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, $F(\alpha, \beta) \subseteq E$.

$\implies$: Given $\alpha \in L$, it is algebraic over $K$ since $L / K$ is algebraic. So there exist $a_{0}, \dots, a_{n-1} \in K$ such that $$ a(\alpha) = \alpha^{n} + a_{n-1} \alpha^{n-1} + \cdots + a_{0} = 0. $$ Hence $\alpha$ is algebraic over $F(a_{0}, \dots, a_{n-1})$, and $$ [F(\alpha, a_{0}, \dots, a_{n-1}) : F(a_{0}, \dots, a_{n-1})] < \infty. $$ Since $K / F$ is algebraic, each $a_{i}$ is algebraic over $F$, so $[F(a_{i}) : F] < \infty$ for all $i$. Thus $$ [F(a_{0}, \dots, a_{n-1}) : F] < \infty. $$ By Theorem 1207, we write $$ [F(\alpha, a_{0}, \dots, a_{n-1}) : F] = [F(\alpha, a_{0}, \dots, a_{n-1}) : F(a_{0}, \dots, a_{n-1})][F(a_{0}, \dots, a_{n-1}) : F], $$ then we can conclude that $[F(\alpha, a_{0}, \dots, a_{n-1}) : F] < \infty$. Since $F \subseteq F(\alpha) \subseteq F(\alpha, a_{0}, \dots, a_{n-1})$, it follows that $[F(\alpha) : F] < \infty$. Then by Proposition 1403, $\alpha$ is algebraic over $F$. Since $\alpha$ is arbitrary, $L / F$ is algebraic.

$\impliedby$: Since $L / F$ is algebraic and $K \subseteq L$, $K / F$ is also algebraic. Moreover, every $\alpha \in L$, there exist $a_{0}, \dots, a_{n-1} \in F$ such that $$ a(\alpha) = \alpha^{n} + a_{n-1}\alpha^{n-1} + \cdots + a_{0} = 0. $$ Since $F \subseteq K$, each $a_{i} \in K$, hence $a(x) \in K[x]$, so $\alpha$ is algebraic over $K$. Therefore, $L / K$ is algebraic.

Clearly, $f(x)$ splits completely in $F(\alpha_{1}, \dots, \alpha_{n})$.

Claim: $F \subseteq E \subseteq K$ and $f(x)$ splits completely over $E$ $\implies$ $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$.

Proof of the Claim: Suppose $f(x) = b(x - \beta_{1})\cdots(x - \beta_{n})$ with $b, \beta_{1}, \dots, \beta_{n} \in E$. Then we have $$ a(x - \alpha_{1}) \cdots (x - \alpha_{n}) = b(x - \beta_{1}) \cdots (x - \beta_{n}) $$ in $K[x]$. Since $K[x]$ is a UFD, after reordering we have $\alpha_{i} = \beta_{i}$ for all $1 \leq i \leq n$. And $F(\beta_{1}, \dots, \beta_{n}) \subseteq E$. Hence $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$.

1. Suppose $F \subseteq E \subseteq F(\alpha_{1}, \dots, \alpha_{n})$ and $f(x)$ splits completely over $E$, then by our claim, $E = F(\alpha_{1}, \dots, \alpha_{n})$. So $F(\alpha_{1}, \dots, \alpha_{n})$ is a splitting field of $f(x)$.
2. Since $f(x)$ splits completely over $E$, by our claim, $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$. Moreover, $E$ is a splitting field of $f(x)$, then by definition, $E = F(\alpha_{1}, \dots, \alpha_{n})$.

Let $\deg f = n$. We prove by induction on $n$.

$n = 1$: it is clear that $F[x]$ is a splitting field of $f(x)$.

Assume it holds for $\leq n - 1$. Suppose $f(x) = p(x)a(x)$ for some irreducible $p(x) \in F[x]$ with $\deg p > 0$ and $a(x) \in F[x]$. Then $p(x)$ is prime in $F[x]$, and by Theorem 1206, let $E_{1} = F[x] / \left( p(x) \right)$, then $E_{1}$ is a field extension of $F$ and $\theta = \overline{x} \in E_{1}$ is a root of $p(x)$. Then $\theta$ is also a root of $f(x)$ since $f(x) \in \left( p(x) \right)$. So $f(x) = (x - \theta)f_{1}(x)$ for some $f_{1}(x) \in E_{1}[x]$ and $\deg f_{1} = n - 1$. By induction, there exists a field extension $E / E_{1}$ such that $f_{1}(x) = a(x - \alpha_{1})\cdots(x - \alpha_{n-1})$ with $\alpha_{1}, \dots, \alpha_{n-1} \in E$. Then $f(x) = a(x - \theta)(x - \alpha_{1}) \cdots (x - \alpha_{n-1})$ in $E[x]$, i.e., splits completely in $E[x]$.

By Proposition 1503, $K = F(\theta, \alpha_{1}, \dots, \alpha_{n-1})$ is a splitting field of $f(x)$ in $F[x]$.

By Proposition 1410, $F(\theta, \alpha_{1}, \dots, \alpha_{n-1})$ is a finite extension of $F$.