MT310 Homework 11

Solution. We know that the minimal polynomial of ζ over Q is 1 + x + · · · + x p−1 , so [Q(ζ) : Q] = p − 1.
Since α = ζ + ζ −1 , it follows that ζ 2 − αζ + 1 = 0, so ζ is a root of the polynomial f(x) = x 2 − αx + 1 ∈
Q(α)[x]. In fact, the roots of f(x) are ζ and ζ −1 , which are complex hence do not lie in Q(α) ⊂ R.
Hence f(x) is irreducible in Q(α)[x]. This means [Q(ζ) : Q(α)] = deg f = 2. Hence
[Q(α) : Q] =
[Q(ζ) : Q]
[Q(ζ) : Q(α)] = 1 2 [Q(ζ) : Q] = 1 2
(p − 1).
Exercise 5. Let α, β ∈ C be two algebraic numbers whose minimal polynomials have degrees m, n
respectively. Assume that gcd(m, n) = 1. Prove that Q(α) ∩ Q(β) = Q.
Solution. Let F = Q(α) ∩ Q(β). Then n = [Q(α) : Q] = [Q(α) : F ] · [F : Q], so [F : Q] divides n.
Likewises [F : Q] divides m. Since gcd(m, n) = 1, we have [F : Q] = 1, that is, F = Q.
Exercise 6. Let ζ = e 2πi/5 and let τ = (1 + √ 5)/2 be the golden ratio.
a) Compute [Q(ζ) : Q].
b) Show that Q(τ) ⊂ Q(ζ).
c) Find the minimal polynomial of ζ over Q(τ).
Hint: Factor x 4 + x 3 + x 2 + x + 1 = (x 2 + ax + 1)(x 2 + bx + 1) with a, b ∈ R.
Solution.
a) The minimal polynomial of ζ over Q is x 4 + x 3 + x 2 + x + 1, so [Q(ζ) : Q] = 4.
b) It suffices to show that τ ∈ Q(ζ). From class, we know that
hence τ = 1 + ζ + ζ −1 ∈ Q(ζ).
ζ + ζ −1 = 2 cos(2π/5) = 1 2 (−1 + √ 5) = τ − 1.
c) There are various ways to do this. One way is to use the previous result: τ = 1 + ζ + ζ −1 , so
ζτ = ζ + ζ 2 + 1, so ζ is a root of x 2 + (1 − τ)ζ + 1.
Another way is to follow the hint and try to factor
x 4 + x 3 + x 2 + x + 1 = (x 2 + ax + 1)(x 2 + bx + 1) = x 4 + (a + b)x 3 + (2 + ab)x 2 + (a + b)x + 1.
We must have b = 1 − a, and 2 + a(1 − a) = 1, or a 2 − a − 1 = 0. This means {a, b} = {τ, 1 − τ}. Now
ζ is root of either x 2 + τx + 1 or x 2 + (1 − τ)x + 1. But ζ 2 + τ + 1 has positive imaginary part, hence
is not zero. So we again find that ζ is a root of x 2 + (1 − τ)ζ + 1.
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