Structure of Simple Transcendental Field Extension

Definition
Let $F/K$ be a field extension and $\alpha \in F$.

Let $K(X)$ be the Field of Rational Functions in an indeterminate $X$.

If $\alpha$ is transcendental over $K$ then $K(\alpha) \simeq K(X)$.

Proof
Let $\phi : K[X] \to K[\alpha]$ be the Evaluation Homomorphism.

Since $\phi(f) = f(\alpha)$, $\phi(f) = 0 \implies f = 0$ by the definition of a trancendental element.

Moreover $\phi$ is surjective by the corollary to Field Adjoined Set.

Therefore by the First Isomorphism Theorem for Rings $K[\alpha] \simeq K[X]$.

Since the construction of the quotient field $K(X)$ uses only the ring axioms it follows that $Q(K[\alpha]) = Q(K[X])$ where $Q$ maps a integral domain to it's quotient field.

That is, $K(\alpha) \simeq K(X)$.