Hurkyl said:
The ordered field axioms imply that x*x > 0 for any nonzero x.
Yikes. This is not the way to go. Drop the order axioms. I just want a way to say that any two complex numbers satisfy trichotomy (<, >, =). I was hoping to do this by mapping C onto R via a one-to-one function and then comparing the real numbers produced by that function.
Strictly speaking, if I have z and q (both complex) and G() is my one-to-one (preferably continuous) function of C onto R, then I want to be able to say that:
1.) G(z)=G(q) or
2.) G(z)>G(q) or
3.) G(z)<G(q)
In this fashion, I could define z>q, z<q, and z=q in a similar fashion to the real numbers.
Ideally (this may be far-fetched), the properties of this complex ordering would be generalizations of order in the real numbers. From the complex orders, the real orders could be derived by simply assuming that Im(z) = Im(q) = 0.
As far as that goes, I don't know what's possible and what has already been done.
I do know that ordering the complex numbers is considered to be a null point. I also think, but cannot prove, that any such ordering as the one I described will not be unique.