TITLE: Show that $\beta $ is algebraic over $F(\alpha)$.
QUESTION [6 upvotes]: I have started reading field theory.
Let $E$ be an extension field of $F$ and let $\alpha,\beta\in E$.Suppose that $\alpha $ is transcendental over $F$ but algebraic over $F(\beta)$.
Show that $\beta$ is algebraic over $F(\alpha)$.
Since $\alpha$ is  algebraic over $F(\beta)\implies \exists p(x)\neq 0$ such that $p(\alpha)=0$ .So $p(x)$ must be a polynomial over $F(\beta)$ and not over $F$.
But these facts are taking me nowhere near the solution.Any help will be appreciated.

REPLY [8 votes]: Consider the polynomial $p$ and write it as
$$
p(x) = a_0 + a_1x + \ldots + a_nx^n
$$
where each $a_i \in F(\beta)$ is of the form
$$
a_i = \frac{q_i(\beta)}{r_i(\beta)}
$$
where $q_i,r_i \in F[x]$ are polynomials. Hence if $r := \prod r_i$, then
$$
r(\beta)[\tilde{q_o}(\beta) + \tilde{q_1}(\beta)\alpha + \ldots + \tilde{q_n}(\beta)\alpha^n] = 0
$$
for some polynomials $\tilde{q_1}, \ldots, \tilde{q_n} \in F[x]$. Collecting like terms, one can write this in the form
$$
b_0 + b_1\beta + \cdots +  b_m\beta^m = 0
$$
where each $b_i$ is a polynomial expression in $\alpha$. This proves that $\beta$ is algebraic over $F(\alpha)$