From: "G. A. Edgar" Subject: Re: Field automorphisms of C - counterexample details needed. Date: Sun, 19 Dec 1999 13:18:50 -0500 Newsgroups: sci.math Keywords: Using axiom of choice to construct field automorphisms of C In article <945617937.18008.0.nnrp-09.9e98a2e3@news.demon.co.uk>, Scotty wrote: > OK, but I'm still not convinced that such an automorphism exists, when I > came across a reference to it in Pertti Lounesto's book I was astounded. It > contradicts a much cherished 'theorem' of mine that field theory was either > boring or trivial. So if I assume the Axiom of Choice how exactly do I prove > the existence of the automophism? kfoster explans this. An automorphism on Q[r,i] that he builds can be extended to Q[r,i] with c independent inteterminates adjoined, then extended to the algebraic closure. Any algebraically closed field of characteristic zero of power c is isomorphic to C (from the Axiom of Choice), so we are done. In <945564646.21525.0.nnrp-11.9e98a2e3@news.demon.co.uk>, Scotty said: > and I'm pretty sure > you can't extend to all of C without [invoking AC] That argument goes like this: Any automorphism of C to itself that is discontinuous fails the property of Baire [inverse image of every open set has the property of Baire, i.e. differs from an open set by a first category set], but Solovay constructs a model of ZF where every set has the property of Baire. So in that model, our automorphism is impossible. -- Gerald A. Edgar edgar@math.ohio-state.edu ============================================================================== From: Kurt Foster Subject: Re: Field automorphisms of C - counterexample details needed. Date: Sun, 19 Dec 1999 16:57:08 GMT Newsgroups: sci.math In <945564646.21525.0.nnrp-11.9e98a2e3@news.demon.co.uk>, Scotty said: . I'm looking for a proof of the following: . There exists a field automorphism of C sending sqrt(2) to -sqrt(2) and . sqrt(sqrt(2)) to i sqrt(sqrt(2)). Also any automorphism which is . neither the identify or complex conjugation permutes an infinity of . subfields all isomorphic to R. If r > 0 is real, and r^4 = 2, then any field automorphism of a field containing Q(r, i), sending r to i*r, will send r*r to (i*r)(i*r), i.e. will send sqrt(2) to -sqrt(2) automatically. You can look at the Galois group G the (Eisenstein) irreducible polynomial p(x) = x^4 - 2, which (I think) is the dihedral group with 8 elements. This is the group of Q-automorphisms of the field K obtained by adjoining the roots of p(x) to Q. It is a transitive subgroup of S_4, so there is an element that sends r to i*r. However, none of the automorphisms in G (apart from the identity and complex conjugation) is defined a priori outside of the field K. You can view them as restrictions of Q-automorphisms of finite normal extensions of Q which contain K; but there are many such fields, defining different extensions of the Q-automorphisms of K. I doubt you can even extend to automorphisms of G to the field of algebraic numbers (a countable extension of Q) without invoking the Axiom of Choice; and I'm pretty sure you can't extend to all of C without doing so.