The typical characterization of points constructible by compass and straightedge is the following:

Let $S\subseteq\mathbb{C}$ with $0,1\in S$, $K_0 = \mathbb{Q}(S\cup \bar{S})$ and $a\in\mathbb{C}$. Then $a$ is constructible from $S$ by compass and straightedge if and only if there is a tower of quadratic field extensions $K_0 \subseteq \ldots \subseteq K_n$ such that $a\in K_n$.

For constructible $a$ it follows that $a$ is algebraic over $K_0$ and $[K_0(a) : K_0]$ is a power of two. However, it is known that this is not sufficient for $a$ to be constructible.

Now I wonder if the constructibility of $a$ is equivalent to the following:

$a$ is algebraic over $K_0$ and the degree of the normal hull of $K_0(a)$ over $K_0$ is a power of two.

The direction ,,$\Leftarrow$'' is true, I think. If N is the normal hull of $K_0(a)$, then $K_0\subseteq N$ is a finite Galois extension, and thus the order of $G = \operatorname{Gal}(K_0 \subseteq N)$ is a power of two. As a $2$-group, it contains a chain of subgroups $\{\operatorname{id}\} = U_n < \ldots < U_0 = G$ of index $2$ each. The respective fixed fields give the needed tower of quadratic field extensions.

But I wasn't able to proof ,,$\Rightarrow$'', nor did I find a counter example.

The typical characterization of points constructible by compass and straightedge is the following:

Let $S\subseteq\mathbb{C}$ with $0,1\in S$, $K_0 = \mathbb{Q}(S\cup \bar{S})$ and $a\in\mathbb{C}$. Then $a$ is constructible from $S$ by compass and straightedge if and only if there is a tower of quadratic field extensions $K_0 \subseteq \ldots \subseteq K_n$ such that $a\in K_n$.

For constructible $a$ it follows that $a$ is algebraic over $K_0$ and $[K_0(a) : K_0]$ is a power of two. However, it is known that this is not sufficient for $a$ to be constructible.

Now I wonder if the constructibility of $a$ is equivalent to the following sharper criterion:

$a$ is algebraic over $K_0$ and the degree of the normal hull of $K_0(a)$ over $K_0$ is a power of two.

The direction ,,$\Leftarrow$'' is true, I think. If $N$ is the normal hull of $K_0(a)$, then $K_0\subseteq N$ is a finite Galois extension, and thus the order of $G = \operatorname{Gal}(K_0 \subseteq N)$ is a power of two. As a $2$-group, it contains a chain of subgroups $\{\operatorname{id}\} = U_n < \ldots < U_0 = G$ of index $2$ each. The respective fixed fields give the needed tower of quadratic field extensions.

But I wasn't able to proof ,,$\Rightarrow$'', nor did I find a counter example.