Exercises on Galois correspondence theorem

1. Calculate the Galois group of the following polynomials in ℚ[x]

   1. x³ + 2x² + 5x + 10

   2. x⁴ − 7x² + 10

   3. x⁵ + 2x⁴ − 5x³ − 10x² + 6x + 12

   4. x³ − 1

   5. x³ − 7

   6. x⁴ − 5

   7. x⁴ − 4x² + 2.

2. Calculate the splitting field K and the Galois group of x⁶ + x⁴ − 4x² − 4 ∈ ℚ[x]. Determine all subfield of K.

3. Let K the splitting field of the polynomial x³ + √3x² − 2x − 2√3 ∈ ℚ(√3)[x]. Calculate G(K, ℚ(√3)).

4. Let K the field ℚ(i, ∛5). Calculate the autormorphism group of K. Check whether K is a Galois extension of ℚ.

5. Let K the field ℚ(i∜2). Calculate the autormorphism group of K. Check whether K is a Galois extension of ℚ.

6. The Galois group of the polynomial x³ − 5 ∈ ℚ(ζ), over ℚ(ζ), with ζ primitive fifth root of unity.

7. Find the Galois groups of x⁴ + 1 and x⁸ −1 ∈ ℚ[x].

8. Find the Galois groups of x⁵ − 1.

Solutions

1. 1. x³ + 2x² + 5x + 10 = (x² + 5) (x + 2). The splitting field is K = ℚ(√5i). One automorphism is the identity and the other σ(√5i) = −√5i; So {id, σ} ≅ ℤ₂. ■

   2. If we replace x² = y we get y² −7y + 10. We find the two roots y = 2 and y = 5, so (x² − 2)(x² − 5), and each of the factor is irreducible over ℚ. Hence the splitting field is ℚ(√2,√5). The four roots of thr original polynomial are √2, −√2, √5, −√5, and an automorhpsms of the splitting field has to send each elemnet in {√2, −√2} to something in there, similarly for {√5, −√5}. If f(a) = −a, then automatically f(−a) = a, because f(−a) = −f(a) = −(−a) = a. Further, since f is an automorphism, we must have f(a)² = f(a²)

      σ1 = id
      σ2:	σ2(√2) = −√2	σ2(√5) = −√5
      σ3:	σ3(√2) = −√2	σ3(√5) = √5
      σ4:	σ4(√2) = √2	σ4(√5) = −√5

      G(ℚ(√2,√5),ℚ) ≅ ℤ₂ x ℤ₂.  ■

   3. A factor is (x + 2). Dividing the polynomial by x + 2 yo find x⁴−5x²+6; by substitution of y = x² I've found two roots +3;+2, so x⁵ + 2x⁴ − 5x³ − 10x² + 6x + 12 = (x + 2) (x² − 3) (x² − 2). So the splitting field is ℚ(√2,√3). The four roots of thr original poly are −2, √2, −√2, √3, −√3. So we have G(ℚ(√2,√3),ℚ) with the following elements

      σ1 = id
      σ2:	σ2(√2) = −√2	σ2(√3) = −√3
      σ3:	σ3(√2) = −√2	σ3(√3) = √3
      σ4:	σ4(√2) = √2	σ4(√3) = −√3

      G(ℚ(√2,√3),ℚ) ≅ ℤ₂ x ℤ₂.  ■

   4. ℚ(ζ) is the splitting field with ζ = e^i2π/3 = cos 2π/3 + i sin 2π; cos value is −1/2, that sin value is √3/2., so ζ = (−1 + √3i)/2. The Galois group G(ℚ(ζ), ℚ) ≅ ℤ₂ since it contains the ideintity automorphism and the automorphism such that ζ ⟶ ζ².  ■

