abstract-algebrafield-theory
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.
$F(\alpha)$ means the smallest field containing both $F$ and $\alpha$.
$\gamma$ algebraic over $F$ means that there is a non-zero polynomial $p(X) \in F[X]$ (i.e., a polynomial with coefficients in $F$) with $p(\gamma) = 0$. (And transcendental means such a polynomial does not exist).
Now the problem itself. The situation is as follows.
Since $\alpha$ is algebraic over $F(\beta)$, there is a non-zero polynomial $f(X) \in F(\beta)[X]$ with $f(\alpha) = 0$. The coefficients are elements of $F(\beta)$, but clearing denominators we may as well assume they are elements of $F[\beta]$.
So, $f(\alpha)$ is a polynomial expression in both $\alpha$ and $\beta$ and we can see it as a polynomial expression $g(\beta)$ in $\beta$ with coefficients in $F[\alpha]$, i.e., $g(Y) \in F[\alpha][Y]$. (To be precise, there is a polynomial $h(X,Y) \in F[X,Y]$ such that $f(X) = h(X,\beta)$ and $g(Y) = h(\alpha,Y)$.) Now $0 = f(\alpha) = g(\beta)$.
What is still left to show is that $g(Y)$ is not the zero polynomial, i.e., that not all its coefficients are $0$. But its coefficients are of the form $c(\alpha)$ with $c(X) \in F[X]$ and because $\alpha$ is transcendental over $F$, $c(\alpha)$ is $0$ only if $c(X) = 0$. So, if $g(Y)$ were the zero polynomial, so would $f(X)$ be.
Example. Take $\alpha = T^2$ and $\beta = T^3$ in the field ${\mathbb Q}(T)$ of rational functions over ${\mathbb Q}$. Then $\alpha$ is transcendental over ${\mathbb Q}$. Also, $\beta$ is algebraic over ${\mathbb Q}(\alpha)$ as it satisfies $\beta^2 - \alpha^3 = 0$ (i.e., $\beta$ is a root of the polynomial $Y^2 - \alpha^3$ over ${\mathbb Q}(\alpha)$). Exactly the same relation shows that $\alpha$ is algebraic over ${\mathbb Q}(\beta)$ (as $\alpha$ is a root of the polynomial $\beta^2 - X^3$ over ${\mathbb Q}(\beta)$).
If $\beta$ would be algebraic over $K$, then $\alpha = f(\beta) \in K(\beta)$ in contradiction with $\alpha$ transcendental over $K$.
Note: remember that any element of a simple extension $K(\gamma)$ of finite degree is algebraic over $K$.
Best Answer
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 $F(\alpha)$