Re: New paper, algebraic integers, Galois Theory

From: W. Dale Hall (mailtowd-hall_at_pacbell.net)
Date: 10/13/04 

Date: Wed, 13 Oct 2004 16:08:27 GMT

James Harris wrote:
> "W. Dale Hall" <mailtowd-hall@pacbell.net> wrote in message news:<t7Tad.2522$6q2.149@newssvr14.news.prodigy.com>...
>
>>James Harris wrote:
>>
>>>"*** T. Winter" <***.Winter@cwi.nl> wrote in message news:<I5G8yE.G0n@cwi.nl>...
>>>
>>>
>>>>In article <3c65f87.0410111713.4decc512@posting.google.com> jstevh@msn.com (James Harris) writes:
>>>>
>>>>>"*** T. Winter" <***.Winter@cwi.nl> wrote in message news:<I5EF41.GtG@cwi.nl>...
>>
>> ...
>>
>>>>>> > would, supposedly, each have to share non-unit factors with 2 and 3.
>>>>>>
>>>>>>Define:
>>>>>>r1 = (-1 + sqrt(-23))/2 and r2 = (-1 - sqrt(-23))/2 (the roots),
>>>>>>p1 = (3 + sqrt(-23))/2 and p2 = (3 - sqrt(-23))/2,
>>>>>>q1 = (-2 - sqrt(-23)) and q2 = (-2 + sqrt(-23)).
>>>>>>
>>>>>>You can verify the following:
>>>>>> p1 * q1 = r1^3
>>>>>> p2 * q2 = r2^3
>>>>>> p1 * p2 = 8
>>>>>> q1 * q2 = 27
>>>>>>So the common factor between r1 and 2 is p1^(1/3), and between r1 and 3
>>>>>>it is q1^(1/3).
>>>>>
>>>>>Yeah, but how do you know that q2 doesn't have 27 itself as a factor?
>>>>
>>>>Perhaps because q2 is a root of x^2 + 4x + 27, and so q2^2 + 4.q2 + 27 = 0?
>>>
>>>
>>>So? It has *two* roots, so how do you know that one doesn't have have
>>>27 as a factor?
>>>
>>
>>You're suggesting that it is possible that one of the roots of the
>>quadratic equation
>>
>> x^2 + x + 27 = 0
>>
>>is a multiple of 27.
>>
>>Note that the product of the roots q1,q2 is
>>
>> q1 q2 = 27.
>>
>>Now, if q1 is divisible by 27, we would have
>>
>> q1 = 27 r
>>
>>where r is an algebraic integer. Let's find a polynomial
>>that r must satisfy.
>>
>>Suppose q1 = 27 r. Since q1 satisfies the equation
>>
>> x^2 + x + 27 = 0,
>>
>>we can substitute
>>
>> x = 27 y
>>
>>to obtain an equation satisfied by r:
>>
>> (27 y)^2 + (27 y) + 27 = 0
>>
>>That is,
>>
>> 27^2 y^2 + 27 y + 27 = 0
>>
>>or, dividing both sides by 27:
>>
>> 27 y^2 + y + 1 = 0.
>>
>>
>>What do we conclude? If q is a root of the equation
>>
>> x^2 + x + 1 = 0,
>>
>>and we suppose q = 27 r, then r is a root of the
>>equation
>>
>> 27 x^2 + x + 1 = 0.
>>
>>Thus, r cannot be an algebraic integer.
>>
>>
>
>
> So? Why does it matter if r is or is not an algebraic integer?
>
> Why do *you* think it matters?
>
> Notice, I'm NOT DISAGREEING with the conclusion that r is not an algebraic integer.
>
> I'm just asking, a simple question, why does it matter?
>
> It's a direct question which I've already asked more than once in this thread.
>
>
> James Harris

By now, you should realize that discussions of divisibility and
existence of common factors requires the context of a ring. In the
present discussion, the natural domain to be considered is the ring
of algebraic integers, since that is where Galois Theory shows one
can find the common factors whose existence you're apparently disputing.

You have maintained that the ring of algebraic integers is somehow
flawed, and that making use of that ring constitutes some sort of error.

