Re: Finishing theorem, algebraic integers

From: David Kastrup (dak_at_gnu.org)
Date: 10/01/04 

Date: Fri, 01 Oct 2004 14:50:11 +0200

jstevh@msn.com (James Harris) writes:

> Bill Dubuque <wgd@nestle.csail.mit.edu> wrote in message news:<y8zacv79pln.fsf@nestle.csail.mit.edu>...
>> James Harris <jstevh@msn.com> wrote:
>> >
>> > In the ring of algebraic integers, given a monic polynomial with
>> > integer coefficients that is irreducible over rationals that has a
>> > non-unit last coefficient C that is a composite natural number,
>> > given a non-unit root A of the polynomial it is IMPOSSIBLE to find a
>> > non-unit algebraic integer factor of A that is coprime to any prime
>> > factor p of C, if any coefficient besides the leading one is coprime to p
>>
>> SIMPLE COUNTEREXAMPLE X^2 - 3 X + 6
>>
>> \3 divides X via 3 | X^2 = 3 X - 6
>>
>> \3 coprime to p = 2 via \3 \3 - 2 = 1
>>
>> non-leading coef -3 is coprime to p = 2
>>
>> --Bill Dubuque
>
> Yup. Oh, here's the fixed one.
>
> In the ring of algebraic integers, given a monic polynomial with
> integer coefficients that is irreducible over rationals that has a
> non-unit last coefficient C that is a composite natural number, given
> a non-unit root A of the polynomial it is IMPOSSIBLE to find a
> non-unit algebraic integer factor of A that is coprime to any prime
> factor p of C, if any coefficient besides the leading one is coprime
> to C.
>
> Note that what I'm showing is a weird failure of decomposability in
> the ring of algebraic integers.

Uh, you are showing nothing. You are wildly claiming at the moment
without showing anything. 

-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
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