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

Question

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.

Answer 1

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.

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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 $Th(\mathbb{C})$.)