Re: JSH: Attacking the conclusion
From: W. Dale Hall (mailtowd_hall_at_pacbell.net)
Date: 06/17/04
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Date: Thu, 17 Jun 2004 19:16:13 GMT
James Harris wrote:
> ANYONE can attack any math paper by just going to the conclusion and
> claiming it's wrong.
>
Anyone can claim anything. This is not a valid argument.
If you mean that a conclusion of any math paper can be proven
to be false, then you have made a specific claim that you
cannot support.
Your best response is simple: select a paper in mathematics
that people generally agree is valid, find any conclusion,
and prove that the conclusion is false.
Make a claim, support that claim. It's pretty simple,
isn't it?
> It's a cheat. It's a dodge. It's specious.
>
> Claims of counterexamples against my work have by me repeatedly been
> shown to be false.
>
Really? You have shown that the polynomial products I've produced are
in error? You have shown that the fact that the products are what I
claim they are does not support my claim that the a's and 5 are never
coprime?
> In every case posters rely on a circular argument which basically
> relies on the ring of algebraic integers not having the problem I've
> proven it has.
>
You have proven no such thing. Let's set some of the details down
for the sake of definiteness:
The claim I am refuting is:
In the factorization
65 x^3 - 12 x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
one of the coefficients a_* is coprime to 5.
I show that each a_* has a non-unit factor in common with
5 in the ring of algebraic integers. I conclude that none
of the a_* can be coprime to 5.
My argument refuting this follows (I have omitted details of the
polynomials q,r,s. They will be provided upon request):
There are polynomials q(x),r(x),s(x) with integer coefficients
that satisfy the following conditions:
q(x)r(x) = P(x)A(x) + 5
r(x)s(x) = P(x)B(x) + x
where P(x) is the minimal polynomial of -a, namely
P(x) = x^3 - 12 x^2 + 65
and where A(x) and B(x) are polynomials with integer
coefficients. Thus, for -a = any root of P(x), we have
the following factorizations:
5 = q(-a)r(-a)
a = r(-a)s(-a)
These factorizations hold in *any* ring containing both the
ring of integers and the number a.
I have further shown that the number r(-a) has as the minimal
polynomial
MP_r = x^3 - 969 x^2 + 315 x + 5
and so (noting that this polynomial is irreducible over Q) is
an algebraic integer which is not a unit in the ring of
algebraic integers.
Thus, in the ring of algebraic integers, the numbers a and 5
have the common, non-unit factor r(-a).
None of the above algebra relies on anything other than simple
manipulation of polynomials (i.e., multiplication and addition), and
the elementary result that a unit of the ring of algebraic integers
must have a minimal polynomial with both the leading term and constant
term equal to (rational) integers.
If you have shown that any of those assumptions fails, then you need to
make that claim explicitly. I claim you haven't found any such result.
You have claimed that the argument is circular because the
condition that u and v be coprime in the ring of algebraic
integers entails that any common factor of u and v, in that
ring, be a unit (in the ring of algebraic integers), while
I have shown that r(-a) cannot be a unit in the ring of
algebraic integers.
There is no reason behind your assertion that this is a circular
argument. Simply saying "it assumes that the algebraic integers
do not have the problem [you have] proven it has" does nothing
to show circularity.
> Here's basically how it goes:
>
> There are numbers that should properly be considered factors of 1.
>
This is a non-mathematical statement. Make it mathematical, and you
will have something to say. That something may be correct or incorrect,
and I would imagine the latter, but so far you've just made an
esthetic or a moral assertion, not a mathematical one.
> However, in the ring of algebraic integers, these numbers are NOT
> factors of 1.
>
> Note: I did not say that they should be considered factors of 1 in
> the ring of algebraic integers.
>
However, your claim is that the numbers 5 and a are coprime
*in the ring of algebraic integers*.
I have found a non-unit common factor *in that ring*, and for
the numbers to be coprime *in that ring* any such factor would
necessarily be a unit *in that ring*. This is a contradiction:
no number can be both a unit and a non-unit of the same ring.
Even you must admit that.
My argument holds in the ring of algebraic integers, and my result
holds in that ring.
