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Let us suppose that the first-order language of ordered fields has symbols for addition, subtraction, multiplication and order, and constant symbols for 0 and 1. An ordered field is said to be archimedian if for any positive elements $x<y$ there is a positive integer $n$ such that $nx>y$.

Question: Is the first-order theory of archimedian ordered fields identical to the first-order theory of ordered fields? The non-trivial part of this question is: Suppose a sentence holds in every archimedian ordered field. Does it hold in every ordered field?

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Can you explain how you axiomatize the "archimedian" part? Do you add a constant $n$ for each natural number together with a sentence for $n = 1+...+1$ ($n$ times)? –  Jason DeVito Sep 11 '12 at 1:02
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@Jason: The question doesn't involve any axiomatization of "archimedian". We have two classes of ordered fields, and the problem is whether their theories are equal. Note that the compactness theorem implies that the class of archimedian ordered fields is not first-order definable. For example, ultrapowers of the rationals are not archimedian, but are elementarily equivalent to the rationals. –  falang Sep 11 '12 at 1:50

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up vote 13 down vote accepted

Here is a statement which is true in archimedian fields but not in general ordered fields:

For all $0 < a < b$, there exists $x$ such that $a < x^2 < b$.

This is true in archimedian fields because every archimedean field contains $\mathbb{Q}$ and is contained in $\mathbb{R}$, and every interval in $\mathbb{R}_{+}$ contains the square of a rational.

Let $K$ be the field $\mathbb{R}(t)$, with $f(t)>g(t)$ if $f-g$ is positive for $t$ sufficiently large. I claim that there are no squares between $t$ and $t+1$. Proof: Define $\deg p(t)/q(t) = \deg p - \deg q$, where $p(t)$ and $q(t)$ are polynomials. Then every square has even degree, but every element between $t$ and $t+1$ has degree $1$.


I usually try to give some clue where I came up with these answers. Vague thought process here: I want a non-archimedean ordered field $R$ and a statement which is true in both $\mathbb{Q}$ and $\mathbb{R}$, but not in $R$. So $R$ better NOT be real closed, because real closed fields have the same first order theory as $\mathbb{R}$, and my statement had better USE that $R$ is not real closed. The axioms of real closed fields are all about existence, so a statement which fails in a non-real-closed field should like something like $\exists x : \cdots$. But it also has to be true in $\mathbb{Q}$, which has very few elements. How can I do that? Maybe something like $\forall y \exists x : \cdots$; then it would be true in $\mathbb{R}$ because there are lots of choices for $x$, and true in $\mathbb{Q}$ because there are few choices for $y$.

So I want an equation, dependent on a parameter, which is always solvable in $\mathbb{Q}$, but for a nontrivial reason. (If it would solvable for a trivial reason, that reason would probably hold in every ordered field.) How about $\forall y>0 \exists x_1, x_2, x_3, x_4 : y = x_1^2+x_2^2+x_3^2+x_4^2$? That's true in $\mathbb{Q}$, for nontrivial reasons, and trivially true in $\mathbb{R}$, but false in $\mathbb{R}(t)$ because sums of squares have even degreee. I don't know if it's true in other archimedean fields, though. Still, the idea of using the fact that squares are reasonably spread out in $\mathbb{Q}$, but only land in discrete lumps of even degree in $\mathbb{R}(t)$ sounds like a good one...

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Thank you, both for the answer and also the heuristics! –  falang Sep 12 '12 at 0:14

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