> I think I've figured out a way to show basically all of you, including
> people who think they don't know any math that mathematicians have
> been lying about my work.  It's so trivial you *should* wonder why
> they thought they could get away with it.
...
> That is, MOST of you will doubt algebra itself rather than consider
> that mathematicians, whom you probably don't even personally know,
> would lie.
...
> Mathematicians *are* a part of society after all.  Why should they
> tell the truth now?  It'd be like Bush owning up.  They can just sit
> tight, and be quiet, like so many American citizens or they can out
> and out lie, like so many other patriots.
James,
What are you?  A mathematician?  That is the name reserved for those
with an understanding of mathematics.  If you are not a mathematician,
you in fact do not understand what exactly you are talking about. 
Instead, you are a magician who pulls numbers out of his hat.
If you are in fact a mathematician, wouldn't people be following your
own advice by not listening to you?  In such case, you are complaining
about yourself.
Deny that you are a mathematician, and admit that you do not
understand what you are talking about.  Otherwise, stop attacking a
yourself.
-Chris Pitman
====
I'm saving this to print it out, later,
in order to lend some credence to my generally non-analytical comments
to your alleged & continuously-improved proof
(of what I refer to as the plausibly first major result
of la noodling de Fermat .-)
    otherwise,
I've come to the conclusion that you are just a big joker,
considering the soap-opera of your collected works,
as recently posted on another thread --
where they threatened you with some hardcore (so-
called) elliptical forms & their semistability ... and
you ran-off with your tail between your legs
(not that I understood too much of it, either .-)...  either that,
or you're some sort of rogue agent or other,
fishing-about for some ridiculous clue of your fancy (or
that of yo'boss); eh?
on the other hand, if you're really serious about this,
the only dysaster taht could conceivably result is, a)
you'd have a temporary loss of self-esteem,
in being denied this woderful attention from a tiny group
of sycophants (why else would we waste our time, herein?), or b)
you'd miraculously find one of the first billion-or-so proofs!
 
> My paper Advanced Polynomial Factorization depends on considering a
> factor of a polynomials that I call g.
> 
> (Paper linked to at http://groups.msn.com/AmateurMath as usual.)
> 
> And in my paper I start by showing that I can write that as
> 
>   g = r + c
> 
> where either r=0, or r changes as the polynomial's value changes,
> while c does not.
> 
> Now you can consider all factors of a given polynomial using g's, with
> something like
> 
>   g_1...g_k = P(x)
> 
> where you have k factors.  For instance, for P(x)=x^2 + 2x + 1,
> 
>   g_1 = x+1, g_2 = x+1, gives you 2 factors.
> 
> Those are polynomial factors, but I'm generalizing in a simple way to
> say that for the factors g, in general, you have an element I call r,
> which changes as the independent variable changes, and you have
> another element I call c, which does not.
> 
> For my example up above it's easy, as with g_1 = x + 1, x varies as x
> varies, while 1 does not.
> 
> Now that's enough that the proof in the paper is straightforward, but
> posters have argued with me anyway, with some trying to argue over the
> definition of polynomial, amazingly enough.
> 
> However, consider that the g's have an important feature, which is
> that when x=0, I have
> 
>   g_1...g_k = P(0).
> 
> For instance, with my simple example, with P(x) = x^2 + 2x + 1, with
> x=0,
> 
>   P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1.
> 
> You see, P(0) gives the constant term, so at x equal 0, the g's must
> multiply to give the constant term.
> 
> So then, maybe you still want to believe the mathematicians and
> question that I can write g = r + c.
> 
> Well consider that substituting gives me
> 
>     g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x),  which gives
> 
>   r_1...r_k +...+c_1...c_k = P(x), which is
> 
>   r_1...r_k +...+P(0) = P(x),  
> 
> which means that if you believe the mathematicians then they've
> convinced you to doubt algebra itself, as then you must believe that
> everything to the left of P(0) above can *maybe* be constant, but also
> *maybe* vary as x varies.
> 
> So why would mathematicians argue against such a simple result?
> 
> Two reasons I suggest.  First because they wish to disagree with me. 
> Second because they probably believe that they can get away with it.
> 
> That is, MOST of you will doubt algebra itself rather than consider
> that mathematicians, whom you probably don't even personally know,
> would lie.
> 
> So where does this lead?
> 
> Well the polynomial I show in the paper is
> 
>   P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf
> 
> which seems to be just complicated enough to give mathematicians room
> to lie.
> 
> For instance, you may be saying, HEY, what's with the 'm' when you
> had 'x' before??!!!
> 
> Well, there's no rule that says that you have to use the letter x as
> the variable label for a polynomial.  Also, there are historical
> reasons for my usage as it goes back to my work with FLT where x, y
> and z are used with x^p + y^p = z^p.
> 
> Finally, the weirder thing is that one poster in particular got a lot
> of mileage out of questioning my finding the constant term with an
> expression like the above by using m=0, as that gives me
> 
>   P(0) = 3xy^2 + y^3
> 
> and he got a lot of mileage for YEARS (before I had discovered proof I
> want to add and after) by emphasizing that two of the ROOTS of such an
> expression considered as a polynomial with respect to x are not
> defined at that point.
> 
> Well that's easy enough to see as the original expression is
> 
>   (v^3+1)x^3 - 3vxy^2 + y^3
> 
> which if you *wish* to see it as a polynomial with respect to x, is of
> degree 3, but when v=-1, it's of degree 1, so if you solve for the
> roots, you'll get funky stuff.
 
> For mathematicians the situation could be considered one of the worst
> possible disasters they can imagine.
--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...
La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto:
(FOSSILISATION [McCainanites?] (TM/sic))/
BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.
Http://www.tarpley.net/bushb.htm (content partiale, below):
  17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81
  23 -- Le FIN d'HISTOIRE
  24 -- L'ORDEUR du MONDE NOUVEAU
  25 -- THYROID STORK !?!
====
Go home, little man...
> I think I've figured out a way to show basically all of you, including
> people who think they don't know any math that mathematicians have
> been lying about my work.  It's so trivial you *should* wonder why
> they thought they could get away with it.
 Here goes.
 My paper Advanced Polynomial Factorization depends on considering a
> factor of a polynomials that I call g.
 (Paper linked to at http://groups.msn.com/AmateurMath as usual.)
 And in my paper I start by showing that I can write that as
   g = r + c
 where either r=0, or r changes as the polynomial's value changes,
> while c does not.
 Now you can consider all factors of a given polynomial using g's, with
> something like
   g_1...g_k = P(x)
 where you have k factors.  For instance, for P(x)=x^2 + 2x + 1,
   g_1 = x+1, g_2 = x+1, gives you 2 factors.
 Those are polynomial factors, but I'm generalizing in a simple way to
> say that for the factors g, in general, you have an element I call r,
> which changes as the independent variable changes, and you have
> another element I call c, which does not.
 For my example up above it's easy, as with g_1 = x + 1, x varies as x
> varies, while 1 does not.
 Now that's enough that the proof in the paper is straightforward, but
> posters have argued with me anyway, with some trying to argue over the
> definition of polynomial, amazingly enough.
 However, consider that the g's have an important feature, which is
> that when x=0, I have
   g_1...g_k = P(0).
 For instance, with my simple example, with P(x) = x^2 + 2x + 1, with
> x=0,
   P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1.
 You see, P(0) gives the constant term, so at x equal 0, the g's must
> multiply to give the constant term.
 So then, maybe you still want to believe the mathematicians and
> question that I can write g = r + c.
 Well consider that substituting gives me
     g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x),  which gives
   r_1...r_k +...+c_1...c_k = P(x), which is
   r_1...r_k +...+P(0) = P(x),
 which means that if you believe the mathematicians then they've
> convinced you to doubt algebra itself, as then you must believe that
> everything to the left of P(0) above can *maybe* be constant, but also
> *maybe* vary as x varies.
 So why would mathematicians argue against such a simple result?
 Two reasons I suggest.  First because they wish to disagree with me.
> Second because they probably believe that they can get away with it.
 That is, MOST of you will doubt algebra itself rather than consider
> that mathematicians, whom you probably don't even personally know,
> would lie.
 So where does this lead?
 Well the polynomial I show in the paper is
   P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf
 which seems to be just complicated enough to give mathematicians room
> to lie.
 For instance, you may be saying, HEY, what's with the 'm' when you
> had 'x' before??!!!
 Well, there's no rule that says that you have to use the letter x as
> the variable label for a polynomial.  Also, there are historical
> reasons for my usage as it goes back to my work with FLT where x, y
> and z are used with x^p + y^p = z^p.
 Finally, the weirder thing is that one poster in particular got a lot
> of mileage out of questioning my finding the constant term with an
> expression like the above by using m=0, as that gives me
   P(0) = 3xy^2 + y^3
 and he got a lot of mileage for YEARS (before I had discovered proof I
> want to add and after) by emphasizing that two of the ROOTS of such an
> expression considered as a polynomial with respect to x are not
> defined at that point.
 Well that's easy enough to see as the original expression is
   (v^3+1)x^3 - 3vxy^2 + y^3
 which if you *wish* to see it as a polynomial with respect to x, is of
> degree 3, but when v=-1, it's of degree 1, so if you solve for the
> roots, you'll get funky stuff.
 Now when I was finding the proof of FLT...remember the process took
> some years...at times I'd talk of polynomials with respect to other
> than m, but I refined my discourse as my understanding improved.
 However, people arguing with me did not.
 You may *choose* to believe that they did not because they don't know
> enough mathematics to follow, but we're talking about actual
> mathematicians here.
 What's more rational?
 I say it's more rational to suppose that they *did* figure out that it
> worked as described, but also noticed that as long as they disagreed,
> no one seemed to call them on making false statements, except me, and
> they knew my credibility wasn't so great.
 For most of you, there's probably the belief that there's some funky
> higher math involved that your pitiful brain can't follow or you
> don't know about, as you may suppose that mathematicians either
> wouldn't lie, or they wouldn't lie in such a dumb way where I could
> catch them so easily.
 But consider what's in the balance:
 1.  I discovered a proof of Fermat's Last Theorem that's more
> available to people in general than most of what mathematicians have
> been producing lately.
 2.  Worse I did so having said I'd find it years ago, and having spent
> years looking for it posting a lot of my ideas, and getting in insult
> battles with posters, quite a few who happened to be mathematicians.
 3.  Then to add insult to injury, I keep questioning mathematicians in
> terms of their ethics, and maybe *extremely importantly* I doubt that
> Wiles found a proof of Fermat's Last Theorem.
 And those are just highlights as there's more but I think that kind of
> gives my point.
 For mathematicians the situation could be considered one of the worst
> possible disasters they can imagine.
 EXCEPT, it looks like all they have to do is either stay quiet, or
> *claim* I'm wrong.
 Many of you simply believe them, and question algebra itself, which is
> rather sad.  I'd think at least some of you valued your educations.
 Others of you may figure it doesn't matter, maybe because Western
> civilization seems to be based on lying anyway, and maybe you figure I
> should just grow up, accept that everybody lies and move on.  And I
> don't have to talk about Enron or pedophile Catholic priests or things
> in that vein.
 I mean, look at George W. Bush and Iraq.  If people can be *killed*
> over lies, without consequences to the liars, then what's with some
> freaking stupid math?
 Good point.
 Mathematicians *are* a part of society after all.  Why should they
> tell the truth now?  It'd be like Bush owning up.  They can just sit
> tight, and be quiet, like so many American citizens or they can out
> and out lie, like so many other patriots.
 After all, that's so easy, now isn't it?
 Which is why you need to understand why I talk about mathematicians
> potentially being prosecuted.  Liars don't just stop because the gig
> is up, as then, they wouldn't necessarily be liars, then eh?
 It'd be against their *true* natures.

> James Harris
====
[snip preamble]
> Well consider that substituting gives me
> 
>     g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x),  which gives
> 
>   r_1...r_k +...+c_1...c_k = P(x), which is
> 
>   r_1...r_k +...+P(0) = P(x),  
> 
> which means that if you believe the mathematicians then they've
> convinced you to doubt algebra itself, as then you must believe that
> everything to the left of P(0) above can *maybe* be constant, but also
> *maybe* vary as x varies.
How do you mean?
The stuff to the left of P(0) appears to be
r_1...r_k + c_1.r_2...r_k + r_1.c_2.r_3...r_k + (lots of other terms
                                        involving r_i and c_i)
In general, this is going to vary with x. That seems reasonable enough,
because the r_i can vary with x. For all x, the value of this expression
will be the same as P(x)-P(0).
It's only going to be constant if P(x) is constant. In which case,
either all the g_i are constant, or you've chosen some non-polynomial
function for at least one of the g_i.
e.g. for P(x)==1, you might have g_1 = x^2+1, g_2 = 1/(x^2+1)
There doesn't seem to be any problem with this.
> So why would mathematicians argue against such a simple result?
> 
> Two reasons I suggest.  First because they wish to disagree with me. 
> Second because they probably believe that they can get away with it.
> 
> That is, MOST of you will doubt algebra itself rather than consider
> that mathematicians, whom you probably don't even personally know,
> would lie.
> 
> So where does this lead?
> 
> Well the polynomial I show in the paper is
> 
>   P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf
> 
> which seems to be just complicated enough to give mathematicians room
> to lie.
I suppose this must have been discussed before, but I don't like your
notation here: you seem to want to talk about a polynomial in m whose
coefficients are functions of x, y and f. To make this clearer maybe
you could write
P(x,y,f)(m) = (v^3+1)x^3 - 3vxy^2 + y^3
or start by defining
Q(m,x,y,f) = (v^3+1)x^3 - 3vxy^2 + y^3
and then say considering Q as a polynomial in m...
> For instance, you may be saying, HEY, what's with the 'm' when you
> had 'x' before??!!!
No, that's fine. Although your post would be easier to read if you
didn't change variable half way through.
> Well, there's no rule that says that you have to use the letter x as
> the variable label for a polynomial.  Also, there are historical
> reasons for my usage as it goes back to my work with FLT where x, y
> and z are used with x^p + y^p = z^p.
> 
> Finally, the weirder thing is that one poster in particular got a lot
> of mileage out of questioning my finding the constant term with an
> expression like the above by using m=0, as that gives me
> 
>   P(0) = 3xy^2 + y^3
> 
> and he got a lot of mileage for YEARS (before I had discovered proof I
> want to add and after) by emphasizing that two of the ROOTS of such an
> expression considered as a polynomial with respect to x are not
> defined at that point.
> 
> Well that's easy enough to see as the original expression is
> 
>   (v^3+1)x^3 - 3vxy^2 + y^3
> 
> which if you *wish* to see it as a polynomial with respect to x, is of
> degree 3, but when v=-1, it's of degree 1, so if you solve for the
> roots, you'll get funky stuff.
Well, goodness knows why you were arguing.
But you shouldn't say as a polynomial with respect to x, [it] is of
degree 3, since (as you go on to explain) there are exceptions.
A true statement is, as a polynomial in three variables (v, x and y),
the highest power of x appearing is 3. It follows that, as a polynomial
with respect to x, it has degree *at most* 3.
More specifically,
When v =/= -1, it has degree 3.
When v = 1 and y =/= 0, it has degree 1.
When v = 1 and y = 0 it is the zero polynomial.
If you're going to be pedantic, which is a virtue, you should
remember the last case as well.
-- 
David Collier
Hampshire, UK
====
> Go home, little man...
This is the only home he has...
V.
-- 
====
> I've looked at your paper... now looking at this I'll bite.  I may be 
> getting in over my head but I'll point out the things that don't make 
> sense to me.  I'm open to being enlightened.
> I think I've figured out a way to show basically all of you, including
> people who think they don't know any math that mathematicians have
> been lying about my work.  It's so trivial you *should* wonder why
> they thought they could get away with it.
> 
> Either you aren't being clear, or they missed something.  It is a strong 
> charge to suggest that ALL mathematicians are in this great conspiracy 
> to teach incorrect math.  Sort of goes against what they stand for.
I say mathematicians but don't specifically say ALL mathematicians.
I'm talking about a pattern of denial that I'm seeing in
mathematicians that I've contacted or observed.
 
> Here goes.
> 
> My paper Advanced Polynomial Factorization depends on considering a
> factor of a polynomials that I call g.
Should be factor of a polynomial, and not polynomials.
> 
> (Paper linked to at http://groups.msn.com/AmateurMath as usual.)
> 
> And in my paper I start by showing that I can write that as
> 
>   g = r + c
> 
> where either r=0, or r changes as the polynomial's value changes,
> while c does not.
> 
> Questions: is g the polynomial to be factored or a factor?  Is r a 
> polynomial, variable, or constant?  Is c a polynomial or constant?  What 
> do you mean about a polynomial's value changing?  A polynomial is in the 
> form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0.  It simply _is_, it 
> doesn't change.
Hmmm...doesn't sound like you actually looked at the paper.
In the paper the ring is algebraic integers; therefore, r and c are
algebraic integers.
In the paper it's noted that I'm generalizing beyond polynomial
factors, that is, factors of a polynomial that are themselves
polynomials.
In the paper I give an example of g = sqrt(x+1), so it's hard to
understand how you could miss the truth.
I state as included in your post, so it's hard to see how you missed
it, but I'll give it again here for emphasis:
>> where either r=0, or r changes as the polynomial's value changes,
> while c does not.
But you state above:
What do you mean about a polynomial's value changing?  A polynomial
is in the
form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0.  It simply _is_,
it doesn't change.
That's called bait-and-switch as you switched from the polynomial's
value to the polynomial.
I didn't say the polynomial changes.  I said its value changes and you
noted that in one sentence and switched in the other.
Now given, say, the polynomial P(x) = x+1, the polynomial's value at
x=0, is 1.
It's not rocket science.
And for the rest of you, I've seen a troubling tendency from posters
and people I know are actual mathematicians to cheat in this way.
Over and over and over again, they'll use deceitful tactics, and from
what I've seen it works.
It's like most of you WANT TO BELIEVE THEM, even when they're trashing
logic, mathematics, and what I'd think are typical ideas about
fairplay.
Mathematicians cheat.  Ok, so I'm not saying ALL mathematicians, but
I'm noting a trend that I've noticed time and time again.
Over and over and over again, mathematicians have cheated.  I point
out that they cheat, and then they just keep doing it!!!
> Note: at this point I'm fairly confused, but I'll read on.
I'll keep reading on as well.
> Now you can consider all factors of a given polynomial using g's, with
> something like
> 
>   g_1...g_k = P(x)
> 
> where you have k factors.  For instance, for P(x)=x^2 + 2x + 1,
> 
>   g_1 = x+1, g_2 = x+1, gives you 2 factors.
> 
> 
> This part made sense, except you'd usually note them as g_1(x), g_2(x).
I gave a simple example with factors of the polynomial that are
themselves polynomials.
But you can't even assume that if P(x) is of degree n that k=n.
In fact k can vary out to infinity.
Just like 2 has an infinite number of factors in the ring of algebraic
integers.
Similarly a polynomial like P(x) = x+1 can have an infinity of
factors.
It's that simple.
Now if that bothers you and you start trying to throw polynomial
factors back at me, I'll look at you the way someone might look at a
person who refuses to believe that numbers like 2 and 3 aren't prime
in higher rings.
Then that person in denial might keep saying things like, but 2(3)=6,
and what do you mean 'bout factors of 3?
They might ask me if I'm not making up my own definition of factor.
Basically, yes, I'm saying that an inability to comprehend that like
integers can have non-integer factors in a ring, and not a field,
polynomials can have non-polynomial factors is primitive.
Are mathematicians math experts or not?
> Those are polynomial factors, but I'm generalizing in a simple way to
> say that for the factors g, in general, you have an element I call r,
> which changes as the independent variable changes, and you have
> another element I call c, which does not.
> 
> So, r is a function, and c is a constant?  Why are you talking about 
> independent variables changing?
I'm not worried about whether or not r fits the formal definition for
function, and when I've talked about functions in this area there's
just been more confusion.
What matters for the argument is that r changes as x changes, and not
specifics about how, while c remains constant.
Polynomials represent single independent variable systems.
For more generality I said independent variable rather than x.
> For my example up above it's easy, as with g_1 = x + 1, x varies as x
> varies, while 1 does not.
> 
> so in this case, r=x, c=1?
Yup.  But it's a trivial case.  The method is general enough to handle
not having an explicit representation for r.  It merely notes that it
changes, not exactly what it is.
> Now that's enough that the proof in the paper is straightforward, but
> posters have argued with me anyway, with some trying to argue over the
> definition of polynomial, amazingly enough.
> 
> Question: does each factor g_i(x) get its own r_i(x)?  I'm not looking 
> at the proof right now, but simply trying to understand what you are 
> claiming to have prooved.  The terminology you are using is not clear to 
> me, and I can't easily comment on the value or validity of your work 
> until that is made clear.  If you read math papers, you will note that 
> people specify what each object is, whether it's an integer, polynomial 
> over the rationals, etc.  I don't see these details.
Yet in the paper it's specifically stated that the ring is algebraic
integers.
Therefore, variables like r and c are in that ring.
When more information is needed beyond that, I provide it.
Again, I see a person saying the damndest things.
Name *one* thing in the paper that isn't defined that's part of the
argument.
The *insinuation* that there's something just pisses me off.
If there's something GIVE THE EVIDENCE.
I'm sick of all these statements without evidence.
MATHEMATICIANS NEED TO QUIT CHEATING!!!!!!!!!!!!!!!
> However, consider that the g's have an important feature, which is
> that when x=0, I have
> 
>   g_1...g_k = P(0).
> 
> Don't you mean (g_1...g_k)(0)=P(0)?
I'm not interested in a form debate.  There's style and there's
substance.
If you want to argue style over substance why don't you switch to
fashion?
Those runway models look nice as they sashay down the catwalk,
swinging those hips.
But if you want to talk mathematics, don't give me style issues.
> For instance, with my simple example, with P(x) = x^2 + 2x + 1, with
> x=0,
> 
>   P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1.
> 
> x+1 is not 1.  x+1 at x=0 is 1.
Yeah, which is why I say that with x=0 above.
Hmmm...you're just trying to piss me off, aren't you?
I mean, you've brought up so much bullshit stuff.
> You see, P(0) gives the constant term, so at x equal 0, the g's must
> multiply to give the constant term.
> 
> So then, maybe you still want to believe the mathematicians and
> question that I can write g = r + c.
> 
> Just questioning what it means.
Yeah right.  I think you're politicking.
> Well consider that substituting gives me
> 
>     g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x),  which gives
> 
> This seems to suggest that each g_i gets it's own r_i, c_i.
Hell yeah!!!
>   r_1...r_k +...+c_1...c_k = P(x), which is
> 
> This step is NOT clear.  What all goes between the product of r's and 
> product of c's?
Now you don't know how to multiply?
Sorry, I'm not giving a class on multiplying out expresssions like
  (r_1 + c_1)...(r_k + c_k)
as I'll let you go to some extent, but not that far.
>   r_1...r_k +...+P(0) = P(x),  
> 
> which means that if you believe the mathematicians then they've
> convinced you to doubt algebra itself, as then you must believe that
> everything to the left of P(0) above can *maybe* be constant, but also
> *maybe* vary as x varies.
> 
> Well, based on how you're defining things it appears that all of your 
> r's should be written as r_i(x), not simply r_i.  What am I missing?  If 
> r depends on x, please clearly indicate it.  If r does not depend on x, 
> then your initial definition of r doesn't make sense.  There is 
> ambiguity in here that appears to be the source of your *maybe*s, but I 
> think it came from you, not mathematicians.
Style questions don't interest me.
 
