Let $z \in \mathbb{C}$. Consider the following statements:

  1. The point $z$ can be constructed with straightedge and compass starting from the points $\{ 0,1\}$.
  2. There is a field extension $K / \mathbb{Q}$ which has a tower of subextensions, each one of degree 2 over the next, and such that $z \in K$
  3. The field extension $\mathbb{Q}(z) / \mathbb{Q}$ has a tower of subextensions, each one of degree 2 over the next.

The usual way to prove that a geometric construction is impossible is to use that 1 and 2 are equivalent. My question is: are 2 and 3 equivalent? At first sight this looked like it was going to be true and elementary, but I could not prove it or find a counterexample.

  • $\begingroup$ Michael Artin covers this in depth in chapter 13 of his Algebra book. $\endgroup$ Feb 10, 2010 at 22:31

1 Answer 1


I think they're equivalent. Clearly 3 ==> 2. To see 2 ==> 3, say an extension E/L of fields is 2-filtered if it has a tower of subextensions each of degree 2 over the next. Then note firstly that if E/L and E'/L are 2-filtered, then so is the compositum of E and E' in any extension of L containing both (induct on the length of the tower of, say, E); and secondly for any map f:E-->E' of extensions of L, if E is 2-filtered then so is f(E). Together these two facts imply that in 2 we may assume that K is Galois over Q, say with group G. Then Q(z)/Q corresponds to a subgroup H, and by Galois theory we just need to prove the following group-theoretic lemma:

Lemma: Let G be a 2-group and H a subgroup. Then there is a chain of subgroups connecting H to G where each is index 2 in the next.

Proof: Recall that we can find a central order two subgroup Z of G. If H\cap Z=1, then HZ gets us one up in the filtration. Otherwise H\cap Z=Z, i.e. Z is in H, and we can use induction on |G|, replacing G by G/Z.

  • $\begingroup$ Thanks! That is it. I had gotten all the way to stating that group-theory lemma, and had failed to prove it, which was the easier part. $\endgroup$ Feb 10, 2010 at 22:58
  • 1
    $\begingroup$ The lemma works for $p$-groups, where $p$ is any prime. $\endgroup$ Feb 11, 2010 at 4:10

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