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