Quartics are solvable

Since the mathematicians have invaded the theory of relativity, I do not understand it myself any more, Albert Einstein […] How do you get a solution? Pray.

Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true, Bertrand Russell.

Counting Formula. Let S be a finite set on which a group G operates and let Stab(α) and Orb(α) be the stabilizer and orbit of an element α of S. Then, |G| = |Orb(α)|·|Stab(α)|.

Theorem. Let F ⊆ ℂ be a field with char(F) ≠ 2, and let f∈ F[x] be an irreducible, separable polynomial of degree 4. Let K be the splitting field of f(x) over F (or K be the splitting field of an irreducible polynomial over ℚ) Then, f is solvable over F.  

We are considering polynomials of degree 4, f(x) = x⁴ + ax³ + bx² + cx +d. Besides, we are assuming that f is irreducible. Otherwise, it is the product of polynomials of lesser degrees that have already been considered and shown to be solvable.

Let f factors as f(x) = (x -α₁)(x -α₂)(x -α₃)(x -α₄) in F, that is, let α₁, α₂, α₃, and α₄ be the distinct roots of f in K (Notice that we are in a characteristic zero field, namely ℚ or f is separable by assumption).

f is irreducible ⇒ [Recall, Let F ⊆ ℂ, f ∈ ℚ[x] be a polynomial of degree n, and let K be the splitting field for f over ℚ. Then, f is irreducible if and only if Gal(f) is a transitive subgroup of S_n. ] G acts on { α₁, α₂, α₃, α₄} transitively and Gal(f) is a transitive subgroup of S₄. S₄ is isomorphic to the octahedral group, a rotation group. Any subgroup will be a rotation group, too. Therefore, the transitive subgroups of S₄ are S₄, A₄, D₄, C₄, and D₂.

By the Orbit Stabilizer or Counting formula, |G| = |Orb(α₁)|·|Stab(α₁)| where Φ the mapping is defined Φ: G → Orb(α₁), g → g * x, where * denotes the group action.

Orb(x) (e.g. Orb(α₁)) denotes the orbit of x, the set of elements in {α₁, α₂, α₃, α₄} to which x (e.g. α₁) can be moved by the elements of G, {g·x: g ∈ G} = [G acts on {α₁, α₂, α₃, α₄} transitively] {α₁, α₂, α₃, α₄}, |Orb(α₁)| = 4.

|G| = 4·|Stab(α₁)| ⇒ 4 divides |G|

Let’s study the transitive subgroups of S₄:

1. S₄, |S₄| = 24
2. A₄, |A₄| = 12.
3. There are three conjugate subgroups isomorphic to D₄, the dihedral group of order 8, ⟨(1324), (12)⟩, ⟨(1234), (13)⟩, and ⟨(1243), (14)⟩. (1324) is odd = (13)(32)(24), so it is not in A₄. |D₄| = 8
4. There are three conjugate subgroups isomorphic to C₄, the cyclic group of order 4, ℤ/4ℤ, namely ⟨(1234)⟩, ⟨(1243)⟩, and ⟨(1324)⟩, e.g., ⟨(1234)⟩ is odd, so it is not in A₄. |C₄| = 4.
5. D₂ is the Klein four group, it consists of the identity and three products of disjoint transpositions = {(), (12)(34), (13)(24), (14)(23)}. All these cycles are even. |D₂| = 4. It is a normal subgroup of S₄.

Proposition. Let G be the Galois group of an irreducible, separable quartic polynomial f. Let D = Disc(f). Then, D is a square in F if and only if G = A₄ or D₂. If D is not a square in F, then G is S₄, D₄, or C₄.

Proof.

Let’s see the conditions:

1. |G| needs to be divisible by 4
2. G needs to be a transitive subgroup of S₄.
3. Finally, the discriminant D of f is a square in F (every element of G fixes δ) ↭ if G contains no odd permutation ↭ G ⊆ A_n ↭ G is A₄ or D₂. Otherwise, the discriminant is not a square in F ↭ G = S₄, D₄, or C₄.

To determine which case occurs, further analysis is required. The key idea is to work with an associated cubic polynomial.

Definition. Let {α₁, α₂, α₃, α₄} be the roots of f, deg(f) = 4, and K be the splitting field of f. Resolvent cubic is a cubic polynomial defined from a monic polynomial of degree four, g(x) = (x -β₁)(x -β₂)(x -β₃) ∈ K[x] where β₁ = α₁α₂ + α₃α₄, β₂ = α₁α₃ + α₂α₄, β₃ = α₁α₄ + α₂α₃, β_i∈ K.

