Is the class of reducts of ordered fields axiomatizable?

Question

Let $(F,+,-,*,0,1,\leq)$ be an ordered field. We know that the class of ordered fields is axiomatizable, by definition. Is the $\{+,-,*,0,1\}$ class of reducts of ordered fields axiomatizable? And if it is, is it finitely axiomatizable?

Answer 1

Yes, a field $F$ admits an ordering iff $-1$ is not a sum of squares in $F$ (see Show that formally real fields can be totally ordered into ordered fields for instance). Such a field is called formally real. This gives an axiomatization, since for each $n$, "$-1$ is not a sum of $n$ squares" can be expressed in the first-order language of rings.

On the other hand, no finitely many of these axioms suffice, so formally real fields are not finitely axiomatizable. Indeed, it is known that for any $n$, there is a field in which $-1$ is not a sum of $n$ squares but it is a sum of $N$ squares for some $N>n$. Explicitly, if $N$ is a power of $2$ and $F$ is a formally real field, let $K$ be the field of fractions of the ring $F[x_1,\dots,x_{N}]/(1+x_1^2+\dots+x_N^2)$. Then clearly $K$ is not formally real (since $-1=x_1^2+\dots+x_N^2$ in $K$), but it can be shown that $-1$ is not a sum of fewer than $N$ squares in $K$. See $-1$ as sum of $k$ squares with $k$ least postive integer for more discussion.