I was remembering some stuff from my calculus course, and I got to the following question:
By a field order, I mean a total order $\leq$ on a field $\mathbb{F}$ which is invariant by addition and by multiplication of elements greater than $0$ (There's an alternative definition using the set of positive elements, given, for example, in Spivak's book Calculus). Every ordered field must have characteristic zero.
A simple result is the following: $\mathbb{Q}$ admits only one field order (which is the usual order $\leq$ ). Let's prove this: Since in any field order $\preceq$ we have $0\preceq 1$, it follows by some inductions that hence $0\preceq q$ whenever $0\leq q$, and it follows easily that $\preceq=\leq$.
So my question is the following:
Is there any field which admits two distinct field orders? Is the archimedian property necessary for uniqueness of the field order?
The simplest example of non-archimedian ordered field I know is the following: Let $\mathbb{Q}(x)$ be the field of rational functions in $\mathbb{R}$. Given $q\in \mathbb{Q}(x)$, we say that $q>0$ if there exists some number $K>0$ such that $q(x)>0$ for $x>K$.
The ordering on $\mathbb{Q}(x)$ corresponding to $x = \pi$ is archimedean.. The ordering on $\mathbb{Q}(x)$ corresponding to $x = e$ is a different archimedean ordering. The one corresponding to $x$ being positive and infinite is a non-archimedean ordering. Yet these all have the same underlying field.
Another example is that we can give two different orderings on $\mathbb{Q}(\sqrt{2})$; the nonstandard one comes from pretending $\sqrt{2}$ is actually $-\sqrt{2}$ instead.