Proof. sketch [1504A]

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 $F$.