Proof. sketch [14011A]

Let $\alpha, \beta$ be any elements of $E$, it suffices to show that $\alpha + \beta, -\alpha, \alpha \beta, \alpha ^{-1} \in E$. Since they are all in $F(\alpha, \beta)$, it suffices to show that $F(\alpha, \beta) \subseteq E$. By Proposition 1410, $F(\alpha, \beta) / F$ is finite and hence is algebraic. Therefore, $F(\alpha, \beta) \subseteq E$.