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?


2 Answers 2


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 $(\mathbb{C},+,\cdot,0,1)$.

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    $\begingroup$ Could we avoid the $f$ part if we say every number can be written as $a+bi$ where $a,b\in R$? $\endgroup$ Commented Feb 25, 2020 at 9:36
  • $\begingroup$ @KeshavSrinivasan Yes! $\endgroup$ Commented Feb 25, 2020 at 14:34
  • $\begingroup$ @AlexKruckman Can you post that as an answer, along with a proof of why that change works? $\endgroup$ Commented Feb 25, 2020 at 14:42
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    $\begingroup$ @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. $\endgroup$ Commented Feb 26, 2020 at 14:34
  • $\begingroup$ @AlexKruckman OK thanks. $\endgroup$ Commented Feb 26, 2020 at 22:17

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 $\lambda$.

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    $\begingroup$ I’m talking about second-order theories here. Just like the second-order theory of real numbers is categorical. $\endgroup$ Commented Feb 25, 2020 at 8:12
  • $\begingroup$ Do you have a reference for the $\lambda$-categoricity of $\mathbb{C}$? Cheers $\endgroup$ Commented Feb 28, 2021 at 18:46
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    $\begingroup$ @aidangallagher4 You'll find this eg in "A Course in Model Theory" by Bruno Poizat $\endgroup$ Commented Mar 10, 2021 at 17:43

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