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?
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2$\begingroup$ Marker's book has this. $\endgroup$– Benjamin SteinbergCommented Jun 29, 2018 at 11:55
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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.
answered Jun 29, 2018 at 11:18
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$\begingroup$ 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. $\endgroup$– tomaszCommented Jun 29, 2018 at 14:35
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3$\begingroup$ @tomasz There is only one algebraically closed field of characteristic $0$ and cardinality $2^\omega$. $\endgroup$ Commented Jun 29, 2018 at 14:38
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$\begingroup$ @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... $\endgroup$ Commented Jun 29, 2018 at 15:26
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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. $\endgroup$ Commented Jun 29, 2018 at 16:49
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1$\begingroup$ @AlexKruckman : Indeed, but it's a standard fact (unless I'm mistaken) that a nontrivial ultraproduct of countable structures has cardinality $2^{\aleph_0}$ $\endgroup$ Commented Jun 29, 2018 at 16:57