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 $f(\alpha, \beta)=0$.

Question

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.

Answer 1

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.