Statement of problem with ring of algebraic integers
James Harris
following explanation for the problem with the ring of algebraic
integers.
Algebraic integers are defined as the roots of monic polynomials with
integer coefficients.
However, it can be shown that in the ring of algebraic integers, given
the algebraic integers a_1, a_2, and a_3, where a_1 a_2 a_3 has the
algebraic integer--non unit and non zero--f^2 as a factor that a_3 is
coprime to f, and it's implied that a_1 and a_2 each have a factor of
f that is f.
But as noted by several posters there is an argument, which proves
that neither a_1 nor a_2 can have f as a factor within the ring of
algebraic integers.
Therefore, it is required that a_1, a_2 and a_3 are each coprime to f
within the ring of algebraic integers, which is the contradiction.
That is a contradiction which shows a problem with the ring.
I have described it as being incomplete.
The mathematical proofs for each segment of the above are readily
available.
The proof that a_3 is coprime to f can be found by going to the link
http://groups.msn.com/AmateurMath to get the paper Advanced Polynomial
Factorization.
James Harris
Dot
Harris <jst...@msn.com> wrote:
> However, it can be shown that in the ring of algebraic integers, given
> the algebraic integers a_1, a_2, and a_3, where a_1 a_2 a_3 has the
> algebraic integer--non unit and non zero--f^2 as a factor that a_3 is
> coprime to f, and it's implied that a_1 and a_2 each have a factor of
> f that is f.
You write "...and it's implied that...". Do you mean to claim
that the following is a theorem?
If a_1, a_2, and a_3 are algebraic integers such that
a_1 * a_2 * a_3 is divisible by f^2 (where f is a non-zero
non-unit algebraic integer), and if a_3 is coprime to f,
then a_1 and a_2 are both divisible by f.
Because that's not a theorem. It's incorrect. Take
a_1 = (2 + i)^2
a_2 = (2 - i)^2
a_3 = 1
f = 5
-- Dot.
Arturo Magidin
>following explanation for the problem with the ring of algebraic
>integers.
>
>Algebraic integers are defined as the roots of monic polynomials with
>integer coefficients.
>
>However, it can be shown that in the ring of algebraic integers, given
>the algebraic integers a_1, a_2, and a_3, where a_1 a_2 a_3 has the
>algebraic integer--non unit and non zero--f^2 as a factor that a_3 is
>coprime to f, and it's implied that a_1 and a_2 each have a factor of
>f that is f.
This "it can be shown" is not established. It is your claim, based on
your arguments, that so far nobody has been able to follow; this
because you insist on using non-standard terminology, you refuse to
define your terms formally, and your arguments are vague are confused.
>But as noted by several posters there is an argument, which proves
>that neither a_1 nor a_2 can have f as a factor within the ring of
>algebraic integers.
In fact, it proves that NONE of a_1, a_2, or a_3 are multiples of f.
>Therefore, it is required that a_1, a_2 and a_3 are each coprime to f
>within the ring of algebraic integers, which is the contradiction.
No, this is you again taking what we have said, and
mis-stating/mis-understanding/mis-representing/lying about
it.
In fact, we noted that NONE of a_1, a_2, or a_3 are coprime to f.
The real key to the step of your FLT proof is NOT the divisibility by
f, but is really that all but two of the coefficients are coprime to
f. But that is precisely what is false. NONE of the coefficients are
coprime to f.
>That is a contradiction which shows a problem with the ring.
Or with that argument of yours that you insist on making as opaque,
complicated, and nonsensical as possible.
[.rest deleted.]
======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
C. Bond
> After considering various discussion with various posters I have the
> following explanation for the problem with the ring of algebraic
> integers.
>
> Algebraic integers are defined as the roots of monic polynomials with
> integer coefficients.
>
> However, it can be shown that in the ring of algebraic integers, given
> the algebraic integers a_1, a_2, and a_3, where a_1 a_2 a_3 has the
> algebraic integer--non unit and non zero--f^2 as a factor that a_3 is
> coprime to f, and it's implied that a_1 and a_2 each have a factor of
> f that is f.
You are attempting to elevate "proof by assertion" to a high art. It
hasn't worked before, and it won't wor now.
> But as noted by several posters there is an argument, which proves
> that neither a_1 nor a_2 can have f as a factor within the ring of
> algebraic integers.
Too bad you haven't paid more attention to other 'posters'. You would have
repaired or abandoned your arguments otherwise.
> Therefore, it is required that a_1, a_2 and a_3 are each coprime to f
> within the ring of algebraic integers, which is the contradiction.
>
> That is a contradiction which shows a problem with the ring.
Any contradiction you have arrived at shows a problem with your arguments.
There is no problem with the ring.
> I have described it as being incomplete.
>
> The mathematical proofs for each segment of the above are readily
> available.
>
> The proof that a_3 is coprime to f can be found by going to the link
> http://groups.msn.com/AmateurMath to get the paper Advanced Polynomial
> Factorization.
>
> James Harris
Go back to your sandbox.
--
It takes a village to raise an idiot.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
Nora Baron
> After considering various discussion with various posters I have the
> following explanation for the problem with the ring of algebraic
> integers.
>
> Algebraic integers are defined as the roots of monic polynomials with
> integer coefficients.
>
> However, it can be shown that in the ring of algebraic integers, given
> the algebraic integers a_1, a_2, and a_3, where a_1 a_2 a_3 has the
> algebraic integer--non unit and non zero--f^2 as a factor that a_3 is
> coprime to f, and it's implied that a_1 and a_2 each have a factor of
> f that is f.
>
> But as noted by several posters there is an argument, which proves
> that neither a_1 nor a_2 can have f as a factor within the ring of
> algebraic integers.
>
In fact "other posters" have noted that the main conclusion of your
Advanced Polynomial Factorization paper is false. I noted this in
a post on June 18 [proof is given below] and W. Dale Hall noted it
in a post on June 22. These were in the thread called "My Work --
Objective Review." Hall's proof that you were wrong was completely
different and independent of my proof, which follows:
===============================================================================
From June 18 post:
The claim is made in "Advanced Polynomial Factorization" that if
P(x) = 65*x^3 - 12*x + 1
is factored in the form
P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1),
where a1, a2, and a3 are algebraic integers, then at least one
of a1, a2, and a3 must be coprime to 5 in the algebraic integers.
Thus in the following, assume that one of the ai,
say a1, is, as claimed, coprime to 5 in the algebraic integers.
That is, there exist algebraic integers s and t such that
s*a1 + t*5 = 1. [1]
Let x = 1/u. Then P(x) = 0 implies that
P(x) = 65/u^3 -12/u + 1 = 0, or
Q(u) = u^3 - 12*u^2 + 65 = 0.
Let r1, r2, and r3 be the roots of this
equation. All of r1, r2, and r3 are algebraic
integers. Note that r1, r2, and r3 happen to
be the negatives a1, a2, and a3 in some order -
say, a1 = -r1, a2 = -r2, a3 = -r3.
Note that Q(u) is irreducible over the rationals.
Let H be the field of algebraic numbers; clearly
r1, r2, and r3 are in H. Let F12 be an automorphism
of H such that F12(r1) = r2, and F12 leaves fixed
the subfield of rational numbers. Note also that
F12(a1) = a2, since a1 = -r1 and a2 = -r2.
{That such an automorphism F12 exists is well-known,
and is described in:
http://www.math.niu.edu/~beachy/aaol/galois.html,
see especially Proposition 8.6.2 on that page. Or
see the excellent textbook, Abstract Algebra, by
John Beachy and William D. Blair.}
Now apply the automorphism F12 to both sides of
equation [1]:
F12(s)*F12(a1) + F12(t)*F12(5) = F12(1).
But F12(1) = 1, F12(5) = 5, and F12(a1) = a2. Thus
we obtain
s' * a2 + t' * 5 = 1, [2]
where s' = F12(s) and F12(t) = t'. Note that the
automorphism F12 preserves the property of being
an algebraic integer; that is, s' and t', like
s and t, are algebraic integers. Therefore
equation [2] implies that a1 is coprime to 5
also.
Similarly one shows that a3 is coprime to 5.
Thus, from the assumption that one of a1, a2,
or a3 is coprime to 5, we deduce that all three
of them must be coprime to 5.
But a1 * a2 * a3 = 65. Therefore at least one
of a1, a2, or a3 is NOT coprime to 5. Therefore
a contradiction. The source of the contradiction
was the initial assumption that one of a1, a2, or
a3 is coprime to 5. Therefore NONE of a1, a2, and
a3 are coprime to 5, contradicting the main result
of "Advanced Polynomial Factorization."
Nora B.
===============================================================================
Nora Baron
as noted.
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03070...@posting.google.com>...
> After considering various discussion with various posters I have the
> following explanation for the problem with the ring of algebraic
> integers.
>
> Algebraic integers are defined as the roots of monic polynomials with
> integer coefficients.
>
> However, it can be shown that in the ring of algebraic integers, given
> the algebraic integers a_1, a_2, and a_3, where a_1 a_2 a_3 has the
> algebraic integer--non unit and non zero--f^2 as a factor that a_3 is
> coprime to f, and it's implied that a_1 and a_2 each have a factor of
> f that is f.
>
> But as noted by several posters there is an argument, which proves
> that neither a_1 nor a_2 can have f as a factor within the ring of
> algebraic integers.
>
In fact "other posters" have noted that the main conclusion of your
http://www.math.niu.edu/~beachy/aaol/galois.html,
equation [2] implies that a2 is coprime to 5
also.
**** misprint was there - I had said " ... a1
is coprime to 5 ..."
Similarly one shows that a3 is coprime to 5.
Thus, from the assumption that one of a1, a2,
or a3 is coprime to 5, we deduce that all three
of them must be coprime to 5.
But a1 * a2 * a3 = 65. Therefore at least one
of a1, a2, or a3 is NOT coprime to 5. Therefore
a contradiction. The source of the contradiction
was the initial assumption that one of a1, a2, or
a3 is coprime to 5. Therefore NONE of a1, a2, and
a3 are coprime to 5, contradicting the main result
of "Advanced Polynomial Factorization."
Nora B.
===============================================================================
> Therefore, it is required that a_1, a_2 and a_3 are each coprime to f
> within the ring of algebraic integers, which is the contradiction.
>
> That is a contradiction which shows a problem with the ring.
>
> I have described it as being incomplete.
>
> The mathematical proofs for each segment of the above are readily
> available.
>
> The proof that a_3 is coprime to f can be found by going to the link
> http://groups.msn.com/AmateurMath to get the paper Advanced Polynomial
> Factorization.
>
>
> James Harris.
James Harris
> In article <3c65f87.03070...@posting.google.com>, James
> Harris <jst...@msn.com> wrote:
>
> > However, it can be shown that in the ring of algebraic integers, given
> > the algebraic integers a_1, a_2, and a_3, where a_1 a_2 a_3 has the
> > algebraic integer--non unit and non zero--f^2 as a factor that a_3 is
> > coprime to f, and it's implied that a_1 and a_2 each have a factor of
> > f that is f.
>
> You write "...and it's implied that...". Do you mean to claim
> that the following is a theorem?
>
> If a_1, a_2, and a_3 are algebraic integers such that
> a_1 * a_2 * a_3 is divisible by f^2 (where f is a non-zero
> non-unit algebraic integer), and if a_3 is coprime to f,
> then a_1 and a_2 are both divisible by f.
Looks like I left it ambiguous. The a's are roots of a monic
polynomial of degree 3 that is irreducible over rationals.
