Re: Algebraic integer mistake, basic

From: KRamsay (kramsay_at_aol.com)
Date: 06/21/04 

Date: 21 Jun 2004 06:22:05 GMT

In article <cb4san$rna$1@agate.berkeley.edu>, magidin@math.berkeley.edu (Arturo
Magidin) writes:
|>Since the algebraic numbers are countable, the axiom of choice
|>is not necessary.
|
|Okay, I'm intrigued...
|
|Does that mean that Zorn's Lemma for subsets of a countable set is a
|theorem of ZF? Or that there are ways of showing the existence of such
|maximal subrings of a countable ring without having to appeal to any
|version of Zorn's Lemma?

I can't see offhand whether Zorn's lemma for a family of subsets of
a countable set is a theorem of ZF. I'd like to know too.

In this case, however, we have a slightly special case, where the
family F is closed under intersection.

Let the algebraic numbers A be {a0, a1, a2, ...}. Define a nested
family of subsets Y_n of A by letting Y_n = the minimal element of
F containing Y_{n-1} and x_n if there is one (i.e., the ring generated
by Y_{n-1} and x_n if it satisfies the condition), and Y_n = Y_{n-1}
otherwise. Denote by Y the union of the Y_n.

No element of F properly contains Y, because if Y' were an element
of F properly containing Y, then there would be some a_n in Y'-Y,
but this contradicts the fact that a_n was not included in Y_n.
The set Y is in F because it's the union of a chain of elements of
F and isn't properly contained in any element of F. Hence it's also
a maximal element of F.

Keith Ramsay

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