Simple explanation of algebraic integer issue
David Kastrup
> It turns out that the definition for algebraic integers where they
> are roots of *monic* polynomials leaves off certain numbers that
> should be included, which is how I found a way to show some wacky
> things.
>
> It's easy enough to explain as consider
>
> x^3 - 3x + 2
>
> which as it turns out has a root that is coprime to 2. That is, it
> doesn't share any non-unit factors with 2 in the ring of algebraic
> integers.
>
> Well, that *should* make it a unit i.e. factor of 1,
Why? 3 is coprime to 2, and it certainly isn't a unit either.
[rest relying on x being a unit snipped]
> It's a "perfect storm" type kind of problem as unless you go
> deliberately looking for the problem or stumble across it, it just
> sits there invisible, as it did for over a hundred years.
[...]
> So is it a triviality? A historical curiosity?
>
> Perhaps, but then again, it's also what mathematicians believed
> until now was impossible--an error in "core".
And so on. The usual delusions of grandeur over a big non-discovery.
> I can trace out the proof step-by-step as it's high school algebra.
Have fun. Start by showing that an algebraic integer coprime to 2
must be a unit. Then show that 3 is a unit. Sounds like a great
project.
> Mathematicians can, of course, continue to fight me on it, but think
> about it.
>
> Maybe, after all, I'm looking for a fight.
With wet noodles for a weapon?
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
C. Bond
> It turns out that the definition for algebraic integers where they are
> roots of *monic* polynomials leaves off certain numbers that should be
> included, which is how I found a way to show some wacky things.
>
> It's easy enough to explain as consider
>
> x^3 - 3x + 2
>
> which as it turns out has a root that is coprime to 2. That is, it
> doesn't share any non-unit factors with 2 in the ring of algebraic
> integers.
>
> Well, that *should* make it a unit i.e. factor of 1, but turn the
> polynomial around using x=1/y and set it to 0 and you get
>
> 1/y^3 - 3/y + 2 = 0
>
> which is
>
> 2y^3 - 3y^2 + 1 = 0
>
> which would force a root to be a unit as well, but there's a quite
> correct theorem proving that the root of a non-monic primitive can NOT
> be an algebraic integer.
I must have missed a key point, so correct me if you will. I understood
that the root of a non-monic, *irreducible* polynomial can not be an
algebraic integer. Is your use of *primitive* the same as *irreducible*?
But wait, 2y^3-3y^2+1 = (2y^2-y-1)(y-1) and is therefore reducible. What's
wrong here?[snip]--
There are two things you must never attempt to prove: the unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
Chip Eastham
"James Harris" <jst...@msn.com> wrote in message
news:3c65f87.03092...@posting.google.com...
> It turns out that the definition for algebraic integers where they are
> roots of *monic* polynomials leaves off certain numbers that should be
> included, which is how I found a way to show some wacky things.
>
> It's easy enough to explain as consider
>
> x^3 - 3x + 2
>
> which as it turns out has a root that is coprime to 2. That is, it
> doesn't share any non-unit factors with 2 in the ring of algebraic
> integers.
>
Hi, James:
I'm not sure how "wacky" you would like to show things,
but let me recommend you work on making sense instead.
The polynomial x^3 - 3x + 2 has a root x = 1, which is
coprime to 2 and also a unit.
After factoring out (x - 1), what's left? A quadratic:
(x^2 + x - 2)(x - 1) = x^3 - 3x + 2
Surely a high school grasp of algebra is adequate to
extract the remaining roots of the quadratic factor.
Hint: The product of those two roots is -2, so both
of them are nontrivial divisors of 2 in the ring of
algebraic integers.
regards, chip
David C. Ullrich
>It turns out that the definition for algebraic integers where they are
>roots of *monic* polynomials leaves off certain numbers that should be
>included,
Before this shows that there's any problem with the math you
need to clarify exactly what "should" means...
But never mind that, it's easy to see the math below is simply
wrong:
>which is how I found a way to show some wacky things.
>
>It's easy enough to explain as consider
>
> x^3 - 3x + 2
>
>which as it turns out has a root that is coprime to 2. That is, it
>doesn't share any non-unit factors with 2 in the ring of algebraic
>integers.
>
>Well, that *should* make it a unit i.e. factor of 1,
As has already been pointed out: Which of the three roots are
you referring to? One root is x = 1, which _is_ a unit. The
other two roots are _not_ coprime to 2, in the sense you're
using the word - they're _factors_ of 2, hence certainly
"share factors with" 2.
>[...]
>
>Here's one interesting oddity that follows from the error:
>
>You can have algebraic integers 'a', 'b', 'c' and differing prime
>numbers p_1 and p_2 such that abc = p_1 p_2 but neither 'a', 'b', nor
>'c' share ANY non-unit factors with p_1 in the ring of algebraic
>integers!!!
First, if this were true then so what? Second, this is one of
those things you keep claiming, although you never give
a _correct_ example. What are a, b, c, p_1, p_2, exactly?
>So far some of you have shown a desire to leave the problem in core,
>and it's my job to disabuse you of that notion as it's a betrayal of
>mathematics, and civilization.
>
>I can trace out the proof step-by-step as it's high school algebra.
>
>Mathematicians can, of course, continue to fight me on it, but think
>about it.
>
>Maybe, after all, I'm looking for a fight.
Could be. Don't forget that you said this, the next time you
claim that all you're interested in is the math.
>James Harris
David C. Ullrich
**************************
As far as I'm concerend you're trying to wait until I die, so I figure
maybe you should die instead. How about that, eh? Wouldn't that be a
better twist?
You refuse to follow the math, so the great Powers that control
reality and *speak* in mathematics decide to kill you instead of me.
So what do you think about that, eh? Oh, can't hear Them talking?
Well, I guess that's because you don't really understand Mathematics,
the true language, which is THE language.
They're talking about you now, and They agree with my assessment, and
will not penalize me as They allowed the others like Galois and Abel
to be penalized.
They will kill you instead.
James Harris speaking on "Weird factorization, genius"
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>It turns out that the definition for algebraic integers where they are
>roots of *monic* polynomials leaves off certain numbers that should be
>
>It's easy enough to explain as consider
>
> x^3 - 3x + 2
>
>which as it turns out has a root that is coprime to 2.
No, it does not. All the roots are factors of two, and none of them
are units in the ring of all algebraic integers.
> That is, it
>doesn't share any non-unit factors with 2 in the ring of algebraic
>integers.
No, it is not. Let r1, r2, r3 be the three complex roots. Then
(-1)*r1*r2*r3 = 2, since
x^3 - 3x + 2 = (x-r1)(x-r2)(x-r3).
Therefore, each root is a factor of 2 in the ring of all algebraic
integers. r1 is a factor of 2 because when multiplied by the algebraic
integer (-1)*r2*r3 you get 2; r2 is a factor of 2 because when
multiplied by the algebraic integer (-1)*r1*r3 you get 2; and r3 is a
factor of 2 because when multiplied by the algebraic integer
(-1)*r2*r3 you get 2.
And none of them are units in the rings of algebraic integers, as you
do manage to show correctly below.
>
>Well, that *should* make it a unit i.e. factor of 1, but turn the
>polynomial around using x=1/y and set it to 0 and you get
>
> 1/y^3 - 3/y + 2 = 0
>
>which is
>
> 2y^3 - 3y^2 + 1 = 0
>
>which would force a root to be a unit as well, but there's a quite
>correct theorem proving that the root of a non-monic primitive can NOT
>be an algebraic integer.
[.rest deleted.]
======================================================================
"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
Arturo Magidin
C. Bond <cb...@ix.netcom.com> wrote:
>James Harris wrote:
>
>> It turns out that the definition for algebraic integers where they are
>> roots of *monic* polynomials leaves off certain numbers that should be
>> included, which is how I found a way to show some wacky things.
>>
>> It's easy enough to explain as consider
>>
>> x^3 - 3x + 2
>>
>> which as it turns out has a root that is coprime to 2. That is, it
>> doesn't share any non-unit factors with 2 in the ring of algebraic
>> integers.
>>
>> Well, that *should* make it a unit i.e. factor of 1, but turn the
>> polynomial around using x=1/y and set it to 0 and you get
>>
>> 1/y^3 - 3/y + 2 = 0
>>
>> which is
>>
>> 2y^3 - 3y^2 + 1 = 0
>>
>> which would force a root to be a unit as well, but there's a quite
>> correct theorem proving that the root of a non-monic primitive can NOT
>> be an algebraic integer.
>
>I must have missed a key point, so correct me if you will. I understood
>that the root of a non-monic, *irreducible* polynomial can not be an
>algebraic integer.
>Is your use of *primitive* the same as *irreducible*?
First, we should include "with integer coefficients" somewhere along
the line. If you talk about irreducibility over Q, then you need to
specify primitive, which means the gcd of the coefficients is 1. This
is to avoid things like
2x^2+2
which is irreducible over Q, nonmonic, but its roots (i and -i) are
both algebraic integers. James is wrong in leaving out "irreducible",
definitely, which I missed in my reply. For instance,
(2x+1)*(x^2+1) is primitive, non-monic, but has roots that are
algebraic integers.
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>It turns out that the definition for algebraic integers where they are
>roots of *monic* polynomials leaves off certain numbers that should be
>included, which is how I found a way to show some wacky things.
>
>It's easy enough to explain as consider
>
> x^3 - 3x + 2
>
>which as it turns out has a root that is coprime to 2.
Yes, because it is reducible:
x^3-3x+2 = (x-1)(x^2 +x-2)=(x-1)(x-1)(x+2)
The root which is coprime to 2 is 1, twice.
> That is, it
>doesn't share any non-unit factors with 2 in the ring of algebraic
>integers.
>
>Well, that *should* make it a unit i.e. factor of 1,
Of course it is: x=1 is a unit.
> but turn the
>polynomial around using x=1/y and set it to 0 and you get
>
> 1/y^3 - 3/y + 2 = 0
>
>which is
>
> 2y^3 - 3y^2 + 1 = 0
>
>which would force a root to be a unit as well, but there's a quite
>correct theorem proving that the root of a non-monic primitive can NOT
>be an algebraic integer.
No, you are misquoting the theorem. The theorem is that a non-monic,
primitive, IRREDUCIBLE polynomial with integer coefficients cannot
have algebraic integers as roots. The problem with your example is
that the polynomial you get here is just as reducible as the one
above:
2y^3 - 3y^2 + 1 = (2y+1)(y^2-2y+1) = (2y+1)(y-1)(y-1)
so that y=1 is a root of y-1, and the one root of x^3-3x+2 which is
not a unit is the root of x+2, namely -2, whose inverse (-1/2) is a
root of 2y+1.
======================================================================
"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
I suppose the simplest way to distinguish the cases is as follows:
1) Given any monic polynomial with integer coefficients we have the result
that all roots are algebraic integers.
2) Multiply the given polynomial by (2x+1) yielding a new polynomial which is
non-monic but has integer coefficients,
3) All roots of the original monic polynomial are still roots of the new
polynomial with the added root: x = -1/2, which is not an algebraic integer.
Conclusion: if a non-monic polynomial with integer coefficients can be
expressed as the product of a monic polynomial with integer coefficients and
a non-monic polynomial with integer coefficients, this composite polynomial
DOES have some roots which are algebraic integers.
It might be worth adding that it doesn't follow that a *reducible* non-monic
polynomial with integer coefficients WILL have roots that are algebraic
integers, only that it MAY have them (unless shown otherwise).
