# Structure of Simple Transcendental Field Extension

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## Definition

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

Let $\map K X$ be the Field of Rational Functions in an indeterminate $X$.

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

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## Proof

Let $\phi: K \sqbrk X \to K \sqbrk \alpha$ be the Evaluation Homomorphism.

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We have that:

- $\map \phi f = \map f \alpha$

Therefore by definition of transcendental element:

- $\map \phi f = 0 \implies f = 0$

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

Therefore by the First Isomorphism Theorem for Rings:

- $K \sqbrk \alpha \simeq K \sqbrk X$

We have that the construction of the field of quotients $\map K X$ uses only the ring axioms.

Thus it follows that:

- $\map Q {K \sqbrk \alpha} = \map Q {K \sqbrk X}$

where $Q$ maps a integral domain to its field of quotients.

That is:

- $\map K \alpha \simeq \map K X$

$\blacksquare$