# Math 121 Homework 5 Solutions

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1 Math 2 Homework Solutions Problem 2, Section 4.. Let τ : C C be the complex conjugation, defined by τa + bi = a bi. Prove that τ is an automorphism of C. First Solution. Every element of C may be written uniquely as a + bi. To see that τ is a ring homomorphism, we must check that and τa + bi + c + di = τa + bi + τc + di τa + bic + di = τa + biτc + di. The first identity is easy to check. The second means which is also easy to check. ac bd ad + bci = a bic di, Second Solution. Note that C = Ri is the splitting field of the irreducible polynomial x 2 +. This has roots i, i an it follows from Theorem 8 on page 9 that there is an isomorphism τ : C C over R such that τi = i. Now τa + bi = τa + τbτi = a bi, that is, τ is complex conjugation. This proves that complex conjugation is an automorphism. Problem 3, Section 4.. Determine the fixed field of complex conjugation on C. Solution: Any complex number may be written uniquely as a + bi with a, b R. If τa + bi = a + bi then a bi = a + bi which is equivalent to b = 0 and so a + bi = a R. Therefore the fixed field of complex conjugation τ is R.

2 Problem 4, Section 4.. Prove that Q 2 and Q 3 are not isomorphic. Solution. If there is an isomorphism φ : Q 2 Q 3 and then φ 2 = a + b 3 satisfies a + b 3 2 = 2. That is, a 2 + 3b 2 + 2ab 3 = 2. Note that and 3 are linearly independent over Q, so 2ab = 0, that is, either a = 0 or b = 0. Thus either a 2 = 2 or b 2 = 2/3. Neither of these equations may be solved with a or b in Q, which is a contradiction. Problem, Section 4.2. Determine the minimal polynomial over Q for the element 2 +. Solution. Let fx be the minimal polynomial for α = 2 +. If β is any root of fx then by Theorem 8 on page 9 there is an isomorphism σ : Qα Qβ such that σα = β. By Theorem 27 on page 4, we may extend σ to an automorphism of the splitting field E of f over Q. Now consider σ 2. Since 2 is a root of the polynomial x 2 over Q, so is σ 2. Therefore σ 2 = ± 2, and similarly σ = ±. Therefore β = σ 2 + = ± 2 ±. Thus the possible roots of 2+ are among the four element set { ± 2 ± } and so f divides the polynomial 2 + x + x This equals x 4 4x To show that this is the minimal polynomial, it is sufficient to show that it is irreducible over Q. None of its roots are in Q, so if it is reducible, it splits as two quadratic factors. One of these will have 2 + as a root. Hence one of the quadratic factors is one of 2 + = x 2 2 3, x + = x 2 x + 3, x = x It is easy to see that none of these are in Q[x], so x 4 4x is irreducible over Q, and this is the minimal polynomial. 2

3 Section 4.2, Problem. Prove that the Galois group of x p 2 over Q a b for p a prime is isomorphic to the group of matrices for a, b F 0 p, a 0. Solution. The polynomial is irreducible by Eisenstein s criterion. So its splitting field is generated by the roots of x p 2 which are αζ i 0 i < p where α = 2 /p and ζ = e 2πi/p. Thus the splitting field is Qα, i. Note that Qζ/Q has degree φp = p, while Qα/Q has degree p. These degrees are coprime, so [Qα, i : Q] = pp by Corollary 22 on page 29. Let m Z be prime to p. We will show that there exists τ α GalK/Q such that τ m α = α τ m ζ = ζ m. Since K/Q is Galois, K/Qα is also Galois and has degree p. Indeed, K = Qα, ζ so the effect of θ GalK/Qα is determined by θζ which must be a primitive p-th root of unity. There are p elements of the Galois group and exactly p such p-th roots of unity, so GalK/Qα must be transitive on these. Therefore we can find τ m as required. Now we will show that GalK/Q contains an automorphism σ such that σα = αζ, σζ = ζ. 2 Indeed, α and αζ are roots of the same irreducible polynomial x p 2, so we may find an automorphism σ such that σ α = αζ. Now σ ζ is a primitive p-th root of unity, say σ ζ = ζ k, p k. Then τ m ζ k = ζ for some α. So if k is the image of k in F p, m = /k. Then τ m σ has the desired effect 2. The elements τ m σ k and σ km τ m have the same effect, α αζ km, ζ ζ m. So τ m σ k τ m = σ km. now the affine group G of matrices a, b F p, a 0 also has generators m t m =, s = a b 0 k, so s k = subject to the same relations t m s k t m = s km. Therefore there is an isomorphism GalK/Q G in which τ m t m and σ s.. for 3