C. Bond
> "Chip Eastham" <eas...@bellsouth.net> wrote in message news:<hradnXylRMa...@comcast.com>...
> > "James Harris" <jst...@msn.com> wrote in message
> > news:3c65f87.03092...@posting.google.com...
> > > It turns out that the definition for algebraic integers where they are
> > > roots of *monic* polynomials leaves off certain numbers that should be
> > > included, which is how I found a way to show some wacky things.
> > >
> > > It's easy enough to explain as consider
> > >
> > > x^3 - 3x + 2
> > >
> > > which as it turns out has a root that is coprime to 2. That is, it
> > > doesn't share any non-unit factors with 2 in the ring of algebraic
> > > integers.
> > >
> > > Well, that *should* make it a unit i.e. factor of 1, but...
> >
> > Hi, James:
> >
> > I'm not sure how "wacky" you would like to show things,
> > but let me recommend you work on making sense instead.
> >
> > The polynomial x^3 - 3x + 2 has a root x = 1, which is
> > coprime to 2 and also a unit.
>
> Oh yeah, you're right. That should be x^3 + 3x - 2.
>
> > After factoring out (x - 1), what's left? A quadratic:
> >
> > (x^2 + x - 2)(x - 1) = x^3 - 3x + 2
> >
> > Surely a high school grasp of algebra is adequate to
> > extract the remaining roots of the quadratic factor.
>
> Ok, so now you want to be insulting. Ok, Chip Eastham, what's with you people?
>
> Oh, I know, you're FREAKING EVIL.
>
> James Harris
On the contrary. He pointed out a false statement you had made that you asserted to be true. That makes
YOU the evil one, not Chip. You are the promoter of lies, he is a defender of truth. Refute that!
C. Bond
[snip]
> Brave words but I'm still making silly mistakes like posting the WRONG
> FREAKING POLYNOMIAL!!!
>
> James Harris
How come when you make false assertions they are "silly mistakes", and anyone who points them is "evil"?
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>Oh yeah, you're right. That should be x^3 + 3x - 2.
Okay, this one is irreducible over Q: if it were reducible it would
have a root, and the only possible rational roots are 1, -1, 2, and
-2. Checking each shows that it does not have rational roots.
So, do you agree or disagree with each of the following?
1. If r1, r2, r3 are the three roots of x^3+3x-2, then
x^3+3x-2 = (x-r1)(x-r2)(x-r3).
2. Therefore, r1*r2*r3*(-1) = 2.
3. Therefore, r1 is a factor of 2 in the ring of all algebraic
integers, since (-1)*r2*r3 is an algebraic integer, and
r1* ( (-1)*r2*r3 ) = 2.
4. Likewise, r2 is a factor of 2 in the ring of all algebraic
integers, since (-1)*r1*r3 is an algebraic integer, and
r2* ( (-1)*r1*r3 ) = 2.
5. Likewise, r3 is a factor of 2 in the ring of all algebraic
integers, since (-1)*r2*r3 is an algebraic integer, and
r3* ( (-1)*r2*r3 ) = 2.
6. In the ring of all algebraic integers, if x and y are algebraic
integers, then x is a common factor of x and x*y.
7. In the ring of all algebraic integers, if x and y are algebraic
integers and x is not an algebraic integer unit, then x is a
non-unit common factor of x and x*y.
8. r1 is a common factor of r1 and 2; r2 is a common factor of r2 and
2; r3 is a common factor of r3 and 2.
9. r1 is not a unit in the ring of algebraic integers, because 1/r1
satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
irreducible, primitive polynomial with integer coefficients, and
therefore its roots are not algebraic integers.
10. Therefore, r1 is a non-unit common factor of r1 and 2. (From 7, 8,
and 9).
11. r2 is not a unit in the ring of algebraic integers. Because 1/r2
satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
irreducible, primitive polynomial with integer coefficients, and
therefore its roots are not algebraic integers.
12. Therefore, r2 is a non-unit common factor of r2 and 2. (From 7, 8,
and 11).
13. r3 is not a unit in the ring of algebraic integers. Because 1/r3
satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
irreducible, primitive polynomial with integer coefficients, and
therefore its roots are not algebraic integers.
14. Therefore, r3 is a non-unit common factor of r3 and 2. (From 7, 8,
and 13).
15. In the ring of all algebraic integers, x and y are not coprime if
and only if there exists a non-unit common factor of x and y (in
the ring of all algebraic integers).
[This is what you use as a definition; it is equivalent to the
standard definition for the ring of all algebraic integers, so it does
not matter in ->that<- context]
16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprime
to 3. (From 10, 12, 14, and 15).
17. The claim that x^3+3x-2 has a root which is coprime to 2 is false.
Now, you claimed that x^3+3x-2 has a root which is coprime to 2. So
you must think that at least one of 1-17 is wrong. Which one, and why?
======================================================================
"[Gabriele Rossetti] has left a vast body of writings... in which
he has attempted to prove the truth of his unorthodox interpre-
tation of medieval literature. They present a formidable
record of unsystematic research in which we see an enthusiast
plunging farther and farther and farther from the logic of facts
and good sense until truth is lost in the dreadful nightmare
of an idee fixe. There is no real evolution of the Theory
although it grows and expands until it embraces ever wider
horizons. The numerous inaccuracies of deduction, mis-statements
of historical fact, and self-contradictions...have caused critics
to turn away from them in disgust... [...] It is impossible to
read far... without realizing that we have to deal with a work of
faith and imagination rather than of reasoning. There is an
appearance of reason, for the author is set on proving by logic
the truth of what he already believes by intuition. The truth
is plain to him and he cannot comprehend why others do not
immediately accept it, but as they desire demonstration he has
multiplied his proofs. It is the redundancy and confusion of a
prophet expounding by a familiar method the truth revealed to his
own simple soul in a flash of inspiration... In such work as
this... it is idle to look for the calm reasoning of a scholar;
we do not find it, and there is little or no advantage in
attacking the obvious inconsistencies and absurdities that abound."
-- E.R. Vincent, _Gabriele Rossetti in England_, quoted in
_The Shakespearan Ciphers Examined_, by William F.
Friedman and Elizebeth S. Friedman
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Christopher J. Henrich
Harris <jst...@msn.com> wrote:
> It turns out that the definition for algebraic integers where they are
> roots of *monic* polynomials leaves off certain numbers that should be
> included, which is how I found a way to show some wacky things.
>
> It's easy enough to explain as consider
>
> x^3 - 3x + 2
>
> which as it turns out has a root that is coprime to 2. That is, it
> doesn't share any non-unit factors with 2 in the ring of algebraic
> integers.
The roots of this cubic are three algebraic integers, and their product
is -2. (High school algebra.) Therefore each one is a factor of 2.
For this reason, I am surprised by the statement that one of them is
"coprime to 2."
--
Chris Henrich
Arturo Magidin
The three roots are 1 (twice) and -2; 1 is coprime to 2, in the
trivial way. The reason this is not a problem and not something wacky
is that the polynomial is reducible over Q:
x^3-3x+2 = (x-1)(x-1)(x+2).
James later stated that he should have written x^3+3x-2, which is
irreducible over Q. In that case, none of the roots are units. all of
them are factors of 2 and of themselves, and therefore you would be
well to be surprised at the statement that one of them is "coprime to
2."
======================================================================
Chip Eastham
> > news:3c65f87.03092...@posting.google.com...
> > > It turns out that the definition for algebraic integers where they are
> > > roots of *monic* polynomials leaves off certain numbers that should be
> > > included, which is how I found a way to show some wacky things.
> > >
> > > It's easy enough to explain as consider
> > >
> > > x^3 - 3x + 2
> > >
> > > which as it turns out has a root that is coprime to 2. That is, it
> > > doesn't share any non-unit factors with 2 in the ring of algebraic
> > > integers.
> > >
> > > Well, that *should* make it a unit i.e. factor of 1, but...
> >
> > Hi, James:
> >
> > I'm not sure how "wacky" you would like to show things,
> > but let me recommend you work on making sense instead.
> >
> > The polynomial x^3 - 3x + 2 has a root x = 1, which is
> > coprime to 2 and also a unit.
>
>
> >
> > (x^2 + x - 2)(x - 1) = x^3 - 3x + 2
> >
> > Surely a high school grasp of algebra is adequate to
> > extract the remaining roots of the quadratic factor.
>
people?
>
> Oh, I know, you're FREAKING EVIL.
>
>
> James Harris
Hi, James. Perhaps I am "FREAKING EVIL" in some sense, but
probably this would be a theological rather than mathematical
issue.
I beg you not to sully your own posts with ad hominem arguments,
either attacking me personally or all mathematicians as a group. It
may be frustrating to have one's errors pointed out publically, but
I rather think this is purpose of public newsgroups.
No insult is intended in pointing out that the high school level of
algebra, a frame of reference to which you often refer as standard,
is sufficient to refute one of your claims. You and I both frequently
post here, and I think it to my advantage to be gracious when an
error of mine is pointed out.
Here, for example, I said the other two roots were "nontrivial"
divisors of 2. As Arturo points out, the other two roots were
1 (a repeat) and -2 (an associate of 2), so in fact both factors
are "trivial" divisors of 2.
Although naturally embarrassed at my own sloppy thinking,
I'm happy to see the statement corrected.
Thanks, Dr. Magidin! You're a credit to your profession.
regards, chip
Gib Bogle
> In article <3c65f87.03093...@posting.google.com>,
> James Harris <jst...@msn.com> wrote:
>
>
>>Oh yeah, you're right. That should be x^3 + 3x - 2.
>
>
> Okay, this one is irreducible over Q: if it were reducible it would
> have a root, and the only possible rational roots are 1, -1, 2, and
> -2. Checking each shows that it does not have rational roots.
>
> So, do you agree or disagree with each of the following?
>
> 1. If r1, r2, r3 are the three roots of x^3+3x-2, then
> x^3+3x-2 = (x-r1)(x-r2)(x-r3).
<steps 2 - 17 snipped>
Hope springs eternal. But you should know by now that JSH does not
answer questions.
Gib
Bill Dubuque
>
> It turns out that the definition for algebraic integers where they are
> roots of *monic* polynomials leaves off certain numbers that should be
> included, which is how I found a way to show some wacky things.
As I already explained, as far as your purpose is concerned, any attempt
to generalize the algebraic integers is necessarily doomed to failure.
http://google.com/groups?selm=y8zllty36hr.fsf%40nestle.ai.mit.edu
Indeed, below you fail spectacularly, refuting your own hypothesis:
> It's easy enough to explain as consider
>
> f(x) = x^3 + 3x + 2 [corrected here and below per later JSH post -wgd]
>
> which as it turns out has a root coprime to 2. That is, it doesn't
> share any non-unit factors with 2 in the ring of algebraic integers.
>
> Well, that *should* make it a unit i.e. factor of 1, but turn the
> polynomial around using x = 1/y and set it to 0 and you get
>
> 1/y^3 + 3/y + 2 = 0
>
> which is
>
> 2y^3 + 3y^2 + 1 = 0
>
> which would force a root to be a unit as well, but there's a quite
> correct theorem proving that the root of a non-monic primitive
> [irreducible polynomial -wgd] can NOT be an algebraic integer.
Summarizing your argument: if you assume by hypothesis that
f(x) has a root coprime to f(0) in the ring of algebraic integers
then this contradicts a known theorem (which you agree is correct).
Hence your hypothesis is false.
