TITLE: How do I prove that $\forall \beta\in F(\alpha)\setminus F$ is transcendental?
QUESTION [5 upvotes]: Let $E/F$ be a field extension.
Let $\alpha\in E$ be transcendental over $F$.
Let $\beta\in F(\alpha)\setminus F$.
Then, how do I prove that $\beta$ is transcendental over $F$?
Here's how I tried:
Since $\beta\in F(\alpha)$, there exist $f(X),g(X)\in F[X]$ such that $\beta=\frac{f(\alpha)}{g(\alpha)}$.
Suppos $\beta$ is algebraic over $F$.
Then, there exists $h\in F[Y]\setminus\{0\}$ such that $h(\beta)=0$. Set $n=deg(h)$.
Then, $\sum_{i=0}^n h_i (\frac{f(\alpha)}{g(\alpha)})^i=0$
Hence, $(\sum_{i=0}^n h_i f^i g^{n-i})(\alpha)=0$.
To conclude, it should be proven that $\sum_{i=0}^n h_i f^i g^{n-i}$ is not a zero polynomial.
How do I prove this? Or is there another way to prove that $\beta$ is transcendental?

REPLY [5 votes]: The relation $\beta=f(\alpha)/g(\alpha)$ can be written as
$$
f(\alpha)-\beta g(\alpha)=0
$$
which means that $\alpha$ satisfies the polynomial
$$
f(X)-\beta g(X)
$$
which has coefficients in $K(\beta)$. Since $\beta\ne0$ and $g(X)\ne0$, this polynomial is nonzero, hence $\alpha$ is algebraic over $K(\beta)$.