Re: Automorphisms of complex numbers

From: G. A. Edgar (edgar_at_math.ohio-state.edu.invalid)
Date: 07/24/04


Date: Sat, 24 Jul 2004 08:35:36 -0400

In article <cds69r$40i$1@daisy.noc.ucla.edu>, H. Enderton <hbe@sonia.math.ucla.edu> wrote:

> Tim Mellor <timm@amsta.leeds.ac.uk> wrote:
> >What is the cardinality of the automorphism group of the complex
> >numbers, C?
> >I say 2^c, where c:=|C|
>
> Right, you get full marks for this. I assume that you mean the
> structure (C; 0, 1, +, x), i.e., the complex field. This field is
> built up by starting with the rational field as its prime subfield.
> Then we add c transcendentals. Then we take the algebraic closure.
>
> Any permutation of those c transcendentals (the Hamel basis)
> generates an automorphism of the field. So we get 2^c automorphisms.
>
> But only *two* of those automorphisms are continuous, viz. the
> identity and the complex conjugates.
>

And those are the only two whose existence can be proved in ZF.

-- 
G. A. Edgar                               http://www.math.ohio-state.edu/~edgar/


Relevant Pages

  • Re: Automorphisms of complex numbers
    ... you get full marks for this. ... > built up by starting with the rational field as its prime subfield. ... > Any permutation of those c transcendentals ... > generates an automorphism of the field. ...
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  • Re: Automorphisms of complex numbers
    ... you get full marks for this. ... built up by starting with the rational field as its prime subfield. ... Any permutation of those c transcendentals ... generates an automorphism of the field. ...
    (sci.math)
  • Re: Automorphisms of complex numbers
    ... you get full marks for this. ... built up by starting with the rational field as its prime subfield. ... Any permutation of those c transcendentals ... generates an automorphism of the field. ...
    (sci.logic)