-Bill Dubuque
Andrzej Kolowski
> In article <3c65f87.03093...@posting.google.com>,
> James Harris <jst...@msn.com> wrote:
>
> >Oh yeah, you're right. That should be x^3 + 3x - 2.
>
> Okay, this one is irreducible over Q: if it were reducible it would
> have a root, and the only possible rational roots are 1, -1, 2, and
> -2. Checking each shows that it does not have rational roots.
>
The following set of 17 statements makes the issue as clear
as it could possibly be, with the exception of a misprint ...
> So, do you agree or disagree with each of the following?
>
> 1. If r1, r2, r3 are the three roots of x^3+3x-2, then
> x^3+3x-2 = (x-r1)(x-r2)(x-r3).
>
> 2. Therefore, r1*r2*r3*(-1) = 2.
>
> 3. Therefore, r1 is a factor of 2 in the ring of all algebraic
> integers, since (-1)*r2*r3 is an algebraic integer, and
> r1* ( (-1)*r2*r3 ) = 2.
>
> 4. Likewise, r2 is a factor of 2 in the ring of all algebraic
> integers, since (-1)*r1*r3 is an algebraic integer, and
> r2* ( (-1)*r1*r3 ) = 2.
>
> 5. Likewise, r3 is a factor of 2 in the ring of all algebraic
> integers, since (-1)*r2*r3 is an algebraic integer, and
> r3* ( (-1)*r2*r3 ) = 2.
>
> 6. In the ring of all algebraic integers, if x and y are algebraic
> integers, then x is a common factor of x and x*y.
>
> 7. In the ring of all algebraic integers, if x and y are algebraic
> integers and x is not an algebraic integer unit, then x is a
> non-unit common factor of x and x*y.
>
> 8. r1 is a common factor of r1 and 2; r2 is a common factor of r2 and
> 2; r3 is a common factor of r3 and 2.
>
> 9. r1 is not a unit in the ring of algebraic integers, because 1/r1
> satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
> irreducible, primitive polynomial with integer coefficients, and
> therefore its roots are not algebraic integers.
>
You might just note here that if A is a unit in the ring of algebraic
integers, then 1/A is an algebraic integer. But I think even Harris
knows that.
> 10. Therefore, r1 is a non-unit common factor of r1 and 2. (From 7, 8,
> and 9).
>
> 11. r2 is not a unit in the ring of algebraic integers. Because 1/r2
> satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
> irreducible, primitive polynomial with integer coefficients, and
> therefore its roots are not algebraic integers.
>
> 12. Therefore, r2 is a non-unit common factor of r2 and 2. (From 7, 8,
> and 11).
>
> 13. r3 is not a unit in the ring of algebraic integers. Because 1/r3
> satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
> irreducible, primitive polynomial with integer coefficients, and
> therefore its roots are not algebraic integers.
>
> 14. Therefore, r3 is a non-unit common factor of r3 and 2. (From 7, 8,
> and 13).
>
> 15. In the ring of all algebraic integers, x and y are not coprime if
> and only if there exists a non-unit common factor of x and y (in
> the ring of all algebraic integers).
>
> [This is what you use as a definition; it is equivalent to the
> standard definition for the ring of all algebraic integers, so it does
> not matter in ->that<- context]
>
> 16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprime
> to 3. (From 10, 12, 14, and 15).
>
You mean r3 is not coprime to 2.
> 17. The claim that x^3+3x-2 has a root which is coprime to 2 is false.
>
>
> Now, you claimed that x^3+3x-2 has a root which is coprime to 2. So
> you must think that at least one of 1-17 is wrong. Which one, and why?
>
Really, it couldn't be clearer. To most of us.
- Andrzej
Arturo Magidin
Andrzej Kolowski <akol...@hotmail.com> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<blcah3$20jb$1...@agate.berkeley.edu>...
>> In article <3c65f87.03093...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> >Oh yeah, you're right. That should be x^3 + 3x - 2.
>>
>> Okay, this one is irreducible over Q: if it were reducible it would
>> have a root, and the only possible rational roots are 1, -1, 2, and
>> -2. Checking each shows that it does not have rational roots.
>>
>
> The following set of 17 statements makes the issue as clear
>as it could possibly be, with the exception of a misprint ...
Couple of misprints, actually; I was still working with x^3-3x+2:
>> 1. If r1, r2, r3 are the three roots of x^3+3x-2, then
>> x^3+3x-2 = (x-r1)(x-r2)(x-r3).
>>
>> 2. Therefore, r1*r2*r3*(-1) = 2.
That should be r1*r2*r3 = 2.
>> 3. Therefore, r1 is a factor of 2 in the ring of all algebraic
>> integers, since (-1)*r2*r3 is an algebraic integer, and
>> r1* ( (-1)*r2*r3 ) = 2.
And this should be r1* (r2*r3) = 2, and "r2*r3 is an algebraic
integer". Same with 4, and 5 below.
>> 16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprime
>> to 3. (From 10, 12, 14, and 15).
>>
>
> You mean r3 is not coprime to 2.
Yes. Thank you and sorry for the missed sign.
Brian Quincy Hutchings
so, given the "high school level" of your latest treatise,
which I don't dare to look at -- hey, I've made *some* progress --
that is probably the audience that would be so unencumbered
by the hypothetically evil math, as to be able to vet your hypotheses
in a way that you might concur with. well?
however, how can you prove a proof,
just by the tautology of "it be math, QED?"
some time ago, on http://listserv.acsu.buffalo.edu/
on its "geodesic" list, a guy was belaboring his efforts
to prove the Kepler conjecture about 12 balls around one;
recently, I realized taht he was merely digesting,
down to identical arbitrary factors, a book
that went through Hale's alleged proof. now,
having perused taht book at a bookstore, I realized, why:
the book shows taht the proof was not complete,
in spite of it's having been called, "prooven."
since you've never said taht you've ever proven any thing
*but* le derniere theorem de Fermat, as it is speciously (or
popularly) called, you might have to explain why it is that,
"Class, we're just going to jump directly into The Math; after all,
it is mathematics!" even if you were wrong,
they'd still learn some thing; eouldn't they?
so, you dystrust professional math folks, as well
as anyone who deigns to take them seriously
in a sci.math forum. so,
seek your proof, else where, amongst those who are receptive
(no matter how "high" their IQs .-)
well?
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03093...@posting.google.com>...
> Hmmm...are mathematicians as a group evil? I question; you decide.
>
> I realized that it's simpler to refer people to a Hong Kong website
> where they allow posting with LaTeX so it looks prettier anyway.
>
> http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759
>
> The math IS rather basic and I still think it's mostly at high school
> level, so it's amazing that so many of you keep fighting it to the
> extent that I have to keep pushing and pushing.
> It IS math after all. It's not like I'm making things up.
--les ducs de Buffet!
http://larouchepub.com
Will Twentyman
> I realized that it's simpler to refer people to a Hong Kong website
> where they allow posting with LaTeX so it looks prettier anyway.
>
> http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759
Do you want the responses posted there or here? It might be easier to
simply copy the relevant material here so everyone can easily look at it
in context.
>
> The math IS rather basic and I still think it's mostly at high school
> level, so it's amazing that so many of you keep fighting it to the
> extent that I have to keep pushing and pushing.
So are the counter-arguments.
>
> But it's not amazing if mathematicians are willing to lie about
> mathematics for *social* reasons, or then again, you might be ashamed
> at such an error, while I think it's fascinating.
How about sticking to math instead of social analysis?
--
Will Twentyman
email: wtwentyman at copper dot net
W. Dale Hall
James Harris wrote:
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<blcah3$20jb$1...@agate.berkeley.edu>...
>
>>In article <3c65f87.03093...@posting.google.com>,
>>James Harris <jst...@msn.com> wrote:
>>
>>
>>>Oh yeah, you're right. That should be x^3 + 3x - 2.
>>
>>Okay, this one is irreducible over Q: if it were reducible it would
>>have a root, and the only possible rational roots are 1, -1, 2, and
>>-2. Checking each shows that it does not have rational roots.
>
>
> Well here we go again. I noticed a poster trying to make a big deal
> about the lateness of my reply but hey, I reply through Google and it
> takes it a while to put up messages.
>
> The math proving my point is basically high school stuff, so it's
> easy, which makes it fascinating to me that so many of you *trust*
> Magidin. Now I understand, I think, why he keeps lying, as what
> choice does he have?
>
> But why do any of you believe him and his voodoo math?
>
Why do you continually claim that people are lying? Is your world
so delicate that there is no possibility that you are in error?
I, for one, find Magidin's arguments to be carefully formulated,
with none of the sham informality that your arguments cloak themselves
in, and when I study them carefully, I find them to be correct. The
times I've done the same with regard to your arguments, I find them
festooned with holes, in fact I find them comprising more hole than
substance.
>
>>So, do you agree or disagree with each of the following?
>>
>>1. If r1, r2, r3 are the three roots of x^3+3x-2, then
>> x^3+3x-2 = (x-r1)(x-r2)(x-r3).
>
>
> Agree.
>
>
>>2. Therefore, r1*r2*r3*(-1) = 2.
>
>
> Agree.
>
>
>>3. Therefore, r1 is a factor of 2 in the ring of all algebraic
>> integers, since (-1)*r2*r3 is an algebraic integer, and
>> r1* ( (-1)*r2*r3 ) = 2.
>
>
> Trivially, yes, so I agree.
>
>
>>4. Likewise, r2 is a factor of 2 in the ring of all algebraic
>> integers, since (-1)*r1*r3 is an algebraic integer, and
>> r2* ( (-1)*r1*r3 ) = 2.
>
>
> Again, trivially, yes, so I agree.
>
>
>>5. Likewise, r3 is a factor of 2 in the ring of all algebraic
>> integers, since (-1)*r2*r3 is an algebraic integer, and
>> r3* ( (-1)*r2*r3 ) = 2.
>
>
> Yet again, trivially, yes, so I agree.
>
>
>>6. In the ring of all algebraic integers, if x and y are algebraic
>> integers, then x is a common factor of x and x*y.
>
>
> Agree.
>
>
>>7. In the ring of all algebraic integers, if x and y are algebraic
>> integers and x is not an algebraic integer unit, then x is a
>> non-unit common factor of x and x*y.
>
>
>
> Hmmm...what does being a unit have to do with anything?
>
> If I have xy = 2, where y=2, isn't x STILL a factor of 2?
>
> Then again, yeah, sounds ok to me.
>
> Agree.
>
>
>>8. r1 is a common factor of r1 and 2; r2 is a common factor of r2 and
>> 2; r3 is a common factor of r3 and 2.
>
>
> Agree.
>
>
>>9. r1 is not a unit in the ring of algebraic integers, because 1/r1
>> satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
>> irreducible, primitive polynomial with integer coefficients, and
>> therefore its roots are not algebraic integers.
>
>
> Agree.
>
>
>>10. Therefore, r1 is a non-unit common factor of r1 and 2. (From 7, 8,
>> and 9).
>
>
> Agree.
>
>
>>11. r2 is not a unit in the ring of algebraic integers. Because 1/r2
>> satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
>> irreducible, primitive polynomial with integer coefficients, and
>> therefore its roots are not algebraic integers.
>
>
> Agree.
>
>
>>12. Therefore, r2 is a non-unit common factor of r2 and 2. (From 7, 8,
>> and 11).
>
>
> Agree.
>
>
>>13. r3 is not a unit in the ring of algebraic integers. Because 1/r3
>> satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
>> irreducible, primitive polynomial with integer coefficients, and
>> therefore its roots are not algebraic integers.
