Re: JSH: Understanding some crucial math

Rotwang wrote:
James Harris wrote:


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.


What kind of functions are f(x) and g(x)? Are they eg. polynomials, continuous, analytic, or are they any functions with domain and codomain the algebraic integers?


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

is the only constraint so they can be any functions that can fit.



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.


"Not sufficient" for what purpose?


I step through some basic algebra, which cannot work in the ring of
algebraic integers.

So if you start with

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

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

in the ring of algebraic integers, you can't get to

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

and still have a_1(x) and a_2(x) be algebraic integers, if f(x) and
g(x) are not rational for a rational algebraic integer x.

Weird result, so I'll point out that for rational algebraic integer
values of x, that give rational values for f(x) and g(x), there is no
problem and you can do it all in the ring of algebraic integers.


But now let a_1(x) = g(x) + 6 or a_1(x) = 7*f(x)
[...]
which gives 7*P(x) = (a_1(x) + 7)(a_2(x) + 7)


I think you mean g(x)-6


Yup. My mistake.

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


I'm afraid I don't understand this at all. What is necessary? Do you mean that the definitions you give for the a's are the only ones that will make them algebraic integers?


No. I'm showing that you cannot do certain basic algebra and remain in
the ring of algebraic integers.

So starting with

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

where g(0) = f(0) = 0, doing some basic things pushes you out of the
ring, showing that the ring of algebraic integers just cannot do the
full job.

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!!!


I don't understand why this is the case... can you explain in more detail please?


It's the argument that supposedly counters my paper on this subject
that has been put forward for years now by posters like W. Dale Hall,
*** Winter and others.

That argument which so many posters have gone on and on about, shows
the result.

Note that what I normally do is start at the end with something like

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

and work back to show that you're pushed out of the ring.

This time I just started at the other end.

The supposed counterexample arguments of Hall, Winter and others now
can be used at this point to show the weird result.


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:


What is the flaw? Exactly what established belief about the algebraic integers is contradicted by what you have written?


Well, you can't do some basic algebra and stay in the ring.


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.


This definition makes no sense. You start by saying that the ring consists of all numbers such that conditions 1 and 2 hold, but the conditions given are properties of rings, not single numbers. By way of analogy, suppose I present you with a bag of wooden bricks of various shapes and colours, and I define a set of bricks as follows:


The set S of bricks consists of all bricks such that the following condition holds:

1. All bricks are the same colour.


This definition makes no sense. One can modify it so that it does in several ways; one way would be to say that S is the union of all sets T of bricks such that all elements of T are the same colour, however this definition has the problem that S no longer has the property that all its elements are the same colour (in fact S consists of ALL bricks). Another way would be to define S to be a maximal set of bricks which all have the same colour, however this definition has the property that it is not unique (for example the set Sr of all red bricks and the set St of all yellow bricks both satisfy the definition).

The object ring contains all the numbers you can fit into it that don't
contradict those properties.

For instance, 1/2 can't fit, because 2*(1/2) = 1, making 2 a unit, when
-1 and 1 are the only rationals that are units in the ring.

So it's an obvious definition that is easily shown to work.

It is an exclusionary definition, blocking out numbers that contradict
with those properties, but including any that do not.

What is so complicated about that?


James Harris

.