MT310 Homework 11

MT310 Homework 11
Solutions
Due Friday, April 30 by 5:00
Exercise 1. Let F be a field, let V be a vector space over F and let V 0 be the set of nonzero vectors
in V . For u, v ∈ V 0 , say that u ∼ v if there exists a ∈ F such that au = v. Prove that this is an
equivalence relation on V 0 .
Comment. If [u] is an equivalence class under this relation, the set [u] ∪ {0} is called a line in V .
Proof. Reflexivity: If u ∼ v then there exists a ∈ F such that au = v, so a −1 v = u so v ∼ u.
Symmetry: u = 1 · u so u ∼ u.
Transitivity: If u ∼ v and v ∼ w then there exist a, b ∈ F with au = v and bv = w.
bau = bv = w, so u ∼ w.
Hence
Optional Exercise for Extra Credit (20 points total)
[Solutions Posted Separately]
Exercise 2. Let K ⊂ F ⊂ E be three fields. Suppose that {α 1 , . . . , α m } is a K-basis of F and
{β 1 , . . . , β n } is an F -basis of E. Prove that {α i β j : 1 ≤ i ≤ m, 1 ≤ j ≤ n} is a K-basis of E.
Comment. This proves that the degree of a field extension is multiplicative: [E : K] = [E : F ][F : K].
Proof. Let γ ∈ E. Since {β 1 , . . . , β n } spans E over F , there are f j ∈ F such that γ = ∑ n
j=1 f jβ j .
Since {α 1 , . . . , α m } spans F over Kd, there are, for each j, elements c ij ∈ K such that f j = ∑ m
i=1 c ijα i .
Then we have
n∑
n∑ m∑
γ = f j β j = γ = c ij α i β j ,
j=1
so {α i β j : 1 ≤ i ≤ m, 1 ≤ j ≤ n} spans E over K.
Suppose we have scalars c ij ∈ K such that
n∑
j=1 i=1
j=1 i=1
m∑
c ij α i β j = 0.
Since {β 1 , . . . , β n } is linearly independent, we have ∑ m
i=1 c ijα i = 0 for each i. Since {α 1 , . . . , α m } is
linearly independent, we have c ij = 0 for each i, j. Hence the set {α i β j : 1 ≤ i ≤ m, 1 ≤ j ≤ n} is
linearly independent. Since this set spans E and is linearly lindependent over F , it is a an F -basis of
E.
Exercise 3. Let α = 3√ 2, ζ = e 2πi/3 , F = Q(α) and E = F (ζ). Compute [E : Q].
Comment. On the previous homework, you showed that the roots of x 3 − 2 are α, αζ, αζ 2 . The field E is
smallest subfield of C containing these roots.
Solution. We have [E : Q] = [E : Q(α)][Q(α) : Q] = 3[E : F ]. The minimal polynomial of ζ over Q is
x 2 + x + 1, whose roots are complex, and do not lie in F . So x 2 + x + 1 is irreducible over F , and is
also the minimal polynomial of α over F . It follows that E = F (ζ) = 2, so [E : Q] = 3 · 2 = 6.
Exercise 4. Let p > 2 be a prime, let α = 2 cos(2π/p) and let p α (x) be the minimal polynomial of α
over Q. Compute the degree of p α (x).
(A formula for p α (x) was stated in class without proof. Do not use this formula.)
Hint: Note that α = ζ + ζ −1 , where ζ = e 2πi/p . Compute [Q(ζ) : Q] and [Q(ζ) : Q(α)], to deduce [Q(α) : Q].

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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