Harristotelian Logic

I want to step quickly through some very basic algebra to help people undestand an important result.

I will try to keep the operation within the ring of algebraic integers.

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. The point of this exercise is to show how you can find that the ring of algebraic integers is NOT sufficient, even with very basic algebra.

Now let

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

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

Now multiply by 7, so I have

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

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) + 1).

But now let

a_1(x) = g(x) + 6 or a_1(x) = 7*f(x)

and

a_2(x) = 7*f(x) or a_2(x) = g(x) + 6

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.

Notice at this point, you have P(x) is a polynomial, and I didn't even need the condition that it have integer coefficients.

Mathematicians who think they can show me wrong here need only demonstrate a counter-example.

So what is the dramatic conclusion?

Remember, I started with

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

so the conclusion from everything above is that if f(x) and g(x) are NOT rational, then they cannot be algebraic integers if a_1(x) and a_2(x) are, so you CANNOT STAY IN THE RING OF ALGEBRAIC INTEGERS.

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

Freaky, eh? I'm wondering if I need P(x) to have integer coefficients but I don't think I do.

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.