The proof presented below, giving an (almost) complete characterisation of constructible regular polygons, is such a beautiful gem of a proof that I can’t help but record it here so that I might not forget it. I’ll go over far more details than is usually done, simply because it amuses me how many different areas of undergraduate mathematics are touched upon.

Theorem 1The regular -gon is constructible by ruler and compass if and only if has the form , where are distinct primes of the form .

Primes of this form are called Fermat primes. The only known Fermat primes are , and heuristics suggest that there may be no others, though this is an open problem.

We must say what we mean by “constructible”. We assume we are given a plane (a sheet of paper, say), with two points already labelled. We parameterise the plane by points of , and we rotate and scale so that the two labelled points are and . There are three things we are allowed to do:

- Whenever we have two labelled points we can draw the straight line through them.
- Whenever we have two labelled points we can draw the circle with one as centre and the other on the circumference.
- We can label any point of intersection (of two lines, two circles, or a line and a circle).

Any point of the plane (considered as an element of ) that can be labelled through a sequence of the above operations is called constructible. Denote by the set of constructible numbers.

Lemma 2The set is the smallest subfield of closed under taking square roots.

The lemma can be proved as follows.

- Show closure under by constructing the parallelogram with vertices .
- Show closure under the rotation .
- Show closure under taking real parts. Hence , where .
- Show closure under for by constructing the right triangle with leg lengths and the similar right triangle with base . Deduce that is a ring.
- Show that if then by doing so first for (another right triangle construction) and then using . Thus is a field.
- Show that is closed under taking square roots (yet another right triangle construction). Thus show that is closed under taking square roots by bisecting an appropriate angle.
- Finally show that any subfield of closed under taking square roots is closed under the “constructing” rules we listed when defining . The idea here is that one never has to solve an equation worse than a quadratic (when intersecting two circles, first construct the perpendicular bisector to the segment connecting the two centres, and then compute the intersection of this line with one of the circles).

The following is our algebraic characterisation of constructibility.

Lemma 3We have if and only if the degree of the normal closure of is a power of .

*Proof:* By the previous lemma, can be described as the union of all fields such that for all . Let . Thus lies in some such field . But note by the tower law that

where each factor . Thus is a power , and by another application of the tower law so is .

Conversely suppose that is a power of . Then the Galois group is a -group. Recall that any -group (except the trivial group) has nontrivial centre: by counting conjugacy class sizes, the size of must be divisible by . Thus by induction (and classifying abelian -groups) it follows that in every -group there is a sequence of normal subgroups such that , , and for each . Specialising to , Galois theory now implies that is the top of a tower of quadratic extensions starting from , so .

By saying “the -gon is constructible” we mean that the vertices of some regular -gon are constructible. Clearly this is equivalent to saying is an element of . Note that , where the th cyclotomic polynomial is defined by

Note that , so by the uniqueness of polynomial division and induction we have that .

Lemma 4is irreducible over .

*Proof:* Suppose where is irreducible and . We must show that each is a root of . Since some is a root of , it suffices to show that implies whenever is a prime not dividing . Towards this end, suppose that is such a prime and but . Then since . Since is irreducible this implies that divides . Denoting reductions modulo by a tilde and using Freshman’s dream, divides . Thus and are not coprime, so , and thus , has a multiple root over . But this is a contradiction since and its formal derivative are coprime.

Thus has minimal polynomial , and moreover is a normal extension. We have therefore reduced our task to finding those for which is a power of .

Writing as a product of primes with each , we have (OK, the details must stop somewhere)

This is a power of iff every which occurs has and a power of . But if is prime, then itself must be a power of , since for every odd we have