Does there exist an ordered valued field $K$ whose value group is the additive group of $K$?

Question

The title says it all. Can one prove the existence of an ordered valued field $K$ whose value group is the additive group of $K$? Would or could the valuation respect the order on $K$?

Answer 1

You can obtain such a field by iterating the following construction:

Given an ordered group $G$, take the fraction field $k(G)$ of polynomials with coefficients in the ordered field $k$ and indeterminates $X^g,g \in G$, with the product extending $X^g X^h=X^{g+h}$. The valuation is $v \frac{P}{Q} = v P - v Q$ where $v P$ is the least $g \in G$ such that $X^g$ occurs in the polynomial $P$. This valued field has value group $G$ and residue field $k$. Say that a polynomial $P$ is positive if it is non zero and the coefficient of $X^{v P}$ in $P$ is positive. Extending this to fractions in the usual way, you get an ordered field extension of $k$ whose valuation ring is convex (which is the standard way to say that the valuation is compatible with the ordering).

Note that $X^G:=\{ X^g \ : \ g\in G\}$ is a copy within $(k(G)^{>0},\times,<)$ of the ordered group $G$. So setting $F_0=k(G)$ and $F_{n+1}=k(F_n)$ for all $n \in \mathbb{N}$, we have natural inclusions $F_n \hookrightarrow F_{n+1}$. Set $F:= \bigcup \limits_{n \in \mathbb{N}} F_n$ with the induced valuation $v$. Then there is a natural isomorphism between $(F,+,<)$ and the value group of $F$, given by $F_n \ni f \longmapsto v X^f$.

One can do the same thing by iterating the Hahn series field construction.

---

There are natural examples of ordered valued fields satisfying this. This is the case for instance of logarithmic-exponential transseries, surreal numbers, and hyperreal numbers under the continuum hypothesis.