> Because that's not a theorem. It's incorrect. Take
>
> a_1 = (2 + i)^2
> a_2 = (2 - i)^2
> a_3 = 1
> f = 5
>
>
> -- Dot.
Looks like I need to do a rewrite.
James Harris
James Harris
> My previous reply to this had a misprint, corrected below
> as noted.
>
>
> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03070...@posting.google.com>...
> > After considering various discussion with various posters I have the
> > following explanation for the problem with the ring of algebraic
> > integers.
> >
> > Algebraic integers are defined as the roots of monic polynomials with
> > integer coefficients.
> >
> > However, it can be shown that in the ring of algebraic integers, given
> > the algebraic integers a_1, a_2, and a_3, where a_1 a_2 a_3 has the
> > algebraic integer--non unit and non zero--f^2 as a factor that a_3 is
> > coprime to f, and it's implied that a_1 and a_2 each have a factor of
> > f that is f.
I left out that the a's are roots of a cubic with integer coefficients
irreducible over Q.
> > But as noted by several posters there is an argument, which proves
> > that neither a_1 nor a_2 can have f as a factor within the ring of
> > algebraic integers.
> >
>
> In fact "other posters" have noted that the main conclusion of your
> Advanced Polynomial Factorization paper is false. I noted this in
> a post on June 18 [proof is given below] and W. Dale Hall noted it
> in a post on June 22. These were in the thread called "My Work --
> Objective Review." Hall's proof that you were wrong was completely
> different and independent of my proof, which follows:
You're being remarkably dense Nora Baron.
Think about it.
And you didn't read my original post, now did you?
> **** misprint was there - I had said " ... a1
> is coprime to 5 ..."
>
>
> Similarly one shows that a3 is coprime to 5.
>
> Thus, from the assumption that one of a1, a2,
> or a3 is coprime to 5, we deduce that all three
> of them must be coprime to 5.
Now I'm not agreeing with your argument, but as you can see it doesn't
matter.
> But a1 * a2 * a3 = 65. Therefore at least one
> of a1, a2, or a3 is NOT coprime to 5. Therefore
> a contradiction. The source of the contradiction
> was the initial assumption that one of a1, a2, or
> a3 is coprime to 5. Therefore NONE of a1, a2, and
> a3 are coprime to 5, contradicting the main result
> of "Advanced Polynomial Factorization."
Nope.
> Nora B.
Sorry Nora Baron, but you need to actually read posts to which you
reply.
> >
> > The proof that a_3 is coprime to f can be found by going to the link
> > http://groups.msn.com/AmateurMath to get the paper Advanced Polynomial
> > Factorization.
Readers who look at the paper will see the *implication* that two of
the a's have a factor of f in the ring of algebraic integers, as that
follows from the assumption that algebraic integers are a complete
ring, and it's proven that one of them must be coprime to f.
However, if you move to rationals and divide f off from the a's that
apparently have it as a factor--remember one has been proven to be
coprime to f--you end up with a non monic primitive irreducible over
Q, so the result is NOT an algebraic integer.
Given that the definition of "factor" would require that the result be
an algebraic integer, it is clear that all of the a's must be coprime
to f in the ring of algebraic integers.
It's a bizarre and fascinating result that comes from that definition
of algebraic integers as roots of monic polynomials with integer
coefficients.
James Harris
James Harris
some replies. ___JSH
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03070...@posting.google.com>...
> After considering various discussion with various posters I have the
> following explanation for the problem with the ring of algebraic
> integers.
>
> Algebraic integers are defined as the roots of monic polynomials with
> integer coefficients.
>
> However, it can be shown that in the ring of algebraic integers, given
> the algebraic integers a_1, a_2, and a_3, where a_1 a_2 a_3 has the
> algebraic integer--non unit and non zero--f^2 as a factor that a_3 is
> coprime to f, and it's implied that a_1 and a_2 each have a factor of
> f that is f.
Also the a's are roots of a cubic with integer coefficients that is
irreducible over Q.
> But as noted by several posters there is an argument, which proves
> that neither a_1 nor a_2 can have f as a factor within the ring of
> algebraic integers.
That's where it gets fascinating. It turns out that the *definition*
arguments are correct in that by the definition of "factor" it cannot
be true that ANY of the a's have a factor of f within the ring of
algebraic integers.
I'll repeat how that conclusion is reached:
1. In my paper Advanced Polynomial Factorization, I prove that one of
the a's must be coprime to f, where it appears that two of the a's
have a factor of f that is f, within the ring of algebraic integers.
2. But you can show that if either of the a's have a factor of f that
is f, within the ring of algebraic integers, if you move to the field
of rationals and divide it off, the result is NOT an algebraic
integer.
> Therefore, it is required that a_1, a_2 and a_3 are each coprime to f
> within the ring of algebraic integers, which is the contradiction.
>
> That is a contradiction which shows a problem with the ring.
> I have described it as being incomplete.
>
> The mathematical proofs for each segment of the above are readily
> available.
>
> The proof that a_3 is coprime to f can be found by going to the link
> http://groups.msn.com/AmateurMath to get the paper Advanced Polynomial
> Factorization.
Read the paper.
James Harris
Nora Baron
Right. At this point you think I have ignored your main point.
Not so. See if you can follow this. Let "APF" denote the main
argument in "Advanced Polynomial Factorization". If APF is
correct, it implies that at least one of a1, a2, or a3 is
coprime to 5. Abbreviate this by "A123C5". Thus
APF ==> A123C5.
What I showed below is that A123C5 is false.
Therefore APF is false.
But you need APF for your main argument here: Remember what you
said at the end of the post to which I was replying?
"The proof that a_3 is coprime to f can be found by going to the
link http://groups.msn.com/AmateurMath to get the paper
Advanced Polynomial Factorization."
So, right. I *had* thought about it. Now you think about it.
Oh yes. See above.
> > **** misprint was there - I had said " ... a1
> > is coprime to 5 ..."
> >
> >
> > Similarly one shows that a3 is coprime to 5.
> >
> > Thus, from the assumption that one of a1, a2,
> > or a3 is coprime to 5, we deduce that all three
> > of them must be coprime to 5.
>
> Now I'm not agreeing with your argument, but as you can see it doesn't
> matter.
>
Wrong. You've forgotten your own logic.
> > But a1 * a2 * a3 = 65. Therefore at least one
> > of a1, a2, or a3 is NOT coprime to 5. Therefore
> > a contradiction. The source of the contradiction
> > was the initial assumption that one of a1, a2, or
> > a3 is coprime to 5. Therefore NONE of a1, a2, and
> > a3 are coprime to 5, contradicting the main result
> > of "Advanced Polynomial Factorization."
>
> Nope.
>
Yep.
> > Nora B.
>
> Sorry Nora Baron, but you need to actually read posts to which you
> reply.
>
Sorry James Harris, I did. Too bad you're not able to follow
the thinking. Incidentally, other than "Nope" I don't see
that you have much of a counterargument here, eh?
> > >
> > > The proof that a_3 is coprime to f can be found by going to the link
> > > http://groups.msn.com/AmateurMath to get the paper Advanced Polynomial
> > > Factorization.
>
> Readers who look at the paper will see the *implication* that two of
> the a's have a factor of f in the ring of algebraic integers, as that
> follows from the assumption that algebraic integers are a complete
> ring, and it's proven that one of them must be coprime to f.
>
You are right that two of the a's have a factor of f. However,
the "proof" that the third one must be coprime to f is wrong.
I have pointed out where you made your error previously, not that
it matters. The argument I gave above shows there is no hope
of fixing it anyway: the APF conclusion is wrong. APF is dead meat.
Nora B.
C. Bond
The exact values of a1, a2 and a3 for the polynomial problem you cite are given below.
Let w = (1/3)ArcTan[Sqrt[12415]/63], a number whose value is approx. 0.352065133...
Then,
a1 = -4-8Cos[w] (approx. -11.50930064..)
a2 = -4+4Cos[w]+4Sqrt[3]Sin[w] (approx. 2.143751171..)
a3 = -4+4Cos[w]-4Sqrt[3]Sin[w] (approx. -2.634450526..)
not necessarily in any preferred order.
Which of these numbers is coprime to 5, and which are not?
--
There are two things you must never attempt to prove: the unprovable -- and the obvious.
James Harris
> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03070...@posting.google.com>...
<deleted>
> > > In fact "other posters" have noted that the main conclusion of your
> > > Advanced Polynomial Factorization paper is false. I noted this in
> > > a post on June 18 [proof is given below] and W. Dale Hall noted it
> > > in a post on June 22. These were in the thread called "My Work --
> > > Objective Review." Hall's proof that you were wrong was completely
> > > different and independent of my proof, which follows:
> >
> > You're being remarkably dense Nora Baron.
> >
> > Think about it.
> >
>
> Right. At this point you think I have ignored your main point.
> Not so. See if you can follow this. Let "APF" denote the main
> argument in "Advanced Polynomial Factorization". If APF is
> correct, it implies that at least one of a1, a2, or a3 is
> coprime to 5. Abbreviate this by "A123C5". Thus
>
> APF ==> A123C5.
>
> What I showed below is that A123C5 is false.
But Nora Baron you have not shown it to be false.
Your own statement is that if one is coprime they all are.
I've noted that yes, that weird conclusion is the proper one, as in
the ring of algebraic integers they ARE all coprime to f, which is the
problem.
> Therefore APF is false.
But that does not follow from what you've given.
> But you need APF for your main argument here: Remember what you
> said at the end of the post to which I was replying?
>
> "The proof that a_3 is coprime to f can be found by going to the
> link http://groups.msn.com/AmateurMath to get the paper
> Advanced Polynomial Factorization."
>
> So, right. I *had* thought about it. Now you think about it.
<deleted>
As I said before you're being remarkably dense Nora Baron.
The facts are that you claim to have proven that if one of the a's is
coprime to f, then they all are. I say that the problem is that given
the definition of algebraic integers it is true that all of the a's
are coprime to f in the ring of algebraic integers.
That's the fact Nora Baron.
Let me repeat it: Given the definition for algebraic integers it is
true that all of the a's *must* be coprime to f in the ring of
algebraic integers.
However, that is what shows a problem with the ring.
What I've done is prove that one of the a's is coprime to f in my
paper Advanced Polynomial Factorization. As noted by me, there are
ways to show that the remaining a's must then be coprime to f. You
are someone who claims to be able to show that conclusion.
As I've said before, proofs don't duel. If you have a proof Nora
Baron, it could NOT conflict with my proof.
James Harris
Nora Baron
> nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03070...@posting.google.com>...
> > jst...@msn.com (James Harris) wrote in message news:<3c65f87.03070...@posting.google.com>...
>
> <deleted>
>
> > > > In fact "other posters" have noted that the main conclusion of your
> > > > Advanced Polynomial Factorization paper is false. I noted this in
> > > > a post on June 18 [proof is given below] and W. Dale Hall noted it
> > > > in a post on June 22. These were in the thread called "My Work --
> > > > Objective Review." Hall's proof that you were wrong was completely
> > > > different and independent of my proof, which follows:
> > >
> > > You're being remarkably dense Nora Baron.
> > >
> > > Think about it.
> > >
> >
> > Right. At this point you think I have ignored your main point.
> > Not so. See if you can follow this. Let "APF" denote the main
> > argument in "Advanced Polynomial Factorization". If APF is
> > correct, it implies that at least one of a1, a2, or a3 is
> > coprime to 5. Abbreviate this by "A123C5". Thus
> >
> > APF ==> A123C5.
