# Computing Galois groups. Degree 2 and 3.

To raise new questions, new possibilities, to regard old problems from a new angle, requires creative imagination and marks real advance in science, Albert Einstein.

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

• Degree 2. The Galois group is a subgroup of S2, Gal(f) ≤ S2. Therefore, it is ℤ/2ℤ when it is irreducible, e.g., x2 + 1, Gal(f) ≋ ℤ/2ℤ; or the trivial subgroup, {id} when the polynomial is not irreducible, e.g., x2-1, Gal(f) ≋ {id}.
• Degree 3.

Proposition. Let f ∈ ℚ[x] be a polynomial of degree 3. Let δ = $\sqrt{Δ}$ be the square root of the discriminant of f. Then, the Galois group of f can be found using the following classification,

$Gal(f) = \begin{cases} \mathbb{ℤ/2ℤ}, \sqrt{Disc(f)} ∉ ℚ, f~ reducible \\ {id}, \sqrt{Disc(f)} ∈ ℚ, f~ reducible \\ \mathbb{A_3} ≋ \mathbb{ℤ/3ℤ}, \sqrt{Disc(f)} ∈ ℚ, f~ irreducible \\ \mathbb{S_3}, \sqrt{Disc(f)} ∉ ℚ, f~ irreducible \end{cases}$

f = x3 +a2x2 + a1x + a0. Disc(f) = (α12)213)223)2 = -4a23a0 + a22a12 + 18a2a1a0 -4a13 -27a02.

In particular, Disc(x3 +ax + b) = -4a3 -27b2.

1. x3-x-1 ∈ ℚ[x] is irreducible by Rational root test because any rational root r/s must have r = ±1, s = ±1, but these values are not roots. It has a discriminant equals to -23 (it is not a square in ℚ), Gal(f)=S3
2. Let f = x3 -3 is irreducible over ℚ by Eisenstein’s criterion (p = 3). It has a discriminant equals to -92·3, so it is not a square in ℚ, thus Gal(f) = S3.
3. x3-3x+1 ∈ ℚ[x] is irreducible by Rational root test because any rational root r/s must have r = ±1, s = ±1, but these values are not roots. It has a discriminant equals to 81 = 92 (9 ∈ ℚ) ⇒ Gal(f) = A3.
4. x3 -x2 -2x +1 ∈ ℚ[x] is irreducible. It has a discriminant equals to 49 = 72 (9 ∈ ℚ) ⇒ Gal(f) = A3.
5. x4 + x2 -4x +1 ∈ ℚ[x] is irreducible over ℚ. It has a discriminant equals to 169 = 132 (9 ∈ ℚ) ⇒ Gal(f) = A3.
6. Let f = x3 +2x2 -3x is reducible over ℚ because f = x(x2 +2x -3), so x = 0 is a solution. Its discriminant is 122 and $\sqrt{Δ}=12∈ \mathbb{Q}$ ⇒ Gal(f) = {id}.
7. x3 -2x -1 is reducible over ℚ because x = -1 is a solution. Its discriminant is Δ = 3, so $\sqrt{3} ∉ ℚ$ ⇒ Gal(f) ≋ ℤ/2ℤ.
8. x3 +5x +5 is irreducible by Eisenstein’s criterion (p = 5). Its discriminant is Δ = -4·53 -27·52 = 52·-47 = -1175, so $\sqrt{-1175} ∉ ℚ$ ⇒ Gal(f) = S3.

Futhermore, we can see that $\lim_{x \to \infin}f(x)=\infin, \lim_{x \to -\infin}f(x)=-\infin$, so by the Intermediate Value Theorem has at least one real root. f’(x) = 3x2 +5 > 0 ⇒ f is always increasing ⇒ f has exactly one real root and two conjugate complex roots, say α ∈ ℝ, β ∈ ℂ, $\bar β$. Then, its splitting field is K = ℚ(α, β). Let’s explore its Galois group. σ ∈ Gal(K/ℚ), σ = (α β $\bar β$), σ-1 = (α $\bar β$ β), |σ| = 3. Let’s consider the conjugate automorphism, too, say τ: β → $\bar β$, α → α, |τ| = 2.

Proposition. Let f ∈ ℚ[x] be an irreducible quadratic which has exactly two real roots. Then, Gal(f) = S4 or D4. Proof.

Let K be the splitting field of f, α ∈ ℝ, a root of f, ℚ(α) ≠ K, K ⊄ ℝ (there are four roots, α1; α2 and two complex roots, say z = a + ib, $\bar z = a -ib, b ≠ 0$)

[K : ℚ] ≥ 8 ⇒ |Gal(f)| ≥ 8 ⇒ G = S4, D4 or A4.

D = Disc(f) = $\prod_{i < j} (α_i-α_j)^2 = (α_1-α_2)^2(α_1-a-ib)^2(α_1-a+ib)^2(α_2-a-ib)^2(α_2-a+ib)^2(2ib)^2$ ⇒ [$(α_1-α_2)^2 > 0, (α_1-a-ib)^2(α_1-a+ib)^2 > 0, (α_2-a-ib)^2(α_2-a+ib)^2 > 0, (2ib)^2 < 0$] D = Disc(f) < 0 ⇒ Disc(f) is not a square of ℚ ⇒ Gal(f) ⊄ A4 ⇒ Gal(f) = S4 or D4

Proposition. Let F be a subfield of the complex numbers, F ⊆ ℂ Let f ∈ F[x] be an irreducible polynomial. If one root of f is solvable over F, then f is solvable, i.e., all roots of f are solvable.

Proof.

f is solvable ↭ all roots of f are solvable. Let α ∈ ℂ be a solvable root of f.

F ⊆s.r. F1s.r. ··· ⊆s.r. Fr, α ∈ Fr ⇒ [Let L/F be a radical extension. Then, there exist a K/L extension such that K/F is both Galois and radical.] F ⊆ Fr ⊆ L that is both radical and Galois.

L/F Galois ⇒ L/F is normal. Since f is irreducible and has a root α in L ⇒ [L/F is normal] f splits completely in L ⇒ All roots of f are solvable because they “live” in a radical extension, so they are all solvable. ∎

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