TITLE: Transcendental Extensions. $F(\alpha)$ isomorphic to $F(x)$
QUESTION [5 upvotes]: Let $E$ be an extension field of $F$ and $\alpha \in E$. Then $\alpha$ is transcendental over $F$ if and only if $F(\alpha)$ is isomorphic to $F(x)$, the field of fractions of $F[x]$. 
This was a theorem in an abstract algebra textbook with a very brief proof. Can someone please explain why this theorem holds? I'm having difficulty grasping the concepts at hand.
Thanks!

REPLY [2 votes]: Evaluating every polynomial $f(x)\in F[x]$ at at the fixed point $\alpha\in E$ is a ring homomorphism from the integral domain $F[x]\to $E that is identity on the constants (the polynomials of degree 0).
This evaluation homomorphism not being injective is the same as saying $\alpha\in E$ is algebraic over $F$. (Any polynomial in the kernel gives a relation for $\alpha$, making it algebraic).
That means for transcendental elements it is injective, and hence this homomorphism can be extended to its field of fractions.
EDIT:
So by basic theorems of ring isomorphism the domain ring, in fact a field $F(x)$ is isomorphic to the image the subfield of $E$ generated by $F$ and $\alpha$.