2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Feb 1 at 6:14
  • $\begingroup$ 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. $\endgroup$ – Alex Kruckman Feb 1 at 15:12
6
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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