Proposition. necessary and sufficient conditions for $K / F$ to be finite or algebraic [14010]

    Let $K / F$ be a field extension.

    1. $K / F$ is finite if and only if there exist $\alpha_{1}, \dots, \alpha_{n} \in K$ algebraic over $F$ such that $K = F(\alpha_{1}, \dots, \alpha_{n})$.
    2. $K / F$ is algebraic if and only if there exists $S \subseteq K$ such that every $\alpha \in S$ is algebraic over $F$ and $K = F(S)$.

    A finite extension requires a finite number of algebraic generators ($\alpha_{1}, \dots, \alpha_{n}$).

    An algebraic extension requires a set $S$ of algebraic generators, but $S$ can be infinite.

    Not all algebraic extensions are finite. For example, the algebraic closure of $\mathbb{Q}$, denoted as $\overline{\mathbb{Q}}$, is an algebraic extension over $\mathbb{Q}$, but $[\overline{\mathbb{Q}} : \mathbb{Q}] = \infty$.

    Proof. sketch [14010A]

      1. $\implies$: 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 $[F(\alpha_{i}) : F] < \infty$.

      2. $\implies$: 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.