If a field has unique ordering, must its subfields also have unique ordering? Let $L$ be a field and suppose that there exists a unique total order $\leq$ on $L$ with respect to which $(L,\leq)$ is an ordered field.
Now let $K$ be a subfield of $L$. Clearly $K$ is an ordered field with $\leq$ restricted to $K$. Must this be the unique order structure on $K$ or could there be another field structure on $K$? I guess if there is another order $\preceq$ on $K$ then it cannot be extended to $L$ since that would contradict uniqueness for $L$. Any ideas?
 A: No, unique orderability need not be "downwards-preserved."
The most immediate (in my opinion) example of a uniquely orderable field is $\mathbb{R}$; this is a convenient example to think about since it has lots of subfields. One easy way to get "multiple orderability" would be to find a subfield with nontrivial automorphisms. A natural place to look for such a subfield is around transcendental reals. And indeed, we can find such an example this way:

The field $\mathbb{Q}(\pi)$ (for example) has nontrivial automorphisms and hence is not uniquely orderable.

(Consider $\pi\mapsto -\pi$.)

The key point is that, in an ordered field, only nonnegative elements can have square roots. This means in particular that if $F$ is a field where, for every $x\in F$, either $x$ or $-x$ has a square root in $F$, then $F$ is uniquely orderable. This is the case with $\mathbb{R}$ but not with $\mathbb{Q}(\pi)$ since in the latter neither $\pi$ nor $-\pi$ has a square root.
So in $\mathbb{Q}(\pi)$ we have some freedom and can make either $\pi$ or $-\pi$ positive as we prefer. However, once we "see" $\sqrt{\pi}$, we're forced to make $\pi$ positive - and this is why we cannot extend an "anomolous" ordering of $\mathbb{Q}(\pi)$ which puts $\pi\trianglelefteq-\pi$ to an ordering of $\mathbb{R}$.
