Can we construct $\Bbb C$ without first identifying $\Bbb R$? Sometimes it is useful to consider $\Bbb C$ as our primitive and identify $\Bbb R$ as a subset of $\Bbb C$.  Thus we can define $\Bbb R$ (or at least a set with all of the interesting properties of $\Bbb R$) from $\Bbb C$.
This suggests to me that there is some way of constructing $\Bbb C$ without first constructing (or taking as a primitive) $\Bbb R$.  However, I've never seen such a construction of $\Bbb C$ (a quick Google search didn't provide me one, either).  I've the Cayley-Dickson construction and the matrix construction many times, but are they the only known ways of constructing $\Bbb C$?
My question:

Is there a way to construct the set of complex numbers without already having (or first constructing) the real numbers?

 A: Any algebraically closed field of characteristic $0$ having the same cardinality as $\mathbb C$ is isomorphic (non-canonically) to $\mathbb C$. This allows one to construct a lot of fields which are abstractly isomorphic to $\mathbb C$ without ever looking at the real numbers. 
For example, the algebraic closure of $\mathbb Q_p$ is isomorphic to $\mathbb C$. However, no isomorphism between them is continuous, so this is not very interesting if you're interested in $\mathbb C$ as a topological field.
A: If you want to avoid $\Bbb R$ and just use general machinery, one way to do it is to use $\Bbb Q(i)$ or any finite extension of $\Bbb Q$ which has zero real embeddings. You can ensure this by taking the extension to be cyclotomic, for example. Then you know there is a norm on the vector space $\Bbb Q(i)$ given by
$$\lVert a+bi\rVert=|a|+|b|.$$
It's easily verified that it is archimedean--this is handy because it will give you a copy of $\Bbb R$ as a subset when you're finished making $\Bbb C$. Now, you can verify that addition, subtraction, multiplication, and inversion of non-zero elements is continuous so that you have a topological field.
Then by forming the metric completion and declaring it to be $\Bbb C$, you automatically have that this is a field because of continuity of the field operations. It is not otherwise obvious that your set of equivalence classes should form such a thing.
A: Another approach: start with the commutative semiring $\mathbb{R}_{\geq 0}.$ Write:


*

*$\mathbf{CSem}$ for the category of commutative semirings

*$\mathbf{CRing}$ for the category of commutative semirings with a distinguished element $(-1)$ such that $(-1)+1=0$, also known as the category of commutative rings.

*$\mathbf{CRing}[i]$ for the category of commutative semirings with a distinguished element $i$ such that $i^2+1=0$.


(All my semirings have a $0$ and a $1$.)
Now there is a forgetful functor $U_\mathrm{I} : \mathbf{CRing} \rightarrow \mathbf{CSem}$ and another $U_\mathrm{II} : \mathbf{CRing}[i] \rightarrow \mathbf{CSem}$. Both have left adjoints. Just like $\mathbb{R}$ is the image of $\mathbb{R}_{\geq 0}$ under the left-adjoint to $U_\mathrm{I}$, similarly $\mathbb{C}$ is the image of $\mathbb{R}_{\geq 0}$ under the left-adjoint to $U_\mathrm{II}$.
(You should check that the details play out as expected, since I have not.)
