of algebraic integers is also a factorzation in
      the ring of objects, and

  (2) There does exist a factorization of

          65*x^3 - 12*x + 1

      of the form you claimed,

          (a1*x + 1)*(a2*x + 1)*(a3*x + 1)

      where each of a1, a2, and a3 is an algebraic integer.

      This follows from the fact that if 65*x^3 - 12*x + 1
      is factored as above, then -a1, -a2, and -a3 are
      roots of

                y^3 - 12*y^2 - 65

      which by definition means they are algebraic integers.
      That the latter polynomial does have a factorization is
      guaranteed by the Fundamental Theorem of Algebra.

  Now put (1) and (2) together, and you
conclude that there is a factorization of the polynomial
 
          65*x^3 - 12*x + 1

of the form

          (a1*x + 1)*(a2*x + 1)*(a3*x + 1)

where a1, a2, and a3 are OBJECTS. None of a1,
a2, and a3 are coprime to 5 in the ring of
algebraic integers. Are any of them coprime to 5
in the ring of OBJECTS?

  Take a1, for example. It is not coprime to 5
in the ring of algebraic integers. Therefore there
exist nonunit algebraic integers s, t, and w such that

        a1 = s * w
         5 = t * w.

  Of course w is the common factor of a1 and 5 in the
algebraic integers.

  But since s, t, and w are algebraic integers, they are
also OBJECTS. Therefore w is also a common factor of
a1 and 5 in the ring of OBJECTS. That is, a1 and 5 are
also not coprime in the ring of OBJECTS.

  The same holds true for a2 and a3.

  This contradicts your statement in APF, *even if you were
talking about coprimeness in the ring of Objects.*

  Again, I must point out: your statements in APF are crystal
clear. You were NOT talking about any "ring of objects". The
term objects was not even mentioned. You were doing all your
algebra in the ring of algebraic integers. The paper was incoherent
unless that is the ring in which your conclusions applied.

 
>
> James Harris

Relevant Pages

  • Re: JSH: The "Published" paper he dosent what you to know about.
    ... I'm not sure what he wants from his "object ring", ... so the other factors needed to be coprime to f --- in the ring ... happen in the ring of algebraic integers. ... Dedekind's work and the theory of ideals. ...
    (sci.math)
  • Re: JSH: Resolution now possible
    ... My position is that the definition of the ring of algebraic ... >properly units that are excluded on the technicality that they are ... The ring of algebraic integers was defined not arbitrarily, ... From that you conclude that a factorization of the desired ...
    (sci.math)
  • Re: Attacking my algebraic integer work
    ... > The paper Advanced Polynomial Factorization has been retired. ... >> his arithmetic in the ring of algebraic integers. ... not coprime to 5 in some yet-to-be-well-defined ring of objects, ...
    (sci.math)
  • Re: JSH: Brainstorming over, for now
    ... You requested a proof that one could prove that an algebraic number could be expressed as the quotient of two coprime algebraic integers. ... >> So consider what follows from the ring of algebraic numbers, ... You're right, factorization can be done by infants using ordinary household materials, but we're really save, and the NSA doesn't need to get in on the game? ...
    (sci.math)
  • Re: JSH: Blocking and ring of algebraic integers
    ... with the factorization that IS blocked you also get ... algebraic integers as well, when the g's are non-rational with non- ... in the ring of algebraic integers with the f's not polynomials nor ...
    (sci.math)