Re: Some math, algebraic integers
From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 09/18/04
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Date: Sat, 18 Sep 2004 23:15:53 +0000 (UTC)
In article <1zU2d.17114$QJ3.13457@newssvr21.news.prodigy.com>,
W. Dale Hall <mailtowd-hall@pacbell.net> wrote:
[.snip.]
>Can I find an extension of Q(z), for which there *is* a common, non-unit
>integral factor of q and 2? Here's where I'm well out of my depth on the
>topic (and I would invite anyone who is better versed than I am to chime
>in at this point).
Yes, you can.
> I'll give it a blind stab:
>
>Let's take the quadratic extension K of Q(z) that makes the ring of
>integers O_K into a PID (principal ideal domain: every ideal is
>generated by a single element .
I think this is not always possible; it is possible to have a
number field K with ring of integers A such that (i) A is not a PID;
(ii) For every finite extension L or K, the ring of integers of L is
not a PID. (A famous result of, I think, Borevic and Shafaraveich on
infinite towers of Hilbert Class Fields). However, it may be possible
when the original extension is a quadratic extension, for all I know,
or at least when the class number is 2.
However, the following two things are always possible:
(i) Way too powerful for what we want here: Given a number field K,
with ring of integers A, there is always a finite extension L
of K, with ring of integers R, in which every extension of an
ideal of A is principal. That is, for every ideal I of A, the
ideal IR of R is principal. This is the Hilbert Class Field of
K.
(ii) Just right for what is needed here: given a number field K,
with ring of integers A, and an ideal I of A, there exists a
finite extension L of K with ring of integers R in which IR is
principal.
(ii) is a consequence of the oft-mentioned finiteness of the class
number. Since the ideal generated by q and 2 is not the trivial ideal,
but is not principal, then by the finiteness of the class number we
must have that the ideal (q,2)^2 is principal, say (q,2)^2 = (w). Then
extend K by taking w^{1/2}, and then (w^{1/2})=(q,2) so w^{1/2} is the
desired element.
This is basically what you are doing. The question of whether w^{1/2}
can be a unit is of course relevant, but easy to deal with: since then
w^2 would itself be a unit, which would mean that (q,2)^2 is the
trivial ideal. By Unique Factorization of ideals, that would mean that
(q,2) itself is trivial, which means q and 2 are coprime, which
contradicts our original conclusion that they were not (despite having
no nonunit common factors).
In any number field K, any ideal of its ring of integers has a finite
power which is principal. So the same argument holds: there is a
finite extension of K where any two elements that do not generate the
trivial ideal have a non-unit common factor. In particular, in the
ring of all algebraic integers, any two elements which have no
non-unit common factor are in fact coprime (the converse always
holds).
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
- Next message: Andrzej Kolowski: "Re: My paper, and the cheaters"
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