Let $F$ be the field of algebraic numbers over $\mathbb{Q}$. I do remember that this means $F$ is a field extension of rationals over $\mathbb{Q}$.
How do I show that no field extension of $F$ is algebraic? I am stuck on this and I would like a hint or two.
A field extension $E/F$ is algebraic if for every $\alpha \in E$, the field extension $F(\alpha)/F$ is finite. Equivalently, $\alpha$ is the root of some polynomial in $F[x]$. This says nothing about whether $E/F$ is finite or infinite. Indeed, both possibilities can occur.
As mentioned in the comments above, when $F$ is the set of all elements that are algebraic over $\mathbb{Q}$, $F/\mathbb{Q}$ is not a finite extension. Can you see why?
Keeping with $F$ the field of algebraic numbers over $\mathbb{Q}$, suppose $E/F$ is a nontrivial field extension. In other words, $F \subset E$ and $F \neq E$. Then there exists an $\alpha \in E \setminus F$. Can $\alpha$ be algebraic over $\mathbb{Q}$? Can $\alpha$ be algebraic over $F$?