> That's the problem. That is, the problem is that these numbers
> provably should be considered to be factors of 1, and finding a ring
> where they are is not even hard.
>
No. As long as you are sticking with the esthetic "should be considered"
there is no "provably" to be had. It's not so hard to climb out of that
hole: just give a mathematical definition of the phrase
"should be considered to be a factor of 1",
and you'll have something (correct or not) to say. Until then, you're
not making sense.
> They are factors of 1 in a ring made up of numbers such that only 1
> and -1 are integers units, which includes algebraic integers, and
> other numbers besides them.
>
Didn't someone show that your desired ring had to have all integers
invertible (and thus to lead to the full field of algebraic numbers)?
> That's it. You have a ring where only 1 and -1 are integer units,
> where that's the principal defining characteristic and you can show
> that some irrational units in that ring, are not algebraic integers.
>
But you haven't demonstrated what that ring is. No one will argue that
there aren't plenty of rings between the ring of algebraic integers
and the ring of algebraic numbers. They are incredibly easy to find:
Take A, the ring of algebraic numbers, and {a1,a2,...,aN}
a set of N non-integral algebraic numbers. Then the ring
A[a1,a2,...,aN] = the set of all values P(a1,a2,...,aN)
where P(x1, x2,..., xN) is a polynomial in N variables
with coefficients in A.
What you haven't shown is that your notion (a) exists, and (b) has
the other properties you think it has. You can postulate all you like,
but until you prove some existence lemma, and demonstrate the
characteristics of this alleged ring, no one will pay it any mind.
> Simple.
>
> But, when attacking my work, posters like Hall, Magidin, and Decker
> assert that because these numbers are not units in the ring of
> algebraic integers i.e. factors of 1 in that ring, then I must be
> wrong.
>
You yourself have admitted that the numbers r(-a) are not units
in the ring of algebraic integers, haven't you?
I have thus found a non-unit common factor of a and 5, within that
ring, right?
It then follows, does it not, that a and 5 cannot be coprime in
the ring of algebraic integers?
You have called the following argument circular. Please
show precisely where the circularity lies, keeping in mind
that it does not specify the algebraic integers, yet applies
to that ring:
Let a,b,m,n,p be elements of R, a commutative ring
with identity, and suppose a and b have a common
factor within R, as follows:
a = mn
b = mp
Suppose also that a and b are coprime in R.
I will show that m must be a unit of R.
By definition, "a and b coprime in R" means
that every ideal of R containing both a and b
is R itself; by noting that the sum of the
principal ideals Ra + Rb is such an ideal
(i.e., it contains both a and b), we have
R = Ra + Rb
Specializing to the identity element 1 of R, we
find that there must be elements t,u of R with:
1 = ta + ub
Substituting:
1 = tmn + ump
= m(tn + up)
And we find that the common factor m is a unit.
Thus, if a,b are coprime elements of R, then
any common factor is a unit of R.
Equivalently (via the contrapositive): if any
two elements of R have a common factor that
is not a unit then they cannot be coprime.
> But I've proven that I'm right using rather basic algebra.
>
You have tricked yourself into believing you've proven something,
by virtue of having imprecise definitions and overly vague arguments.
If you were correct, then your result would have been correct. Your
result is flawed, and that in itself indicates your method is flawed.
> That algebra shows that there is a ring of numbers where -1 and 1 are
> the only integer units that is indeed LARGER than the ring of
> algebraic integers.
>
You say there is a LARGER ring. There are many LARGER rings. The
existence of a LARGER ring is absolutely irrelevant for the validity
of a proof within the smaller ring of algebraic integers.
Yet you're saying that a and 5 are coprime in this alleged LARGER ring?
I don't address any LARGER ring. My result holds in the ring of
algebraic integers; yours doesn't. Simply saying that the ring
of algebraic integers has some problem, therefore no proof in that
context is valid, is no argument at all, unless you can prove that the
set of algebraic integers fails to be a commutative ring or fails
to be integrally closed (neither of which is likely, given the number
of times the relevant proofs have been revisited).
> That's it. That's the amazing conclusion that mathematicians have
> been running from for so long.
>
> It's actually kind of a neat fact.
>
>
> James Harris
Whatever.
Dale.
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