Re: Basic answer, and refutation, end of arguments

From: *** T. Winter (***.Winter_at_cwi.nl)
Date: 10/01/04 

Date: Fri, 1 Oct 2004 12:33:39 GMT

In article <3c65f87.0409301418.6921f79a@posting.google.com> jstevh@msn.com (James Harris) writes:
> Now consider x^2 + x + 6.
>
> It's has the roots
>
> x = (-1 + sqrt(-23))/2
>
> and
>
> x = (-1 - sqrt(-23))/2
>
> and, intriguingly enough, no non-unit algebraic integer can be found
> that is a factor of either one of those roots that is coprime to 2 or
> 3.
>
> It's a stunning and argument killing result as no mathematician on the
> planet can deliver the factor, and if challenged they may do odd
> things.

Nonsens. Let's make some definitions:
    r1 = (-1 + sqrt(-23))/2; r2 = (-1 - sqrt(-23))/2;
    p1 = (-3 - sqrt(-23))/2; p2 = (-3 + sqrt(-23))/2;
    q1 = ( 2 + sqrt(-23)) ; q2 = ( 2 - sqrt(-23));
The following is easily verified:
    p1 * p2 = 2^3;
    q1 * q2 = 3^3;
    p1 * q1 = r1^3;
    p2 * q2 = r2^3;
    p1 and p2 coprime to q1 and q2;
and also all of p1, p2, q1, q2, r1 and r2 are algebraic integers.
So the factors you are seeking are:
    r1 = ( p1^(1/3) )( q1^(1/3) );
    r2 = ( p2^(1/3) )( q2^(1/3) );
where
    p1^(1/3) and p2^(1/3) are coprime to 3 and
    q1^(1/3) and q2^(1/3) are coprime to 2.

> The full result--more formally so that there's no room for
> confusion--is that in the ring of algebraic integers, given a non-unit
> irrational algebraic integer A, that is a factor of a composite
> natural C it is IMPOSSIBLE to find a non-unit algebraic integer factor
> of A that is coprime to any prime factor of C.
>
> To refute, someone needs to just demonstrate the contrary, by
> delivering a factor.

See above.

> At least some one of you should apologize to me.

The other way around, methinks. 

-- 
*** t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~***/
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