>
>
> Agree.
>
>
>>14. Therefore, r3 is a non-unit common factor of r3 and 2. (From 7, 8,
>> and 13).
>
>
> Agree.
>
>
>>15. In the ring of all algebraic integers, x and y are not coprime if
>> and only if there exists a non-unit common factor of x and y (in
>> the ring of all algebraic integers).
>>
>>[This is what you use as a definition; it is equivalent to the
>>standard definition for the ring of all algebraic integers, so it does
>>not matter in ->that<- context]
>
>
> Agree.
>
>
>>16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprime
>> to 3. (From 10, 12, 14, and 15).
>
>
> Disagree.
>
> Actually I'm glad you presented it that way Magidin as it shows how
> you've fooled so many by getting them to take an illogical step.
>
> Here it IS true that x and y are NOT coprime if there exists a
> non-unit factor common to both x and y, but the converse, is NOT true
> in the ring of algebraic integers as it is screwy.
>
You state:
x and y are NOT coprime if there exists a non-unit
factor common to x and y
Magidin has shown, and you have agreed, to the following:
r1 is a non-unit factor common to r1 and 2. (step 10)
r2 is a non-unit factor common to r2 and 2. (step 12)
r3 is a non-unit factor common to r3 and 2. (step 14).
Yet, in step 16, modulo a typo (Magidin surely intended to type
"r3 is not coprime to 2"), you disagree with the combined statement.
Or, perhaps you didn't read his statement as containing the clear typo-
graphical error (typing "2" when he clearly [referencing step 14] meant
to type "3").
> You have a gap.
>
> For those readers confused consider that in the ring of *evens* 2 and
> 6 don't share non-unit factors, but by Magidin's definition they are
> NOT coprime.
>
No. Magidin uses the definition that r and s are coprime in the ring R
if the minimal ideal in R containing r and s is R itself. If the ring
contains a multiplicative identity, this is equivalent to the existence
of members m and n for which
mr + ns = 1.
Given the fact that 2Z doesn't have a multiplicative identity, it is
necessary to deal with the previous form of the definition (involving
ideals, rather than a decomposition of 1 into the prospective coprime
elements).
Given that definition, and the ring 2Z, the members 2 and 6 are
coprime, since the minimal ideal containing 2 and 6 is 2Z. How do we
see this?
Let I be an ideal (an additive subgroup of the ring, which is
closed under multiplication by ring elements) of 2Z, containing
both 2 and 6.
Since I is closed under multiplication by elements of 2Z, and
2 is already in I, then -2*2 = -4 is also in I. Since 6 is
also in I, and since I is an additive subgroup of 2Z, we then
know that -4 + 6 = 2 is in I (I know, I already knew that, but
we continue). Multiplying by ring elements, we obtain all other
elements of 2Z. Actually, the minimal ideal of 2Z containing
2 is already 2Z. Thus, 2 is coprime to every other element,
including 6, of the ring 2Z.
> Why?
>
> Because 3 is not even, but 2(3) = 6, so by Magidin's definition of
> coprime they are NOT coprime. Now the problem with algebraic integers
> is more complicated, but that example should help you to see that
> Magidin is lying to you.
>
No, you're mis-representing his position. Part of honest argumentation
involves attempting to attack the position that the person *actually*
holds, rather than purposely misreading it and attacking that mistaken
position.
> His mathematical presentations have gaps which cannot be filled.
>
No, you only produce gaps by taking advantage of typographical or
other inconsequential errors. When you attempt to address his *actual*
arguments, you consistently fail.
> Further, since his position has been repudiated by Professor McKenzie
> who shot down his key objection he is against the discipline of
> mathematics as are his supporters.
>
Hey, if you want to keep holding Professor McKenzie up as a person who
supports your claim, you should at least offer Professor McKenzie the
courtesy of speaking for himself. It is a sign of profound disrespect
to this person who has befriended you, without any apparent recompense
to himself, to act otherwise.
> Again see http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759
>
>
> James Harris
Dale.
Jack Rudd
> It turns out that the definition for algebraic integers where they are
> roots of *monic* polynomials leaves off certain numbers that should be
> included, which is how I found a way to show some wacky things.
>
>
>
> which as it turns out has a root that is coprime to 2.
True enough. I make its roots x=1 (twice) and x=-2; 1 is comprime to 2
in the Algebraic Integers.
> Well, that *should* make it a unit i.e. factor of 1, but turn the
> polynomial around using x=1/y and set it to 0 and you get
>
> 1/y^3 - 3/y + 2 = 0
>
> which is
>
> 2y^3 - 3y^2 + 1 = 0
Which has a repeated root of y=1 and a root of y=-1/2.
> which would force a root to be a unit as well, but there's a quite
> correct theorem proving that the root of a non-monic primitive can NOT
> be an algebraic integer.
Bollocks. Absolute crap. I think the correct formulation of that
theorem states that such a primitive must have at least one root that
isn't an algebraic integer, not that all its roots must be. Consider
the set of polynomial equations of the form 2x^2-(2n+1)x+n=0, where n
is an integer. Then they are all non-monic primitives, but they all
have the algebraic integer solution x=n.
> Maybe, after all, I'm looking for a fight.
It took you that long to work that out?
Jack Rudd
Arturo Magidin
[.personal attack removed.]
>> So, do you agree or disagree with each of the following?
>>
>> 1. If r1, r2, r3 are the three roots of x^3+3x-2, then
>> x^3+3x-2 = (x-r1)(x-r2)(x-r3).
>
>Agree.
>
>> 2. Therefore, r1*r2*r3*(-1) = 2.
>
>Agree.
>
>> 3. Therefore, r1 is a factor of 2 in the ring of all algebraic
>> integers, since (-1)*r2*r3 is an algebraic integer, and
>> r1* ( (-1)*r2*r3 ) = 2.
>
>Trivially, yes, so I agree.
>
>> 4. Likewise, r2 is a factor of 2 in the ring of all algebraic
>> integers, since (-1)*r1*r3 is an algebraic integer, and
>> r2* ( (-1)*r1*r3 ) = 2.
>
>Again, trivially, yes, so I agree.
>
>> 5. Likewise, r3 is a factor of 2 in the ring of all algebraic
>> integers, since (-1)*r2*r3 is an algebraic integer, and
>> r3* ( (-1)*r2*r3 ) = 2.
>
>Yet again, trivially, yes, so I agree.
>
>> 6. In the ring of all algebraic integers, if x and y are algebraic
>> integers, then x is a common factor of x and x*y.
>
>Agree.
>
>> 7. In the ring of all algebraic integers, if x and y are algebraic
>> integers and x is not an algebraic integer unit, then x is a
>> non-unit common factor of x and x*y.
>
>
>Hmmm...what does being a unit have to do with anything?
The last statement is "x is a NON-UNIT common factor of x and
x*y". Clearly, we need x to be not a unit for that to be true.
>If I have xy = 2, where y=2, isn't x STILL a factor of 2?
Yes, but it is not a "non-unit factor of 2", because in that case x=1,
so x is a unit.
>Then again, yeah, sounds ok to me.
>
>Agree.
>
>> 8. r1 is a common factor of r1 and 2; r2 is a common factor of r2 and
>> 2; r3 is a common factor of r3 and 2.
>
>Agree.
>
>> 9. r1 is not a unit in the ring of algebraic integers, because 1/r1
>> satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
>> irreducible, primitive polynomial with integer coefficients, and
>> therefore its roots are not algebraic integers.
>
>Agree.
>
>> 10. Therefore, r1 is a non-unit common factor of r1 and 2. (From 7, 8,
>> and 9).
>
>Agree.
>
>> 11. r2 is not a unit in the ring of algebraic integers. Because 1/r2
>> satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
>> irreducible, primitive polynomial with integer coefficients, and
>> therefore its roots are not algebraic integers.
>
>Agree.
>
>> 12. Therefore, r2 is a non-unit common factor of r2 and 2. (From 7, 8,
>> and 11).
>
>Agree.
>
>> 13. r3 is not a unit in the ring of algebraic integers. Because 1/r3
>> satisfies the polynomial 2x^2 - 3x + 1, which is non-monic,
>> irreducible, primitive polynomial with integer coefficients, and
>> therefore its roots are not algebraic integers.
>
>Agree.
>
>> 14. Therefore, r3 is a non-unit common factor of r3 and 2. (From 7, 8,
>> and 13).
>
>Agree.
>
>> 15. In the ring of all algebraic integers, x and y are not coprime if
>> and only if there exists a non-unit common factor of x and y (in
>> the ring of all algebraic integers).
>>
>> [This is what you use as a definition; it is equivalent to the
>> standard definition for the ring of all algebraic integers, so it does
>> not matter in ->that<- context]
>
>Agree.
>
>> 16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprime
>> to 3. (From 10, 12, 14, and 15).
>
>Disagree.
Why?
By 10, which you agree to, r1 is a non-unit common factor of r1 and
2. Therefore, r1 and 2 have a non-unit common factor (namely, r1). By
15 that means that r1 and 2 are not coprime (since there exists a
non-unit common factor of x and y, namely r1).
Same thing with r2 and 2, and r3 and 2.
The 3 is a typo.
>Actually I'm glad you presented it that way Magidin as it shows how
>you've fooled so many by getting them to take an illogical step.
What step is illogical and y?
>Here it IS true that x and y are NOT coprime if there exists a
>non-unit factor common to both x and y, but the converse, is NOT true
>in the ring of algebraic integers as it is screwy.
Point the zeroth. so in fact you are contesting point 15, not point
16, even though you wrote that you "agreed" with point 15. You should
really read more carefully. Point 15 was:
>> 15. In the ring of all algebraic integers, x and y are not coprime if
>> and only if there exists a non-unit common factor of x and y (in
>> the ring of all algebraic integers).
So you are saying that this is not true. It is not "if and only if",
you claim that only "only if" holds. Fine.
Replace 15 with:
15'. In the ring of all algebraic integers, if there exists a non-unit
common factor of x and y (in the ring of all algebraic integers) then
x and y are not coprime.
That one you claim to agree to.
Point the first: I never used the converse: there exists a non-unit
factor common to both r1 and 2 (point 10), and so I conclude that r1
and 2 are not coprime. There exists a non-unit factor common to both
r2 and 2 (point 12), so I conclude that r2 and 2 are not
coprime. There exists a non-unit factor common to both r3 and 2 (point
14), so I conclude that r3 and 2 are not coprime.
Where did I supposedly use the converse? The converse would be "If r1
and 2 are not coprime, then there exists a non-unit factor common to
both r1 and 2." I never invoked that.
Point the second: Actually, the converse IS true in the ring of
algebraic integers. I already gave you a proof:
http://groups.google.com/groups?selm=bhbev8%24m80%241%40agate.berkeley.edu
(Aug 12 2003) and the only step which is not fully proven in it is
Dedekind's Theorem on the finiteness of the class number.
>You have a gap.
Nope. I never used the converse.
>For those readers confused consider that in the ring of *evens* 2 and
>6 don't share non-unit factors, but by Magidin's definition they are
>NOT coprime.
This is a red herring. We are working in the ring of all algebraic
integers, and I did NOT use the converse of the statement, I only used
the statement in its direct form.
>His mathematical presentations have gaps which cannot be filled.
You have not correctly pointed any gap. You INSINUATE that I use the
converse of a result, when I used the result itself.
