Let $E$ be an extension of field $F$, and let $\alpha, \beta \in E$. Suppose $\alpha$ is transcendental over $F$ but algebraic over $F(\beta)$.
Show that $\beta$ is algebraic over $F(\alpha)$.
Okay, first questions: What does the notation $F(\alpha)$ and $F(\beta)$ mean? And being transcendental means it solves no equations with rational coefficients, but what does it mean for a field?
Best Answer
$F(\alpha)$ 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 ${\mathbb Q}(\beta)$).