Computing Galois groups II. Quartics and beyond.

When you have eliminated the impossible, whatever remains, however improbable, must be the truth, Sherlock Holmes

Theorem: Let f(x) be the general nth degree polynomial in F[x] of deg(f) = n ≤ 4, then f is solvable (by radicals).

Mod p Irreducible Test. Let p be a prime and suppose that f(x)∈ ℤ[x] with deg f(x) ≥ 1. Let f′(x) be the polynomial in ℤ_p[x] obtained from f(x) by reducing all the coefficients of f(x) modulo p. If f′(x) is irreducible over ℤ_p and deg f(x)= deg f′(x), then f(x) is irreducible over ℚ.

Resolvent cubic is a cubic polynomial defined from a monic polynomial of degree four, g(x) = (x -β₁)(x -β₂)(x -β₃) ∈ K[x]. We define β₁ = α₁α₂ + α₃α₄, β₂ = α₁α₃ + α₂α₄, β₃ = α₁α₄ + α₂α₃, β_i∈ K.

Corollary. If g splits completely in F ↭ G = D₂. If g has exactly one root in F ↭ G = D₄ or C₄. g is irreducible ↭ G = S₄ or A₄

Disc(x⁴ +cx +d) = 256d³ -27c⁴. Resolvent cubic of x⁴ +cx + d = x³ -4dx -c²

• f = x⁴ -2. It is irreducible by Eisenstein criteria (p = 2). f = $(x^2 - \sqrt{2})(x^2 + \sqrt{2}) = (x - \sqrt[4]{2})(x + \sqrt[4]{2})(x - i\sqrt[4]{2})(x + i\sqrt[4]{2})$. Roots: $α_1= \sqrt[4]{2}, α_2 = iα_1, α_3=-α_1, α_4=-α_2$

K = $ℚ(\sqrt[4]{2}, \sqrt[4]{2}i) = ℚ(\sqrt[4]{2}, i)$, and this can be easily seen because i = $\frac{i\sqrt[4]{2}}{\sqrt[4]{2}}$ so $ℚ(\sqrt[4]{2}, i) ⊆ ℚ(\sqrt[4]{2}, \sqrt[4]{2}i)$, but obviously it is also the case that $\sqrt[4]{2}i = \sqrt[4]{2}·i$ so $ℚ(\sqrt[4]{2}, \sqrt[4]{2}i) ⊆ ℚ(\sqrt[4]{2}, i)$.

$β_1 = 2i\sqrt{2}, β_2=0, β_3=-2i\sqrt{2}, g = (x -2i\sqrt{2})(x + 2i\sqrt{2})x = x^3+8x$, g is reducible, and has exactly one root in ℚ ⇒ Gal(f) = D₂ (it is not possible because it is not completely reducible -Corollary-) D₄ or C₄. To discriminate between these two choices, further analysis is required.

Therefore, [K: ℚ] = [K : ℚ(i, ∜2)]·[ℚ(∜2) : ℚ] = 8 ⇒ Gal(f) = D₄.

The dihedral group D₄ is the symmetry group of the square. D₄ = ⟨r, s : r⁴ = s² = id, rs = sr^-1⟩.

1. r = $(\sqrt[4]{2},i\sqrt[4]{2},-\sqrt[4]{2},-i\sqrt[4]{2})$, |r| = 4, ℤ₄ ≋ ⟨r⟩. Notice that $r(i)=r(\frac{i\sqrt[4]{2}}{\sqrt[4]{2}}) = \frac{r(i\sqrt[4]{2})}{r(\sqrt[4]{2})} = \frac{-\sqrt[4]{2}}{i\sqrt[4]{2}} = \frac{-1}{i} = i$, so it could be easily seen that its fixed field is ℚ(i).
2. r² = $(\sqrt[4]{2},-\sqrt[4]{2})(i\sqrt[4]{2},-i\sqrt[4]{2})$. The center of D₄ is given by ℤ(D₄) = {e, r²} = ⟨r²⟩ ≋ ℤ₂. What is its fixed field? i is fixed by r², but also $r^2(\sqrt{2})=r^2(\sqrt[4]{2}·\sqrt[4]{2})=r^2(\sqrt[4]{2})·r^2(\sqrt[4]{2})=-\sqrt[4]{2}·-\sqrt[4]{2}=\sqrt{2}$, and its fixed field is ℚ(i, $\sqrt{2}$).
3. Of course, s is meant to be complex conjugation, say s = (i, -i), |s| = 2, and $ℚ(\sqrt[4]{2})$ is fixed by s. Futhermore, $\sqrt{2}$ is fixed by both s and r², so the fixed field of V₄ = ⟨r², s⟩ is $\mathbb{Q}(\sqrt{2})$
4. rs: $\sqrt[4]{2}→\sqrt[4]{2}→i\sqrt[4]{2},i\sqrt[4]{2}→-i\sqrt[4]{2}→\sqrt[4]{2}, i→-i→-i, -i→i→i, -\sqrt[4]{2}→-\sqrt[4]{2}→-i\sqrt[4]{2}, -i\sqrt[4]{2}→i\sqrt[4]{2}→-\sqrt[4]{2}$. Therefore, rs = $(\sqrt[4]{2},i\sqrt[4]{2})(i,-i)(-\sqrt[4]{2},-\sqrt[4]{2}i)$ and the reader can also verify that r³s = $(\sqrt[4]{2},-i\sqrt[4]{2})(i,-i)(-\sqrt[4]{2},\sqrt[4]{2}i)$

