Theorem. isomorphism induced in the middle [1408]

Let $K_{1} / F_{1}$ and $K_{2} / F_{2}$ be field extensions, $\varphi : F_{1} \to F_{2}$ be a field isomorphism and let $p_{1}(x) = a_{n}x^{n} + \cdots + a_{1}x + a_{0} \in F_{1}[x]$ be an irreducible polynomial, and let $p_{2}(x) = \varphi(a_{n})x^{n} + \cdots + \varphi(a_{1})x + \varphi(a_{0}) \in F_{2}[x]$. If $\alpha \in K_{1}$ is a root of $p_{1}(x)$ and $\beta \in K_{2}$ is a root of $p_{2}(x)$, then $$ \Psi: F_{1}(\alpha) \to F_{2}(\beta), \quad c_{m}\alpha^{m} + \cdots + c_{1}\alpha + c_{0} \mapsto \varphi(c_{m})\beta^{m} + \cdots + \varphi(c_{1})\beta + \varphi(c_{0}) $$ for $c_{i} \in F_{1}$, is well-defined and is a field isomorphism.

Exegesis 1. diagram illustrations and a special case [1408A]

$F_{1} = F_{2} = F, K_{1} = K_{2} = K$ implies that if $\alpha, \beta \in K$ are both roots of an irreducible polynomial $p(x) \in F[x]$, then $$ F(\alpha) \cong F(\beta), \quad a(\alpha) \mapsto a(\beta), \quad a(x) \in F[x] $$ is well-defined and is an isomorphism.

Example 1.1. An example for $F=\mathbb{Q}$ and $K=\mathbb{C}$ [1408AC]

$$ \mathbb{Q}\left( \sqrt[3]{ 2 } \right) \cong \mathbb{Q}\left( \sqrt[3]{ 2 } e^{ 2 \pi i / 3 } \right). $$

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 $\Psi = \varphi_{2} \circ \sigma \circ \varphi_{1}^{-1}$.