Newsgroups: sci.math
Subject: Re: Algebraic characterization of reals
From: mann@vms.huji.ac.il (Avinoam Mann)
Date: 16 Mar 1995 13:43 IDT

The other day, I've posted an 'answer' to the title problem. My thanks to Tim
Chow for pointing out that this answer was totally wrong. Indeed it has been
pointed out in this thread already that there are 'zillions' of non-isomorphic
real closed fields whose algebraic closure is isomorphic to the complex numbers
(proof below). What I think is a correct characterisation is by the three
properties below. First, let me recall that a field F is real-closed if, first,
for each non-zero element x either x or -x, but not both, is a square, and,
next, each polynomial of odd degree over F has a root in F. Such a field has
characteristic 0, and adding to it the square root of any non-square yields an
algebraically closed field. Moreover, if F is any field whose alg. closure is
of finite dimension over it, then F is alg. closed or real closed.
Now the real field can be characterised by the following:
1. It's real closed.
2. If we order F by declaring each square to be positive, the ordering is
archimedean. This means that given any x in F, there exists an integer n
such that n - x is a square in F.
3. F has no proper extension satisfying 1. and 2.
An alternative to 3. is
3.' F is Dedekind complete (wrt the ordering in 2.).

Now for the calculation of the number of real closed fields in C (the complex
numbers). Choose a transcendence basis B for the reals. Then B has the power c
of the continuum. Let A be a subset of B of the same power, and let F be the
alg. closure of Q(A) in R (Q - rationals; R - reals). Then F is real closed,
its alg. closure is an alg. closed field of char. 0 and power c, hence this
alg. closure is isomorphic to C, so F is isomorphic to a field of codimension
2 in C. Suppose F and K are two isomorphic real closed subfields of R. The
isomorphism between them must fix all rationals. Also squares go to squares,
so the ordering is preserved. Now real closed fields have a unique ordering
(by the assumption that x or -x is square), so their ordering must be the one
induced by their being subsets of R. But each real number is determined by
the set of rationals below it, so the isomorphism from F to K send each
element of F to itself, and thus F = K. It follows that the number of non-
isomorphic such fields is at least 2¬c, and it cannot be more, this being
the number of all subsets of C.
Since every field of char. 0 and power c is isomorphic to a subfield of C,
it follows that 2¬c is the number of all isomorphism types of real closed
fields of power c.