Characteristic $0$ non-algebraically closed field which does not admit an ordering

Question

Are there any characteristic $0$ fields which are not algebraically closed and do not admit an ordering making them an ordered field?

Answer 1

Take $\mathbb{C}(X)$ (the field of rational functions over the complex numbers). This field has characteristic zero and does not admit an ordering, otherwise also $\mathbb{C}$ would. It is not algebraically closed as the polynomial $T^2 - X\in \mathbb{C}(X)[T]$ has no zero in $\mathbb{C}(X)$ (the square root is not a rational function).

Answer 2

Yes: the field $\mathbb{Q}[i]$ . It has an element whose square is $-1$ and that's enough to prove it.

Answer 3

How about $\mathbb{Q}[i]$?

Not algebraically closed for similar reasons to $\mathbb{Q}$.

Not an ordered field for similar reasons to $\mathbb{C}$.

Answer 4

Also, for $p \in \mathbb{N}$ prime, $\mathbb{Q}_p$, the $p$-adic numbers, are an unorderable field of characteristic zero that is not algebraically closed. (Perhaps also interesting, their algebraic closure is a field extension of infinite degree, unlike $\mathbb{R}$.)