# An analogue of rational functions for Hahn series

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?

• If $\Gamma=\mathbb{Q}$, the ring of Hahn polynomials with finitely many terms is $\varinjlim_n k[t^{1/n}]$ and its fraction field is $\varinjlim_n k(t^{1/n})$, and they all sit inside the Puiseux field $\varinjlim_n k((t^{1/n}))$ which is still far from the Hahn series. I don't think your question has an answer: Hahn series are intrinsically local to a place, there is no corresponding global object. Aug 13 '17 at 1:51
• @Gro-Tsen I don't understand your second sentence. "local" and "global" are geometric terms; I'm just talking algebraically. Aug 13 '17 at 4:38
• I think it helps to understand comm. algebra to think geometrically, but if you don't like that, I can restate it as follows: you can embed $k(t)$ in the Laurent series field in many ways by expanding w.r.t. $t-a$ for $a\in k$ (let's say $k$ is alg.ly closed) or $1/t$ ("at infinity"), each one gives you an embedding or "place": $k(t)$ treats these places equally, hence the "global", whereas $k((t))$ is relative to the choice of one place for completion, hence "local". I.e.: a "global" object should have the $t\mapsto t+a$ and $t\mapsto 1/t$ automorphisms that $k(t)$ has. (contd.) Aug 13 '17 at 10:23
• (contd.) So I would say a good object that "is to $k((t^{\mathbb{Q}}))$ what $k(t)$ is to $k((t))$" should not discriminate between $t$ and $t-a$ or $1/t$, i.e., it should have automorphisms taking $t$ to $t+a$ or $1/t$. As such, I think the best answer you can get (at least for $k$ algebraically closed and $\Gamma=\mathbb{Q}$) is simply "the algebraic closure of $k(t)$, and the way it embeds in $k((u^{\mathbb{Q}}))$ for every $u=t-a$ and $u=1/t$". • But of course, the real question if you want to define an object is: what do you want to do with it? (This determines which properties matter.) Aug 13 '17 at 10:32
• @Gro-Tsen I have some concrete examples of "Hahn rational functions" that are formal quotients of finite sums of non-integral powers of a variable $t$, and I want to know where they live. I care about the particular variable $t$ (as opposed to $t-a$ or $1/t$ or whatever) because I want to be able to evaluate such functions at particular values of $t$. Aug 13 '17 at 14:13

The field $k(t^\Gamma)$ is sometimes called "the field of generalized rational functions". It is covered in section 2.9 of I. Efrat, "Valuations, Orderings, and Milnor $K$-Theory", AMS, 2006.