For any field $k$, we have both the field $k(t)$ of rational functions (formal quotients of polynomials, i.e. the field of fractions of $k[t]$) and the field $k((t))$ of formal Laurent series (which is also the field of fractions of the formal power series ring $k[[t]]$). The composite embedding $k[t] \hookrightarrow k[[t]] \hookrightarrow k((t))$ induces, by the universal property of the field of fractions, a ring homomorphism $k(t) \to k((t))$, which is injective since any homomorphism of fields is injective. In other words, rational functions can be regarded as certain formal power series without loss of information. This is all well-known.

Now if $\Gamma$ is an ordered abelian group, we also have the field $k[[t^\Gamma]]$ of Hahn series over $k$ with value group $\Gamma$. Picking any positive element of $\Gamma$ to call $1$ yields a unique map $\mathbb{Z} \to \Gamma$, which is injective since $\Gamma$ is ordered; thus any such field of Hahn series properly contains $k((t))$ as the Hahn series with all integer exponents. My question is:

What is the analogue of $k(t)$, sitting inside $k[[t^\Gamma]]$ the same way that $k(t)$ sits inside $k((t))$?

It seems to me that the above argument generalizes fairly easily to the Hahn context. That is, let "$k[t^\Gamma]$" be the ring of "Hahn polynomials", i.e. Hahn series with finitely many terms, and let "$k(t^\Gamma)$" be its field of fractions ("Hahn rational functions"). Then for the same reasons we should get an embedding $k(t^\Gamma)\hookrightarrow k[[t^\Gamma]]$. So unless I've made a mistake here, my question boils down to

What is $k(t^\Gamma)$ called, and where can I read about it?

5more comments