• x⁴ -4x³ -4x² + 8x -2 is irreducible by Eisenstein criteria (p = 2). Its roots are $x_{1,3} = 1 +\sqrt{2} ± \sqrt{3+\sqrt{2}}, x_{2,4} = 1 -\sqrt{2} ± \sqrt{3-\sqrt{2}}$. Δ = 64512 is not a square in ℚ.

Since [K : ℚ] = 8, and Δ is not a square in F (G = D₄, C₄ -|C₄| = 4, or S₄ -|S₄| = 24), the only option with order 8 is D₄ = ⟨a, b: a⁴ = b² = e, ab = ba^-1⟩.

It is obviously solvable by radicals: $\mathbb{Q}⊆\mathbb{ℚ}(\sqrt{2})⊆{ℚ}(\sqrt{2},\sqrt{3+\sqrt{2}})⊆{ℚ}(\sqrt{2},\sqrt{3+\sqrt{2}},\sqrt{3-\sqrt{2}})$

• f(x) = x⁴ +1, roots: ±1, ±i, g = x³ -4x. g is completely reducible in ℚ, its roots are -2, 0, and 2 ⇒ Gal(f) = D₂. Another way of obtaining the same result is seeing that K = ℚ(i), [K : ℚ] = 2, K/ℚ ≋ D₂.

• f(x) = x⁴ -16x² + 4. It is irreducible modulo 3 (or over ℤ₃, Mod p Irreducible Test) ⇒ f is irreducible in ℚ[x]. Its discriminant (Disc(f) = Disc(g) = ) Δ = 1920², so $\sqrt{Δ} = 1920 ∈ \mathbb{Q}$. Futhermore, the resolvent is g = x³ -128x² +3840x, so it is reducible over ℚ ⇒ Gal(K/Q) ≋ D₂.

• f(x) = x⁴ -4x² +2, is irreducible by Eisenstein, p = 2 | -4, 2 |2, but 2 ɫ a_n = 1, p² = 2² = 4 ɫ a₀ = a₂

x⁴ -4x² +2 = (x⁴ -4x +4) -2 = (x²-2)² -2 = 0 ⇒ (x²-2)² = 2 ⇒ $(x^2-2)=±(\sqrt{2}) ⇒ x^2 = 2±(\sqrt{2})$ ⇒ Its roots are $±α = ±\sqrt{2+\sqrt{2}}, ±β = ±\sqrt{2-\sqrt{2}}.$ Notice that β = $\frac{α^2-2}{α}∈ℚ(α)$, so K = [🚀] $ℚ(α)=ℚ(\sqrt{2+\sqrt{2}})$. An automorphism Φ: ℚ(α) → ℚ(α) is defined by permuting the four roots, e.g., σ: α → β, β → -α, Gal(K/ℚ) ≋ [Since [K : ℚ] = 4, the options with order 4 are Gal(K/ℚ) ≋ D₂ or C₄] D₂ or C₄

🚀 ℚ(α) = $ℚ(\sqrt{2+\sqrt{2}}), β = \sqrt{2-\sqrt{2}} = \frac{(\sqrt{2-\sqrt{2}})·(\sqrt{2+\sqrt{2}})}{(\sqrt{2+\sqrt{2}})}=\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}$. It is obvious that α ∈ ℚ(α) = $ℚ(\sqrt{2+\sqrt{2}}) ⇒ α^2 = 2 + \sqrt{2} ∈ ℚ(α) ⇒ \sqrt{2} ∈ ℚ(α) ⇒ β = \frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} ∈ ℚ(α)$ ⇒ K = ℚ(±α, ±β) = ℚ(α).

