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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.

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If some advertisement is authorized here, this can be found as Theorem 5.1.1 of my book, A field guide to Algebra (Undergraduate Texts in Mathematics, Springer-Verlag, 2005). – ACL Jan 20 '13 at 18:03
To followers of certain sports, ACL stands for anterior cruciate ligament --- see – Gerry Myerson Jan 20 '13 at 22:21
up vote 2 down vote accepted

Your question is about showing that the normal hull of $K_n$ over $K_0$ has $2$-power degree, if $[K_i:K_{i-1}]=2$ for all $i$. But that follows be induction: Let $L$ be the normal hull of $K_{n-1}$ over $K_0$, so $[L:K_0]$ is a $2$-power.

The normal hull $N$ of $K_n$ over $K_0$ is the composite of the conjugates of $K_n$ over $K_0$. But all these conjugates are extensions of $L$ of degree $2$ (or $1$, if $K_n\subseteq L$), and from that the claim follows.

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I got it, thank you! – azimut Jan 20 '13 at 18:44

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