Mathematics 352 – Abstract Algebra II Problem Set 9 Due: April 11, 2008

From Hungerford

Additional Problem. In this problem, following on the material from Chapter 15 we discussed before Easter, you will derive a complete characterization of the regular polygons in ℝ^{2} that are constructible by straightedge and compass. This question was another famous and long-standing problem that was solved by Gauss in the early 19th century. (He was so proud of this, in fact, that he had a regular 17-gon, which is constructible, carved on his tombstone!)

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

2^{ℓ−1}p_{1}^{e_{1}−1}(p_{1} − 1) ⋯ p_{k}^{e_{k}−1}(p_{k} − 1).

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 odd primes here 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 know about factorizations of the Fermat numbers.