6
$\begingroup$

This is a reference request. Let $A$ be a formula in the language of rings which is of the form $\forall_{x_1}\dots\forall_{x_n}\exists_{y_1}\dots\exists_{y_m} F$, where $F$ is quantifier-free. I once read that if $A$ is valid for all finite fields, then it is valid for $\mathbb C$. Where would I find a proof of this statement?

$\endgroup$
1
  • 2
    $\begingroup$ Marker's book has this. $\endgroup$ – Benjamin Steinberg Jun 29 '18 at 11:55
11
$\begingroup$

I don't know a reference but it's not that hard to prove.

Let's call your formula $\varphi$. Then by classical arguments, the models of $\varphi$ are closed under directed union, in particular since $\varphi$ holds in any finite field, it holds in any $\overline{\mathbb{F}_p} = \displaystyle\bigcup_{n<\omega}\mathbb{F}_{p^n}$, the algebraic closure of the field with $p$ elements, which is the directed union of its finite subfields.

But then by Los's theorem, $\varphi$ holds in $\displaystyle\prod_{p\in \mathbb{P}}\overline{\mathbb{F}_p}/\mathcal{U}$ for any $\mathcal{U}$ ultrafilter on $\mathbb{P}$. Picking a non-principal ultrafilter yields a field isomorphic to $\mathbb{C}$, so the formula holds in $\mathbb{C}$ too.

$\endgroup$
11
  • $\begingroup$ I don't see why the resulting field would be isomorphic to the complex field (in absence of CH). But it will be an algebraically closed field of characteristic zero, which is what is important here. $\endgroup$ – tomasz Jun 29 '18 at 14:35
  • 3
    $\begingroup$ @tomasz There is only one algebraically closed field of characteristic $0$ and cardinality $2^\omega$. $\endgroup$ – Emil Jeřábek Jun 29 '18 at 14:38
  • $\begingroup$ @tomasz your argument is perhaps simpler, but the proof that the theory algebraically closed fields of characteristic zero is complete goes through (at least the proof I know) its $\kappa$-categoricity, for $\kappa$ uncountable, and since this field and $\mathbb{C}$ have the same cardinal... $\endgroup$ – Maxime Ramzi Jun 29 '18 at 15:26
  • 1
    $\begingroup$ The fact that the ultraproduct has size $2^{\aleph_0}$ is not completely trivial. The trick is finding a continuum family of sequences $(x_p)_{p\in \mathbb{P}}$ which you know to be distinct in the ultraproduct, i.e. such that $\{p\in \mathbb{P}\mid x_p = y_p\}$ is finite whenever $(x_p)$ and $(y_p)$ are distinct sequences in the family. $\endgroup$ – Alex Kruckman Jun 29 '18 at 16:49
  • 1
    $\begingroup$ @AlexKruckman : Indeed, but it's a standard fact (unless I'm mistaken) that a nontrivial ultraproduct of countable structures has cardinality $2^{\aleph_0}$ $\endgroup$ – Maxime Ramzi Jun 29 '18 at 16:57

Your Answer

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