> So why would mathematicians argue against such a simple result?
> 
> Because it's not clearly stated.
Bullshit.
> Two reasons I suggest.  First because they wish to disagree with me. 
> Second because they probably believe that they can get away with it.
> 
> Right now, I don't understand what you're trying to say because there is 
> too much that you have not defined.  I cannot disagree with you because 
> I do not understand you.  I cannot accept your claims until I understand 
> them, however.  I've worked through enough math to have headaches from 
> the strain of maintaining precision in language.  Your paper lacks that 
> precision.
Stated without supporting evidence.
Yes more of what I've found to be typical behavior from
mathematicians.
> That is, MOST of you will doubt algebra itself rather than consider
> that mathematicians, whom you probably don't even personally know,
> would lie.
> 
> No, I just want to know what you're trying to say.
Like how you asked about things stated in the paper despite saying at
the top that you read the paper?
I've demonstrated in several places that you can't be taken at your
word.
> So where does this lead?
> 
> Well the polynomial I show in the paper is
> 
>   P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf
> 
> which seems to be just complicated enough to give mathematicians room
> to lie.
> 
> I'd like to know why you introduced v, then defined it in terms of m and 
> f, rather than do the substitution yourself.  Also, why are you 
> apparently leaving x,y,f as unaccounted for variables?  What type of 
> polynomial is P supposed to be?
IT'S ALL IN THE PAPER YOU CLAIM TO HAVE READ!!!!!!!!!!!!!!
That's it.  I'm tired of your game.
Yes undergrads, if you thought that mathematicians played by rules,
notice how they're trying to change rules of fairplay in dealing with
me, a non-mathematician.
====
> I've looked at your paper... now looking at this I'll bite.  I may be
> getting in over my head but I'll point out the things that don't make
> sense to me.  I'm open to being enlightened.
[snip]
> Notes: I am a mathematician.  I have only recently started reading this
> board, so I am not familiar with your previous difficulties with
> mathematicians.
It looks like you and James have been tricked into wasting
a few hours of each others time.
I really enjoyed James' opening line
and how it gradually evolved into
      IT'S ALL IN THE PAPER YOU CLAIM TO
          HAVE READ!!!!!!!!!!!!!!
      That's it.  I'm tired of your game.
If this was your intention, very well done! If not, don't take it
personal - it's a question of medication dosage. There must
be a balance somewhere. We know it will be unstable, but
it will be a balance - for a few weeks.
Welcome to the playground of James Harrass.
Dirk Vdm
====
> 
>>I've looked at your paper... now looking at this I'll bite.  I may be
>>getting in over my head but I'll point out the things that don't make
>>sense to me.  I'm open to being enlightened.
> 
> 
> [snip]
> 
> 
>>Notes: I am a mathematician.  I have only recently started reading this
>>board, so I am not familiar with your previous difficulties with
>>mathematicians.
> 
> 
> It looks like you and James have been tricked into wasting
> a few hours of each others time.
> 
> I really enjoyed James' opening line
> and how it gradually evolved into
>       IT'S ALL IN THE PAPER YOU CLAIM TO
>           HAVE READ!!!!!!!!!!!!!!
>       That's it.  I'm tired of your game.
> 
> If this was your intention, very well done! If not, don't take it
> personal - it's a question of medication dosage. There must
> be a balance somewhere. We know it will be unstable, but
> it will be a balance - for a few weeks.
> 
> Welcome to the playground of James Harrass.
> 
> Dirk Vdm
> 
> 
I noticed it.  I've lurked long enough that his response isn't 
*entirely* surprising.  I suppose I should go try to make sense of his 
paper now, to see where I'm missing things.  It would have been helpful 
if he had started by posting the paper here instead of posting a link to 
a pdf.  It makes quoting his paper... more challenging.  And makes it 
less accessible.
I looked at his paper long enough to get confused by it.  I didn't try 
to make sense of it.  I suppose I must defend my honor as a 
mathematician and start looking for the definitions in his paper.
-- 
Will Twentyman
====
> 
>>I've looked at your paper... now looking at this I'll bite.  I may be 
>>getting in over my head but I'll point out the things that don't make 
>>sense to me.  I'm open to being enlightened.
> 
Note: looked at is not the same as reading for understanding.
>I think I've figured out a way to show basically all of you, including
>people who think they don't know any math that mathematicians have
>been lying about my work.  It's so trivial you *should* wonder why
>they thought they could get away with it.
>>Either you aren't being clear, or they missed something.  It is a strong 
>>charge to suggest that ALL mathematicians are in this great conspiracy 
>>to teach incorrect math.  Sort of goes against what they stand for.
> 
> I say mathematicians but don't specifically say ALL mathematicians.
> 
Distinction noted.  My mistake.
> I'm talking about a pattern of denial that I'm seeing in
> mathematicians that I've contacted or observed.
>  
> 
>(Paper linked to at http://groups.msn.com/AmateurMath as usual.)
And in my paper I start by showing that I can write that as
  g = r + c
where either r=0, or r changes as the polynomial's value changes,
>while c does not.
>>Questions: is g the polynomial to be factored or a factor?  Is r a 
>>polynomial, variable, or constant?  Is c a polynomial or constant?  What 
>>do you mean about a polynomial's value changing?  A polynomial is in the 
>>form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0.  It simply _is_, it 
>>doesn't change.
> 
> Hmmm...doesn't sound like you actually looked at the paper.
I'll be going over it in detail this weekend.  Do you mind if I quote it 
directly in my responses?
> In the paper the ring is algebraic integers; therefore, r and c are
> algebraic integers.
An algebraic integer is a constant.  It does not vary.  Period.  Note: 
I'm taking my definitions of terms from Wolfram's Mathworld.  If you are 
using a different definition, please provide it.
> In the paper it's noted that I'm generalizing beyond polynomial
> factors, that is, factors of a polynomial that are themselves
> polynomials.
> 
> In the paper I give an example of g = sqrt(x+1), so it's hard to
> understand how you could miss the truth.
To be honest, I didn't get past the statement of your lemma.  I will 
attempt to correct that.
> I state as included in your post, so it's hard to see how you missed
> it, but I'll give it again here for emphasis:
> 
> 
>where either r=0, or r changes as the polynomial's value changes,
>while c does not.
> 
> 
> But you state above:
> 
> What do you mean about a polynomial's value changing?  A polynomial
> is in the
> form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0.  It simply _is_,
> it doesn't change.
> 
> That's called bait-and-switch as you switched from the polynomial's
> value to the polynomial.
That's called using standard terminology.  Precise use of terms is a 
hallmark of mathematics.  A polynomial is an expression.  For different 
values of its variable(s) it evaluates to different values, but that is 
not the same as saying the polynomial's value changes.
> 
> I didn't say the polynomial changes.  I said its value changes and you
> noted that in one sentence and switched in the other.
> 
> Now given, say, the polynomial P(x) = x+1, the polynomial's value at
> x=0, is 1.
> 
> It's not rocket science.
Correct.  But you did NOT change the value of the polynomial, you 
changed x and you changed P at x.  P(x)=x+1, P(0)=1.  P's value didn't 
change, you simply evaluated it for a given value of x.
> And for the rest of you, I've seen a troubling tendency from posters
> and people I know are actual mathematicians to cheat in this way.
If insisting on precise use of terminology is cheating, you might as 
well give up.  We've been trained to be very precise in how words are 
using, because many (if not all) of us got proofs shredded when we 
forgot a technicality.
[rant deleted]
> 
>>Note: at this point I'm fairly confused, but I'll read on.
> 
> 
> I'll keep reading on as well.
> 
> 
>Now you can consider all factors of a given polynomial using g's, with
>something like
  g_1...g_k = P(x)
where you have k factors.  For instance, for P(x)=x^2 + 2x + 1,
  g_1 = x+1, g_2 = x+1, gives you 2 factors.
>This part made sense, except you'd usually note them as g_1(x), g_2(x).
> 
> 
> I gave a simple example with factors of the polynomial that are
> themselves polynomials.
> 
> But you can't even assume that if P(x) is of degree n that k=n.
> 
> In fact k can vary out to infinity.
> 
> Just like 2 has an infinite number of factors in the ring of algebraic
> integers.
> 
> Similarly a polynomial like P(x) = x+1 can have an infinity of
> factors.
> 
> It's that simple.
That's not a problem, but your notation was inconsistent.
> Now if that bothers you and you start trying to throw polynomial
> factors back at me, I'll look at you the way someone might look at a
> person who refuses to believe that numbers like 2 and 3 aren't prime
> in higher rings.
> 
> Then that person in denial might keep saying things like, but 2(3)=6,
> and what do you mean 'bout factors of 3?
> 
> They might ask me if I'm not making up my own definition of factor.
> 
> Basically, yes, I'm saying that an inability to comprehend that like
> integers can have non-integer factors in a ring, and not a field,
> polynomials can have non-polynomial factors is primitive.
> 
> Are mathematicians math experts or not?
Theoretically, yes.  On the other hand, my dad is a mathematician by 
degree only.  He has forgotten much of what he knew through lack of use.
>Those are polynomial factors, but I'm generalizing in a simple way to
>say that for the factors g, in general, you have an element I call r,
>which changes as the independent variable changes, and you have
>another element I call c, which does not.
>>So, r is a function, and c is a constant?  Why are you talking about 
>>independent variables changing?
> 
> 
> I'm not worried about whether or not r fits the formal definition for
> function, and when I've talked about functions in this area there's
> just been more confusion.
> 
> What matters for the argument is that r changes as x changes, and not
> specifics about how, while c remains constant.
If r changes as x changes, regardless of how, then either r is a 
function of x, a relation on x, or not well-defined.  I'll accept 
function of x.  The reason I ask is that it would be nice if you noted 
it as r(x), rather than simply as r.
> 
> Polynomials represent single independent variable systems.
> 
> For more generality I said independent variable rather than x.
> 
For generality, I'll assume whatever you stick in () is the independent 
variable.
> 
>For my example up above it's easy, as with g_1 = x + 1, x varies as x
>varies, while 1 does not.
>>so in this case, r=x, c=1?
> 
> 
> Yup.  But it's a trivial case.  The method is general enough to handle
> not having an explicit representation for r.  It merely notes that it
> changes, not exactly what it is.
I can deal with this being a special case, I just want the details 
nailed down.
>Now that's enough that the proof in the paper is straightforward, but
>posters have argued with me anyway, with some trying to argue over the
>definition of polynomial, amazingly enough.
>>Question: does each factor g_i(x) get its own r_i(x)?  I'm not looking 
>>at the proof right now, but simply trying to understand what you are 
>>claiming to have prooved.  The terminology you are using is not clear to 
>>me, and I can't easily comment on the value or validity of your work 
>>until that is made clear.  If you read math papers, you will note that 
>>people specify what each object is, whether it's an integer, polynomial 
>>over the rationals, etc.  I don't see these details.
> 
> 
> Yet in the paper it's specifically stated that the ring is algebraic
> integers.
> 
> Therefore, variables like r and c are in that ring.
If r varies with x, then it's not an algebraic integer.  It might be 
something you could call an algebraic function, but it's NOT a constant.
> 
> When more information is needed beyond that, I provide it.
> 
> Again, I see a person saying the damndest things.
> 
> Name *one* thing in the paper that isn't defined that's part of the
> argument.
I'll go over your work this weekend and see what I can conclude.  I was 
responding the the content of the post, not what it refers to.
> The *insinuation* that there's something just pisses me off.
> 
> If there's something GIVE THE EVIDENCE
> I'm sick of all these statements without evidence.
> 
> MATHEMATICIANS NEED TO QUIT CHEATING!!!!!!!!!!!!!!!
> 
Just be sure we ARE cheating before accusing.  It does nothing for your 
credibility.  If we are confused, explain.  If we have made an error, 
point it out *clearly*.  If we respond to the content of a post that 
refers to an outside document, don't be surprised if there is a limited 
amount of cross-referencing.
>However, consider that the g's have an important feature, which is
>that when x=0, I have
  g_1...g_k = P(0).
>>Don't you mean (g_1...g_k)(0)=P(0)?
> 
> 
> I'm not interested in a form debate.  There's style and there's
> substance.
In mathematics, the line between style and substance can be *VERY* fine. 
  Don't be so quick to dismiss it.  Also, don't be surprised if failure 
to use the accepted style generates additional confusion.  I've seen it 
too often.
> 
> If you want to argue style over substance why don't you switch to
> fashion?
> 
I need too much help shopping for clothes to switch to fashion.
> But if you want to talk mathematics, don't give me style issues.
What looks like style may be substance.  Also, consistent style makes it 
easier to see the substance.
>For instance, with my simple example, with P(x) = x^2 + 2x + 1, with
>x=0,
  P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1.
>>x+1 is not 1.  x+1 at x=0 is 1.
> 
> 
> Yeah, which is why I say that with x=0 above.
> 
> Hmmm...you're just trying to piss me off, aren't you?
No, I'm trying to get you to be precise.
Try writing it as P(x)=g_1(x)*g_2(x) with x=0,
P(0)=1=g_1(0)*g_2(0), g_1(0) = g_2(0) = 0+1 = 1
The added level of precision will prevent confusion on the part of your 
reader and make it easier to see the content, and comment intelligently 
on it.  We might even all decide that you're a genius.
> 
> I mean, you've brought up so much bullshit stuff.
You haven't seen how bad it can get.  Trust me, the level of detail can 
get a LOT worse.
>You see, P(0) gives the constant term, so at x equal 0, the g's must
>multiply to give the constant term.
So then, maybe you still want to believe the mathematicians and
>question that I can write g = r + c.
>>Just questioning what it means.
> 
> Yeah right.  I think you're politicking.
If you want us to accept your proof as valid, then we need to understand 
it.  If you don't want to conform your writing to the conventions, don't 
be surprised if you are scorned.  Can you name a field of study that 
does NOT have rules for presentation of material?
>Well consider that substituting gives me
    g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x),  which gives
>>This seems to suggest that each g_i gets it's own r_i, c_i.
> 
> Hell yeah!!!
Good!  I understood something :)
>  r_1...r_k +...+c_1...c_k = P(x), which is
>>This step is NOT clear.  What all goes between the product of r's and 
>>product of c's?
> 
> Now you don't know how to multiply?
No, I'm saying writing the first and last terms can easily lead to 
erroneous conclusions.
> Sorry, I'm not giving a class on multiplying out expresssions like
> 
>   (r_1 + c_1)...(r_k + c_k)
> 
> as I'll let you go to some extent, but not that far.
Summation and product notations can make it compact without missing 
anything.
>  r_1...r_k +...+P(0) = P(x),  
which means that if you believe the mathematicians then they've
>convinced you to doubt algebra itself, as then you must believe that
>everything to the left of P(0) above can *maybe* be constant, but also
>*maybe* vary as x varies.
>>Well, based on how you're defining things it appears that all of your 
>>r's should be written as r_i(x), not simply r_i.  What am I missing?  If 
>>r depends on x, please clearly indicate it.  If r does not depend on x, 
>>then your initial definition of r doesn't make sense.  There is 
>>ambiguity in here that appears to be the source of your *maybe*s, but I 
>>think it came from you, not mathematicians.
> 
> Style questions don't interest me.
That is probably why you are having so many conflicts.  If your style is 
not within the accepted norm, you will be viewed as babbling.  Have you 
ever had an instructor that speaks broken English or has a thick accent? 
  Some students fight through that to understand, others hear only 
babbling.  The same thing can happen with style.  Some will attempt to 
decipher, others will dismiss until you adapt to the accepted standards.
>So why would mathematicians argue against such a simple result?
>>Because it's not clearly stated.
> 
> Bullshit.
Actually, that was one of the first responses I saw to your paper. 
Something to the effect of It's not right, it's not wrong, it's 
unintelligible.  If you are not communicating clearly, your arguments 
will hold little weight.
>Two reasons I suggest.  First because they wish to disagree with me. 
>Second because they probably believe that they can get away with it.
>>Right now, I don't understand what you're trying to say because there is 
>>too much that you have not defined.  I cannot disagree with you because 
>>I do not understand you.  I cannot accept your claims until I understand 
>>them, however.  I've worked through enough math to have headaches from 
>>the strain of maintaining precision in language.  Your paper lacks that 
>>precision.
> 
> Stated without supporting evidence.
That's because what I call precision, you denegrate as style.
> Yes more of what I've found to be typical behavior from
> mathematicians.
So why not play our game by our rules?  If you can't win within our 
rules for communication, it may mean you need to convince us of the 
problem in HOW we communicate, not WHAT we communicate.
>That is, MOST of you will doubt algebra itself rather than consider
>that mathematicians, whom you probably don't even personally know,
>would lie.
>>No, I just want to know what you're trying to say.
> 
> Like how you asked about things stated in the paper despite saying at
> the top that you read the paper?
Looking at a paper is NOT understanding it.  I've had to look up 3 terms 
in your paper and I'm hoping that you are using the same definitions I 
found.  If you don't appear to be, I'll note it in my response to your 
paper.
> I've demonstrated in several places that you can't be taken at your
> word.
I'll leave that statement for others to judge.
>So where does this lead?
Well the polynomial I show in the paper is
  P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf
which seems to be just complicated enough to give mathematicians room
>to lie.
>>I'd like to know why you introduced v, then defined it in terms of m and 
>>f, rather than do the substitution yourself.  Also, why are you 
>>apparently leaving x,y,f as unaccounted for variables?  What type of 
>>polynomial is P supposed to be?
> 
> 
> IT'S ALL IN THE PAPER YOU CLAIM TO HAVE READ!!!!!!!!!!!!!!
See notes above.
> 
> That's it.  I'm tired of your game.
> 
> Yes undergrads, if you thought that mathematicians played by rules,
> notice how they're trying to change rules of fairplay in dealing with
> me, a non-mathematician.
> 
> who claim to be logicians or scientists.
> 
> They're not; they're bullshit artists.
> 
> 
> James Harris
And you'll notice that he never addressed my last question.  James: if 
you are unwilling to answer questions asking for clarification, you are 
not likely to persuade anyone.  It would be more helpful to cite the 
find what addresses what.  May I suggest that in the future you answer 
questions, rather than insult the person asking them?  If I was in some 
way rude, I appologize.  I will do my best to not insult you.  However, 
I *will* ask questions when I don't understand what is being said.  It 
is safer to do that than assume you mean something other than what you 
intended.
I believe I asked it above, but I will ask again: do I have your 
-- 
Will Twentyman
====
> 
> I've looked at your paper... now looking at this I'll bite.  I may be
> getting in over my head but I'll point out the things that don't make
> sense to me.  I'm open to being enlightened.
>> [snip]
> Notes: I am a mathematician.  I have only recently started reading this
> board, so I am not familiar with your previous difficulties with
> mathematicians.
>> It looks like you and James have been tricked into wasting
>> a few hours of each others time.
>> I really enjoyed James' opening line
>> and how it gradually evolved into
>>       IT'S ALL IN THE PAPER YOU CLAIM TO
>>           HAVE READ!!!!!!!!!!!!!!
>>       That's it.  I'm tired of your game.
>> If this was your intention, very well done! If not, don't take it
>> personal - it's a question of medication dosage. There must
>> be a balance somewhere. We know it will be unstable, but
>> it will be a balance - for a few weeks.
>> Welcome to the playground of James Harrass.
>> Dirk Vdm
> 
> I noticed it.  I've lurked long enough that his response isn't 
> *entirely* surprising.  I suppose I should go try to make sense of his 
> paper now, to see where I'm missing things.  It would have been helpful 
> if he had started by posting the paper here instead of posting a link to 
> a pdf.  It makes quoting his paper... more challenging.  And makes it 
> less accessible.
> 
> I looked at his paper long enough to get confused by it.  I didn't try 
> to make sense of it.  I suppose I must defend my honor as a 
> mathematician and start looking for the definitions in his paper.
I've been lurking long enough to see several well-meaning mathematicians 
take a lot of trouble to try to understand what JSH is doing, and to 
help him to clarify it.  The outcome has become quite predictable. 
Rather than follow the advice that he define his terms and follow 
accepted standards of proof, JSH turns on his would-be helpers and 
abuses them as liars and cheats.  You are asking for the same treatment.
Gib
====
> Deny that you are a mathematician, and admit that you do not
> understand what you are talking about.  Otherwise, stop attacking a
> yourself.
He has frequently proclaimed that he is not a mathematician.  For a 
rational person this would disqualify him from attempting to do 
mathematics, but JSH is immune to such logic.
Gib
====
> 
>>I've looked at your paper... now looking at this I'll bite.  I may be
>>getting in over my head but I'll point out the things that don't make
>>sense to me.  I'm open to being enlightened.
> 
> 
> [snip]
> 
> 
>>Notes: I am a mathematician.  I have only recently started reading this
>>board, so I am not familiar with your previous difficulties with
>>mathematicians.
That's just sad.  It means that maybe you picked up bad habits which
as I've feared are typical for mathematicians.
In any event, if you get serious I'll work through it carefully as
long as you don't try any tricks, no bait-and-switch, nor anything
else not aboveboard.
 
> 
> 
> It looks like you and James have been tricked into wasting
> a few hours of each others time.
Nope.  I don't necessarily post for one person.  Undergrads who *can*
follow and see what's going on can also get out of mathematics while
they can.
Biology is a good major.
> 
> I really enjoyed James' opening line
> and how it gradually evolved into
>       IT'S ALL IN THE PAPER YOU CLAIM TO
>           HAVE READ!!!!!!!!!!!!!!
>       That's it.  I'm tired of your game.
Yeah, my elation turned to frustration as first I was repeating what
was in the paper, then dealing with underhanded things like
bait-and-switch and finally I just got tired of it.
> 
> If this was your intention, very well done! If not, don't take it
> personal - it's a question of medication dosage. There must
> be a balance somewhere. We know it will be unstable, but
> it will be a balance - for a few weeks.
> 
> Welcome to the playground of James Harrass.
> 
> Dirk Vdm
> 
> 
> 
> I noticed it.  I've lurked long enough that his response isn't 
> *entirely* surprising.  I suppose I should go try to make sense of his 
> paper now, to see where I'm missing things.  It would have been helpful 
> if he had started by posting the paper here instead of posting a link to 
> a pdf.  It makes quoting his paper... more challenging.  And makes it 
> less accessible.
The paper started on the newsgroup.
Undergrads the warning is clear.  Mathematicians don't play fair, they
seem to run from logic if it doesn't suit them, and when you catch
them, they just come up with more *stuff*.
> I looked at his paper long enough to get confused by it.  I didn't try 
> to make sense of it.  I suppose I must defend my honor as a 
> mathematician and start looking for the definitions in his paper.
What I'm doing is showing that apparently there is little honor as a
mathematician, which is a modern thing.
Mathematicians like Gauss, Euler, Dedekind, and so many others
wouldn't play these bullshit games, I'm sure.
But I think a lot of you got in under a system where lies are not only
tolerated, but expected, where mathematics is just a club, so you can
pay your bills, get married, and teach poor unsuspecting students who
get caught up in the grandeur of mathematics past.
Those students need to remember that the future keeps happening
despite the past.
Mathematicians today *can* be as contemptibly bad as I've been showing
despite the grand history of the discipline.
It apparently is going through a bad spell.
James Harris
====
>   g_1...g_k = P(0).
> 
> Don't you mean (g_1...g_k)(0)=P(0)?
> 
> I'm not interested in a form debate.  There's style and there's
> substance.
It's not style, it's precision. Your g_1...g_k = P(0) admits of the
interpretation g_1...g_k is a function that is identially P(0), i.e.,
for all x: (g_1....g_k)(x) = P(0).
V.
-- 
====
No kidding!!
 > Go home, little man...
 This is the only home he has...
 V.
 --
====
You're an admirable person to try and make sense of his drither. Good luck,
Will Twentyman!!!
>>I've looked at your paper... now looking at this I'll bite.  I may be
>>getting in over my head but I'll point out the things that don't make
>>sense to me.  I'm open to being enlightened.
>   > [snip]
>   >>Notes: I am a mathematician.  I have only recently started reading this
>>board, so I am not familiar with your previous difficulties with
>>mathematicians.
>   > It looks like you and James have been tricked into wasting
> a few hours of each others time.
> I really enjoyed James' opening line
> and how it gradually evolved into
>       IT'S ALL IN THE PAPER YOU CLAIM TO
>           HAVE READ!!!!!!!!!!!!!!
>       That's it.  I'm tired of your game.
> If this was your intention, very well done! If not, don't take it
> personal - it's a question of medication dosage. There must
> be a balance somewhere. We know it will be unstable, but
> it will be a balance - for a few weeks.
> Welcome to the playground of James Harrass.
> Dirk Vdm
>  
> I noticed it.  I've lurked long enough that his response isn't
> *entirely* surprising.  I suppose I should go try to make sense of his
> paper now, to see where I'm missing things.  It would have been helpful
> if he had started by posting the paper here instead of posting a link to
> a pdf.  It makes quoting his paper... more challenging.  And makes it
> less accessible.
 I looked at his paper long enough to get confused by it.  I didn't try
> to make sense of it.  I suppose I must defend my honor as a
> mathematician and start looking for the definitions in his paper.
 --
> Will Twentyman
>
====
> But I think a lot of you got in under a system where lies are not only
> tolerated, but expected, where mathematics is just a club, so you can
> pay your bills, get married, and teach poor unsuspecting students who
> get caught up in the grandeur of mathematics past.
My lecturers actively encourage us to find flaws in the stuff presented to
us. Frankly, they would be rather disappointed if we believed something 
just
because they told us it is true.
Jon
====
>We've been trained to be very precise in how words are 
>using, ...
And words using goodly we is...   ;-)
-- 
Thomas Wasell     | Does the name Pavlov ring a bell? 
wasell@bahnhof.se | 
====
> 
> 
>>We've been trained to be very precise in how words are 
>>using, ...
> 
> 
> And words using goodly we is...   ;-)
> 
*sigh*  I knew I shouldn't have studied under Yoda and Jar-Jar Binks.
-- 
Will Twentyman
====
> 
> I've been lurking long enough to see several well-meaning mathematicians 
> take a lot of trouble to try to understand what JSH is doing, and to 
> help him to clarify it.  The outcome has become quite predictable. 
> Rather than follow the advice that he define his terms and follow 
> accepted standards of proof, JSH turns on his would-be helpers and 
> abuses them as liars and cheats.  You are asking for the same treatment.
> 
> Gib
> 
> 
I seem to have received it already.  I can hope he will listen, and if 
not, I can be know that I attempted to help someone.  It is up to him 
whether he accepts help.
-- 
Will Twentyman
====
> 
> 
> 
> Yeah, my elation turned to frustration as first I was repeating what
> was in the paper, then dealing with underhanded things like
> bait-and-switch and finally I just got tired of it.
Notice that you assumed
1) I made a bait-and-switch and
2) If I did, it was intentional rather than a result of not understanding.
>If this was your intention, very well done! If not, don't take it
>personal - it's a question of medication dosage. There must
>be a balance somewhere. We know it will be unstable, but
>it will be a balance - for a few weeks.
Welcome to the playground of James Harrass.
Dirk Vdm

>>I noticed it.  I've lurked long enough that his response isn't 
>>*entirely* surprising.  I suppose I should go try to make sense of his 
>>paper now, to see where I'm missing things.  It would have been helpful 
>>if he had started by posting the paper here instead of posting a link to 
>>a pdf.  It makes quoting his paper... more challenging.  And makes it 
>>less accessible.
> 
> 
> The paper started on the newsgroup.
In sci.math, where I picked this up, you posted a link to the paper that 
do not recall ever seeing it in sci.math, but that could easily be an 
oversite.
-- 
Will Twentyman
====
> 
>  I can be know that I attempted to help someone.  
> 
Note to self: I have GOT to proofread before hitting <send>
-- 
Will Twentyman
====
> 
>>I've looked at your paper... now looking at this I'll bite.  I may be 
>>getting in over my head but I'll point out the things that don't make 
>>sense to me.  I'm open to being enlightened.
> 
> 
> Note: looked at is not the same as reading for understanding.
Ok.
 
>I think I've figured out a way to show basically all of you, including
>people who think they don't know any math that mathematicians have
>been lying about my work.  It's so trivial you *should* wonder why
>they thought they could get away with it.
>>Either you aren't being clear, or they missed something.  It is a strong 

>>charge to suggest that ALL mathematicians are in this great conspiracy 
>>to teach incorrect math.  Sort of goes against what they stand for.
> 
> I say mathematicians but don't specifically say ALL mathematicians.
> 
> 
> Distinction noted.  My mistake.
Note that I've been careful to actually have working mathematicians in
mind when I say that mathematicians have been lying about my work.
Two that readily come to mind are David Ullrich, a professor at
Oklahoma State University, and Arturo Magidin, whose university I'm
not sure of at this time as I'd have to check a header in a post of
his, but he obtained his Ph.D from Berkeley University.
I've verified that each is associated as given by checking at the
websites of the respective universities.
They are mathematicians as noted.
> I'm talking about a pattern of denial that I'm seeing in
> mathematicians that I've contacted or observed.
>  
> 
>(Paper linked to at http://groups.msn.com/AmateurMath as usual.)
>And in my paper I start by showing that I can write that as
>  g = r + c
>where either r=0, or r changes as the polynomial's value changes,
>while c does not.
>>Questions: is g the polynomial to be factored or a factor?  Is r a 
>>polynomial, variable, or constant?  Is c a polynomial or constant?  What 

>>do you mean about a polynomial's value changing?  A polynomial is in the 

>>form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0.  It simply _is_, it 

>>doesn't change.
> 
> Hmmm...doesn't sound like you actually looked at the paper.
> 
> I'll be going over it in detail this weekend.  Do you mind if I quote it 
> directly in my responses?
No.
> In the paper the ring is algebraic integers; therefore, r and c are
> algebraic integers.
> 
> An algebraic integer is a constant.  It does not vary.  Period.  Note: 
> I'm taking my definitions of terms from Wolfram's Mathworld.  If you are 
> using a different definition, please provide it.
You deleted context.  My guess is that you were asking what r and c
were and I noted that because the ring is algebraic integers, they are
algebraic integers.
For instance, given P(x)=x+1 in the ring of integers, if you come back
to me and ask me what is x, I'll tell you that it is an integer.
If you reply that integers do not vary I'll doubt your veracity and
assume that you are deliberately trying to deceive.
I suggest that you be careful about what you delete out when you make
such statements.
My assessment at this time is that you have been *repeatedly* caught.
My own view is that your behavior is typicaly of what I've seen from
other mathematicians.
Clearly I see an advantage in pointing out to others--especially
undergrads--that behavior like yours is, in my experience, typical of
the mathematical community.
Mathematicians cheat.
> In the paper it's noted that I'm generalizing beyond polynomial
> factors, that is, factors of a polynomial that are themselves
> polynomials.
> 
> In the paper I give an example of g = sqrt(x+1), so it's hard to
> understand how you could miss the truth.
> 
> To be honest, I didn't get past the statement of your lemma.  I will 
> attempt to correct that.
Well you can say To be honest but I think you need to work harder at
showing that you're being honest.
> I state as included in your post, so it's hard to see how you missed
> it, but I'll give it again here for emphasis:
> 
> 
>where either r=0, or r changes as the polynomial's value changes,
>while c does not.
> 
> 
> But you state above:
> 
> What do you mean about a polynomial's value changing?  A polynomial
> is in the
> form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0.  It simply _is_,
> it doesn't change.
> 
> That's called bait-and-switch as you switched from the polynomial's
> value to the polynomial.
> 
> That's called using standard terminology.  Precise use of terms is a 
> hallmark of mathematics.  A polynomial is an expression.  For different 
> values of its variable(s) it evaluates to different values, but that is 
> not the same as saying the polynomial's value changes.
Posts on newsgroups are not formal, and there's always a possibility
of misunderstandings.
I'm not a mathematician.  There are likely to be any number of times
when I end up using something that's not exactly the way you may be
used to seeing it, but if you're a mathematician, by definition you're
a math expert.
If I say the value of P(x) = x + 1 varies as x varies, then are you
suggesting you can't figure out what I mean, even if it's not exactly
what you're used to seeing?
> 
> I didn't say the polynomial changes.  I said its value changes and you
> noted that in one sentence and switched in the other.
> 
> Now given, say, the polynomial P(x) = x+1, the polynomial's value at
> x=0, is 1.
> 
> It's not rocket science.
> 
> Correct.  But you did NOT change the value of the polynomial, you 
> changed x and you changed P at x.  P(x)=x+1, P(0)=1.  P's value didn't 
> change, you simply evaluated it for a given value of x.
Then what changed?  You say, ...you changed P at x.  Yet above you
have:
> What do you mean about a polynomial's value changing?  A polynomial
> is in the
> form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0.  It simply _is_,
> it doesn't change.
Now I take your claim to be that the value of a polynomial is its
explicit expression.
While I said things like the polynomial's value at x=0.
Your claim is that my usage is nonstandard.
Ok, I don't care to hold on to a nonstandard usage as my aim is to
communicate.
So then, however it's stated so that I can talk of a polynomial, like
x+1, and also talk of the value for some particular x.
> And for the rest of you, I've seen a troubling tendency from posters
> and people I know are actual mathematicians to cheat in this way.
> 
> If insisting on precise use of terminology is cheating, you might as 
> well give up.  We've been trained to be very precise in how words are 
> using, because many (if not all) of us got proofs shredded when we 
> forgot a technicality.
> 
> [rant deleted]
<deleted>
Again this is a *post* on a newsgroup and not a formal paper.  I'm
trying to explain in a medium that allows questions to be asked in
case there is a misunderstanding.
Here there seems to be a misunderstanding as to how one talks of the
values for the explicit expression of a polynomial P(x) for a given x.
Now I'm suspicious of you still, but again, I see your post as typical
of what I've come to expect from mathematicians.
Which is also why I say you're like lying English professors.
I also say you're like bad lawyers who argue over technicalities, as
if the point or the truth doesn't matter.
I'm not interested in reading further as you've repeatedly
demonstrated behavior I find deceitful and annoying.
James Harris
====
[cut]
> Ok, I don't care to hold on to a nonstandard usage as my aim is to
> communicate.
> 
> So then, however it's stated so that I can talk of a polynomial, like
> x+1, and also talk of the value for some particular x.
There are probably many ways. Below is one way.
   Let A be the ring of algebraic integers.
   Let P(X) = X+1 be a polynomial over A.
   Let x be an algebraic integer.
   Consider the algebraic integer P(x).
   etc.
Another example is your lemma being discussed:
   Let A be the ring of algebraic integers.
   Let P(X) be a polynomial in A[X].
   Let x be an algebraic integer.
   Let g be an algebraic integer that is a factor of P(x).
   Then, I can find algebraic integers r and c such that
   g = r + c where c is a factor of P(0) and
   either r = 0 or else x and r are not coprime.
Note that the above implies that c depends upon x,
whereas you say that c is independent of x. Thus,
the above phrasing is not quite what you want.
-- Bill Hale
====
> 
>I think I've figured out a way to show basically all of you, including
>people who think they don't know any math that mathematicians have
>been lying about my work.  It's so trivial you *should* wonder why
>they thought they could get away with it.
>>Either you aren't being clear, or they missed something.  It is a strong 

>>charge to suggest that ALL mathematicians are in this great conspiracy 
>>to teach incorrect math.  Sort of goes against what they stand for.
I say mathematicians but don't specifically say ALL mathematicians.
>Distinction noted.  My mistake.
> 
> 
> Note that I've been careful to actually have working mathematicians in
> mind when I say that mathematicians have been lying about my work.
> 
> Two that readily come to mind are David Ullrich, a professor at
> Oklahoma State University, and Arturo Magidin, whose university I'm
> not sure of at this time as I'd have to check a header in a post of
> his, but he obtained his Ph.D from Berkeley University.
> 
> I've verified that each is associated as given by checking at the
> websites of the respective universities.
> 
> They are mathematicians as noted.
For the record, I have a Master's Degree in math from the University of 
Illinois, Urbana-Champaign.  My emphasis was in logic, not algebra, so 
I'll be doing a fair bit of refreshing as I work through this.
>Hmmm...doesn't sound like you actually looked at the paper.
>>I'll be going over it in detail this weekend.  Do you mind if I quote it 
>>directly in my responses?
> 
> No.
>In the paper the ring is algebraic integers; therefore, r and c are
>algebraic integers.
>>An algebraic integer is a constant.  It does not vary.  Period.  Note: 
>>I'm taking my definitions of terms from Wolfram's Mathworld.  If you are 
>>using a different definition, please provide it.
> 
> 
> You deleted context.  My guess is that you were asking what r and c
> were and I noted that because the ring is algebraic integers, they are
> algebraic integers.
I will attempt to deal with this carefully in my comments about your paper.
> For instance, given P(x)=x+1 in the ring of integers, if you come back
> to me and ask me what is x, I'll tell you that it is an integer.
> 
> If you reply that integers do not vary I'll doubt your veracity and
> assume that you are deliberately trying to deceive.
> 
> I suggest that you be careful about what you delete out when you make
> such statements.
I will keep that in mind.
>I state as included in your post, so it's hard to see how you missed
>it, but I'll give it again here for emphasis:
where either r=0, or r changes as the polynomial's value changes,
>while c does not.

