# How much choice is needed to show that formally real fields can be ordered?

Background: a field is formally real if -1 is not a sum of squares of elements in that field. An ordering on a field is a linear ordering which is (in exactly the sense that you would guess if you haven't seen this before) compatible with the field operations.

It is immediate to see that a field which can be ordered is formally real. The converse is a famous result of Artin-Schreier. (For a graceful exposition, see Jacobson's Basic Algebra. For a not particularly graceful exposition which is freely available online, see http://math.uga.edu/~pete/realspectrum.pdf.)

The proof is neither long nor difficult, but it appeals to Zorn's Lemma. One suspects that the reliance on the Axiom of Choice is crucial, because a field which is formally real can have many different orderings (loc. cit. gives a brief introduction to the real spectrum of a field, the set of all orderings endowed with a certain topology making it a compact, totally disconnected topological space).

Can someone give a reference or an argument that AC is required in the technical sense (i.e., there are models of ZF in which it is false)? Does assuming that formally real fields can be ordered recover some weak version of AC, e.g. that any Boolean algebra has a prime ideal? (Or, what seems less likely to me, is it equivalent to AC?)

This is equivalent (in ZF) to the Boolean Prime Ideal Theorem (which is equivalent to the Ultrafilter Lemma).

Reference: R. Berr, F. Delon, J. Schmid, Ordered ﬁelds and the ultraﬁlter theorem, Fund Math 159 (1999), 231-241. online

• Well, that's perfect, thanks. It's amazing to me that this paper is so recent. – Pete L. Clark Dec 11 '09 at 20:11
• Yes, one would have thought that all standard theorems that invoke Zorn's Lemma would have been "done" long ago. – Konrad Swanepoel Dec 11 '09 at 20:35

At the very least, it implies the following:

Let $f: A \to B$ be a map of sets where every fiber has cardinality $2$. Then there is a section of $f$.

Proof: Let $V$ be the $\mathbb{R}$-vector space spanned by the elements of $A$, subject to the relation $a_1 = - a_2$ whenever $f(a_1)=f(a_2)$. Let $S$ be $\mathrm{Sym}(V)$ and $K$ be $\mathrm{Frac}(S)$.

Then $K$ is formally real (any proposed sum of squares only uses finitely many elements of $A$, so we can reduce to the case of a finite dimensional vector space, where this is obvious). But to choose an ordering, we need to decide which element of each fiber of $f$ will be positive and which will be negative. The positive elements form a section.