> >
> > What I showed below is that A123C5 is false.
>
> But Nora Baron you have not shown it to be false.
>
> Your own statement is that if one is coprime they all are.
>
> I've noted that yes, that weird conclusion is the proper one, as in
> the ring of algebraic integers they ARE all coprime to f, which is the
> problem.
>
Thanks for conceding "that weird conclusion."
But you didn't go quite far enough. Yes, I show that if one
of a1, a2, or a3 is coprime to 5, then they all are. But if
you had read a little farther into the proof, you would have
seen this:
"But a1 * a2 * a3 = 65. Therefore at least one of a1,
a2, or a3 is NOT coprime to 5. Therefore a contradiction.
The source of the contradiction was the assumption that
one of a1, a2, or a3 is coprime to 5."
You must therefore conclude that NONE of a1, a2, or a3 is
coprime to 5. The conclusion of "Advanced Polynomial
Factorization" is therefore false.
> > Therefore APF is false.
>
> But that does not follow from what you've given.
>
Oh yes. That is exactly what follows from what I've given.
> > But you need APF for your main argument here: Remember what you
> > said at the end of the post to which I was replying?
> >
> > "The proof that a_3 is coprime to f can be found by going to the
> > link http://groups.msn.com/AmateurMath to get the paper
> > Advanced Polynomial Factorization."
> >
> > So, right. I *had* thought about it. Now you think about it.
>
> <deleted>
>
> As I said before you're being remarkably dense Nora Baron.
>
> The facts are that you claim to have proven that if one of the a's is
> coprime to f, then they all are. I say that the problem is that given
> the definition of algebraic integers it is true that all of the a's
> are coprime to f in the ring of algebraic integers.
>
> That's the fact Nora Baron.
>
No. You were right when you agreed that I had shown that
if one of a1, a2, and a3 were coprime to 5, then they all
are. You did not get the conclusion of my argument. Why?
Because you are desperately reluctant to admit what it
implies: that your main argument in "Advanced Polynomial
Factorization" is incorrect.
There are really only two choices here. There is nothing
wrong with the definition of algebraic integers, or the proofs
of their properties going back to Gauss and Dedekind. The algebraic
integers are not "incomplete". Either your argument in "APF"
is wrong, or mine is. You have essentially conceded above that
the heart of my argument is not wrong. Therefore yours *is*.
I have told you previously exactly where your reasoning went
off track. No, it has nothing to do with your stubborn misuse
of standard terminology. That is just froth and semantics.
It has to do with assuming that in factoring a polynomial, you
can generalize from a degenerate case, where the polynomial
has degree 1, to the usual case where it has degree three. It
is an error you have made many, many times over the past 1-2
years. It is a real hard unavoidable kernel of error, not a
semantic misunderstanding, and, though several people have
tried, no one has succeeded in getting you to see it.
Perhaps for you it is a kind of 'pons asinorum.' Do you
remember that from geometry? Anyway, it leaves you stuck thinking
you have a valid proof, not understanding why you don't;
thinking mathematicians are all stupid liars, avoiding what has
cursed you for 8 years.
8 years: it's quite sad. You have learned a little math in
that time. You have learned what a ring is, what algebraic
numbers are, what algebraic integers are, what fields are, a tiny
bit of Galois theory. You have learned about congruences, Fermat's
Little Theorem, the fact that there are no primes in the algebraic
integers, the fact that the AIs are infinitely divisible. You
have learned a bit about permutations. Maybe you know a little
about conjugates. You know what units are.
It's sad that I can summarize what you know in one paragraph,
8+ lines.
In 8 years, you have learned about as much as an average student
would in 2-3 weeks in an algebraic number theory course. You don't
know or understand the most basic nontrivial tool of algebraic
number theory, the Eucldean algorithm. You don't know Gauss's
Lemma, many other basic theorems. This is all Chapter 1 stuff.
If you were really a math genius you would have discovered all
these for yourself. That hasn't happened. You don't have a
single nontrivial lemma to your name in algebraic number theory,
no general methods, nothing anyone else wants to use. Your
"Advanced Polynomial Factorization" is not worth the paper it's
written on. It's built on a simple error that you yourself
cannot understand.
That's the fact, James Harris.
Nora B.
[snip]
Virgil
jst...@msn.com wrote on 7/6/03 09:24:
> As I've said before, proofs don't duel. If you have a proof Nora
> Baron, it could NOT conflict with my proof.
>
>
> James Harris
Valid proofs may not duel, but JSH has such a long and reliable history for
providing invalid arguments labeled as proofs, that I would bet against him
in any such "duel".
Wayne Brown
<snip>
> But Nora Baron you have not shown it to be false.
<snip>
> As I said before you're being remarkably dense Nora Baron.
<snip>
> That's the fact Nora Baron.
<snip>
> As I've said before, proofs don't duel. If you have a proof Nora
> Baron, it could NOT conflict with my proof.
You have an odd fixation about using people's full names. It's but one of
the many things that make your articles so ludicrous. (Calling everyone
else liars every few sentences is another.)
Did your momma use your full name every time she scolded you? "You shut
your mouth, James Steven Harris!" You must have heard that many,
many times. Is that why you're obsessed with other people's names?
--
Wayne Brown | "When your tail's in a crack, you improvise
fwb...@bellsouth.net | if you're good enough. Otherwise you give
| your pelt to the trapper."
"e^(i*pi) = -1" -- Euler | -- John Myers Myers, "Silverlock"
W. Dale Hall
James Harris wrote:
> nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03070...@posting.google.com>...
>
>>jst...@msn.com (James Harris) wrote in message news:<3c65f87.03070...@posting.google.com>...
>
>
> <deleted>
>
>>>> In fact "other posters" have noted that the main conclusion of your
>>>>Advanced Polynomial Factorization paper is false. I noted this in
>>>>a post on June 18 [proof is given below] and W. Dale Hall noted it
>>>>in a post on June 22. These were in the thread called "My Work --
>>>>Objective Review." Hall's proof that you were wrong was completely
>>>>different and independent of my proof, which follows:
>>>
>>>You're being remarkably dense Nora Baron.
>>>
>>>Think about it.
>>>
>>
>> Right. At this point you think I have ignored your main point.
>>Not so. See if you can follow this. Let "APF" denote the main
>>argument in "Advanced Polynomial Factorization". If APF is
>>correct, it implies that at least one of a1, a2, or a3 is
>>coprime to 5. Abbreviate this by "A123C5". Thus
>>
>> APF ==> A123C5.
>>
>> What I showed below is that A123C5 is false.
>
>
> But Nora Baron you have not shown it to be false.
>
> Your own statement is that if one is coprime they all are.
>
> I've noted that yes, that weird conclusion is the proper one, as in
> the ring of algebraic integers they ARE all coprime to f, which is the
> problem.
>
>
The polynomial JSH claims leads to this weird conclusion is this:
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
I will show that the a's cannot be coprime to 5. Here's how:
Consider the polynomial
p(x) = x^3 - 12 x^2 + 65,
and note that the roots of p(x) are precisely the negatives of the a's.
Define the polynomials
q(x) = 8 x^2 - 76 x -185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104
The following is then true:
For z any root of the polynomial
p(x) = x^3 - 12 x^2 + 65,
we have
q(z)*r(z) = 5
r(z)*s(z) = z.
By virtue of the polynomials q,r,s having integer coefficients, and
of z being a root of p(x), we know that z, q(z), r(z), and s(z) are all
algebraic integers.
Thus, r(z) is a common factor of z and 5, in A, the ring of algebraic
integers. Since it is guaranteed that JSH will claim r(z) is a unit
of A, here is the minimal polynomial for r:
MinPoly(r(z)) = x^3 - 969 x^2 + 315 x + 5
The products were found via KASH, and have been verified in Macsyma.
>> Therefore APF is false.
>
>
> But that does not follow from what you've given.
>
>
>> But you need APF for your main argument here: Remember what you
>>said at the end of the post to which I was replying?
>>
>> "The proof that a_3 is coprime to f can be found by going to the
>> link http://groups.msn.com/AmateurMath to get the paper
>> Advanced Polynomial Factorization."
>>
>> So, right. I *had* thought about it. Now you think about it.
>
>
> <deleted>
>
> As I said before you're being remarkably dense Nora Baron.
>
> The facts are that you claim to have proven that if one of the a's is
> coprime to f, then they all are. I say that the problem is that given
> the definition of algebraic integers it is true that all of the a's
> are coprime to f in the ring of algebraic integers.
>
So, you must have some disagreement with the work I've shown above?
> That's the fact Nora Baron.
>
> Let me repeat it: Given the definition for algebraic integers it is
> true that all of the a's *must* be coprime to f in the ring of
> algebraic integers.
>
No, you're incorrect, and your methods for producing what it is you
refer to as proof are flawed. Flawed methods produce flawed arguments,
and lead you to make incorrect conclusions.
> However, that is what shows a problem with the ring.
>
> What I've done is prove that one of the a's is coprime to f in my
> paper Advanced Polynomial Factorization. As noted by me, there are
> ways to show that the remaining a's must then be coprime to f. You
> are someone who claims to be able to show that conclusion.
>
> As I've said before, proofs don't duel. If you have a proof Nora
> Baron, it could NOT conflict with my proof.
>
But what about my example of a common factor between 5 and each of
the a's?
>
> James Harris
Dale
James Harris
There's a problem in mathematics, which is fascinating one, and yes,
it also is rather wacky.
> But you didn't go quite far enough. Yes, I show that if one
> of a1, a2, or a3 is coprime to 5, then they all are. But if
> you had read a little farther into the proof, you would have
> seen this:
>
> "But a1 * a2 * a3 = 65. Therefore at least one of a1,
> a2, or a3 is NOT coprime to 5. Therefore a contradiction.
> The source of the contradiction was the assumption that
> one of a1, a2, or a3 is coprime to 5."
But that's wrong. Now you need to try and prove that given that
a1 a2 a3 = 65
in the ring of algebraic integers, a1, a2 or a3 must each have some
non unit factor of 5.
That should be interesting.
> You must therefore conclude that NONE of a1, a2, or a3 is
> coprime to 5. The conclusion of "Advanced Polynomial
> Factorization" is therefore false.
Which is false. It IS true that all of the a's are coprime to 5 in
the ring of algebraic integers, as that's been proven mathematically.
So the conclusion that none of them is coprime to 5 is false.
> > > Therefore APF is false.
> >
> > But that does not follow from what you've given.
> >
>
>
> Oh yes. That is exactly what follows from what I've given.
You're still being remarkably dense Nora Baron.
> > > But you need APF for your main argument here: Remember what you
> > > said at the end of the post to which I was replying?
> > >
> > > "The proof that a_3 is coprime to f can be found by going to the
> > > link http://groups.msn.com/AmateurMath to get the paper
> > > Advanced Polynomial Factorization."
> > >
> > > So, right. I *had* thought about it. Now you think about it.
> >
> > <deleted>
> >
> > As I said before you're being remarkably dense Nora Baron.
> >
> > The facts are that you claim to have proven that if one of the a's is
> > coprime to f, then they all are. I say that the problem is that given
> > the definition of algebraic integers it is true that all of the a's
> > are coprime to f in the ring of algebraic integers.
> >
> > That's the fact Nora Baron.
> >
>
>
> No. You were right when you agreed that I had shown that
> if one of a1, a2, and a3 were coprime to 5, then they all
> are. You did not get the conclusion of my argument. Why?