   5. The roots are ∛7, ∛7ζ, ∛7ζ². So the splitting field is ℚ(∛7, ζ). There are six elements in the Galoois group

      σ1 = id
      σ2:	σ2(ζ) = ζ2	σ2(∛7) = ∛7
      σ3:	σ3(ζ) = ζ	σ3(∛7) = ζ ∛7
      σ4 = σ32:	σ4(ζ) = ζ	σ4(∛7) = ζ2 ∛7
      σ5 = σ2 ∘ σ3:	σ5(ζ) = ζ2	σ5(∛7) = ζ2 ∛7
      σ6 = σ2 ∘ σ32:	σ(ζ) = ζ2	σ6(∛7) = ζ ∛7

      So haveing the Galois groups 6 elements and the only subgroup of S₃ that has order 6 is S₃ itself, so G(ℚ(∛7, ζ), ℚ) ≃ S₃.  ■

   6. The roots are ∜5, ∜5i, −∜5, −∜5i. Splitting field ℚ(∜5, i). If f is an automorphism then f(b) ∈ {b,bi,-b,-bi} and f(i) ∈ {i,−i} with b =∜5. So we have G(ℚ(∜5, i),ℚ) with the following eight elements

      σ1 = id
      σ2:	σ2(∜5) = −∜5	σ2(i) = i
      σ3:	σ3(∜5) = ∜5	σ3(i) = −i
      σ4:	σ4(∜5) = −∜5	σ3(i) = −i
      σ5:	σ5(∜5) = ∜5i	σ4(i) = i
      σ6:	σ6(∜5) = ∜5i	σ4(i) = −i
      σ7:	σ7(∜5) = −∜5i	σ5(i) = i
      σ8:	σ8(∜5) = −∜5i	σ6(i) = −i

      so G(ℚ(∜5, i),ℚ) ≃ D₄, the group of symmetries of a square.  ■

   7. With the substitution y = x² we obtain y² − 4y +2 with roots 2 ± √2. The polynomial splits as (x² − 2 + 2) (x² + 2 + 2), so the splitting field is ℚ(√(2 + √2) and we have the four root ±√(2 + √2) and √(2 - √2). The Galois group is the following

      σ1 = id
      σ2:	σ2(√(2 + √2)) = −(2 + √2)
      σ3:	σ3(2 + √2) = −(2 + √2)	σ3(i) = −i
      σ4:	σ4(2 + √2) = (2 + √2)	σ4(i) = −i

      so G(ℚ(√(2 + √2), ℚ) ≅ ℤ₄.  ■

2. By substitution y = x² we have y³ + y² − 4y − 4 = (y + 1) (y² − 4) = (x² + 1) (x² − 2) (x² + 2). Its roots are ± √2, ± i√2. Its splitting field is K = ℚ(√2, i). We have

   [ℚ(√2, i) : ℚ] = [ℚ(√2, i) : ℚ(√2)]: [ℚ(√2) : ℚ]

   Thus K = [ℚ(√2, i) : ℚ] = 4. A basis of K over ℚ, is given by 1, i, √2, i√2. Every σ ∈ G(K, ℚ), is given by the images of i and √2. σ(i) can be in {±i} and σ(√2) in {±√2}. We have four automorphisms:

   σ1 = id
   σ2:	σ2(√2) = √2	σ2(i) = −i
   σ3:	σ3(√2) = −√2	σ3(i) = i
   σ4:	σ4(√2) = −√2	σ3(i) = −i

   As in the Klein group, every element other than the identity has order 2, si G(K, ℚ) ~ V. By Galois Correspondence theorem, the subfields of K are all and only the fields fixed by the subgroup of G(K, ℚ) .
   σ₃(a + bi + c√2 + di √2) = a −bi −c √2 + di√2. We see that σ₃ fixes all elements a + di √2, which are part of ℚ(i√2).
   σ₂(a + bi + c√2 + di √2) = a − bi + c √2 − di√2. So σ₂ fixes the elements a + c√2, which are part of ℚ(√2).
   σ₄(a + bi + c√2 + di √2) = a −bi −c √2 − di√2, so σ₄ fixes ℚ and σ₁ fixes K. ■

