Re: The actual math, advanced polynomial factorization
From: W. Dale Hall (mailtowd-hall_at_pacbell.net)
Date: 09/26/04
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Date: Sun, 26 Sep 2004 09:18:24 GMT
James Harris wrote:
> The following factorization is useful:
>
> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> Note that x, m, f, and u are all independent variables. It's a lot of
> variables, yes, but they're independent of each other.
>
> Also note that a_1, a_2 and a_3 are dependent variables.
>
> Now divide both sides by f^2 to get
>
>
> (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2
>
> and all the arguing for over TWO YEARS NOW has been over what that
> means.
>
> Let m=0, and you get
>
>
> (0*f^4 - 3* 0*f^2 + 3*0) x^3 - 3(-1 + 0*f^2 )x u^2 + u^3 f =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2
>
> which is
>
> -3(-1)x u^2 + u^3 f = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2
>
> which is
>
> u^2(3x + uf) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2
>
> and it's clear that two of the a's have to equal 0, while one equals
> 3.
>
> But that means that the f^2 divides through two of the factors
>
> (a_1 x + uf), (a_2 x + uf), or (a_3 x + uf)
>
> and, remember, x, m, f and u are all independent variables.
>
> Therefore, two of the factors have f as a factor in general without
> regard to the value of m, but it turns out that IN THE RING OF
> ALGEBRAIC INTEGERS they do not always have f as a factor. Notice that
> "not always", it's important.
>
> Some posters rather than accept that x, m, f and u are independent
> variables so their value doesn't matter to each other, claim my result
> only holds if m=0, as a special case, but if variables are independent
> of each other, how can a particular value be a special case?
>
> So why fight the result? Well Evariste Galois came up with the theory
> for how you can *represent* roots of irrational polynomials.
>
> What I've shown is an analysis tool for actually figuring out factors
> with irrationals, but no one had tools of this power before, and with
> the results of Galois they made a leap: They decided that the work of
> Galois was not just about representation, but also about actual
> factors.
>
What you say is not true.
Galois theory does indeed talk about factors, but you have to understand
what Galois theory actually says; it helps to know a little algebra, but
your grasp of algebra is insufficient.
For instance, Galois theory says that, given any two roots r and s of an
irreducible polynomial P(x) over a field F, for any arithmetic statement
in the elements of F(r) (The field F with r attached) that is true, must
also be true with r replaced by s. This forms a general symmetry
principle that applies for roots of polymomials.
Note that it is *NOT* just a statement about representing roots. For
instance, as an example of an arithmetic statement in F(r), one
could imagine finding a factorization (Q,R,S: polynomials in F)
Q(r) = R(r)S(r)
then one would be able to deduce that this factorization is also
valid:
Q(s) = R(s)S(s).
Aside: Actually, these facts don't really require the full
power of Galois theory. They are a result of the fact
that the field F(r) is canonically isomorphic to the
quotient of the polynomial ring F[x] by the ideal
enerated by P(x).
However you might note that this quotient, written:
F[x]/<P(x)>
makes no mention of the root r at all. In this model
of F(r), it is the class of x modulo the ideal P(x)
that serves as our abstracted quantity r.
In considering the other root s, we also have this:
F(s) is canonically isomorphic to
F[x]/<P(x)>,
with the class of x modulo <P(x)> being
identified with the root s.
As a corollary, one finds that F(r) is canonically
isomorphic to F(s), with r corresponding to s under
the canonical isomorphism.
Now, suppose the polynomial in question has integer coefficients, and
is monic. Then if w is any factor of r in the ring of algebraic
integers, that fact can be expressed via a true arithmetic statement
involving only the value r and the ring of integers. By the general
symmetry principle I mentioned in the preceding paragraph, this
same statement with r replaced by s will also be true.
The result is that the number s will *necessarily* be divisible by
an algebraic integer w'. The two numbers w and w' are not totally
independent of each other, for they have the same irreducible
polynomial (that is, for any polynomial P(x) having P(w) = 0, we
will also have P(w') = 0). In other words, if you find *any*
algebraic integer that divides one root r of an irreducible monic
polynomial P(x), you'll find an algebraic integer w' *with the same
irreducible polynomial* that divides any other root s of P(x).
