Non-archimedean matrix fields

Question

Are there any examples of sets $A\subseteq{\Bbb C}^{n\times n}$ for which you can find operations so that $(A,+,\,\cdot\ ,<)$ is a non-archimedean ordered field? I feel like the answer is probably no, because to get an ordered field out the structure would have to be very similar to ${\Bbb R}$, but then again matrix groups seem to have a lot of degrees of freedom, so I'm not so sure. (If the answer is no, what about for $A\subseteq{\Bbb C}^{\omega\times\omega}$ or even larger bases?)

Answer 1

It is not necessary to go to matrix fields to find non-archimedean ordered fields, even if we restrict to the standard $+$ and $\cdot$ operation on ${\Bbb R}$. One simple example of a non-archimedean ordered field is the field of rational functions ${\Bbb R}(x)$ or ${\Bbb Q}(x)$, where $f\prec g$ if the leading coefficient of the numerator of $g-f$ is positive (i.e. $\lim_{x\to\infty}e^x(g(x)-f(x))=\infty$, although the functions are usually treated only formally and not as functions).

We can reduce this to a subset of ${\Bbb R}$ by looking at ${\Bbb Q}(\pi)$ with the inherited order (which is to say, $f(\pi)\prec g(\pi)$ iff $f\prec g$). To prove that this map is injective (so that $\prec$ is well-defined on ${\Bbb Q}(\pi)$), we need only note that if $f(\pi)=g(\pi)$, then the numerator of $g-f$ has a root at $\pi$, so since $\pi$ is transcendental and the numerator of $g-f$ is a rational polynomial, it must be that $g-f=0$, i.e. $f=g$.

Thus, $({\Bbb Q}(\pi),+,\cdot,\prec)$ is isomorphic to $({\Bbb Q}(x),+,\cdot,\prec)$, so ${\Bbb Q}(\pi)\subseteq{\Bbb R}\subseteq{\Bbb C}^{1\times1}$ is a non-archimedean ordered field since ${\Bbb Q}(x)$ is.