If $F$ is an ordered field and $G$ is an ordered abelian group, one can define the *Hahn product* $F \boxtimes G$ to be the set of formal Laurent series with coefficients in $F$ and exponents in $G$. It is easy to see that this is a ring that derives a linear order from the orders on $F$ and $G$ lexicographically. Via the usual division algorithm it can be shown that $F \boxtimes G$ is also a field.


Just as orders on $F$ and $G$ induces an order on $F \boxtimes G$, I would like for it to be true that if $G$ is a field, and $F$ and $G$ both admit exponential maps -- meaning a group homomorphism from the additive group to the multiplicative group -- then there is an induced exponential map on $F \boxtimes G$. This exponential map $\textrm{exp}$ would ideally have the (loosely stated) properties that


* $\textrm{exp}$ maps the whole field $F \boxtimes G$ bijectively to the positive elements $(F \boxtimes G)\_{> 0}$
* in some fashion $\textrm{exp}$ respects the individual orders -- perhaps strictly respecting the order on $F$ and reversing the order on $G$
* in some fashion $\textrm{exp}$ respects the exponential maps on $F$ and $G$


I've been looking for such a map in $\mathbb{R} \boxtimes \mathbb{R}$ and haven't been able to find one that matches the criteria, nor prove that a map with those properties can't exist.



Is there an exponential map on (Hahn) ordered fields?
There is no such exponential map. This was demonstrated in:


F.-V. Kuhlmann, S. Kuhlmann, S. Shelah, Exponentiation in power series fields, Proc. Amer. Math. Soc. 125 (1997) 3177–3183.


