TITLE: Quotients and polynomial?
QUESTION [3 upvotes]: I'm studying abstract algebra, and I'm confused in quotients and polynomials. 
The exercise that I feel confused was this;
Let $E$ be an extension field of a field $F$ and let $\alpha\in E$ be transcendental over $F$. Show that every element of $F(\alpha)$ that is not in $F$ is also transcendental over $F$.
(A First Course in Abstract Algebra, Exercise 33 in Section 29)
I'm trying to prove using contradiction supposing $\beta\in F(\alpha)\setminus F$ and $f(\beta) = 0$. 
When seeing a solution, I cannot understand a statement that "Since $\beta\in F(\alpha)$, $\beta$ can be expressed as a quotient of polynomials in $\alpha$ with coefficient in $F$."
Later processes are clear for me, but that statement doesn't make sense.
Could you help me?

REPLY [1 votes]: Elements of $F(\alpha)$ have an explicit form. They can be written as rational functions of alpha (quotients of polynomials) with coefficients in $F$. If such a rational expression satisfies a polynomial equation, derive that so does $\alpha$ itself.

REPLY [1 votes]: The statement you mention is simply the fact that $F(\alpha) \cong F(X)$, the field of rational functions in one variable, because $\alpha$ is transcendental over $F$.
To make things concrete, take $F=\mathbb Q$ and $\alpha=\pi$.
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