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


*

*the axioms of fields,

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

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

*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.
A: 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. 
