Proof. sketch [1408B]

WLOG, assume $p_{1}(x)$ and $p_{2}(x)$ are monic. Then by Proposition 1407, $$ \begin{align*} &\varphi_{1} : F_{1}[x] / \left( p_{1}(x) \right) \to F_{1}(\alpha), \qquad a(x) \mapsto a(\alpha), \\ &\varphi_{2} : F_{2}[x] / \left( p_{2}(x) \right) \to F_{2}(\beta), \qquad a(x) \mapsto a(\beta) \end{align*} $$ are field isomorphisms. Also, define $$ \sigma : F_{1}[x] / \left( p_{1}(x) \right) \to F_{2}[x] / \left( p_{2}(x) \right) , \quad c_{m}x^{m} + \cdots + c_{0} \mapsto \varphi(c_{m}) x^{m} + \cdots + \varphi(c_{0}), $$ which is well-defined and is an isomorphism induced by the isomorphism $\varphi$.

Then $\varphi_{2} \circ \sigma \circ \varphi_{1} : F_{1}(\alpha) \to F_{2}(\beta)$ is a field isomorphism and we have $\Psi = \varphi_{2} \circ \sigma \circ \varphi_{1}^{-1}$.