The degree of a constructible number is a power of two

Statement

Let $z \in \mathbb{C}$ be a constructible number. Then $z$ is algebraic over $\mathbb{Q}$, and $[\mathbb{Q}(z) : \mathbb{Q}] = 2^r$ for some integer $r \ge 0$.

Remarks

The proof: there is a tower $\mathbb{Q} = F_0 \subseteq F_1 \subseteq \dots \subseteq F_n$ with $z \in F_n$ and $[F_j : F_{j-1}] = 2$ for each $j$. By the tower law, $[F_n : \mathbb{Q}] = 2^n$, and $[\mathbb{Q}(z) : \mathbb{Q}]$ divides it. So the minimal polynomial of $z$ over $\mathbb{Q}$ has degree a power of two. The converse is false: some algebraic numbers whose degree is a power of two are not constructible. A sufficient condition: the splitting field of the minimal polynomial of $z$ has degree a power of two over $\mathbb{Q}$.

Explanations

L1 human

Each construction step at most doubles the degree of the field. So a number whose minimal polynomial has degree $3$, like $\sqrt[3]{2}$, can never be constructed. This simple test settles several problems that were open for two thousand years.