Let σ be defined by σ: α → β, β → -α. σ²(α) = σ(β) = -α, σ³(α) = σ(-α) = -β, σ⁴(α) = σ(-β) = α. σ²(β) = σ(-α) = -β, σ³(β) = σ(-β) = α, σ⁴(β) = σ(α) = β ⇒ σ⁴ = id ⇒ σ has order 4 ⇒ [D₂ has all elements of order 2 except the identity, but C₄ has an element of order 4] Gal(K/Q) ≋ C₄.

• f(x) = x⁴ +4x² +2 ∈ ℚ[x] is irreducible by Eisenstein’s criteria (p = 2) ⇒ Gal(f) is a transitive subgroup of S₄ = {S₄, A₄, D₈, V₄, C₄}.

x⁴ +4x² +2 = 0 ↭ $x^4 +4x^2 +4 -2 = 0 ↭ (x+2)^2 = 2 ↭ x = ±\sqrt{-2 ±\sqrt{ 2 }}, α_1 = \sqrt{-2 +\sqrt{ 2 }}, α_2 = -\sqrt{-2 +\sqrt{ 2 }}, α_3 = \sqrt{-2 -\sqrt{ 2 }}, α_4 = -\sqrt{-2 -\sqrt{ 2 }}$

By a similar reasoning than before K = ℚ(α₁) because $α_1^2 = -2 + \sqrt{2} ∈ K ⇒ \sqrt{2} ∈ K$. Futhermore, α₁α₃ = $(\sqrt{-2 +\sqrt{ 2 }})(\sqrt{-2 -\sqrt{ 2 }}) = i(\sqrt{2 -\sqrt{ 2 }})i(\sqrt{2 +\sqrt{ 2 }}) = -\sqrt{2} ⇒ α_3 = \frac{-\sqrt{2}}{α_1} ∈ K, α_4 = -α_3 ∈ K$, so K is the splitting field of the irreducible polynomial f over ℚ ⇒ [K : ℚ] = deg(f) = 4 = |Gal(f)|. Since [K : ℚ] = 4, the options with order 4 are Gal(K/ℚ) ≋ D₂ or C₄

Next, σ ∈ Gal(f), σ(α₁) = α₃ is an element of order 4 ⇒ [D₂ has all elements of order 2 except the identity, but C₄ has an element of order 4] Gal(K/Q) ≋ C₄.

Besides, the discriminant is Δ = 2048, so $\sqrt{Δ}∉\mathbb{Q}$, and the resolvent is g = x³ +32x² + 128x which is reducible over ℚ ⇒ Gal(K/ℚ) ≋ D₂ or C₄.

• f(x) = x⁴ -x -1 ∈ ℚ[x], f(x) = x⁴ -x -1 is irreducible modulo 2 (Mod p Irreducible Test, 0-0-1=-1≠0, 1-1-1=-1≠0) ⇒ f is irreducible in ℚ[x]. [Disc(x⁴+ cx +d) = 256d³ -27d⁴. Resolvent cubic(x⁴+ cx +d) = x³ -4dx -d²] Disc(f) = -283 is not a square in ℚ, g = x³ +4x -1 ∈ ℚ[x] is irreducible [By rational root test, any root must be ±1, but none of them are roots] ⇒ g is irreducible, Disc(f) is not a square ⇒ Gal(f) ≋ S₄.

• f(x) = x⁴ + x³ + 1. It is irreducible modulo 2 or over ℤ₂ (Mod p Irreducible Test) ⇒ f is irreducible over ℚ. Disc(f) = 229 is not a square in ℚ, g = x³ -3x² -61x -1 ∈ ℚ[x] is irreducible [By rational root test, any root must be ±1, but none of them are roots] ⇒ g is irreducible, Disc(f) is not a square ⇒ Gal(f) ≋ S₄.

