Does there exist a metrizable non-Archimedean ordered field $\mathbb{F}$ that is Cauchy-complete under some metric $d$, where $\mathbb{F}$ is not isomorphic to any field extension of $\mathbb{R}$? I ask because I am thinking about whether the generalization of Hilbert spaces as "vector spaces over such fields, equipped with a positive-definite sesquilinear form" would allow for fields that are not isomorphic to a field extension of $\mathbb{R}$. I have done very little field theory so far, so I would appreciate detailed explanations.
If this is not possible, I would be interested in what happens if you take out the order requirement as well. Thanks.
First of all, let me comment on the "correct" notion of Cauchy completeness for an ordered field $F$. Given an ordinal $\xi$, say a $\xi$-indexed sequence $(x_\alpha)_{\alpha<\xi}$ is Cauchy if for each $\epsilon>0$ in $F$ there exists $\alpha<\xi$ such that $|x_\beta-x_\gamma|<\epsilon$ for all $\beta,\gamma\geq \alpha$. Similarly, say $(x_\alpha)$ converges to $x\in F$ if for each $\epsilon>0$ there exists $\alpha<\xi$ such that $|x_\beta-x|<\epsilon$ for all $\beta\geq \alpha$. Then $F$ can be called Cauchy-complete if all Cauchy ordinal-indexed sequences in $F$ converge to some element of $F$. Any ordered field has a Cauchy completion obtained by formally adjoining limits to all Cauchy ordinal-indexed sequences (which can in fact be taken to all be indexed by the cofinality of $F$).
(There are many equivalent formulations of this notion, such as using nets or filters instead of ordinal-indexed sequences. Another equivalent formulation is that a Dedekind cut $(L,R)$ in $F$ is always filled by an element of $F$ as long as either for all $\epsilon>0$ there exists $x\in L$ such that $y<x+\epsilon$ for all $y\in L$ or for all $\epsilon>0$ there exists $x\in R$ such that $y>x-\epsilon$ for all $y\in R$.)
In particular, for instance, consider the field $\mathbb{Q}(x)$ ordered such that $x$ is infinitely large. This field is not Cauchy-complete, but let $F$ be its Cauchy completion. (Since $\mathbb{Q}(x)$ is countable, you can construct $F$ by just taking ordinary $\mathbb{N}$-indexed Cauchy sequences.) Note that every element of $\mathbb{Q}(x)$ is either infinitely large or is infinitesimally close to some element of $\mathbb{Q}$: a rational function $f(x)/g(x)$ is infinitely large if $\deg f>\deg g$, infinitesimal if $\deg f<\deg g$, and if $\deg f=\deg g$ then it is infinitesimally close to $a/b$ where $a$ and $b$ are the leading coefficients of $f$ and $g$. Any element in the completion $F$ can be approximated arbitrarily closely by elements of $\mathbb{Q}(x)$, which means any element of $F$ is infinitesimally close to an element of $\mathbb{Q}(x)$ since $\mathbb{Q}(x)$ has infinitesimals. Thus every element of $F$ is either infinitely large or infinitesimally close to an element of $\mathbb{Q}$. In particular, for instance, $F$ cannot contain a square root of $2$, so $\mathbb{R}$ does not embed in $F$.
(If you insist on talking about completeness in terms of a metric, this example can also be given a metric: say $d(f,g)=2^n$ where $n\in\mathbb{Z}$ is the difference between the degrees of the numerator and denominator of $f-g$. This defines a metric on $\mathbb{Q}(x)$ compatible with its order topology and the metric completion with respect to this metric can naturally be identified with the Cauchy completion.)