3. A root of the polynomial is −√3 so a factor is (x + √3); dividing the polynomial by this factor we find the other (x² − 2), so

   x³ + √3x² − 2x − 2√3 = (x + √3) (x² − 2)

   So the splitting field is K = ℚ(√2,√3)) The roots of the polynomial are −√3 and ±√2. An element σ ∈ G(K, ℚ(√3)) sends each root in {√2, −√2}, to the same set; The only automorphism, in addition to the identity is σ(√2) = √2. Hence G(K, ℚ(√3)) ≅ ℤ₂. ■

4. The minimal polynomial of i over ℚ is x² + 1, σ(i) = ± i. The minimal polynomial of ∛5 is x³ −5, σ(∛5) must be a root of x³ − 5: the roots are ∛5 and the complex roots ∛5ζ, ∛5ζ², with ζ primitive 3rd root, so 1/2(−1 ± i √3)∛5. These last two roots do not belong to K, otherwise √3 as well, would be in K. So K is not a splitting field for the polynomial and it cannot be a Galois extension of ℚ. So we must have σ(∛5) = ∛5. Thus there are two autormorphisms: the identity automorphism, and the automorphism σ that sends i to −i and ∛5 to ∛5: every element of K = {q₀, q₁ ∛5 + q₂ (∛5)² + q₃i + q₄ i + q₅i(∛5)²} is sent to its complex conjugate and [K:ℚ] = 6. Hence G(K, ℚ(√3)) ≅ ℤ₂. This is not a Galois extension since |Aut(K/ℚ)| = 2 and by theorem 9.3.4 we should have |G(K,F)| = [K : F].  ■

5. The minimal polynomial is x⁴ − 2. An element of K is a₀ + a₁ (i∜2) + a₂(i∜2)² + a₃ (i∜2)³, with a_i ∈ ℚ. The splitting field of x⁴ − 2 is ℚ(i,∜2). In a Galois group, G(E, ℚ), E must be the splitting field of the polynomial; So we must show that ℚ(i,∜2) and ℚ(i∜2) are not the same extension. ℚ(∜2) and ℚ(i∜2) are isomorphic: composing the isomorphism φ : ℚ(i∜2) ⟶ ℚ(∜2) with the embedding Q(∜2) ⟶ ℝ, we have an embedding ℚ(i∜2) ⟶ ℝ. However, the splitting field of x⁴ − 2 cannot be embedded in ℝ, because it contains i. Thus K is not a Galois extension of ℚ. Therefore, the fixed field of Aut(K/ℚ) is larger than just ℚ. Let's find out what this field is. Notice that an automorphism σ : K ⟶ K must send “i∜2” to a root of x⁴ − 2. The possibilities are “i∜2” and “−i∜2”, (Since the other two roots of x⁴ − 2, namely, “∜2” and “−∜2”, are missing from K.). We thus have only two automorphisms: the identity and σ(i∜2) = −i∜2. To calculate the fixed field of Aut(K/ℚ), we let σ act on an element of K

   σ(a₀ + a₁ (i∜2) + a₂(i∜2)² + a₃ (i∜2)³) = a₀ − a₁ (i∜2) + a₂(i∜2)² − a₃ (i∜2)³

   So it fixes only a₀ + a₂ (i∜2)² = a₀ − √2. So the fixed field by Aut(K/ℚ) is ℚ(√2).  ■

6. Let b = 3^1/5. The splitting fiels of the polynomial is ℚ(b, ζ). Let's call σ_i the automorphism defined by σ_i(b) = bζⁱ. Then σ_i(bζ^j) = σ_i(b) ⋅ σ_i(ζ)^j = bζⁱ ⋅ ζ^j = bζ^i+j, since σ_i ∈ Gal(ℚ(b,ζ) / ℚ(ζ)), we must have σ_i(ζ) = ζ.