It goes even farther than that, in fact. If this factor w is also
a factor of an ordinary integer n in the ring of algebraic integers,
w' will *also* be a factor of n in the ring of algebraic integers.
> Now the algebra supporting my case is simple enough, but I've been
> arguing for over two years on the same points and posters have
> routinely rejected basic algebra, like that independent variables are
> independent, while making up their own kind of weird mathematics I've
> at times called voodoo mathematics.
>
There is no doubt in my mind that you deny what I've written above.
I'll make you a deal: pose a factorization that you wish to claim
causes the above general principle (w divides r <==> w' divides s)
or the minimal extension (w divides r and n <==> w' divides s and n)
to fail. If it fails, then I'm wrong and will retract the above
statements. If it passes, you may wish to try again. And again.
And again.
My claim is that it won't fail, because I claim the symmetry
principle stated above is a valid method.
> Let me show you how it works. Like they'll say there exists algebraic
> integer functions w_1(m), w_2(m), and w_3(m), where
>
> w_1(m) w_2(m) w_3(m) = f, and
>
> a_1 = b_1 w_1(m), a_2 = b_2 w_2(m), and a_3 = b_3 w_3(m)
>
> where the b's are algebraic integers.
>
> Now with
>
> (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2
>
> you then have
>
> (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f =
>
> (b_1 x + u w_2(m)w_3(m))(b_2 + uw_1(m)w_3(m))(b_3 x + uw_1(m)w_2(m))
>
> where all might seem well and good, like why can't you just have these
> functions that somehow relate INDEPENDENT VARIABLES but I've said
> that before and the sci.math'ers make fun of me.
>
> There's just one small little problem with that setup, f^2 was
> multiplied times both sides before, but what if I play a trick on the
> voodoo math?
>
> Let's say that rather than multiply both sides by f^2, I multiply
> times 13^2?
>
> You see, the point of have some constant multiple like f^2 is that if
> you divide it off it's gone, like if I have
>
> 4(x^2 + 3x + 2) = 4(x+1)(x+2)
>
> and divide 4 off, it's gone. But not in the voodoo math of sci.math
> posters...if the factorization is irrational. They always come back
> to say, yeah, if you're talking about polynomial factors like in my
> simple example, then yeah, that's how constant multiples behave!!!
>
> It gets weirder. My result holds for a family of polynomials.
>
And just what is that result? Please state the result that
you're claiming. So far, I see many references to an
alleged result, but no statement of what it is.
> To see that let's make some adjustments:
>
> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)
>
> and multiplying through by f^2 gives
>
> (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x u^2 f^2 + u^3 f^3
>
> where notice in the last two terms you have u and f paired, so let
> y=uf, so you get
>
>
> (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x y^2 + y^3
>
> and the factorization now is
>
> (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x y^2 + y^3 =
>
> (a_1 x + y)(a_2 x + y)(a_3 x + y)
>
> and my result from before still applies because the a's are not
> dependent on the value of y, while still having the same values as
> before, so now I can change y, as yet another independent variable, so
> let's put some numbers in.
>
> Let f=sqrt(2), m=1, y=1, which gives
>
> (4 - 3(2) + 3)(2)x^3 - 3(-1 + 2)x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x
> + 1)
>
> which is
>
> 2x^3 - 3x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
>
> and you can just look at it and see that x=1, works as a root of the
> polynomial 2x^3 - 3x + 1.
>
> That leaves room for only two of the a's to have sqrt(2) as a factor,
> and in this case you can see it work IN the ring of algebraic
> integers.
>
> Notice that if you wish you can try your own numbers.
>
As in the polynomial factorization:
65 x^3 - 12 x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
for which you maintain that one of the a_i is coprime to 5 in the ring
of algebraic integers?
You know that each of the a_i's has a common non-unit factor with 5,
in the ring of algebraic integers. You have seen the proof, which
involves only ordinary arithmetic.
How can any one of them be coprime to 5 in that ring?
Answer that one.
No, you won't, will you? You continue to sell the same pile
of irrelevant junk, over and over again, but instead of producing
a cogent argument, you fill each and every article of yours with
the following sort of nonsense:
> Mathematically the results are always the same without regard to how
> reducible over rationals the primary cubic may be.
