Download MATH30010: Field Theory Homework 5: Solutions

MATH30010: Field Theory
Homework 5: Solutions
1. Let L/F be an extension of fields, and suppose that α, β ∈ L are
algebraic over F of degree r and s respectively. Let K = F (α, β).
(a) Give an explicit example of F, L, α and β where [K : F ] < rs.
(b) Prove that if (r, s) = 1, then [K : F ] = rs.
Solution:
√
√
3
3
(a) Take F = Q, L = C,
α
=
2
and
β
=
ω
2 where
√
1
3i
.
ω = e2πi/3 = − +
2
2
√
√
Then degQ (α) = degQ (β) = 3, but K = Q(α, β) = Q( 3 2, ω 3 2) =
√
Q( 3 2, ω) is of degree 6 (not 9) over Q.
(b) Firstly, K = F (α, β) = F (α)(β) =⇒
[K : F ] = [F (α) : F ][F (α)(β) : F (α)] = degF (α)degF (α) (β) ≤ rs
since degF (α) (β) ≤ degF (β) = s.
On the other hand F (α) ⊂ K =⇒ r = [F (α) : F ]|[K : F ] by
the Tower Law, and similarly F (β) ⊂ K =⇒ s|[K : F ].
Since (r, s) = 1 it follows that rs|[K : F ]. Since [K : F ] ≤ rs it
now follows that rs = [K : F ].
2. Let F be a field of characteristic p > 0. Let Ψ : F → F be the
Frobenius homomorphism, a 7→ ap . Recall that Fp ⊂ Fix(Ψ). Prove
that in fact Fp = Fix(Ψ). [Hint: You may wish to use the fact that,
over any field, a polynomial of degree d with coefficients in the field
has at most d roots in the field.]
Solution: Recall, from class, that the elements of Fp are precisely the
p roots of the polynomial xp − x ∈ Fp [x].
Let a ∈ F . Then a ∈ Fix(Ψ) ⇐⇒ Ψ(a) = a ⇐⇒ ap = a ⇐⇒
ap − a = 0 ⇐⇒ a is a root of xp − x ⇐⇒ a ∈ Fp .
3. Let n > 2. Recall that ζn := e2πi/n = cos(2π/n) + i sin(2π/n) and
that
ζn + ζn−1
2π
ζn + ζ¯n
=
= cos
.
<(ζn ) =
2
2
n
Let F = Q(ζn ) and K = Q(cos(2π/n)) ⊂ F .
(a) Show that the map σ : F → F, z 7→ z̄ (complex conjugation) is
a field automorphism.
(b) Show that K ⊂ Fix(σ) and deduce that K 6= F .
(c) Prove that ζn is the root of a quadratic polynomial with coefficients in K.
(d) Explain why it follows that
2π
1
degQ (cos
) = degQ (ζn ).
n
2
(e) Show that degQ (cos(2π/5)) = 2. Can you find the minimal polynomial of cos(2π/5)?
(f) Show that degQ (cos(2π/7)) = 3.
Solution: (Observe that ζn ζ¯n = |ζn |2 = 12 = 1 =⇒ ζn−1 = ζ¯n .)
(a) F = Q(ζn ) = Q(ζn−1 ) = Q(ζ¯n ). But σ(Q) = Q and σ(ζn ) = ζ¯n .
It follows that σ is a bijective map from F to F . Furthermore
σ(1) = 1, and for all a, b ∈ F σ(a + b) = σ(a) + σ(b) and σ(ab) =
σ(a)σ(b), so σ is an automorphism of F .
(b) Clearly Fix(σ) = F ∩ R and ζ 6∈ Fix(σ). Thus Fix(σ) is a proper
subfield of F .
However Q, cos(2π/n) ∈ R =⇒ K = Q(cos(2π/n)) ⊂ Fix(σ). It
thus follows that K is a proper subfield of F and thus [F : K] ≥ 2.
(c) ζn is a root of
2
x − 2 cos
2π
n
x + 1 ∈ K[x].
