Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
To followers of certain sports, ACL stands for anterior cruciate ligament --- see en.wikipedia.org/wiki/Anterior_cruciate_ligament –  Gerry Myerson Jan 20 '13 at 22:21

1 Answer 1

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.

share|improve this answer
    
I got it, thank you! –  azimut Jan 20 '13 at 18:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.