Theorem. [1506]

    1. All splitting fields of $f(x) \in F[x]$ are isomorphic.
    2. More generally, let $\varphi : F_{1} \xrightarrow{\cong} F_{2}$ be an isomorphism of fields. Given $f_{1}(x) = a_{n}x^{n} + \cdots + a_{1}x + a_{0} \in F_{1}[x]$, let $f_{2}(x) = \varphi(a_{n})x^{n} + \cdots + \varphi(a_{1})x + \varphi(a_{0}) \in F_{2}[x]$. Suppose that $E_{1}$ is a splitting field of $f_{1}(x)$ over $F_{1}$ and $E_{2}$ is a splitting field of $f_{2}(x)$ over $F_{2}$. Then the isomorphism $\varphi$ extends to an isomorphism $\sigma : E_{1} \xrightarrow{\cong} E_{2}$, i.e., there exists $\sigma$ such that the diagram

      The induction argument requires the more general statement.

      The strategy $F \subseteq F(\alpha) \subseteq K$ with directly handling simple extension $F(\alpha) / F$ and apply induction to $E / F(\alpha)$ will be repeatedly used in several proofs.

    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 $E_{2}$.