Please read again points 15 and 16 carefully. Replace point 15 by
point 15', and you will see that point 16 follows from using ONLY
point 15', not the converse.
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<blcc0a$213e$1...@agate.berkeley.edu>...
[.snip.]
>> x^3-3x+2 = (x-1)(x-1)(x+2).
>>
>> James later stated that he should have written x^3+3x-2, which is
>> irreducible over Q. In that case, none of the roots are units. all of
>> them are factors of 2 and of themselves, and therefore you would be
>> well to be surprised at the statement that one of them is "coprime to
>> 2."
>
>That's actually a useful statement to show how the wacky error with
>algebraic integers can lead to a false "proof" for those of you who
>think it doesn't matter to have this particular error in "core".
>
>Here Magidin IS correct in that none of the roots can be units, but
>that's because *in the ring of algebraic integers* it's provably true
>that any unit is a root of a monic polynomial with integer
>coefficients and a last coefficient of 1 or -1.
Excellent! Last time I claimed (and proved) this, you stated this was
false.
http://groups.google.com/groups?selm=b8rsue%2429gh%241%40agate.berkeley.edu
(May 1st 2003). Quoting from that:
THEOREM. Let f(x) be a monic polynomial with integer coefficients,
which is irreducible over Q. Then a root of f(x) is an algebraic
integer unit if and only if the constant term of f(x) is 1 or -1.
Proof. The definition of "algebraic integer unit" is "an algebraic
integer u such that 1/u is also an algebraic integer."
Let r be a root. If the constant term is 1 or -1, then r divides 1 or
-1 (in the ring of algebraic integers), so it is a unit.
Conversely, suppose that r is an algebraic integer unit. Let
s=1/r. Then s is an algebraic integer. Write
f(x) = x^n + ... + a_0.
0 = s^nf(r) = s^n(r^n+...+a_0)
= 1 + a_{n-1}s + ... + a_0s^n.
So s is a root of g(y) = a_0x^n+...+1; since g(y) is the polynomial we
get by taking f(1/y) and multiplying through by y^n, it follows from
irreducibility of f(x) that g(x) is irreducible.
Since s is the root of the irreducible primitive polynomial g(y) with
integer coefficients, it follows that the leading coefficient of g(y)
is either 1 or -1. So a_0=1 or a_0=-1. Therefore, if r is an algebraic
integer unit, then the constant term of f(x) is either 1 or -1. QED
Your reply was:
http://groups.google.com/groups?selm=3c65f87.0305020604.5984d6a1%40posting.google.com
(May 2 2003)
quoting from it:
-- Begin Insert --
> THEOREM. Let f(x) be a monic polynomial with integer coefficients,
> which is irreducible over Q. Then a root of f(x) is an algebraic
> integer unit if and only if the constant term of f(x) is 1 or -1.
>
> Proof. The definition of "algebraic integer unit" is "an algebraic
> integer u such that 1/u is also an algebraic integer."
>
> Let r be a root. If the constant term is 1 or -1, then r divides 1
> or
> -1 (in the ring of algebraic integers), so it is a unit.
>
> Conversely, suppose that r is an algebraic integer unit. Let
> s=1/r. Then s is an algebraic integer. Write
<deleted>
As I guessed you use the *definition* which means your argument IS
circular as it has to be because what you have can't be a theorem,
when I've proven it false.
-- End Insert --
So you now agree that the theorem is correct. Good.
>But that does NOT prove that none of the roots are coprime to 2.
Nobody said that by itself proved it. What PROVES that none of the
roots are coprime to 2 is:
1. If there is a non-unit common factor of x and y, then x and y are
not coprime.
2. r1 is not a unit; r2 is not a unit; r3 is not a unit.
3. r1 is a factor of r1 and of 2. r2 is a factor of r2 and 2. r3 is a
factor of r3 and 2.
4. Therefore, r1 is a non-unit common factor of r1 and 2. (2 and 3)
5. Therefore, r1 and 2 have a non-unit common factor. (4)
6. Therefore, r1 and 2 are not coprime. (1 and 4)
7. r2 is a non-unit common factor of r2 and 2 (2 and 3).
8. Therefore, r2 and 2 have a non-unit common factor (7)
9. Therefore, r2 is not coprime to 2. (1 and 8)
10. r3 is a non-unit common factor of r3 and 2 (2 and 3)
11. Therefore, r3 and 2 have a non-unit common factor (10)
12 Therefore, r3 and 2 are not coprime (1 and 11).
13. Therefore, none of the roots are coprime to 2 (6, 9, and 12).
>Here Magidin is using one thing, but switching to another, as if you
>just believe that to be coprime to 2, one of the roots has to be a
>unit, then you accept his claim as true.
That would be an incorrect statement as to fact. I never assume that
if a root is coprime to 2 then it is not a unit.
I never used what you CLAIM I am using. I dare you to point out which
of the steps above "uses" what you claim I am using.
>To catch Magidin here you have to force him to *prove*
>non-coprimeness, and he cannot.
I just did.
>Various posters have tried this
>particular trick, and I'd catch them on it, and they'd start going in
>circles.
Actually, you are. You keep claiming that we are using "the converse"
of "if x and y have a non-unit common factor then they are not
coprime", but we do not.
Also, as long as we're at it:
You've said that the DEFINITION of being "coprime" is
"x and y are coprime" means that any common factor of x and y is a
unit.
Is that not the same as stating:
1. If x and y are coprime then any common factor of x and y is a
unit;
and
2. If every common factor of x and y is a unit, then x and y are
coprime.
?
Why not? If the "definition" does not mean both things, then it is not
a definition, it is just a ->property<- of being coprime.
If it is the same, then surely their contrapositives are valid as
well:
"x and y are not coprime" means that there is a common factor of x
and y which is not a unit
and that means BOTH
I. If x and y are not coprime, then there is a common factor of x and
y which is not a unit.
AND
II. If there is a common factor of x and y which is not a unit, then x
and y are not coprime.
But that is immaterial. You admit that II is true, and that is the
only thing I am using.
Rick Decker
James Harris wrote:
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<blcc0a$213e$1...@agate.berkeley.edu>...
>
<snip>
>>
>>James later stated that he should have written x^3+3x-2, which is
>>irreducible over Q. In that case, none of the roots are units. all of
>>them are factors of 2 and of themselves, and therefore you would be
>>well to be surprised at the statement that one of them is "coprime to
>>2."
>>
<snip>
>
> Here Magidin IS correct in that none of the roots can be units, but
> that's because *in the ring of algebraic integers* it's provably true
> that any unit is a root of a monic polynomial with integer
> coefficients and a last coefficient of 1 or -1.
>
> But that does NOT prove that none of the roots are coprime to 2.
Of course not, nor did he claim that as a proof. However it's trivial
to see that none of the roots are coprime to 2. Denote the roots by r1,
r2, and r3, so that
x^3 + 3x - 2 = (x - r1)(x - r2)(x - r3)
Then r1*r2*r3 = 2, so r1 divides 2 in the ring of algebraic integers
(in fact 2 / r1 = r2*r3). Thus r1 and 2 share a common nonunit factor,
namely r1. The same holds for r2 and r3, so none of the roots are
coprime to 2.
Regards,
Rick
Complete Moron
----------------------------------------------------
I do not claim, that James Harris made any errors in his "Advanced
polynomial factorization" paper, because it is really hard to figure
out what he really means, so it's even harder to decide whether it is
correct.
One reason it maybe hard to convince him that he is making mistakes
all over is that the main statement of his "paper" is actually true.
His proof is not likely to be correct, but that's a really hard
concept to grasp that a proof of a true statement may be itself
incorrect.
So this is just to let you all know that what he claims in that paper
is actually a triviality:
It is actually not obvious first what his main statement is, because
James does not believe in labeling things as "theorem" or "proof".
Nonetheless the following is a quote from his paper:
"This paper will show, using basic algebraic methods, that given the
factorization, in the ring of algebraic integers, 65x^3-12x + 1 =
(a1*x + 1)(a2*x + 1)(a3*x + 1) one of the a's is coprime to 5"
And here is why this is trivial. In other words, here is a proof:
(this is no longer a quotation)
Since (a1*x + 1)(a2*x + 1)(a3*x + 1)=a1*a2*a3*x^3+...+(a1+a2+a3)*x+1,
one obtains, that a1+a2+a3=-12, and therefore if none of the a's is
coprime to 5, then 12 is not coprime to 5 either. But then 1=5*5-2*12
is also not coprime to 5. This is a contradiction.
Complete
Arturo Magidin
Complete Moron <complet...@comcast.net> wrote:
>It is actually not obvious first what his main statement is, because
>James does not believe in labeling things as "theorem" or "proof".
>Nonetheless the following is a quote from his paper:
>
>"This paper will show, using basic algebraic methods, that given the
>factorization, in the ring of algebraic integers, 65x^3-12x + 1 =
>(a1*x + 1)(a2*x + 1)(a3*x + 1) one of the a's is coprime to 5"
>
>And here is why this is trivial. In other words, here is a proof:
>(this is no longer a quotation)
>Since (a1*x + 1)(a2*x + 1)(a3*x + 1)=a1*a2*a3*x^3+...+(a1+a2+a3)*x+1,
>one obtains, that a1+a2+a3=-12, and therefore if none of the a's is
>coprime to 5, then 12 is not coprime to 5 either.
6 is not coprime to 30.
10 is not coprime to 30.
15 is not coprime to 30.
Therefore, 6+10+15 = 31 is not coprime to 30.
======================================================================
"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
William Hale
[cut]
> > 15. In the ring of all algebraic integers, x and y are not coprime if
> > and only if there exists a non-unit common factor of x and y (in
> > the ring of all algebraic integers).
> >
> > [This is what you use as a definition; it is equivalent to the
> > standard definition for the ring of all algebraic integers, so it does
> > not matter in ->that<- context]
>
Actually, you say below that you do not agree with this.
>
> > 16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprime
> > to 3. (From 10, 12, 14, and 15).
>
> Disagree.
>
> Actually I'm glad you presented it that way Magidin as it shows how
> you've fooled so many by getting them to take an illogical step.
>
> Here it IS true that x and y are NOT coprime if there exists a
> non-unit factor common to both x and y, but the converse, is NOT true
> in the ring of algebraic integers as it is screwy.
Step 15 is the definition that is being used for the term "coprime".
For a definition, the converse is always true. It does not require
a proof: there is no gap.
Otherwise, when someone says "Suppose r1 is not coprime to 2", we
would not be able to know what was meant by that statement. The
statement "Suppose r1 is not coprime to 2" can be unwind using
the converse of the definition of "coprime" into a statement
involving non-unit factors common to r1 and 2, thereby eliminating
the term "coprime". That is what definitions do: they give names
to new relationships involving previous defined or primitive terms.
> You have a gap.
>
> For those readers confused consider that in the ring of *evens* 2 and
> 6 don't share non-unit factors, but by Magidin's definition they are
> NOT coprime.
No, by Magidin's definition, they are coprime.
> Why?
>
> Because 3 is not even, but 2(3) = 6, so by Magidin's definition of
> coprime they are NOT coprime.
But, 3 is not an even integer. So, you have not exhibited a non-unit
even integer that is a common factor of 2 and 6. You have not shown
that 2 and 6 are not coprime.
-- Bill Hale
Arturo Magidin
W. Dale Hall <wd_...@pacbell.net> wrote:
Some corrections:
[.snip.]
>No. Magidin uses the definition that r and s are coprime in the ring R
>if the minimal ideal in R containing r and s is R itself.
