From: ah170@FreeNet.Carleton.CA (David Libert)
Subject: Re: existence of an algebraic closure of a field
Date: 3 Jun 1999 03:05:55 GMT
Newsgroups: sci.math
Keywords: dependence on Axiom of Choice
Luc Bourhis (luc.bourhis@durham.ac.uk) writes:
> What conditions a field K must satisfy in order to have an algebraic
> closure L ? (L contains K and any algebraic equation of degree n in L
> has n solutions in L : I precise this only because I did not learn
> mathematic in english and I am not sure to use the correct terms)
> --
> Luc Bourhis
>
> Center for Particle Theory
> University of Durham, UK
With the Axiom of Choice every field has an algebraic closure.
Furthermore it is unique up to isomorphism in the following sense. If F
is a field and i1: F -> A1 & i2: F -> A2 are two algebraic
closures in the sense that i1 and i2 are homeomorphisms of fields and
A1 and A2 are algebraically closed and A1 and A2 are alebraic
extensions of the respective images of F under i1 and i2, then there
exists an isomorphism of fields g making the following diagram commute:
g
A1 ------> A2 The proof of these claims can be done by a Zorn's
Lemma argument. Consider all algebraic field
^ ^ extensions of F having the same cardinality as F.
| / Consider all inclusions of these as a poset. Argue
| i1 / that a union of an increasing chain of algebraic
| / i2 extensions is an algebraic extension. By Zorn's
| / Lemma take a maximal such field extension of F, A.
| / If A is not algebraically closed consider a proper
| / algebraic extension of A by one root of some polynomial
| / over A. This contradicts the maximality of A.
| /
F For the uniqueness claim, given i1, i2, A1, A2 as above
consider A'1 A'2 g' for various choices of A'1 and
A'2 respective subfields of A1 and A2, and superfields of the F images of
i1 and i2 respectively. Consider i1 : F -> A'1 as acting by the obvious
restriction. Consider only those isomorphisms g' : A1' -> A2' which
make the corresponding diagram commute, using the notion for i1, i2 just
indicated. Order triples A'1, A'2, g' by inclusion. Argue a union of
a chain is in this set. By Zorn's Lemma find a maximal element A''1,
A''2, g''. If A''1 is a proper subset of A1 then consider the extension
of A''1 by an single element, this is an algebraic extension (since
contained in A1 which was an algebraic extension of i1 image of F), so
g'' can be lifted to an isomorphism of a corresponding algebraic
extension of A''2 inside A2 (since A2 is algebraically closed). This
contradicts maximality.
There is no way to get alternate proofs of the above claims using no
AC. Jech's book _The Axiom of Choice_ has two very nice independence
results relating to each of these. If ZFC is consistent then there
exists a model of M1 ZF with a element F1 in the model where M1 satisfies
"F1 is a field having no algebraic closure".
Also if ZFC is consistent there exists a model M2 of ZF with an element
F2 of the model so M2 satisfies "F2 is a field having at least two
algebraic closures not isomorphic in the sense above".
Jech has some other similar independence results relating to vector
spaces. In ZFC every vector space has a basis. Also in ZFC all bases
of a vector space have the same cardinality (ie dimension is well
defined).
In ZF alone, if a vector space V has at least one finite basis then all
bases have that same cardinality. But to get uniqueness of cardinality
in the case of infinite bases the ZFC proof uses AC.
These two ZFC results about existence and uniqueness up to cardinality
of bases can be done by Zorn's Lemma arguents, similar to those above.
Jech has two independence results in his book relating to these, namely
a ZF model with a vector space having no basis, and another ZF model with
a vector space having at least two bases of distinct cardinality.
These four independence results in Jech's book are not due to Jech.
Also related there is a stronger result, that easily implies the
independence result about no basis. As I indicated above, ZF proves
AC -> every vector space has a basis. Jech in his book cites without
proof a result on Andreas Blass:
ZF proves: (every vector space has a basis) -> AC.
Yet another reason to believe in AC I say.
--
David Libert (ah170@freenet.carleton.ca)
1. I used to be conceited but now I am perfect.
2. "So self-quoting doesn't seem so bad." -- David Libert
3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig