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A theory is categorical if it has a unique model up to isomorphism. First-order Peano arithmetic is not categorical, but second-order Peano arithmetic is categorical, with the natural numbers as its unique model. The first-order theory of real closed fields is not categorical, but the second-order theory of Dedekind-complete ordered fields is categorical, with the real numbers as its unique model. ZFC is not categorical, but Morse-Kelley Set Theory with an appropriate axiom about inaccessible cardinals is categorical.

My question is, what theory of the complex numbers is categorical? The first order theory of algebraically closed fields of characteristic zero is not categorical, because both the field of algebraic complex numbers and the field of complex numbers satisfy it. So is there some second-order axiom we can add to this theory to make it categorical?

Keshav Srinivasan
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In full second-order logic you can characterize $\mathbb{C}$ in the language $(+,\cdot,0,1)$ up to isomorphism using the following axioms:

  1. First-order axioms stating that $(M,+,\cdot,0,1)$ forms an algebraically closed field of characteristic zero.
  2. A second-order axiom stating that there is a subset $R$, a function $f$ and a relation $<$ such that $(R,+,\cdot,0,1,<)$ forms a Dedekind-complete ordered field and $f$ is a bijection between $R$ and the whole structure $M$.

This theory is categorical. Why? As Olivier Roche alluded to in his answer, the theory of algebraically closed fields of characteristic zero has a unique model in each cardinality $\lambda > \aleph_0$. Moreover, every Dedekind-complete ordered field has the cardinality of $|\mathbb{R}|$, so the models of the theory above are precisely the algebraically closed fields of characteristic zero of cardinality $|\mathbb{R}|$, so they are all isomorphic to $(\mathbb{C},+,\cdot,0,1)$.

Z. A. K.
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    Could we avoid the $f$ part if we say every number can be written as $a+bi$ where $a,b\in R$? – Keshav Srinivasan Feb 25 '20 at 09:36
  • @KeshavSrinivasan Yes! – Alex Kruckman Feb 25 '20 at 14:34
  • @AlexKruckman Can you post that as an answer, along with a proof of why that change works? – Keshav Srinivasan Feb 25 '20 at 14:42
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    @KeshavSrinivasan I don't think it's worth writing a separate answer, since Z.A.K.'s answer contains the key ideas. Note that you can't write down "every number can be written as $a+bi$ with $a,b\in R$", since you don't have a constant symbol for $i$. But already the axiom $\exists i, \forall z, \exists x, \exists y, (R(x)\land R(y)\land (z = x + yi))$ implies that in any model $M$, there is an injective function $M\to R\times R$, so $|M|\leq |R|^2 = |R|$ so by Cantor-Schröder-Bernstein, $|M| = |R|$, and then the last paragraph of Z.A.K.'s answer applies. – Alex Kruckman Feb 26 '20 at 14:34
  • @AlexKruckman OK thanks. – Keshav Srinivasan Feb 26 '20 at 22:17
  • Is $f$ a bijection between the set of complex numbers and the set of real numbers? I thought such a function cannot exist. – Filippo Oct 24 '21 at 11:02
  • @Filippo: Why did you think that? The set of complex numbers and the set of real numbers both have the cardinality of the continuum, so they are both in bijection with the set $\mathcal{P}(\mathbb{N})$ consisting of all subsets of the naturals. Since the composition of two bijections is a bijection, there is a bijection between $\mathbb{C}$ and $\mathbb{R}$ – Z. A. K. Oct 24 '21 at 11:47
  • @Z.A.K. Thank you! Okay, so we require the existence of such a function $f$ solely because we want $M$ to have the same cardinality as the real numbers? – Filippo Oct 24 '21 at 15:35
  • I.e. this function does not have a special interpretation (e.g. as the function assigning to each complex number its real number), but you only wrote out the definition of "$R$ and $M$ have the same cardinality"? I have very little knowledge in this topics, that's why I'm asking. – Filippo Oct 24 '21 at 15:38
  • Yes, the point of the second-order axiom is to force R and M to have the same cardinality. Since R has the cardinality of the continuum, so will M. And there is only one algebraically closed field of cardinality continuum. – Z. A. K. Oct 24 '21 at 19:52
  • @Z.A.K. Thank you for the clarification! "There is only one algebraically closed field of cardinality continuum" - Is the isomorphism between two such fields unique? Or can there be several isomorphisms? – Filippo Oct 24 '21 at 20:24
  • @Filippo: Alas, no: $\mathbb{C}$ is isomorphic to itself in a lot of non-obvious ways. And then there's complex conjugation. – Z. A. K. Oct 24 '21 at 22:00
  • @Z.A.K. Thank you for the link! However, I not sure what you are trying to say with the second sentence, "And then there's complex conjugation". Are you referring to the fact that given an algebraically closed field of cardinality continuum, there is not a preferred automorphism we can call conjugation? – Filippo Oct 27 '21 at 06:25
  • @Filippo: I am referring to the fact that $\mathbb{C}$ is isomorphic to itself in many non-obvious ways, and in one obvious way, which is complex conjugation. The latter already ensures that algebraically closed fields of cardinality continuum are not uniquely isomorphic. (Btw we'll have to stop this exchange at this point. I think your original question was clarified, and this is now starting to turn into a discussion, which is not allowed according to the Math.SE rules) – Z. A. K. Oct 27 '21 at 06:56
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Since $\mathbb{C}$ is infinite, there's no first order theory whose only model is supported by $\mathbb{C}$. This is a consequence of Löwenheim–Skolem theorem.

On the other hand, it is well known that the theory of $(\mathbb{C}, +, \cdot, 0, 1)$ is $\lambda$-categorical for every uncountable $\lambda$.

Olivier Roche
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