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 S_{2}, Gal(f) ≤ S_{2}. Therefore, it is ℤ/2ℤ when it is irreducible, e.g., x^{2}+ 1, Gal(f) ≋ ℤ/2ℤ; or the trivial subgroup, {id} when the polynomial is not irreducible, e.g., x^{2}-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 = x^{3} +a_{2}x^{2} + a_{1}x + a_{0}. Disc(f) = (α_{1} -α_{2})^{2}(α_{1} -α_{3})^{2}(α_{2} -α_{3})^{2} = -4a_{2}^{3}a_{0} + a_{2}^{2}a_{1}^{2} + 18a_{2}a_{1}a_{0} -4a_{1}^{3} -27a_{0}^{2}.

In particular, Disc(x^{3} +ax + b) = -4a^{3} -27b^{2}.

- x
^{3}-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)=S_{3} - Let f = x
^{3}-3 is irreducible over ℚ by Eisenstein’s criterion (p = 3). It has a discriminant equals to -9^{2}·3, so it is not a square in ℚ, thus Gal(f) = S_{3}. - x
^{3}-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 = 9^{2}(9 ∈ ℚ) ⇒ Gal(f) = A_{3}. - x
^{3}-x^{2}-2x +1 ∈ ℚ[x] is irreducible. It has a discriminant equals to 49 = 7^{2}(9 ∈ ℚ) ⇒ Gal(f) = A_{3}. - x
^{4}+ x^{2}-4x +1 ∈ ℚ[x] is irreducible over ℚ. It has a discriminant equals to 169 = 13^{2}(9 ∈ ℚ) ⇒ Gal(f) = A_{3}. - Let f = x
^{3}+2x^{2}-3x is**reducible**over ℚ because f = x(x^{2}+2x -3), so x = 0 is a solution. Its discriminant is 12^{2}and $\sqrt{Δ}=12∈ \mathbb{Q}$ ⇒ Gal(f) = {id}. - x
^{3}-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. - x
^{3}+5x +5 is**irreducible**by Eisenstein’s criterion (p = 5). Its discriminant is Δ = -4·5^{3}-27·5^{2}= 5^{2}·-47 = -1175, so $\sqrt{-1175} ∉ ℚ$ ⇒**Gal(f) = S**._{3}

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) = 3x^{2} +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) = S_{4} or D_{4}.
Proof.

Let K be the splitting field of f, α ∈ ℝ, a root of f, ℚ(α) ≠ K, K ⊄ ℝ (there are four roots, α_{1}; α_{2} and two complex roots, say z = a + ib, $\bar z = a -ib, b ≠ 0$)

[K : ℚ] ≥ 8 ⇒ |Gal(f)| ≥ 8 ⇒ G = S_{4}, D_{4} or A_{4}.

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) ⊄ A _{4} ⇒ Gal(f) = S_{4} or D_{4}** ∎

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.} F_{1} ⊆_{s.r.} ··· ⊆_{s.r.} F_{r}, α ∈ F_{r} ⇒ [Let L/F be a radical extension. Then, there exist a K/L extension such that K/F is both Galois and radical.] F ⊆ F_{r} ⊆ 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. ∎