# 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 Sn. ] 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 ⊆ An ↭ 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, βi∈ 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}] βi for some i

∀σ ∈ G, σg = g ⇒ [Every permutation of the roots αi permutes the element βj, 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 ∈ KG[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) = α123) -α423) = (α14)·(α23) ⇒ [By assumption, αi are all distinct, and therefore α14 ≠ 0, α23 ≠ 0] β1 - β2 ≠ 0 ⇒ β1 ≠ β2.

β1 - β3 = (α1α2 + α3α4) - (α1α4 + α2α3) = α124) -α324) = (α13)·(α24) ≠ 0 ⇒ β1 ≠ β3.

Analogously, β2 - β3 = (α12)·(α34) ≠ 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:

D square ∈ F D square ∉ F
g is reducible over F G = D2 G = D4 or G = C4
g is irreducible over F G = A4 G = S4

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 αi defines a transitive operation on B (βj), 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 Πi is one way of partitioning a set of 4 elements into two subsets of equal cardinality. S4 acts on a Πi 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(βi) = βi ∀i -it fixes βi or σ acts trivially on βi-} = [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(βi) = βi ↭ G ⊆ Ker(Φ) = D2G = 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=eG ⇒ eH=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|)∎

D square ∈ F D square ∉ F
g is reducible over F D2 G = D4 or G = C4
g is irreducible over F G = A4 G = S4

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.