Mathematics 352 – Abstract Algebra II Discussion 3 – Which Regular Polygons are Constructible? Due: April 30, 2008

Background

Building on the material from Chapter 15 we discussed before Easter, in this discussion, you will derive a complete answer to the question: which regular polygons in ℝ^{2} are constructible by straightedge and compass?

Discussion Questions

A) Show that the regular n-gon is constructible if and only if [ℚ(ζ_{n}) : ℚ] is equal to a power of 2, where ζ_{n} = e^{2πi/n}. (Hint: ζ_{n} = cos (\frac{2π}{n} ) + i sin (\frac{2π}{n} ).)

B) Prove that ℚ(ζ_{n}) is a Galois extension of ℚ.

C) By looking at the Galois group Gal_{ℚ}ℚ(ζ_{n}), deduce that the degree [ℚ(ζ_{n}) : ℚ] is equal to the number of integers m, 1 ≤ m < n, that are relatively prime to n. (Hint: You can use without proof the fact we mentioned in class that if ϕ_{k}(x) is the minimal polynomial of ζ_{k} over ℚ) (the kth cyclotomic polynomial), then

x^{n} − 1 = ∏_{d|n} ϕ_{d}(x).

Any element of the Galois group Gal_{ℚ}ℚ(ζ_{n}) must take ζ_{n} to an nth root of 1 that is not a dth root of 1 for any d < n; do you see why?)

D) Show that if an integer of the form 2^{k} + 1 is prime, then k must be a power of 2: k = 2^{r} for some r ≥ 0.

E) If the prime factorization of n in ℤ is

n = 2^{ℓ}p_{1}^{e_{1}} ⋯ p_{k}^{e_{k}} ,

where the p_{i} are odd, then the number of integers m, 1 ≤ m < n, relatively prime to n is given by

ϕ(n) = 2^{ℓ−1}p_{1}^{e_{1}−1}(p_{1} − 1) ⋯ p_{k}^{e_{k}−1}(p_{k} − 1),

(the Euler ϕ-function). Deduce from this statement and parts A) and D) that a regular n-gon is constructible by straightedge and compass if and only if n is equal to 2^{ℓ} times a product of distinct odd primes of the form p = 2^{2^{r}} + 1. This says, for instance, that the regular 11−gon, the regular 13−gon, the regular 19−gon, etc. are not constructible by straightedge and compass.

Historical Note

The question of which regular polygons are constructible with straightedge and compass was another famous and long-standing problem going back to Euclid (or earlier). In the Elements, Proposition 1 is a construction of the equilateral triangle. Squares, regular pentagons, hexagons, octagons, decagons, dodecagons, etc. all have well-known constructions. The general question, though, remained unsolved until a complete answer was found by Gauss in the early 19th century. He was so proud of this work, in fact, that he reportedly asked to have a regular 17-gon, which is constructible, carved on his tombstone! The stonemason he approached “talked him out of it,” though, apparently saying that the design would be indistinguishable from a circle(!)

The odd primes in part E are called Fermat primes, after the 17th century French mathematician (and lawyer) Pierre de Fermat (of “Fermat’s Last Theorem” fame). Only five of them are known to exist: 3 = 2 + 1 (hence the equilateral triangle is constructible – Euclid’s Proposition 1), 5 = 2^{2} +1, 17 = 2^{4} +1 (hence Gauss’s 17-gon construction mentioned above!), 257 = 2^{8} + 1, and 65537 = 2^{16} + 1. Fermat apparently thought that all the numbers 2^{2^{r}} + 1 would be prime. But in fact, in all other cases where the prime factorization of 2^{2^{r}} + 1 has been determined, it is a composite number! As of 2007, only the first 12 Fermat numbers have been completely factored, but some prime factors are known for many more of them. For instance,

2^{2^{5}} + 1 = 4294967297 = 641 · 6700417,

(where both factors are prime) as was found by hand by Euler in 1732(!). Then

2^{2^{6}} + 1 = 67280421310721 · 274177,

(both factors are prime), and

2^{2^{7}} + 1 = 5704689200685129054721 · 59649589127497217

(both factors prime again). There can be more than two distinct prime factors (in fact this

seems to be true for all 2^{2^{r}} +1 with r ≥ 9). See http://www.prothsearch.net/fermat.html

for what is known about factorizations of the Fermat numbers.