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I know that $\mathbb R$ is - up to a unique isomorphism of ordered fields - the unique complete ordered field. Is there a similar characterization of the complex numbers? I guess that $\mathbb C$ can not simply be defined as a field with particular properties, but that additional structure must be part of the definition.$^1$

Anyways, I am sure that many people have thought about this issue and that there are several possible approaches. I would like to learn about them.

$^1$ E.g. two functions from $\mathbb C$ onto $\mathbb R$ (the projection onto the real part and the projection onto the imaginary part).

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    $\begingroup$ I don't think you can get away from the fact that complex conjugation is an automorphism of $\mathbb C$. That means you'll never have a unique isomorphism. $\endgroup$
    – eyeballfrog
    Mar 27 at 22:37
  • $\begingroup$ @eyeballfrog In this paper $\mathbb C$ is introduced as a field $F$ together with an automorphism $F\ni z\mapsto \overline{z}\in F$ and an ordering on the set $\{z:\overline{z}=z\}$. And indeed this does only define $\mathbb C$ up to "duique" isomorphism (see theorem 6). $\endgroup$
    – Filippo
    Mar 28 at 9:53

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Up to "duique" isomorphism (there are two of them), it is the only algebraic closure of $\Bbb R$. I don't think you can do unique, as complex conjugation will always be present.

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  • $\begingroup$ We can only count the fields up to isomorphism anyway, since we don't want $\mathbb{R}(i)$ to be different from $\mathbb{R}(\square)$ just using the symbol $\square$ in place of $i$. So the automorphism doesn't matter - there's just the one field. $\endgroup$
    – aschepler
    Mar 27 at 13:50
  • $\begingroup$ @aschepler Sure, but even though there are plenty of candidate objects, the uniqueness of any isomorphism between them would still often be a desirable property. For instance, a product object in a given category (if it exists) is very much non-unique, but for any two objects that satisfy the role of $X\times Y$, there is only a single isomorphism between them that work with the projection maps. Not so in this case. $\endgroup$
    – Arthur
    Mar 27 at 13:53
  • $\begingroup$ By that logic, doesn't $\mathbb{Q}[x]/(x^2-2)$ describe two fields? $\endgroup$
    – aschepler
    Mar 27 at 13:58
  • $\begingroup$ @aschepler No. I have no idea what that even means. But it means that if you have another field $F$, and $F\cong \Bbb Q[x]/(x^2-2)$, then there are two isomorphisms $F \to Q[x]/(x^2-2)$, as $x\mapsto -x$ yields an automorphism of $\Bbb Q[x]/(x^2-2)$. Since the question specifically asked about unique isomorphisms, I felt it apt to address the fact that no, you can't get unique isomorphisms this way, there will always be two. $\endgroup$
    – Arthur
    Mar 27 at 14:02
  • $\begingroup$ So the statement is essentially "all fields with property $P$ are isomorphic to each other, and do not have non-trivial automorphisms"? $\endgroup$
    – aschepler
    Mar 27 at 14:04
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This answer notes that up to isomorphism there is only one field that is a vector space of dimension 2 over $\mathbb R$. Does that count as characterizing $\mathbb C$ up to isomorphism?

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  • $\begingroup$ Thank you for your answer. To be precise, it characterizes $\mathbb C$ up to "duique" isomorphism (not unique), right? So my answer to your question would be "no", but I still consider your answer useful (+1), just like Arthur's answer. $\endgroup$
    – Filippo
    Mar 27 at 13:54
  • $\begingroup$ In fact it looks like some people think that $\mathbb C$ can not be defined up to a unique isomorphism, so maybe I have to be content with this. $\endgroup$
    – Filippo
    Mar 27 at 13:59

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