Re: JSH: The "Published" paper he dosen't what you to know about.


"marcus_b" <marcus_bruckner@xxxxxxxxx> wrote in message
news:1158630475.008443.214370@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Sue wrote:
Paper published and withdrawn paper by crank/crackpot James Harris
exposes
the "Over Iinterpertations of Galois Theory" by mathematicians.
http://www.ne-plus-ultra.net/pubs/beckwith_factorization.pdf

ADVANCED POLYNOMIAL FACTORIZATION
James Harris
A. W. Beckwith
Department of Physics and
Texas Center for Superconductivity
University of Houston,
Houston, Texas 77204-5005, USA,
Fermi National Laboratory
Batavia, IL 60510-050

Abstract.
Algebraic method for determining distribution of factors within a
polynomial
factorization, which breaks through what was seen as a barrier from over
interpretations of Galois Theory

Correspondence:
A. W. Beckwith: projectbeckwith2@xxxxxxxxx
PACS numbers : 02.10.-v, 02.10.De, 02.10.Ox , 02.20.Bb

A.M.S. (MOS) Subject Classification Codes. 11R04,11R09
KeyWords 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,

65x^3 - 12x+ 1 = (a1x + 1)(a2x + 1)(a3x+ 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- c.
_

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 - 3*m^2*f^2 + 3*m)*x^3 ? 3(?1 + m*f^2)*x*u^2 + u^3f)

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 f2.

That expression comes from expanding (v^3+1)*x^3-3v*x*y^2+y^3, using the
substitutions

v = -1+m*f^2, and y = u*f,

where additional variables provide an additional degree of freedom.

Now consider the factorization

P(m) = (a1*x + u*f)(a2*x + u*f)(a3*x + u*f)

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 + 3*m)

so

a1*a2*a3 = mf^2*(m^2*f^4 - 3m*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*(3*x*u^2 +
u^3*f)
and also that P(0)/f^2 = 3*x*u^2 + u^3*f, which is coprime to f.

Then I have the factor of P(m), g1, where g1 = a1*x+u*f, where here I
also
have that a1 is not coprime to m.

From my factorization lemma, I have that, when m =0, g1 = c = u*f,
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 + u*f 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^6-3*m^2*f^4 + 3*m)*x^3-3*(?1+ m*f^2)*x*u^2 + u^3 = 65*x^3-2*x+
1

which may be more easily seen from using

v = -1 + m*f^2 = 4, y=1 with (v^3 +1)*x^3 ? 3*v*x*y^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.

Another version, different in detail, not co-authored by Beckwith,
appears in a webzine called 'Chiaroscuro':

http://www.etienne.nu/isis/Chiaroscuro8.pdf

The date is July 2005. The first sentence is: "Over a hundred years
ago subtle error crept into number theory."

A variant of Harris's deeply flawed argument involving constant
terms of functions is still present.

Is this the version that was sent to the Annals of Mathematics?

Marcus.


It has a conclusion the other one does not, so your reference may be it.


.

Relevant Pages

  • Re: Overinterpertations of Galois Theory Exposed
    ... ADVANCED POLYNOMIAL FACTORIZATION ... While you know that the algebraic integer factors are themselves factors ... that are themselves polynomials. ... Who is this A. W. Beckwith ...
    (sci.math)
  • Re: Overinterpertations of Galois Theory Exposed
    ... ADVANCED POLYNOMIAL FACTORIZATION ... While you know that the algebraic integer factors are themselves factors ... that are themselves polynomials. ... Here f is a non unit, non zero algebraic integer coprime to 3 and x, and u ...
    (sci.math)
  • Re: Secret paper that was published, then withdrawn
    ... ADVANCED POLYNOMIAL FACTORIZATION ... While you know that the algebraic integer factors are themselves factors ... polynomial that are themselves polynomials. ... Here f is a non unit, non zero algebraic integer coprime to 3 and x, and u ...
    (sci.math)
  • Re: Galois Theory ok, weird lie from sci.math
    ... > I looked over Galois Theory as I hadn't really considered it before, ... > and noted that it's a field theory. ... However,the result applies to polynomials ... algebraic integer coefficients, and then claims a year later ...
    (sci.math)
  • Re: JSH: The "Published" paper he dosent what you to know about.
    ... the "Over Iinterpertations of Galois Theory" by mathematicians. ... ADVANCED POLYNOMIAL FACTORIZATION ... Determining the distribution of factors within irrational algebraic integers ... that are themselves polynomials. ...
    (sci.math)
Loading