I have been thinking the following question for several days:
Let $\mathcal{L}$ be the first-order language of theory of rings. Also let $\exists x \forall y \varphi(x,y)$ be a sentence of $\mathcal{L}$ with $\varphi(x,y)$ quantifier-free. Consider the infinite algebraic extension L of $\mathbb{Q}$ and finite subextensions $\mathbb{Q} \subseteq K_i \subset L$ for all i .
From field theory we know
L is algebraic extension of $\mathbb{Q}$ if and only if L is direct limit of its finite subextensions.
Then L=$\varinjlim K_i=\bigcup_iK_i$. Now let $K_i \models \exists x \forall y \varphi(x,y)$ for all i and $\mathbb{Q} \models \exists x \forall y \varphi(x,y)$.
Is it necessary that $L \models \exists x \forall y \varphi(x,y)$ ?
I try to give a counterexample but failed.
Any hints or answers are welcomed. Thank you!