Field having an archimedean ordering and a non archimedean ordering
Let $F$ be a formally real field, so that $F$ has at least one ordering (compatible with the operations of $F$).
Question. is it possible for $F$ to have an archimedean ordering $\leq_1$ and a non-archimedean ordering $\leq_2$ ?
I know that the answer is NO in several cases:
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formally real number fields (because of the description of orderings of finite extensions of $\mathbb{Q}$)
-
real closed fields (since such fields have a unique ordering).
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formally real pythagorean fields (pythagorean: every sum of two squares is a square), but the arguments I have are extremely complicated and indirect... To be quick, I think I can show that if $F$ is pythagorean and admits an archimedean ordering , $SO_3(F)$ is a simple group, while if $F$ is pythagorean and admits a non archimedean ordering, I can show that $SO_3(F)$ is not a simple group.
I am pretty sure the answer is NO in general, but I cannot come up with simple arguments ( I am probably missing something obvious...)
If needed, I am mainly interested to have a simple proof in the case of formally real pythagorean fields.
EDIT. Apparently, there are fields $F$ for which the answer is YES (see GEdgar's comment).
The real question would then be:
Question (V2). Let $F$ be a formally real pythagorean field. Find a direct proof of the fact that $F$ cannot have an archimedean ordering $\leq_1$ and a non-archimedean ordering $\leq_2$.
Take $\mathbb Q$
together with one transcendental $𝑥$.
Then the countable field $\mathbb Q(x)$
can be ordered:
(a) so that $x=\pi$,
but it can also be ordered
(b) so that $𝑥 > r$ for all $r \in \mathbb Q$.