# Formally real fields with unique non-Archimedean ordering

My question is rather simple. Do there exist a formally real field that admits a unique ordering (so sums of squares are the positive elements) and such that this ordering is not archimedean?

Oh, I have forgotten to add that the field I am looking for cannot be euclidean (in particular it cannot be real closed)!

The question I have in mind is the following. I proved many years ago that if K is a field that admits a unique ordering and this ordering is archimedean, then every automorphism f of K(x), where x is an indeterminate, maps K onto K. I wonder if archimedeanity can be dropped in this statement, and I realize that I know very few fields K whose unique ordering is not archimedean; essentially, real closed fields that cannot be embedded into the reals. And for such fields I know that f(K) = K!!

• You should have a more explicit title and maybe add a bit of context to your question. – M. Dus Mar 20 '18 at 9:50
• I guess the confusion is that the box named "Title" does not refer to your professional title, but to the subject of your posting. – Carlo Beenakker Mar 20 '18 at 9:51
• Every real closed field has a unique ordering (the positive elements are precisely the squares). Just pick a non-archimedean one (e.g. Puiseaux series over $\mathbb{R}$). – Denis Nardin Mar 20 '18 at 9:53
• (Linguistic note @DenisNardin: the correct spelling is Puiseux series.) – Peter Heinig Mar 20 '18 at 10:24
• @PeterHeinig Ugh, this means I've been mispronouncing them for the last couple of years... Thanks! – Denis Nardin Mar 20 '18 at 10:32

Yes, such fields exist. Let $T$ be the first-order theory in the language of fields with an extra constant $c$, axiomatized by

1. the axioms of fields,

2. every element or its negative is a sum of 4 squares,

3. some element which is a sum of two squares is not a square,

4. the axioms $c>n$ for each standard natural number constant $n$.

Every finite subtheory of $T$ has a model: indeed, the subtheory of $T$ with axioms 4. only for $n<N$ is satisfied in $(\mathbb Q,N)$. Thus, by the compactness theorem, $T$ has a model, and this is a non-archimedean non-euclidean uniquely orderable field.

• This is a very beautiful answer. Many thanks! I will try to find an automorphism of K(x) that does not maps K onto K, where K denotes the field you have found. – JOSE MANUEL GAMBOA MUTUBERRIA Mar 22 '18 at 10:56

In addition to the above very nice model-theoretic construction, it seems interesting to describe such fields Galois-theoretically. Here is one such construction:

By a result of Ershov, the free profinite product of absolute Galois groups of fields is again an absolute Galois group of some field, and in fact such a field can have an arbitrarily large cardinality. Let $K$ be a field of cardinality $>\aleph$ whose absolute Galois group is the free profinite product of a group of order $2$ and of some absolute Galois group of a non-ordered field (e.g. of $\hat{\mathbb{Z}}$). By a result of Herfort and Ribes https://eudml.org/doc/152727, every torsion element in a free profinite product is conjugate to an element of one of the free factors. The Artin-Schreier theorem therefore implies that $K$ has exactly one ordering. Since a field with an Archimedean ordering embeds in $\mathbb{R}$, the ordering on $K$ is non-Archimedean.