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