Do there exist some results on the theory of $\mathbb{C}$ in the language $\{0,1+, \times, \overline{\cdot}\}$, where $\overline{\cdot}$ is the conjugation map $\overline{a+ib} = a - ib$?

I'm wondering in particular if it has QE (or if a sensible expansion of the language has QE), and if it's decidable. But properties like model-completeness, (bi)-interpretability with standard, well-understood structures etc. would be helpful to know too.


1 Answer 1


It is a decidable theory, because it is interpretable in the real-closed field $\langle\mathbb{R},+,\cdot,0,1\rangle$, which has a decidable theory. We can interpret complex numbers $a+bi$ as pairs of real numbers $(a,b)$, and the complex structure, including conjugation, is definable in the reals. (Indeed, this is easily seen to be a bi-interpretation, since we can define $\mathbb{R}$ via conjugation in $\mathbb{C}$.) By Tarski's theorem on real-closed fields, that theory is decidable, and so we can decide the complex theory also.

Basically, for any given statement in the complex field with conjugation, we can translate it to a question in the real-closed field. By Tarski's result, that question is equivalent to a quantifier-free assertion in the real-closed field $\langle\mathbb{R},+,\cdot,<\rangle$, which we can then easily decide.

Since the interpretation doesn't involve any quantifiers (note that we can clear the use of unary minus in conjugation by moving negatives to the other side of any equation), it follows as Alex Kruckman notes in the comments that we will get model-completeness of the complex field with conjugation.

My earlier claim that we get full QE for $\mathbb{C}$ stumbles on the fact that $\langle \mathbb{R},+,\cdot\rangle$ doesn't have QE, since you need the order in Tarski's result. As Alex mentions, you can define the positive real line in $\langle\mathbb{C},+,\cdot,\bar{}\rangle$, but this will not be quantifier-free definable.

If we add the real and imaginary part operators and the order for the real line as a relation on $\mathbb{C}$, however, then we will get QE in the corresponding expansion, and this expansion will also be bi-interpretable with the real-closed field.

  • 7
    $\begingroup$ There's a little bit of a hitch with QE, because the theory of $\langle\mathbb{R},+,\cdot,0,1\rangle$ does not have QE! You have to throw in the ordering $\leq$. So for example the formula $\exists x\, (x = \overline{x}\land x^2 = y)$ which defines the non-negative real numbers in $\mathbb{C}$ is not equivalent to a quantifier-free formula. On the other hand, model-completeness transfers from $\mathbb{R}$ to $\mathbb{C}$ with conjugation for exactly the reason you give: the interpretation involves only quantifier-free formulas. $\endgroup$ Nov 6, 2019 at 22:40
  • $\begingroup$ Ah, you are right. I have edited. $\endgroup$ Nov 6, 2019 at 23:25
  • $\begingroup$ @JoelDavidHamkins When you mention bi-interpretation above, do you mean that it's a biinterpretation of the theories, or merely of the models? I've asked a question about this here. $\endgroup$ Sep 7, 2021 at 15:56
  • $\begingroup$ Yes, it is a (parametric) bi-interpretation of theories. $\endgroup$ Sep 8, 2021 at 7:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.