Re: JSH: Resolution now possible
From: Brian Smith (brianscsmith_at_yahoo.com)
Date: 10/19/04
- Next message: Brad Cooper: "Q on mapping in complex plane"
- Previous message: NHL: "problem from coddington and levinson"
- In reply to: James Harris: "JSH: Resolution now possible"
- Next in thread: Jesse F. Hughes: "Re: JSH: Resolution now possible"
- Messages sorted by: [ date ] [ thread ]
Date: 18 Oct 2004 19:54:17 -0700
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0410170613.507bf41f@posting.google.com>...
> I have outlined a complete ring I call the object ring based on
> two primary requirements:
>
> a. No rational unit other than 1 or -1 is in the ring.
>
> b. No non-unit member of the ring is a factor of any two integers
> that are coprime in the ring of integers.
James, I took a look through the Google archives and all your
arguments about algebraic integers and factorization seem to boil down
to this:
Start with the equation
P = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f)
and factor it with respect to x into three terms as follows:
P = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
The conjecture is that there is some ring R (the 'object ring') in
which
for all m and f that exactly two of the a_i are multiples of f and
the
other is coprime to f.
I will call this the 'Harris Conjecture'.
I have determined that 3 must be a unit in your ring as follows:
By setting the two expressions for P equal to each other, the
following facts about a_1, a_2 and a_3 can be found:
a_1*a_2*a_3 = m^3*f^6 - 3m^2*f^4 + 3m*f^2
a_1*a_2 + a_1*a_3 + a_2*a_3 = 0
a_1 + a_2 + a_3 = 3 - 3mf^2
A monic equation in x whose solutions are the a_i's is as follows:
A = (x - a_1)*(x - a_2)*(x - a_3) = 0
A = x^3 - (a_1+a_2+a_3)x^2 + (a_1*a_2+a_1*a_3+a_2*a_3)x -
(a_1*a_2*a_3) = 0
A = x^3 - (3 - 3mf^2)x^2 + (0)x - (m^3*f^6 - 3m^2*f^4 + 3m*f^2) = 0
Now consider a greatest common factor of a_1, a_2, a_3, lets call it
g. g is a unit if the set {a_1, a_2, a_3} has no non-unit common
factor or g is not a unit if there is a nontrivial common factor.
Let a_1=g*b_1, a_2=g*b_2 and a_3=g*b_3 then A can be written as:
A = (x - g*b_1)(x - g*b_2)(x - g*b_3) = 0
A = x^3 - g*(b_1+b_2+b_3)*x^2 + g^2*(b_1*b_2+b_1*b_3+b_2*b_3)*x -
g^3*(b_1*b_2*b_3) = 0
We can set the coefficients equal to each other to get:
3(mf^2 - 1) = g*(b_1 + b_2 + b_3)
0 = g^2*(b_1*b_2 + b_1*b_3 + b_2*b_3)
mf^2*(m^2f^4 - mf^2 + 1) = g^3*(b_1*b_2*b_3)
Let h be the greatest common factor 3 and f. Then g is a multiple of
h^(2/3) since h^(2/3) can be factored out of 3, (h^(2/3))^3=h^2 can be
factored out of f^2, and trivially (h^(2/3))^2 can be factored out of
0. The largest value of h is 3. By largest I mean that all possible
values of h are factors 3.
If h is not a unit in ring R then h^(2/3) is also not a unit and
contradicts the Harris Conjecture since all the a_i's would then have
a non-unit factor of f, that factor being h^(2/3). Therefore 3 must
be a unit in ring R for the Harris Conjecture to be true.
- Next message: Brad Cooper: "Q on mapping in complex plane"
- Previous message: NHL: "problem from coddington and levinson"
- In reply to: James Harris: "JSH: Resolution now possible"
- Next in thread: Jesse F. Hughes: "Re: JSH: Resolution now possible"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