>But you state above:
What do you mean about a polynomial's value changing?  A polynomial
>is in the
>form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0.  It simply _is_,
>it doesn't change.
That's called bait-and-switch as you switched from the polynomial's
>value to the polynomial.
>>That's called using standard terminology.  Precise use of terms is a 
>>hallmark of mathematics.  A polynomial is an expression.  For different 
>>values of its variable(s) it evaluates to different values, but that is 
>>not the same as saying the polynomial's value changes.
> 
> Posts on newsgroups are not formal, and there's always a possibility
> of misunderstandings.
> 
> I'm not a mathematician.  There are likely to be any number of times
> when I end up using something that's not exactly the way you may be
> used to seeing it, but if you're a mathematician, by definition you're
> a math expert
> 
> If I say the value of P(x) = x + 1 varies as x varies, then are you
> suggesting you can't figure out what I mean, even if it's not exactly
> what you're used to seeing?
As stated here, I can understand it.  I just wish to be careful to not 
place assumptions on what you mean that may not be accurate.  If it 
makes me sound like I'm being nitpicky, I'll accept that.  I prefer that 
to blasting you only to find there has been a misunderstanding.  The 
exchange below is a good example of trying to clearly understand what 
you're saying, even if it sounds like I'm asking annoying and/or stupid 
questions.
>I didn't say the polynomial changes.  I said its value changes and you
>noted that in one sentence and switched in the other.
Now given, say, the polynomial P(x) = x+1, the polynomial's value at
>x=0, is 1.
It's not rocket science.
>>Correct.  But you did NOT change the value of the polynomial, you 
>>changed x and you changed P at x.  P(x)=x+1, P(0)=1.  P's value didn't 
>>change, you simply evaluated it for a given value of x.
> 
> 
> Then what changed?  You say, ...you changed P at x.  Yet above you
> have:
> 
> 
>What do you mean about a polynomial's value changing?  A polynomial
>is in the
>form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0.  It simply _is_,
>it doesn't change.
> 
> 
> Now I take your claim to be that the value of a polynomial is its
> explicit expression.
> 
> While I said things like the polynomial's value at x=0.
> 
> Your claim is that my usage is nonstandard.
> 
> Ok, I don't care to hold on to a nonstandard usage as my aim is to
> communicate.
> 
> So then, however it's stated so that I can talk of a polynomial, like
> x+1, and also talk of the value for some particular x.
> 
> 
[deletia]
> 
> Again this is a *post* on a newsgroup and not a formal paper.  I'm
> trying to explain in a medium that allows questions to be asked in
> case there is a misunderstanding.
Understood.
> Here there seems to be a misunderstanding as to how one talks of the
> values for the explicit expression of a polynomial P(x) for a given x.
> 
> Now I'm suspicious of you still, but again, I see your post as typical
> of what I've come to expect from mathematicians.
> 
> Which is also why I say you're like lying English professors.
> 
> I also say you're like bad lawyers who argue over technicalities, as
> if the point or the truth doesn't matter.
That's because I've found that technicalities can make the difference 
between a solid proof and a very pretty piece of garbage.  When doing 
proofs, technicalities frequently matter.
> I'm not interested in reading further as you've repeatedly
> demonstrated behavior I find deceitful and annoying.
I can't promise that I'll say anything in the future that will not be 
similarly annoying.  It is the nature of asking questions that they can 
become annoying.  As for deceit, it suggests I have an agenda of some 
sort.  I'll clearly state my agenda: to understand what you are saying, 
and determine if what you have said is valid.  This could result in 
demonstrating that you are correct and misunderstod, or that you have 
made some mistakes that may or may not be fixable.  Either way is fine 
with me.  If you believe I have some other agenda that is not 
appropriate, please state it and I will do my best to avoid it.
-- 
Will Twentyman
====
> I think I've figured out a way to show basically all of you, including
> people who think they don't know any math that mathematicians have
> been lying about my work.  It's so trivial you *should* wonder why
> they thought they could get away with it.
> 
> Here goes.
> 
> My paper Advanced Polynomial Factorization depends on considering a
> factor of a polynomials that I call g.
> 
> (Paper linked to at http://groups.msn.com/AmateurMath as usual.)
> 
soon to be published as Mein Kampf. Twenty years of struggle against
the world's mathematicians and their lies.
====
> Two that readily come to mind are David Ullrich, a professor at
> Oklahoma State University, and Arturo Magidin, whose university I'm
> not sure of at this time as I'd have to check a header in a post of
> his, but he obtained his Ph.D from Berkeley University.
Sheesh, you can't even get the name of the University right!  It's The
University of California at Berkeley, or just Berkeley, but *not*
Berkeley University!
> Posts on newsgroups are not formal, and there's always a possibility
> of misunderstandings.
If you're going to talk about mathematics you need to use the proper
conversation in a coffee shop.
(BTW, speaking of terminology, I really wish you'd refer to newsgroup
Terminology matters in more places than mathematics, you know.)
> I'm not a mathematician.  There are likely to be any number of times
> when I end up using something that's not exactly the way you may be
> used to seeing it, but if you're a mathematician, by definition you're
> a math expert.
Idiot.  If two mathematicians (or math experts as you prefer to
call them) try to discuss mathematics using non-standard terminology,
and without clearly defining their non-standard terms *in relation to*
standard terminology, then they'll probably confuse *each other*.  It's
even *more* important for someone like you to be very careful and precise
in defining and using your terms.  That's why standard terminology exists
in the first place!
> I'm not interested in reading further as you've repeatedly
> demonstrated behavior I find deceitful and annoying.
This is James Harris code-speak for I'm completely out of my depth and
can't understand your argument well enough to dispute it intelligently,
so I'll just ignore you and hope you go away.  Whenever James starts
accusing people of lying or cheating or being deceitful, it's a sure
sign he knows he's outclassed.
-- 
Wayne Brown                | When your tail's in a crack, you improvise
fwbrown@bellsouth.net      |  if you're good enough.  Otherwise you give
                           |  your pelt to the trapper.
e^(i*pi) = -1  -- Euler  |           -- John Myers Myers, 
Silverlock
====
You know, Harris, you seem to think that anyone who has a degree in 
ANYTHING
is a liar. What's your pathetic excuse?? Why do you, who have NO FORMAL
DEGREE IN MATHEMATICS, feel you can critique REAL mathematicians???
You are such a loser, Harris!! The time you waste on these forums is
amazing. If you could put forth 1/8th the effort into having a REAL life
instead of this pathetic research you're doing, you may actually make a
friend or two. Until then, you continue to be a waste of human flesh...
Don't forget your doomsday predictions from about a month or two ago to
those who doubted you... Weren't you planning on killing us all and you 
were
going to be the only one left around???
Can you really expect anyone to take you seriously??? You're such a joke...
>>I've looked at your paper... now looking at this I'll bite.  I may be
>>getting in over my head but I'll point out the things that don't make
>>sense to me.  I'm open to being enlightened.
> [snip]
>>Notes: I am a mathematician.  I have only recently started reading
this
>>board, so I am not familiar with your previous difficulties with
>>mathematicians.
 That's just sad.  It means that maybe you picked up bad habits which
> as I've feared are typical for mathematicians.
 In any event, if you get serious I'll work through it carefully as
> long as you don't try any tricks, no bait-and-switch, nor anything
> else not aboveboard.
 > It looks like you and James have been tricked into wasting
> a few hours of each others time.
 Nope.  I don't necessarily post for one person.  Undergrads who *can*
> follow and see what's going on can also get out of mathematics while
> they can.
 Biology is a good major.
 > I really enjoyed James' opening line
> and how it gradually evolved into
>       IT'S ALL IN THE PAPER YOU CLAIM TO
>           HAVE READ!!!!!!!!!!!!!!
>       That's it.  I'm tired of your game.
 Yeah, my elation turned to frustration as first I was repeating what
> was in the paper, then dealing with underhanded things like
> bait-and-switch and finally I just got tired of it.
 > If this was your intention, very well done! If not, don't take it
> personal - it's a question of medication dosage. There must
> be a balance somewhere. We know it will be unstable, but
> it will be a balance - for a few weeks.
> Welcome to the playground of James Harrass.
> Dirk Vdm
>   > I noticed it.  I've lurked long enough that his response isn't
> *entirely* surprising.  I suppose I should go try to make sense of his
> paper now, to see where I'm missing things.  It would have been helpful
> if he had started by posting the paper here instead of posting a link 
to
> a pdf.  It makes quoting his paper... more challenging.  And makes it
> less accessible.
 The paper started on the newsgroup.
 Undergrads the warning is clear.  Mathematicians don't play fair, they
> seem to run from logic if it doesn't suit them, and when you catch
> them, they just come up with more *stuff*.
 > I looked at his paper long enough to get confused by it.  I didn't try
> to make sense of it.  I suppose I must defend my honor as a
> mathematician and start looking for the definitions in his paper.
 What I'm doing is showing that apparently there is little honor as a
> mathematician, which is a modern thing.
 Mathematicians like Gauss, Euler, Dedekind, and so many others
> wouldn't play these bullshit games, I'm sure.
 But I think a lot of you got in under a system where lies are not only
> tolerated, but expected, where mathematics is just a club, so you can
> pay your bills, get married, and teach poor unsuspecting students who
> get caught up in the grandeur of mathematics past.
 Those students need to remember that the future keeps happening
> despite the past.
 Mathematicians today *can* be as contemptibly bad as I've been showing
> despite the grand history of the discipline.
 It apparently is going through a bad spell.

> James Harris
====
> Don't forget your doomsday predictions from about a month or two ago to
> those who doubted you... Weren't you planning on killing us all and you 
were
> going to be the only one left around???
I don't remember that. There was the thing about how he
has Army generals just waiting for word from him to invade
the halls of academia. And the thing about how he's going
to press charges and everyone in the math establishment is
going to jail. And occasionally the thing about how the
human race is doomed to extinction if we don't accept his
proof as correct (haven't seen that one in a while). But I
don't remember that particular threat or prediction.
         - Randy
====
> 
>Hmmm...doesn't sound like you actually looked at the paper.
>>I'll be going over it in detail this weekend.  Do you mind if I quote it 

>>directly in my responses?
> 
> No.
> 
Having spent the weekend going through musty books on algebra...  here
is my analysis:
BTW, I'm using Google to post this since my usual newsfeed is acting
up.
Original paper will be quoted, the rest is my comments.  There may be
formatting issues with the quoted paper since it was a .pdf rather
than plain text.
>ADVANCED POLYNOMIAL FACTORIZATION
>JAMES HARRIS
>Abstract. Algebraic method for determining distribution of fac-
>tors within a polynomial factorization, which breaks through what
>was seen as a barrier from overinterpretations of Galois Theory.
>1. Advanced Polynomial Factorization Approached
>Determining the distribution of factors within irrational algebraic
>integers has long been considered impossible as it is not possible to
do
>using Galois Theory. However a simple technique through the intro-
>duction of more variables makes it possible. To highlight the
standard
>belief consider the algebraic integers (1 &#8722; sqrt(21))/2 and
(1+sqrt(21))/2 >which are roots of x^2 + x &#8722; 5.
This is a minor note, but it should be -1 for both roots.  While this
has little impact on the validity of the rest of the work, it should,
at the very least, be corrected in your final draft.
>While you know that the algebraic integer factors are themselves
>factors of 5, in what way is each a factor of 5? Can either not have
non
>unit factors of 5? How do you know?
These questions don't make a lot of sense to me, but I'll skip over
that to get to what I believe is the true problem.
>In looking to consider distribution of algebraic integer factors
within
>a factorization I'll be using a more complicated example than x 2 +x
&#8722; 5.
>The paper will show that given the factorization, in the ring of
alge-
>braic integers,
>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>one of the a's is coprime to 5, using basic algebraic methods.
And here we run into two problems.  First, while it is true that the
a's are algebraic integers, it is something that at least deserves a
comment.  Second, it is far from obvious that coprime is meaningful
on the algebraic integers.  Here's the problem: coprime is based on
the definition of the least common divisor.  The LCD is only
guaranteed to exist for Principal Ideal Domains and Unique
algebraic integers are certainly NOT a UFD, and I've seen nothing to
indicate anything stronger than that they are a commutative ring. 
Until someone can point to work that indicates that LCD is defined on
the algebraic integers, this paper cannot move forward.  The remainder
of the paper (which I have NOT checked for accuracy but am including
for completeness) depends on the concept of coprime being meaningful
and defined on the algebraic integers.
>First I'll need a simple lemma to generalize beyond factors of a
poly-
>nomial that are themselves polynomials.
>Lemma 1.1. Factorization Lemma:
>Given a factor g of a polynomial P(x), further defined as a factor
>for all x, which means that the value of g for a value 'a' of x is a
factor
>of P(a), within the ring of algebraic integers, there exists r and c
such
>that
>g = r + c
>where r=0, or is not coprime to x, and c is a factor of the constant
>term P(0).
The notation here could be improved.  g should be g(x), r should be
r(x).
>Proof. Let x=0, then g must be a factor of P(0), so at that point c =
g.
>(1) If when x does not equal 0, g=c, r=0.
>(2) If when x does not equal 0, g =/= c there must exist r which
varies
>with x, and as r equals 0 when x equals 0 it is not coprime to
>x.
>
I'm not sure step 2 makes sense.  I don't see why it makes r coprime
to x.
>As an example consider sqrt(x + 1) which is a non polynomial factor
of
>x+1, and while there are an infinity of irrational solutions consider
the
>rational solution at x=35.
Are you sure you want to talk about non-polynomial roots?  While
legitimate, it changes the topic of discussion from polynomials over
the algebraic integers to algebraic functions over the algebraic
integers.
>Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,
>and c=1. But for different values of x, g and r will vary, while r
will
>not be coprime to x.
Again, coprime needs to be defined for the particular ring under
consideration.
>2. Primary Argument
>Given
>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>in the ring of algebraic integers. Let
>P(m) = f^2 ((m^3 f^4 &#8722; 3m^2 f^2 + 3m)x^3 &#8722; 3( &#8722; 1 +
mf^2 )xu^2 + u^3 f)
>Here f is a non unit, non zero algebraic integer coprime to 3 and x,
>and u a non unit, non zero algebraic integer coprime to f. Note P(m)
>has a factor that is f^2 .
Again, coprime must be defined on the algebraic integers.  Also, the
change of variable and the introduction of additional variables seems
odd.  Since it appears that you mean for f and u to be constants, it
appears that you actually have P(m,x).
>That expression comes from expanding (v^3 +1)x^3 &#8722; 3vxy^2 + y^3
, using
>the substitutions v = &#8722; 1+mf^2 , and y = uf, where additional
variables
>provide an additional degree of freedom.
>Consider that a similar idea can be used to factor 3, prime in
integers,
>as x^2 +7x+10 = 3, allows you to find factors of 3 in the ring of
algebraic
>integers.
I don't follow what you mean by this at all.  Some additional
explanation would be helpful.  What are the factors and how did you
compute them?
>Now consider the factorization
>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)
>where multiplying out shows that
>a1a2a3 = m^3 f^6 &#8722; 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 &#8722; 3m^2
f^2 + 3m)
>so a1a2a3 = mf^2 (m^2 f^4 &#8722; 3mf^2 + 3).
>Therefore, at least one of the a's cannot be coprime to m, and at
>least one of the a's must equal 0 when m=0.
>(Note: The a's are roots of a monic polynomial with algebraic integer
>coefficients so they are algebraic integers.)
Maybe.  Did P(m) have a constant term of 1?  if so, uf=1.  The logic
of your note could stand some clarification.  (note: coprime shows up
again)
>Notice that the constant term P(0) is given by
>P(0) = f^2 (3xu^2 + u^3 f)
>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.
>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1
>is not coprime to m.
>g1 = c = uf
>meaning f is a factor of the constant term.
>Therefore, exactly two of the a's equal 0, when m=0, to get the
factor
>f^2 in the constant term P(0), while one must not equal 0, or f^3
would
>be the factor.
This statement doesn't make a lot of sense either... but I'm not
trying as hard at this point.  If the other problems can be fixed,
I'll be happy to examine it more carefully.
>Now as noted before in general P(m) has a factor that is f^2 , and
>separating that factor off, gives a constant term coprime to f;
therefore,
>given g1 = a1x + uf
>where with m = 0, g1 gives a factor of f it must have that same
factor
>in general, proving that two of the a's have a factor that is f.
>Therefore, one factor is coprime to f.
>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't
>change the a's, I have
>(m^3 f^6 &#8722; 3m^2 f^4 + 3m)x^3 &#8722; 3( &#8722; 1 + mf^2 )xu^2
+ u^3 = 65x^3 &#8722; 12x + 1
>which may be more easily seen from using v = &#8722; 1 + mf^2 = 4,
y=1
>with (v^3 + 1)x^3 &#8722; 3vxy^2 + y^3 .
>Therefore, with the factorization
>65x^3 &#8722; 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)
>one of the a's is coprime to 5, which shows where some of the alge-
>braic integer factors distribute despite the factors being
irrational.
Overall analysis: if you can clarify the notation and the logical
connections, along with your result about factors not being factors,
what you probably have is a proof that coprime is not always defined
on the algebraic integers.
====
   [.sci.skeptic removed from newsgroups.]
   [.snip.]
>>In looking to consider distribution of algebraic integer factors
>within
>>a factorization I'll be using a more complicated example than x 2 +x
>&#8722; 5.
>>The paper will show that given the factorization, in the ring of
>alge-
>>braic integers,
>>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>>one of the a's is coprime to 5, using basic algebraic methods.
And here we run into two problems.  First, while it is true that the
>a's are algebraic integers, it is something that at least deserves a
>comment.  Second, it is far from obvious that coprime is meaningful
>on the algebraic integers.  Here's the problem: coprime is based on
>the definition of the least common divisor.  
The usual notion is not of a least common divisor, but of a
->greatest<- common divisor, GCD.
Coprime also makes sense for ideals: two ideals are coprime if and
only if there is no prime ideal containing both. If we assume the
Axiom of Choice and that our rings have a 1, then this is equivalent
to comaximal, no maximal ideal contains both ideals. As such, it is
possible to define coprime for arbitrary commutative rings with 1:
two elements, x and y in R, are coprime if and only if (x) and (y) are
coprime; this occurs if and only if (x,y)=(1); this occurs if and only
if there exist a and b in R such that ax+by=1.
>The LCD is only
>guaranteed to exist for Principal Ideal Domains and Unique
>algebraic integers are certainly NOT a UFD, and I've seen nothing to
>indicate anything stronger than that they are a commutative ring.
They are a Bezout Domain: every finitely generated ideal is principal:
so given any two elements x and y, they DO have a GCD, given by any
generator of the ideal (x,y); however, it is not unique, and there is
no natural choice among all of them. On the plus side, any two differ
by a constant.
   [.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
magidin@math.berkeley.edu
====
...
 >The paper will show that given the factorization, in the ring of
 > alge-
 >braic integers,
 >65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
 >one of the a's is coprime to 5, using basic algebraic methods.
 > 
 > And here we run into two problems.  First, while it is true that the
 > a's are algebraic integers, it is something that at least deserves a
 > comment.  Second, it is far from obvious that coprime is meaningful
 > on the algebraic integers.  Here's the problem: coprime is based on
 > the definition of the least common divisor.
The common definition for coprime is that within a ring, two numbers
a and b are called co-prime if there are numbers x and y such that
a*x + b*y = 1.  This is a change from the former definition where
a and b are coprime if LCD(a, b) = 1 but the theorem that there are
numbers x and y such that a*x + b*y = LCD(a, b) assures that the
definitions are the same when both could be applied.  So in the
algebraic integers 2 and 3 are coprime because 2 * 2 + 3 * (-1) = 1.
 >Lemma 1.1. Factorization Lemma:
 >Given a factor g of a polynomial P(x), further defined as a factor
 >for all x, which means that the value of g for a value 'a' of x is a
 > factor
 >of P(a), within the ring of algebraic integers, there exists r and c
 > such
 >that
 >g = r + c
 >where r=0, or is not coprime to x, and c is a factor of the constant
 >term P(0).
 > 
 > The notation here could be improved.  g should be g(x), r should be
 > r(x).
More could be improved, but as written it is wrong.  Let me attempt a
better wording (as I think JSH intends it):
  Lemma 1.1.  Factorisation lemma;
  Let's have a ring R (JSH intends the algebraic integers, but I 
generalise)
  and a polynomial P(X) in R[X].  A function g: R => R is defined to be
  a factor of P(X) if for all X in R, g(X) is a divisor of P(X).
  In that case there is a function r: R => R and a constant c in R such
  that either r(X) == 0, or r(X) is not coprime to X for all X in R
  and c divides P(0) in R.
However with this formulation the lemma is wrong (this could be done in
*all* rings that contain the integers):
  Let R be the algebraic integers.
  Let P(X) be the polynomial X + 5.
  Let g(X) be the function:
    g(X) = if X != 7: X + 5
         = if X = 7: 3
this satisfies the conditions: g(7) = 3 and P(7) = 12, for all other
X, g(X) = P(X), so g(X) is a factor of P(X).
Because of the definition of g(X) for X != 7 we find:
  c = 5,
  r(X) = if X != 7: X
       = if X = 7: -2.
But r(7) = -2 is coprime to X = 7.
 > Proof. Let x=0, then g must be a factor of P(0), so at that point c = 
g.
Agreed.
 > (1) If when x does not equal 0, g=c, r=0.
And so P(X) is a multiple of c.  The formulation is awkward...
 > (2) If when x does not equal 0, g =/= c there must exist r which 
varies
 > with x, and as r equals 0 when x equals 0 it is not coprime to x.
 > 
 > I'm not sure step 2 makes sense.  I don't see why it makes r coprime
 > to x.
(You mean *not* coprime.)  Indeed, it makes no sense.
 > Are you sure you want to talk about non-polynomial roots?  While
 > legitimate, it changes the topic of discussion from polynomials over
 > the algebraic integers to algebraic functions over the algebraic
 > integers.
But indeed, that is what he wants to do, opening a can of worms.
 > Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,
 > and c=1. But for different values of x, g and r will vary, while r 
will
 > not be coprime to x.
 > 
 > Again, coprime needs to be defined for the particular ring under
 > consideration.
As it is defined, that is no problem.  And indeed, the Lemma holds
for sqrt(x + 1) as factor of x + 1.  But I am not sure for what
kind of functions g(x) the lemma holds.  Step 2 in the proof is no
clarification for that, and there are (as shown above) function for
which it does not hold.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
> [...]
[The algebraic integers] are a Bezout Domain: every finitely generated 
ideal is principal:
>so given any two elements x and y, they DO have a GCD, given by any
>generator of the ideal (x,y); however, it is not unique, and there is
>no natural choice among all of them. On the plus side, any two differ
>by a constant.
That was a slight Harrisism for differ by a unit, right?
************************
David C. Ullrich
====
> [...]
>>[The algebraic integers] are a Bezout Domain: every finitely generated 
ideal is principal:
>>so given any two elements x and y, they DO have a GCD, given by any
>>generator of the ideal (x,y); however, it is not unique, and there is
>>no natural choice among all of them. On the plus side, any two differ
>>by a constant.
That was a slight Harrisism for differ by a unit, right?
Well, it was certainly a lapsus and I meant differ by a unit.
But just what is a Harrisism? It is not a simple error like the
above, certainly. It should encompass some complete missing the point
of a definition, theorem, or something, no?
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
   [.snip.]
>More could be improved, but as written it is wrong.  Let me attempt a
>better wording (as I think JSH intends it):
>  Lemma 1.1.  Factorisation lemma;
>  Let's have a ring R (JSH intends the algebraic integers, but I 
generalise)
>  and a polynomial P(X) in R[X].  A function g: R => R is defined to be
>  a factor of P(X) if for all X in R, g(X) is a divisor of P(X).
>  In that case there is a function r: R => R and a constant c in R such
>  that either r(X) == 0, or r(X) is not coprime to X for all X in R
>  and c divides P(0) in R.
You forgot and g(X)=r(X)+c as functions...
>However with this formulation the lemma is wrong (this could be done in
>*all* rings that contain the integers):
>  Let R be the algebraic integers.
>  Let P(X) be the polynomial X + 5.
>  Let g(X) be the function:
>    g(X) = if X != 7: X + 5
>         = if X = 7: 3
>this satisfies the conditions: g(7) = 3 and P(7) = 12, for all other
>X, g(X) = P(X), so g(X) is a factor of P(X).
>Because of the definition of g(X) for X != 7 we find:
>  c = 5,
>  r(X) = if X != 7: X
>       = if X = 7: -2.
>But r(7) = -2 is coprime to X = 7.
A much easier way to see that the lemma is false in a ring with 1
would be to note that if X is a unit, then r(X) cannot be not coprime
to X, and so that if X is a unit that would necessarily imply that
r(X) is identically zero. That means that r(X) would always have to be
the constant zero function, which means that g(X) would have to be
constant. This is clearly false for pretty much most rings.
======================================================================
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
magidin@math.berkeley.edu
====
> [...]
[The algebraic integers] are a Bezout Domain: every finitely generated 
ideal is principal:
>so given any two elements x and y, they DO have a GCD, given by any
>generator of the ideal (x,y); however, it is not unique, and there is
>no natural choice among all of them. On the plus side, any two differ
>by a constant.
>>That was a slight Harrisism for differ by a unit, right?
Well, it was certainly a lapsus and I meant differ by a unit.
But just what is a Harrisism? It is not a simple error like the
>above, certainly. It should encompass some complete missing the point
>of a definition, theorem, or something, no?
Calling it a Harrisism wasn't really appropriate, because it doesn't
involve any sort of awesome cluelessness, just writing one word
where you meant another, in an informal context. The only reason
it reminded me of JSH was because of perennial confusion over
the difference between constants and non-constants (no, not
that you were actually confused about anything like that.) Sorry.
>======================================================================
>It's not denial. I'm just very selective about
> what I accept as reality.
>    --- Calvin (Calvin and Hobbes)
>======================================================================
Arturo Magidin
>magidin@math.berkeley.edu
************************
David C. Ullrich
====
>
   [.snip.]
>> On the plus side, any two differ by a constant.
That was a slight Harrisism for differ by a unit, right?
>>Well, it was certainly a lapsus and I meant differ by a unit.
>>But just what is a Harrisism? It is not a simple error like the
>>above, certainly. It should encompass some complete missing the point
>>of a definition, theorem, or something, no?
Calling it a Harrisism wasn't really appropriate, because it doesn't
>involve any sort of awesome cluelessness, just writing one word
>where you meant another, in an informal context. The only reason
>it reminded me of JSH was because of perennial confusion over
>the difference between constants and non-constants (no, not
>that you were actually confused about anything like that.) Sorry.
Actually, I was more interested in coming up with a good notion of
Harrisism we can use on the newsgroup... (-: You know, like
spoonerism. 
It seems a bit hard to encompass all the behaviors that could be thus
named, but we should we able to hash out something. I mean, had I
replied attacking you for taking me to task on what was clearly just a
simple error instead of just noting the error and leaving it at that
or something, that would be harris-like behavior...
Perhaps:
Harrisism: the repeated and dogged blatantly incorrect use of a
technical term, caused by either ignorance or confusion.
(So the recent discussion on 'well-ordering the reals' would be a 
Harrisism...)
Any other suggestions?
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
thank you.  that's how bad I am, that I was always puzzled
by that ax + by = 1 definition of coprimeness,
as recondite as it is ... I think!
 
