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Is it true that every real closed field can be elementarily embedded in some other real closed filed with the same Archimedean classes (I mean in a proper extension)? Can for example real numbers be elementarily embedded in another real closed field with the same Archimedean classes? R (real numbers) is not the only Archimedean field, is it right?

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If you're asking whether the field of reals has a proper, Archimedean extension, the answer is no, and the question is not at the level appropriate for MO. So I'll vote to close. If you intended to ask something else, please clarify. –  Andreas Blass Dec 12 '11 at 21:05
    
Do you mean Archimedean classes of the additive group or of the multiplicative group? –  Emil Jeřábek Dec 13 '11 at 12:59

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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 $$a\sim b\Leftrightarrow\mathrm{sign}(a)=\mathrm{sign}(b)\land\exists n\in\omega\smallsetminus\{0\}\\,(n^{-1}|a|\le|b|\le n|a|).$$ 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 $R$.

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