Structure of Simple Transcendental Field Extension

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

Let $K \left({X}\right)$ be the Field of Rational Functions in an indeterminate $X$.

If $\alpha$ is transcendental over $K$ then $K \left({\alpha}\right) \simeq K \left({X}\right)$.

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

Since $\phi \left({f}\right) = f \left({\alpha}\right)$, $\phi \left({f}\right) = 0 \implies f = 0$ 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 $K \left[{\alpha}\right] \simeq K \left[{X}\right]$.

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

That is, $K \left({\alpha}\right) \simeq K \left({X}\right)$.