Axiomatic definition of the complex numbers

Question

The real numbers may be defined axiomatically as a complete ordered field. This description characterises them up to isomorphism.

Question: is there a similar way to define the field of complex numbers?

In contrast to the difficulty of constructing $\mathbb R$ from $\mathbb Q$, the construction of $\mathbb C$ from $\mathbb R$ is straightforward. For instance, $\mathbb C$ may be defined as $\mathbb R^2$ with the operations $(a,b)+(c,d)=(a+b,c+d)$ and $(a,b)\cdot(c,d)=(ac-bd,ad+bc)$. Then, $i$ can be defined as $(0,1)$, and we may identify $\mathbb R$ with the subfield $\{(a,0):a\in\mathbb R\}$. Any model of $\mathbb C$ is isomorphic to this model. However, it feels somewhat artificial to describe a model of $\mathbb C$ as a field $F$ which is isomorphic to $\mathbb R^2$ with the aforementioned operations; it would be like describing a model of $\mathbb R$ as a field which is isomorphic to the collection of Dedekind cuts.

What I'm looking for is a list of properties that are satisfied by a field $K$ if and only if $K$ is a model of $\mathbb C$. This answer uses the Upward Lowenheim-Skolem theorem to conclude that no first-order theory will do the job. I have read that $\mathbb C$ can be described as the algebraic closure of $\mathbb R$, but I don't have the requisite algebraic knowledge to understand this description; moreover, this description still references $\mathbb R$, and so it still feels artificial in some sense. By contrast, the axiomatic description of $\mathbb R$ makes no reference to other fields such as $\mathbb Q$. I'm looking for properties more along the lines of: $K$ is a field, there is an $x\in K$ such that $x^2=-1$, etc.

Answer 1

As a topological field (=field with a topology such that the field operations are continuous) $\Bbb C$ is the only algebraically closed topological field that is Hausdorff, not discrete and locally compact.

Not sure if this is what you want because it refers to a topology, but this doesn't refer to $\Bbb R$ in any way.

Edit: Here's why this holds. (I'll use "locally compact" to mean locally compact and Hausdorff. Non-Hausdorff topological fields always carry the trivial topology, so they are not very interesting anyway.) Actually, it's possible to completely classify all non-discrete locally compact topological fields. (Such a field is also called a local field in number theory.)

A complete list is given as follows:

• The fields of formal Laurent series over a finite field $\Bbb F_q((T))$ with the $T$-adic topology.
• Finite extensions of $\Bbb Q_p$ for a prime $p$.
• $\Bbb R$
• $\Bbb C$

As you can see, the only algebraically closed one on the list is $\Bbb C$, which proves the characterization.

How does one prove such a thing? You can find the details in Markus Stroppel - Locally Compact Groups.

Here's an idea. Let $K$ be a locally compact non-discrete field. The first step is to construct an absolute value on $K$. This is done as follows: Choose a Haar measure $\mu$ for the additive group of $K$. Then for any $a \in K$, the map $\nu:X \mapsto \mu(aX)$ defines another Haar measure. So by uniqueness of Haar measures, you get that $\nu=\alpha \mu$ for some $\alpha \in \Bbb R_{>0}$. Call that scalar $\alpha = |a|$. Then one can show that $a \mapsto |a|$ (extended such that $|0|=0$) defines almost an absolute value on $K$. I'm saying almost because in the case of $\Bbb C$, it's actually the square of an absolute value. (There are more lenient defintions of absolute values that allow this).

The next step is to show that the valued field we get is complete, this is a simple exercise.

Now if $K$ has characteristic $0$, it contains $\Bbb Q$ and hence a completion of $\Bbb Q$. By Ostrowski's theorem, it must contain $\Bbb R$ or $\Bbb Q_p$ for some $p$.

If $K$ has characteristic $p$, then it contains $\Bbb F_p$. Since $\overline{\Bbb F_p}$ has no nontrivial absolute value (everything is zero or a root of unity), it must contain an element $T$ transcendental over $\Bbb F_p$. Thus it contains $\Bbb F_p(T)$ and hence the completion of $\Bbb F_p(T)$. One can show that every such completion is isomorphic to $\Bbb F_q((T))$ for some $q=p^n$.

So now we're in the situation that we have a subfield isomorphic to $\Bbb Q_p, \Bbb R$ or $\Bbb F_q((T))$. All these fields are themselves locally compact. One can show that any extension of locally compact fields $L/F$ must be of finite degree. Then we're almost done. To simplify the finite characteristic case, one can show any finite extension of $\Bbb F_q((T))$ is itself isomorphic to $\Bbb F_{q^n}((X))$ for some $n$.

Answer 2

Not an answer, just an idea...

There exists an element $i\in K$ with $i^2 + 1 = 0$, where $0$ and $1$ are the additive and the multiplicative identities of $K$, respectively.

We've already made good progress by adding this statement to our list of properties/axioms for $K$, since the existence of such an element is essential to the structure of $\mathbb C$. We're still far from characterizing $\mathbb C$, though: the fields $\mathbb Q (i)$ and $\mathbb F_2$ both have this property, but neither is isomorphic to $\mathbb C$. Notice that $\mathbb Q (i)$ is not complete, i.e. "it has holes", and $\mathbb F_2$ is discrete. This suggests that the addition of another axiom stipulating the "completeness" of $K$ might be enough to force $K\cong\mathbb C$. The natural approach would be to introduce a suitable metric $d$ on $K$ and postulate that $K$ be complete with respect to $d$, but this implicitly assumes the reals have been constructed because metrics are, by definition, functions into the non-negative reals.

Thus, the question arises: can we characterize the completeness of a space without the use of metrics? One idea is to stipulate that $K$ be a topological space, i.e. can be endowed with a topology, and see if there's any topological characterizations of the completeness of $\mathbb C$. It is known that such a characterization exists for $\mathbb R$ (see this MathOverflow post), so by analogy, there might be one for $\mathbb C$.

Edit

Have a look at Lukas Heger's answer. It shows the topological characterization of $\mathbb C$ we seek does exist.