• f(x) = x⁴ +8x + 12 ∈ ℚ[x] is irreducible (not easy). Disc(f) = 576, $\sqrt{576}=24∈ ℚ$. The resolvent is g = x³ -48x -64, g’= [g modulo 5] x³ -3x -4 = x³ + 2x +1 is irreducible modulo 5: 0+0+1=1≠0, 1+2+1=4≠0, 8+2+1=1≠0, 27+6+1=34=4≠0, 64+8+1=73=3≠0 ⇒ g is irreducible in ℚ and Disc(f) ∈ ℚ ⇒ Gal(f) ≋ A₄.

• f(x) = x⁴ +24x +36 ∈ ℚ[x] is irreducible modulo 17 or over ℤ₁₇ (Mod p Irreducible Test, tedious 😞). Disc(f) = 1728², $\sqrt{Δ}=1728 ∈ ℚ$. The resolvent g is g = x³ -2304x -36864 is irreducible modulo 3 or over ℤ₃ ⇒ g is irreducible in ℚ and Disc(f) ∈ ℚ ⇒ Gal(f) ≋ A₄.

• f(x) = x⁶ -2x⁵ -x⁴ +4x² +2x -4 = (x³-2)(x -1)(x +1)(x -2) so it has three roots in ℚ (namely 1, -1, and 2) and two non-real-roots ⇒ [Insolvability of quintics II Let f ∈ ℚ[x] be an irreducible polynomial of prime degree p with exactly one pair of non-real complex conjugate roots. Then, f is not solvable by radicals, G ≋ ℤ_p] Gal(f) = S₃

• f(x) = x⁵ -7 is irreducible over ℚ (Eisenstein criteria). This is based on an exercise on Fields and Galois Theory, John. M. Howie, Springer. The primitive root w = $e^{\frac{2πi}{5}}$ has minimum polynomial x⁴ +x³ + x² +x +1 ⇒ |ℚ(w) : ℚ| = 4.

Futhermore, x⁵ -7 is irreducible over ℚ(w). Suppose for the sake of contradiction, x⁵ -7 is reducible over ℚ(w) ⇒ [Abel’s Theorem. Let F be a field of characteristic 0, let p be a prime number, and a ∈ F. If x^p -a is reducible over F, then it has a linear factor x -c in F[x].] ∃b∈ ℚ(w): b = $\sqrt[5]{7}$. However, x⁵ -7 is irreducible over ℚ ⇒ [ℚ($\sqrt[5]{7})$: ℚ] ≥ 5, but [ℚ($\sqrt[5]{7})$: ℚ] ≤_{Since b ∈ ℚ(w)} [ℚ(w) : ℚ] = 4⊥

The roots of x⁵ -7 in ℂ are α, αw, αw², αw³, αw⁴, where α = $\sqrt[5]{7}$ and w = $e^{\frac{2πi}{5}}$. The Galois group consist of σ_{p, q}: α → αw^p, w → w^q, 0 ≤ p, q ≤ 4, σ_0,1 = id.

σ_p,q∘σ_r,s(α) = σ_p,q(αw^r) = σ_p,q(α)(σ_p,q(w))^r = αw^pw^qr = αw^p+qr. σ_p,q∘σ_r,s(w) = σ_p,q(w^s) = w^qs. Therefore, σ_p,q∘σ_r,s = σ_p+qr,qs where the addition and multiplication p+qr,qs is in mod 5 ⇒ σ_1,1² = σ_{2, 1}. More generally, σ_{1, 1}^p = σ_{p, 1}. Similarly, σ_{0, 2}² = σ_{0, 2²}, and in general, σ_{0, 2}^q = σ_{0, 2^q}, and σ_{p, 1}σ_{0, 2^q} = σ_{p, 2^q}.

The Galois group is generated by β = σ_{1, 1} and γ = σ_{0, 2} with presentation ⟨β, γ| β⁵ = γ⁴ = β²γβ^-1γ^-1 = id⟩

• β⁵ = (σ_1,1)⁵ = σ_{5, 1} = σ_0,1 = id.
• γ⁴ = (σ_0,2)⁴ = σ_{0, 2⁴} = σ_0,1 = id.
• γβ = σ_0,2σ_1,1 = [σ_p,q∘σ_r,s = σ_p+qr,qs] σ_{2, 2}. β²γ = σ_2,1σ_0,2 = σ_{2, 2}. Therefore, γβ = β²γ ⇒ β²γβ^-1γ^-1 = id.