(d) It follows that [F : K] = [K(ζn ) : K] = degK (ζn ) ≤ 2. Thus
[F : K] = 2. But then, by the Tower Law,
2π
1
1
) = [K : Q] = [F : Q] = degQ (ζn ).
degQ (cos
n
2
2
(e) Let ζ = ζ5 .
Since 5 is prime, degQ (ζ) = 5 − 1 = 4 by a previous exercise.
Thus degQ (cos(2π/5)) = 12 · 4 = 2.
Recall that
2π
ζ + ζ −1
2π
cos
=
and hence 2 cos
= ζ + ζ −1
5
2
5
and
0 = 1 + ζ + ζ 2 + ζ 3 + ζ 4.
Multiplying this latter by ζ −2 gives
0 = ζ −2 + ζ −1 + 1 + ζ + ζ 2 .
Now using ζ 2 + ζ −2 = (ζ + ζ −1 )2 − 2, we find that cos(2π/n)
satisfies the quadratic
4x2 + 2x − 1 = 0.
Since it has degree 2 over Q, its minimal polynomial is thus
1
1
x2 + x −
2
4
(Observe that it follows, solving the quadratic, that
√
2π
−1 + 5
cos
.)
=
5
4
(f) Since 7 is prime, degQ (ζ7 ) = 6 and hence degQ (cos(2π/7)) =
1
· 6 = 3.
2
4. Prove that the roots of the polynomial q(x) = x4 − 6x2 + 6 are constructible.
Solution: Let a be a root of q(x) and let b = a2 . Thus b2 − 6b + 6 = 0
and hence
√
√
√
6±2 3
6 ± 12
=
= 3 ± 3.
b=
2
2
Either √way, b p
is a positive real number and thus all four values of
√
a = ± b = ± 3 ± 3 are real numbers. Since these real numbers
are obtained from rational numbers by field operations and the taking
of square roots, they are constructible numbers.
5. Prove that the regular n-gon can be constructed by ruler and compass
if and only if cos(2π/n) is constructible.
Solutions: The vertices of the regular n-gon in the complex plane are
the numbers ζnk , k = 0, 1, . . . , n − 1. The regular n-gon is constructible
if and only if the side-length is constructible. Since 1 and ζn are
neighbouring vertices, the side-length is β := |1 − ζn |.
Now β is constructible if and only if β 2 = |1−ζn |2 = (1−ζn )(1− ζ¯n ) =
2 − (ζn + ζ¯n ) = 2 − 2 cos(2π/n) is constructible. So β is constructible
if and only if
2π
2 − β2
cos
=
n
2
is constructible.
(Observe that it follows that
s
β=
2 − 2 cos
2π
n
= 2 sin
π n
.)
6. Let F8 = F2 (x3 + x + 1) be the field with 8 elements.
(a) Find a primitive root of F8 ; i.e. a nonzero element of order 7.
Prove all assertions.
(b) Does F8 contain any subfield of order 4? Explain.
Solution:
(a) As it turns out, any element other than 0 and 1 is a primitive
root for F8 . For instance, let θ ∈ F8 be a root of x3 + x + 1.
Recall that each element of F8 is uniquely expressible in the form
a0 + a1 θ + a2 θ2 with ai ∈ F2 .
So θ3 = 1 + θ and the successive powers of θ are:
θ1
θ2
θ3
θ4
θ5
θ6
θ7
=
=
=
=
=
=
=
θ
θ2
1+θ
θ + θ2
θ2 + θ3 = 1 + θ + θ2
θ + θ2 + θ3 = 1 + θ2
θ + θ3 = 1.
So θ is a primitive root for F8 .
(b) No. If F ⊂ F8 is a subfield then F2 ⊂ F ⊂ F8 =⇒
3 = [F8 : F2 ] = [F8 : F ][F : F2 ] =⇒ [F : F2 ] = 1 or 3 =⇒ |F | =
2 or 8.