$\forall\exists$-sentence true in all finite fields is true in algebraically closed fields? Let $\phi$ be a $\forall\exists$-sentence in the language of rings that is true in every finite field. I want to show that $\phi$ is true in every algebraically closed field.
I know that ACF is strongly minimal, model-complete, and has QE, but is not complete. Since it is model-complete, it is inductive and therefore equivalent to an $\forall\exists$-theory $T$, so I am trying to show that $\phi\in T$, but I haven't had much luck in doing so.
I think where I am getting lost is that I have no idea how to use the assumption on finite fields- is there some detail in the relation between finite fields and algebraically closed fields I am missing?
 A: Expand $\varphi$ out as $\forall\bar{x}\exists\bar{y}\psi(\bar{x},\bar{y})$, where $\psi$ is atomic, and fix a prime $p$; we'll first show that $\varphi$ holds in every algebraically closed field of characteristic $p$. Since the theory of algebraically closed fields of characteristic $p$ is complete, it suffices to show that $\varphi$ holds in a single model of this theory, so let $F$ be the algebraic closure of $\mathbb{F}_p$.
To show $F\models\varphi$, let $\bar{a}$ be any $|\bar{x}|$-tuple from $F$; we want to show that $F\models\exists\bar{y}\psi(\bar{a},\bar{y})$. The field $E:=\mathbb{F}_p(\bar{a})$ is a finitely generated algebraic extension of $\mathbb{F}_p$, hence finite; by hypothesis thus $E\models\varphi$, and so in particular $E\models\exists\bar{y}\psi(\bar{a},\bar{y})$. Thus we may find a tuple $\bar{b}$ from $E$ such that $E\models\psi(\bar{a},\bar{b})$, and then $F\models\psi(\bar{a},\bar{b})$, as needed.

Okay, so this shows that $\varphi$ holds in every algebraically closed field of positive characteristic. But now by compactness and completeness of $\mathrm{ACF}_0$, this means that $\varphi$ also holds in every algebraically closed field of characteristic $0$ (why?), and so we are done.
