| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
02 Nov 2006 12:05:05 AM |
| Object: |
Factorization of Anal Explosions |
If you understand how rules create behavior in exclusionary rings then
it is easy to understand the coverage problem of the ring of algebraic
integers by considering what certain things proven in that ring mean
mathematically.
For instance, consider a number N which is proven to NOT be a unit in
the ring of algebraic integers.
What does that tell you?
Well, first a number is an algebraic integer by the rule that numbers
which are roots of monic polynomials with integer coefficients are
algebraic integers.
It can be further shown that to be a unit in the ring of algebraic
integers a number must be the root of a monic polynomial with integer
coefficients and a last coefficient of 1 or -1.
So there is ANOTHER deeper rule created by the first parent rule, which
is that to be a unit in the ring of algebraic integers certain
additional conditions must be met.
Then what do you know about a number N that is proven to not be a unit
in the ring of algebraic integers?
You know that it is NOT the root of a monic polynomial with integer
coefficients where the last coefficient is 1 or -1, and nothing else.
That's all you get from determining that a number N is not a unit in
the ring of algebraic integers, as you have the exclusion, based on the
rule, so by the rule the exclusion is the result of N not fitting
necessary conditions.
BUT, so what? Who cares if N is not the root of a monic polynomial
with integer coefficients and a last coefficient of 1 or -1? Does that
indicate anything else mathematically?
Provably, it does not.
I like an analogy where to start consider
x^2 + 3x + 2 = (x+2)(x+1)
where you have 2 and 1 paired, so you have a unit with a non-unit and
can remain in the ring of integers.
But BY THE EXCLUSIONARY RULES with the ring of algebraic integers you
cannot have a non-rational non-unit paired with unit if you have a
monic polynomial with integer coefficients!!!
So a parallel to a simple case in the ring of integers is excluded from
the ring of algebraic integers.
But so what? Maybe non-rationals are just different?
Maybe, but, you can prove that they are not and that the exclusion has
no algebraic meaning but is just a quirk of a special exclusionary
definition.
Importantly, abstracting out key properties of rings like the ring of
integers and the ring of algebraic integers I found I could focus on
two key conditions to define a ring which I call the object 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.
And those conditions do not exclude a parallel to
x^2 + 3x + 2 = (x+2)(x+1)
with non-rationals.
May not seem a big deal, but anywhere you see a mathematical argument
where the conclusion rests on proving that some number is not a unit,
then it has a very limited scope in terms of what has actually been
found.
But it gets worse.
Dedekind in working out information about the ring of algebraic
integers worked quite hard at trying to prove it does not have the
coverage problem that it DOES have.
He found he couldn't prove what he wanted, but finally he used a fairly
new approach and thought he finally got it.
That approach was the theory of ideals.
James Harris
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| User: "=?ISO-8859-1?Q?Jos=E9_Carlos_Santos?=" |
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| Title: Re: Factorization of Anal Explosions |
02 Nov 2006 03:23:00 AM |
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On 02-11-2006 6:05, wrote:
If you understand how rules create behavior in exclusionary rings then
it is easy to understand the coverage problem of the ring of algebraic
integers by considering what certain things proven in that ring mean
mathematically.
To whom it may concern: this post is just a re-post of this one:
http://groups.google.com/group/alt.math.undergrad/msg/e94f432b8748de4d
Of course, it wasn't really James who posted it for the second time.
Best regards,
Jose Carlos Santos
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