This is not the definition I was using, either. That would be
"comaximal" in a ring with 1, and in rings without 1 (or in the
absence of the Axiom of Choice) it need not be equivalent to coprime.
I'll give the definitions in a second, but let me just point out that
James's "objection" is moot. I used only HIS definition in the
argument I gave (which is really just copying some results that you,
Dale, had gotten and posted, with a few more multiplications of
polynomials made explicitly.
Anyway, the definitions:
Let R be a ring, which may or may not have a multiplicative
identity. An ideal I of R is a "prime ideal" if and only if it is not
equal to R, and whenever x and y are elements of R such that
x*y is in I, then either x is in I or y is in I.
An ideal J of R is a "maximal ideal" if and only if it is an ideal,
properly contained in R, and there does not exist any ideal I which
properly contains J and is properly contained in R.
Two ideals A and B are coprime if and only if there does not exist a
prime ideal containing both.
Two elements r and s of R are "coprime" if and only if the principal
ideals they generate, (r) and (s), are coprime.
Two ideals A and B are coprime if and only if there does not exist a
maximal ideal containing both.
Two elements r and s of R are "comaximal" if and only if the principal
ideals they generate, (r) and (s), are comaximal. For rings with 1,
this is equivalent to the minimal ideal in R that contains r and s
being R itself.
If the ring is commutative and has a 1, then every maximal ideal
is prime. So coprime -> comaximal.
If the ring has a 1, and you assume the Axiom of Choice (Zorn's
Lemma), then every proper ideal is contained in a maximal ideal, so
comaximal->coprime.
So, if your ring is commutative, has a 1, and you assume the Axiom of
Choice, then coprime is equivalent to comaximal. Your definition is
the definition of comaximal.
> If the ring
>contains a multiplicative identity, this is equivalent to the existence
>of members m and n for which
>
> mr + ns = 1.
This is true for "comaximal".
>Given the fact that 2Z doesn't have a multiplicative identity, it is
>necessary to deal with the previous form of the definition (involving
>ideals, rather than a decomposition of 1 into the prospective coprime
>elements).
>
>Given that definition, and the ring 2Z, the members 2 and 6 are
>coprime, since the minimal ideal containing 2 and 6 is 2Z. How do we
>see this?
>
> Let I be an ideal (an additive subgroup of the ring, which is
> closed under multiplication by ring elements) of 2Z, containing
> both 2 and 6.
>
> Since I is closed under multiplication by elements of 2Z, and
> 2 is already in I, then -2*2 = -4 is also in I. Since 6 is
> also in I, and since I is an additive subgroup of 2Z, we then
> know that -4 + 6 = 2 is in I (I know, I already knew that, but
> we continue). Multiplying by ring elements, we obtain all other
> elements of 2Z.
Ehr, no. Multiplying by ring elements once you have 2 gives you all
the multiples of 4. However, an ideal is also closed under differences,
since you have 2 you have 0=2-2; and you have -4 so you have
4=0-(-4); and you have 6 so you have 4-6 = -2.
It is also closed under sums, so you have 2+2, 2+2+2, 2+2+2+2,....,,
so you get all positive multiples of 2; and you also have -2, -2+(-2),
-2+(-2)+(-2),.... so you get all negative multiples of 2, so you get
all of 2Z.
> Actually, the minimal ideal of 2Z containing
> 2 is already 2Z.
Yes. The minimal ideal containing 2n is {2n*k : k an integer}. And the
ideal generated by 2n is {4n*k: k an integer}.
Using the definition on principal ideals, note that (2)={4k: k is an
integer} and (6) = {12k: k is an integer}. The former contains the
latter, but (2) is not prime: 2*2 is in (2), but 2 is not. Therefore,
if there were a prime ideal containing both, then it would properly
contain {4k: k an integer}. That means that there is an even integer a
with a = 2 (mod 4) in the ideal. But that means that a-2=4k, so a-4k =
2; therefore, the ideal contains 2, and as you show above, that means
that the ideal is all of 2Z, and hence is not prime. Thus, (2) and (6)
are coprime, so 2 and 6 are coprime.
They are not, however, comaximal: (2) = {4k: k is an integer} is a
maximal ideal, as just shown above. And it contains (6); so (2) and
(6) are both contained in the maximal ideal (2), hence they are not
comaximal.
James is using the definition "a and b are coprime in R if and only if
every common divisor of a and b in R is a unit." For this definition
to even ->begin<- to make sense, you need to be able to talk about
units in R. The only way you can even begin to talk about units in R
is if R has a multiplicative identity. So James's definition does not
even ->begin<- to make sense in 2Z, making each and every mention of
2Z nothing but a red herring.
It is easy to verify that if a and b are coprime under the definition
above involving prime ideals (assuming the Axiom of Choice and that R
is commutative and with 1), then they are coprime under this
definition. The two are not equivalent in general, as the old example
Z[sqrt(-5)] shows, since 2 and 1+sqrt(-5) are "coprime" under the
'common divisor' definition, but not under the definition as ideals.
However, in the case of the ring of all algebraic integers, one can
show that the two definitions are equivalent (again, assuming the
Axiom of Choice to be able to get the 'coprime' as linear combination
property). It follows from the Finiteness of the Class Number. I
posted a proof some time ago, glossing only over the last fact:
http://groups.google.com/groups?selm=bhbev8%24m80%241%40agate.berkeley.edu
(August 12, 2003)
C. Bond
[snip]
> Mistakes in core mathematics can be great fun to play with, so posters
> like Rick Decker can run themselves in knots, if they don't follow the
> math.
>
> Remember an error in core is a big deal. People can "prove" all kinds
> of things with such an error, which is why it should be handled.
>
> James Harris
The error is in your argument, not in core mathematics. The only big deal here is the magnitude of your ego
and its reciprocal: your vanishingly small intelligence.
Will Twentyman
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<blcc0a$213e$1...@agate.berkeley.edu>...
> That's actually a useful statement to show how the wacky error with
> algebraic integers can lead to a false "proof" for those of you who
> think it doesn't matter to have this particular error in "core".
>
> Here Magidin IS correct in that none of the roots can be units, but
> that's because *in the ring of algebraic integers* it's provably true
> that any unit is a root of a monic polynomial with integer
> coefficients and a last coefficient of 1 or -1.
>
> But that does NOT prove that none of the roots are coprime to 2.
>
> Here Magidin is using one thing, but switching to another, as if you
> just believe that to be coprime to 2, one of the roots has to be a
> unit, then you accept his claim as true. Notice he says "well to be
> surprised".
>
> To catch Magidin here you have to force him to *prove*
> non-coprimeness, and he cannot. Various posters have tried this
> particular trick, and I'd catch them on it, and they'd start going in
> circles.
>
Okay, I'll take a stab at it:
x^3+3x-2 factors as (x-a)(x-b)(x-c) where a,b,c are roots.
Multiplying this out, we get that:
-a-b-c=0
ab+ac+bc=3
abc=2
Looking at the last equation, we see three factors of 2: a,b,c since
2=a(bc),
2=b(ac),
2=c(ab).
These factors are each factors of themselves: a=1*a, b=1*b, c=1*c
So, a is a factor of a and 2,
b is a factor of b and 2,
c is a factor of c and 2.
Since you admit that a,b,c are not units, and you have defined two
numbers to be coprime if they have no non-unit factors in common,
a is not coprime to 2, because they have a common non-unit factor: a
b is not coprime to 2, because they have a common non-unit factor: b
c is not coprime to 2, because they have a common non-unit factor: c
Therefor, *none of the roots are coprime to 2*. QED.
Now, having said that, I expect you will find an objection somewhere.
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>Rick Decker <rde...@hamilton.edu> wrote in message news:<3F7B20A7...@hamilton.edu>...
>and 'b' such that r_1 a + 2b = 1 in the ring of algebraic integers,
Yes, it does. For rewriting 2 as r1*(r2*r3), we have that if such an a
and b existed, then:
1 = r_1*a +
= r_1*a + r1*r2*r3*b
= r_1*(a+r2*r3*b)
Since r2, r3, and b are algebraic integers, so is r2*r3*b. Since a is
also an algebraic integer by assumption, so is a+r2*r2*b.
So if a and b existed, then there would be an algebraic integer x
(namely, x=a+r2*r3*b) such that r_1*x = 1.
Therefore, r_1 would be an algebraic integer unit.
But the ONLY complex number x with the property that r_1*x = 1 is
x=1/r_1.
Therefore, you are claiming that a+r2*r3*b = 1/r1.
But we know that a+r2*r3*b is an algebraic integer. And we know that
1/r1 is NOT an algebraic integer.
This is a contradiciton. The contradiction arises from assuming that a
and b exist. Therefore, no such a and no such b exist.
The same argument, word for word, exchanging r1 and r2, shows the same
for r2; and exchanging r1 and r3, shows it for r3.
But this is all a red herring.
>and in fact for one of the r's it is true that they *do* exist as both
>you and Magidin are wrong.
Please state explicilty
(a) which r;
(b) what is the value of a; and
(c) what is the value of b.
Otherwise, your claim is unfounded. But this is all a red herring.
>
Red herring.
We used YOUR definition of coprime. It is using YOUR definition of
"coprime" that you conclude that one of r1, r2, or r3 is coprime to
2. But using THAT definition, it turns out that NONE of r1, r2, or r3
are coprime to 2. So your claim is wrong.
>Again, the trick is to try and use the result that it is not a unit,
>when in fact, my point is that the ring of algebraic integers is
>screwed up in that it is not a unit.
So: the trick is to use that it is not a unit, when in fact it is not
a unit.
Why is it a trick to use something that is true?
>
>Mistakes in core mathematics can be great fun to play with, so posters
>like Rick Decker can run themselves in knots, if they don't follow the
>math.
You seem to be tying yourself in knots. You claim it is a trick (or an
error) to use a true fact. Why?
Brian Quincy Hutchings
just can't keep up with endless begging of the question.
why do I persist?
wouldn't some thing like his "nonpolynomial roots" be the subject
of Galois theory?... as I recall,
it's the same thing as "exploring the radical."
jackru...@hotmail.com (Jack Rudd) wrote in message news:<b64489b2.03100...@posting.google.com>...
> isn't an algebraic integer, not that all its roots must be. Consider
> the set of polynomial equations of the form 2x^2-(2n+1)x+n=0, where n
> is an integer. Then they are all non-monic primitives, but they all
> have the algebraic integer solution x=n.
--les ducs d'Enron!
http://members.tripod.com/~american_almanac/mozart.htm
W. Dale Hall
Arturo Magidin wrote:
> In article <3F7A900...@pacbell.net>,
> W. Dale Hall <wd_...@pacbell.net> wrote:
>
> Some corrections:
>
> [.snip.]
>
>
>
>>No. Magidin uses the definition that r and s are coprime in the ring R
>>if the minimal ideal in R containing r and s is R itself.
>
>
> This is not the definition I was using, either. That would be
> "comaximal" in a ring with 1, and in rings without 1 (or in the
> absence of the Axiom of Choice) it need not be equivalent to coprime.
>
Oops, you're right. I should have thought of that.
> I'll give the definitions in a second, but let me just point out that
> James's "objection" is moot. I used only HIS definition in the
> argument I gave (which is really just copying some results that you,
> Dale, had gotten and posted, with a few more multiplications of
> polynomials made explicitly.
Yes, I appreciate that; perhaps it is more to the point to indicate
in these discussions just where and when JSH props up strawmen in
a vain pretense of addressing the issues his critics have raised.
