Proof. sketch [1407A]

    1. By the proof of Proposition 1405, the kernel of the following ring homomorphism $$\varphi : F[x] \to K, \qquad a(x) \mapsto a(\alpha)$$ is $\left( m_{\alpha, F}(x) \right)$. By the First Isomorphism Theorem, $$F[x] / \left( m_{\alpha, F}(x) \right) \to \mathrm{Im}\varphi, \qquad \overline{a(x)} \mapsto a(\alpha)$$ is a well-defined isomorphism. Moreover, in the proof of Proposition 1403, we have shown $\mathrm{Im}\varphi = F(\alpha)$. And by Theorem 1206, $[F(\alpha) : F] = \deg m_{\alpha, F}(x)$.

    2. Suppose $\frac{a(x)}{b(x)} = \frac{c(x)}{d(x)}$, then $$a(x)d(x) = b(x)c(x) \implies a(\alpha)d(\alpha) = b(\alpha)c(\alpha) \implies a(\alpha)b(\alpha)^{-1} = c(\alpha)d(\alpha)^{-1}.$$ So the map is well-defined. Also, it is clearly a field homomorphism. By definition of $F(\alpha)$, it is surjective. Denote the map by $\phi$. If $\phi$ is not injective, then there exists $\frac{a(x)}{b(x)}\in F(x)$ such that $a(\alpha) b(\alpha)^{-1} = 0 \implies a(\alpha) = 0$. Then $\alpha$ is algebraic, contradicting $\alpha$ is transcendental. So $\phi$ is an isomorphism.

      Now, if $[F(\alpha) : F] < \infty$, then by Proposition 1403, $\alpha$ is algebraic over $F$, a contradiction. So $[F(\alpha) : F] = \infty$.