We all know that the complex field structure $\langle\mathbb{C},+,\cdot,0,1\rangle$ is interpretable in the real field $\langle\mathbb{R},+,\cdot,0,1\rangle$, by encoding $a+bi$ with the real-number pair $(a,b)$. The corresponding complex field operations are expressible entirely within the real field.

Meanwhile, many mathematicians are surprised to learn that the converse is not true — we cannot define a copy of the real field inside the complex field. (Of course, the reals $\mathbb{R}$ are a *subfield* of $\mathbb{C}$, but this subfield is not a definable subset of $\mathbb{C}$, and the surprising fact is that there is no definable copy of $\mathbb{R}$ in $\mathbb{C}$.) Model theorists often prove this using the core ideas of stability theory, but I made a blog post last year providing a comparatively accessible argument:

The argument there makes use in part of the abundance of automorphisms of the complex field.

In a comment on that blog post, Ali Enayat pointed out that the argument therefore uses the axiom of choice, since one requires AC to get these automorphisms of the complex field. I pointed out in a reply comment that the conclusion can be made in ZF+DC, simply by going to a forcing extension, without adding reals, where the real numbers are well-orderable.

My question is whether one can eliminate all choice principles, getting it all the way down to ZF.

**Question.** Does ZF prove that the real field is not interpretable in the complex field?

I would find it incredible if the answer were negative, for then there would be a model of ZF in which the real number field was interpretable in its complex numbers.

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