>   Lemma 1.1.  Factorisation lemma;
>   Let's have a ring R (JSH intends the algebraic integers, but I 
generalise)
>   and a polynomial P(X) in R[X].  A function g: R => R is defined to be
>   a factor of P(X) if for all X in R, g(X) is a divisor of P(X).
>   In that case there is a function r: R => R and a constant c in R such
>   that either r(X) == 0, or r(X) is not coprime to X for all X in R
>   and c divides P(0) in R.
> However with this formulation the lemma is wrong (this could be done in
> *all* rings that contain the integers):
>   Let R be the algebraic integers.
>   Let P(X) be the polynomial X + 5.
>   Let g(X) be the function:
>     g(X) = if X != 7: X + 5
>          = if X = 7: 3
> this satisfies the conditions: g(7) = 3 and P(7) = 12, for all other
> X, g(X) = P(X), so g(X) is a factor of P(X).
> Because of the definition of g(X) for X != 7 we find:
>   c = 5,
>   r(X) = if X != 7: X
>        = if X = 7: -2.
> But r(7) = -2 is coprime to X = 7.
--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...
La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto:
(FOSSILISATION [McCainanites?] (TM/sic))/
BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.
Http://www.tarpley.net/bushb.htm (content partiale, below):
  17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81
  23 -- Le FIN d'HISTOIRE
  24 -- L'ORDEUR du MONDE NOUVEAU
  25 -- THYROID STORK !?!
====
        ... stuff deleted ...
>>That was a slight Harrisism for differ by a unit, right?
Well, it was certainly a lapsus and I meant differ by a unit.
But just what is a Harrisism? It is not a simple error like the
>above, certainly. It should encompass some complete missing the point
>of a definition, theorem, or something, no?
>>Calling it a Harrisism wasn't really appropriate, because it doesn't
>>involve any sort of awesome cluelessness, just writing one word
>>where you meant another, in an informal context. The only reason
>>it reminded me of JSH was because of perennial confusion over
>>the difference between constants and non-constants (no, not
>>that you were actually confused about anything like that.) Sorry.
> 
> 
> Actually, I was more interested in coming up with a good notion of
> Harrisism we can use on the newsgroup... (-: You know, like
> spoonerism. 
> 
> It seems a bit hard to encompass all the behaviors that could be thus
> named, but we should we able to hash out something. I mean, had I
> replied attacking you for taking me to task on what was clearly just a
> simple error instead of just noting the error and leaving it at that
> or something, that would be harris-like behavior...
> 
> Perhaps:
> 
> Harrisism: the repeated and dogged blatantly incorrect use of a
> technical term, caused by either ignorance or confusion.
> 
Harrisment? I think it fits [as well as having a fitting homophone].
> (So the recent discussion on 'well-ordering the reals' would be a 
Harrisism...)
> 
Oh, you meant Harrisy!
> Any other suggestions?
> 
Harrisism would be included in the next edition of the Diagnostic and
Statistical Manual of Mental Disorders, DSM-V. A subtype of Borderline
Personality Disorder. Occasionally compounded with Narcissistic
Personality Disorder.
> ======================================================================
> It's not denial. I'm just very selective about
>  what I accept as reality.
>     --- Calvin (Calvin and Hobbes)
> ======================================================================
> 
> Arturo Magidin
> magidin@math.berkeley.edu
> 
> 
> 
There are also
        Harristotelian logic (n), a contradiction in terms if ever
        there was one. It would involve a major effort to catalogue
        all the variants of this. Virtually any JSH thread will
        contain prime examples of this, that will, ideally, factor
        into any serious definition.
        Harrisite (n), a moral relative of the leech, one who depends
        critically on the assistance of others, but who actively
        subjects his helpers to abuse. Also, a member of the
        legion of imagined JSH supporters: JSH is addressing the
        Harrisites when he makes his appeals to the masses.
        Harriside (n), the so-called majority of one, although
        some may imagine a throng on the Harriside. The opposite
        of JSH's usage of some in the classic some may disagree
        with this argument. Not to be confused with:
        Harricide (n), an occasional fantasy of some of the less well
        self-controlled participants in these threads (several readers
        have suggested Harricide as a solution for sci.math).
        Harristic (n), an informal argument (that can only be
        expressed using utterly trivial (and non-informative)
        examples. Also (adj) an epithet painting someone's
        argument as belonging to this category. (The proof
        of the Advanced Factorization Lemma by JSH was via a
        Harristic argument). By extension, this usage may be
        applied to any object that fails to meet even minimal
        standards for its type, as in That sieve sure makes
        a Harristic bucket!
        Harrispective (adj): paying absolutely no regard to the
        arguments of others (usage: Harrispective of Magidin's
        argument, we find q to be an algebraic integer unit with
        its inverse not integral over the rationals). Also (n): a
        peculiar form of insight leading to the propensity to pay
        no attention to others' arguments.
        Harision (n): fierce, often comically so, verbal attack on
        the provider of technical assistance. All attempts to
        correct the argument were met with insult and Harrision.
OK, that's all I could come up with. Sorry I couldn't go with the
suggested meaning of Harrisism, however.
Dale.
====
>
>   [.snip.]
 On the plus side, any two differ by a constant.
>>That was a slight Harrisism for differ by a unit, right?
Well, it was certainly a lapsus and I meant differ by a unit.
But just what is a Harrisism? It is not a simple error like the
>above, certainly. It should encompass some complete missing the point
>of a definition, theorem, or something, no?
>>Calling it a Harrisism wasn't really appropriate, because it doesn't
>>involve any sort of awesome cluelessness, just writing one word
>>where you meant another, in an informal context. The only reason
>>it reminded me of JSH was because of perennial confusion over
>>the difference between constants and non-constants (no, not
>>that you were actually confused about anything like that.) Sorry.
Actually, I was more interested in coming up with a good notion of
>Harrisism we can use on the newsgroup... (-: 
That's a lie. (hth)
>You know, like spoonerism. 
>It seems a bit hard to encompass all the behaviors that could be thus
>named, but we should we able to hash out something. I mean, had I
>replied attacking you for taking me to task on what was clearly just a
>simple error instead of just noting the error and leaving it at that
>or something, that would be harris-like behavior...
Perhaps:
Harrisism: the repeated and dogged blatantly incorrect use of a
>technical term, caused by either ignorance or confusion.
(So the recent discussion on 'well-ordering the reals' would be a 
Harrisism...)
Any other suggestions?
======================================================================
>It's not denial. I'm just very selective about
> what I accept as reality.
>    --- Calvin (Calvin and Hobbes)
>======================================================================
Arturo Magidin
>magidin@math.berkeley.edu

************************
David C. Ullrich
====
> Will Twentyman

>Hmmm...doesn't sound like you actually looked at the paper.
>>I'll be going over it in detail this weekend.  Do you mind if I
quote it
>>directly in my responses?

> No.

Having spent the weekend going through musty books on algebra...
here is my analysis:
Original paper will be quoted, the rest is my comments.  There may be
formatting issues with the quoted paper since it was a .pdf rather
than plain text.
>ADVANCED POLYNOMIAL FACTORIZATION
>JAMES HARRIS
>Abstract. Algebraic method for determining distribution of fac-
>tors within a polynomial factorization, which breaks through what
>was seen as a barrier from overinterpretations of Galois Theory.
>1. Advanced Polynomial Factorization Approached
>Determining the distribution of factors within irrational algebraic
>integers has long been considered impossible as it is not possible to >do
using Galois Theory. However a simple technique through the intro-
>duction of more variables makes it possible. To highlight the standard
>belief consider the algebraic integers (1 - sqrt(21))/2 and
>(1+sqrt(21))/2 >which are roots of x^2 + x - 5.
This is a minor note, but it should be -1 for both roots.  While this
has little impact on the validity of the rest of the work, it should,
at the very least, be corrected in your final draft.
>While you know that the algebraic integer factors are themselves
>factors of 5, in what way is each a factor of 5? Can either not
>have non unit factors of 5? How do you know?
These questions don't make a lot of sense to me, but I'll skip over
that to get to what I believe is the true problem.
>In looking to consider distribution of algebraic integer factors
>within a factorization I'll be using a more complicated example than
>x^2 +x - 5.
>The paper will show 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, using basic algebraic methods.
And here we run into two problems.  First, while it is true that the
a's are algebraic integers, it is something that at least deserves a
comment.  Second, it is far from obvious that coprime is meaningful
on the algebraic integers.  Here's the problem: coprime is based on
the definition of the least common divisor.  The LCD is only
guaranteed to exist for Principal Ideal Domains and Unique
algebraic integers are certainly NOT a UFD, and I've seen nothing to
indicate anything stronger than that they are a commutative ring.
Until someone can point to work that indicates that LCD is defined on
the algebraic integers, this paper cannot move forward.  The remainder
of the paper (which I have NOT checked for accuracy but am including
for completeness) depends on the concept of coprime being meaningful
and defined on the algebraic integers.
>First I'll need a simple lemma to generalize beyond factors of a
>polynomial that are themselves polynomials.
>Lemma 1.1. Factorization Lemma:
>Given a factor g of a polynomial P(x), further defined as a factor
>for all x, which means that the value of g for a value 'a' of x is
>a factor of P(a), within the ring of algebraic integers, there exists
>r and c such that
>g = r + c
>where r=0, or is not coprime to x, and c is a factor of the constant
>term P(0).
The notation here could be improved.  g should be g(x), r should be
r(x).
>Proof. Let x=0, then g must be a factor of P(0), so at that point
>c = g.
>(1) If when x does not equal 0, g=c, r=0.
>(2) If when x does not equal 0, g =/= c there must exist r which
>varies with x, and as r equals 0 when x equals 0 it is not coprime to
>x.
>
I'm not sure step 2 makes sense.  I don't see why it makes r coprime
to x.
>As an example consider sqrt(x + 1) which is a non polynomial factor
>of x+1, and while there are an infinity of irrational solutions
>consider the rational solution at x=35.
Are you sure you want to talk about non-polynomial roots?  While
legitimate, it changes the topic of discussion from polynomials over
the algebraic integers to algebraic functions over the algebraic
integers.
>Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,
>and c=1. But for different values of x, g and r will vary, while r
>will not be coprime to x.
Again, coprime needs to be defined for the particular ring under
consideration.
>2. Primary Argument
>Given
>65x^3 - 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>in the ring of algebraic integers. Let
>P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 +
u^3 f)
>Here f is a non unit, non zero algebraic integer coprime to 3 and x,
>and u a non unit, non zero algebraic integer coprime to f. Note P(m)
>has a factor that is f^2 .
Again, coprime must be defined on the algebraic integers.  Also, the
change of variable and the introduction of additional variables seems
odd.  Since it appears that you mean for f and u to be constants, it
appears that you actually have P(m,x).
>That expression comes from expanding (v^3 +1)x^3 - 3vxy^2 + y^3 ,
>using the substitutions v = - 1+mf^2 , and y = uf, where additional
>variables provide an additional degree of freedom.
>Consider that a similar idea can be used to factor 3, prime in
>integers, as x^2 +7x+10 = 3, allows you to find factors of 3 in the
>ring of algebraic integers.
I don't follow what you mean by this at all.  Some additional
explanation would be helpful.  What are the factors and how did you
compute them?
>Now consider the factorization
>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)
>where multiplying out shows that
>a1a2a3 = m^3 f^6 - 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 - 3m^2 f^2 + 3m)
>so a1a2a3 = mf^2 (m^2 f^4 - 3mf^2 + 3).
>Therefore, at least one of the a's cannot be coprime to m, and at
>least one of the a's must equal 0 when m=0.
>(Note: The a's are roots of a monic polynomial with algebraic
>integer coefficients so they are algebraic integers.)
Maybe.  Did P(m) have a constant term of 1?  if so, uf=1.  The logic
of your note could stand some clarification.  (note: coprime shows up
again)
>Notice that the constant term P(0) is given by
>P(0) = f^2 (3xu^2 + u^3 f)
>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.
>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1
>is not coprime to m.
>g1 = c = uf
>meaning f is a factor of the constant term.
>Therefore, exactly two of the a's equal 0, when m=0, to get the
>factor f^2 in the constant term P(0), while one must not equal 0, or >f^3
would be the factor.
This statement doesn't make a lot of sense either... but I'm not
trying as hard at this point.  If the other problems can be fixed,
I'll be happy to examine it more carefully.
>Now as noted before in general P(m) has a factor that is f^2 , and
>separating that factor off, gives a constant term coprime to f;
>therefore, given g1 = a1x + uf
>where with m = 0, g1 gives a factor of f it must have that same
>factor in general, proving that two of the a's have a factor that is f.
>Therefore, one factor is coprime to f.
>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't
>change the a's, I have
>(m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 + u^3 = 65x^3 -
12x + 1
>which may be more easily seen from using v = - 1 + mf^2 = 4, y=1
>with (v^3 + 1)x^3 - 3vxy^2 + y^3 .
>Therefore, with the factorization
>65x^3 - 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)
>one of the a's is coprime to 5, which shows where some of the alge-
>braic integer factors distribute despite the factors being irrational.
Overall analysis: if you can clarify the notation and the logical
connections, along with your result about factors not being factors,
what you probably have is a proof that coprime is not always defined
on the algebraic integers.
-- 
Will Twentyman
====
Admire you, Will.... wait to see what his reply is.....
> Will Twentyman
>   >Hmmm...doesn't sound like you actually looked at the paper.
>>I'll be going over it in detail this weekend.  Do you mind if I
> quote it
>>directly in my responses?
>   > No.
>   Having spent the weekend going through musty books on algebra...
> here is my analysis:

> Original paper will be quoted, the rest is my comments.  There may be
> formatting issues with the quoted paper since it was a .pdf rather
> than plain text.
 >ADVANCED POLYNOMIAL FACTORIZATION
>JAMES HARRIS
>Abstract. Algebraic method for determining distribution of fac-
>tors within a polynomial factorization, which breaks through what
>was seen as a barrier from overinterpretations of Galois Theory.
>1. Advanced Polynomial Factorization Approached
>Determining the distribution of factors within irrational algebraic
>integers has long been considered impossible as it is not possible to 
>do
> using Galois Theory. However a simple technique through the intro-
>duction of more variables makes it possible. To highlight the standard
>belief consider the algebraic integers (1 - sqrt(21))/2 and
>(1+sqrt(21))/2 >which are roots of x^2 + x - 5.
 This is a minor note, but it should be -1 for both roots.  While this
> has little impact on the validity of the rest of the work, it should,
> at the very least, be corrected in your final draft.
 >While you know that the algebraic integer factors are themselves
>factors of 5, in what way is each a factor of 5? Can either not
>have non unit factors of 5? How do you know?
 These questions don't make a lot of sense to me, but I'll skip over
> that to get to what I believe is the true problem.
 >In looking to consider distribution of algebraic integer factors
>within a factorization I'll be using a more complicated example than
>x^2 +x - 5.
>The paper will show 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, using basic algebraic methods.
 And here we run into two problems.  First, while it is true that the
> a's are algebraic integers, it is something that at least deserves a
> comment.  Second, it is far from obvious that coprime is meaningful
> on the algebraic integers.  Here's the problem: coprime is based on
> the definition of the least common divisor.  The LCD is only
> guaranteed to exist for Principal Ideal Domains and Unique
> algebraic integers are certainly NOT a UFD, and I've seen nothing to
> indicate anything stronger than that they are a commutative ring.
> Until someone can point to work that indicates that LCD is defined on
> the algebraic integers, this paper cannot move forward.  The remainder
> of the paper (which I have NOT checked for accuracy but am including
> for completeness) depends on the concept of coprime being meaningful
> and defined on the algebraic integers.
 >First I'll need a simple lemma to generalize beyond factors of a
>polynomial that are themselves polynomials.
>Lemma 1.1. Factorization Lemma:
>Given a factor g of a polynomial P(x), further defined as a factor
>for all x, which means that the value of g for a value 'a' of x is
>a factor of P(a), within the ring of algebraic integers, there exists
>r and c such that
>g = r + c
>where r=0, or is not coprime to x, and c is a factor of the constant
>term P(0).
 The notation here could be improved.  g should be g(x), r should be
> r(x).
 >Proof. Let x=0, then g must be a factor of P(0), so at that point
>c = g.
>(1) If when x does not equal 0, g=c, r=0.
>(2) If when x does not equal 0, g =/= c there must exist r which
>varies with x, and as r equals 0 when x equals 0 it is not coprime to
>x.
> 
> I'm not sure step 2 makes sense.  I don't see why it makes r coprime
> to x.
 >As an example consider sqrt(x + 1) which is a non polynomial factor
>of x+1, and while there are an infinity of irrational solutions
>consider the rational solution at x=35.
 Are you sure you want to talk about non-polynomial roots?  While
> legitimate, it changes the topic of discussion from polynomials over
> the algebraic integers to algebraic functions over the algebraic
> integers.
 >Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,
>and c=1. But for different values of x, g and r will vary, while r
>will not be coprime to x.
 Again, coprime needs to be defined for the particular ring under
> consideration.
 >2. Primary Argument
>Given
>65x^3 - 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>in the ring of algebraic integers. Let
>P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 +
> u^3 f)
>Here f is a non unit, non zero algebraic integer coprime to 3 and x,
>and u a non unit, non zero algebraic integer coprime to f. Note P(m)
>has a factor that is f^2 .
 Again, coprime must be defined on the algebraic integers.  Also, the
> change of variable and the introduction of additional variables seems
> odd.  Since it appears that you mean for f and u to be constants, it
> appears that you actually have P(m,x).
 >That expression comes from expanding (v^3 +1)x^3 - 3vxy^2 + y^3 ,
>using the substitutions v = - 1+mf^2 , and y = uf, where additional
>variables provide an additional degree of freedom.
>Consider that a similar idea can be used to factor 3, prime in
>integers, as x^2 +7x+10 = 3, allows you to find factors of 3 in the
>ring of algebraic integers.
 I don't follow what you mean by this at all.  Some additional
> explanation would be helpful.  What are the factors and how did you
> compute them?
 >Now consider the factorization
>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)
>where multiplying out shows that
>a1a2a3 = m^3 f^6 - 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 - 3m^2 f^2 + 3m)
>so a1a2a3 = mf^2 (m^2 f^4 - 3mf^2 + 3).
>Therefore, at least one of the a's cannot be coprime to m, and at
>least one of the a's must equal 0 when m=0.
>(Note: The a's are roots of a monic polynomial with algebraic
>integer coefficients so they are algebraic integers.)
 Maybe.  Did P(m) have a constant term of 1?  if so, uf=1.  The logic
> of your note could stand some clarification.  (note: coprime shows up
> again)
 >Notice that the constant term P(0) is given by
>P(0) = f^2 (3xu^2 + u^3 f)
>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.
>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1
>is not coprime to m.
>g1 = c = uf
>meaning f is a factor of the constant term.
>Therefore, exactly two of the a's equal 0, when m=0, to get the
>factor f^2 in the constant term P(0), while one must not equal 0, or 
>f^3
> would be the factor.
 This statement doesn't make a lot of sense either... but I'm not
> trying as hard at this point.  If the other problems can be fixed,
> I'll be happy to examine it more carefully.
 >Now as noted before in general P(m) has a factor that is f^2 , and
>separating that factor off, gives a constant term coprime to f;
>therefore, given g1 = a1x + uf
>where with m = 0, g1 gives a factor of f it must have that same
>factor in general, proving that two of the a's have a factor that is f.
>Therefore, one factor is coprime to f.
>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't
>change the a's, I have
>(m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 + u^3 = 65x^3 -
> 12x + 1
>which may be more easily seen from using v = - 1 + mf^2 = 4, y=1
>with (v^3 + 1)x^3 - 3vxy^2 + y^3 .
>Therefore, with the factorization
>65x^3 - 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)
>one of the a's is coprime to 5, which shows where some of the alge-
>braic integer factors distribute despite the factors being irrational.
 Overall analysis: if you can clarify the notation and the logical
> connections, along with your result about factors not being factors,
> what you probably have is a proof that coprime is not always defined
> on the algebraic integers.
 --
> Will Twentyman

>
====
> ...
>>The paper will show that given the factorization, in the ring of
>alge-
>>braic integers,
>>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>>one of the a's is coprime to 5, using basic algebraic methods.
And here we run into two problems.  First, while it is true that the
>a's are algebraic integers, it is something that at least deserves a
>comment.  Second, it is far from obvious that coprime is 
meaningful
>on the algebraic integers.  Here's the problem: coprime is based on
>the definition of the least common divisor.
> 
> The common definition for coprime is that within a ring, two numbers
> a and b are called co-prime if there are numbers x and y such that
> a*x + b*y = 1.  This is a change from the former definition where
> a and b are coprime if LCD(a, b) = 1 but the theorem that there are
> numbers x and y such that a*x + b*y = LCD(a, b) assures that the
> definitions are the same when both could be applied.  So in the
> algebraic integers 2 and 3 are coprime because 2 * 2 + 3 * (-1) = 1.
> 
I saw that as one definition, but I'm not sure that's what JSH had in 
mind.  Either way, we now have a definition, can he make it work?
>Are you sure you want to talk about non-polynomial roots?  While
>legitimate, it changes the topic of discussion from polynomials over
>the algebraic integers to algebraic functions over the algebraic
>integers.
> 
> But indeed, that is what he wants to do, opening a can of worms.
Not to mention completely changing the topic.
>> Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,
>> and c=1. But for different values of x, g and r will vary, while r 
will
>> not be coprime to x.
Again, coprime needs to be defined for the particular ring under
>consideration.
> 
> As it is defined, that is no problem.  And indeed, the Lemma holds
> for sqrt(x + 1) as factor of x + 1.  But I am not sure for what
> kind of functions g(x) the lemma holds.  Step 2 in the proof is no
> clarification for that, and there are (as shown above) function for
> which it does not hold.
I've noticed that he stopped replying to us.  Have we perhaps convinced 
him?  It would be nice to know.
-- 
Will Twentyman
====
> I've noticed that he stopped replying to us.  Have we perhaps convinced 
> him?  It would be nice to know.
Oh, don't worry, he comes and goes in cycles... perhaps he circulates
around to other newsgroups in a great migration, I'm not sure.
In any case, probably in about another week he'll start posting a new
series of rants and will treat these threads as if they never happened.
====
...
 > I've noticed that he stopped replying to us.  Have we perhaps convinced 
 > him?  It would be nice to know.
No chance.  He just stopped reading this thread and will start anew
in another thread.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
> Will Twentyman
>   >Hmmm...doesn't sound like you actually looked at the paper.
>>I'll be going over it in detail this weekend.  Do you mind if I
>  quote it
>>directly in my responses?
>   > No.
>   Having spent the weekend going through musty books on algebra...
> here is my analysis:
> 
> 
  Harris's paper has already been proven to be incorrect
by two posters independently.  See the Harris thread in sci.math
entitled My Work -- Objective Review, and the post
on June 18 by Nora Baron and that on June 22 by W. Dale Hall.
[snip]
> 
> Overall analysis: if you can clarify the notation and the logical
> connections, along with your result about factors not being factors,
> what you probably have is a proof that coprime is not always defined
> on the algebraic integers.
  This is not accurate - the algebraic integers form a Bezout
domain - suggest you consult some good texts on algebraic 
number theory.
  Andrzej
====
> 
>Will Twentyman
>Hmmm...doesn't sound like you actually looked at the paper.
>>I'll be going over it in detail this weekend.  Do you mind if I
>> quote it
>>directly in my responses?

>No.

>>Having spent the weekend going through musty books on algebra...
>>here is my analysis:
> 
> 
>   Harris's paper has already been proven to be incorrect
> by two posters independently.  See the Harris thread in sci.math
> entitled My Work -- Objective Review, and the post
> on June 18 by Nora Baron and that on June 22 by W. Dale Hall.
> 
> 
> [snip]
Noted.
>>Overall analysis: if you can clarify the notation and the logical
>>connections, along with your result about factors not being factors,
>>what you probably have is a proof that coprime is not always defined
>>on the algebraic integers.
> 
> 
> 
>   This is not accurate - the algebraic integers form a Bezout
> domain - suggest you consult some good texts on algebraic 
> number theory.
> 
> 
>   Andrzej
The key word was if.  I missed a better definition of coprime, but had 
a strong suspicion that the holes were huge.
-- 
Will Twentyman
====
> 
>  
> 
> 
> Yeah, my elation turned to frustration as first I was repeating what
> was in the paper, then dealing with underhanded things like
> bait-and-switch and finally I just got tired of it.
> 
> Notice that you assumed
> 1) I made a bait-and-switch and
> 2) If I did, it was intentional rather than a result of not 
understanding.
What I've noticed is that mathematicians tend to follow a pattern when
they're trying to deny the correctness of my work.
 