> Because you are desperately reluctant to admit what it
> implies: that your main argument in "Advanced Polynomial
> Factorization" is incorrect.
Really? I think you're lying because if you *believe* what you're
saying then you must believe there's a flaw in the paper. If there's
a flaw in the paper then someone can point it out.
If you're not lying Nora Baron, I challenge you to give a flaw in my
paper Advanced Polynomial Factorization.
You see, enough with all the assertions without proof when because
it's mathematics there's such an *easy* path to take.
> There are really only two choices here. There is nothing
> wrong with the definition of algebraic integers, or the proofs
> of their properties going back to Gauss and Dedekind. The algebraic
> integers are not "incomplete". Either your argument in "APF"
> is wrong, or mine is. You have essentially conceded above that
> the heart of my argument is not wrong. Therefore yours *is*.
>
> I have told you previously exactly where your reasoning went
> off track. No, it has nothing to do with your stubborn misuse
> of standard terminology. That is just froth and semantics.
> It has to do with assuming that in factoring a polynomial, you
> can generalize from a degenerate case, where the polynomial
> has degree 1, to the usual case where it has degree three. It
> is an error you have made many, many times over the past 1-2
> years. It is a real hard unavoidable kernel of error, not a
> semantic misunderstanding, and, though several people have
> tried, no one has succeeded in getting you to see it.
I've split up a long section to focus on what is apparently the
assertion that Nora Baron is holding on to desperately. I'll give an
example to explain the assertion.
Consider P(x) = x^2 y^2 + 2x + 3y + 2,
where yes, it's important that there are two variables, but the
polynomial is with respect to x.
Now I say that the constant term of the polynomial is given by P(0)
and is
P(0) = 3y+2
and yes, if you wish, you can stick in something like y=5, and get
actual numbers, but the point is that for P(x) the constant term is
P(0), which you get with x=0.
However, Nora Baron is citing a problem with that as I've dropped the
degree of the polynomial. Now with something like P(x) = x^2 + 4x +
4, you can look and see the constant term, but it's also true that
P(0) gives the constant term, so what would you think if someone
claimed that it was a "degenerate case" that didn't prove what you
said?
Now the actual expression in the paper is just a little more
sophisticated, but that's basically what's done in the paper.
I point out that Nora Baron buried her assertion in quite a bit of
verbiage.
> Perhaps for you it is a kind of 'pons asinorum.' Do you
> remember that from geometry? Anyway, it leaves you stuck thinking
> you have a valid proof, not understanding why you don't;
> thinking mathematicians are all stupid liars, avoiding what has
> cursed you for 8 years.
Looks to me like you're desperate Nora Baron or you wouldn't have
buried your supposed objection in so many words where it looks like
you're trying to psyche me out.
> 8 years: it's quite sad. You have learned a little math in
> that time. You have learned what a ring is, what algebraic
> numbers are, what algebraic integers are, what fields are, a tiny
> bit of Galois theory. You have learned about congruences, Fermat's
> Little Theorem, the fact that there are no primes in the algebraic
> integers, the fact that the AIs are infinitely divisible. You
> have learned a bit about permutations. Maybe you know a little
> about conjugates. You know what units are.
Now if you had an actual error in my paper, what's with all the
rhetoric?
> It's sad that I can summarize what you know in one paragraph,
> 8+ lines.
>
> In 8 years, you have learned about as much as an average student
> would in 2-3 weeks in an algebraic number theory course. You don't
> know or understand the most basic nontrivial tool of algebraic
> number theory, the Eucldean algorithm. You don't know Gauss's
> Lemma, many other basic theorems. This is all Chapter 1 stuff.
> If you were really a math genius you would have discovered all
> these for yourself. That hasn't happened. You don't have a
> single nontrivial lemma to your name in algebraic number theory,
> no general methods, nothing anyone else wants to use. Your
> "Advanced Polynomial Factorization" is not worth the paper it's
> written on. It's built on a simple error that you yourself
> cannot understand.
>
>
> That's the fact, James Harris.
Mathematics is a hard discipline. It doesn't care about social status
or who says what as it's about truth. Maybe for someone like Nora
Baron mathematical truth is too hard to take, so you get chattering
when mathematical logic is needed. Maybe Nora Baron is desperate to
hold on to a math world where she understands, where a math professor
is at the top of the heap, and mathematicians are the ones who win
arguments like the one being had.
But the fact is that sticking to the math, my conclusion is the only
one that's possible, which is that in the situation discussed the a's
are ALL coprime to 5 in the ring of algebraic integers.
It's bizarre and wacky, but also fascinating.
James Harris
C. Bond
[snip]
> The polynomial JSH claims leads to this weird conclusion is this:
>
> 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>
> I will show that the a's cannot be coprime to 5. Here's how:
>
> Consider the polynomial
>
> p(x) = x^3 - 12 x^2 + 65,
>
> and note that the roots of p(x) are precisely the negatives of the a's.
>
> Define the polynomials
>
> q(x) = 8 x^2 - 76 x -185
> r(x) = 8 x^2 - 4 x - 45
> s(x) = 4 x^2 - 37 x - 104
>
> The following is then true:
> For z any root of the polynomial
>
> p(x) = x^3 - 12 x^2 + 65,
>
> we have
>
> q(z)*r(z) = 5
> r(z)*s(z) = z.
>
> By virtue of the polynomials q,r,s having integer coefficients, and
> of z being a root of p(x), we know that z, q(z), r(z), and s(z) are all
> algebraic integers.
>
> Thus, r(z) is a common factor of z and 5, in A, the ring of algebraic
> integers. Since it is guaranteed that JSH will claim r(z) is a unit
> of A, here is the minimal polynomial for r:
>
> MinPoly(r(z)) = x^3 - 969 x^2 + 315 x + 5
>
> The products were found via KASH, and have been verified in Macsyma.
[snip]
> Dale
Hmmm. Elsewhere, in this same thread, James is still confronting Nora with the claim that one or all of the a's
*must* be coprime to 5. Is there any way to bring the facts together at one point? I tried posting the exact values
of the a's with the request that James identify those which are coprime to 5 and those which are not so he would
have the benefit of illustrating his point, but he has so far declined.
Nora Baron
Warning: I am going to use the definition of
"coprime."
Suppose A and B are algebraic integers that are coprime to 5.
This means that there exist r, s, r' and s' such that
r*A + s*5 = 1 and
r'*B + s'*5 = 1.
Multiply the first equation by B:
r*A*B + s*5*B = B.
Substitute into the second equation:
r'(r*A*B + s*5*B) + s'*5 = 1.
Rearrange a bit:
r'*r*A*B + (s*B + s')*5 = 1.
Let r" = r'*r and s" = s*B + s'. Then
r"*(A*B) + s"*5 = 1.
Therefore A*B is also coprime to 5.
Thus if two numbers are coprime to 5,
their product is also coprime to 5.
I daresay you can see how to get the same
conclusion for 3 numbers, eh? If not, let
me know.
> > You must therefore conclude that NONE of a1, a2, or a3 is
> > coprime to 5. The conclusion of "Advanced Polynomial
> > Factorization" is therefore false.
>
> Which is false. It IS true that all of the a's are coprime to 5 in
> the ring of algebraic integers, as that's been proven mathematically.
>
See above.
I am not a liar.
Of course I believe there is a flaw in your paper. I specify
exactly where it is and what it is. See below for some more
details.
> If you're not lying Nora Baron, I challenge you to give a flaw in my
> paper Advanced Polynomial Factorization.
>
I did. I said explicitly where the error is and what it is.
But there is also value in giving the proof that I gave, and
in Dale Hall's proof. Why? Because if you make some trivial
change in your argument, but come to same conclusion, we will
still know it is wrong.
You love saying that proofs can't duel. I accept that. It
is a logical consequence of the definition of the word "proof".
But arguments can duel. You don't have a proof. You have an argument.
Your argument contains an error. I know where it is and what
it is, and I have pointed it out. It is simply too bad if
you cannot understand what I am saying. Your only recourse is
to find an error in MY argument AND an error in Dale Hall's
separate argument.
> You see, enough with all the assertions without proof when because
> it's mathematics there's such an *easy* path to take.
>
My principle assertion is that I can prove your main conclusion
in "Advanced Polynomial Factorization" is wrong. I gave a proof
in detail. I note that you have not identified an error in it.
W. Dale Hall, in this same thread (as well as in an earlier
thread) has given a different, totally independent proof of the
same thing.
I also specified exactly where your reasoning went wrong. There
is some more detail on this below.
Good example, I like it. Note that when x = 0, the
degree of x^2 + 4*x + 4 drops from 2 to 0. Note also
that P(0) is divisible by 4. Can I conclude from that
that P(x), for any other integer x, is divisible by 4?
No: let x = 1, for example. Yet this is exactly parallel
to what you are doing in your proof: concluding from
a degenerate case something which you want to be true
more generally: claiming it without proof, and obviously,
in view of my proof and Dale Hall's proof, claiming it
*incorrectly*.
Here is another example, a little closer to what you
are doing:
Let P(m) = m*x^2 + 7*x + 7.
Note that when m = 0, P(m) = 7*x + 7. As a polynomial
in x, this is divisible by 7.
Can I therefore conclude that P(m) is divisible by 7
for all values of m?
Again, this is analogous to what you are doing.
I know you will not agree. Your polynomial is
*different*. Of course it is. It is more complicated.
But the principle is the same. Most importantly,
you have said something about factoring that polynomial
in the degenerate case and concluded that that same
something must be true more generally. You do see
that that is what you are doing, eh? And you do see
that you are doing it without justification, eh?
That is, without proof?
And you do see that your unproven result disagrees
with the my proof and that of Dale Hall, eh? So
it *must be wrong*. Remember? Proofs cannot duel,
but your argument has a big fat gap. It is not a proof.
> Now the actual expression in the paper is just a little more
> sophisticated, but that's basically what's done in the paper.
>
> I point out that Nora Baron buried her assertion in quite a bit of
> verbiage.
>
It's not buried. It's right there in plain English and math.
> > Perhaps for you it is a kind of 'pons asinorum.' Do you
> > remember that from geometry? Anyway, it leaves you stuck thinking
> > you have a valid proof, not understanding why you don't;
> > thinking mathematicians are all stupid liars, avoiding what has
> > cursed you for 8 years.
>
> Looks to me like you're desperate Nora Baron or you wouldn't have
> buried your supposed objection in so many words where it looks like
> you're trying to psyche me out.
>
> > 8 years: it's quite sad. You have learned a little math in
> > that time. You have learned what a ring is, what algebraic
> > numbers are, what algebraic integers are, what fields are, a tiny
> > bit of Galois theory. You have learned about congruences, Fermat's
> > Little Theorem, the fact that there are no primes in the algebraic
> > integers, the fact that the AIs are infinitely divisible. You
> > have learned a bit about permutations. Maybe you know a little
> > about conjugates. You know what units are.
>
> Now if you had an actual error in my paper, what's with all the
> rhetoric?
>
See above.
Nora B.
[blather deleted]
Brian Quincy Hutchings
you nasty ingrate -- sorry!:
let's say we have triangle ABC with |AB| = |AC| now, if you have
already proved the SSS theorem, then this theorem is easy to prove.
Just draw the median from A to side BC, and you're pretty much home.
If not, then you have to do some things with isometries to show that
if two triangles have three corresponding sides of equal length, then
they are congruent.
