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 $Th(\mathbb{R}, 0, 1, +, *, <)$?

  • $\begingroup$ Any two real-closed fields are elementarily equivalent. $\endgroup$ – André Nicolas Nov 16 '15 at 1:45
  • $\begingroup$ You mean that every positive element has a square root. $\endgroup$ – Asaf Karagila Nov 16 '15 at 1:46
  • $\begingroup$ (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.) $\endgroup$ – Asaf Karagila Nov 16 '15 at 1:52
  • $\begingroup$ The OP wrote "is a square", not "is a square root" :) $\endgroup$ – Alex Kruckman Nov 16 '15 at 5:52
  • $\begingroup$ @Alex: Ah, I blame the late hour for that, then! :-) $\endgroup$ – 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 $p$.

In the other direction, we first need to prove that given a real-closed field every definable bounded set has a least upper bound. But this is easy once you prove that every definable set is a finite union of intervals (which may be closed, and thus singletons).

  • $\begingroup$ I think you mean "every definable set is a finite union of intervals" $\endgroup$ – nombre Nov 16 '15 at 12:35
  • $\begingroup$ You're absolutely right. Thanks! $\endgroup$ – Asaf Karagila Nov 16 '15 at 13:34
  • $\begingroup$ I think you misunderstand. I do not allow parameters in the axiom schema. I meant, definable without parameters. Would the answer to the question, change, then? This is why I have not accepted your answer. $\endgroup$ – user107952 Jun 22 '17 at 19:29
  • 1
    $\begingroup$ Remind me again in two years. $\endgroup$ – Asaf Karagila Jun 22 '17 at 20:06

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