Re: Some number theory

From: W. Dale Hall (mailtowd-hall_at_pacbell.net)
Date: 09/24/04 

Date: Fri, 24 Sep 2004 05:30:04 GMT

James Harris wrote:
> It turns out that my work on advanced polynomial factorization
> techniques can actually be seen to work, but that's something that
> often gets lost in all the arguing!
>
> Besides I usually end up hearing at least one person claim a single
> example doesn't mean anything!
>
> Consider the advanced factorization:
>
> (v^3 + 1)x^3 - 3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)
>
> with v = -1 + mf^2, and for now let y=1, m = 1 and f = sqrt(2),
>
> which gives
>
> v = -1 + (1)(2) = 1,
>
> 2x^3 - 3x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
>
> which *does* factor over Q.
>

As in

        2x^3 - 3x + 1 = (x - 1)(2x^2 + 2x - 1)

So what?

The a's would be the roots of the equation

        x^3 + 3x^2 - 2 = (x + 1)(x^2 + 2x - 2)

> Notice that in that case exactly two of the a's have sqrt(2) as a
> factor, while one is coprime to 2, in the ring of algebraic integers.
>

So what? Virtually everyone has told you that there is no necessary
relation among roots of non-irreducible polynomials.

> That ring of algebraic integers is quirky, but the algebra is not.
>
> You see, iff the cubic is irreducible over Q then you can't see what's
> happening directly, like you can above, and in THAT case, the problem
> with the ring of algebraic integers kicks in.
>
> Notice my example actually isn't fully reducible into integer roots,
> but still it goes far enough for you to see what's happening because
> one of the roots is 1.
>
> Neat.
>
> The algebra applies without regard to the ring, or reducibility, which
> should tell the more mathematically astute of you something:
>

Oh, really? You mean you could do this if the ring had positive
characteristic, or were not commutative, or had zero divisors up the
wazoo, or failed to have unique factorization?

> If I were wrong, then a result with reducibles could be found to prove
> me wrong directly without any room for argument.
>

Look. You've been proven wrong plenty of times. As far as "a result
with reducibles could be found to prove [you] wrong", WTF is that
supposed to mean?

> What my work shows is what happens for a family of polynomials that
> fit into
>
> (v^3 + 1)x^3 - 3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)
>
> and the techniques are advanced because they're new to mathematicians
> (amazingly enough) and because of what they allow you to do.
>
> Basically I found a powerful analysis tool.
>
> Look at the full paper and see how simple it is
>
> http://www.ne-plus-ultra.net/index.php?option=content&task=view&id=46&Itemid=26
>
> and then ask yourself, why fight algebra?
>

Indeed. Note, for example, your so-called "Primary Argument", that
concludes with this claim:

        Therefore, with the factorization

                65 x^3 - 12 x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)

        one of the a's is coprime to 5.

Understanding you to use "coprime" to mean "coprime in the ring of
algebraic integers", you'll have no problem addressing the following
factorization.

Why fight algebra?

Hell, why fight simple arithmetic?

I have, on several occasions, requested that you point out which
of the following steps (1-5 below) is invalid:

The correctness of all of the following statements is sufficient
to prove that none of the a_i is coprime to 5 in the ring of algebraic
integers, so for your argument to stand, at least one of the following
must fail.

Here is an outline of the argument; if you could just point out which
of the statements relies on the alleged false assumption then you will
assist in furthering this discussion.

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.

I claim that not one of these individual points depends on the
assumption you say I've made. The above constitutes the entire
argument and all commentary should be restricted to the argument,
so let's see where your objection lies.

>
> James Harris

I'm not going to give up on getting this answered. Every time I see you
put this bogus claim on sci.math or any other newsgroup that I frequent,
you can expect to see this or a similar message.

I'll note that not one person (except you, who is the only one here
who is craving recognition) has found an error in the above outline.
It can be verified by using simple arithmetic, and if you have access
to a computer algebra (or even a flexible computer arithmetic) routine,
it's pretty darned simple to do.

If it's wrong, then YOU can find out that it's wrong, and I've given
you all the ammunition you need: just find the incorrect statement.
You can take my word for it that if the implication that I say is true
is in fact true, then at least one of the statements (1-5 above) will
turn out to be incorrect.

Feel free to consult any sources of expertise that you trust; I'm in no
hurry here, and certainly don't mind having anyone look over my shoulder
on this work. Use any help you want.

Feel free to accept any of the above statements provisionally, with the
right to rescind your acceptance. I'm not interested in tricking you or
in setting up any sort of trap. I don't care for you to say you accept
what you don't really understand, and if you decide you don't quite
believe any of the steps, I don't think you need to be held to some
contractual obligation to believe what you don't really believe.

I trust that you will behave honestly; I know, for instance, that if
you understand each step, you will see that it is true. I also am
willing to credit you with the ability to follow the logic that
says that the truth of the full set of statements 1-5 above actually
implies that each a_i is not coprime to 5.

I don't know why you're so frightened. After all, if you were as
confident as you pretend, you'd step right up to the plate and have
a swing at this slow lob straight over the center.

On the other hand, I guess I do know why you're so frightened,
after all.

Dale.

Relevant Pages

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