# Is this an axiomatization of real closed fields?

I know that real closed fields are defined as ordered fields where every positive element is a square and every odd polynomial has a root. But can they also be axiomatized as totally ordered fields which satisfy the first-order completeness axiom schema? In other words, I am asking whether those axioms give the same theory. Also, my second question is, is either of those theories the same as $Th(\mathbb{R}, 0, 1, +, *, <)$?

• Any two real-closed fields are elementarily equivalent. – André Nicolas Nov 16 '15 at 1:45
• You mean that every positive element has a square root. – Asaf Karagila Nov 16 '15 at 1:46
• (In every semiring every element is a square root. Simply because $x\cdot x$ is a well-defined object. The difficulty is to assess when something has a square root.) – Asaf Karagila Nov 16 '15 at 1:52
• The OP wrote "is a square", not "is a square root" :) – Alex Kruckman Nov 16 '15 at 5:52
• @Alex: Ah, I blame the late hour for that, then! :-) – Asaf Karagila Nov 16 '15 at 8:34

If by first-order schema for completeness you mean that you also allow parameters, then the answer to both your questions is yes. But first let us clarify that real-closed field are often taken as the first axiomatization you propose, and they are in fact elementarily equivalent to $\Bbb R$.
Assume the schema holds, then for every positive $r$, consider the set $\{x\mid 0<x\land x^2<r\}$, this is a bounded set and definable from $r$, therefore it has a least upper bound. But it is not hard to check that this least upper bound is in fact $\sqrt r$. So every positive element is a square root.
Next, suppose that $p$ is a polynomial of an odd degree, without loss of generality $\lim_{x\to\infty}p(x)=\infty$, so $\lim_{x\to -\infty}p(x)=-\infty$. Let $s<r$ be two real numbers such that $p(s)<0<p(r)$, and consider the set $\{x\mid s<x<r\mid p(x)<0\}$. This is a bounded set, as it is bounded by $r$, so it has a least upper bound. And again it is not hard to check that this upper bound is in fact a root for $p$.