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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://alpha.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?)

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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

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    $\begingroup$ Well, that's perfect, thanks. It's amazing to me that this paper is so recent. $\endgroup$ Commented Dec 11, 2009 at 20:11
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    $\begingroup$ Yes, one would have thought that all standard theorems that invoke Zorn's Lemma would have been "done" long ago. $\endgroup$ Commented Dec 11, 2009 at 20:35
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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.

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Let me give an answer from a different perspective. Konrad Swanepoel's accepted answer shows what happens in the general case, for formally real fields of any cardinality. However, it is possible to consider only countable formally real fields, in the setting of second-order arithmetic (i.e. reverse mathematics). The definition of a formally real field makes sense in this setting since the condition of being formally real only quantifies over finite objects, namely finite sums of field elements (in fact it is a $\Sigma^0_1$ condition).

The statement that every countable formally real field is orderable is equivalent over $\mathsf{RCA}_0$ to weak König's lemma, the statement that every infinite tree $T \subseteq 2^{<\mathbb{N}}$ has an infinite path. This is a result of Friedman, Simpson, and Smith (1983), 'Countable algebra and set existence axioms', Annals of Pure and Applied Logic 25(2):141–181. A textbook treatment of this and other results concerning countable formally real and ordered fields can be found in sections II.9 and IV.4 of Simpson's book Subsystems of Second Order Arithmetic.

Another result of Friedman, Simpson, and Smith is that the existence and uniqueness of the real closure of a countable ordered field is provable in $\mathsf{RCA}_0$. The proof draws on Tarski's result that the theory of real closed fields admits elimination of quantifiers. By the previous equivalence between WKL and the orderability of countable formally real fields, the statement that every countable formally real field has a real closure is thus also equivalent to WKL over $\mathsf{RCA}_0$.

As noted in the paper of Berr, Delon, and Schmid (1999) linked in Konrad Swanepoel's answer, Lombardi and Roy (1991) and Sander (1991) independently showed that the existence and uniqueness of the real closure of an ordered field is provable in ZF without the axiom of choice. The results for countable fields are thus analogous to those for fields of arbitrary cardinality, with weak König's lemma playing the role of the ultrafilter lemma and the base theory $\mathsf{RCA}_0$ playing the role of ZF. (This analogy is not preserved in general: the classification of equivalences over $\mathsf{RCA}_0$ for countable structures is quite different to the one over ZF for structures of arbitrary cardinality.)

(Apologies for posting an answer on such an old and previously answered question, but I saw it brought to the front page by an edit today…)

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    $\begingroup$ No apologies necessary, also because all these old questions regularly turn up in Google search results. Every addition is valuable in my opinion. $\endgroup$ Commented Oct 8, 2023 at 23:49
  • $\begingroup$ It might be worth pointing out that WKL is, in a sense, an SOA version of UF. In ZF the Ultrafilter Lemma is equivalent to the statement that $2^X$ is compact in the product topology for all $X$. WKL is exactly the statement that $2^\omega$ is compact in the product topology (in ZF this is just provable). $\endgroup$
    – Asaf Karagila
    Commented Nov 10, 2023 at 16:54

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