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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.