4 Section 4.2, Problem 6. Let K = Q 8 2, i and let F = Qi, F 2 = Q 2, F 3 = Q 2. Prove that GalK/F = Z 8, GalK/F 2 = D 8 and GalK/F 3 = Q 8. The book devotes several pages to this example. However we will solve this from scratch. Solution. First let us observe that K is Galois and of degree 6 over Q. To begin with, α = 8 2 is the root of the polynomial x 2 which is irreducible by Eisenstein s criterion, so [Qα : Q] = 8. Now i / Qα since Qα R. Therefore [Qα, i : Q] = 2 and so [Qα, i : Q] = [Qα, i : Qα][Qα : Q] = 6. The field Qα, i contains the primitive 8-th roots of unity since it contains 2 = α 4 and hence ε 8 = 2 + i = e 2πi/8. Hence it contains all the roots of x 8 2 = 7 αε k 8. It is thus a splitting field for this polynomial, and Galois over Q. k=0 Lemma. Let β = αε k 8 be any root of x 8 2. Then there exist elements σ, σ GalK/Q such that σα = β, σi = i, and σ α = β, σ i = i. Proof. There exists φ : Qα Qβ such that α β by Theorem 8 on page 9. By Theorem 27 on page 4, this extends to an automorphism of the splitting field K. We show that i / Qβ. Otherwise φ i Qα R and φ i is a root of x 2 + = 0, that is, φ i = ±i which is not real, so this is a contradiction. Since Qβ has degree 8 over Q and i / Qβ, Qβ, i has degree 6 over Q and hence Qβ, i = K. Since K is Galois over Q, it is Galois over Qβ and GalK/Qβ = 2. Let ψ be the generator, so that ψβ = β and ψi = i. Then φ and ψφ both take α to β, and one takes i to i, the other i to i. This gives us σ and σ. We may now handle the three cases. Each of GalK/Qi, Gal K/Q 2 and Gal K/Q 2 has order 8, which is the degree of the extension. 4

5 First, the Lemma gives us an element θ GalK/Q such that θα = ε 8 α and θi = i. This is then in GalK/Qi. We find that θ 2 = 2 since 2 = α 4 and 2 = ε 8 α 4. Therefore θε 8 = θ 2 + i = 2 + i = ε 8. We may now calculate that θ 2 α = ε 2 8α = iα and θ 2 ε 8 = ε 8 so θ 4 α = α and θ 4 i = i. We see that θ has order 8 and therefore GalK/Qi is cyclic of order 8. Next, let us show that Gal K/Q 2 = D8. The Lemma gives us two elements ρ, τ GalK/Q such that ρα = iα, ρi = i, and τα = α, τi = i. Since 2 = α 4, both of these fix 2 and so they are in Gal K/Q 2. It is easy to see that ρ has order 4 and τ has order 2, and τρτ = ρ, so they generate a dihedral group of order 8. Finally, let us show that Gal K/Q 2 = Q8. The Lemma gives us two elements µ and ν such that µα = ε 8 α µi = i We will show that µ, ν satisfy, and να = ε 8 α νi = i. µ 2 = ν 2, µ 4 =, νµν = µ. These relations define the quaternion group Q 8. Since 2 = iα 4, µ sends 2 to iε8 α 4 = 2 and is in Gal K/Q 2 ; similarly ν is also in this Galois group. Using ε 8 = 2 + i we also calculate Using this we calculate µε 8 = νε 8 = ε 3 8. µ 2 α = α µ 2 i = i and ν has the same effect. Thus µ 2 = ν 2 and µ, ν have order 4. The relation νµν = µ may now be checked.

6 Section 4.2, Problem 3. Show that Q is a cyclic quartic field, that is, it is Galois over Q of degree 4 with cyclic Galois group. Consider the polynomial fx = x x + 2. In every square root, the argument is a positive real number and as usual we are taking the positive square root. This polynomial equals x x 2 = x 2 2 = x 4 4x It is Eisenstein, hence irreducible. Let α = and β = 2. Then 2 = α 2 Qα, and since αβ = = = 2 we also have 2 β = α Qα. The other two roots are α, β and so Qα is the splitting field of f. Since f is separable Qα is Galois over Q. Its degree is 4, so the Galois group has degree 4. Now to compute the Galois group, Theorem 8 on page 9 guarantees that there is an isomorphism σ : Qα Qβ such that σα = β. Then σ 2 = σα 2 = β 2 = 2. Remembering that β = 2/α we then have σβ = σ 2 /σα = 2/β = α. This σ permutes the roots cyclicly: α β α β α. This proves that the Galois group is cyclic of order 4. Section 4.3, Problem. Exhibit an explicit isomorphism between the splitting fields of x 3 x + and x 3 over F 3. Solution. If x 3 x + = 0 then y 3 y = 0 where y = x. 6

7 Section 4.4, Problem 8. Determine the splitting field of the polynomial x p a where a F p, a 0. Solution. The polynomial is irreducible and separable by Problem in Section 3., which was assigned in Homework 3. It has degree p, so every root must lie in the unique field with p p elements. Thus F p p is the splitting field. 7

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