My geometry book states:
"The theorem is known as the ass's bridge, probably because of the
drawing used in Euclid's proof and because of the dimness of anyone
who cannot grasp the theorem."
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03070...@posting.google.com>...
> > Perhaps for you it is a kind of 'pons asinorum.' Do you
> > remember that from geometry? Anyway, it leaves you stuck thinking
> > you have a valid proof, not understanding why you don't;
> > thinking mathematicians are all stupid liars, avoiding what has
> > cursed you for 8 years.
> > 8 years: it's quite sad. You have learned a little math in
> > that time. You have learned what a ring is, what algebraic
> > numbers are, what algebraic integers are, what fields are, a tiny
> > bit of Galois theory. You have learned about congruences, Fermat's
> > Little Theorem, the fact that there are no primes in the algebraic
> > integers, the fact that the AIs are infinitely divisible. You
> > have learned a bit about permutations. Maybe you know a little
> > about conjugates. You know what units are.
> > It's sad that I can summarize what you know in one paragraph,
> > 8+ lines.
> >
> > In 8 years, you have learned about as much as an average student
> > would in 2-3 weeks in an algebraic number theory course. You don't
--Dec.2000 'WAND' Chairman Paul O'Neill, reelected
to Board. Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
James Harris
> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03070...@posting.google.com>...
> > nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03070...@posting.google.com>...
<deleted>
> > > Thanks for conceding "that weird conclusion."
> >
> > There's a problem in mathematics, which is fascinating one, and yes,
> > it also is rather wacky.
> >
> > > But you didn't go quite far enough. Yes, I show that if one
> > > of a1, a2, or a3 is coprime to 5, then they all are. But if
> > > you had read a little farther into the proof, you would have
> > > seen this:
> > >
> > > "But a1 * a2 * a3 = 65. Therefore at least one of a1,
> > > a2, or a3 is NOT coprime to 5. Therefore a contradiction.
> > > The source of the contradiction was the assumption that
> > > one of a1, a2, or a3 is coprime to 5."
> >
> > But that's wrong. Now you need to try and prove that given that
> >
> > a1 a2 a3 = 65
> >
> > in the ring of algebraic integers, a1, a2 or a3 must each have some
> > non unit factor of 5.
> >
> > That should be interesting.
> >
>
> Warning: I am going to use the definition of
> "coprime."
>
> Suppose A and B are algebraic integers that are coprime to 5.
>
> This means that there exist r, s, r' and s' such that
>
> r*A + s*5 = 1 and
Let's consider that in the ring of evens with 2 and 6, as 2 is coprime
to 6 in the ring of evens.
Then by your reasoning I'd have to have something like
2r + 6s = 1
which is not true, as 1 isn't even in the ring.
Yet *somehow* 2 is coprime to 6 in the ring of evens.
<deleted>
> Thus if two numbers are coprime to 5,
> their product is also coprime to 5.
Your approach is faulty in context as it doesn't handle if one number
isn't in the ring.
You've moved outside the ring.
The problem requires that you concentrate Nora Baron. You need to be
very careful with the mathematics.
> I daresay you can see how to get the same
> conclusion for 3 numbers, eh? If not, let
> me know.
Why don't you think about 2 and 6 in the ring of evens, where 2 is
coprime to 6?
Also, remember that you can't talk of a product in the ring of
algebraic integers, where some of your wouldbe factors are outside the
ring.
> > > You must therefore conclude that NONE of a1, a2, or a3 is
> > > coprime to 5. The conclusion of "Advanced Polynomial
> > > Factorization" is therefore false.
> >
> > Which is false. It IS true that all of the a's are coprime to 5 in
> > the ring of algebraic integers, as that's been proven mathematically.
> >
>
>
> See above.
All the a's must be coprime to 5 in the ring of algebraic integers.
For readers who find this confusing you need to focus on that "in the
ring of algebraic integers" or you may run in circles never figuring
out exactly what's wrong.
<deleted>
> > >
> > > No. You were right when you agreed that I had shown that
> > > if one of a1, a2, and a3 were coprime to 5, then they all
> > > are. You did not get the conclusion of my argument. Why?
> > > Because you are desperately reluctant to admit what it
> > > implies: that your main argument in "Advanced Polynomial
> > > Factorization" is incorrect.
> >
> > Really? I think you're lying because if you *believe* what you're
> > saying then you must believe there's a flaw in the paper. If there's
> > a flaw in the paper then someone can point it out.
> >
>
>
> I am not a liar.
How are readers to judge? I suggest to you Nora Baron that readers
must judge by what you *say* mathematically, not what you say in your
defense when accused.
Shorten your posts, focus on the math, and find an actual error or
mathematical support for your position, and prove that you're not a
liar.
> Of course I believe there is a flaw in your paper. I specify
> exactly where it is and what it is. See below for some more
> details.
>
Yet your claim is false, and as I explained before, I'll explain again
now.
> > If you're not lying Nora Baron, I challenge you to give a flaw in my
> > paper Advanced Polynomial Factorization.
> >
>
>
> I did. I said explicitly where the error is and what it is.
You did not, and have not.
> But there is also value in giving the proof that I gave, and
> in Dale Hall's proof. Why? Because if you make some trivial
> change in your argument, but come to same conclusion, we will
> still know it is wrong.
Why would I make any changes?
I strongly suggest to the reader that Nora Baron is lying.
> You love saying that proofs can't duel. I accept that. It
> is a logical consequence of the definition of the word "proof".
> But arguments can duel. You don't have a proof. You have an argument.
> Your argument contains an error. I know where it is and what
> it is, and I have pointed it out. It is simply too bad if
> you cannot understand what I am saying. Your only recourse is
> to find an error in MY argument AND an error in Dale Hall's
> separate argument.
>
If you actually pointed out an error, why would you have to talk so
much?
I suggest to readers that Nora Baron is trying to convince them, not
with mathematics, but with rhetoric.
> > You see, enough with all the assertions without proof when because
> > it's mathematics there's such an *easy* path to take.
> >
>
> My principle assertion is that I can prove your main conclusion
> in "Advanced Polynomial Factorization" is wrong. I gave a proof
> in detail. I note that you have not identified an error in it.
That is a lie. As you claim that the contradiction raised by all of
the a's being coprime to 5 proves that none of them can be coprime to
5 in the ring of algebraic integers, I've pointed out that is false,
as in fact they all *are* coprime to 5 in that ring.
To date you have NOT given a proof that none of them can be coprime to
5 in the ring of algebraic integers.
You have asserted. When challenged you gave what I've already
commented on in this post.
> W. Dale Hall, in this same thread (as well as in an earlier
> thread) has given a different, totally independent proof of the
> same thing.
He has not proven any such thing as what he posted doesn't rule out
*all* of the a's being coprime to 5 in the ring of algebraic integers.
> I also specified exactly where your reasoning went wrong. There
> is some more detail on this below.
>
You made a false assertion, which I answered.
Why would you? The assertion I made is that P(0) is the constant term
of the polynomial.
> No: let x = 1, for example. Yet this is exactly parallel
> to what you are doing in your proof: concluding from
> a degenerate case something which you want to be true
> more generally: claiming it without proof, and obviously,
> in view of my proof and Dale Hall's proof, claiming it
> *incorrectly*.
That statement is false.
Again consider P(x) = x^2 y^2 + 2x + 3y + 2, where if you wish I can
say y=2, so that you have P(x) = 4x^2 + 2x + 8.
But notice that with the original P(0) = 3y + 2 = 8, with y=2.
That's the constant term of the polynomial P(x).
So why have the extra symbol y?
Because I can factor the polynomial differently because of it.
That is, having the symbol y instead of the number 2, means that I can
factor the polynomial P(x) in ways that I couldn't otherwise.
What I do in the paper is consider a more sophisticated example, where
the extra symbols allow me to get non polynomial factors of the
polynomial which in the paper is P(m).
>
> Here is another example, a little closer to what you
> are doing:
>
> Let P(m) = m*x^2 + 7*x + 7.
>
>
> Note that when m = 0, P(m) = 7*x + 7. As a polynomial
> in x, this is divisible by 7.
And in context, the constant term of the polynomial is P(0) = 7x+7.
> Can I therefore conclude that P(m) is divisible by 7
> for all values of m?
Nope.
> Again, this is analogous to what you are doing.
You are lying.
> I know you will not agree. Your polynomial is
> *different*. Of course it is. It is more complicated.
> But the principle is the same. Most importantly,
> you have said something about factoring that polynomial
> in the degenerate case and concluded that that same
> something must be true more generally. You do see
> that that is what you are doing, eh? And you do see
> that you are doing it without justification, eh?
> That is, without proof?
Well the polynomial in the paper is
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)
and you'll notice that it has a factor that is f^2.
I consider P(0) to get the constant term, but factor with respect to
x, and that's it.
Notice in context now Nora Baron's example above where she talked of
factors of the constant term and then questioned them being factors of
the polynomial.
However here you can see the factor f^2 is clearly not just a factor
of the constant term.
> And you do see that your unproven result disagrees
> with the my proof and that of Dale Hall, eh? So
> it *must be wrong*. Remember? Proofs cannot duel,
> but your argument has a big fat gap. It is not a proof.
>
What's with all the verbiage? You talk a lot, and I see it as trying
to convince people of a lie.
> > Now the actual expression in the paper is just a little more
> > sophisticated, but that's basically what's done in the paper.
And I've given the expression this time.
> > I point out that Nora Baron buried her assertion in quite a bit of
> > verbiage.
> >
>
> It's not buried. It's right there in plain English and math.
>
The readers can judge for themselves.
> > > Perhaps for you it is a kind of 'pons asinorum.' Do you
> > > remember that from geometry? Anyway, it leaves you stuck thinking
> > > you have a valid proof, not understanding why you don't;
> > > thinking mathematicians are all stupid liars, avoiding what has
> > > cursed you for 8 years.
> >
> > Looks to me like you're desperate Nora Baron or you wouldn't have
> > buried your supposed objection in so many words where it looks like
> > you're trying to psyche me out.
> >
> > > 8 years: it's quite sad. You have learned a little math in
> > > that time. You have learned what a ring is, what algebraic
> > > numbers are, what algebraic integers are, what fields are, a tiny
> > > bit of Galois theory. You have learned about congruences, Fermat's
> > > Little Theorem, the fact that there are no primes in the algebraic
> > > integers, the fact that the AIs are infinitely divisible. You
> > > have learned a bit about permutations. Maybe you know a little
> > > about conjugates. You know what units are.
> >
> > Now if you had an actual error in my paper, what's with all the
> > rhetoric?
> >
>
> See above.
>
>
> Nora B.
>
> [blather deleted]
Oh and in that "blather" I noted that Nora Baron may not care about
the truth, but about social issues. That is, her brain may be quite
dead set against the truth, as it'd upset the status quo.
However I hope that some of you prize that quality of mathematics
where truth is independent of social issues. If a proof being true
meant that all of humanity would die, you'd just have fun with the
time you have left because you'd realize that no matter what all of
humanity thought, no matter what they did, they couldn't change the
truth.
Proofs don't give a damn about society, social issues, or whether or
not someone can handle the truth.
Proofs are just true.
And for some of you, I thought that was a large part of why you liked
mathematics.
