# Forall exists formula

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?

• Marker's book has this. – Benjamin Steinberg Jun 29 '18 at 11:55

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.
• @tomasz There is only one algebraically closed field of characteristic $0$ and cardinality $2^\omega$. – Emil Jeřábek Jun 29 '18 at 14:38
• @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... – Maxime Ramzi Jun 29 '18 at 15:26
• 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. – Alex Kruckman Jun 29 '18 at 16:49
• @AlexKruckman : Indeed, but it's a standard fact (unless I'm mistaken) that a nontrivial ultraproduct of countable structures has cardinality $2^{\aleph_0}$ – Maxime Ramzi Jun 29 '18 at 16:57