Proof. sketch [14010A]

$\implies$: Suppose $[K : F] = n < \infty$. Let $\alpha_{1}, \dots, \alpha_{n}$ be a basis of $K$ as an $F$-vector space. Then $K = F(\alpha_{1}, \dots, \alpha_{n})$. Moreover, $F(\alpha_{i}) \subseteq K$ implies $[F(\alpha_{i}) : F] \leq [K : F] < \infty$, then by Proposition 1403, $\alpha_{j}$ is algebraic over $F$.

$\impliedby$: By Theorem 1207, $$[K : F] = [F(\alpha_{1}, \dots, \alpha_{n}) : F] \leq [F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] [F(\alpha_{1}, \dots, \alpha_{n-1}) : F].$$ By HW2 Prob 1, $[F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] \leq [F(\alpha_{n}) : F]$. So $$[K : F] \leq [F(\alpha_{n}) : F][F(\alpha_{1}, \dots, \alpha_{n-1}) : F]$$ Therefore, by induction we have $[K : F] \leq [F(\alpha_{n}) : F][F(\alpha_{n-1}) : F] \cdots [F(\alpha_{1}) : F]$, in which every $[F(\alpha_{i}) : F] < \infty$.

$\implies$: Take $S = K$.

$\impliedby$: Given $\beta \in K = F(S)$, $\beta \in F(\alpha_{1}, \dots, \alpha_{n})$ for some $\alpha_{i} \in S$. The $\alpha_{i}$’s are algebraic, so $F(\alpha_{1}, \dots, \alpha_{n}) / F$ is finite. Hence $[F(\beta) : F]$ is finite since $F(\beta) \subseteq F(\alpha_{1}, \dots, \alpha_{n})$. Then by Proposition 1403, $\beta$ is algebraic over $F$. Therefore $K / F$ is algebraic.