James Harris
Michael Brown
> Nora Baron wrote:
> > James Harris wrote:
> > > Nora Baron wrote:
<snip>
> > > > But you didn't go quite far enough. Yes, I show that if one
> > > > of a1, a2, or a3 is coprime to 5, then they all are. But if
> > > > you had read a little farther into the proof, you would have
> > > > seen this:
> > > >
> > > > "But a1 * a2 * a3 = 65. Therefore at least one of a1,
> > > > a2, or a3 is NOT coprime to 5. Therefore a contradiction.
> > > > The source of the contradiction was the assumption that
> > > > one of a1, a2, or a3 is coprime to 5."
> > >
> > > But that's wrong. Now you need to try and prove that given that
> > >
> > > a1 a2 a3 = 65
> > >
> > > in the ring of algebraic integers, a1, a2 or a3 must each have some
> > > non unit factor of 5.
> > >
> > > That should be interesting.
> > >
> >
> > Warning: I am going to use the definition of
> > "coprime."
> >
> > Suppose A and B are algebraic integers that are coprime to 5.
> >
> > This means that there exist r, s, r' and s' such that
> >
> > r*A + s*5 = 1 and
>
> Let's consider that in the ring of evens with 2 and 6, as 2 is coprime
> to 6 in the ring of evens.
This is a dodge, since that the ring of evens does not have a multiplicitive
identity, yet the ring of algebraic integers does. Since Nora is talking
about a unitary ring, you should as well. However, you are quite right that
the above equation does not work in non-unitary rings.
OK, back to lurking ...
<snip>
--
Michael Brown
Add michael@ to emboss.co.nz - My inbox is always open
fishfry
"Michael Brown" <s...@signature.below> wrote:
Are you sure you're just lurking? Or are you another closet JSH addict?
Virgil
s...@signature.below wrote on 7/7/03 21:46:
Even worse, James Harris didn't even notice that coprimeness cannot even be
defined in rings without a unity, so his "evens" example fails before it
starts.
James Harris
It's not a dodge, it's a fact. My point is that the assertion that
coprimeness is defined by just that *one* way is false, as I can
easily show with 2 and 6.
That's important because algebraic integers have a quirky problem,
which people might approach by considering 2 being coprime to 6 in the
ring of evens.
That is, you can have an algebraic integer z, where z=xy, and x is an
algebraic integer, but y is not, so z is coprime to x, in the ring of
algebraic integers.
I like the example with 6=2(3) where in the ring of evens 2 is coprime
to 6 because it shows how that's possible without fractions or
anything "fractional".
I remind that the problem has been in established mathematics for over
a hundred years. If you even *consider* the possibility, the
mathematics is almost trivial. It's easy for me to prove the flaw,
but so far, from what I've seen in posts many of you have simply
denied the algebra.
Those of you who are mathematicians know that the *idea* of a problem
in "core" would be resisted. Just the idea of a problem there is
probably something you may dogmatically resist. But here's a test of
what kind of mathematician are you: Are you one in name only, or can
you go with mathematical logic, no matter how much it hurts?
James Harris
David C. Ullrich
It's an irrelevant fact. What words mean depends on the
context. The _definition_ of "x and y are coprime" _in_
the algebraic integers is that there exist a and b with
ax + by = 1. (From which it follows easily that if x and y
are both coprime to z then xy is coprime to z, as Nora
proved in detail above.) That's _not_ the definition of
coprimness in the ring of even integers.
The fact that the definition makes no sense in one
context doesn't mean that it's not the definition in
another context. It _is_ the standard definition for
coprimeness in the algebraic integers.
Of course the fact that it's the standard definition
doesn't mean you can't use the word "coprime" to
mean something else. You can use the word
"coprime" to mean anything you want, as long
as you _say_ what your definition _is_. This
is one of the many things you've never done,
say exactly what you mean by "coprime". If
you gave a coherent definition then people would
use that definition in discussing your work - since
you repeatedly refuse to say what definition you're
using the standard definition is the only one people
have to go by.
>My point is that the assertion that
>coprimeness is defined by just that *one* way is false, as I can
>easily show with 2 and 6.
>
>That's important because algebraic integers have a quirky problem,
>which people might approach by considering 2 being coprime to 6 in the
>ring of evens.
>
>That is, you can have an algebraic integer z, where z=xy, and x is an
>algebraic integer, but y is not, so z is coprime to x, in the ring of
>algebraic integers.
No, that doesn't follow. At least it doesn't follow with the standard
definition of "coprime" - if you have some other definition in mind
you need to _state_ the definition!
(Why it doesn't follow with the standard definition: Let x = 4,
y = 1/2, z = 2. Then x, z are algebraic integers, y is not an
algebraic integer and z = xy. But x and z are not coprime:
Suppose to the contrary that ax + bz = 1 for some algebraic
integers a and b. That says 4a + 2b = 1. Hence 2a + b = 1/2,
which shows that 1/2 is an algebraic integer, which it's not.)
>I like the example with 6=2(3) where in the ring of evens 2 is coprime
>to 6 because it shows how that's possible without fractions or
>anything "fractional".
>
>I remind that the problem has been in established mathematics for over
>a hundred years. If you even *consider* the possibility, the
>mathematics is almost trivial. It's easy for me to prove the flaw,
>but so far, from what I've seen in posts many of you have simply
>denied the algebra.
>
>Those of you who are mathematicians know that the *idea* of a problem
>in "core" would be resisted. Just the idea of a problem there is
>probably something you may dogmatically resist. But here's a test of
>what kind of mathematician are you: Are you one in name only, or can
>you go with mathematical logic, no matter how much it hurts?
>
>
>James Harris
************************
David C. Ullrich
Nora Baron
This seems rather silly, since obviously the algebra
here was in the ring of algebraic integers, not in the
ring of even numbers mod 2. The universally accepted
definition of coprime that I am using *requires* that the
ring be a ring with unit, which the AI's obviously
are and the even integers obviously are not.
You really don't want to accept this, do you?
> <deleted>
>
> > Thus if two numbers are coprime to 5,
> > their product is also coprime to 5.
>
> Your approach is faulty in context as it doesn't handle if one number
> isn't in the ring.
>
> You've moved outside the ring.
>
Nonsense. See above. 1 is in the ring of
algebraic integers. What on earth are you
thinking ???
> The problem requires that you concentrate Nora Baron. You need to be
> very careful with the mathematics.
>
Agreed!
> > I daresay you can see how to get the same
> > conclusion for 3 numbers, eh? If not, let
> > me know.
>
> Why don't you think about 2 and 6 in the ring of evens, where 2 is
> coprime to 6?
>
See above. This is just smoke and blather, irrelevant to
my argument as you well know.
> Also, remember that you can't talk of a product in the ring of
> algebraic integers, where some of your wouldbe factors are outside the
> ring.
>
Like what? A, B, r, s, r', s', 5, and 1 are all algebraic
integers, and the algebraic integers are closed under sums
and products (i.e., they form a ring). Again, what the heck
are you talking about ???
> > > > You must therefore conclude that NONE of a1, a2, or a3 is
> > > > coprime to 5. The conclusion of "Advanced Polynomial
> > > > Factorization" is therefore false.
> > >
> > > Which is false. It IS true that all of the a's are coprime to 5 in
> > > the ring of algebraic integers, as that's been proven mathematically.
> > >
> >
> >
> > See above.
>
> All the a's must be coprime to 5 in the ring of algebraic integers.
>
Assuming your argument is right. Which it isn't.
> For readers who find this confusing you need to focus on that "in the
> ring of algebraic integers" or you may run in circles never figuring
> out exactly what's wrong.
>
> <deleted>
>
> > > >
> > > > No. You were right when you agreed that I had shown that
> > > > if one of a1, a2, and a3 were coprime to 5, then they all
> > > > are. You did not get the conclusion of my argument. Why?
> > > > Because you are desperately reluctant to admit what it
> > > > implies: that your main argument in "Advanced Polynomial
> > > > Factorization" is incorrect.
> > >
> > > Really? I think you're lying because if you *believe* what you're
> > > saying then you must believe there's a flaw in the paper. If there's
> > > a flaw in the paper then someone can point it out.
> > >
> >
> >
> > I am not a liar.
>
> How are readers to judge? I suggest to you Nora Baron that readers
> must judge by what you *say* mathematically, not what you say in your
> defense when accused.
>
Couldn't agree more. Readers should not take anything I say
on faith. Judge the math for yourselves, readers!
> Shorten your posts, focus on the math, and find an actual error or
> mathematical support for your position, and prove that you're not a
> liar.
>
> > Of course I believe there is a flaw in your paper. I specify
> > exactly where it is and what it is. See below for some more
> > details.
> >
>
> Yet your claim is false, and as I explained before, I'll explain again
> now.
>
> > > If you're not lying Nora Baron, I challenge you to give a flaw in my
> > > paper Advanced Polynomial Factorization.
> > >
> >
> >
> > I did. I said explicitly where the error is and what it is.
>
> You did not, and have not.
>
I did. I said it was in the part where you consider a
degenerate 1-degree case of your polynomial, which ordinarily
has degree 3. I said you assumed that a factorization
property which applies to the degree 1 case (when m = 0)
can be generalized to apply when m is nonzero. Here is the
explicit statement in your argument:
" ... 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."
The part I capitalized is the part you are
asserting without any proof. You are assuming that
a pattern you observe in a degenerate case can be
assumed to hold in all nondegenerate cases. There
is no algebraic theorem or underlying principle
which justifies that key phrase, "IN GENERAL".
This is the big fat gap in your argument.
Is this explicit enough for you? I must say I
am rather annoyed at having to point this out. If
you weren't so heavily into denial you would have
seen it for yourself. And even if you didn't, you
still have my proof and Dale Hall's to contend
with. What that means is that not only do you
have a gap, but also you are not going to be
able to fill it. The main conclusion of APF
is *false* and you cannot fix it.
> > But there is also value in giving the proof that I gave, and
> > in Dale Hall's proof. Why? Because if you make some trivial
> > change in your argument, but come to same conclusion, we will
> > still know it is wrong.
>
> Why would I make any changes?
>
> I strongly suggest to the reader that Nora Baron is lying.
>
The reader should judge for him or herself on that.
> > You love saying that proofs can't duel. I accept that. It
> > is a logical consequence of the definition of the word "proof".
> > But arguments can duel. You don't have a proof. You have an argument.
> > Your argument contains an error. I know where it is and what
> > it is, and I have pointed it out. It is simply too bad if
> > you cannot understand what I am saying. Your only recourse is
> > to find an error in MY argument AND an error in Dale Hall's
> > separate argument.
> >
>
> If you actually pointed out an error, why would you have to talk so
> much?
>
> I suggest to readers that Nora Baron is trying to convince them, not
> with mathematics, but with rhetoric.
>
Whatever works. I provided both.
> > > You see, enough with all the assertions without proof when because
> > > it's mathematics there's such an *easy* path to take.
> > >
> >
> > My principle assertion is that I can prove your main conclusion
> > in "Advanced Polynomial Factorization" is wrong. I gave a proof
> > in detail. I note that you have not identified an error in it.
>
> That is a lie. As you claim that the contradiction raised by all of
> the a's being coprime to 5 proves that none of them can be coprime to
> 5 in the ring of algebraic integers, I've pointed out that is false,
> as in fact they all *are* coprime to 5 in that ring.
>
No - see above. You ring-of-even-numbers objection is
irrelevant and specious, and you know it.
> To date you have NOT given a proof that none of them can be coprime to
> 5 in the ring of algebraic integers.
>
See above.
> You have asserted. When challenged you gave what I've already
> commented on in this post.
