Re: JSH: Same old crap

jstevh@xxxxxxx wrote: 
I'm stuck with few options with you journal killers having alerted the
math world to me, while getting one of my papers yanked.


Didn't you submit a version of your SWJPAM article to the Annals?
Did anyone interfere with that (or any) submission?
Why do you imagine you have few options?

So I'm stuck with Usenet, and I see the same old crap, but you don't
understand, I'm going to push towards resolution and note areas of
agreement.


By "the same old crap", you seem to mean that the people who engage
in these threads refuse to accept your arguments as being valid. It
does your case no service to engage in provocative language:

	1. To my understanding, no one is lying about mathematics.

	2. To claim that Barry Mazur's career is soon to be
	   ended in disgrace (as you did in a recent article
	   on sci.crypt) is to invite ridicule. His career may
	   be completed soon (he is close to 70 years old), but
	   the claim that he is engaged in a conspiracy to hide
	   long-standing errors in mathematics that you have
	   exposed is the stuff of delusion on your part.

I'll point out that the provocative language that comes your way
is also objectionable (to me); it is in large part provoked.

For instance, in the ring of algebraic integers for non-rational roots
f is not coprime to them in that ring.

That's not an area of disagreement.

Claiming that it disproves my examples is false as well as my argument
shows logically how results in the ring of algebraic integers
necessarily lead to an apparent contradiction with what follows
algebraically.


You have said "two roots divisible by f, the other coprime to f", and
that simply will not hold for an irreducible cubic. The fact is that
every number coprime (in the ring of algebraic integers) to any of the
roots will be coprime to all the roots. This cannot be violated in any
integral extension of Z that contains the roots, since once numbers are
coprime in some ring they remain coprime in any extension of that ring.

The so-called contradiction is apparent only to you. Your arguments
are sufficiently vague as to mask the evident errors that make the
situation in the ring of algebraic integers appear to you to be contradictory. Make your arguments more complete, and the errors
will surely emerge.

Like you can prove a number has f as a factor, and also prove that
number does not have f as a factor in the ring of algebraic integers,
so when I say prove a number has f as a factor, then, in what ring?


Proving that one number is divisible by another is a matter of showing
membership in the ideal generated by that other number; showing that
it isn't divisible by that number is similarly a matter of showing
that the first is not in that same ideal.

Your question suggests your problem. It's up to you to provide a suitable ring (or to provide a family of rings defined by some
general property, and to prove that this family is nonempty),
and then to prove that your results hold in any member of this
family. Your methods don't seem to be adequate to prove this.

What is evident from your assertions of an "object ring" is that
any such ring must fail to include all roots of integral polynomials
for which it contains one root (such rings surely exist). It is
puzzling to imagine why such a ring should be preferable to a ring
in which the presence of all such roots can be shown.

First though, it's best to consider how a number can be algebraically
proven to have f as a factor.


OK, then go to the definition, and prove the existence of the number
that yields the complementary factor, in the ring in question. That's
how to do what you're considering. So far, I haven't seen any such
proof.

What do I mean by "algebraically proven" to have a factor?


No one really knows what you mean. That's why you need to be
clear in exposition. For example, the recent claim that some
fact about a rational root of a monic polynomial with integer
coefficients can in any way contradict Ideal theory or algebraic
number theory or Galois theory shows that you fail to understand
even the basic elements of these areas.

Your recent remark:

	SEE how the mistake works and why it's so important
	for interpretations of Galois Theory to have irreducible
	cases where the ambiguity of the radicals hides the
	solutions somewhat?

again shows your ignorance. No mathematician who knows algebra is
at all surprised that the roots of reducible polynomials have different
properties than the roots of irreducible polynomials. At the very least,
different irreducible factors of a reducible polynomial are completely
independent: there is, in general, NO relationship among the roots of
different irreducible factors. After all, I can take arbitrary
irreducible polynomials p,q,r and produce reducible polynomials

	pq, pr, qr

and since there is no general relationship among the roots of p,q,r
separately, there can be no general relationship that connects the
roots of pq (or pr or qr) as a single entity. However, there is a
tight relationship among all the roots of an irreducible polynomial;
this tight relationship is made explicit by Galois theory.

In fact, your interpretation of ambiguity of radicals is a naive
attempt to speak to the fact that the Galois group permutes roots.
Galois theory is far more general, in that it is not restricted to
even-order radicals (as your comments have been), but relates to
roots of arbitrary irreducible polynomials. To be sure, one can
treat reducible polynomials, but the action of the Galois group
is necessarily one that acts independently on the roots of the
separate reducible factors.


James Harris


Dale.
.

Relevant Pages

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