Let $R$ be a real closed field. Recall that the *ladder* of $R$ is the divisible, ordered abelian group obtained by quotienting $R$ by a certain equivalence relation.


Note that $R$ has trivial ladder iff $R$ is a subfield of $\mathbb R$. If $R$ has trivial ladder and $L$ is a divisible ordered abelian group, let $R\langle\!\langle x^L\rangle \!\rangle$ be the field of Puiseux series. Then $R\langle \!\langle x^L\rangle \!\rangle$ has ladder $L$. Conversely, from any real closed field, we may extract a maximal subfield of $\mathbb R$ and a ladder. I'm wondering whether that's all there is to it, i.e., whether every real closed field is of the form $R\langle\! \langle x^L \rangle\! \rangle$. Let me state this more formally:


**Questions:** Let $R$ be a real closed field with ladder $L$, and let $R\_0 \subseteq R$ be a maximal subfield with trivial ladder.


1. Is $R$ characterized up to isomorphism by $R\_0$ and $L$?
2. Is $R$ characterized up to isomorphism over $R\_0$ by $L$?
3. The relative version: for any extension $R \to S$ of real closed fields such that $R\_0 = S\_0$, is $S$ characterized up to isomorphism over $R$ by the induced map on ladders?


As indicated in the title question, I'm happy to assume that $R\_0 = \mathbb R$ if that simplifies things.


EDIT: The first two paragraphs of this question were edited from the version to which @EmilJeřábek [responded](https://mathoverflow.net/a/308058). In order that that answer can still make sense, the original first two paragraphs appear below.



> 
> Let $R$ be a real closed field. Recall that $x,y \in R$ are *comparable* if there are $m,n \in \mathbb Z$ such that $mx > y$ and $ny > z$. Recall that the *ladder* of $R$ is the linear order obtained by quotienting by comparability.
> 
> 
> Note that $R$ has trivial ladder iff $R$ is a subfield of $\mathbb R$. If $R$ has trivial ladder and $L$ is a linear order, let $R^L$ be the real closure of the purely transcendental extension $R(L)$ with transcendence basis $L$, ordered as in $L$. Then $R^L$ has ladder $L$. Conversely, from any real closed field, we may extract a maximal subfield of $\mathbb R$ and a ladder. I'm wondering whether that's all there is to it, i.e. whether every real closed field is of the form $R^L$.
> 
> 
> 



Is any real closed extension of $\mathbb R$ characterized up to isomorphism by its ladder?
First, the initial claims in the question are false: the ladder of $R^L$ is *not* $L$. For example, for any $x\in L$, the field $R^L$ contains the element $\sqrt x$, which is not “comparable” to any element of $L$.


Not every linear order is isomorphic to a ladder of a real-closed field; for example, any such ladder is a dense order. In fact, a linear order is isomorphic to a ladder of a real-closed field if and only if it can be expanded to a divisible ordered group. This follows from the fact that any ordered field $(F,\le)$ carries a canonical structure of a valued field, with valuation ring $\{x\in F:\exists n\in\mathbb N\,(-n<x<n)\}$. Then the ladder of $F$ is exactly the underlying order of the value group of $F$. If $F$ is real-closed, its value group is divisible. Conversely, if $L$ is a divisible ordered group, the field of Puiseux series $\mathbb R\langle\!
\langle x^L\rangle\!\rangle$ is a real-closed field with value group $L$.


Now, the answer to all the questions is *no, not at all*. For instance, if $K$ is a dense subfield of $L$, then $K$ and $L$ have the same ladder; in particular, any real-closed field has the same ladder (in fact, the same value group and the same residue field) as its completion.