>
For most people what I gave previously would have been
enough. You however have lost all your objectivity. You
continue to throw out silly objections (e.g. the even number
argument). As in previous arguments you have to be
dragged kicking and screaming to what is true. That is
what is happening now.
> > W. Dale Hall, in this same thread (as well as in an earlier
> > thread) has given a different, totally independent proof of the
> > same thing.
>
> He has not proven any such thing as what he posted doesn't rule out
> *all* of the a's being coprime to 5 in the ring of algebraic integers.
>
I am pretty sure he assumed the following: that if
5 is a divisor of a1 * a2 * a3, then at least one of
a1, a2, or a3 shares algebraic integer factors with 5.
You yourself have freely assumed that kind of thing in
the past. It makes intuitive sense. I gave an explicit,
complete proof of it in the post to which you are
replying. For most people that proof would not have
been necessary. I guess given what is at stake for you,
it is not totally unreasonable for you to ask for it.
> > I also specified exactly where your reasoning went wrong. There
> > is some more detail on this below.
> >
>
> You made a false assertion, which I answered.
>
Your answer, involving a ring without unit,
is irrelevant.
Why would I? Because it is extremely similar to what
you do with P(0) in your argument. Certainly not because
I believe it.
> > No: let x = 1, for example. Yet this is exactly parallel
> > to what you are doing in your proof: concluding from
> > a degenerate case something which you want to be true
> > more generally: claiming it without proof, and obviously,
> > in view of my proof and Dale Hall's proof, claiming it
> > *incorrectly*.
>
> That statement is false.
>
Better be explicit on that. What part is false,
and why? Note that saying that you have a "proof"
is not sufficient, since the whole issue here is,
your (non) "proof" has a big fat gap and two of us have
independent counterproofs.
> Again consider P(x) = x^2 y^2 + 2x + 3y + 2, where if you wish I can
> say y=2, so that you have P(x) = 4x^2 + 2x + 8.
>
> But notice that with the original P(0) = 3y + 2 = 8, with y=2.
>
> That's the constant term of the polynomial P(x).
>
> So why have the extra symbol y?
>
> Because I can factor the polynomial differently because of it.
>
> That is, having the symbol y instead of the number 2, means that I can
> factor the polynomial P(x) in ways that I couldn't otherwise.
>
> What I do in the paper is consider a more sophisticated example, where
> the extra symbols allow me to get non polynomial factors of the
> polynomial which in the paper is P(m).
>
> >
> > Here is another example, a little closer to what you
> > are doing:
> >
> > Let P(m) = m*x^2 + 7*x + 7.
> >
> >
> > Note that when m = 0, P(m) = 7*x + 7. As a polynomial
> > in x, this is divisible by 7.
>
> And in context, the constant term of the polynomial is P(0) = 7x+7.
>
Yep.
> > Can I therefore conclude that P(m) is divisible by 7
> > for all values of m?
>
> Nope.
>
Right again!
> > Again, this is analogous to what you are doing.
>
> You are lying.
>
Nope.
You have yet to correctly identify any error in my
proof or Dale Hall's. I, on the other hand, have
zeroed in on the error in your argument.
>
> > > Now the actual expression in the paper is just a little more
> > > sophisticated, but that's basically what's done in the paper.
>
> And I've given the expression this time.
>
>
> > > I point out that Nora Baron buried her assertion in quite a bit of
> > > verbiage.
> > >
> >
> > It's not buried. It's right there in plain English and math.
> >
>
> The readers can judge for themselves.
>
Absolutely.
> > > > Perhaps for you it is a kind of 'pons asinorum.' Do you
> > > > remember that from geometry? Anyway, it leaves you stuck thinking
> > > > you have a valid proof, not understanding why you don't;
> > > > thinking mathematicians are all stupid liars, avoiding what has
> > > > cursed you for 8 years.
> > >
> > > Looks to me like you're desperate Nora Baron or you wouldn't have
> > > buried your supposed objection in so many words where it looks like
> > > you're trying to psyche me out.
> > >
> > > > 8 years: it's quite sad. You have learned a little math in
> > > > that time. You have learned what a ring is, what algebraic
> > > > numbers are, what algebraic integers are, what fields are, a tiny
> > > > bit of Galois theory. You have learned about congruences, Fermat's
> > > > Little Theorem, the fact that there are no primes in the algebraic
> > > > integers, the fact that the AIs are infinitely divisible. You
> > > > have learned a bit about permutations. Maybe you know a little
> > > > about conjugates. You know what units are.
> > >
> > > Now if you had an actual error in my paper, what's with all the
> > > rhetoric?
> > >
> >
> > See above.
> >
> >
> > Nora B.
> >
> > [blather deleted]
>
> Oh and in that "blather" I noted that Nora Baron may not care about
> the truth, but about social issues. That is, her brain may be quite
> dead set against the truth, as it'd upset the status quo.
>
More blather.
> However I hope that some of you prize that quality of mathematics
> where truth is independent of social issues. If a proof being true
> meant that all of humanity would die, you'd just have fun with the
> time you have left because you'd realize that no matter what all of
> humanity thought, no matter what they did, they couldn't change the
> truth.
>
More blather. When someone starts injecting phrases like
"social issues" and "all of humanity" into a mathematical
discussion, you know the manure level is rising over the top
of your boots, and you know that person doesn't really want
to deal with the math.
> Proofs don't give a damn about society, social issues, or whether or
> not someone can handle the truth.
>
> Proofs are just true.
>
Not if they have big fat gaps or errors, like
yours.
Nora B.
W. Dale Hall
James Harris wrote:
> nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03070...@posting.google.com>...
... stuff deleted ...
>> W. Dale Hall, in this same thread (as well as in an earlier
>>thread) has given a different, totally independent proof of the
>>same thing.
>
>
> He has not proven any such thing as what he posted doesn't rule out
> *all* of the a's being coprime to 5 in the ring of algebraic integers.
>
Baloney. I have given, for *each* root z of the polynomial
p(x) = x^3 - 12 x^2 + 65,
a common factor r(z)= 8 z^2 - 4 z - 45
between z and 5. Let q,r,s be defined as follows:
q(x) = 8 x^2 - 76 x -185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104
Then
q(z)*r(z) = 5
r(z)*s(z) = z.
Further, r(z) is not a unit. The splitting field K of the polynomial
r(x) over the rationals Q has a multiplicative map called the norm:
N: K ----> Q
The norm has these properties:
1. N( integer of K ) = integer of Q (= ordinary integer)
2. N( *** ) = integer <==> *** is an integer
3. N( *** ) = unit of Z (that is, +/- 1)
<==> *** is a unit in the ring of integers.
4. N( r(z) ) = -5, for each of the roots z defined above.
Thus, r(z) is a factor in common between z and 5. In every case, r(z) is
an algebraic integer, and in no case (for these roots z) is r(z) a unit.
Thus, NONE of the a's is coprime to 5.
You are wrong.
Your method of argument is flawed, leading you to draw incorrect
conclusions.
... stuff deleted ...
However, I couldn't resist including this:
> However I hope that some of you prize that quality of mathematics
> where truth is independent of social issues. If a proof being true
> meant that all of humanity would die, you'd just have fun with the
> time you have left because you'd realize that no matter what all of
> humanity thought, no matter what they did, they couldn't change the
> truth.
>
How does that figure into the fact that you can't abide the existence
of the above factorizations? None of the a's is comprime to 5. You now
have the actual factors (well, you might have to do some arithmetic).
> Proofs don't give a damn about society, social issues, or whether or
> not someone can handle the truth.
>
Apparently, that person who can't handle the truth is you.
> Proofs are just true.
>
Problem is, your methodology does not yield correct arguments. It never
has, and, until you straighten up and give up on that sorry-ass line of
argument you have, it never will.
The mathematics doesn't lie. There are the factors. You are ignoring
them. You are the one in error.
> And for some of you, I thought that was a large part of why you liked
> mathematics.
>
You think people believe you're correct, and yet argue against your
position? Think again. Or, just never mind. Go watch another action
movie and get your agression rocks off in your own mind.
>
> James Harris
Dale.
C. Bond
defend the truth. He will not confront or acknowledge his errors unless someone succeeds in rubbing his nose in
them, and sometimes not even then.
Some of the more capable and literate of the contributors to these threads have succeeded in meeting the challenge
of identifying, analyzing and disproving James' fundamental claims, but that will never be enough to shut him up.
Even a simple, concrete counterexample will not serve. The long history of promoting errors would be sad if James
were simply a would-be technologist who found himself in way over his head. But this is not the case. He is an
abusive, arrogant megalomaniac with no evidence of talent, cognitive skills or humanity.
Perhaps he disagrees with my assessment. If so, I call on him to correct his errors and see where his arguments
lead when they are not diverted by mistakes.
--
A man of integrity will identify, acknowledge and correct his errors. A man without it will deny, ignore or defend
them.
Arturo Magidin
No, 2 is not coprime to 6 in the ring 2Z.
Your error here is apparently thinking that "coprime" means that they
have no common divisors other than units. This is not the case in
general. In an arbitrary ring, "a and b are coprime" is STRONGER than
the statement "any common divisor of a and b is a unit."
For example, in the ring Z[sqrt(-5)], 2 and 1+sqrt(-5) have no common
divisors other than units, but they are NOT coprime.
As it happens, IN THE RING OF ALL ALGEBRAIC INTEGERS, one can prove
that "a and b are coprime" is equivalent to "any common divisor of a
and b is a unit", but that is a property that only SOME rings have.
>Then by your reasoning I'd have to have something like
>
> 2r + 6s = 1
>
>which is not true, as 1 isn't even in the ring.
>
>Yet *somehow* 2 is coprime to 6 in the ring of evens.
It is not. You are once again misusing standard terminology.
Small wonder your arguments are so... solid.
[.snip.]
======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
Arturo Magidin <mag...@math.berkeley.edu> wrote:
[.snip.]
(Nora Baron, then James Harris, then me)
>>> Warning: I am going to use the definition of
>>> "coprime."
>>>
>>> Suppose A and B are algebraic integers that are coprime to 5.
>>>
>>> This means that there exist r, s, r' and s' such that
>>>
>>> r*A + s*5 = 1 and
>>
>>Let's consider that in the ring of evens with 2 and 6, as 2 is coprime
>>to 6 in the ring of evens.
>
>No, 2 is not coprime to 6 in the ring 2Z.
Oops. This is incorrect. 2 and 6 are coprime in the ring 2Z. The
problem is that James is not using the definition, but a property
which is equivalent to it in rings with 1, which 2Z most certainly is
not. That is, he is taking a theorem out of context and trying to
apply it where it does not belong. A typical error of his.
>Your error here is apparently thinking that "coprime" means that they
>have no common divisors other than units. This is not the case in
>general. In an arbitrary ring, "a and b are coprime" is STRONGER than
>the statement "any common divisor of a and b is a unit."
>
>For example, in the ring Z[sqrt(-5)], 2 and 1+sqrt(-5) have no common
>divisors other than units, but they are NOT coprime.
>
>As it happens, IN THE RING OF ALL ALGEBRAIC INTEGERS, one can prove
>that "a and b are coprime" is equivalent to "any common divisor of a
>and b is a unit", but that is a property that only SOME rings have.
Perhaps a slightly more expansive explanation is in order,
particularly given my error above.
Given a ring R, two ideals I and J are "coprime" if and only if there
is no prime ideal of R which contains both I and J; they are
"comaximal" if and only if there is no maximal ideal of R which
contains both I and J.
