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. Hence
bau = bv = w, so u ∼ w.
Optional Exercise for Extra Credit (20 points total)
[Solutions Posted Separately]
Exercise 2. Let K ⊂ F ⊂ E be three fields. Suppose that {α1 , . . . , αm } is a K-basis of F and
{β1 , . . . , βn } is an F -basis of E. Prove that {αi βj : 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].
Pn
Proof. Let γ ∈ E. Since {β1 , . . . , βn } spans E over F , there are fj ∈ F such that γ = j=1 fj βj .
Pm
Since {α1 , . . . , αm } spans F over Kd, there are, for each j, elements cij ∈ K such that fj = i=1 cij αi .
Then we have
m
n
n X
X
X
cij αi βj ,
γ=
fj βj = γ =
j=1 i=1
j=1
so {αi βj : 1 ≤ i ≤ m, 1 ≤ j ≤ n} spans E over K.
Suppose we have scalars cij ∈ K such that
n X
m
X
cij αi βj = 0.
j=1 i=1
Pm
Since {β1 , . . . , βn } is linearly independent, we have i=1 cij αi = 0 for each i. Since {α1 , . . . , αm } is
linearly independent, we have cij = 0 for each i, j. Hence the set {αi βj : 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 = e2πi/3 , F = Q(α) and E = F (ζ). Compute [E : Q].
Comment. On the previous homework, you showed that the roots of x3 − 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
x2 + x + 1, whose roots are complex, and do not lie in F . So x2 + 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 pα (x) be the minimal polynomial of α
over Q. Compute the degree of pα (x).
(A formula for pα (x) was stated in class without proof. Do not use this formula.)
Hint: Note that α = ζ + ζ −1 , where ζ = e2πi/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 + · · · + xp−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) = x2 − 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]
= 21 [Q(ζ) : Q] = 12 (p − 1).
[Q(ζ) : Q(α)]
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 = e2πi/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 x4 + x3 + x2 + x + 1 = (x2 + ax + 1)(x2 + bx + 1) with a, b ∈ R.
Solution.
a) The minimal polynomial of ζ over Q is x4 + x3 + x2 + x + 1, so [Q(ζ) : Q] = 4.
b) It suffices to show that τ ∈ Q(ζ). From class, we know that
√
ζ + ζ −1 = 2 cos(2π/5) = 12 (−1 + 5) = τ − 1.
hence τ = 1 + ζ + ζ −1 ∈ Q(ζ).
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 x2 + (1 − τ )ζ + 1.
Another way is to follow the hint and try to factor
x4 + x3 + x2 + x + 1 = (x2 + ax + 1)(x2 + bx + 1) = x4 + (a + b)x3 + (2 + ab)x2 + (a + b)x + 1.
We must have b = 1 − a, and 2 + a(1 − a) = 1, or a2 − a − 1 = 0. This means {a, b} = {τ, 1 − τ }. Now
ζ is root of either x2 + τ x + 1 or x2 + (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 x2 + (1 − τ )ζ + 1.
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