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.


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 $$ b_0 + b_1\beta + \ldots b_m\beta^m = 0 $$ where each $b_i$ is a polynomial expression in $\alpha$. This proves that $\beta$ is algebraic over $F(\alpha)$

  • $\begingroup$ Just want to ask that if $\beta$ is algebraic over $F$, then $r_i(\beta)=1$ right? And also I wonder that where the fact "$\alpha$ is transcendental over $F$" is applied in the proof. $\endgroup$ – Alan Wang Jul 25 '18 at 5:00

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