Show that $\beta $ is algebraic over $F(\alpha)$.

Solution 1:

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 $F(\alpha)$