# First-order properties of Euclidean fields (instead of real closed fields)

Let $$\mathbb{F}$$ be an ordered field.

• $$\mathbb{F}$$ is called Euclidean if $$\forall x>0 \in \mathbb{F}\ \exists y \in \mathbb{F} : y^2=x$$, i.e. a square root exists.
• $$\mathbb{F}$$ is called real closed if it is Euclidean and every polynomial of odd degree has at least one zero in $$\mathbb{F}$$.

Real closed fields have the same first-order properties as the field of real numbers, i.e. statements that involve symbols like $$+, \cdot, =,\leq,\dots$$ are true for a real closed field $$\mathbb{F}$$ if and only if they are true over the real numbers $$\mathbb{R}$$.

If real closed fields are generalisations of the real numbers, then the Euclidean fields are generalisations of the field of constructible numbers. It is wrong to say that Euclidean fields have the same first-order properties as the constructible numbers, since all real closed fields are Euclidean, so the property of having a $$n$$th root may be different. But is the following true?

A first-order-logic statement holds over the constructible numbers, but not over the real numbers, if and only if it holds for any Euclidean field $$\mathbb{F}$$, but not its real closure $$\overline{\mathbb{F}}$$.

Is there a counter example? Does one direction hold? Are there references to this? More generally, is there any way in which real closed fields are better behaved under "logical generalisations" than the Euclidean fields?

I am interested in this question because in my research the Euclidean property seems to suffice most of the time, but people just use real closed fields since they have a well established theory.

• I think your forward direction is trivially false. Constructible number don't have cube root of 2, but Euclidean field can have cube root of 2, simply by constructing an Euclidean field over $Q(2^{1/3})$. As for the backward direction, what does it mean to hold over all Euclidean field but not real closure? Real closed field are Euclidean. – needanstyvm Oct 28 '19 at 13:50
• Thanks, you are right. Having a cube root, having a fifth root, having a severth root etc. are all statements that are not true for the constructible numbers, but they are true for some Euclidean non real closed fields. I wonder if these examples are "the only" ones. Backward direction: There might be a property in a Euclidean $\mathbb{F}$, witch does not hold in it's real closure $\overline{\mathbb{F}}$. – Strichcoder Oct 28 '19 at 14:16