Re: JSH: Understanding some crucial math

James Harris wrote:


P(x) = (f(x) + 1)*(g(x) + 1) is the only constraint so they can be any functions
that can fit.


OK


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


Try f(x)=sqrt(x) and g(x)=-sqrt(x), so P(x)=1-x. Take x=2, so that x is rational but f(x) and g(x) aren't. Then a_1(x)=7sqrt(2) and a_2(x)=-sqrt(2)-6, which are both algebraic integers. Is there something wrong with this counterexample?


[...]doing some basic things pushes you out of the ring,


What does this mean? Do you mean that the set of algebraic integers are not closed under addition, multiplication or additive inverses (i.e. that it is not a ring)? If so, then can you provide an explicit example of algebraic integers whose sum, product or additive inverses are not algebraic integers?


showing that the ring of algebraic integers just cannot do the
full job.


What "job" is it that they cannot do?

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.


Can you give a citation please?


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.


I would like you to be more specific please... tell me which algebraic operations push you out of the ring, and give me a reference to some existing literature which claims that they shouldn't.


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.


OK, so you are defining it to be a maximal ring that satisfies conditions 1 and 2. Then how do you show that it is unique?


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


Let's go back to my brick analogy. Suppost that you start with the empty set, then you can add a red brick, and condition 1 (which was that all elements of S be the same colour) is satisfied. After that you may add all the remaining red bricks and condition 1 is still satisfied, but adding any yellow bricks will violate condition 1. However, suppose that you start with the empty set and add a yellow brick: again condition 1 is satisfied, and you can add all the remaining yellow bricks, but no red ones. Therefore the definition of S as a maximal set satisfying condition 1 is not unique. How can you guarantee that your object ring definition does not run into the same problem?
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