The field of real numbers $\mathbb{R}$ is total-ordered and not algebraicaly closed, the field of complex numbers $\mathbb{C}$ is not ordered but is algebraicaly closed.
Intuitively how these two properties are related?
For example:
what about other fields, how the two properties relate?
is there any algebraicaly closed field that is ordered?
why ordering breaks closure? (if indeed so) (some intuitive argument will do)
why closure breaks ordering? (if indeed so) (some intuitive argument will do)
is it true that $\mathbb{C}$ cannot have any kind of order? (what about ordering by magnitude and (sub-ordering by) phase, for example)
does this have any relation to automorphisms of $\mathbb{C}$ over $\mathbb{R}$? (i.e conjugate numbers)
...
This follows on a related question
Adding Gerry Myerson's comments as an answer to this question so any other people that might see this, see an actual answer. (so please dont thank or upvote me for the answer)
Any algebraically closed filed will have an element whose square is minus one, so being an ordered field isn't going to happen
Of course, the complex numbers can have "an order", but an ordered field is one in which the order and the field operations work together, and that's the kind of order the complex numbers can't have.
There are only 2 field automorphisms of the complex numbers over the reals: the identity, and complex conjugation. Similarly, there are 2 field automorphisms of the rationals adjoin 2√ over the rationals, but those fields are both ordered, so I don't see much joy there.
PS As Gerry Myerson said, "lets wait and see if sth better comes along"..