Re: JSH: Critique means slow, and thorough

From: William Hughes (wpihughes_at_hotmail.com)
Date: 03/29/05 

Date: 29 Mar 2005 06:06:48 -0800

jstevh@msn.com wrote:

              <snip>
>
> Consider a ring where the following two properties hold:
>
> 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.
>

As mentioned before, if an algebraic number m is in a ring R containing
the integers, then there exists n in R such that mn is an integer.

As mentioned before, if n is a unit, then mn is a factor of m.

Therefore, as mentioned before, condition 2 is not needed.

                          - "William Hughes"

> Those properties hold with the ring of integers, the ring of gaussian
> integers, and the ring of algebraic integers.
>
> But, it cannot be shown that ANY of those rings exhaust all
> possibilities.
>
> That is, maybe there is some other ring, where those properties hold
> that has members that are not integer, gaussian integers, nor
algebraic
> integers.
>
> If so, then that means that you might have an algebraic number x/y
> where x and y are members of that ring, but not algebraic integers.
>
> So consider what follows from the ring of algebraic numbers, which is
> also a field, as every algebraic number can be written as a ratio of
> coprime algebraic integers so I can show what happens in detail, so
in
> algebraic numbers I can have
>
> a/b
>
> where a and b are algebraic integers, so you have
>
> a/b = x/y
>
> where in the ring where x and y are members, you may also have
>
> cx/cy = a/b
>
> where cx and cy are algebraic integers, while x and y are not
algebraic
> integers.
>
> Then coprimeness in the ring of algebraic integers does not mean
> coprimeness in the more inclusive ring.
>
> But if you don't realize that possibility, and worse, assume that
> you've included all rings where the two key properties hold, you can
> have an odd thing, where you can prove "coprimeness" by relying on
> coprimeness in the ring of algebraic integers, and algebraically find
> that two numbers are not coprime, and thus have the appearance of
> proving two different and opposite things.
>
> A simple analogy that I've given before is to consider 6 and 2 in the
> ring of evens.
>
> Imagine someone doesn't know about 1 or odds, and they have 6/2
coprime
> in the ring of evens and declare that 2 is coprime to 6, in general,
> but, of course, in the ring of integers, 6/2 = 3, so they are wrong.
>
> Notice the problem isn't coprimeness in the ring of evens here, as it
> exists and 2 and 6 are in fact coprime in that ring, but the
assertion
> of what you can call global coprimeness, where in any ring where the
> two key properties hold, 6 and 2 are coprime.
>
> Does that make sense?
>
> Can you see why having a bigger ring that includes the ring of
> algebraic integers, but has members outside of it where those key
> properties hold, would cause problems?
>
>
> James Harris

Relevant Pages

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