Ytukyg

Let $kappa$ be a regular uncountable ordinal. Let $No(kappa)$ denote the field of surreal numbers of birthdate $ < kappa$.

In Fields of surreal numbers and exponentiation (2000), P. Ehrlich and L.v.d. Dries proved that $No(kappa)$ is isomorphic to $mathbb{R}((x))^{No(kappa)}_{<kappa}$ which is the subfield of Hahn series of length $< kappa$ over $mathbb{R}$ with value group $No(kappa)$.

Let $S$ be the subset of $mathbb{R}((x))^{No(kappa)} = mathbb{R}((x))^{No(kappa)}_{<{kappa}^+}$ of Hahn series of either length $<kappa$ or of length $kappa$ whose $kappa$-sequence of exponents is cofinal in $No(kappa)$.

It is not too difficult to see that $S$ is stable under $+$ and $-$.

I wonder if it is a subfield of $mathbb{R}((x))^{No(kappa)}$, in which case it would be an example of completion of $No(kappa)$. Does anyone know how to prove/disprove this?

So as said in the comments, this follows from the following general result in valuation theory:

Let $F$ be a field, let $(G,+)$ be a non trivial ordered group, and let $F[[varepsilon^G]]$ be the field of Hahn series with value group $G$ and coefficients in $F$. This is equipped with the uniform structure and topology induced by the standard valuation $v$ whose valuation ring is the set of series whose support is a subset of $G^{geq 0}$.

For $f in F[[varepsilon^G]]$ and $g in G$, we let $f|_g$ denote the series with support $operatorname{supp} f cap G^{> g}$ which coincides with $f$ on this set. This is a truncation of $f$ as a series.

Corollary 3.2.18 in ADH:

If $T$ is a subfield of $F[[varepsilon^G]]$ which is stable under truncation and contains $varepsilon^G$, then the set $overline{T}:={f in F[[varepsilon^G]]:forall g in G, f|_g in T}$ is a completion of $T$, that is, a maximal dense valued field extension of $T$.

In the case when $F$ is ordered, the same definition gives a maximal dense ordered field extension of $T$, that is, the Cauchy-completion of $T$ as an ordered field.

Let's apply this to the case $F=mathbb{R}$, $G=mathbf{No}(kappa)$ and $T=mathbf{No}(kappa)$. We have $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$ which is stable under truncation in $mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa^+}$ and contains $varepsilon^{mathbf{No}(kappa)}$.

Let $f in overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}$. If $operatorname{supp} f$ is cofinal in $mathbf{No}(kappa)$, then let $alpha$ denote its order type and assume $kappa< alpha$. The ordinal $alpha$ must be limit so $kappa+1<alpha$. By the definition of the completion, we have $f|_{x_{kappa+1}} in mathbf{No}(kappa)$ where $x_{kappa+1}$ is the $(kappa+1)$-th element of $operatorname{supp} f$. But the order type of $operatorname{supp} f|_{x_{kappa+1}}$ is $kappa$, which contradicts the identity $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$. So $alphaleqkappa$, and thus $f in S$. Else $operatorname{supp} f$ is has an upper bound $x$ in $mathbf{No}(kappa)$ and then $f|_{x+1}=f in mathbf{No}(kappa)$ so $f in S$. This proves the inclusion $overline{mathbf{No}(kappa)} subseteq S$.

Conversely we have $mathbf{No}(kappa) subset overline{mathbf{No}(kappa)}$. For $f in mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{leq kappa}$ whose support is cofinal in $mathbf{No}$, the order type $alpha$ of $operatorname{supp} f$ satisfies $kappa=operatorname{cof}(kappa)leqalphaleq kappa$ so $alpha=kappa$. For $x in mathbf{No}(kappa)$, there is $y in operatorname{supp} f$ with $x<y$, so $operatorname{supp} f|_x<operatorname{supp} f|_yleq kappa$. Thus $f|_x$ lies in $mathbf{No}(kappa)$. We deduce that $f in overline{mathbf{No}(kappa)}$.

Hence $S=overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}=overline{mathbf{No}(kappa)}$ as desired.