Constructible regular polygons

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 1 The regular {n}-gon is constructible by ruler and compass if and only if {n} has the form {2^k p_1 \cdots p_l}, where {p_1,\ldots,p_l} are distinct primes of the form {2^{2^m} + 1}.

Primes of this form are called Fermat primes. The only known Fermat primes are {3,5,17,257,65537}, 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 {\mathbf{C}}, and we rotate and scale so that the two labelled points are {0} and {1}. There are three things we are allowed to do:

  1. Whenever we have two labelled points we can draw the straight line through them.
  2. Whenever we have two labelled points we can draw the circle with one as centre and the other on the circumference.
  3. 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 {\mathbf{C}}) that can be labelled through a sequence of the above operations is called constructible. Denote by {\mathbf{E} \subset \mathbf{C}} the set of constructible numbers.

Lemma 2 The set {\mathbf{E}} is the smallest subfield of {\mathbf{C}} closed under taking square roots.

The lemma can be proved as follows.

  1. Show closure under {(x,y)\mapsto x+y} by constructing the parallelogram with vertices {0,x,y,x+y}.
  2. Show closure under the rotation {x\mapsto ix}.
  3. Show closure under taking real parts. Hence {\mathbf{E} = \mathcal{R}\mathbf{E} + i \mathcal{R}\mathbf{E}}, where {\mathcal{R}\mathbf{E} = \mathbf{E}\cap\mathbf{R}}.
  4. Show closure under {(x,y)\mapsto xy} for {x,y\in\mathcal{R}\mathbf{E}} by constructing the right triangle with leg lengths {1,x} and the similar right triangle with base {y}. Deduce that {\mathbf{E}} is a ring.
  5. Show that if {0\neq x\in\mathbf{E}} then {1/x\in\mathbf{E}} by doing so first for {x\in\mathcal{R}\mathbf{E}} (another right triangle construction) and then using {1/x = \bar{x}/|x|^2}. Thus {\mathbf{E}} is a field.
  6. Show that {\mathcal{R}\mathbf{E}} is closed under taking square roots (yet another right triangle construction). Thus show that {\mathbf{E}} is closed under taking square roots by bisecting an appropriate angle.
  7. Finally show that any subfield {K} of {\mathbf{C}} closed under taking square roots is closed under the “constructing” rules we listed when defining {\mathbf{E}}. 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 3 We have {x\in\mathbf{E}} if and only if the degree {[K:\mathbf{Q}]} of the normal closure {K} of {\mathbf{Q}(x)} is a power of {2}.

Proof: By the previous lemma, {\mathbf{E}} can be described as the union of all fields {\mathbf{Q}(\alpha_1,\ldots,\alpha_m)} such that {\alpha_k^2\in\mathbf{Q}(\alpha_1,\ldots,\alpha_{k-1})} for all {k=1,\ldots,m}. Let {x\in\mathbf{E}}. Thus {x} lies in some such field {L = \mathbf{Q}(\alpha_1,\ldots,\alpha_m)}. But note by the tower law that

\displaystyle [L:\mathbf{Q}] = \prod_{k=1}^{m} [\mathbf{Q}(\alpha_1,\ldots,\alpha_k):\mathbf{Q}(\alpha_1,\ldots,\alpha_{k-1})],

where each factor {[\mathbf{Q}(\alpha_1,\ldots,\alpha_k):\mathbf{Q}(\alpha_1,\ldots,\alpha_{k-1})]\leq 2}. Thus {[L:\mathbf{Q}]} is a power {2}, and by another application of the tower law so is {[K:\mathbf{Q}]}. \Box

Conversely suppose that {[K:\mathbf{Q}]} is a power of {2}. Then the Galois group {\text{Gal}(K/\mathbf{Q})} is a {2}-group. Recall that any {p}-group {G} (except the trivial group) has nontrivial centre: by counting conjugacy class sizes, the size of {Z(G)} must be divisible by {p}. Thus by induction (and classifying abelian {p}-groups) it follows that in every {p}-group {G} there is a sequence of normal subgroups {G_i \triangleleft G} such that {G_0 = \{e\}}, {G_n = G}, and {|G_k/G_{k-1}| = p} for each {k=1,\ldots,n}. Specialising to {p=2}, Galois theory now implies that {K} is the top of a tower of quadratic extensions starting from {\mathbf{Q}}, so {x\in K\subset\mathbf{E}}.

By saying “the {n}-gon is constructible” we mean that the vertices of some regular {n}-gon are constructible. Clearly this is equivalent to saying {\zeta_n = \exp(i2\pi/n)} is an element of {\mathbf{E}}. Note that {\Phi_n(\zeta_n) = 0}, where the {n}th cyclotomic polynomial {\Phi_n} is defined by

\displaystyle \Phi_n(X) = \prod_{(k,n)=1} (X-\zeta_n^k).

Note that {\prod_{d|n} \Phi_d(X) = X^n-1 \in \mathbf{Z}[X]}, so by the uniqueness of polynomial division and induction we have that {\Phi_n(X) \in \mathbf{Z}[X]}.

Lemma 4 {\Phi_n} is irreducible over {\mathbf{Q}}.