>If this was your intention, very well done! If not, don't take it
>personal - it's a question of medication dosage. There must
>be a balance somewhere. We know it will be unstable, but
>it will be a balance - for a few weeks.
>Welcome to the playground of James Harrass.
>Dirk Vdm
>   >>I noticed it.  I've lurked long enough that his response isn't 
>>*entirely* surprising.  I suppose I should go try to make sense of his 
>>paper now, to see where I'm missing things.  It would have been helpful 

>>if he had started by posting the paper here instead of posting a link to 

>>a pdf.  It makes quoting his paper... more challenging.  And makes it 
>>less accessible.
> 
> 
> The paper started on the newsgroup.
> 
> In sci.math, where I picked this up, you posted a link to the paper that 
> do not recall ever seeing it in sci.math, but that could easily be an 
> oversite.
It started on the newsgroup.
Now let's see about your other posts.
James Harris
====
>> 
>>>  
>> 
>> 
>> Yeah, my elation turned to frustration as first I was repeating what
>> was in the paper, then dealing with underhanded things like
>> bait-and-switch and finally I just got tired of it.
>> 
>> Notice that you assumed
>> 1) I made a bait-and-switch and
>> 2) If I did, it was intentional rather than a result of not 
understanding.
What I've noticed is that mathematicians tend to follow a pattern when
>they're trying to deny the correctness of my work.
This is the pattern they tend to follow:
1. This doesn't make any sense
2. You didn't define this term
2a. You keep using the term factor.  I do not think it means what
    you think it means.
3. Are you going to stop ranting at mathematicians and actually write
   some math?
4. I guess not.
5. Are these algebraic integers, integers, polynomials, or what?
6. If you numbered your equations it would make life a lot easier.
7. No one is objecting to that specific example, it's the general
   case that we have a problem with. 
8. To hell with this.  I'm going to get a beer.
Alan
-- 
Defendit numerus
====
> 
> >Hmmm...doesn't sound like you actually looked at the paper.
> >>I'll be going over it in detail this weekend.  Do you mind if I quote 
it 
>>directly in my responses?
> 
> No.
> 
> 
> Having spent the weekend going through musty books on algebra...  here
> is my analysis:
Hmmm...so my short paper sent you to books for an entire weekend?
This post should be interesting.
> BTW, I'm using Google to post this since my usual newsfeed is acting
> up.
> 
> Original paper will be quoted, the rest is my comments.  There may be
> formatting issues with the quoted paper since it was a .pdf rather
> than plain text.
> 
>ADVANCED POLYNOMIAL FACTORIZATION
>JAMES HARRIS
>Abstract. Algebraic method for determining distribution of fac-
>tors within a polynomial factorization, which breaks through what
>was seen as a barrier from overinterpretations of Galois Theory.
>1. Advanced Polynomial Factorization Approached
>Determining the distribution of factors within irrational algebraic
>integers has long been considered impossible as it is not possible to
>  do
>using Galois Theory. However a simple technique through the intro-
>duction of more variables makes it possible. To highlight the
>  standard
>belief consider the algebraic integers (1 &#8722; sqrt(21))/2 and
> (1+sqrt(21))/2 >which are roots of x^2 + x &#8722; 5.
> 
> This is a minor note, but it should be -1 for both roots.  While this
> has little impact on the validity of the rest of the work, it should,
> at the very least, be corrected in your final draft.
That was pointed out by someone else, I've acknowledged it as a
mistake, and I've changed it in the paper.
>While you know that the algebraic integer factors are themselves
>factors of 5, in what way is each a factor of 5? Can either not have
>  non
>unit factors of 5? How do you know?
> 
> These questions don't make a lot of sense to me, but I'll skip over
> that to get to what I believe is the true problem.
Well, it looks to me like your comment might be taken as somewhat
negative, but the preamble to the paper can be skipped.
>In looking to consider distribution of algebraic integer factors
>  within
>a factorization I'll be using a more complicated example than x 2 +x
>  &#8722; 5.
>The paper will show that given the factorization, in the ring of
>  alge-
>braic integers,
>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>one of the a's is coprime to 5, using basic algebraic methods.
 
> And here we run into two problems.  First, while it is true that the
> a's are algebraic integers, it is something that at least deserves a
> comment.  
Why should that be commented on here?
>Second, it is far from obvious that coprime is meaningful
> on the algebraic integers.  
That was discussed by others in reply to your post.
However I'll add that I switched to coprime rather than use factor
in this situation because algebraic integers is an incomplete ring so
there is a problem with the term factor if I want to stay in the ring.
 That is, rather than say that one of the a's is coprime, I might have
said that only two of the a's have a factor of f, but the ring is
incomplete, so that's problematic.
That is, the problem with algebraic integers being incomplete means
that I'd have to knowingly use factor in a way not exactly correct,
but coprime still works, so I used it.
>Here's the problem: coprime is based on
> the definition of the least common divisor.  The LCD is only
> guaranteed to exist for Principal Ideal Domains and Unique
> algebraic integers are certainly NOT a UFD, and I've seen nothing to
> indicate anything stronger than that they are a commutative ring. 
> Until someone can point to work that indicates that LCD is defined on
> the algebraic integers, this paper cannot move forward.  The remainder
> of the paper (which I have NOT checked for accuracy but am including
> for completeness) depends on the concept of coprime being meaningful
> and defined on the algebraic integers.
The problem is that algebraic integers are incomplete as a ring.  What
the paper shows is that two of the a's should have f as a factor, as
the paper shows them to have f as a factor, but technically, it's not
a factor within the ring of algebraic integers.
>First I'll need a simple lemma to generalize beyond factors of a
>  poly-
>nomial that are themselves polynomials.
>Lemma 1.1. Factorization Lemma:
>Given a factor g of a polynomial P(x), further defined as a factor
>for all x, which means that the value of g for a value 'a' of x is a
>  factor
>of P(a), within the ring of algebraic integers, there exists r and c
>  such
>that
>g = r + c
>where r=0, or is not coprime to x, and c is a factor of the constant
>term P(0).
> 
> The notation here could be improved.  g should be g(x), r should be
> r(x).
I'm not interested in style issues.
You have claimed to be a mathematician, and mathematicians are defined
to be math experts.  The fundamental question about the paper is not
style but correctness.  Minor issues aren't relevant, but so far
that's all you have.
For instance, you talked about the sign problem in the preamble, and
then you questioned the use of coprime, and now you talk about
notation.
These type of issues are the kinds of things I see brought up by
people I deem aren't interested in the truth.
The discussion is not a game.  It's not about winning or losing.  It's
about the truth.
If I'm correct then there's a problem in mathematics which needs to be
fixed.
If I'm wrong then you should be able to find more than a style issue.
>Proof. Let x=0, then g must be a factor of P(0), so at that point c =
>  g.
>(1) If when x does not equal 0, g=c, r=0.
>(2) If when x does not equal 0, g =/= c there must exist r which
>  varies
>with x, and as r equals 0 when x equals 0 it is not coprime to
>x.
> 
> 
> I'm not sure step 2 makes sense.  I don't see why it makes r coprime
> to x.
The argument isn't complicated but I simplified that section rather
than deal with it, after I realized it wasn't correct.
That is, r is not coprime to x or contains a unit factor of x.
With that unit factor things get messy as I'm sure some would jump on
that and claim that since r would always have a unit factor of x, it's
meaningless.
Rather than get into an involved explanation with more room for
confusion, I realized that I really was just using the fact that any
factor of a polynomial can be split up between what's constant and
what's varying.
That is, r changes, but c does not, as x varies.
The proof of that is actually trivial as all you do is take g at 0,
which gives you c, then r = g-c.
Given that c is constant, while g is not, obviously r is not either.
>As an example consider sqrt(x + 1) which is a non polynomial factor
>  of
>x+1, and while there are an infinity of irrational solutions consider
>  the
>rational solution at x=35.
> 
> Are you sure you want to talk about non-polynomial roots?  While
> legitimate, it changes the topic of discussion from polynomials over
> the algebraic integers to algebraic functions over the algebraic
> integers.
Which would seem to indicate that you haven't read further down in the
paper, or haven't bothered to consider what I said before this
section:
First I'll need a simple lemma to generalize beyond factors of a
polynomial that are themselves polynomials.
Now I'd think that says what you need to know, and besides, what's
with the talk of roots?
What are non-polynomial roots?
Are you sure you're a mathematician?
>Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,
>and c=1. But for different values of x, g and r will vary, while r
>  will
>not be coprime to x.
> 
> Again, coprime needs to be defined for the particular ring under
> consideration.
And it sounds to me like maybe you didn't spend a weekend going over
math texts as you repeatedly brought up as an issue something other
posters dismissed immediately.
>2. Primary Argument
>Given
>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>in the ring of algebraic integers. Let
>P(m) = f^2 ((m^3 f^4 &#8722; 3m^2 f^2 + 3m)x^3 &#8722; 3( &#8722; 1 +
>  mf^2 )xu^2 + u^3 f)
>Here f is a non unit, non zero algebraic integer coprime to 3 and x,
>and u a non unit, non zero algebraic integer coprime to f. Note P(m)
>has a factor that is f^2 .
> 
> Again, coprime must be defined on the algebraic integers.  Also, the
> change of variable and the introduction of additional variables seems
> odd.  Since it appears that you mean for f and u to be constants, it
> appears that you actually have P(m,x).
Nope.  And if you're a mathematician changing variables as I did
shouldn't be a problem for you.
Again, the question is to the correctness of the paper, and not style
issues.
So far, just like before when I got upset, you've spent a lot of time
on nonessential issues, or ones that others quickly refuted.
It seems to me that your behavior is consistent with what I've seen
from posters trying to hide the truth, rather than get to the bottom
of things.
But the issue is taught mathematics which is wrong.
>That expression comes from expanding (v^3 +1)x^3 &#8722; 3vxy^2 + y^3
>  , using
>the substitutions v = &#8722; 1+mf^2 , and y = uf, where additional
>  variables
>provide an additional degree of freedom.
>Consider that a similar idea can be used to factor 3, prime in
>  integers,
>as x^2 +7x+10 = 3, allows you to find factors of 3 in the ring of
>  algebraic
>integers.
> 
> I don't follow what you mean by this at all.  Some additional
> explanation would be helpful.  What are the factors and how did you
> compute them?
I explain to some extent where the expression I use comes from, and
then give a quick analogy.  If the analogy confuses you, you can skip
it, as you did with the other section that confused you.
It's not part of the argument, but merely there to help.
>Now consider the factorization
>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)
>where multiplying out shows that
>a1a2a3 = m^3 f^6 &#8722; 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 &#8722; 3m^2
>  f^2 + 3m)
>so a1a2a3 = mf^2 (m^2 f^4 &#8722; 3mf^2 + 3).
>Therefore, at least one of the a's cannot be coprime to m, and at
>least one of the a's must equal 0 when m=0.
>(Note: The a's are roots of a monic polynomial with algebraic integer
>coefficients so they are algebraic integers.)
> 
> Maybe.  Did P(m) have a constant term of 1?  if so, uf=1.  The logic
> of your note could stand some clarification.  (note: coprime shows up
> again)
Strange question, given what follows:
>Notice that the constant term P(0) is given by
>P(0) = f^2 (3xu^2 + u^3 f)
>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.
>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1
>is not coprime to m.
>g1 = c = uf
>meaning f is a factor of the constant term.
>Therefore, exactly two of the a's equal 0, when m=0, to get the
>  factor
>f^2 in the constant term P(0), while one must not equal 0, or f^3
>  would
>be the factor.
> 
> This statement doesn't make a lot of sense either... but I'm not
> trying as hard at this point.  If the other problems can be fixed,
> I'll be happy to examine it more carefully.
In the paper I take a rather complicated expression, consider it as a
polynomial with respect to m, meaning that I let only m vary, and then
find that I can factor it in a not surprising way, given the
expression.
That factorization gives me non-polynomial factors, with which I use
my lemma to show that only one of the a's is coprime to f.
The factorization I use, isn't terribly surprising, and the argument
follows rather simply from a rather basic lemma, and the distributive
property.
>Now as noted before in general P(m) has a factor that is f^2 , and
>separating that factor off, gives a constant term coprime to f;
>  therefore,
>given g1 = a1x + uf
>where with m = 0, g1 gives a factor of f it must have that same
>  factor
>in general, proving that two of the a's have a factor that is f.
>Therefore, one factor is coprime to f.
>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't
>change the a's, I have
>(m^3 f^6 &#8722; 3m^2 f^4 + 3m)x^3 &#8722; 3( &#8722; 1 + mf^2 )xu^2
>  + u^3 = 65x^3 &#8722; 12x + 1
>which may be more easily seen from using v = &#8722; 1 + mf^2 = 4,
>  y=1
>with (v^3 + 1)x^3 &#8722; 3vxy^2 + y^3 .
>Therefore, with the factorization
>65x^3 &#8722; 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)
>one of the a's is coprime to 5, which shows where some of the alge-
>braic integer factors distribute despite the factors being
>  irrational.
> 
> Overall analysis: if you can clarify the notation and the logical
> connections, along with your result about factors not being factors,
> what you probably have is a proof that coprime is not always defined
> on the algebraic integers.
Well I refer you to the statements by others about coprime in the
ring of algebraic integers.
What I can do is take out the use of the term in the paper, and use
factor in the ring of algebraic integers, and note that will lead to
a contradiction as the point of the paper is that the ring is
incomplete.
What I want to impress upon readers is that I'm quite willing to work
with mathematicians to explain the mistake that they're teaching.
The ring of algebraic integers is incomplete, it's easy to show, and
I've shown it with my paper.
If mathematicians are having trouble understanding any part of it, I'm
able to explain further.
James Harris
====
[...]
| What I can do is take out the use of the term in the paper, and use
| factor in the ring of algebraic integers, and note that will lead to
| a contradiction as the point of the paper is that the ring is
| incomplete.
| 
| What I want to impress upon readers is that I'm quite willing to work
| with mathematicians to explain the mistake that they're teaching.
If you want to show that a mistake is being taught, you should quote it
specifically.
| The ring of algebraic integers is incomplete, it's easy to show, and
| I've shown it with my paper.
| 
| If mathematicians are having trouble understanding any part of it, I'm
| able to explain further.
You have a big problem with terminology. You use terms that aren't
used by others in the context in which you use them, such as counting
factors of 5 in algebraic integers. I haven't seen anything I would
consider an adequate definition of those terms. You don't explain
why your terms should be considered relevant to the ones mathematicians
are using.
I don't know what you think is a fair way for things to work, but the
way things actually work, this kind of problem with terminology will
absolutely prevent you from making any headway.
In an experiment in communication, I once tried elaborating on such a
concept for you. You usually write as though the number of factors of
5 in an algebraic number satisfied certain axioms which are familiar
to me: being a rational number v(x)>=0 associated with each algebraic
integer x <> 0, where v(5)=1, v(xy)=v(x)+v(y) for any algebraic
integers x<>0 and y<>0, and v(x+y)>=min{v(x),v(y)} for algebriac
integers x<>0, y<>0, and x+y<>0. Assuming that we have such a function
v, then we can show that v(r1)=0 and v(r2)=v(r3)=1/2 for the roots
r1, r2, and r3 of your cubic, if we take them in the right order.
I think this is the best way to try to make sense of your argument
in your advanced polynomial factorization thing.
Unfortunately, this known concept is not so directly related to
divisibility in the algebraic integers, so even if you did manage to
produce a definition on these lines, it wouldn't show mathematicians
are doing anything wrong when they make claims about divisibility in
the algebraic integers as they teach about it in classrooms. They're
just apples and oranges.
Say we define the number of factors of 5 in an algebraic integer x to be
the highest rational number r such that 5^r divides x in the algebraic
integers, i.e. such that there exists an algebraic integer y such that
5^r*y = x. I think this definition works (i.e., I think there is such a
highest rational power for each algebraic number, although I haven't
tried to write out a proof). If we define it that way, though, then it
simply doesn't have one of the properties I listed above.
Keith Ramsay
====
> The ring of algebraic integers is incomplete, it's easy to show, and
> I've shown it with my paper.
> 
> If mathematicians are having trouble understanding any part of it, I'm
> able to explain further.
I don't understand what it means to say, The ring of algebraic
integers is incomplete.
I don't claim to be a mathematician (I'm a jigsaw puzzle retailer,
actually), but there's considerable evidence that most mathematicians
who've looked can't understand it either. Could you please (either or
both):
(a) Post a clear definition, of the sort one sees in maths books of a
complete ring
(b) Say of some simple rings whether they are complete or not -
The ring of integers (Z)
The ring of integers modulo 12 (clock arithmetic)
The ring of integers modulo 2 (even/odd arithmetic)
The ring of integers modulo 7
(b') If all of these are complete, can you name another ring other
than the ring of algebraic integers which is also incomplete?
Brian Chandler
----------------
geo://Sano.Japan.Planet_3
Jigsaw puzzles from Japan at:
http://imaginatorium.org/shop/
====
> [...]
> | What I can do is take out the use of the term in the paper, and use
> | factor in the ring of algebraic integers, and note that will lead 
to
> | a contradiction as the point of the paper is that the ring is
> | incomplete.
> | 
> | What I want to impress upon readers is that I'm quite willing to work
> | with mathematicians to explain the mistake that they're teaching.
> 
> If you want to show that a mistake is being taught, you should quote it
> specifically.
That's not very sensible if you expect that for a mistake to be taught
for any length of time it'd have some subtlety, unless you're also
questioning the competence of mathematicians.
It seems to me that rather than deal with the important question,
which is whether or not I'm correct, people often try to introduce
weird ad hoc conditions.
If you disagree with that assessment, can you please explain why you
believe there should be something simple and short enough for me to
quote?
 
> | The ring of algebraic integers is incomplete, it's easy to show, and
> | I've shown it with my paper.
> | 
> | If mathematicians are having trouble understanding any part of it, I'm
> | able to explain further.
> 
> You have a big problem with terminology. You use terms that aren't
> used by others in the context in which you use them, such as counting
> factors of 5 in algebraic integers. I haven't seen anything I would
> consider an adequate definition of those terms. You don't explain
> why your terms should be considered relevant to the ones mathematicians
> are using.
That statement possibly has something to do with what I now call the
preamble in the paper as it is there to help explain context, but
isn't part of the actual argument.
Here's what I say, copied from the paper (some editing for format):
To highlight the standard belief consider the algebraic integers
(-1-sqrt(21))/2 and (-1+sqrt(21))/2 which are roots of x^2 + x - 5.
While you know that the algebraic integer factors are themselves
factors of 5, in what way is each a factor of 5? Can either not have
non unit factors of 5? How do you know?
There I'm talking about algebraic integers, so one can assume I'm
talking about algebraic integer factors.  Given that 5 has algebraic
integer factors, how is what I say nonstandard Keith Ramsay?
I'd like you to carefully explain your assertion.  The question I'm
raising is the possibility that you lied.
> I don't know what you think is a fair way for things to work, but the
> way things actually work, this kind of problem with terminology will
> absolutely prevent you from making any headway.
What will prevent me from making headway is if mathematicians
continually lie.
Now then in what way is it improper terminology to talk about
algebraic integer factors of 5?
 
