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!)

1. 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} ).)

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

3. 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?)

4. 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.

5. 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 ruler 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. (Note: The odd primes here are called Fermat primes. 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!)