Monday, September 11, 2006

 

Problem with algebraic integers

The results on prime numbers are telling but nothing compared to the sheer reach of my most controversial result, which shows a problem with what are called algebraic integers, as in, they can't do the full job.

Algebraic integers are kind of simple in that they are roots of monic polynomials with integer coefficients.

But now I'm going to step through a very short bit of algebra:

I will use a polynomial P(x), functions f(x), g(x), a_1(x) and a_2(x), where x is an algebraic integer.

Now let

P(x) = 175x^2 - 15x + 2 and consider the factorization

P(x) = (f(x) + 1)*(g(x) + 2)

where f(0) = g(0) = 0.

Now multiply by 7, so I have

7*P(x) = 7*(f(x) + 1)*(g(x) + 2)

and there are an infinite number of ways that you can distribute 7 through the factors on the right side, but we don't have time to check them all, so let me just pick one, and have

7*P(x) = (7*f(x) + 7)*(g(x) + 2).

But now let

a_1(x) = g(x) - 5 or a_1(x) = 7*f(x)

and

a_2(x) = 7*f(x) or a_2(x) = g(x) - 5

which gives

7*P(x) = (a_1(x) + 7)(a_2(x) + 7)

which is necessary for the a's to be algebraic integers if f(x) and g(x) are not rational.

And, get this, if a_1(x) and a_2(x) are non-rational algebraic integers—that is, given an algebraic integer x you get an algebraic integer result—then f(x) and g(x) cannot be algebraic integers!!!

That is the crucial point.

Remember f(0) = g(0) = 0.

Seem minor? Well, that result is big enough for a math journal to die in a fight over a paper in this area because if you pull the thread, so to speak, that is, consider the necessary mathematical conclusions that follow, then ideal theory can't work.

Run yourselves around on that one for a while, and hopefully some of you will understand the flaw with current views on the ring of algebraic integers and why my object ring is necessary:

The object ring is defined by two conditions, and includes all numbers such that these conditions are true:
  1. 1 and -1 are the only rationals that are units in the ring.
  2. Given a member m of the ring there must exist a non-zero member n such that mn is an integer, and if mn is not a factor of m, then n cannot be a unit in the ring.
Faced with the inability of the ring of algebraic integers to cover all the ground necessary, I figured out the key properties needed for a
ring that does.





<< Home

This page is powered by Blogger. Isn't yours?