Re: Attacking my algebraic integer work

From: Nora Baron (norabaron_at_hotmail.com)
Date: 03/29/05 

Date: 28 Mar 2005 17:33:46 -0800

jstevh@msn.com wrote:
> William Hughes wrote:
> > jstevh@msn.com wrote:
> >
> > <snip>
> >
> > > That a new paper is at the Annals is not overwhelming unless they
> > > accept it.
> > >
> > > And several months with them doesn't necessarily mean much, as
> > journals
> > > can keep papers for quite some time, only to reject.
> > >
> >
> >
> > You can take this a little further. By now several people at the
> > Annals
> > must have seen the paper. A short simple paper that purports to
> > overturn
> > huge amounts of modern mathematics. And yet there have been no
> > comments,
> > not expedited publishing, no rumors, no didley squat. Doesn't
> > look good.
> >
>
>
> I agree.
>
> > <snip>
> >
> > > Yup, you heard me right. I want to consider the position that I
am
> > > wrong, with the assumption that posters claiming I'm wrong are
> right,
> > > in order to go through their positions carefully to see if that
> > > assumption is true.
> > >
> > > This thread will be for that purpose.
> > >
> > >
> >
> > A key part of your arguments has been that dividing through
> > by a non-constant factor cannot lead to a counterexample as
> > this would involve changing the value of the constant term. In
this
>
> I'm not sure about your claim here about that being a key part of my
> argument.
>
> > context it has been noted that the question
> >
> > Given a(0) = 0, w(x) non-constant and w(0) = 1, what
> > is the constant term of
> >
> > g(x) = a(x)/w(x) + 7/w(x) ?
> >
> > is critical . However, you have ignored this question.
>
> If g(x) is a continuous function, lies in the ring of algebraic
> integers, and w(x) is a continuous function in that ring, which is a
> factor of g(x) in that ring, then
>
> g(0) = 7.
>
> I use the ring of algebraic integers here as it's simpler.
>
> You can move outside of that ring to consider constant terms, but
I'll
> start simply and see how it goes.
>

  Mr Hughes is right that this is close to the central issue
in your paper "Advanced Polynomial Factorization".

  However, some of the context has been lost. You were
considering a *function* of the integer variable m, but
factored as a *polynomial* in x. Explicitly,

[1] P(m) = (a1(m)*x + uf)*(a2(m)*x + uf)*(a3(m)*x + uf).

The functions a1(m), etc., are not polynomial functions. You
repeatedly noted at the time how remarkable it was that you were
pioneering a method of factoring a polynomial with nonpolynomial
functions.

  You specified that the function P(m) was divisible by f^2. We
noted that one must assume that when f^2 is divided out of the right-
hand side of [1], it may divide out in a *variable* way. This is a
reasonable assumption, because when m = 0, two of a1(m), a2(m),
and a3(m) were 0 [P(0) was a first-degree polynomial in x] but
when m = 1, P(m) was a 3rd degree polynomial in x. Therefore
a1(m), a2(m), and a3(m) *cannot be constant functions*.

  Now say f^2 = v1(m)*v2(m)*v3(m), and that this decomposition
of f^2 is such that

         (ai(m)*x + uf)/vi(m) = (ai(m)/vi(m))*x + u f/vi(m)

is a factorization with algebraic integer coefficients for
i = 1, 2, 3. That is , ai(m)/vi(m) and f/vi(m) are all
algebraic integers. The existence of such functions vi(m) is
guaranteed by a theorem of Dedekind.

  Here is what you said about this on Oct 10, 2004:

----------------------------------------------------------------------------
<Quote from Oct 10 post:>

 http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1086485&
 messageID=3408408#3408408

  Well, let's put that to the test "Nora Baron" and see if the algebra
  reaches you.

  Consider a_1 x + uf, and you claim it has a factor that is v_1(m)
  that varies with m, now just divide through by that factor and
  you get

  a_1 x/v_1(m) + uf/v_1(m)

  and you can see that the term constant with respect to m is now
  uf/v_1(m), which is a basic contradiction, as now it *varies*
  with m.

Understand?

<End of quote>
-----------------------------------------------------------------------------

  It was clear right there where you were making your mistake: You
had forgotten your own definition of *constant term*. You reasonably
defined the constant term of a function h(m) to be h(0). A nice,
simple definition.

  Now if you *correctly* apply *your own* definition to

      a_1 x/v_1(m) + uf/v_1(m)

then, because a_1(0) = 0, you obtain

      uf/v_1(0)

NOT, as you said on Oct 10 [above]

      uf/v_1(m).

  Once you realize that, all the rest of your "constant terms
must remain constant" argument - which, after all, reduces just to
that trivial tautology which buys you nothing - comes unravelled.

  But apparently you never got it.

  The major conclusion was, you CAN factor f^2 out of the right side
of [1] and still have all the coefficients be algebraic integers.
The factorization *itself* is a function of m. You do NOT need to
invent a new ring, as you incorrectly inferred, to achieve such
a factorization.

  Understand?

  Want to try again?

  Nora B.

> > In your new critical phase will you answer the question?
> >
> > - "William Hughes"
>
> Well, I'll try.
>
>
> James Harris

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