7. x⁴ + 1 is irreducible over ℚ, see Example 5.7.5, in which we already found the roots. The splitting field of x⁴ + 1 is K(ω) with ω = (1 + i)/√2, we've already calculated in Example 4.13.4 the other three roots: ω³, ω⁵, ω⁷. Thus [K: ℚ] = 4. Let σ₁,σ₂,σ₃ be the non-identity elements of Gal(ℚ(b) / ℚ). That is, σ₁(ω) = ω³, σ₂(ω) = ω⁵ and σ₃(ω) = ω⁷. First of all, notice that σ₁²(ω) = σ₁(ω³) = ω⁹ = ω. Similarly, σ₂²(ω) = ω and ω₃²(ω) = ω Therefore, Gal(ℚ[b] / ℚ) has to be (isomorphic to) the Klein group. Since ω⁴ = −1 then (ω⁴)² = ω⁸ = 1. So x4 + 1 and x⁸ − 1 have the same splitting field. We know from 9.3.9 Proposition that the Galois group of x⁸ − 1 must be isomorphicto U(ℤ_n) = {1,3,5,7}. ■

8. The splitting field of x⁵ − 1 over ℚ is K = ℚ(ζ) with ζ primitive 5-th root of unity. Since x⁵ − 1 is reducible over ℚ a factor is (x−1), doing the division of x⁵ − 1 by (x − 1) we obtain x⁴ + x³ + x² +1, which is the miniaml polynomial of ζ. So [K : ℚ] = 4 and the four elements of Gal(K / ℚ) are

   σ0 = id
   σ1:	σ1(ζ) = ζ2
   σ2 = σ1 ∘ σ1∘ σ1:	σ2(ζ) = ζ3
   σ3 = σ1 ∘ σ1:	σ3(ζ) = ζ4

   σ₁ is the generator of the group which is isomorphic to ℤ₄. In particular Gal(K / ℚ) has three subgroups of order 2: (a) the whole group, (b) the subgroup {id, σ₁²} = {id, σ₃}, and (c) the trivial.
   Notice that σ₃(ζ) = ζ⁴ and σ₃(ζ⁴) = (ζ⁴)⁴ = ζ¹⁶ = (ζ⁵)³ ⋅ ζ = 1³ ⋅ ζ = ζ. Since σ₃ swaps ζ and ζ⁴, it follows that σ₃ fixesζ + ζ⁴. Notice that σ₃ swaps ζ² and ζ³ as well, and therefore fixes ζ² + ζ³ as well. We need to find the minimal polynomial of either ζ + ζ⁴ or ζ² + ζ³, to show that ℚ(ζ + ζ⁴) = ℚ(ζ² + ζ³) is a proper intermediate field between ℚ and ℚ(ζ). We know that ζ⁴ + ζ³ + ζ² + ζ + 1 = 0 and noticing that (ζ² + ζ³)² = ζ⁴ + ζ⁶ + 2 = ζ⁴ + ζ + 2 since that ζ⁵ = 1 and ζ⁶ = z, the minimal polynomial is

   x² + x − 1

   Using the quadratic formula or completing the square, we get the roots (−1 ± √5)/2, thus K_H = ℚ(√5). Chosing ζ = e^{(i⋅ 2π/5)} we have ζ² + ζ³ = (−1 − √5)/2 and ζ + ζ⁴ = (−1 + √5)/2. The minimal polynomial of ζ over K_H is x² + (ζ³ + ζ² + 1)x + 1. By solving this quadratic equation by the quadratic formula: ζ = (−b ± √(b² − 4)) / 2, with b = ζ³ + ζ² + 1. From the above value for ζ² + ζ³ we have

   b = ζ² + ζ³ + 1 = (−1 − √5)/2 + 1 = (−√5+1)/2
   b² = [(−√5+1)/2]² = (3−√5)/2.

   So

   $$\zeta =\frac{(-1+\sqrt{5})}{4}+\frac{1}{2}i\sqrt{(5+\sqrt{5})/2}$$