>
> Anyway, the definitions:
>
> Let R be a ring, which may or may not have a multiplicative
> identity. An ideal I of R is a "prime ideal" if and only if it is not
> equal to R, and whenever x and y are elements of R such that
> x*y is in I, then either x is in I or y is in I.
>
> An ideal J of R is a "maximal ideal" if and only if it is an ideal,
> properly contained in R, and there does not exist any ideal I which
> properly contains J and is properly contained in R.
>
> Two ideals A and B are coprime if and only if there does not exist a
> prime ideal containing both.
>
> Two elements r and s of R are "coprime" if and only if the principal
> ideals they generate, (r) and (s), are coprime.
>
Ah, I had assumed one used the minimal ideals containing r and s,
respectively. Plus, ignored the difference between prime and maximal.
... the 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
>
Thanks for the corrections.
Dale.
Arturo Magidin
W. Dale Hall <wd_...@pacbell.net> wrote:
[.snip.]
>> I'll give the definitions in a second, but let me just point out that
>> James's "objection" is moot. I used only HIS definition in the
>> argument I gave (which is really just copying some results that you,
>> Dale, had gotten and posted, with a few more multiplications of
>> polynomials made explicitly.
>
>Yes, I appreciate that; perhaps it is more to the point to indicate
>in these discussions just where and when JSH props up strawmen in
>a vain pretense of addressing the issues his critics have raised.
Fair enough... But then it perhaps also important to indicate not only
the strawmen, but also the red herrings. In this case, it was both. (A
red herring made of straw? How odd... (-: )
>
>Ah, I had assumed one used the minimal ideals containing r and s,
>respectively. Plus, ignored the difference between prime and maximal.
For rings with 1, the principal ideal (x) and the minimal ideal
containing x are of course the same. I guess you could define them
via the minimal ideal in general, it's just that I've never seen it
done that way and I ->have<- seen it done via the principal ideal. You
get weird things, like the fact that 2 is comaximal to every element
in 2Z even though it is not a unit, but then, rings without 1 are
weird anyway, so I don't think it matters; having two elements be
comaximal but not coprime is weird as well, after all.
Arturo Magidin
Yes, that was a typo.
Suppose that there are a and b such that
1 = r_1*a + 2*b.
Then using the fact taht 2=r_1*(r_2*r_3) we have:
1 = r_1*a + 2*b
= r_1*a + (r_1*(r_2*r_3))*b
= r_1*a + r_1*r_2*r_3* b
and the rest follows as before.
>On the next line you use "r_1" in one place and then have "r1", and
>what's with introducing "r2" and "r3"?
>
>If you try Galois Theory,
I did not. [...]
But if you could find a and b with 1=r_1*a + 2*b, then you would be
able to EXHIBIT an algebraic integer which, when multiplied by r_1,
would equal 1. That means that you would be able to EXHIBIT that r_1
is a unit.
>> Since r2, r3, and b are algebraic integers, so is r2*r3*b. Since a is
>> also an algebraic integer by assumption, so is a+r2*r2*b.
>>
>> So if a and b existed, then there would be an algebraic integer x
>> (namely, x=a+r2*r3*b) such that r_1*x = 1.
>>
>> Therefore, r_1 would be an algebraic integer unit.
>
>And here's the real point I think, trying to find any way to claim
>that r_1 would have to be a unit if coprime to 2, but voodoo math
>isn't satisfying.
Then you should stop using voodoo math, with expressions like "should
factor", "divides off", and "just like".
Now, I made a simple typo, and you go off and make a big deal out of
it. Isn't that a perfect example of your hypocrisy?
Complete Moron
Oops! Of course, in Harrisian fashion, I should call you an evil
liar...except that you are right. That is embarrassing. Fortunately I
am only a Moron.
Anyway, this gave me the idea, that a variation of above argument
actually proves that JH's statement is false. I realize that you or
someone else may have posted this already, but let me do it to correct
my mistake.
I also feel better to be on the opposite side than JH.
So, here is the quote from James Harris' "Advanced polynomial
factorization":
> >
> >"This paper will show, using basic algebraic methods, that given the
> >factorization, in the ring of algebraic integers, 65x^3-12x + 1 =
> >(a1*x + 1)(a2*x + 1)(a3*x + 1) one of the a's is coprime to 5"
> >
Note: it should probably be "a" ring and not "the" ring.
Claim (CM)
This is false
Proof
Let K be the splitting field of the above polynomial and let A be the
ring of integers in K. Further let p in A be a prime divisor of 5.
Since a1*a2*a3=65, p divides one of the ai's, let's say a1. Hence a1
and 5 are not coprime.
Next consider an element s of Gal(K/Q), such that s(a1)=a2. Then s(p)
divides a2 and hence a2 and 5 are not coprime either. Repeat for a t
in Gal(K/Q) such that t(a1)=a3 to conclude that none of the a's are
coprime to 5.
Q.E.D.
OK, is this correct, or did I make a mistake again?
To be honest, I am not sure why one would care about a statement like
this. James?
Brian Quincy Hutchings
Bill Dubuque <w...@nestle.ai.mit.edu> wrote in message news:<y8zk77q...@nestle.ai.mit.edu>...
> Summarizing your argument: if you assume by hypothesis that
> f(x) has a root coprime to f(0) in the ring of algebraic integers
> then this contradicts a known theorem (which you agree is correct).
>
> Hence your hypothesis is false.
--les ducs d'Enron!
http://www.schillerinstitute.org/fid_02-06/033_cervantes.html
Herman Jurjus
"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message news:blhntd$lu6$1...@agate.berkeley.edu...
[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
Can this book be found online, somewhere? I looked for it, but with no success.
Many thanks in advance,
Herman Jurjus
Arturo Magidin
No, this is correct.
http://groups.google.com/groups?selm=d6af759.0205061750.67cdf17%40posting.google.com
(May 6, 2002)
http://groups.google.com/groups?selm=all16e%24dro%241%40agate.berkeley.edu
(Sept. 10 2002)
Part 6b of
http://groups.google.com/groups?selm=b89tbu%242pi8%241%40agate.berkeley.edu
(Apr 24, 2003)
among other times when the argument has been presented.
Arturo Magidin
Herman Jurjus <h.ju...@hetnet.nl> wrote:
[re: Budget of Paradoxes]
>Can this book be found online, somewhere? I looked for it, but with no success.
I doubt it; it's not, as far as I know, even in print anymore.
Many university libraries have copies, either of the annotated 1915
edition, or the original 1872 edition, or the Dover reprint from 1954.
I bought my copy of the 1915 edition from Bookfinder,
www.bookfinder.com, after keeping an eye out for several months until
a good copy at a price I was willing to pay showed up (as I recall,
around US$100). Right now, they seem to have a copy of the Dover
edition for $60 through Abebooks:
http://dogbert.abebooks.com/abe/BookDetails?bi=228256424
Arturo Magidin
>> Yes, it does. For rewriting 2 as r1*(r2*r3), we have that if such an a
>> and b existed, then:
>>
>> 1 = r_1*a +
>
>Ok, now I see it, as above he probably missed putting in 2b and has an
>inconsistency using r_1 in one place but r1 in another below.
Yes. Two typos. One more below that you mercifully forgave:
>> = r_1*a + r1*r2*r3*b
>> = r_1*(a+r2*r3*b)
>>
>> Since r2, r3, and b are algebraic integers, so is r2*r3*b. Since a is
>> also an algebraic integer by assumption, so is a+r2*r2*b.
>
>Ok.
Should be, of course, a+r2*r3*b.
>> So if a and b existed, then there would be an algebraic integer x
>> (namely, x=a+r2*r3*b) such that r_1*x = 1.
>>
>> Therefore, r_1 would be an algebraic integer unit.
>
>Hmmm...yeah, you're right. Interesting. I'll need to think about
>this carefully.
It's the same as with integers, and the Euclidean Algorithm.
If you have two integers a and b, and you can find integers x and y
such that ax+by = 1, then a and b are coprime. Exactly the same
argument: if there were some factor, say z, that divides both a and b,
a=z*r, b=z*s, then
1 = ax + by
= (z*r)x + (z*s)y
= z* (rx + sy)
so z must be a unit, with inverse rx+sy.
That's why I said that the statement: "there exist x and y such that
ax+by=1" is ->stronger<- for general rings than the statement "x and y
have no common factor other than units." For an arbitrary ring, if the
first holds, then the second must also holds. There are rings for
which the latter holds, but the former does not (Z[sqrt(-5)] with a=2
and b=1+sqrt(-5)).
In the ring of all algebraic integers, the two statements happen to be
equivalent. But proving they are equivalent (that is, proving that the
latter statement implies the first) is something that Dedekind said
was ->hard<-. It's a consequence of one of his famous theorems, the
Finiteness of the Class Number.
======================================================================
Arturo Magidin
>> Readers should notice how *easy* it is to show that Arturo Magidin is
>> doing wacky stuff!!! The tolerance of the sci.math newsgroup for his
>> anti-math is rather amazing.
>
>YUCK! I hate it when I screw-up. It turns out that I was wrong and
>'a' and 'b' can't both be algebraic integers, which--when cleaned
>up--Arturo Magidin's argument shows.
So I wasn't a damn liar trying to confuse everyone?
Fancy that.
I assume, then, that in addition to noting here that I was correct,
you will be apologizing for any personal attack you launched against
me, at least with respect to this argument? For instance, will you
explicitly apologize for claiming that I was "doing wacky stuff",
"anti-math", or trying to fool readers with this argument?
W. Dale Hall
James Harris wrote:
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<blfl77$1ni$1...@agate.berkeley.edu>...
... stuff deleted ...
>
>>1 = r_1*a +
>
>
> Ok, now I see it, as above he probably missed putting in 2b and has an
> inconsistency using r_1 in one place but r1 in another below.
>
>
>> = r_1*a + r1*r2*r3*b
>> = r_1*(a+r2*r3*b)
>>
>>Since r2, r3, and b are algebraic integers, so is r2*r3*b. Since a is
>>also an algebraic integer by assumption, so is a+r2*r2*b.
>
>
> Ok.
>
>
>>So if a and b existed, then there would be an algebraic integer x
>>(namely, x=a+r2*r3*b) such that r_1*x = 1.
>>
>>Therefore, r_1 would be an algebraic integer unit.
>
>
> Hmmm...yeah, you're right. Interesting. I'll need to think about
> this carefully.
>
>
> James Harris
So, after countless people (Magidin here (and on many previoius
occasions), and myself (on multiple occasions as well)) have
pointed this exact argument out to you, only to be told rudely
that we "have been chasing [our] tails" and "gone in a circle",
and been the subject of much unwarranted gloating, *now* you
finally get the fact that a non-unit divisor of the number k
cannot be coprime to k, in whatever ring?
Are you at last prepared to recognize that the sequence of
steps refutes your claim regarding the following factorization?
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
Here are the steps:
1. The following factorizations exist in the ring of algebraic integers:
q(ai) r(ai) = 5
r(ai) s(ai) = ai
for each coefficient ai in the factorization
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
where ths ai's are algebraic integers. (I have previously
exhibited the polynomials q,r,s explicitly, and have shown,
also explicitly, that the factorizations of 5 and ai hold.
I do not duplicate that effort here.)
2. r(ai) is non-unit algebraic integer for each ai. ( I have
previously exhibited the minimal polynomial for r(ai), and
have explicitly shown that r(ai) satisfies that polynomial.
