Proof. sketch [1506A]

    1. Proposition 1503 show that all splitting fields of $f(x)$ contained in a common field extension $K$ of $F$ are the same.
    2. We prove by induction on $\deg f_{1}(x) = n$.

      1. Case $n = 1$: in this case, $E_{1} = F_{1}$ and $E_{2} = F_{2}$. The isomorphism $\sigma$ is simply $\varphi$.

      2. Case $n$: assume the theorem holds for all $\deg f_{1}(x) < n$.

        Let $\alpha$ be a root of $f_{1}(x)$ over $E_{1}$, we first find its corresponding root $\beta$ of $f_{2}(x)$ over $E_{2}$ such that $\sigma(\alpha) = \beta$. After that, we establish $F_{1}(\alpha) \cong F_{2}(\beta)$ and write $f_{1}(x) = (x - \alpha)g_{1}(x)$ and $f_{2}(x) = (x - \beta)g_{2}(x)$ with $\deg g_{1}(x) < n$ and $\deg g_{2}(x) < n$, and then use assumption on $g_{1}, g_{2}$ over $F_{1}(\alpha), F_{2}(\beta)$ to get $F_{1}(\alpha)(\alpha_{1}, \dots, \alpha_{n-1}) \cong F_{2}(\beta)(\beta_{1}, \dots, \beta_{n-1})$ where $\alpha_{1}, \dots, \alpha_{n-1}$ are the roots of $g_{1}(x)$ and $\beta_{1}, \dots, \beta_{n-1}$ are the roots of $g_{2}(x)$. And finally claim that $E_{1} = F_{1}(\alpha, \alpha_{1}, \dots, \alpha_{n-1}) = F_{1}(\alpha, \dots, \alpha_{n-1})$ and the same for $E_{2}$.