# Show that the set of real numbers is not definable in the field of complex number

Like what the title suggest, I wish to show that $$\mathbb{R}$$ cannot be define in $$\mathbb{C}$$. I want to make use of the following proposition ;

(David Marker) Fix a structure $$M$$, if $$X \subset M$$ is definable over a set of parameters $$A$$, then every automorphisms of $$M$$ that fixes $$A$$ pointwise must fixes $$X$$ setwise.

So if we assume the contrary that $$\mathbb{R}$$ is definable in $$\mathbb{C}$$, then it is definable over a finite set of parameters, $$A$$, from $$\mathbb{C}$$. Now all that remains is to find an automorphism $$\sigma$$ that fixes everypoint in $$A$$ but maps some real numbeers into complex numbers.

For whatever reason, I'm having some difficulties coming up with such an explicit $$\sigma$$. Any help or insights is deeply appreciated.

• You can't write one down explicitly, you'll have to use something like transfinite induction. See math.stackexchange.com/a/2092817/86856 for an example of the sort of argument involved. – Eric Wofsey Feb 1 at 6:14
• See this similar question for a stronger result: $\mathbb{R}$ is not even interpretable in $\mathbb{C}$. The question gives an argument using the notion of stability, and my answer gives a more elementary argument using automorphisms. – Alex Kruckman Feb 1 at 15:12

As Eric Wofsey said, actually building an automorphism of $$\mathbb{C}$$ which moves a real is difficult: while we can prove that for every transcendental real $$r$$ there is a field automorphism of $$\mathbb{C}$$ which moves $$r$$, this proof is $$(i)$$ a transfinite recursion argument and $$(ii)$$ relies on the axiom of choice.

There is, however, an automorphism argument which does work:

We'll show a stronger result - that for any transcendental real number $$r$$ and any formula $$\varphi(x)$$ such that $$\mathbb{C}\models\varphi(r)$$, there is a non-real complex number $$s$$ such that $$\mathbb{C}\models\varphi(s)$$. The argument uses automorphisms and elementary submodels, and goes as follows (outlined only - it's a good exercise to fill it all in):

• Fix a transcendental real $$r$$, a transcendental non-real complex number $$s$$, and a formula $$\varphi(x)$$ such that $$\mathbb{C}\models\varphi(r)$$. Let $$M$$ be a countable elementary submodel of $$\mathbb{C}$$ which contains $$r$$ and $$s$$ - we know such an $$M$$ exists by the downward Lowenheim-Skolem theorem.

• Note that we've replacing the not-obviously-well-orderable field $$\mathbb{C}$$ by a countable, and hence well-orderable, field $$M$$. Intuitively, this should (and does) make the axiom of choice irrelevant here.
• Now show that $$M$$ is a countable algebraically closed field of characteristic zero and infinite transcendence degree.

• By a standard back-and-forth argument, there is an automorphism of $$M$$ swapping $$r$$ and $$s$$, so $$M\models\varphi(s)$$. Back-and-forth arguments are still recursive constructions, but they don't use transfinite recursion; also, this is probably a result you've already proved.

• Remembering how $$M$$ and $$\mathbb{C}$$ are related, conclude that $$\mathbb{C}\models\varphi(s)$$ as desired.

Remark $$1$$. We can avoid elementary submodels entirely by working with Ehrenfeucht-Fraisse games instead; however, elementary submodel juggling is an important technique to learn.

Remark $$2$$. An even stronger fact is true: $$\mathbb{C}$$ is strongly minimal, meaning that for any model $$N$$ of $$Th(\mathbb{C})$$, every definable-with-parameters subset of $$N$$ is either finite or cofinite. (First, prove that $$\mathbb{C}$$ has quantifier elimination, and then show that every quantifier-free formula with parameters defines either a finite or a cofinite set; then lift this to every model of $$Th(\mathbb{C})$$.)