A question inspired by the concept of constructible numbers.
Let $K$ be a field, $\Omega$ some algebraic closure of $K$ and $p$ a prime number. Let $L$ be the set of all elements $\alpha \in \Omega$ for which $\dim_K(K[\alpha])$ is a power of $p$. Is $L$ a subring/subfield of $\Omega$ in general? If not, under what conditions on $p$ or $K$ is it?
The answer is positive for the constructible numbers ($p = 2, K = \mathbb{Q}$).
$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\C}{\mathbb{C}}$$\newcommand{\Q}{\mathbb{Q}}$Let $K = \Q$, $\Omega = \C$, $p = 3$, and $\omega \in \C$ a primitive third root of unity.
Then $\sqrt[3]{2}, \omega \sqrt[3]{2} \in L$, but $\omega = (\sqrt[3]{2})^{-1} \cdot \omega \sqrt[3]{2} \notin L$, as $\Size{\Q[\omega] : \Q} = 2$
So $L$ is not a subfield in this case, nor a subring, as $(\sqrt[3]{2})^{-1} \in \Q[\sqrt[3]{2}]$.
You can do the same, mutatis mutandis, for any odd prime $p$. The reason the argument fails for $p = 2$ is of course that $\Size{\Q[-1] : \Q} = 1 = 2^{0}$.