From: ah170@FreeNet.Carleton.CA (David Libert)
Subject: Re: Are the "experts" lying? re: Simple FLT proof
Date: 17 Oct 2000 07:55:21 GMT
Newsgroups: sci.math
Summary: [missing]

  I have just written a couple of articles showing strange examples of
polynomials over rings of coefficients with zero divisors.

  These examples led me to reconsider my claimed proof of Eisenstein's
Criterion as written a couple of articles ago in this thread.

  I noted there that both Lang and Hungerford stated this as being over
a UFD of coefficients, and that in restating it I only seemed to need
commutative coefficients for the proof.

  I can expand on this point.  Hungerford states the criterion in terms
of having an irreducible element in the coefficient ring with all the
assumed divisibility properties with respect to coefficients of the
polynomial.  Lang states it as a prime element.

  The details of the proof actually use a primeness property of this
element.  In a UFD every irreducible is prime, so that is no problem in
Hungerford's proof.

  I stated Eisenstein's criterion using primeness instead of
irreducibility, so I didn't need to use the UFD premise to pass from the
irreducibility stated to the primeness to be actually used.

  So I thought this way I was ok just assuming commutative coefficients.

  But I have just noticed another point.  Given polynomial f  and an
assumed factoring   f =  g*h,  late in the proof, having made if
necessary a case dependent interchange of g and h, I proved from using
the assumed divisibilities that  degree f = degree g. 

  So far that is fine.  But then I next concluded that  h  is therefore
a constant polynomial, and then went on to argue that case.

  Why must  h  be a constant polynomial?  Suppose  h  has non-zero
coefficient of power >0.  Then this would seem to multiply by the
leading coefficient of  g  to give a high coefficient to f, beyond
degree of f.  But with zero divisors  we could have such a high
multiplication still yielding a high zero coefficient for f, hence
degree f  manages to stay as low as  degree g,  even with non-constant
h.

  Both Lang and Hungerford have this covered.  Hungerford assumes the
coefficients are a UFD, and Hungerford's definition of UFD includes
being an integral domain, which includes no zero divisors.

  Lang does the same thing, but with different names.  Lang calls UFD 
factorial  and calls integral domain entire.

  So my statement of  Eisenstein's Criterion should include that S  the
ring of coefficients has no zero divisors.

  With this  degree g = degree f  ->  h is constant and the proof is
saved.

  Note all the uses I made of Eisenstein's Criterion were for rings of
coefficients assumed to have no zero divisors, so the applications are
not affected by this.


--
David Libert     (ah170@freenet.carleton.ca)
1. I used to be conceited but now I am perfect.
2. "So self-quoting doesn't seem so bad."  -- David Libert
3. "So don't be a morron."  -- Marek Drobnik bd308  rhetorical salvo IRC sig