Field extensions and algebraic/transcendental elements 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?
 A: $F(\alpha)$ 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 ${\mathbb Q}(\beta)$).
A: $F(\alpha)$ is the smallest field that contains both $F$ and $\alpha$ inside $E$.. 
Being transcendental means, it solves no equations with coefficients in the given field, i.e. $\beta \in E$ is transcendental over  $F(\alpha)$, if no polynomial with coefficients in $F(\alpha)$ has $\beta$ as root.
A: This is an old question, but I would like to share an alternative solution I came up with when I did this exercise.
The conditions in the question is equivalent to the following diagram:

Now we claim $F(\beta)/F$ is transcendental. Suppose not, then $F(\beta)/F$ is algebraic. Since composition of algebraic extensions is algebraic, $F(\alpha, \beta)/F$ is algebraic, which contradicts $\alpha$ being transcendental over $F$.

Hence $F(\beta)$ and $F(\alpha)$ are both transcendental extensions.

This means $\alpha, \beta$ are both transcendental over $F$. So $F(\alpha), F(\beta) \cong F(x)$ for some intermediate $x$ $\implies F(\alpha) \cong F(x) \cong F(\beta)$

Since $F(\alpha) \cong F(\beta)$, we can regard $F(\beta)$ as degree $1$ extension of $F(\alpha)$. $F(\beta)/F(\alpha)$ is algebraic since all finite extensions are algebraic.

Finally, we use again the fact that composition of algebraic extensions is algebraic to conclude $F(\alpha, \beta)/F(\alpha)$ is algebraic.