Proof: Suppose {\Phi_n(X) = f(X) g(X)} where {f} is irreducible and {\deg f\geq 1}. We must show that each {\zeta_n^k} is a root of {f}. Since some {\zeta_n^k} is a root of {f}, it suffices to show that {f(z)=0} implies {f(z^p)=0} whenever {p} is a prime not dividing {n}. Towards this end, suppose that {p} is such a prime and {f(z)=0} but {f(z^p)\neq 0}. Then {g(z^p) = 0} since {\Phi_n(z^p)=0}. Since {f} is irreducible this implies that {f(X)} divides {g(X^p)}. Denoting reductions modulo {p} by a tilde and using Freshman’s dream, {\tilde{f}} divides {\tilde{g}(X^p) = \tilde{g}(X)^p}. Thus {\tilde{f}} and {\tilde{g}} are not coprime, so {\tilde{\Phi_n}}, and thus {X^n-1}, has a multiple root over {\mathbf{F}_p}. But this is a contradiction since {X^n-1} and its formal derivative {nX^{n-1}} are coprime. \Box

Thus {\zeta_n} has minimal polynomial {\Phi_n}, and moreover {\mathbf{Q}(\zeta_n)/\mathbf{Q}} is a normal extension. We have therefore reduced our task to finding those {n} for which {[\mathbf{Q}(\zeta_n):\mathbf{Q}] = \deg\Phi_n = \varphi(n)} is a power of {2}.

Writing {n} as a product of primes {n=p_1^{e_1}\cdots p_m^{e_m}} with each {e_i>0}, we have (OK, the details must stop somewhere)

\displaystyle \varphi(n) = p_1^{e_1 - 1}(p_1 - 1) \cdots p_m^{e_m-1} (p_m-1).

This is a power of {2} iff every {p_i\neq 2} which occurs has {e_i = 1} and {p_i - 1} a power of {2}. But if {p = 2^k + 1} is prime, then {k} itself must be a power of {2}, since for every odd {q} we have

\displaystyle x^q + 1 = (x + 1)(x^{q-1} - x^{q-2} + - \cdots + 1).

The probability of coprimality

A famous result: Two positive integers chosen at random will be coprime with probability {6/\pi^2}.

One has to say what one means by this, since there is no uniform distribution on the positive integers. What we mean is that if we choose {x} and {y} uniformly and independently at random from the integers {\{1,\ldots,N\}}, then {x} and {y} will be coprime with probability tending to {6/\pi^2} as {N\rightarrow\infty}.

Here is one classic proof: Consider the points {(x,y)\in\{1,\ldots,N\}^2} such that {\gcd(x,y)=1}. If we scale this picture by {M}, we will have exactly the points {(X,Y)\in\{1,\ldots,MN\}^2} such that {\gcd(X,Y)=M}. Therefore, whatever the limiting probability {P} of being coprime is, the probability of having {\gcd = M} is exactly {P/M^2}. The law of total probability then implies

\displaystyle  1 = \lim_{N\rightarrow\infty}\mathbf{P}(\gcd(x,y)\in\mathbf{N}) = P \sum_{M=1} \frac{1}{M^2} = P\,\zeta(2).

Another way of imagining a random element on a noncompact group {G} is to look at the profinite completion {\hat{G}}, which is always compact so always possesses a uniform distribution. Now by the Chinese Remainder Theorem,

\displaystyle  \hat{\mathbf{Z}} = \prod_p \mathbf{Z}_p,

so choosing an element at random from {\hat{\mathbf{Z}}} is the same as choosing an element at random from {\mathbf{Z}_p} for each {p}. The probability that a random element of {\mathbf{Z}_p} is divisible by {p} is precisely {1/p}, so the probably that two elements from {\hat{\mathbf{Z}}} are coprime is precisely

\displaystyle  \prod_p (1-1/p^2) = 1/\zeta(2).

This does not imply the asymtotic result in {\mathbf{Z}} mentioned above, but it is a nice way of thinking about it.

EDIT (28/4/12): As pointed out to me by Freddie Manners, the “classic proof” suggested above suffers from the fatal defect of not proving that the limit exists. Here is an alternative proof which does prove that the limit exists. More generally, let {d>1} and consider positive integers {x_1,\ldots,x_d\leq X} chosen uniformly at random. Let {G(X)} be the number of {(x_1,\ldots,x_d)\in[1,X]^d} such that {\gcd(x_1,\ldots,x_d)=1}, and observe that the number of {(x_1,\ldots,x_d)\in[1,X]^d} such that {\gcd(x_1,\ldots,x_d) = n} is precisely {G(X/n)}. Then

\displaystyle  \lfloor X\rfloor^d = \sum_{n\geq1} G(X/n)

for all {X}. Inverting this relationship,

\displaystyle  G(X) = \sum_{n\geq1} \mu(n) \lfloor X/n\rfloor^d.

Thus, the probability that {d} randomly and uniformly chosen integers {x_1,\ldots,x_d\leq X} are all coprime is exactly

\displaystyle P(X) = \sum_{n\geq1} \mu(n) \left(\frac{\lfloor X/n\rfloor}{\lfloor X\rfloor}\right)^d.

The {n}th term here is bounded by {n^{-d}}, so by the dominated convergence theorem {P(X)\rightarrow\sum_{n\geq1} \mu(n) n^{-d} = \zeta(d)^{-1}}.