How do I prove that $\forall \beta\in F(\alpha)\setminus F$ is transcendental?

Question

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?

Answer 1

You've got the right idea, but I think the way you are going about things might be a bit difficult. Instead, consider $F(\beta)$ as an intermediate subfield between $F$ and $F(\alpha)$. If $[F(\beta):F]$ and $[F(\alpha):F(\beta)]$ are both finite, then by the multiplicativity of field extension degree, $[F(\alpha):F]$ is finite, a contradiction. Hence, if we show $[F(\alpha):F(\beta)]$ is finite, it is clear that $[F(\beta):F]$ cannot be finite. See if you can show this; I am happy to post more.

Answer 2

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 $K(\beta)$.