>
> That's how math works.
Problem is, your so-called "math" doesn't even work. All you can put
together are toy problems in which the numbers that are supposed to
be coprime to something are the roots of an irreducible polynomial,
and the other root is not related to that polynomial.
Your claims that it doesn't matter, well, they don't matter. All roots
of an irreducible polynomial are indistinguishable using only arithmetic
in the field of coefficients together with the root in question. Any
relation held in that domain by one root will hold for any other, but
in the analogous domain (i.e., the field of coefficients together with
the new root).
>
> If Galois Theory wasn't such a big deal there wouldn't have been all
> the arguing, which has gone on for over two years now.
>
This sentence doesn't mean anything, does it?
> So, mathematically, my case is solid and absolute, but math society is
> protecting its wrong beliefs about roots of polynomials irreducible
> over Q, and to do that the people engaging in that activity cannot be
> real mathematicians.
Not.
>
> They're social animals. A real mathematician would just accept the
> math as correct and disdain the earlier false assumption as it's just
> wrong.
>
A real mathematician would stop this dancing around and answer the
simple question I've asked for months: why, when you know that each
a_i shares a non-unit algebraic integer factor with 5, in the ring
of algebraic integers, do you continue to claim that one of the a_i
is coprime to 5?
Do you maintain that the ring of algebraic integers is so wacky that
now the ordinary integers no longer work? After all, my proof only
involves ordinary arithmetic.
> Mathematicians don't defend wrong beliefs about mathematics, as they
> actually love mathematics itself.
>
I am not defending any belief. I'm asking a simple question: why do
you claim to be correct, when ordinary arithmetic shows you that you
are just wrong?
> So, if you fight a correct result, to hold on to a false belief, you
> prove you're not really a mathematician, no matter what people call
> you or you call yourself.
Do you claim that my result, which involves only the multiplication of
ordinary integer polynomials, and the application of the simple fact
that the a_i's satisfy a specific polynomial equation of degree 3, is
not correct? My result says that each a_i has a non-unit factor in
the ring of algebraic integers, which is also a factor of 5 in the
same ring. You claim that one of the a_i's is coprime to 5 in that
same ring. The two assertions are patently incompatible, being in
direct contradiction of each other, and my result is backed up by
the five steps that only depend on one using ordinary arithmetic,
and logically entail the result. Your assertion is backed up by
loosely-worded argument, in which nothing is asserted to belong
to the ring of algebraic integers, and nothing actually proven,
and by your continuing libelous* remarks concerning the honesty
of an entire profession (*they would indeed be libelous if there
were the slightest possibility that anyone could believe a word
you write).
If my result or my logic are not correct, what step is incorrect:
In this outline, "a" refers to one of the coefficients that you claim
must be coprime to 5. P(x) refers to the polynomial x^3 - 12 x^2 + 65,
and I note that P(-a) = 0, for each of your values "a".
1. The following formulas are true:
q(x)r(x) = (64 x + 128)P(x) + 5
r(x)s(x) = (32 x + 72)P(x) + x,
where q,r,s are defined as follows:
q(x) = 8 x^2 - 76 x - 185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104
2. Since -a is a root of P(x), we have the following
factorizations:
q(-a)r(-a) = 5
r(-a)s(-a) = -a
3. The minimal polynomial of r(-a) is given as:
MP_r = x^3 - 969 x^2 + 315 x + 5
which is irreducible over Q. This shows that
r(-a) is an algebraic integer, but not a unit
in that ring.
4. From 2 and 3 above, the common factor r(-a)
shared by a and 5 is not a unit in the ring
of algebraic integers.
5. a and 5 are not coprime in the ring of algebraic
integers, since they share a non-unit factor in
common.
>
> That's the problem I think with the modern math world as somehow the
> belief has taken hold that you have a social position, not a calling,
> and that as long as society calls you a mathematician, then you are,
> no matter how false mathematically what you believe is.
>
The problem that I think is that you are so filled with the love of
your own failed attempts at making something of your life that you
cannot change.
If you claim I am stating any mathematical falsehood, you'll stand up
and find the error. The statements (1-5) of my argument are there for
everyone to see, and if you cannot find error in them, then by your
own standards, you must accede to their truth.
>
> James Harris
Dale.
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