> In an experiment in communication, I once tried elaborating on such a
> concept for you. You usually write as though the number of factors of
> 5 in an algebraic number satisfied certain axioms which are familiar
> to me: being a rational number v(x)>=0 associated with each algebraic
> integer x <> 0, where v(5)=1, v(xy)=v(x)+v(y) for any algebraic
> integers x<>0 and y<>0, and v(x+y)>=min{v(x),v(y)} for algebriac
> integers x<>0, y<>0, and x+y<>0. Assuming that we have such a function
> v, then we can show that v(r1)=0 and v(r2)=v(r3)=1/2 for the roots
> r1, r2, and r3 of your cubic, if we take them in the right order.
> I think this is the best way to try to make sense of your argument
> in your advanced polynomial factorization thing.
> 
> Unfortunately, this known concept is not so directly related to
> divisibility in the algebraic integers, so even if you did manage to
> produce a definition on these lines, it wouldn't show mathematicians
> are doing anything wrong when they make claims about divisibility in
> the algebraic integers as they teach about it in classrooms. They're
> just apples and oranges.
There is no divisibility argument within the paper, and I only even
mention roots when the roots are algebraic integers.  I've talked
about x/y in discussing the error in taught mathematics, but did make
a post explaining that was when an argument is considered in the field
of rationals.
Your statement falls flat Keith Ramsay, and I think you're just trying
to sound good enough to fool people, rather than trying to get to the
truth.
What is clear is that the paper does NOT operate over any fields, and
depends only on ring operations.
> Say we define the number of factors of 5 in an algebraic integer x to be
> the highest rational number r such that 5^r divides x in the algebraic
> integers, i.e. such that there exists an algebraic integer y such that
> 5^r*y = x. I think this definition works (i.e., I think there is such a
> highest rational power for each algebraic number, although I haven't
> tried to write out a proof). If we define it that way, though, then it
> simply doesn't have one of the properties I listed above.
> 
> Keith Ramsay
Your post isn't coherent mathematically given what I say in the paper.
 Consider that in the ring of algebraic integers, 5 has algebraic
integer factors, and given algebraic integers r_1 and r_2, where their
product is 5, why are you acting as if it's so difficult to comprehend
that there must be some distribution of factors of 5?
For instance, both could have a factor that is sqrt(5), or one could
have a factor of 1+2i, while the other had a factor of 1-2i.
The question I'm trying to get the reader to explore is, given that
they are roots of this particular polynomial, could one of them only
have unit factors of 5?
How do you know?
James Harris
====
...
 > To highlight the standard belief consider the algebraic integers
 > (-1-sqrt(21))/2 and (-1+sqrt(21))/2 which are roots of x^2 + x - 5.
 > 
 > While you know that the algebraic integer factors are themselves
 > factors of 5, in what way is each a factor of 5? Can either not have
 > non unit factors of 5? How do you know?
Note that the polynomial is primitive.
...
 > For instance, both could have a factor that is sqrt(5), or one could
 > have a factor of 1+2i, while the other had a factor of 1-2i.
 > 
 > The question I'm trying to get the reader to explore is, given that
 > they are roots of this particular polynomial, could one of them only
 > have unit factors of 5?
 > 
 > How do you know?
(Note: with divisor I mean divisor in the ring of algebraic integers.)
1.  Each divisor of r1 is also a divisor of 5 (because r1 is itself a
    divisor of 5).
2.  If all divisors of r1 are units, it is a unit itself (the product
    of units is a unit).
3.  If r1 is a unit, 1/r1 is an algebraic integer (definition of unit).
4.  1/r1 is a root of the primitive polynomial 5x^2 - x - 1, so is not
    an algebraic integer (we have gone ove this before).
So r1 is not a unit and it has non-unit divisors that are also divisors
of 5.
Satisfied?
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
|> If you want to show that a mistake is being taught, you should quote it
|> specifically.
|
|That's not very sensible if you expect that for a mistake to be taught
|for any length of time it'd have some subtlety, unless you're also
|questioning the competence of mathematicians.
|
|It seems to me that rather than deal with the important question,
|which is whether or not I'm correct, people often try to introduce
|weird ad hoc conditions.
|
|If you disagree with that assessment, can you please explain why you
|believe there should be something simple and short enough for me to
|quote?
so it's possible for someone to be wrong, and yet for you to be unable
to point to some one localized place in what they're saying as the
place where the mistake is being made?
-- 
====
[...]
> For instance, both could have a factor that is sqrt(5), or one could
> have a factor of 1+2i, while the other had a factor of 1-2i.
> 
> The question I'm trying to get the reader to explore is, given that
> they are roots of this particular polynomial, could one of them only
> have unit factors of 5?
> 
> How do you know?
may be relevant:
http://www.ams.org/new-in-math/cover/factorization.html
``On the other hand, if D = -5, then R is not a UFD.
David Bernier
====
> 
> The ring of algebraic integers is incomplete, it's easy to show, and
> I've shown it with my paper.
> 
> If mathematicians are having trouble understanding any part of it, I'm
> able to explain further.
> 
> I don't understand what it means to say, The ring of algebraic
> integers is incomplete.
> 
> I don't claim to be a mathematician (I'm a jigsaw puzzle retailer,
> actually), but there's considerable evidence that most mathematicians
> who've looked can't understand it either. Could you please (either or
> both):
You'll have better luck getting Cool Giraffe to explain the apes and
bananas thread than prying any information out of JSH.
> 
> (a) Post a clear definition, of the sort one sees in maths books of a
> complete ring
> 
> (b) Say of some simple rings whether they are complete or not -
> 
> The ring of integers (Z)
> 
> The ring of integers modulo 12 (clock arithmetic)
> 
> The ring of integers modulo 2 (even/odd arithmetic)
> 
> The ring of integers modulo 7
> 
> (b') If all of these are complete, can you name another ring other
> than the ring of algebraic integers which is also incomplete?
> 
> 
> Brian Chandler
> ----------------
> geo://Sano.Japan.Planet_3
> Jigsaw puzzles from Japan at:
> http://imaginatorium.org/shop/
====
>>Having spent the weekend going through musty books on algebra...  here
>>is my analysis:
> 
> Hmmm...so my short paper sent you to books for an entire weekend?
> 
Sorry if I disappointed you.  I tend to do more work in logic/discrete 
math than I do in Algebra.
>ADVANCED POLYNOMIAL FACTORIZATION
>JAMES HARRIS
>Abstract. Algebraic method for determining distribution of fac-
>tors within a polynomial factorization, which breaks through what
>was seen as a barrier from overinterpretations of Galois Theory.
>1. Advanced Polynomial Factorization Approached
>Determining the distribution of factors within irrational algebraic
>integers has long been considered impossible as it is not possible to
>> do
>using Galois Theory. However a simple technique through the intro-
>duction of more variables makes it possible. To highlight the
>> standard
>belief consider the algebraic integers (1 &#8722; sqrt(21))/2 and
>>(1+sqrt(21))/2 >which are roots of x^2 + x &#8722; 5.
>>This is a minor note, but it should be -1 for both roots.  While this
>>has little impact on the validity of the rest of the work, it should,
>>at the very least, be corrected in your final draft.
> 
> 
> That was pointed out by someone else, I've acknowledged it as a
> mistake, and I've changed it in the paper.
Agreed.  I was being complete.
>While you know that the algebraic integer factors are themselves
>factors of 5, in what way is each a factor of 5? Can either not have
>> non
>unit factors of 5? How do you know?
>>These questions don't make a lot of sense to me, but I'll skip over
>>that to get to what I believe is the true problem.
> 
> 
> Well, it looks to me like your comment might be taken as somewhat
> negative, but the preamble to the paper can be skipped.
I just prefer clarity in a paper.  Some things I will take as my own 
ignorance, but I should have a reasonable way to go educate myself. 
These questions appear to be a driving concept behind what you intend to 
do, but it isn't obvious to me what that concept is.  It could be my 
ignorance, but it didn't appear to be clear what you meant.
>In looking to consider distribution of algebraic integer factors
>> within
>a factorization I'll be using a more complicated example than x 2 +x
>> &#8722; 5.
>The paper will show that given the factorization, in the ring of
>> alge-
>braic integers,
>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>one of the a's is coprime to 5, using basic algebraic methods.
>>And here we run into two problems.  First, while it is true that the
>>a's are algebraic integers, it is something that at least deserves a
>>comment.  
> 
> Why should that be commented on here?
> 
Because with a brief comment you can avoid someone ignoring you out of 
hand.  It can easily _appear_ that you have just switched the discussion 
from algebraic integers to algebraic numbers.  _That_ would put 
everything you are doing into question, or possibly cause someone to 
ignore the rest of your work as unrelated to the claimed topic of 
discussion.
Put briefly, you are putting the burden of responsibility for 
understanding you on the reader, rather than taking it on yourself to be 
clear.
> 
>>Second, it is far from obvious that coprime is meaningful
>>on the algebraic integers.  
> 
> 
> That was discussed by others in reply to your post.
It has been pointed out that the definition of coprime I was using is 
not optimal.  A better version states x and y are coprime in a ring if 
there is an a and b in the same ring satisfying ax+by=1.  This simple 
statement negates all my comments about the meaningfulness of coprime.
> However I'll add that I switched to coprime rather than use 
factor
> in this situation because algebraic integers is an incomplete ring so
> there is a problem with the term factor if I want to stay in the ring.
>  That is, rather than say that one of the a's is coprime, I might have
> said that only two of the a's have a factor of f, but the ring is
> incomplete, so that's problematic.
What do you mean by the term incomplete?
> That is, the problem with algebraic integers being incomplete means
> that I'd have to knowingly use factor in a way not exactly correct,
> but coprime still works, so I used it.
If you are using the term factor in a way that is not exactly correct, 
you probably shouldn't use it.  Find another term, or invent one.  If 
you aren't using the words right, then you are not communicating 
effectively.
If I decide to use the word cat to refer to any pet, since the only 
pets I have are cats, then when I say my neighbor's three cats in the 
back yard were barking up a storm last night I should _expect_ to have 
people complain that cats don't bark.
> 
> 
>>Here's the problem: coprime is based on
>>the definition of the least common divisor.  The LCD is only
>>guaranteed to exist for Principal Ideal Domains and Unique
>>algebraic integers are certainly NOT a UFD, and I've seen nothing to
>>indicate anything stronger than that they are a commutative ring. 
>>Until someone can point to work that indicates that LCD is defined on
>>the algebraic integers, this paper cannot move forward.  The remainder
>>of the paper (which I have NOT checked for accuracy but am including
>>for completeness) depends on the concept of coprime being meaningful
>>and defined on the algebraic integers.
> 
> 
> The problem is that algebraic integers are incomplete as a ring.  What
> the paper shows is that two of the a's should have f as a factor, as
> the paper shows them to have f as a factor, but technically, it's not
> a factor within the ring of algebraic integers.
And this is where the terminology becomes important.  If it's a factor, 
fine.  If it's not a factor, then it's something else.  If you don't use 
the term at the right time, you can go in circles trying to make things 
be what they aren't and do what they don't to make the term apply better.
> 
> 
>First I'll need a simple lemma to generalize beyond factors of a
>> poly-
>nomial that are themselves polynomials.
>Lemma 1.1. Factorization Lemma:
>Given a factor g of a polynomial P(x), further defined as a factor
>for all x, which means that the value of g for a value 'a' of x is a
>> factor
>of P(a), within the ring of algebraic integers, there exists r and c
>> such
>that
>g = r + c
>where r=0, or is not coprime to x, and c is a factor of the constant
>term P(0).
>>The notation here could be improved.  g should be g(x), r should be
>>r(x).
> 
> 
> I'm not interested in style issues.
> 
You have yet to learn the difference between style and precision. 
Precision preserves the accuracy of what you say.  Style perserves the 
readability of what you say.  Both are important parts of clear 
communication.
> You have claimed to be a mathematician, and mathematicians are defined
> to be math experts.  The fundamental question about the paper is not
> style but correctness.  Minor issues aren't relevant, but so far
> that's all you have.
I am a mathematician.  My expertise is NOT in algebra or number theory. 
  That is why I was looking up terms in books.  The problem is the 
following: when I look things up and read and can't determine 
correctness because I can't get past the style, the style has become an 
issue and ceased to be minor.
I don't understand what you're trying to say is not a minor issue.
> For instance, you talked about the sign problem in the preamble, and
> then you questioned the use of coprime, and now you talk about
> notation.
The sign problem was minor.  Notation often is not.
> 
> These type of issues are the kinds of things I see brought up by
> people I deem aren't interested in the truth.
Perhaps you are missing part of the truth: namely that these issues are 
significant, not minor.
> 
> The discussion is not a game.  It's not about winning or losing.  It's
> about the truth.
> 
> If I'm correct then there's a problem in mathematics which needs to be
> fixed.
> 
> If I'm wrong then you should be able to find more than a style issue.
I always thought math _was_ a game.  It has rules you must play by.  The 
problem is, the rules don't seem to be what you want them to be.  It 
would be easier to fix the style issues and _then_ deal with the 
substance rather than keep fighting to preserve your style.
>Proof. Let x=0, then g must be a factor of P(0), so at that point c =
>> g.
>(1) If when x does not equal 0, g=c, r=0.
>(2) If when x does not equal 0, g =/= c there must exist r which
>> varies
>with x, and as r equals 0 when x equals 0 it is not coprime to
>x.
>I'm not sure step 2 makes sense.  I don't see why it makes r coprime
>>to x.
> 
> 
> The argument isn't complicated but I simplified that section rather
> than deal with it, after I realized it wasn't correct
> 
> That is, r is not coprime to x or contains a unit factor of x.
> 
> With that unit factor things get messy as I'm sure some would jump on
> that and claim that since r would always have a unit factor of x, it's
> meaningless.
If it gets messy, then you should present it rather than force the 
reader to try to slog through all of that on their own.  Skip simple 
things that are clear cut.  Messy means not obvious a lot of the 
time.
> 
> Rather than get into an involved explanation with more room for
> confusion, I realized that I really was just using the fact that any
> factor of a polynomial can be split up between what's constant and
> what's varying.
> 
> That is, r changes, but c does not, as x varies.
> 
> The proof of that is actually trivial as all you do is take g at 0,
> which gives you c, then r = g-c.
> 
> Given that c is constant, while g is not, obviously r is not either.
This I agree with.
> 
> 
>As an example consider sqrt(x + 1) which is a non polynomial factor
>> of
>x+1, and while there are an infinity of irrational solutions consider
>> the
>rational solution at x=35.
>>Are you sure you want to talk about non-polynomial roots?  While
>>legitimate, it changes the topic of discussion from polynomials over
>>the algebraic integers to algebraic functions over the algebraic
>>integers.
> 
> 
> Which would seem to indicate that you haven't read further down in the
> paper, or haven't bothered to consider what I said before this
> section:
> 
> First I'll need a simple lemma to generalize beyond factors of a
> polynomial that are themselves polynomials.
> 
> Now I'd think that says what you need to know, and besides, what's
> with the talk of roots?
> 
> What are non-polynomial roots?
Ah.  That should be factor.  This is a prime example of saying what 
you _mean_, not just using a word that's related.
> 
> Are you sure you're a mathematician?
Only if a master's degree counts.
> 
> 
>Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,
>and c=1. But for different values of x, g and r will vary, while r
>> will
>not be coprime to x.
>>Again, coprime needs to be defined for the particular ring under
>>consideration.
> 
> 
> And it sounds to me like maybe you didn't spend a weekend going over
> math texts as you repeatedly brought up as an issue something other
> posters dismissed immediately.
> 
> 
>2. Primary Argument
>Given
>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>in the ring of algebraic integers. Let
>P(m) = f^2 ((m^3 f^4 &#8722; 3m^2 f^2 + 3m)x^3 &#8722; 3( &#8722; 1 +
>> mf^2 )xu^2 + u^3 f)
>Here f is a non unit, non zero algebraic integer coprime to 3 and x,
>and u a non unit, non zero algebraic integer coprime to f. Note P(m)
>has a factor that is f^2 .
>>Again, coprime must be defined on the algebraic integers.  Also, the
>>change of variable and the introduction of additional variables seems
>>odd.  Since it appears that you mean for f and u to be constants, it
>>appears that you actually have P(m,x).
> 
> 
> Nope.  And if you're a mathematician changing variables as I did
> shouldn't be a problem for you.
> 
> Again, the question is to the correctness of the paper, and not style
> issues.
Which is easier: fixing the style so there are no style issues, or 
fighting to preserve your style when they obscure the ability to 
determine correctness?
> 
> So far, just like before when I got upset, you've spent a lot of time
> on nonessential issues, or ones that others quickly refuted.
> 
> It seems to me that your behavior is consistent with what I've seen
> from posters trying to hide the truth, rather than get to the bottom
> of things.
> 
> But the issue is taught mathematics which is wrong.
> 
> 
>That expression comes from expanding (v^3 +1)x^3 &#8722; 3vxy^2 + y^3
>> , using
>the substitutions v = &#8722; 1+mf^2 , and y = uf, where additional
>> variables
>provide an additional degree of freedom.
>Consider that a similar idea can be used to factor 3, prime in
>> integers,
>as x^2 +7x+10 = 3, allows you to find factors of 3 in the ring of
>> algebraic
>integers.
>>I don't follow what you mean by this at all.  Some additional
>>explanation would be helpful.  What are the factors and how did you
>>compute them?
> 
> 
> I explain to some extent where the expression I use comes from, and
> then give a quick analogy.  If the analogy confuses you, you can skip
> it, as you did with the other section that confused you.
> 
> It's not part of the argument, but merely there to help.
> 
> 
>Now consider the factorization
>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)
>where multiplying out shows that
>a1a2a3 = m^3 f^6 &#8722; 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 &#8722; 3m^2
>> f^2 + 3m)
>so a1a2a3 = mf^2 (m^2 f^4 &#8722; 3mf^2 + 3).
>Therefore, at least one of the a's cannot be coprime to m, and at
>least one of the a's must equal 0 when m=0.
>(Note: The a's are roots of a monic polynomial with algebraic integer
>coefficients so they are algebraic integers.)
>>Maybe.  Did P(m) have a constant term of 1?  if so, uf=1.  The logic
>>of your note could stand some clarification.  (note: coprime shows up
>>again)
> 
> 
> Strange question, given what follows:
> 
> 
>Notice that the constant term P(0) is given by
>P(0) = f^2 (3xu^2 + u^3 f)
>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.
>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1
>is not coprime to m.
>g1 = c = uf
>meaning f is a factor of the constant term.
>Therefore, exactly two of the a's equal 0, when m=0, to get the
>> factor
>f^2 in the constant term P(0), while one must not equal 0, or f^3
>> would
>be the factor.
>>This statement doesn't make a lot of sense either... but I'm not
>>trying as hard at this point.  If the other problems can be fixed,
>>I'll be happy to examine it more carefully.
> 
> 
> In the paper I take a rather complicated expression, consider it as a
> polynomial with respect to m, meaning that I let only m vary, and then
> find that I can factor it in a not surprising way, given the
> expression.
> 
> That factorization gives me non-polynomial factors, with which I use
> my lemma to show that only one of the a's is coprime to f.
Ok, I just looked at your paper again...  where are the non-polynomial 
factors?  I see a ton of polynomial factors, but no non-polynomial factors.
> 
> The factorization I use, isn't terribly surprising, and the argument
> follows rather simply from a rather basic lemma, and the distributive
> property.
> 
> 
>Now as noted before in general P(m) has a factor that is f^2 , and
>separating that factor off, gives a constant term coprime to f;
>> therefore,
>given g1 = a1x + uf
>where with m = 0, g1 gives a factor of f it must have that same
>> factor
>in general, proving that two of the a's have a factor that is f.
>Therefore, one factor is coprime to f.
>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't
>change the a's, I have
>(m^3 f^6 &#8722; 3m^2 f^4 + 3m)x^3 &#8722; 3( &#8722; 1 + mf^2 )xu^2
>> + u^3 = 65x^3 &#8722; 12x + 1
>which may be more easily seen from using v = &#8722; 1 + mf^2 = 4,
>> y=1
>with (v^3 + 1)x^3 &#8722; 3vxy^2 + y^3 .
>Therefore, with the factorization
>65x^3 &#8722; 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)
>one of the a's is coprime to 5, which shows where some of the alge-
>braic integer factors distribute despite the factors being
>> irrational.
>>Overall analysis: if you can clarify the notation and the logical
>>connections, along with your result about factors not being factors,
>>what you probably have is a proof that coprime is not always defined
>>on the algebraic integers.
> 
> 
> Well I refer you to the statements by others about coprime in the
> ring of algebraic integers.
> 
> What I can do is take out the use of the term in the paper, and use
> factor in the ring of algebraic integers, and note that will lead to
> a contradiction as the point of the paper is that the ring is
> incomplete.
> 
> What I want to impress upon readers is that I'm quite willing to work
> with mathematicians to explain the mistake that they're teaching.
> 
> The ring of algebraic integers is incomplete, it's easy to show, and
> I've shown it with my paper.
> 
> If mathematicians are having trouble understanding any part of it, I'm
> able to explain further.
> 
> 
> James Harris
Sure: what does incomplete mean?
-- 
Will Twentyman
====
>>Having spent the weekend going through musty books on algebra...  here
>>is my analysis:
> 
> Hmmm...so my short paper sent you to books for an entire weekend?
> 
> 
> Sorry if I disappointed you.  I tend to do more work in logic/discrete 
> math than I do in Algebra.
I didn't say I was disappointed.  Actually I *was* intrigued.  Then I
read your post.
>ADVANCED POLYNOMIAL FACTORIZATION
>JAMES HARRIS
>Abstract. Algebraic method for determining distribution of fac-
>tors within a polynomial factorization, which breaks through what
>was seen as a barrier from overinterpretations of Galois Theory.
>1. Advanced Polynomial Factorization Approached
>Determining the distribution of factors within irrational algebraic
>integers has long been considered impossible as it is not possible to
>> do
>using Galois Theory. However a simple technique through the intro-
>duction of more variables makes it possible. To highlight the
>> standard
>belief consider the algebraic integers (1 &#8722; sqrt(21))/2 and
>>(1+sqrt(21))/2 >which are roots of x^2 + x &#8722; 5.
>>This is a minor note, but it should be -1 for both roots.  While this
>>has little impact on the validity of the rest of the work, it should,
>>at the very least, be corrected in your final draft.
> 
> 
> That was pointed out by someone else, I've acknowledged it as a
> mistake, and I've changed it in the paper.
> 
> Agreed.  I was being complete.
Ok.
>While you know that the algebraic integer factors are themselves
>factors of 5, in what way is each a factor of 5? Can either not have
>> non
>unit factors of 5? How do you know?
>>These questions don't make a lot of sense to me, but I'll skip over
>>that to get to what I believe is the true problem.
> 
> 
> Well, it looks to me like your comment might be taken as somewhat
> negative, but the preamble to the paper can be skipped.
> 
> I just prefer clarity in a paper.  Some things I will take as my own 
> ignorance, but I should have a reasonable way to go educate myself. 
> These questions appear to be a driving concept behind what you intend to 
> do, but it isn't obvious to me what that concept is.  It could be my 
> ignorance, but it didn't appear to be clear what you meant.
Well I'm going to explain briefly but I'm not interested in spending a
lot of time explaining things to posters, primarily because I have to
deal with so many, including a lot whom I consider cheats.
Consider r_1 and r_2 algebraic integers where their product is 5.  Now
then, can you imagine different ways in which algebraic integer
factors of 5 might distribute?
As given, one easy way for one to be coprime to 5 is for one of them
to equal 1.
Now both could equal sqrt(5), so neither would be coprime to 5, and
both would have a factor of 5, that is sqrt(5).
Got it?
 
>In looking to consider distribution of algebraic integer factors
>> within
>a factorization I'll be using a more complicated example than x 2 +x
>> &#8722; 5.
>The paper will show that given the factorization, in the ring of
>> alge-
>braic integers,
>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>one of the a's is coprime to 5, using basic algebraic methods.
>>And here we run into two problems.  First, while it is true that the
>>a's are algebraic integers, it is something that at least deserves a
>>comment.  
> 
> Why should that be commented on here?
> 
> 
> Because with a brief comment you can avoid someone ignoring you out of 
> hand.  It can easily _appear_ that you have just switched the discussion 
> from algebraic integers to algebraic numbers.  _That_ would put 
> everything you are doing into question, or possibly cause someone to 
> ignore the rest of your work as unrelated to the claimed topic of 
> discussion.
> 
> Put briefly, you are putting the burden of responsibility for 
> understanding you on the reader, rather than taking it on yourself to be 
> clear.
That's false.
The statement begins with,
The paper will show that given the factorization, in the ring of
algebraic integers...
So the ring is declared to algebraic integers at the start.
So once again, you're shown up, and as I've said repeatedly, your
comments are typical of what I've seen from mathematicians trying to
*hide* the truth.
Undergrads need only note the subject line of this thread to consider
why I keep pointing that out.
I want you to see how clear it is that mathematicians *are* lying.
 
> 
>>Second, it is far from obvious that coprime is meaningful
>>on the algebraic integers.  
> 
> 
> That was discussed by others in reply to your post.
> 
> It has been pointed out that the definition of coprime I was using is 
> not optimal.  A better version states x and y are coprime in a ring if 
> there is an a and b in the same ring satisfying ax+by=1.  This simple 
> statement negates all my comments about the meaningfulness of 
coprime.
> 
> However I'll add that I switched to coprime rather than use 
factor
> in this situation because algebraic integers is an incomplete ring so
> there is a problem with the term factor if I want to stay in the ring.
>  That is, rather than say that one of the a's is coprime, I might have
> said that only two of the a's have a factor of f, but the ring is
> incomplete, so that's problematic.
> 
> What do you mean by the term incomplete?
Because I can take an algebraic integer x, prove that within the ring
of algebraic integers it has the algebraic integer y (I'm being
redundant with the algebraic integers declarations but you claimed a
problem before), and then moving to the field of rationals show that
x/y is NOT an algebraic integer, which is a contradiction.
That is, given that in the ring I can prove x has y as a factor, it
hardly makes sense that in checking, by going to the field of
rationals, I find that x/y is not an algebraic integer.
So x/y *should* be in the ring, as it is not, the ring is incomplete.
> That is, the problem with algebraic integers being incomplete means
> that I'd have to knowingly use factor in a way not exactly correct,
> but coprime still works, so I used it.
> 
> If you are using the term factor in a way that is not exactly correct, 

> you probably shouldn't use it.  Find another term, or invent one.  If 
> you aren't using the words right, then you are not communicating 
> effectively.
Well I used coprime so I wouldn't have to get into discussion on the
subject!!!
 
> If I decide to use the word cat to refer to any pet, since the only 
> pets I have are cats, then when I say my neighbor's three cats in the 
> back yard were barking up a storm last night I should _expect_ to have 
> people complain that cats don't bark.
Your statement is irrelevant, as I've explained in detail why I used
coprime instead of factor and in fact you brought up an issue with
coprime which was shot down by other posters.
My fear is that if it hadn't been, I could have been arguing with you
or any number of other posters for some time.
But thankfully you bent over when some other posters corrected you.
> 
> 
>>Here's the problem: coprime is based on
>>the definition of the least common divisor.  The LCD is only
>>guaranteed to exist for Principal Ideal Domains and Unique
>>algebraic integers are certainly NOT a UFD, and I've seen nothing to
>>indicate anything stronger than that they are a commutative ring. 
>>Until someone can point to work that indicates that LCD is defined on
>>the algebraic integers, this paper cannot move forward.  The remainder
>>of the paper (which I have NOT checked for accuracy but am including
>>for completeness) depends on the concept of coprime being meaningful
>>and defined on the algebraic integers.
> 
> 
> The problem is that algebraic integers are incomplete as a ring.  What
> the paper shows is that two of the a's should have f as a factor, as
> the paper shows them to have f as a factor, but technically, it's not
> a factor within the ring of algebraic integers.
> 
> And this is where the terminology becomes important.  If it's a factor, 
> fine.  If it's not a factor, then it's something else.  If you don't use 
> the term at the right time, you can go in circles trying to make things 
> be what they aren't and do what they don't to make the term apply better.
And I've explained repeatedly, and again I want undergrads to consider
the subject line of this thread.
> 
> 
>First I'll need a simple lemma to generalize beyond factors of a
>> poly-
>nomial that are themselves polynomials.
>Lemma 1.1. Factorization Lemma:
>Given a factor g of a polynomial P(x), further defined as a factor
>for all x, which means that the value of g for a value 'a' of x is a
>> factor
>of P(a), within the ring of algebraic integers, there exists r and c
>> such
>that
>g = r + c
>where r=0, or is not coprime to x, and c is a factor of the constant
>term P(0).
>>The notation here could be improved.  g should be g(x), r should be
>>r(x).
> 
> 
> I'm not interested in style issues.
> 
> 
> You have yet to learn the difference between style and precision. 
> Precision preserves the accuracy of what you say.  Style perserves the 
> readability of what you say.  Both are important parts of clear 
> communication.
Yet if you are a mathematician, then by definition you are a math
expert, so the question will arise, are you bringing up an issue that
removes your ability to comprehend?
Yet you say the notation here could be improved, so I think it
reasonable to suppose that you *do* understand.
I'm not interested in style issues.
> You have claimed to be a mathematician, and mathematicians are defined
> to be math experts.  The fundamental question about the paper is not
> style but correctness.  Minor issues aren't relevant, but so far
> that's all you have.
> 
> I am a mathematician.  My expertise is NOT in algebra or number theory. 
>   That is why I was looking up terms in books.  The problem is the 
> following: when I look things up and read and can't determine 
> correctness because I can't get past the style, the style has become an 
> issue and ceased to be minor.
Then point out such an instance.  So far you've brought up minor
issues which have all turned out to be related to your ignorance, or
your *opinion* that something like notation could be improved.
Remember, my point is that mathematicians work *against* finding the
truth, so it hardly makes sense for you not to be on your best
behavior!!!
And in fact, part of my point is that if you were to behave, you'd
have to admit the truth, which is that there is an error in taught
mathematics.
> I don't understand what you're trying to say is not a minor issue.
Then you need to say that in a relevant portion.  You did say that
about the preamble, but that's just an introduction section.
So far, you've talked a lot, without saying much of anything, while
I've explained quite a bit.
> For instance, you talked about the sign problem in the preamble, and
> then you questioned the use of coprime, and now you talk about
> notation.
> 
> The sign problem was minor.  Notation often is not.
So are you going to just whine and whine about notation?
> 
> These type of issues are the kinds of things I see brought up by
> people I deem aren't interested in the truth.
> 
> Perhaps you are missing part of the truth: namely that these issues are 
> significant, not minor.
Then *show* why the issues are relevant to the paper.
The conclusion of the paper would be a start.  Try mightily to think
of something that might have relevance to the conclusion.
Technicalities will just get ripped by me.
If you're an expert in mathematics, prove it.
> 
> The discussion is not a game.  It's not about winning or losing.  It's
> about the truth.
> 
> If I'm correct then there's a problem in mathematics which needs to be
> fixed.
> 
> If I'm wrong then you should be able to find more than a style issue.
> 
> I always thought math _was_ a game.  It has rules you must play by.  The 
> problem is, the rules don't seem to be what you want them to be.  It 
> would be easier to fix the style issues and _then_ deal with the 
> substance rather than keep fighting to preserve your style.
Yeah right, like I'll move to your turf.  Mathematicians cheat.  If I
start trying to play in areas where they cheat well, I could be
talking about this notation versus that notation for months!!!
The issue is an error in taught mathematics.
I'd think mathematicians would care more about that than notation!!!
But then again, maybe they wouldn't, which is something undergrads
would do well to learn now, better sooner than later.
>Proof. Let x=0, then g must be a factor of P(0), so at that point c =
>> g.
>(1) If when x does not equal 0, g=c, r=0.
>(2) If when x does not equal 0, g =/= c there must exist r which
>> varies
>with x, and as r equals 0 when x equals 0 it is not coprime to
>x.
>>I'm not sure step 2 makes sense.  I don't see why it makes r coprime
>>to x.
> 
> 
> The argument isn't complicated but I simplified that section rather
> than deal with it, after I realized it wasn't correct
> 
> That is, r is not coprime to x or contains a unit factor of x.
> 
> With that unit factor things get messy as I'm sure some would jump on
> that and claim that since r would always have a unit factor of x, it's
> meaningless.
> 
> If it gets messy, then you should present it rather than force the 
> reader to try to slog through all of that on their own.  Skip simple 
> things that are clear cut.  Messy means not obvious a lot of the 
time.
Huh?  What I found was that I could use something simpler than I had
that kept it from getting messy.  I decided to go with simpler.
> 
> Rather than get into an involved explanation with more room for
> confusion, I realized that I really was just using the fact that any
> factor of a polynomial can be split up between what's constant and
> what's varying.
> 
> That is, r changes, but c does not, as x varies.
> 
> The proof of that is actually trivial as all you do is take g at 0,
> which gives you c, then r = g-c.
> 
> Given that c is constant, while g is not, obviously r is not either.
> 
> This I agree with.
Excellent!  Then you just got past the lemma!!!
> 
> 
>As an example consider sqrt(x + 1) which is a non polynomial factor
>> of
>x+1, and while there are an infinity of irrational solutions consider
>> the
>rational solution at x=35.
>>Are you sure you want to talk about non-polynomial roots?  While
>>legitimate, it changes the topic of discussion from polynomials over
>>the algebraic integers to algebraic functions over the algebraic
>>integers.
> 
> 
> Which would seem to indicate that you haven't read further down in the
> paper, or haven't bothered to consider what I said before this
> section:
> 
> First I'll need a simple lemma to generalize beyond factors of a
> polynomial that are themselves polynomials.
> 
> Now I'd think that says what you need to know, and besides, what's
> with the talk of roots?
> 
> What are non-polynomial roots?
> 
> Ah.  That should be factor.  This is a prime example of saying what 
> you _mean_, not just using a word that's related.
Ok.
> 
> Are you sure you're a mathematician?
> 
> Only if a master's degree counts.
Forget that question as I'll admit I was quick on the gun.
The point is that the lemma is there so that I can talk about
non-polynomial factors of a polynomial later.  That's the point of it.
> 
> 
>Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,
>and c=1. But for different values of x, g and r will vary, while r
>> will
>not be coprime to x.
>>Again, coprime needs to be defined for the particular ring under
>>consideration.
> 
> 
> And it sounds to me like maybe you didn't spend a weekend going over
> math texts as you repeatedly brought up as an issue something other
> posters dismissed immediately.
> 
> 
>2. Primary Argument
>Given
>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>in the ring of algebraic integers. Let
>P(m) = f^2 ((m^3 f^4 &#8722; 3m^2 f^2 + 3m)x^3 &#8722; 3( &#8722; 1 +
>> mf^2 )xu^2 + u^3 f)
>Here f is a non unit, non zero algebraic integer coprime to 3 and x,
>and u a non unit, non zero algebraic integer coprime to f. Note P(m)
>has a factor that is f^2 .
>>Again, coprime must be defined on the algebraic integers.  Also, the
>>change of variable and the introduction of additional variables seems
>>odd.  Since it appears that you mean for f and u to be constants, it
>>appears that you actually have P(m,x).
> 
> 
> Nope.  And if you're a mathematician changing variables as I did
> shouldn't be a problem for you.
> 
> Again, the question is to the correctness of the paper, and not style
> issues.
> 
> Which is easier: fixing the style so there are no style issues, or 
> fighting to preserve your style when they obscure the ability to 
> determine correctness?
I changed variables deliberately.  It doesn't change the mathematical
argument at all.
The point is that the lemma is in general, and the use of x with the
I'm not interested in style issues.  The point is that there's a
problem in taught mathematics, and it's not about style.
> 
> So far, just like before when I got upset, you've spent a lot of time
> on nonessential issues, or ones that others quickly refuted.
> 
> It seems to me that your behavior is consistent with what I've seen
> from posters trying to hide the truth, rather than get to the bottom
> of things.
> 
> But the issue is taught mathematics which is wrong.
> 
> 
>That expression comes from expanding (v^3 +1)x^3 &#8722; 3vxy^2 + y^3
>> , using
>the substitutions v = &#8722; 1+mf^2 , and y = uf, where additional
>> variables
>provide an additional degree of freedom.
>Consider that a similar idea can be used to factor 3, prime in
>> integers,
>as x^2 +7x+10 = 3, allows you to find factors of 3 in the ring of
>> algebraic
>integers.
>>I don't follow what you mean by this at all.  Some additional
>>explanation would be helpful.  What are the factors and how did you
>>compute them?
> 
> 
> I explain to some extent where the expression I use comes from, and
> then give a quick analogy.  If the analogy confuses you, you can skip
> it, as you did with the other section that confused you.
> 
> It's not part of the argument, but merely there to help.
> 
> 
>Now consider the factorization
>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)
>where multiplying out shows that
>a1a2a3 = m^3 f^6 &#8722; 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 &#8722; 3m^2
>> f^2 + 3m)
>so a1a2a3 = mf^2 (m^2 f^4 &#8722; 3mf^2 + 3).
>Therefore, at least one of the a's cannot be coprime to m, and at
>least one of the a's must equal 0 when m=0.
>(Note: The a's are roots of a monic polynomial with algebraic integer
>coefficients so they are algebraic integers.)
>>Maybe.  Did P(m) have a constant term of 1?  if so, uf=1.  The logic
>>of your note could stand some clarification.  (note: coprime shows up
>>again)
> 
> 
> Strange question, given what follows:
> 
> 
>Notice that the constant term P(0) is given by
>P(0) = f^2 (3xu^2 + u^3 f)
>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.
>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1
>is not coprime to m.
>  
>  
>g1 = c = uf
>meaning f is a factor of the constant term.
>Therefore, exactly two of the a's equal 0, when m=0, to get the
>> factor
>f^2 in the constant term P(0), while one must not equal 0, or f^3
>> would
>be the factor.
>>This statement doesn't make a lot of sense either... but I'm not
>>trying as hard at this point.  If the other problems can be fixed,
>>I'll be happy to examine it more carefully.
> 
> 
> In the paper I take a rather complicated expression, consider it as a
> polynomial with respect to m, meaning that I let only m vary, and then
> find that I can factor it in a not surprising way, given the
> expression.
> 
> That factorization gives me non-polynomial factors, with which I use
> my lemma to show that only one of the a's is coprime to f.
> 
> Ok, I just looked at your paper again...  where are the non-polynomial 
> factors?  I see a ton of polynomial factors, but no non-polynomial 
factors.
The polynomial is P(m), which is given.  Since m is the only
independent variable, those other factors of P(m) are NOT polynomials.
 They may *look* like polynomials to you, but are in fact
non-polynomial factors of P(m), and cannot be considered polynomials
unless you shift variables.  Which is why I say they may *look* like
polynomials, when they aren't, given that m is the key variable.
> 
> The factorization I use, isn't terribly surprising, and the argument
> follows rather simply from a rather basic lemma, and the distributive
> property.
> 
> 
>Now as noted before in general P(m) has a factor that is f^2 , and
>separating that factor off, gives a constant term coprime to f;
>> therefore,
>given g1 = a1x + uf
>where with m = 0, g1 gives a factor of f it must have that same
>> factor
>in general, proving that two of the a's have a factor that is f.
>Therefore, one factor is coprime to f.
>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't
>change the a's, I have
>(m^3 f^6 &#8722; 3m^2 f^4 + 3m)x^3 &#8722; 3( &#8722; 1 + mf^2 )xu^2
>> + u^3 = 65x^3 &#8722; 12x + 1
>which may be more easily seen from using v = &#8722; 1 + mf^2 = 4,
>> y=1
>with (v^3 + 1)x^3 &#8722; 3vxy^2 + y^3 .
>Therefore, with the factorization
>65x^3 &#8722; 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)
>one of the a's is coprime to 5, which shows where some of the alge-
>braic integer factors distribute despite the factors being
>> irrational.
>>Overall analysis: if you can clarify the notation and the logical
>>connections, along with your result about factors not being factors,
>>what you probably have is a proof that coprime is not always defined
>>on the algebraic integers.
> 
> 
> Well I refer you to the statements by others about coprime in the
> ring of algebraic integers.
> 
> What I can do is take out the use of the term in the paper, and use
> factor in the ring of algebraic integers, and note that will lead 
to
> a contradiction as the point of the paper is that the ring is
> incomplete.
> 
> What I want to impress upon readers is that I'm quite willing to work
> with mathematicians to explain the mistake that they're teaching.
> 
> The ring of algebraic integers is incomplete, it's easy to show, and
> I've shown it with my paper.
> 
> If mathematicians are having trouble understanding any part of it, I'm
> able to explain further.
> 
> 
> James Harris
> 
> Sure: what does incomplete mean?
I've explained above, but I'll explain again.  I want undergrads to
note that I'm going the distance, and that I'm the person who takes a
good deal of trouble to go into detail.
The ring of algebraic integers is incomplete in that I can prove that
an algebraic integer x has an algebraic integer factor y, but then in
the field of rationals, considering x/y I find that it is not an
algebraic integer, which is a contradiction.
The proof is in the paper.  The other with x/y has been discussed at
length in various posts.
Of course the numbers are NOT called x and y in the paper.  They are
called a_1 and a_2, which must have factors of f, since a_3 is proven
to be coprime to f, but a_1 a_2 a_3 has a factor of f, which is f^2.
James Harris
====
> 
> 
>While you know that the algebraic integer factors are themselves
>factors of 5, in what way is each a factor of 5? Can either not have
>>non
>unit factors of 5? How do you know?
>>These questions don't make a lot of sense to me, but I'll skip over
>>that to get to what I believe is the true problem.

