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

jstevh@xxxxxxx wrote:
marcus_b wrote:
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Paper published and withdrawn paper by crank/crackpot James Harris exposes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.

Yes it was.

They told me it was received and was to be reviewed.

I waited and waited, expecting them to follow rules and this would all
be over, but after six months with no word, I contacted Princeton.

I was told that a rejection had been emailed a month after submission.
I never received any rejection. I have still never received a
rejection to this day.

And I was was given no reason for a rejection as the person who told me
this, also told me she just found it noted in the database.

I don't give a lot of details about who I contacted as she is someone I
trust, and she was not the one who put the entry in the database--she
was on vacation when it was done--but was the person who told me the
paper was to be reviewed.

A nice person who may have received a nasty wake-up call about the
Princeton math department with my story. Sometimes even the supposedly
best of the best are just not.

I never received a rejection from Princeton.

Readers who check out that paper can see how I made certain to handle
the technical objections raised by posters by simply noting exactly
what the argument does.

Other readers should note that no mathematical objection to any of my
work has been raised in this thread.


Oh, well, let's remedy that. First, some trivialities -

1. Look at the line

P(0) = f^2 (3xu + u^3 f).

That's an error. From the previous definition of P(m),
it should be

P(0) = f^2 (3xu^2 + u^3 f).

(This is a careless minor error which does not affect
the rest.)

2. In the Conclusion section there is the equation

a^3 + 3^2 - 2 = 0.

This too is an error. It should be

a^3 + 3 a^2 - 2 = 0.

Another sloppy minor error. There are others - e.g.,
font changes in the middle of a paragraph, misplaced
exponents, the "word" 'fitself', etc.. All trivialities,
but they are signs of how carelessly this paper was
written.

3. Now for the main event. You are talking about a polynomial
P(m) [actually the polynomial variable is x, but you
want to emphasize another variable m], degree 3 in x,
which is divisible by f^2 and which is such that

P(0) = u^2 (3x + uf),

where u and f are algebraic integers. You start with
a factorization of P(m) of the form

P(m) = (a1(m)x + uf)(a2(m)x + uf)(a3(m)x + uf),

where a1(m), a2(m), and a3(m) are non-polynomial
functions of m.

You note that since when m = 0, P(m) has degree 1
in x, it must be the case that

a1(0) a2(0) a3(0) = 0, implying

that at least one of these three functions is zero
when m = 0. (In fact, at least two of them must be zero,
since otherwise P(0) would be of degree 2 in x.)

You choose that a1(m) is one of the coefficient
functions which is such that a1(0) = 0. You then argue
that a1(m) must be divisible by f for all m. The
same must be true for another of the coefficient functions
also (choose a3(m)). Since P(m) is divisiible by f^2
but not by any higher power of f, you conclude that
the third coefficient function is not divisible by f
and in fact is coprime to f.

The ring in which this coprimeness condition exists
is not specified. The only ring mentioned is the ring
of algebraic integers, but you do not say explicitly
that the coprimeness occurs in that ring.

However, later in the paper you note that in the ring
of algebraic integers, your result contradicts Galois
theory. If u = 1, the coefficient functions a1(m),
etc., must satisfy the equation

a^3 + 3(-1 + mf^2)a^2 - f^2(m^3f^4 - 3m^2f^2 + 3) = 0

and if the polynomial here is irreducible, Galois
theory says that none of the roots are coprime to f
*** in the ring of algebraic integers. ***

From which you conclude: Galois theory is wrong.

Specifically, the final sentence of your paper is:

"Unfortunately, Galois Theory as often taught would
indicate that the ring of algebraic integers is
correct, and not algebra, requiring that none of
roots be coprime to f, and the alternative - going
with algebra - indicates that Galois Theory as
taught is incorrect, with major implications."

Which must mean that you are claiming that your result
holds *** in the ring of algebraic integers.***
(Actually, since you do not specify a ring, one
might conclude that you are claiming it
holds in ANY ring where the numbers involved can
exist.)

There is another route to proving that none of the
roots of the equation above are coprime to f (if the
polynomial is irreducible). That follows from the
use of a theorem of Dedekind in algebraic number
theory. That theorem also applies to the ring of
algebraic integers.

So (as you have said elsewhere), you are saying that
Dedekind's theorem is wrong also. That is why at the
beginning of your paper, you say "Over a hundred years
ago subtle error crept into number theory."

One of the simpler cases to which your result should
apply is when f = sqrt(5). In that case, the polynomial
equation above becomes

a^3 + 12 a^2 - 65 = 0.

This is a monic polynomial with integer coefficients.
It is irreducible over the rationals. The roots
are algebraic integers. Dedekind's theorem
and Galois theory both say that none of the roots are
coprime to f^2 = 5 in the ring of algebraic integers.
You however infer that one of the roots IS coprime
to 5. And since you are saying that Galois theory is
wrong, you must be saying that it is wrong in the ring
to which it applies: the ring of algebraic integers.
From which we must conclude: you think one of roots is
coprime to 5 *** in the ring of algebraic integers.***
This is inescapable.

So your result does indeed just flatly contradict
two well-known proven classical results in number
theory in this example.

But there is a ANOTHER way to prove that none of
the roots of the simple equation just above are
coprime to f = sqrt(5) (or equivalently, coprime
to 5). Dale Hall provided explicit numerical results
which showed this. He reduced it to just plain
arithmetic. You don't need Galois theory or Dedekind's
theorem to do the computations. All you need
is a pencil.

Which implies in a THIRD, independent way that
your results in the Chiaroscuro paper are wrong.
They contradict elementary arithmetic.

Unless you don't trust your pencil!

It's all goddamn politics. It's so sad, but so effective.


It's effective because it is right and anyone can
check it for themselves. Politics are not involved.


Most of you are so easily manipulated by people who don't give a damn
about mathematics.


You wish.

Mathematicians have decided to hold on to flawed ideas that I easily
and almost trivially prove are wrong because our academic world is
about letting small-minded people live political lives where when the
truth hurts, they can just go on teaching whatever they fee like.


The opposite is true. You prove nothing, almost trivially
or otherwise.


An entire goddamn mathematical journal died over this and it did not
matter.


It died most likely because it was very badly edited.
Its temporary publication of your paper was a symptom of
how incredibly bad a job it was doing. There were a
number of others.


Some of you may delude yourselves that you have hope for your own
ideas, but reality is, if your mathematical research passes the career
test--as in it does not negatively impact the careers of some
"important" mathematicians somewhere in the world--you may be allowed
to have your research accepted.

You MAY be allowed.

My story gives them even more reason to be arbitrary and political as
it tells mathematicians around the world just how much they can get
away with, and how much protection they'll get from people like the
posters who reply to my threads.

You may be allowed to be known for your own research, but hey, think
about it, the game is fixed.


The only "fix" is in the direction of rigorous valid
mathematics. That is the criterion that has repeatedly
tripped you up.

Marcus.


James Harris

.

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