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.

  • $\begingroup$ In case it's helpful, here's an example of an exponential map I was able to construct that does not fit the bill: $$ \textrm{exp}\left( r \cdot t^\eta \right) = e^r \cdot t^{r\eta} $$ where $t$ is the formal indeterminate. This definition is for a single-term Laurent series, but gets extended by linearity; we also take $\textrm{exp}(0) = 1$ of course. Notice in particular how this exponential map assumes that $F \subset G$; I suspect an exponential map that does fit the bill may require something similar. $\endgroup$ – Twiffy Jun 14 '16 at 6:33
  • $\begingroup$ Can you explain how you would extend linearly? Also, please clarify the structure of $G$. Usuallly, additive notations are used for $G$, here you seem to be assuming it is a ring. $\endgroup$ – nombre Jun 14 '16 at 7:21

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.

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  • $\begingroup$ Thanks @PhilipEhrlich, that's perfect. I actually know Franz-Viktor so it's rather embarrassing that I hadn't found that paper myself. $\endgroup$ – Twiffy Jun 14 '16 at 23:24
  • $\begingroup$ It seems that the hyperreals $^*\mathbb{R}$ admit an exponential map in this sense, where we take $\textrm{exp}(x) = e^x$ componentwise on $\mathbb{R}^\mathbb{N}$ and then mod out by an ultrafilter. So the above paper would seem to imply that $^*\mathbb{R}$ cannot be written as (in my notation) $\mathbb{R} \boxtimes G$ for any ordered abelian group $G$. But I would have guessed that the hyperreals (as with the surreals) are a fixed point of the $\mathbb{R} \boxtimes \mbox{--}$ functor. Is it otherwise known that this is not the case? $\endgroup$ – Twiffy Jun 14 '16 at 23:41
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    $\begingroup$ @Twiffy: If $G$ is a set-sized ordered group, then $|\mathbb{R} \boxtimes G| \geq {|\mathbb{R}|}^{\kappa}$ for each cardinal $\kappa$ embedding in $(G,<)$. It is consistent (assuming the continuum hypothesis) that $|^*\mathbb{R}| = |\mathbb{R}|$ is such a cardinal for $G = ^*\mathbb{R}$, and thus that $|\mathbb{R} \boxtimes ^*\mathbb{R}| \geq 2^{|^*\mathbb{R}|} > |^*\mathbb{R}|$. $\endgroup$ – nombre Jun 15 '16 at 9:01
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    $\begingroup$ @Twiffy. Yes, it follows from the K-K-S result that there is no hyperreal ordered field $^*\mathbb{R}$ having an underlying ordered field isomorphic to a Hahn field. On the other hand, like the surreals, many hyperreal ordered fields have underlying ordered fields that are isomorphic to distinguished subfields of Hahn fields. This was pointed out in my: An alternative construction of Conway's ordered field No. Algebra Universalis 25 (1988), no. 1, 7–16. $\endgroup$ – Philip Ehrlich Jun 15 '16 at 15:06
  • $\begingroup$ @Nombre, thanks, that's very clear. $\endgroup$ – Twiffy Jun 17 '16 at 3:38

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