Claim: g(x) ∈ F[x]

Let σ be an arbitrary automorphism of G, α ∈ G, then σg = (x -σβ₁)(x -σβ₂)(x -σβ₃) = g ↭ σ permutes {β₁, β₂, β₃}

And this is the case, e.g., σβ₁ = [β₁ = α₁α₂ + α₃α₄] σ(α₁)σ(α₂) + σ(α₃)σ(α₄) = [It follows from G acting transitively on {α₁, α₂, α₃, α₄}] β_i for some i

∀σ ∈ G, σg = g ⇒ [Every permutation of the roots α_i permutes the element β_j, so the coefficients of g are symmetric functions in the roots -By symmetric functions, we mean that the function remains unchanged when the roots are interchanged-] g ∈ K^G[x] = [K/F is Galois] F[X]∎

Claim. Our irreducible polynomial f and the resolvent have the same discriminant, Disc(f) = Disc(g). Besides, β₁, β₂, and β₃ are all distinct.

Proof.

β₁ - β₂ = (α₁α₂ + α₃α₄) - (α₁α₃ + α₂α₄) = α₁(α₂ -α₃) -α₄(α₂ -α₃) = (α₁ -α₄)·(α₂ -α₃) ⇒ [By assumption, α_i are all distinct, and therefore α₁ -α₄ ≠ 0, α₂ -α₃ ≠ 0] β₁ - β₂ ≠ 0 ⇒ β₁ ≠ β₂.

β₁ - β₃ = (α₁α₂ + α₃α₄) - (α₁α₄ + α₂α₃) = α₁(α₂ -α₄) -α₃(α₂ -α₄) = (α₁ -α₃)·(α₂ -α₄) ≠ 0 ⇒ β₁ ≠ β₃.

Analogously, β₂ - β₃ = (α₁ -α₂)·(α₃ -α₄) ≠ 0 ⇒ β₂ ≠ β₃.

Futhermore, Disc(g) = (β₁ - β₂)(β₁ - β₃)(β₂ - β₃) = $\prod_{i < j} (α_i-α_j)^2$ = Disc(f)∎

Theorem. Let F be a field, F ⊆ ℂ, G be the Galois group of an irreducible, separable quartic polynomial f, G = Gal(K/F) = Gal(f). Let D = Disc(f). Then, we have the following exclusive options:

Proof.

Let B = {β₁, β₂, β₃} ⊆ K, |B| = 3

S₄ permutes {α₁, α₂, α₃, α₄}, S₃ permutes {β₁, β₂, β₃}, so we can consider the mapping Φ: S₄ → S₃. In other words, the operation on the symmetric group S₄ on the roots α_i defines a transitive operation on B (β_j), and the associated permutation representation is a homomorphism from S₄ to S₃.

We get an homomorphism Φ from S₄ to S₃ by considering the action of S₄ on a finite set Y = {Π₁, Π₂, Π₃} where Π₁ = {{1,2}, {3,4}}, Π₂ = {{1,3}, {2,4}} and Π₃ = {{1,4}, {2,3}}. Each Π_i is one way of partitioning a set of 4 elements into two subsets of equal cardinality. S₄ acts on a Π_i by permuting the numbers in each partition, e.g., (1234) interchanges Π₁ and Π₃ while keeping Π₂ fixed, so that Φ((1234)) = (13)(2).

Ker(Φ) = {σ ∈ S₄| σ(β_i) = β_i ∀i -it fixes β_i or σ acts trivially on β_i-} = [β₁ = α₁α₂ + α₃α₄, β₂ = α₁α₃ + α₂α₄, β₃ = α₁α₄ + α₂α₃] {(), (12)(34), (13)(24), (14)(23)} = D₂

(12)(34): β₁ = α₁α₂ + α₃α₄ → α₂α₁ + α₄α₃ = β₁; (12)(34): α₁α₃ + α₂α₄ → α₂α₄ + α₁α₃; (12)(34): α₁α₄ + α₂α₃ → α₂α₃ + α₁α₄; (13)(24): β₁ = α₁α₂ + α₃α₄ → α₃α₄ + α₁α₂ ···

Φ: S₄ → S₃. Let’s consider the restriction of Φ to G. By abuse of our notation, Φ: G → S₃.