>Well, it looks to me like your comment might be taken as somewhat
>negative, but the preamble to the paper can be skipped.
>>I just prefer clarity in a paper.  Some things I will take as my own 
>>ignorance, but I should have a reasonable way to go educate myself. 
>>These questions appear to be a driving concept behind what you intend to 
>>do, but it isn't obvious to me what that concept is.  It could be my 
>>ignorance, but it didn't appear to be clear what you meant.
> 
> 
> Well I'm going to explain briefly but I'm not interested in spending a
> lot of time explaining things to posters, primarily because I have to
> deal with so many, including a lot whom I consider cheats.
> 
> Consider r_1 and r_2 algebraic integers where their product is 5.  Now
> then, can you imagine different ways in which algebraic integer
> factors of 5 might distribute?
> 
> As given, one easy way for one to be coprime to 5 is for one of them
> to equal 1.
> 
> Now both could equal sqrt(5), so neither would be coprime to 5, and
> both would have a factor of 5, that is sqrt(5).
> 
> Got it?
Got it.
>In looking to consider distribution of algebraic integer factors
>>within
>a factorization I'll be using a more complicated example than x 2 +x
>>&#8722; 5.
>The paper will show that given the factorization, in the ring of
>>alge-
>braic integers,
>65x^3 &#8722; 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
>one of the a's is coprime to 5, using basic algebraic methods.
>>And here we run into two problems.  First, while it is true that the
>>a's are algebraic integers, it is something that at least deserves a
>>comment.  
Why should that be commented on here?
>Because with a brief comment you can avoid someone ignoring you out of 
>>hand.  It can easily _appear_ that you have just switched the discussion 
>>from algebraic integers to algebraic numbers.  _That_ would put 
>>everything you are doing into question, or possibly cause someone to 
>>ignore the rest of your work as unrelated to the claimed topic of 
>>discussion.
>>Put briefly, you are putting the burden of responsibility for 
>>understanding you on the reader, rather than taking it on yourself to be 
>>clear.
> 
> 
> That's false.
> 
> The statement begins with,
> 
> The paper will show that given the factorization, in the ring of
> algebraic integers...
> 
> So the ring is declared to algebraic integers at the start.
> 
The definition of an algebraic integer is: the root of a monic 
polynomial with integer coefficients.  You have just introduced a 
non-monic polynomial.  This is likely to make someone pause for a bit 
trying to see the connection.  Why does this seem unlikely?
> So once again, you're shown up, and as I've said repeatedly, your
> comments are typical of what I've seen from mathematicians trying to
> *hide* the truth.
> 
> Undergrads need only note the subject line of this thread to consider
> why I keep pointing that out.
> 
> I want you to see how clear it is that mathematicians *are* lying.
>  
> 
>>Second, it is far from obvious that coprime is meaningful
>>on the algebraic integers.  

>That was discussed by others in reply to your post.
>>It has been pointed out that the definition of coprime I was using is 
>>not optimal.  A better version states x and y are coprime in a ring if 
>>there is an a and b in the same ring satisfying ax+by=1.  This simple 
>>statement negates all my comments about the meaningfulness of 
coprime.
>However I'll add that I switched to coprime rather than use 
factor
>in this situation because algebraic integers is an incomplete ring so
>there is a problem with the term factor if I want to stay in the ring.
> That is, rather than say that one of the a's is coprime, I might have
>said that only two of the a's have a factor of f, but the ring is
>incomplete, so that's problematic.
>>What do you mean by the term incomplete?
> 
> 
> Because I can take an algebraic integer x, prove that within the ring
> of algebraic integers it has the algebraic integer y (I'm being
> redundant with the algebraic integers declarations but you claimed a
> problem before), and then moving to the field of rationals show that
> x/y is NOT an algebraic integer, which is a contradiction.
> 
> That is, given that in the ring I can prove x has y as a factor, it
> hardly makes sense that in checking, by going to the field of
> rationals, I find that x/y is not an algebraic integer.
> 
> So x/y *should* be in the ring, as it is not, the ring is incomplete.
> 
If y is a factor of x, both in the ring, then there is an r in the ring 
which satisfies ry=x.  r is what you are referring to as x/y.  If r is 
NOT in the ring, y is NOT a factor of x.
If you have shown that there is a problem in math, it would be nice if 
you show which theorem that you relied on in your proof has the error. 
Otherwise, your proof has the error.  You cannot redefine what a factor 
is, which is what you appear to be trying to do.  Until you can find 
which theorem from someone else's work that you used has the flaw, all 
you've done is noted that SOMETHING is fishy.  It's up to you to find 
where that something fishy is.  Until then, you haven't discovered 
anything useful.
Note: it is obviously the belief of the people who have responded that 
the something fishy is located in your work.  Until you can locate it 
elsewhere or address all objections (including counterexamples provided 
by others), you will have a hard time convincing anyone of anything.
[ much of rest deleted ]
I'm not interested in style issues.
>You have yet to learn the difference between style and precision. 
>>Precision preserves the accuracy of what you say.  Style perserves the 
>>readability of what you say.  Both are important parts of clear 
>>communication.
> 
> 
> Yet if you are a mathematician, then by definition you are a math
> expert, so the question will arise, are you bringing up an issue that
> removes your ability to comprehend?
> 
> Yet you say the notation here could be improved, so I think it
> reasonable to suppose that you *do* understand.
> 
> I'm not interested in style issues.
Nor am I interested in rewriting your paper with all the style issues 
corrected to see if it still says what you claim it says.  If you find 
that an objectionable trait, I'm sorry.
> 
> 
>You have claimed to be a mathematician, and mathematicians are defined
>to be math experts.  The fundamental question about the paper is not
>style but correctness.  Minor issues aren't relevant, but so far
>that's all you have.
>>I am a mathematician.  My expertise is NOT in algebra or number theory. 
>>  That is why I was looking up terms in books.  The problem is the 
>>following: when I look things up and read and can't determine 
>>correctness because I can't get past the style, the style has become an 
>>issue and ceased to be minor.
> 
> 
> Then point out such an instance.  So far you've brought up minor
> issues which have all turned out to be related to your ignorance, or
> your *opinion* that something like notation could be improved.
When I'm not sure if g is a polynomial or an algebraic integer.  That 
was clarified in another post, but was still an obstacle at the time.
> 
> Remember, my point is that mathematicians work *against* finding the
> truth, so it hardly makes sense for you not to be on your best
> behavior!!!
Have I been insulting to you?  I haven't said what you want to hear, but 
that does not constitute poor behavior.
> 
> And in fact, part of my point is that if you were to behave, you'd
> have to admit the truth, which is that there is an error in taught
> mathematics.
> 
When you fail to address points raised (and mine are minor compared to 
those raised by others), the burden of proof falls back on you.
>Proof. Let x=0, then g must be a factor of P(0), so at that point c =
>>g.
>(1) If when x does not equal 0, g=c, r=0.
>(2) If when x does not equal 0, g =/= c there must exist r which
>>varies
>with x, and as r equals 0 when x equals 0 it is not coprime to
>x.
>I'm not sure step 2 makes sense.  I don't see why it makes r coprime
>>to x.

>The argument isn't complicated but I simplified that section rather
>than deal with it, after I realized it wasn't correct
That is, r is not coprime to x or contains a unit factor of x.
With that unit factor things get messy as I'm sure some would jump on
>that and claim that since r would always have a unit factor of x, it's
>meaningless.
>>If it gets messy, then you should present it rather than force the 
>>reader to try to slog through all of that on their own.  Skip simple 
>>things that are clear cut.  Messy means not obvious a lot of the 
time.
> 
> 
> Huh?  What I found was that I could use something simpler than I had
> that kept it from getting messy.  I decided to go with simpler.
You left a gap in your proof that you didn't address.
>Rather than get into an involved explanation with more room for
>confusion, I realized that I really was just using the fact that any
>factor of a polynomial can be split up between what's constant and
>what's varying.
That is, r changes, but c does not, as x varies.
The proof of that is actually trivial as all you do is take g at 0,
>which gives you c, then r = g-c.
Given that c is constant, while g is not, obviously r is not either.
>>This I agree with.
> 
> 
> Excellent!  Then you just got past the lemma!!!
> 
Wrong, I just got past the statement of the lemma.  Not the proof.
>Are you sure you're a mathematician?
>>Only if a master's degree counts.
> 
> 
> Forget that question as I'll admit I was quick on the gun.
> 
> The point is that the lemma is there so that I can talk about
> non-polynomial factors of a polynomial later.  That's the point of it.
> 
Then finish prooving the lemma.
>Notice that the constant term P(0) is given by
>P(0) = f^2 (3xu^2 + u^3 f)
>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.
>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1
>is not coprime to m.
>> 
>g1 = c = uf
>meaning f is a factor of the constant term.
>Therefore, exactly two of the a's equal 0, when m=0, to get the
>>factor
>f^2 in the constant term P(0), while one must not equal 0, or f^3
>>would
>be the factor.
>>This statement doesn't make a lot of sense either... but I'm not
>>trying as hard at this point.  If the other problems can be fixed,
>>I'll be happy to examine it more carefully.

>In the paper I take a rather complicated expression, consider it as a
>polynomial with respect to m, meaning that I let only m vary, and then
>find that I can factor it in a not surprising way, given the
>expression.
That factorization gives me non-polynomial factors, with which I use
>my lemma to show that only one of the a's is coprime to f.
>>Ok, I just looked at your paper again...  where are the non-polynomial 
>>factors?  I see a ton of polynomial factors, but no non-polynomial 
factors.
> 
> 
> The polynomial is P(m), which is given.  Since m is the only
> independent variable, those other factors of P(m) are NOT polynomials.
>  They may *look* like polynomials to you, but are in fact
> non-polynomial factors of P(m), and cannot be considered polynomials
> unless you shift variables.  Which is why I say they may *look* like
> polynomials, when they aren't, given that m is the key variable.
> 
Since all your factors had integers powers on m, your factors were 
polynomials.  Please show one that isn't.
-- 
Will Twentyman
====
   [.snip.]
>> If you are using the term factor in a way that is not exactly 
correct, 
>> you probably shouldn't use it.  Find another term, or invent one.  If 
>> you aren't using the words right, then you are not communicating 
>> effectively.
Well I used coprime so I wouldn't have to get into discussion on the
>subject!!!
Then you are an idiot. Apparently, you are using a term whose meaning
you do not know in order to avoid arguments and discussions. That's
not how mathematics is done, and that is a very far cry from the
logic you claim to love.
I commented some months ago that you seem to think that terminology
and definitions are like magic spells and incantantions: you do not
need to understand them, you just need to say the words in the right
order with the right inflection, and all your problems will disappear
like magic.
If you do NOT know what coprime means, then either ask, or do not
use it.
If you are NOT willing to give a FORMAL, PRECISE, CLEAR definition of
the terms you use, then you are not doing math.
At this point, you need to provide a FORMAL, PRECISE, CLEAR definition
of complete ring, incomplete ring, and factor. The latter,
because you are clearly NOT using it in the standard meaning.
======================================================================
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
magidin@math.berkeley.edu
====
> Because I can take an algebraic integer x, prove that within the ring
> of algebraic integers it has the algebraic integer y (I'm being
> redundant with the algebraic integers declarations but you claimed a
> problem before), and then moving to the field of rationals show that
> x/y is NOT an algebraic integer, which is a contradiction.
> 
> That is, given that in the ring I can prove x has y as a factor, it
> hardly makes sense that in checking, by going to the field of
> rationals, I find that x/y is not an algebraic integer.
> 
> So x/y *should* be in the ring, as it is not, the ring is incomplete.
This strikes me as a bit nonsensical.
It's kind of like saying that I've proved that in the integers, 2 is
a factor of 3, even though 3/2 isn't an integer.  If you run across a
statement like that, it's not time to declare the integers incomplete,
but rather time to take a close look at why you thought 2 was a factor
to start with, because it simply isn't.
You can't factor a number part-way... unless *both* y and y/x are in
the ring, then y isn't a factor of x, by definition of factor.  You
may be able to multiply together y and y/x to get x, but that's not
the same thing.
====
> You can't factor a number part-way... unless *both* y and y/x are in
> the ring, then y isn't a factor of x, by definition of factor.  You
> may be able to multiply together y and y/x to get x, but that's not
> the same thing.
Sorry, that should have been x/y above in both places.  Typo.
====
James Harris <jstevh@msn.com> spewed all over himself in message:
[most babble deleted for brevity]
 Here I'm right, and you're wrong, as the *definition* of algebraic
> integer leads to an incomplete ring, which I prove mathematically.
 You cannot depend on a *definition* in mathematics to prove something
> that's been disproven by mathematics.

> James Harris
TROLL!!!
Harris, do the world a favor and give it up!! You are attempting to
'educate' math undergrads, when I can almost guarantee they are laughing
their asses off when you post!! You even sound like some spoiled 5-year old
when someone points out that they're wrong. If you want to act like a
mathematician, talk like a mathematician, INCLUDING using their 
definitions,
etc. Otherwise, you're a kook!!
If it's entertainment you're trying to provide, you're doing a great job!!
If it's education you're trying to provide, you suck!!
~Bhuvan
====
I'm putting this bit first to highlight it:
[...]
|  Consider that in the ring of algebraic integers, 5 has algebraic
| integer factors, and given algebraic integers r_1 and r_2, where their
| product is 5, why are you acting as if it's so difficult to comprehend
| that there must be some distribution of factors of 5?
I'm not acting as if I'm having any kind of difficulty. I'm saying
that *you're* having difficulty in realizing that there must be some
distribution of factors of 5 is a meaninglessly vague phrase. You're
saying this as if to say, Shouldn't this be true?, and not as if
you actually had proven it was true.
By the standard meanings of the terms, saying an algebraic integer r_1
has a factor of 5 in the algebraic integers is just the same as
saying it's divisible by 5 in the algebraic integers. That's the same
as saying that there's an algebraic integer t for which r_1=5t.
There isn't, and I think you know there isn't. Since neither r_1 nor
r_2 is divisible by 5, there's no distribution of factors of 5
between them.
I think you use ill-defined terms because on some level you're aware
that using only well-defined terms spoils your fun. It's disillusioning.
It clear away the fog of obfuscation. Suddenly you're back to reality,
and it's the rather ordinary reality that people keep telling you that
you're in, not the amazing story you wish it was.
Back to the top.
| > [...]
| > | What I can do is take out the use of the term in the paper, and use
| > | factor in the ring of algebraic integers, and note that will lead 
to
| > | a contradiction as the point of the paper is that the ring is
| > | incomplete.
| > |
| > | What I want to impress upon readers is that I'm quite willing to work
| > | with mathematicians to explain the mistake that they're teaching.
| >
| > If you want to show that a mistake is being taught, you should quote it
| > specifically.
|
| That's not very sensible if you expect that for a mistake to be taught
| for any length of time it'd have some subtlety, unless you're also
| questioning the competence of mathematicians.
Quite to the contrary-- it's because mathematicians are generally
competent that one can ordinarily say very specifically what is wrong
when they make mistakes.
What makes a mathematical mistake subtle is not what it takes to point
it out once it's been recognized; it's the difficulty in recognizing
it in the first place.
| It seems to me that rather than deal with the important question,
| which is whether or not I'm correct, people often try to introduce
| weird ad hoc conditions.
You created an ad hoc rule that in order to refute a claim of yours,
it's not good enough for someone else to prove that the conclusion is
wrong. You think a proof of yours can show that a proof of a
mathematician is wrong, but you appear unwilling to accept the
possibility that a proof from a mathematician can just as well show
that one of yours is wrong (i.e., not really a proof).
| If you disagree with that assessment, can you please explain why you
| believe there should be something simple and short enough for me to
| quote?
In mathematics, we try to make each *mathematical* statement have its
own well-defined meaning. (There are of course nonmathematical claims
that we sometimes make, which aren't always precise.) If a sequence of
well-defined mathematical statements is wrong, it's wrong because (at
least) one of the statements is wrong.
The key is making sure all the definitions are sound.
Mathematics that is vaguely enough written that it can be wrong
without any individual statements in it being definitely wrong, is
considered very bad. It's bad because an author hasn't defined their
terms well enough.
| > | The ring of algebraic integers is incomplete, it's easy to show, and
| > | I've shown it with my paper.
| > |
| > | If mathematicians are having trouble understanding any part of it, 
I'm
| > | able to explain further.
| >
| > You have a big problem with terminology. You use terms that aren't
| > used by others in the context in which you use them, such as counting
| > factors of 5 in algebraic integers. I haven't seen anything I would
| > consider an adequate definition of those terms. You don't explain
| > why your terms should be considered relevant to the ones mathematicians
| > are using.
|
| That statement possibly has something to do with what I now call the
| preamble in the paper as it is there to help explain context, but
| isn't part of the actual argument.
|
| Here's what I say, copied from the paper (some editing for format):
|
| To highlight the standard belief consider the algebraic integers
| (-1-sqrt(21))/2 and (-1+sqrt(21))/2 which are roots of x^2 + x - 5.
|
| While you know that the algebraic integer factors are themselves
| factors of 5, in what way is each a factor of 5? Can either not have
| non unit factors of 5? How do you know?
|
| There I'm talking about algebraic integers, so one can assume I'm
| talking about algebraic integer factors.  Given that 5 has algebraic
| integer factors, how is what I say nonstandard Keith Ramsay?
|
| I'd like you to carefully explain your assertion.  The question I'm
| raising is the possibility that you lied.
No, I haven't lied. You are not using have non unit factors of 5 in
a standard way.
If at this point in the paper, you meant it in a completely standard
way, here's what the meaning would be. Have a factor of 5 is
synonymous with have 5 as a factor, or be divisible by 5. Also 5
is a non-unit (so that part is redundant).
Now obviously we all know you don't mean that. It's possible you mean
something like, shares non-unit factors with 5, meaning that there
are non-unit algebraic integers that are factors of both numbers. But
this is just guesswork. Any serious writer of mathematics takes
responsibility for taking out the guesswork, NOT forcing the reader to
keep inferring which of various possible meanings is the intended one.
The standard usage is also inconsistent with the rest of the paper.
You write, for example, ...proving that two of the a's have a factor
that is f at a point where you plainly have NOT shown that f is a
factor of any of the a's, in any of the usual ways factor is
defined.
Elsewhere, you refer to an algebraic integer f coprime to x. This
has no standard meaning in the context you use it, where x is a
variable. If you meant coprime in some ring of polynomials, which is
closest to a standard meaning, it would mean that there exists
polynomials P and Q such that P*f+Q*x=1. But if f is an algebraic
integer, that's possible (if and) only if f is a unit, with Q=0, and
you also assume in the same sentence that f is a non unit.
| > I don't know what you think is a fair way for things to work, but 
the
| > way things actually work, this kind of problem with terminology will
| > absolutely prevent you from making any headway.
|
| What will prevent me from making headway is if mathematicians
| continually lie.
Mathematicians don't continually lie. What you keep deciding are
lies are just disagreements with you.
Perhaps the biggest problem of all is that you've reached a point
where you don't have anybody you are willing to trust, who could help
you sort out which of the claims people are making are valid and which
are baloney. So you tend automatically to assume that the things
people say that seem wrong to you are baloney of some kind or another.
This also leaves you without any very good way to learn how to improve
your mathematical tinkering. We keep telling you different things
which _we_ claim would help you avoid pitfalls like you keep falling
into. But (a) you don't trust us enough to believe we've identified a
problem with what you're doing, and (b) you don't trust us to give you
straight advice on how to avoid it; you suspect us of just trying to
waste your time. So you're stuck with only your own inklings of what's
a good way to work with proofs.
| Now then in what way is it improper terminology to talk about
| algebraic integer factors of 5?
I didn't say it was improper to talk about algebraic integer factors
of 5. That has a perfectly well-defined meaning. An algebraic integer
r is a factor of 5 if it divides 5 in the algebraic integers, meaning
that 5/r is also an algebraic integer.
| > In an experiment in communication, I once tried elaborating on such a
| > concept for you. You usually write as though the number of factors of
| > 5 in an algebraic number satisfied certain axioms which are familiar
| > to me: being a rational number v(x)>=0 associated with each algebraic
| > integer x <> 0, where v(5)=1, v(xy)=v(x)+v(y) for any algebraic
| > integers x<>0 and y<>0, and v(x+y)>=min{v(x),v(y)} for algebriac
| > integers x<>0, y<>0, and x+y<>0. Assuming that we have such a function
| > v, then we can show that v(r1)=0 and v(r2)=v(r3)=1/2 for the roots
| > r1, r2, and r3 of your cubic, if we take them in the right order.
| > I think this is the best way to try to make sense of your argument
| > in your advanced polynomial factorization thing.
| >
| > Unfortunately, this known concept is not so directly related to
| > divisibility in the algebraic integers, so even if you did manage to
| > produce a definition on these lines, it wouldn't show mathematicians
| > are doing anything wrong when they make claims about divisibility in
| > the algebraic integers as they teach about it in classrooms. They're
| > just apples and oranges.
|
| There is no divisibility argument within the paper, and I only even
| mention roots when the roots are algebraic integers.
I guess you think of factorization as thoroughly unrelated to
divisibility. Note P(m) has a factor that is f^2 has nothing to do
with divisibility of P(m) by f^2, then. This just highlights how far
your terminology departs from standard. Yes, I automatically take []
is a factor of [] as a synonym for [] divides [], because that's
the way it is with any normal usage of the terms. It's only a mistake
if one is dealing with someone like you who departs so far from
standard usage.
|  I've talked
| about x/y in discussing the error in taught mathematics, but did make
| a post explaining that was when an argument is considered in the field
| of rationals.
|
| Your statement falls flat Keith Ramsay, and I think you're just trying
| to sound good enough to fool people, rather than trying to get to the
| truth.
People have seldom had a hard time getting at the mathematical truth
in these discussions, whenever the actual mathematical question has
been well-defined. (There was *once* a question about polynomial
factorization over the algebraic integers which required some effort,
but that was a rare exception.)
The problem is with getting well-define claims. Take your term
incomplete ring for example. Your definition of incomplete ring
for example amounts essentially to a ring in which one element doesn't
divide another one, but it *should* divide it. You only imagine that
the difficulties created by using this meaninglessly vague term are
someone else's fault.
| What is clear is that the paper does NOT operate over any fields, and
| depends only on ring operations.
You seem to have an odd idea of what one is allowed to say after
having declared an argument to be in a ring. The whole point is to
make all of your mathematical statements be well-defined, not to obey
arbitrary conventions.
When you say that you are working in the algebraic integers, that
provides us with a certain kind of context, which allows us to
understand the meaning of certain terms. For example, after you've
said that you're working in the algebraic integers, if you say y
divides x, it means y divides x in the algebraic integers.
Otherwise it might be ambiguous what it meant, or mean something
different from what you intended.
Saying you're doing this does NOT mean that it magically becomes
forbidden to refer to x/y. Certainly it would be incorrect to assume
that x/y is defined in the algebraic integers. But there's an obvious
meaning to the expression x/y, so it's legitimate to refer to it (just
so long as one realizes that it isn't necessarily an algebraic integer
itself). If we're trying to figure out whether there exists an
algebraic integer z with the property that x=yz, it makes perfect
sense for us to consider first the fact that there exists one and only
one z that satisfies x=yz, and THEN ask whether that z (which exists
in the algebraic numbers) is a member of the algebraic integers (which
is the ring we're considering primarily). Saying that x/y is an
algebraic integer means just the same thing as saying that there
exists an algebraic integer z such that yz=x.
| > Say we define the number of factors of 5 in an algebraic integer x to 
be
| > the highest rational number r such that 5^r divides x in the algebraic
| > integers, i.e. such that there exists an algebraic integer y such that
| > 5^r*y = x. I think this definition works (i.e., I think there is such a
| > highest rational power for each algebraic number, although I haven't
| > tried to write out a proof). If we define it that way, though, then it
| > simply doesn't have one of the properties I listed above.
|
| Your post isn't coherent mathematically given what I say in the paper.
|  Consider that in the ring of algebraic integers, 5 has algebraic
| integer factors, and given algebraic integers r_1 and r_2, where their
| product is 5, why are you acting as if it's so difficult to comprehend
| that there must be some distribution of factors of 5?
Just to repeat, it's not a difficulty I have; it's your difficulty in
saying what you actually mean by that. And I don't think you have a
clear idea of what you mean by that yourself.
| For instance, both could have a factor that is sqrt(5), or one could
| have a factor of 1+2i, while the other had a factor of 1-2i.
|
| The question I'm trying to get the reader to explore is, given that
| they are roots of this particular polynomial, could one of them only
| have unit factors of 5?
|
| How do you know?
If you were using the terms in the standard way, this question would
be, How can you tell when one algebraic integer divides another
algebraic integer?
If you actually defined your question, it would become clear either
that it's the same as the standard one (and that the answer
mathematicians give is correct), or that it's a different question
from the standard one (so that even if you get a different answer to
your question, it's just irrelevant to whether mathematicians are
being truthful). I think this has a lot to do with why you simply keep
the question vague; this kind of clarity would take the drama out of it.
Keith Ramsay
====
[snip for brevity]
>> You can't factor a number part-way... unless *both* y and y/x are in
>> the ring, then y isn't a factor of x, by definition of factor.  You
>> may be able to multiply together y and y/x to get x, but that's not
>> the same thing.
I note your follow-up correction post.  My answer still is that you're
>*assuming* that algebraic integers are complete, but then I say it's
>not, and you try to fault me on your assumption, which I've proven
>false.
How about this?
Let's say that someone claims that they have a complete ring, which is
>a subset of the ring of integers, but they give you 6 as a member, ...
Ok. There are exactly four such rings: Z, 2Z, 3Z, and 6Z (with usual 
addition and multiplication). Of those, only Z is a ring with unity. 
> ... and
>*say* that every member of the ring has a factor of 2.
Why would anyone say that? 
>Now you say, no, that's false.  And they argue with you.
You say, but 6/2 is not in the ring then, as it doesn't have a factor
>of 2, and then they give an argument similar to what you gave above,
>saying that of course as 2 and 6 are in the ring, and by the
>definition every member has 2 as a factor!!!
In Z, 2 is a factor of 6, because 2, 6, and 6/2 are _all_ in Z. 
In 2Z, 2 is a _not_ factor of 6, because 6/2 is not in 2Z.
In 3Z, 2 is a _not_ factor of 6, because 2 is not in 3Z (even though 6/2 
is).
In 6Z, 2 is a _not_ factor of 6, because 2 is not in 6Z.
>The arguing goes back and forth, back and forth, back and forth.
But of course 6/2 = 3, so 3 is not in the ring.  So you're right, and
>they're wrong.
You appear to be arguing that 3 _should_ be in the ring, otherwise the ring 

