Let $K=\mathbb{Q}(\sqrt[3]{5})$. How to show that there is no element $a$ in $K \setminus \mathbb{Q}$ satisfying $a^2 \in \mathbb{Q}$?
My thoughts: Assume for a contradiction that such $a$ exists. Then $a^2 = \frac{p}{q}$, where both $p,q$ are integers $\neq 0$ (otherwise $a \in K \cap \mathbb{Q}$), so $a$ is a root of $qx^2 -p$. Now this has to be irreducible, otherwise it splits in $\mathbb{Q}$ (since it is quadratic and has a root $a$) and therefore $a \in \mathbb{Q}$, a contradiction. Therefore $L=\mathbb{Q}(a)$ defines an intermediate quadratic extension $\mathbb{Q} \leq L \leq K$, which is impossible by the tower law. Is this correct?