Let's retrace the discussion:

Here's what you said on 10/10/04:

        The basic claim is that *each* of the roots of a monic
        polynomial with integer coefficients that is irreducible
        over rationals must share non-unit factors with ALL of
        the prime factors of the last coefficient.

        For instance, with P(x) = x^2 + x + 6, the claim would be
        that the two roots:

                (-1 + sqrt(32))/2 and (-1 - sqrt(32))/2

        would, supposedly, each have to share non-unit factors with 2
        and 3.

        My works shows that it's possible that actually neither does
        and you have to check using advanced polynomial factorization
        techniques.

The natural location where one would make such a claim (and the place
where Galois Theory applies to *prove* the existence of the non-unit
factors) is the ring of algebraic integers. It is in that ring where
your final statement is shown to be in error, once the correct formula
for the roots is found, as I wrote:

        So the roots are

                x = (-1 ± sqrt(-23))/2

And I went on, in that article (also on 10/10/04), to find common
factors between each of the roots and the numbers 2 and 3.

However, you decided to act as though such common factors "may not"
exist, in this exchange with *** Winter (who also found the common
factors), from this article of yours on 10/11/04

>> would, supposedly, each have to share non-unit factors with
>> 2 and 3.
>>
>> Define:
>> r1 = (-1 + sqrt(-23))/2 and
>> r2 = (-1 - sqrt(-23))/2 (the roots),
>> p1 = (3 + sqrt(-23))/2 and p2 = (3 - sqrt(-23))/2,
>> q1 = (-2 - sqrt(-23)) and q2 = (-2 + sqrt(-23)).
>>
>> You can verify the following:
>> p1 * q1 = r13
>> p2 * q2 = r23
>> p1 * p2 = 8
>> q1 * q2 = 27
>> So the common factor between r1 and 2 is p1^(1/3), and
>> between r1 and 3
>> it is q1^(1/3).

        Yeah, but how do you know that q2 doesn't have 27 itself as a
        factor?

What is clear from the context is that the common factors between the
roots and the numbers 2 and 3 are available in the ring of algebraic
integers. Remember, you are ostensibly taking exception to the position
from Galois Theory, predicting a non-unit common factor between each
root of a monic, irreducible polynomial over Z, and each prime factor
of the constant term of that polynomial. Since the factors are *proven*
to exist in the ring of algebraic integers, *that* is where the argument
belongs. Extending the ring beyond that of the algebraic integers, does
*not* refute that position. Finding an extension ring in which common
factor is a unit does *not* refute that position. The *only* way to
refute the following statement:

        Given a monic polynomial P with integer coefficients,
        irreducible over the rationals, a root r of P, and
        a prime factor q of the constant term of P, there
        is a non-unit algebraic integer t which divides both
        r and q, in that there are non-unit algebraic integers
        s,t,u for which

                r = s t
                q = t u.

Is to find such a combination of P, r, and q, for which the
common factor (here, it's t) fails to exist *in the ring
of algebraic integers*.

So, to repeat the question:

> So? Why does it matter if r is or is not an algebraic integer?
>
> Why do *you* think it matters?
>
> Notice, I'm NOT DISAGREEING with the conclusion that r is not an
algebraic integer.
>
> I'm just asking, a simple question, why does it matter?
>
> It's a direct question which I've already asked more than once in
this thread.

That's why. The common factors are shown to exist in the ring of
algebraic integers, that's what one naturally assumes what you're
disputing, and thus your question leads one to assume that that's
what you're talking about.

If you want to change the domain under consideration, it is not proper
to do it without saying so. Mathematical discussion is above all about
the communication of mathematical content and mathematical argument.
It is most definitely *not* about switching context or meaning of
terms behind one's back. When I write an argument, I make a special
effort to make sure it is *clear* what I am talking about, and what
I mean about every term. My argument is intended to be clear on these
matters, when read by someone familiar with mathematical terminology.

Your insistence on going outside the ring of algebraic integers
is silly, when the discussion has focussed on the existence of
quantities within that ring, whose existence you have disputed.

Here's an analogy:

Someone says there are seven days of the week, and lists
them: Sunday, Monday, Tuesday, Wednesday, ..., Saturday.

You respond by saying, "What about Chocolate Sundae?"

It is non-responsive to the argument.

Dale.

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