In the case of a ring with 1, assuming the Axiom of Choice, two
ideals are coprime if and only if they are comaximal, if and only if
the smallest ideal containing both I and J, namely I+J, is the unit
ideal; if and only if there exist i in I and j in J such that i+j = 1.
Two elements, by extension, are said to be "coprime"
(resp. "comaximal") if and only if the principal ideals they generate
are coprime (resp. comaximal). It is then easy to verify that in the
case of a ring with 1, assuming the Axiom of Choice, two elements a
and b are coprime if and only if there exist elements x and y in the
ring such that ax + by = 1 (ax being the element of (a), by the
element of (b)). This happens if and only if the ideal (a,b) generated
by a and b is the unit ideal.
THEOREM. Let R be a ring with 1. Let a, b in R. If a and b are
coprime, then any common divisor of a and b is a unit.
Proof: Let u be a common divisor of a and b; then u divides all
elements of the form ax+by, with x,y in R. In particular, u divides 1
hence is a unit. QED
The converse is false, as the oft-mentioned (in threads involving
James) example of Z[sqrt(-5)], a=2, b=1+sqrt(-5), testifies.
The ring of all algebraic integers is a Bezout domain; this means that
any finitely generated ideal is principal. Therefore, a and b in the
ring of all algebraic integers are coprime if and only (a,b)=(1).
THEOREM. Let R be the ring of all algebraic integers, and let a and b
in R. If and b have no common divisor in R other than units, then a
and b are coprime.
Proof. Assume a and b have no common divisor in R other than units;
let x be a generator of (a,b). It is easy to verify that x must be a
common divisor of a and b: a lies in (a,b), hence a lies in (x), hence
there exists y in R such that a = xy; likewise with b. Therefore, x is
a unit by assumption; since an ideal generated by a unit is the unit
ideal, (1), this proves that (a,b)=(1), hence a and b are coprime in
R. QED.
The case of rings without 1 is much more difficult. For one, it is no
longer necessarily true that being coprime and being comaximal is
equivalent: it could be that there are no maximal ideals containing
both elements, yet there are prime ideals that do.
Let's consider the case of the ring 2Z of even integers. What is the
ideal (2), generated by 2? It contains all elements of the form 2a
with a in 2Z. So
(2) = { 4m : m in Z}, the multiples of 4.
Likewise,
(6) = { 12m: m in Z}, the multiples of 12.
Clearly, (6) is contained in (2). So 6 and 2 are coprime in 2Z if and
only if (2) is not contained in any prime ideal of 2Z.
(2) is not a prime ideal: 2*2 lies in (2), but 2 does not. Now let I
be any ideal of 2Z which properly contains (2); therefore, it contains
an even integer which is not a multiple of 4, hence an element of the
form 4m+2, with m an integer. But then, I contains 4m, hence contains
2. Since it contains 2 and all multiples of 4, it contains all even
integers, hence I=2Z. Therefore, the only ideal that properly contains
(2) is the total ideal, which is not prime either; so (2) and (6) are
indeed coprime in 2Z, hence 2 and 6 are coprime in Z.
Of course, they are NOT comaximal: we have just shown that (2) is
maximal in 2Z, and it clearly contains both (2) and (6). So here we
have an example of two ideals in a ring without 1 which are coprime
but not comaximal, something which is impossible in rings with 1 (even
if we do not assume the Axiom of Choice, which is needed to prove that
comaximal->coprime, the implication coprime->comaximal always holds in
rings with 1, since any maximal ideal which contains both will
necessarily be a prime ideal which contains both ideals).
So, in summary, the error is that James has confused the theorem:
THEOREM: If R is a ring with 1, then two elements a and b are coprime
if and only if there exists x and y in R such that ax+by =
1. (Assuming the Axiom of Choice)
with the definition
DEFINITION. Let R be a ring. Two elements a and b are coprime if and
only if there is no prime ideal that contains both the ideal (a) and
the ideal (b).
The confusion almost certainly arises from the fact that James is
using words whose meaning he does not know, pretending that he does.
Arturo Magidin
[.snip.]
(Nora Baron, then James Harris, then me)
>>> Warning: I am going to use the definition of
>>> "coprime."
>>>
>>> Suppose A and B are algebraic integers that are coprime to 5.
>>>
>>> This means that there exist r, s, r' and s' such that
>>>
>>> r*A + s*5 = 1 and
>>
>>Let's consider that in the ring of evens with 2 and 6, as 2 is coprime
>>to 6 in the ring of evens.
>
>No, 2 is not coprime to 6 in the ring 2Z.
Oops. This is incorrect. 2 and 6 are coprime in the ring 2Z. The
problem is that James is not using the definition, but a property
which is equivalent to it in rings with 1, which 2Z most certainly is
not. That is, he is taking a theorem out of context and trying to
apply it where it does not belong. A typical error of his.
>Your error here is apparently thinking that "coprime" means that they
>have no common divisors other than units. This is not the case in
>general. In an arbitrary ring, "a and b are coprime" is STRONGER than
>the statement "any common divisor of a and b is a unit."
>
>For example, in the ring Z[sqrt(-5)], 2 and 1+sqrt(-5) have no common
>divisors other than units, but they are NOT coprime.
>
>As it happens, IN THE RING OF ALL ALGEBRAIC INTEGERS, one can prove
>that "a and b are coprime" is equivalent to "any common divisor of a
>and b is a unit", but that is a property that only SOME rings have.
Perhaps a slightly more expansive explanation is in order,
equivalent; in fact, the example provided by James gives two elements
which are coprime but not comaximal. It could also be that two
elements are comaximal but not coprime, for example if the ring does
not contain maximal ideals.
Let's consider the case of the ring 2Z of even integers. What is the
ideal (2), generated by 2? It contains all elements of the form 2a
with a in 2Z. So
(2) = { 4m : m in Z}, the multiples of 4.
Likewise,
(6) = { 12m: m in Z}, the multiples of 12.
Clearly, (6) is contained in (2). So 6 and 2 are coprime in 2Z if and
only if (2) is not contained in any prime ideal of 2Z.
(2) is not a prime ideal: 2*2 lies in (2), but 2 does not. Now let I
be any ideal of 2Z which properly contains (2); therefore, it contains
an even integer which is not a multiple of 4, hence an element of the
form 4m+2, with m an integer. But then, I contains 4m, hence contains
2. Since it contains 2 and all multiples of 4, it contains all even
integers, hence I=2Z. Therefore, the only ideal that properly contains
(2) is the total ideal, which is not prime either; so (2) and (6) are
indeed coprime in 2Z, hence 2 and 6 are coprime in 2Z.
Of course, they are NOT comaximal: we have just shown that (2) is
maximal in 2Z, and it clearly contains both (2) and (6). So here we
have an example of two ideals in a ring without 1 which are coprime
but not comaximal, something which is impossible in rings with 1 (even
if we do not assume the Axiom of Choice, which is needed to prove that
comaximal->coprime, the implication coprime->comaximal always holds in
rings with 1, since any maximal ideal which contains both will
necessarily be a prime ideal which contains both ideals).
So, in summary, the error is that James has confused the theorem:
THEOREM: If R is a ring with 1, then two elements a and b are coprime
if and only if there exists x and y in R such that ax+by =
1. (Assuming the Axiom of Choice)
with the definition
DEFINITION. Let R be a ring. Two elements a and b are coprime if and
only if there is no prime ideal that contains both the ideal (a) and
the ideal (b).
The confusion almost certainly arises from the fact that James is
using words whose meaning he does not know, pretending that he does.
======================================================================
Matthew Baker
<bef8oc$1vi7$1...@agate.berkeley.edu>:
>have no common divisors other than units. This is not the case in
>general. In an arbitrary ring, "a and b are coprime" is STRONGER than
>the statement "any common divisor of a and b is a unit."
Now that's interesting - I've been quietly wondering about this while
following this conversation. Could you explain what the definition is
exactly? I can't find it any of my (admittedly limited set of) textbooks.
>As it happens, IN THE RING OF ALL ALGEBRAIC INTEGERS, one can prove
>that "a and b are coprime" is equivalent to "any common divisor of a
>and b is a unit", but that is a property that only SOME rings have.
This question may be redundant once I've seen the definition, but which
rings have this property?
Arturo Magidin
Matthew Baker <m...@haar.clara.co.uk> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) wrote in
><bef8oc$1vi7$1...@agate.berkeley.edu>:
>>Your error here is apparently thinking that "coprime" means that they
>>have no common divisors other than units. This is not the case in
>>general. In an arbitrary ring, "a and b are coprime" is STRONGER than
>>the statement "any common divisor of a and b is a unit."
>
>Now that's interesting - I've been quietly wondering about this while
>following this conversation. Could you explain what the definition is
>exactly? I can't find it any of my (admittedly limited set of) textbooks.
I think I posted the answer a bit later (since I incorrectly said that
2 and 6 are not coprime in the ring 2Z); but:
DEFINITION. Let R be a ring, I and J two ideals of R. I and J are
COPRIME (resp. COMAXIMAL) if and only if there is no prime
(resp. maximal) ideal of R which contains both I and J.
DEFINITION. Let R be a ring, a and b two elements of R. We say that a
and b are coprime (resp. comaximal) if and only if the principal
ideals (a) and (b) of R are coprime (resp. comaximal).
>>As it happens, IN THE RING OF ALL ALGEBRAIC INTEGERS, one can prove
>>that "a and b are coprime" is equivalent to "any common divisor of a
>>and b is a unit", but that is a property that only SOME rings have.
>
>This question may be redundant once I've seen the definition, but which
>rings have this property?
The answer in general may be difficult (that is, if we include rings
without 1 or noncommutative); even if I restrict myself to commutative
rings with 1, and assume the Axiom of Choice so that comaximal and
coprime are equivalent, I do not know the answer off-hand.
Clearly, rings in which every finitely generated ideal is principal
will have the property. It may fail in gcd domains (domains in which
every pair of elements have a gcd), and it may fail in UFDs (e.g., in
Z[x], x and 2 have no common divisors other than units, but they are
not coprime, since (2,x) is a proper ideal).
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Matthew Baker
@agate.berkeley.edu>:
>In article <93B2D0AC0mpb...@217.158.240.11>,
>Matthew Baker <m...@haar.clara.co.uk> wrote:
>>mag...@math.berkeley.edu (Arturo Magidin) wrote in
>><bef8oc$1vi7$1...@agate.berkeley.edu>:
>>>Your error here is apparently thinking that "coprime" means that they
>>>have no common divisors other than units. This is not the case in
>>>general. In an arbitrary ring, "a and b are coprime" is STRONGER than
>>>the statement "any common divisor of a and b is a unit."
>>
>>Now that's interesting - I've been quietly wondering about this while
>>following this conversation. Could you explain what the definition is
>>exactly? I can't find it any of my (admittedly limited set of) textbooks.
>
>I think I posted the answer a bit later (since I incorrectly said that
>2 and 6 are not coprime in the ring 2Z); but:
>
>DEFINITION. Let R be a ring, I and J two ideals of R. I and J are
>COPRIME (resp. COMAXIMAL) if and only if there is no prime
>(resp. maximal) ideal of R which contains both I and J.
>
>DEFINITION. Let R be a ring, a and b two elements of R. We say that a
>and b are coprime (resp. comaximal) if and only if the principal
>ideals (a) and (b) of R are coprime (resp. comaximal).
Yes - many thanks, your full answer in your other post is exactly what I
needed - and makes a lot of sense to me.
<matt