is incomplete. If so, which of the rings 2Z, 3Z and 6Z do you think is 
incomplete? They all contain 0; they're all closed under (usual) addition 
and multiplication. What's the problem?
>Here I'm right, and you're wrong, as the *definition* of algebraic
>integer leads to an incomplete ring, which I prove mathematically.
You cannot depend on a *definition* in mathematics to prove something
>that's been disproven by mathematics.
Try this:
  (*) Definition: The set of contradictory numbers is the set of all
      odd positive integers, that are divisible by 2.
There is nothing wrong with this definition. The set of contradictory 
numbers is perfectly well defined. As any definition it can't be true or 
false, right or wrong. The definition itself doesn't say anything about the 

existence of such numbers, so we need a theorem:
  (*) Theorem: The set of contradictory numbers is empty.
_This_ can be right or wrong.
It the same with algebraic numbers:
  (*) Definition: The _set_ of algebraic integers is the set of all complex 

      roots of monic polynomials with integer coefficients.
Note that the algebraic integers aren't _defined_ to be a ring. The 
definition doesn't even say anything about the existence of algebraic 
integers, so we need:
  (*) Theorem: The set of algebraic integers is not empty.
Now, _this_ can be true or false, and so can this:
  (*) Theorem: The algebraic numbers form a ring under usual addition
      and multiplication.
You (apparently) claim that this theorem is false. The only thing that can 
possibly mean, is that the algebraic integers are either not closed under 
addition or not closed under multiplication. 
When you try to prove that it is false, you appeal to factors. The normal 
definition of factor is:
  (*) Definition: y is a factor of x in the ring R, if and only if, *all* of 

      x, y, and x/y are in R.
This is what is meant by the word factor. The _only_ way to prove that y is 

factor of x in the ring of algebraic integers, is to prove that x, y AND x/y 

are algebraic integers, because that is what the word factor means. 
_That_ 
is why 2 is NOT a factor of 6 in 2Z, 3Z and 6Z.
-- 
Thomas Wasell     | The heart has its reasons which reason knows nothing
wasell@bahnhof.se | of. 
                  |             -- Blaise Pascal
====
...
 > First consider that the definition of algebraic integers as the
 > *roots* of monic polynomials with integer coefficients does not
 > specifically give you any protection from contradiction.
That is true.
 > The definition is over the field of rationals, and it's not clear in
 > just looking at it, if it will give a complete ring.
Indeed, nor is it claimed that it does.
 > Mathematicians apparently *assumed* it did, and have gone on that
 > assumption for some time.
You have that wrong.  It has been *proven* that with this definition
the set of algebraic integers is closed under addition and multiplication.
That is, it has been proven that given two algebraic integers, their
sum and their product is also an algebraic integer.  (See for instance
Van der Waerden, Algebra.)  As the set contains also 0, 1 and the additive
inverse of each element, the set is (from this) actually a ring.
 > People seem fixed on the *definition* with the assumption, when
 > there's no proof based on the definition that the ring is complete.
What do you *mean* with an incomplete ring?  The algebraic integers
are closed under addition and multiplication and satisfy the following
axioms:
1: a + (b + c) = (a + b) + c
2: a + b = b + a
3: there is a 0 such that 0 + a = a
4: for each a there is a (-a) such that a + (-a) = 0
5: a * (b + c) = a * b + a * c
6: (a * b) * c = a * (b * c)
7: a * b = b * a
8: there is a 1 such that 1 * a = a
and so they form a ring (a commutative ring to be precise).
 > And there can't be such a proof because I've *proven* it's not
 > complete, as that definition leads to a contradiction.
In what way not complete?  What of the above axioms does it not
satisfy?  Or is it not closed under addition or multiplication?
 > Go to the following link, find and read my paper Advanced Polynomial
 > Factorization:
 >   http://groups.msn.com/AmateurMath
 > and you will see how you can prove that x has y as a factor.
I might do that when you do not put it behind some doors.
 > Consider c=ab, where 'c' is an algebraic integer, and 'a' is an
 > algebraic integer, but 'b' is not.
Yup, entirely possible.  So what?  So 'a' is not a factor of 'c'.
Your proof that for some reason 'a' must be a factor of 'c'
contained flaws.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
> 
> I'm putting this bit first to highlight it:
Note:  Keith Ramsay is pointing out that he has edited my post in
terms of placement of text, as he has moved a section.
 
> [...]
> |  Consider that in the ring of algebraic integers, 5 has algebraic
> | integer factors, and given algebraic integers r_1 and r_2, where their
> | product is 5, why are you acting as if it's so difficult to comprehend
> | that there must be some distribution of factors of 5?
> 
> I'm not acting as if I'm having any kind of difficulty. I'm saying
> that *you're* having difficulty in realizing that there must be some
> distribution of factors of 5 is a meaninglessly vague phrase. You're
> saying this as if to say, Shouldn't this be true?, and not as if
> you actually had proven it was true.
Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in
some way between r_1 and r_2.
For instance, if r_1 = r_2 = sqrt(5), you have the same factor of 5 in
both r_1 and r_2, while with r_1=5, r_2 = 1, you have all the non unit
factors of 5 with r_1.
It hardly seems like rocket science.
Can you explain why you as a mathematician are having difficulty in
the area Keith Ramsay?
> By the standard meanings of the terms, saying an algebraic integer r_1
> has a factor of 5 in the algebraic integers is just the same as
> saying it's divisible by 5 in the algebraic integers. That's the same
> as saying that there's an algebraic integer t for which r_1=5t.
Well I thought by standard usage a factor of 12 is 3.
Or I could say that a factor of 5 is 1+2i.
Are you saying that is nonstandard usage?
> There isn't, and I think you know there isn't. Since neither r_1 nor
> r_2 is divisible by 5, there's no distribution of factors of 5
> between them.
Well, if r_1 has a factor of 5 that is 1+2i, and r_2 has a factor of 5
that is 1-2i, then I'd say there it is clear there is a distribution
of factors of 5.
I've noted before how I talk of factors of and when I mean a
specific factor I say that is, for instance, a factor of 5 that is
1+2i.
> I think you use ill-defined terms because on some level you're aware
> that using only well-defined terms spoils your fun. It's disillusioning.
> It clear away the fog of obfuscation. Suddenly you're back to reality,
> and it's the rather ordinary reality that people keep telling you that
> you're in, not the amazing story you wish it was.
You're having problems Keith Ramsay with standard usage, as it looks
like you may be sticking in the word multiple where I use the word
factor.
So it's your fault, not mine.
> Back to the top.
> 
> | > [...]
> | > | What I can do is take out the use of the term in the paper, and use
> | > | factor in the ring of algebraic integers, and note that will 
lead to
> | > | a contradiction as the point of the paper is that the ring is
> | > | incomplete.
> | > |
> | > | What I want to impress upon readers is that I'm quite willing to 
work
> | > | with mathematicians to explain the mistake that they're teaching.
> |  | > If you want to show that a mistake is being taught, you should quote 
it
> | > specifically.
> |
> | That's not very sensible if you expect that for a mistake to be taught
> | for any length of time it'd have some subtlety, unless you're also
> | questioning the competence of mathematicians.
> 
> Quite to the contrary-- it's because mathematicians are generally
> competent that one can ordinarily say very specifically what is wrong
> when they make mistakes.
The mistake is in assuming that definition of the ring of algebraic
integers does not lead to contradictions.
> What makes a mathematical mistake subtle is not what it takes to point
> it out once it's been recognized; it's the difficulty in recognizing
> it in the first place.
You don't give a proof.  I contend that you can have a subtle math
mistake that is difficult to point out when it's been recognized, and
indeed it can also be difficult to recognize in the first place.
> | It seems to me that rather than deal with the important question,
> | which is whether or not I'm correct, people often try to introduce
> | weird ad hoc conditions.
> 
> You created an ad hoc rule that in order to refute a claim of yours,
> it's not good enough for someone else to prove that the conclusion is
> wrong. You think a proof of yours can show that a proof of a
> mathematician is wrong, but you appear unwilling to accept the
> possibility that a proof from a mathematician can just as well show
> that one of yours is wrong (i.e., not really a proof).
That's not true.  I've noted that proofs don't duel, which is that
math proofs don't contradict each other.
Therefore, you cannot claim that a proof of mine is wrong based on
another proof, if neither has an error, as they will NOT contradict.
Therefore, if a proof of mine is not a proof, then it must have an
error.
Several people have claimed that they have proofs that refute mine,
but I've also pointed out that there are other interpretations, even
assuming they have proofs.
I've also noted that it's odd that mathematicians would rather try to
duel with proofs than simply find an error in my proof.  But, of
course, as it is a proof, there is no error, which is why I think they
try to find other means.
That's not mathematics.  That's cheating.
> | If you disagree with that assessment, can you please explain why you
> | believe there should be something simple and short enough for me to
> | quote?
> 
> In mathematics, we try to make each *mathematical* statement have its
> own well-defined meaning. (There are of course nonmathematical claims
> that we sometimes make, which aren't always precise.) If a sequence of
> well-defined mathematical statements is wrong, it's wrong because (at
> least) one of the statements is wrong.
> 
> The key is making sure all the definitions are sound.
As I've stated before a proof begins with a truth and proceeds by
logical statements to a conclusion that then must be true.
Therefore it follows that definitions must be sound, or the steps will
not be logical.
> Mathematics that is vaguely enough written that it can be wrong
> without any individual statements in it being definitely wrong, is
> considered very bad. It's bad because an author hasn't defined their
> terms well enough.
The term bad is a human term inapplicable in context as mathematics
is about truth.
That is, a proof is neither bad nor good, it is true or false.
 
> | > | The ring of algebraic integers is incomplete, it's easy to show, 
and
> | > | I've shown it with my paper.
> | > |
> | > | If mathematicians are having trouble understanding any part of it, 
I'm
> | > | able to explain further.
> |  | > You have a big problem with terminology. You use terms that aren't
> | > used by others in the context in which you use them, such as counting
> | > factors of 5 in algebraic integers. I haven't seen anything I would
> | > consider an adequate definition of those terms. You don't explain
> | > why your terms should be considered relevant to the ones 
mathematicians
> | > are using.
> |
> | That statement possibly has something to do with what I now call the
> | preamble in the paper as it is there to help explain context, but
> | isn't part of the actual argument.
> |
> | Here's what I say, copied from the paper (some editing for format):
> |
> | To highlight the standard belief consider the algebraic integers
> | (-1-sqrt(21))/2 and (-1+sqrt(21))/2 which are roots of x^2 + x - 5.
> |
> | While you know that the algebraic integer factors are themselves
> | factors of 5, in what way is each a factor of 5? Can either not have
> | non unit factors of 5? How do you know?
> |
> | There I'm talking about algebraic integers, so one can assume I'm
> | talking about algebraic integer factors.  Given that 5 has algebraic
> | integer factors, how is what I say nonstandard Keith Ramsay?
> |
> | I'd like you to carefully explain your assertion.  The question I'm
> | raising is the possibility that you lied.
> 
> No, I haven't lied. You are not using have non unit factors of 5 in
> a standard way.
Can a number have non unit factors of 5 Keith Ramsay?
I simply applied a not to the front.
If you claim that a number cannot have non unit factors of 5, then I'd
like to know why.
Here your confusion may come from thinking that non unit factors of 5
must be 5.  However another non unit factor of 5 is 1+21.  Yes, 5 is a
factor of itself, but there is no reason to assume that the statement
non unit factors of 5 forces one of the the factors to be 5, though
it may be 5.
Now let's say I have something like 25, which has factors of 5. 
Notice that it *also* has 1+2i as a factor, while it also has 5*5 as a
factor.
You seem to be fixed on a more limiting case where factors of 5
means *multiples* of 5.
But because you read multiples when I say factors of you assume
that I'm wrong.
I use the definitions precisely, while you interpret.
Are you a math expert Keith Ramsay?
If so, why am I demonstrably more precise than you?
> If at this point in the paper, you meant it in a completely standard
> way, here's what the meaning would be. Have a factor of 5 is
> synonymous with have 5 as a factor, or be divisible by 5. Also 5
> is a non-unit (so that part is redundant).
You are incorrect.
 
> Now obviously we all know you don't mean that. It's possible you mean
> something like, shares non-unit factors with 5, meaning that there
> are non-unit algebraic integers that are factors of both numbers. But
> this is just guesswork. Any serious writer of mathematics takes
> responsibility for taking out the guesswork, NOT forcing the reader to
> keep inferring which of various possible meanings is the intended one.
Yet your problem apparently is in adding the term multiple which is
not my fault.
It's your failure Keith Ramsay.
> The standard usage is also inconsistent with the rest of the paper.
> You write, for example, ...proving that two of the a's have a factor
> that is f at a point where you plainly have NOT shown that f is a
> factor of any of the a's, in any of the usual ways factor is
> defined.
Hmmm...that's an interesting point.  What has happened at that point
in the paper is that I've considered the constant term P(0), and the
constant term with f^2 separated off, which is P(0)/f^2 = 3x u^2 + u^3
f, and noted that it is coprime to f.
Now I then consider g_1 at m=0, where c=g_1, and notice it has a
factor of f, and then based on P(0)/f^2 being coprime to f, I have
that r + c, must have a factor of f, which proves that r must have a
factor of f that is f.
> Elsewhere, you refer to an algebraic integer f coprime to x. This
> has no standard meaning in the context you use it, where x is a
> variable. If you meant coprime in some ring of polynomials, which is
> closest to a standard meaning, it would mean that there exists
> polynomials P and Q such that P*f+Q*x=1. But if f is an algebraic
> integer, that's possible (if and) only if f is a unit, with Q=0, and
> you also assume in the same sentence that f is a non unit.
But f while a variable is constant, and so is x.  So it is not in any
ring of polynomials.
Unfortunately, you seem to be fixated on the *letters* as in seeing an
x you may assume that it's varying.  Nope.  It's constant.
That's troubling Keith Ramsay as that's basic algebra.  The symbols
have to be understood as defined, not based on your experience of how
you may be used to seeing them.
 
> | > I don't know what you think is a fair way for things to work, but 
the
> | > way things actually work, this kind of problem with terminology will
> | > absolutely prevent you from making any headway.
> |
> | What will prevent me from making headway is if mathematicians
> | continually lie.
> 
> Mathematicians don't continually lie. What you keep deciding are
> lies are just disagreements with you.
That's not true.  What I've done is explain clearly and in detail.
In reply I find people making specious issues, like your claims about
factor when you apparently are sticking in multiple.
> Perhaps the biggest problem of all is that you've reached a point
> where you don't have anybody you are willing to trust, who could help
> you sort out which of the claims people are making are valid and which
> are baloney. So you tend automatically to assume that the things
> people say that seem wrong to you are baloney of some kind or another.
How can I take people like you seriously when you have trouble with
such simple things as factor of?
> This also leaves you without any very good way to learn how to improve
> your mathematical tinkering. We keep telling you different things
> which _we_ claim would help you avoid pitfalls like you keep falling
> into. But (a) you don't trust us enough to believe we've identified a
> problem with what you're doing, and (b) you don't trust us to give you
> straight advice on how to avoid it; you suspect us of just trying to
> waste your time. So you're stuck with only your own inklings of what's
> a good way to work with proofs.
Yet I've caught you in a contradiction with yourself, where you
apparently have been inserting the word multiple when I say
something like factor of 5, so that you read multiples of 5.  If
that's not what you're doing then, who knows what's going through that
brain of yours.
Readers should consider what follows if they think that's too harsh.
> | Now then in what way is it improper terminology to talk about
> | algebraic integer factors of 5?
> 
> I didn't say it was improper to talk about algebraic integer factors
> of 5. That has a perfectly well-defined meaning. An algebraic integer
> r is a factor of 5 if it divides 5 in the algebraic integers, meaning
> that 5/r is also an algebraic integer.
<deleted>
You're contradicting yourself.  Go back and read over what you said at
the top of this post.  I'm tiring of this exercise in pointing out
your errors.
James Harris
====
> Nope.  There is no claim that y is a factor as it's *provably* a
> factor.
*sigh*  Could you give me your definition of a factor, then?  I suspect
it's different from what mathematicians usually mean, because the way to
prove y is a factor of x is to find some z such that y*z=x, with suitable
properties for all three of them (i.e. all in the algebraic integers,
say).  If you haven't done this, how can you prove it's a factor?
> Rather than deal with the term factor as people get confused, in my
> paper Advanced Polynomial Factorization, I have three numbers a_1,
> a_2, and a_3, and I prove that a_3 is coprime to a factor I call f.
Argh, you profess to want to get away from the term 'factor', and then
you go and use it right again.  Please define your terms rigorously.
> Consider c=ab, where 'c' is an algebraic integer, and 'a' is an
> algebraic integer, but 'b' is not.
> 
> Now if you include fractions or move to a field like algebraic numbers
> that's ok.  But I'm talking about 'b' that's not in any way a fraction
> or fractional.
 It's like with the ring of evens, and given 6 = 2(3), you have that 3
> is outside the ring, as in the ring of evens, 2 is NOT a factor of 6.
> 
> The situation is analogous.
> 
> Do you understand?
No, I don't.  You seem to be operating under the assumption that all
are you claiming that the set of evens is incomplete, because 2
should be a factor of 6, but isn't?
How is your situation with the algebraic integers distinct from this,
if it is distinct?
====
> 
> 
> |
> I'm putting this bit first to highlight it:
> 
> Note:  Keith Ramsay is pointing out that he has edited my post 
> in terms of placement of text, as he has moved a section.
Note that James Harris has drawn attention to Keith Ramsay
Jim Burns
> [...]
> | Consider that in the ring of algebraic integers, 5 has 
> | algebraic integer factors, and given algebraic integers 
> | r_1 and r_2, where their product is 5, why are you acting 
> | as if it's so difficult to comprehend that there must be 
> | some distribution of factors of 5?
> I'm not acting as if I'm having any kind of difficulty. I'm 
> saying that *you're* having difficulty in realizing that 
> there must be some distribution of factors of 5 is a 
> meaninglessly vague phrase. You're saying this as if to say, 
> Shouldn't this be true?, and not as if you actually had 
> proven it was true.
[ etc. etc. etc.]
====
>> 
>> I'm putting this bit first to highlight it:
Note:  Keith Ramsay is pointing out that he has edited my post in
>terms of placement of text, as he has moved a section.
> 
>> [...]
>> |  Consider that in the ring of algebraic integers, 5 has algebraic
>> | integer factors, and given algebraic integers r_1 and r_2, where their
>> | product is 5, why are you acting as if it's so difficult to comprehend
>> | that there must be some distribution of factors of 5?
>> 
>> I'm not acting as if I'm having any kind of difficulty. I'm saying
>> that *you're* having difficulty in realizing that there must be some
>> distribution of factors of 5 is a meaninglessly vague phrase. You're
>> saying this as if to say, Shouldn't this be true?, and not as if
>> you actually had proven it was true.
Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in
>some way between r_1 and r_2.
It makes sense, but (assuming I know what you mean - yes you
_are_ using the language in nonstandard ways) it's simply
_not_ _true_.
Suppose that r_1 = 1 + 2i and r_2 = 1 - 2i. Now sqrt(5) is a factor
of 5. Does it go with r_1 or r_2?
A: Neither (again assuming go with means what I conjecture
it does). In the algebraic integers sqrt(5) is _not_ a factor of r_1
and it is also _not_ a factor of r_2.
Proof for r_1: The minimal polynomial of (1+2i)/sqrt(5) is
5x^4 + 6x^2 + 5. So (1+2i)/sqrt(5) is not an algebraic integer.
Proof for r_2: Exactly the same.
>For instance, if r_1 = r_2 = sqrt(5), you have the same factor of 5 in
>both r_1 and r_2, while with r_1=5, r_2 = 1, you have all the non unit
>factors of 5 with r_1.
It hardly seems like rocket science.
Can you explain why you as a mathematician are having difficulty in
>the area Keith Ramsay?
> By the standard meanings of the terms, saying an algebraic integer r_1
>> has a factor of 5 in the algebraic integers is just the same as
>> saying it's divisible by 5 in the algebraic integers. That's the same
>> as saying that there's an algebraic integer t for which r_1=5t.
Well I thought by standard usage a factor of 12 is 3.
Or I could say that a factor of 5 is 1+2i.
Are you saying that is nonstandard usage?
That's not the usage he's talking about. The problem is
this has a factor of thing. When you say a has a factor
of b that _means_ that b _is_ a factor of a. But when
you say it that's not what you mean, what you mean
is that there _is_ a factor of b which is also a factor of a.
>> There isn't, and I think you know there isn't. Since neither r_1 nor
>> r_2 is divisible by 5, there's no distribution of factors of 5
>> between them.
Well, if r_1 has a factor of 5 that is 1+2i, and r_2 has a factor of 5
>that is 1-2i, then I'd say there it is clear there is a distribution
>of factors of 5.
I've noted before how I talk of factors of and when I mean a
>specific factor I say that is, for instance, a factor of 5 that is
>1+2i.
And many other people have noted before that the way you
use these terms is simply wrong. Nobody knows why you
insist on speaking your own language - if you were actually
trying to communicate you'd use words in their standard
ways instead of requiring the rest of the world to remember
that when _you_ say a has a factor of b it means something
different from what it means when everyone else says it.
>> I think you use ill-defined terms because on some level you're aware
>> that using only well-defined terms spoils your fun. It's disillusioning.
>> It clear away the fog of obfuscation. Suddenly you're back to reality,
>> and it's the rather ordinary reality that people keep telling you that
>> you're in, not the amazing story you wish it was.
You're having problems Keith Ramsay with standard usage, as it looks
>like you may be sticking in the word multiple where I use the word
>factor.
So it's your fault, not mine.
No, his interpretation is _exactly_ what what you said _actually_
means. It's _your_ fault that you mean something else, not his.
>> Back to the top.
>> 
>>[...]
But because you read multiples when I say factors of you assume
>that I'm wrong.
I use the definitions precisely, while you interpret.
Just the opposite. Saying a has a factor of b _does_ mean
that a is a multiple of b. To say what you mean by the phrase
you should say instead a shares a factor with b or a has a
factor in common with b.
>Are you a math expert Keith Ramsay?
If so, why am I demonstrably more precise than you?
It's exactly like when you were being demonstrably more
precise than all the fools who didn't realize that integers
were irrational.
>> If at this point in the paper, you meant it in a completely standard
>> way, here's what the meaning would be. Have a factor of 5 is
>> synonymous with have 5 as a factor, or be divisible by 5. Also 5
>> is a non-unit (so that part is redundant).
You are incorrect.
> 
>> Now obviously we all know you don't mean that. It's possible you mean
>> something like, shares non-unit factors with 5, meaning that there
>> are non-unit algebraic integers that are factors of both numbers. But
>> this is just guesswork. Any serious writer of mathematics takes
>> responsibility for taking out the guesswork, NOT forcing the reader to
>> keep inferring which of various possible meanings is the intended one.
Yet your problem apparently is in adding the term multiple which is
>not my fault.
It's your failure Keith Ramsay.
>[...]
That's not true.  What I've done is explain clearly and in detail.
In reply I find people making specious issues, like your claims about
>factor when you apparently are sticking in multiple.
> [...]
Yet I've caught you in a contradiction with yourself, where you
>apparently have been inserting the word multiple when I say
>something like factor of 5, so that you read multiples of 5.  If
>that's not what you're doing then, who knows what's going through that
>brain of yours.
Readers should consider what follows if they think that's too harsh.
> | Now then in what way is it improper terminology to talk about
>> | algebraic integer factors of 5?
>> 
>> I didn't say it was improper to talk about algebraic integer factors
>> of 5. That has a perfectly well-defined meaning. An algebraic integer
>> r is a factor of 5 if it divides 5 in the algebraic integers, meaning
>> that 5/r is also an algebraic integer.
<deleted
>You're contradicting yourself. 
No he's not. It's not improper to say that an algebraic integer has
a factor of 5, and he never said it was improper. What _is_
improper is to say it has a factor of 5 _when_ that's not what
you actually _mean_.
> Go back and read over what you said at
>the top of this post.  I'm tiring of this exercise in pointing out
>your errors.

>James Harris
************************
David C. Ullrich
====
On Fri, 04 Jul 2003 11:33:25 -0400, Jim Burns <burns.87@osu.edu
>> 
>> 
>> |
>> I'm putting this bit first to highlight it:
>> 
>> Note:  Keith Ramsay is pointing out that he has edited my post 
>> in terms of placement of text, as he has moved a section.
Note that James Harris has drawn attention to Keith Ramsay
Say, I'm glad you pointed that out.
>Jim Burns
> [...]
>> | Consider that in the ring of algebraic integers, 5 has 
>> | algebraic integer factors, and given algebraic integers 
>> | r_1 and r_2, where their product is 5, why are you acting 
>> | as if it's so difficult to comprehend that there must be 
>> | some distribution of factors of 5?
>> I'm not acting as if I'm having any kind of difficulty. I'm 
>> saying that *you're* having difficulty in realizing that 
>> there must be some distribution of factors of 5 is a 
>> meaninglessly vague phrase. You're saying this as if to say, 
>> Shouldn't this be true?, and not as if you actually had 
>> proven it was true.
>[ etc. etc. etc.]
************************
David C. Ullrich
====
> 
>> 
>> I'm putting this bit first to highlight it:
>Note:  Keith Ramsay is pointing out that he has edited my post in
>terms of placement of text, as he has moved a section.
> 
>> [...]
>> |  Consider that in the ring of algebraic integers, 5 has algebraic
>> | integer factors, and given algebraic integers r_1 and r_2, where 
their
>> | product is 5, why are you acting as if it's so difficult to 
comprehend
>> | that there must be some distribution of factors of 5?
>> 
>> I'm not acting as if I'm having any kind of difficulty. I'm saying
>> that *you're* having difficulty in realizing that there must be some
>> distribution of factors of 5 is a meaninglessly vague phrase. You're
>> saying this as if to say, Shouldn't this be true?, and not as if
>> you actually had proven it was true.
>Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in
>some way between r_1 and r_2.
> 
> It makes sense, but (assuming I know what you mean - yes you
> _are_ using the language in nonstandard ways) it's simply
> _not_ _true_.
> 
> Suppose that r_1 = 1 + 2i and r_2 = 1 - 2i. Now sqrt(5) is a factor
> of 5. Does it go with r_1 or r_2?
<deleted>
See what I mean sci.skeptic?  The issue here is that I'm actually
using standard mathematical usage, while several posters apparently
keep reading factors of 5 as multiples of 5.
The problem has to do with their misuse of mathematical terminology.
For instance 2 is a factor of 6, but so is 3, and in fact, so is 6.
When you say, factor of, it means something that is a factor of the
given number.  And similarly factors of would mean factors of the
given number.
But you have these posters, like David Ullrich who is a math professor
at Oklahoma State University, who are lost with rather basic
mathematical language to the point that they *keep* arguing *after*
I've corrected them.  You see their *belief* system apparently is that
as mathematicians they can't be the ones with the error, so amazingly,
they simply keep almost mindlessly repeating it.
You may guess that they say factors of when they mean multiples
of, but I'm using the proper terminology, in the correct way.
So when I say factors of 5, I'm NOT saying multiples of 5.  For
instance, 2 and 3 are factors of 6.  So when I say factors of 6 it
doesn't mean multiples of 6.  And I emphasize that there is a
mathematical term multiples of which applies.
And in fact, using factors of when you mean multiples of while
common, is technically incorrect.
But people like Ramsay and Ullrich are unlikely to reply to *correct*
their mistakes because the society of sci.math lets people it
considers part of the society get away with the dumbest mistakes.
That's how mathematicians operate as demonstrated before your eyes.
And that's how they can have errors in their discipline for years, and
years because they seek to by definition have a society that is
perfect, when in fact, mathematicians are just people, and people make
mistakes.
James Harris