Mathematics 352 – Abstract Algebra II Solutions for Discussion 3 April 28, 2008

A) The regular n-gon is constructible if and only if the point (cos (\frac{2π}{n} ) , sin (\frac{2π}{n} )) is constructible, which is true in turn if and only if

[ℚ (cos (\frac{2π}{n} )) : ℚ] = 2^{m} and [ℚ (sin (\frac{2π}{n} )) : ℚ] = 2^{m′}  (1)

for some m, m′ ≥ 0.

Assume that (1) holds. Note that x = i sin (\frac{2π}{n} ) is a root of the polynomial equation

x^{2} + (1 − cos^{2} (\frac{2π}{n} )) = 0.

For all n > 2, this polynomial is irreducible in ℚ (cos (\frac{2π}{n} )) [x] since it has no real roots but ℚ (cos (\frac{2π}{n} )) ⊂ ℝ. Hence

[ℚ (cos (\frac{2π}{n} ) , i sin (\frac{2π}{n} )) : ℚ (cos (\frac{2π}{n} ))] = 2,

so

[ℚ (cos (\frac{2π}{n} ) , i sin (\frac{2π}{n} )) : ℚ] = 2^{m+1}.

Now ℚ(ζ_{n}) is a subfield of ℚ (cos (\frac{2π}{n} ) , i sin (\frac{2π}{n} )) so [ℚ(ζ_{n}) : ℚ] is also a power of 2, since that degree must divide 2^{m+1}.

Conversely, assume that [ℚ(ζ_{n}) : ℚ] = 2^{ℓ} for some l ≥ 0. Note that

(ζ_{n})^{−1} = cos (\frac{2π}{n} ) − i sin (\frac{2π}{n} ) ,

so

cos (\frac{2π}{n} ) = \frac{1}{2} (ζ_{n} + (ζ_{n})^{−1})

is an element of ℚ(ζ_{n}). It follows that [ℚ (cos (\frac{2π}{n} )) : ℚ] divides 2^{ℓ}, hence is also a power of 2. Hence cos (\frac{2π}{n} ) is constructible. Hence the point (cos (\frac{2π}{n} ) , sin (\frac{2π}{n} )), and the regular n-gon are constructible.

B) The roots of x^{n} − 1 = 0 in ℂ are 1, ζ_{n}, ζ_{n}^{2}, …, ζ_{n}^{n−1}. Hence ℚ(ζ_{n}) = ℚ(1, ζ_{n}, ζ_{n}^{2}, …, ζ_{n}^{n−1}) is the splitting field of x^{n} −1 ∈ ℚ[x]. By Theorem 10.15 in Hungerford, ℚ(ζ_{n}) is a normal extension of ℚ. It is also a separable extension of ℚ, since ℚ has characteristic zero (Theorem 10.17). It is finite dimensional since [ℚ(ζ_{n}) : ℚ] < n. Therefore, ℚ(ζ_{n}) is a Galois extension of ℚ by definition.

C) By the observations in part (B), we know that the roots of x^{n} − 1 = 0 in ℂ are the numbers ζ_{n}^{k} for k with 0 ≤ k < n. Since ℚ(ζ_{n}) is a Galois extension of ℚ, it follows by the Fundamental Theorem of Galois Theory (Theorem 11.11) that

[ℚ(ζ_{n}) : ℚ] = |Gal_{ℚ}ℚ(ζ_{n})|.

However, by Theorem 11.3, given k with 0 ≤ k < n, there exists σ ∈ Gal_{ℚ}ℚ(ζ_{n}) with σ(ζ_{n}) = ζ_{n}^{k} if and only if ζ and ζ_{n}^{k} have the same minimal polynomial, and then by Theorem 11.4 there is exactly one element of the Galois group for each such k. By the equation

x^{n} − 1 = ∏_{d|n} ϕ_{d}(x)

we see that σ ∈ Gal_{ℚ}ℚ(ζ_{n}) can only map ζ_{n} to other roots of the nth cyclotomic polynomial ϕ_{n}(x), since the roots of the other factors are all dth roots of unity for d < n. If c = gcd(k, n) > 1, then (ζ_{n}^{k})^{c} = 1, so ζ_{n}^{k} is not a primitive nth root of unity. Hence the roots of ϕ_{n}(x) are precisely the ζ_{n}^{k} where gcd(n, k) = 1, which implies what we wanted to show.

D) We will show the contrapositive: If k > 1 is not a power of 2, then 2^{k} + 1 is not prime. If an integer k > 1 is not a power of 2, then it must be divisible by some odd prime p ≥ 3. Say k = pq where q is some other integer. But then we have, by the factorization of a sum of like odd powers:

2^{pq} + 1 = (2^{q})^{p} + 1^{p} = (2^{q} + 1)((2^{q})^{p−1} − (2^{q})^{p−2} + ⋯ − 2^{q} + 1).

(Since p is odd, the last term in the second factor here is always +1.) Since p is odd and ≥ 3, 2^{q} + 1 < 2^{pq} + 1. Hence 2^{pq} + 1 cannot be a prime because it is divisible by 2^{q} + 1.

E) By the given formula for the Euler ϕ-function, if n = 2^{ℓ}p_{1}^{e_{1}} ⋯ p_{k}^{e_{k}}, where p_{i} are odd primes, the degree [ℚ(ζ_{n}) : ℚ], which equals the number of integers k with 1 ≤ m < n that satisfy gcd(k, n) = 1 by part (C), is given by the equation:

[ℚ(ζ_{n}) : ℚ] = 2^{ℓ−1}p_{1}^{e_{1}−1}(p_{1} − 1) ⋯ p_{k}^{e_{k}−1}(p_{k} − 1).  (2)

If the regular n-gon is constructible then [ℚ(ζ_{n}) : ℚ] = 2^{m} for some m ≥ 0. However (2) gives a factorization of [ℚ(ζ_{n}) : ℚ] and from this and the Unique Factorization Theorem in ℤ we see that all e_{i} = 1 (or else p_{i} divides the degree [ℚ(ζ_{n}) : ℚ]) and each p_{i} − 1 must be a power of 2, so p_{i} = 2^{2^{r_{i}}} + 1 by part (D). This says that n must be a power of 2 times a product of distinct primes of the form 2^{2^{r}} + 1.

Conversely, if n is a power of 2 times a product of distinct odd primes of the form 2^{2^{r}} + 1, then (2) shows that [ℚ(ζ_{n}) : ℚ] equals a power of 2 since all the factors on the right hand side of (2) are powers of 2 (note that our assumptions ensure e_{i} = 1 for all i). Therefore the regular n-gon is constructible by part (A).