I do not duplicate that effort here).
3. r(ai) is a non-unit factor of both ai and 5, for each i. (This
follows from 1 and 2 above).
4. If ai and 5 were coprime, then there would be algebraic integers
m and n for which
m ai + n 5 = 1
and so by substituting:
m ( r(ai) s(ai) ) + n ( q(ai) r(ai) ) = 1
further, by rearranging factors:
r(ai) m s(ai) + r(ai) n q(ai) = 1
or (after factoring)
r(ai) ( m s(ai) + n q(ai) ) = 1.
5. If ai and 5 were coprime, then r(ai), which we know independently
(from step 2) cannot be a unit, would be shown to be a unit, with
multiplicative inverse given by
1/r(ai) = ( m s(ai) + n q(ai) ).
(The proof is step 4, above, immediately preceding this statement)
Thus, your claim is incorrect.
No doubt you still disagree. Please display the first statement you find
fault with.
Dale.
David C. Ullrich
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bli9p0$rtl$1...@agate.berkeley.edu>...
>> In article <3c65f87.03100...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> >> Readers should notice how *easy* it is to show that Arturo Magidin is
>> >> doing wacky stuff!!! The tolerance of the sci.math newsgroup for his
>> >> anti-math is rather amazing.
>> >
>> >YUCK! I hate it when I screw-up. It turns out that I was wrong and
>> >'a' and 'b' can't both be algebraic integers, which--when cleaned
>> >up--Arturo Magidin's argument shows.
>>
>> So I wasn't a damn liar trying to confuse everyone?
>>
>> Fancy that.
>>
>> I assume, then, that in addition to noting here that I was correct,
>> you will be apologizing for any personal attack you launched against
>> me, at least with respect to this argument? For instance, will you
>> explicitly apologize for claiming that I was "doing wacky stuff",
>> "anti-math", or trying to fool readers with this argument?
>
>Well, if you hadn't spent so much time and effort trying to convince
>others that correct mathematics was wrong...
Has to be said: what a fucking asshole you are! I mean really -
you just _admitted_ that he was wrong and you screwed up
_again_. But no apology for calling him names when he was
correcting what you (finally) agree was an error, just more
complaints about his supposed lies.
>aybe...but your actions
>have had an influence on the future of humanity, as you've been an
>integral part of its failure.
A pompous demented asshole to boot.
>Your postings became part of the test of people's ability to see the
>truth beyond social pressure, and beyond their *belief* in a social
>hierarchy as final arbiter.
>
>They failed and believed you, when so often you were wrong.
>
>I've demonstrated by my acknowledgement what it means to follow the
>math, as it's not about pride or how you will look; it's about truth.
>
>
>James Harris
David C. Ullrich
**************************
As far as I'm concerend you're trying to wait until I die, so I figure
maybe you should die instead. How about that, eh? Wouldn't that be a
better twist?
You refuse to follow the math, so the great Powers that control
reality and *speak* in mathematics decide to kill you instead of me.
So what do you think about that, eh? Oh, can't hear Them talking?
Well, I guess that's because you don't really understand Mathematics,
the true language, which is THE language.
They're talking about you now, and They agree with my assessment, and
will not penalize me as They allowed the others like Galois and Abel
to be penalized.
They will kill you instead.
James Harris speaking on "Weird factorization, genius"
David Moran
"David C. Ullrich" <ull...@math.okstate.edu> wrote in message
news:lakqnvop761tde781...@4ax.com...
> On 2 Oct 2003 19:25:50 -0700, jst...@msn.com (James Harris) wrote:
>
> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message
news:<bli9p0$rtl$1...@agate.berkeley.edu>...
> >> In article <3c65f87.03100...@posting.google.com>,
> >> James Harris <jst...@msn.com> wrote:
> >>
> >> >> Readers should notice how *easy* it is to show that Arturo Magidin
is
> >> >> doing wacky stuff!!! The tolerance of the sci.math newsgroup for
his
> >> >> anti-math is rather amazing.
> >> >
> >> >YUCK! I hate it when I screw-up. It turns out that I was wrong and
> >> >'a' and 'b' can't both be algebraic integers, which--when cleaned
> >> >up--Arturo Magidin's argument shows.
> >>
> >> So I wasn't a damn liar trying to confuse everyone?
> >>
> >> Fancy that.
> >>
> >> I assume, then, that in addition to noting here that I was correct,
> >> you will be apologizing for any personal attack you launched against
> >> me, at least with respect to this argument? For instance, will you
> >> explicitly apologize for claiming that I was "doing wacky stuff",
> >> "anti-math", or trying to fool readers with this argument?
He won't apologize. He is too pompous and arrogant. If I were doing the
things he's doing, I could find a MUCH better way to do it.
David Moran
Arturo Magidin
>> James Harris <jst...@msn.com> wrote:
>>
>> >> Readers should notice how *easy* it is to show that Arturo Magidin is
>> >> doing wacky stuff!!! The tolerance of the sci.math newsgroup for his
>> >> anti-math is rather amazing.
>> >
>> >YUCK! I hate it when I screw-up. It turns out that I was wrong and
>> >'a' and 'b' can't both be algebraic integers, which--when cleaned
>> >up--Arturo Magidin's argument shows.
>>
>> So I wasn't a damn liar trying to confuse everyone?
>>
>> Fancy that.
>>
>> I assume, then, that in addition to noting here that I was correct,
>> you will be apologizing for any personal attack you launched against
>> me, at least with respect to this argument? For instance, will you
>> explicitly apologize for claiming that I was "doing wacky stuff",
>> "anti-math", or trying to fool readers with this argument?
>
>have had an influence on the future of humanity, as you've been an
>integral part of its failure.
I see. So, even though I was right, you will not withdraw your
accusations that I was trying to deceive.
That's why you can make statements like the false statement below:
>Your postings became part of the test of people's ability to see the
>truth beyond social pressure, and beyond their *belief* in a social
>hierarchy as final arbiter.
>
>They failed and believed you, when so often you were wrong.
I have ->never<- been wrong on this, James. It is only that you THINK
I am, and that whenever you finally admit I am right, you do not think
it is necessary for you to withdraw your accusations.
In short, it is only because you lack honesty and integrity that you
continue to think I was wrong even once, let alone "so often."
Arturo Magidin
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bli9p0$rtl$1...@agate.berkeley.edu>...
> > In article <3c65f87.03100...@posting.google.com>,
> > James Harris <jst...@msn.com> wrote:
> >
> > >> Readers should notice how *easy* it is to show that Arturo Magidin is
> > >> doing wacky stuff!!! The tolerance of the sci.math newsgroup for his
> > >> anti-math is rather amazing.
> > >
> > >YUCK! I hate it when I screw-up. It turns out that I was wrong and
> > >'a' and 'b' can't both be algebraic integers, which--when cleaned
> > >up--Arturo Magidin's argument shows.
> >
> > So I wasn't a damn liar trying to confuse everyone?
> >
> > Fancy that.
> >
> > I assume, then, that in addition to noting here that I was correct,
> > you will be apologizing for any personal attack you launched against
> > me, at least with respect to this argument? For instance, will you
> > explicitly apologize for claiming that I was "doing wacky stuff",
> > "anti-math", or trying to fool readers with this argument?
>
> others that correct mathematics was wrong...maybe...but your actions
> have had an influence on the future of humanity, as you've been an
> integral part of its failure.
>
> truth beyond social pressure, and beyond their *belief* in a social
> hierarchy as final arbiter.
>
> They failed and believed you, when so often you were wrong.
>
> math, as it's not about pride or how you will look; it's about truth.
Then the truth is: You did not need to launch personal attacks against
me in response to a 100% pure, nothing but math, argument. You did it
because of your pride. And you were wrong in your personal attacks and
in your estimation of the mathematics presented.
So when you called me a liar, you were telling falsehoods.
Your refusal to explicitly withdraw that lie shows that you are NOT,
in fact, interested in the truth at all.
Which means that you are still being a liar.
Remember that, next time you try to get on your high horse.
Arturo Magidin, from google, sans .sig
Chip Eastham
> In article <3F7BD0D0...@pacbell.net>,
> W. Dale Hall <wd_...@pacbell.net> wrote:
>
> [.snip.]
>
> >> I'll give the definitions in a second, but let me just point out that
> >> James's "objection" is moot. I used only HIS definition in the
> >> argument I gave (which is really just copying some results that you,
> >> Dale, had gotten and posted, with a few more multiplications of
> >> polynomials made explicitly.
> >
> >Yes, I appreciate that; perhaps it is more to the point to indicate
> >in these discussions just where and when JSH props up strawmen in
> >a vain pretense of addressing the issues his critics have raised.
>
> Fair enough... But then it perhaps also important to indicate not only
> the strawmen, but also the red herrings. In this case, it was both. (A
> red herring made of straw? How odd... (-: )
>
> >
> >Ah, I had assumed one used the minimal ideals containing r and s,
> >respectively. Plus, ignored the difference between prime and maximal.
>
> For rings with 1, the principal ideal (x) and the minimal ideal
> containing x are of course the same. I guess you could define them
> via the minimal ideal in general, it's just that I've never seen it
> done that way and I ->have<- seen it done via the principal ideal. You
> get weird things, like the fact that 2 is comaximal to every element
> in 2Z even though it is not a unit, but then, rings without 1 are
> weird anyway, so I don't think it matters; having two elements be
> comaximal but not coprime is weird as well, after all.
>
Other definitions are possible and probably desirable, in order
to have greater "separation" between comaximal and coprime.
I would, for example, want to treat indeterminates X,Y in the
ring of polynomials f[X,Y] over a field f as "coprime" or (at a
minimum if that terminology were "taken") as relatively prime.
Note that the ideal (X,Y) in that ring is maximal, but X,Y have
no (nontrivial) common divisors.
By parallelism with the definition of prime element, one might
consider the property of elements r,s in ring R, that:
for all t in R, r|t and s|t implies rs|t
This property obtains for X,Y in f[X,Y], yet of course it's not
possible to express 1 as uX + vY for any coefficients u,v in
this ring.
regards, chip
Bill Dubuque
>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>
>> Two ideals A and B are coprime if and only if
>> there does not exist a prime ideal containing both.
>>
>> Two elements r and s of R are "coprime" if and only if
>> the principal ideals they generate, (r) and (s), are coprime.
>
> to have greater "separation" between comaximal and coprime.
Indeed, there are various different notions of "coprime" in use.
Thus unless one is a context where the most common alternatives
coincide, e.g. a PID or Bezout domain, one should always define
what "coprime" means. E.g. below are various possible definitions
of a is coprime to b
-----------------
(1) c|a,b => c|1
(2) a,b|c => ab|c
(3) a|bc => a|c
(4) (a) /\ (b) = (ab)
(5) a + b X is a primitive polynomial of degree 1
(6) (a, b) is a primitive ideal
(7) (a, b)^-1 = (1)
(8) (a, b) = (1)
Exercise: what implications hold among these statements
in common classes of domains? Esp. consider these domains:
http://google.com/groups?selm=y8z8yto8k85.fsf%40nestle.ai.mit.edu
-Bill Dubuque
Arturo Magidin
"Relatively prime" is a perfectly fine name. It works beautifully for
UFDs, such as F[X,Y], and has the usual divisibility properties. In
fact, that's the definition I know for "relatively prime" (as opposed
to "coprime"): in a UFD, two elements are relatively prime if and only
if there is no irreducible that divides both.
Arturo Magidin, sans .sig