Let’s assume g splits in F ↭ [g is a product of linear polynomials, g(x) = (x -β₁)(x -β₂)(x -β₃)] so β₁, β₂, β₃ ∈ F ↭ [σ is an automorphism of K that fixes F] ∀σ ∈ G: σ(β_i) = β_i ↭ G ⊆ Ker(Φ) = D₂ ↭ G = D₂.

Let’s assume g = (x-β₁)(x-β₂)(x-β₃) is irreducible in F[x] ⇒ [Using the same argument as before G acts transitively on B, so |G|=|Orb(β₁)||Stab(β₁)| = 3|Stab(β₁)|] and therefore 3 divides |G| ⇒ [Possibilities are: |S⁴| = 24, |A⁴| = 12, |D₄| = 8⊥, |C₄| = 4⊥, |D₂| = 4⊥] ⇒ G = S₄ or A₄ ⇒ [Let G be the Galois group of an irreducible, separable quartic polynomial f. Let D = Disc(f). Then, D is a square in F if and only if G = A₄ or D₂. If D is not a square in F, then G is S₄, D₄, or C₄.] G = S₄ (D is not a square in F) or D = A₄ (D is a square in F).

Suppose 3 divides |G| and therefore, G = S₄ or A₄. Besides, 3 is a prime ⇒ [Cauchy’s theorem: Let G be a finite group and p is a prime number dividing the order of G ⇒ G contains an element of order p] then G contains an element of order 3, say σ ⇒ [Let ϕ: G→ H be a group homomorphism, and let g be an element of G. Then, the order of ϕ(g) divides the order of g: gⁿ=e_G ⇒ e_H=Φ(gⁿ)=Φ(g)ⁿ] Φ: G → S₃ = {(), (12), (13), (23), (123), (132)} ⇒ Φ(σ) has order 1 (this is not possible*) or 3.

(*) Φ(σ) cannot have order 1. Suppose for the sake of contradiction, Φ(σ) = id ⇒ σ ∈ Ker(Φ) = D₂ = {(), (12)(34), (13)(24), (14)(23)}. However, |D₂| = 4, but |σ| = 3 ɫ 4 = |D₂| ⊥

Φ(σ) has order 3 [S₃ = {() -identity-, (12), (13), (23) -three transpositions with order 2-, (123) and (132) order 3}] Φ(σ) is a 3-cycle, i.e., permutations that move all elements and have order three, σ does not operate trivially on B (Ker(Φ)=D₂), it permutes the three elements cyclically, and therefore G acts or operates transitively on B ⇒ g = (x-β₁)(x-β₂)(x-β₃) is irreducible. So we have demonstrated that g = (x-β₁)(x-β₂)(x-β₃) is irreducible ↭ G = S₄ or G = A₄ (3 divides |G|)∎

Check our exercises for examples

Corollary. Let F be a field, F ⊆ ℂ, G be the Galois group of an irreducible, separable quartic polynomial f, G = Gal(K/F) = Gal(f).

1. If g splits completely in F ↭ G = [G ⊆ Ker(Φ) = D₂ ⇒] G = D₂
2. If g has exactly one root in F [If g has two roots in G, then it would split completely in F] ↭ G = D₄ or C₄.
3. g is irreducible if and only if G = S₄ or A₄.

Theorem. Let F be a field, F ⊆ ℂ, G be the Galois group of an irreducible, separable quartic polynomial f, G = Gal(K/F) = Gal(f). Then, f is solvable over F.

Proof.

Let δ = $\sqrt{D}$. The extension F ⊆ F(δ) is degree 1 (δ ∈ F) or 2 (δ ∉ F), i.e., solvable.

Consider f over F(δ), then discriminant of f, say D, is square now in F(δ). Let L be the splitting field of g over F(δ), F ⊆ F(δ) ⊆ L, D = Disc(f) = Disc(g), deg(g) = 3 ⇒ F(δ) ⊆ L is solvable (Recall, cubic polynomials are solvable, too).

Now consider f over L, g splits completely over L ↭ [Previous corollary] Gal(K/L) = D₂. So we have a tower of fields, F ⊆ F(δ) ⊆ L ⊆ K ⇒ [Gal(K/L) = D₂]. Hence f is solvable over F.

Note: D₂ is the symmetry group of the rectangle, the various symmetry mappings are: the identity mapping id, the rotation r (in either direction) of 180°, and the reflections h and v in the vertical and horizontal axes. D₂ is solvable follows from the sequence of subgroups {id} ≤₂ H ≤₂ D₄ where H is the subgroup of symmetries of a square that are orientation preserving, i.e., the rotation and the identity.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.