Show that if $\beta \in E$ is algebraic over $F(\alpha)$, then there is a nonzero polynomial $f(x,y) \in F[x,y]$ such that $f(\alpha, \beta)=0$. Let $E$ be an extension field of F, and let $\alpha \in E$ be transcendental over $F$.
Show that if $\beta \in E$ is algebraic over $F(\alpha)$, then there is a nonzero polynomial $f(x,y) \in F[x,y]$ such that $f(\alpha, \beta)=0$.
I am prepping for an exam and decided to attempt previous years'.
This is one question I cannot seem to find a solution of, and not quite sure how to tackle.
 A: Since
$\beta \in E \tag 1$
is algebraic over $F(\alpha)$, where $\alpha$ is transcendental over $F$, there exists a polynomial
$\theta(x) = \displaystyle \sum_0^m \theta_k x^k \in F(\alpha)[x] \tag 2$
with
$\theta(\beta) = 0; \tag 3$
since the coefficients of $\theta$,
$\theta_k \in F(\alpha), \tag 4$
are each a rational function in $\alpha$ with coefficients from $F$, we may write
$\theta_k(\alpha) = \dfrac{p_k(\alpha)}{q_k(\alpha)}, \; p_k(x), q_k(x) \in F[x]; \tag 5$
if we substitute the $\theta_k(\alpha)$ from (5) into (2), and concurrently take $x = \beta$, then (3) yields
$\displaystyle \sum_0^m  \dfrac{p_k(\alpha)}{q_k(\alpha)}\beta^k = 0; \tag 6$
we may clear the denominators in this expression by multiplying through by the polynomial
$q(\alpha) = \displaystyle \prod_0^m q_k(\alpha); \tag 7$
we obtain
$\displaystyle \sum_0^m \left (\prod_{j = 0, j \ne k}^m q_k(\alpha) \right ) p_k(\alpha) \beta^k = 0; \tag 8$
we may now take
$f(x, y) \in F[x, y] \tag 9$
to be
$f(x, y) = \displaystyle \sum_0^m \left (\prod_{j = 0, j \ne k}^m q_k(x) \right ) p_k(x) y^k, \tag{10}$
and we are done.
