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2 fix the definition

The question is somewhat ambiguous, it’s not clear whether the Archimedean classes are meant to be additive or multiplicative. I will assume the former, i.e., equivalence classes of the relation $$a\sim b\Leftrightarrow\exists n,m\in\omega\smallsetminus\{0\}\,(na\le b\le ma).$$ b\Leftrightarrow\mathrm{sign}(a)=\mathrm{sign}(b)\land\exists n\in\omega\smallsetminus\{0\}\,(n^{-1}|a|\le|b|\le n|a|).$$First, since real-closed fields (rcf) have elimination of quantifiers, any embedding between them is automatically elementary. Thus the question is whether every rcf R has a proper rcf extension S with the same Archimedean classes (i.e., every s\in S is \sim to some r\in R). As Andreas noted above, this property does not hold in general, and in particular, \mathbb R has no proper Archimedean extension. On the other hand, it holds for many other real-closed fields: for example, any Archimedean rcf different from \mathbb R has a proper Archimedean rcf extension (namely, \mathbb R). I think the following characterization holds: Proposition: If R is a rcf, the following are equivalent: 1. R has a proper rcf extension with the same Archimedean classes. 2. There is a Dedekind cut \langle A,B\rangle on the interval (0,1)_R such that$$\tag{$*$}\forall a\in A\,\exists b\in B\,\frac{a+b}2\in A\qquad\text{and}\qquad\forall b\in B\,\exists a\in A\,\frac{a+b}2\in B.$$On the one hand, let S\supseteq R be a rcf with the same Archimedean classes and x\in S\smallsetminus R. We can assume x>1. There exists c\in R such that c\sim x; WLOG c< x< 2c. Then 0< x/c-1< 1, and the Dedekind cut on R determined by x/c-1 is easily seen to satisfy (*). On the other hand, assume the cut \langle A,B\rangle is given. We define an ordering on the rational function field F=R(x) as follows. Using the fact that every nonzero polynomial is a product of linear polynomials and polynomials of the form (x-a)^2+b, where b>0, we see that for every f(x)/g(x)\in F, there are a\in A and b\in B such that f and g have constant sign on (a,b)_R; we define the sign of f(x)/g(x) to be the sign it assumes on (a,b)_R. This makes F an ordered field. Let S be its real closure. For a given \alpha\in S, there exists c\in R such that \alpha\sim c whenever: 1. \alpha=x-a, a\in R. This follows from (*). 2. \alpha=(x-a)^2+b, a,b\in R, b>0. This follows from 1: if u\sim u', v\sim v', and u,v>0, then u+v\sim u'+v'. 3. \alpha\in F. Every such \alpha is a product of an element of R and elements of the form 1 or 2 or their inverses, and u\sim u' and v\sim v' imply uv\sim u'v' and u/v\sim u'/v'. 4. \alpha\in S is such that \alpha^k\in F for some integer k>0. We have \alpha^k\sim c by 3, hence \alpha\sim\sqrt[k]c\in R. 5. \alpha\in S. We have \sum_{i\le d}u_i\alpha^i=0 for some u_i\in F, u_d\ne0. Let i be such that the Archimedean class of u_i\alpha^i is maximal. Since the sum above is 0, there exists j\ne i such that u_j\alpha^j\sim-u_i\alpha^i. Then \alpha^{j-i}\sim-u_i/u_j, hence \alpha\sim c for some c\in R by 4. Thus S is a proper rcf extensions of R with the same Archimedean classes as R. 1 The question is somewhat ambiguous, it’s not clear whether the Archimedean classes are meant to be additive or multiplicative. I will assume the former, i.e., equivalence classes of the relation$$a\sim b\Leftrightarrow\exists n,m\in\omega\smallsetminus\{0\}\,(na\le b\le ma).$$First, since real-closed fields (rcf) have elimination of quantifiers, any embedding between them is automatically elementary. Thus the question is whether every rcf R has a proper rcf extension S with the same Archimedean classes (i.e., every s\in S is \sim to some r\in R). As Andreas noted above, this property does not hold in general, and in particular, \mathbb R has no proper Archimedean extension. On the other hand, it holds for many other real-closed fields: for example, any Archimedean rcf different from \mathbb R has a proper Archimedean rcf extension (namely, \mathbb R). I think the following characterization holds: Proposition: If R is a rcf, the following are equivalent: 1. R has a proper rcf extension with the same Archimedean classes. 2. There is a Dedekind cut \langle A,B\rangle on the interval (0,1)_R such that$$\tag{$*$}\forall a\in A\,\exists b\in B\,\frac{a+b}2\in A\qquad\text{and}\qquad\forall b\in B\,\exists a\in A\,\frac{a+b}2\in B.

On the one hand, let $S\supseteq R$ be a rcf with the same Archimedean classes and $x\in S\smallsetminus R$. We can assume $x>1$. There exists $c\in R$ such that $c\sim x$; WLOG $c< x< 2c$. Then $0< x/c-1< 1$, and the Dedekind cut on $R$ determined by $x/c-1$ is easily seen to satisfy $(*)$.

On the other hand, assume the cut $\langle A,B\rangle$ is given. We define an ordering on the rational function field $F=R(x)$ as follows. Using the fact that every nonzero polynomial is a product of linear polynomials and polynomials of the form $(x-a)^2+b$, where $b>0$, we see that for every $f(x)/g(x)\in F$, there are $a\in A$ and $b\in B$ such that $f$ and $g$ have constant sign on $(a,b)_R$; we define the sign of $f(x)/g(x)$ to be the sign it assumes on $(a,b)_R$. This makes $F$ an ordered field. Let $S$ be its real closure. For a given $\alpha\in S$, there exists $c\in R$ such that $\alpha\sim c$ whenever:

1. $\alpha=x-a$, $a\in R$. This follows from $(*)$.

2. $\alpha=(x-a)^2+b$, $a,b\in R$, $b>0$. This follows from 1: if $u\sim u'$, $v\sim v'$, and $u,v>0$, then $u+v\sim u'+v'$.

3. $\alpha\in F$. Every such $\alpha$ is a product of an element of $R$ and elements of the form 1 or 2 or their inverses, and $u\sim u'$ and $v\sim v'$ imply $uv\sim u'v'$ and $u/v\sim u'/v'$.

4. $\alpha\in S$ is such that $\alpha^k\in F$ for some integer $k>0$. We have $\alpha^k\sim c$ by 3, hence $\alpha\sim\sqrt[k]c\in R$.

5. $\alpha\in S$. We have $\sum_{i\le d}u_i\alpha^i=0$ for some $u_i\in F$, $u_d\ne0$. Let $i$ be such that the Archimedean class of $u_i\alpha^i$ is maximal. Since the sum above is $0$, there exists $j\ne i$ such that $u_j\alpha^j\sim-u_i\alpha^i$. Then $\alpha^{j-i}\sim-u_i/u_j$, hence $\alpha\sim c$ for some $c\in R$ by 4.

Thus $S$ is a proper rcf extensions of $R$ with the same Archimedean classes as $R$.