>>>>A general isomorphism cannot exist since R is an ordered field but C is not...
>>>
>>>Why isn't C ordered?
>>>
>>The order axioms aren't satisfied. For example, is i greater than -i?
>
>We can set an order defined as follows:
>
>z1 <= z2 iif ((y1 < y2) or ((y1 = y2) and (x1 <= x2)))
>
>Where z=x+y*i (or zk=xk+yk*i). (k is an indice)
>
>This relation is reflexive, tranzitive and antisymetric. So, C can be ordered.
>
>Anyway, the fact that we don't have an order on a set doesn't mean it cannot be isomorphic with a set for which we have an order.
>
>As an example, we can have an isomorphism between a circle and R. R is ordered, but we don't have (usual) an order on a circle.
>
I'll buy that, but...
Ordered fields must obey a few certain properties. There are several reasons why C is not an ordered field, but basically the problem is that i is not defined in R (though i**2 is).

One of the properties for an ordered field that results from order axioms is that for any number, say x, in a set, one of these statements is true: 'x = 0', 'x is > 0', '-x > 0'. Clearly, this is not true for i. Thus C is not an ordered field.

On another and perhaps easier track, it's my understanding that an isomorphism must maintain a 1-1 correspondence between sets. Intuitively, you can see that C and R are not 1-1. Proof left as exercise :~)

My old math profs might not like this explanation for its informality, but do you get what I'm driving at?