# Archimedean completeness of some fields

I need a reference (different from Hahn's 1907 paper) for the following result.

Theorem: If $G$ is a totally ordered abelian group, then the field $\mathbb{R}((G))$ is archimedean complete.

• $\mathbb{R}((G))$ consists of all the functions $f:G\to\mathbb{R}$ such that $\{g\in G:f(g)\neq0\}$ is well-ordered.
• Let $E$ be an ordered field. Two nonzero elements $x,y\in E$ are comparable if there exist $m,n\in\mathbb{N}$ such that $|x|<m|y|$ and $|y|<n|x|$ (where $|a|$ is defined as $\max\{a,-a\}$).
• Let $E/ K$ be an extension of ordered fields, where the order on $E$ restricted to $K$ coincides with that of $K$. We say that $E$ is an Archimedean extension of $K$ if for every $x\in E$, there exists $y\in K$ such that $x$ and $y$ are comparable in $E$.
• A field $K$ is Archimedean complete if it has no proper archimedean extension fields.

Does anyone know a good reference for the proof of this result? The more the better.

• I understand the embedding theorem, but in the references there is not mention about the notion of Archimedean completeness. If $K$ is an archimedean extension of $\mathbb{R}((G))$, then $K$ is of type $G$ (the group of Archimedean classes of $K$ is isomorphic to $G$). Thus, by the embedding theorem there exists an embedding $\theta$ from $K$ to $\mathbb{R}((G))$. But still can happen that $K$ is a proper extension of $\mathbb{R}((G))$. How can we discard that possibility? In general, this situation is possible. For example $K((x))$ can be embedded in $K((x^2))$ being a proper extension of it Nov 3 '16 at 22:10
• @Chilote. You are mistaken. Look at the last paragraph on p. 862. It begins: "Theorem 3.2 (i.e. Hahn's Embedding Theorem) is beautifully adapted to the proof of Hahn's completeness theorem.... Hahn defines a complete ordered group to be an ordered group $G$ which does not admit a proper extension $H$ which is Archimedean relative to $G$....." Nov 4 '16 at 1:57