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) = x^{4} + ax^{3} + bx^{2} + 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 -α_{1})(x -α_{2})(x -α_{3})(x -α_{4}) in F, that is, let α_{1}, α_{2}, α_{3}, and α_{4} 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 { α_{1}, α_{2}, α_{3}, α_{4}} transitively and Gal(f) is a transitive subgroup of S_{4}. S_{4} is isomorphic to the octahedral group, a rotation group. Any subgroup will be a rotation group, too. Therefore, the transitive subgroups of S_{4} are S_{4}, A_{4}, D_{4}, C_{4}, and D_{2}.

By the Orbit Stabilizer or Counting formula, |G| = |Orb(α_{1})|·|Stab(α_{1})| where Φ the mapping is defined Φ: G → Orb(α_{1}), g → g * x, where * denotes the group action.

Orb(x) (e.g. Orb(α_{1})) denotes the orbit of x, the set of elements in {α_{1}, α_{2}, α_{3}, α_{4}} to which x (e.g. α_{1}) can be moved by the elements of G, {g·x: g ∈ G} = [G acts on {α_{1}, α_{2}, α_{3}, α_{4}} transitively] {α_{1}, α_{2}, α_{3}, α_{4}}, |Orb(α_{1})| = 4.

|G| = 4·|Stab(α_{1})| ⇒ **4 divides |G|**

Let’s study the transitive subgroups of S_{4}:

**S**, |S_{4}_{4}| = 24**A**, |A_{4}_{4}| = 12.- There are three conjugate subgroups isomorphic to D
_{4}, the dihedral group of order 8, ⟨(1324), (12)⟩, ⟨(1234), (13)⟩, and ⟨(1243), (14)⟩. (1324) is odd = (13)(32)(24), so it is not in A_{4}. |D_{4}| = 8 - There are three conjugate subgroups isomorphic to C
_{4}, the cyclic group of order 4, ℤ/4ℤ, namely ⟨(1234)⟩, ⟨(1243)⟩, and ⟨(1324)⟩, e.g., ⟨(1234)⟩ is odd, so it is not in A_{4}. |C_{4}| = 4. **D**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. |D_{2}_{2}| = 4. It is a normal subgroup of S_{4}.

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 = A_{4} or D_{2}. If D is not a square in F, then G is S_{4}, D_{4}, or C_{4}.

Proof.

Let’s see the conditions:

- |G| needs to be
**divisible by 4** - G needs to be a
**transitive subgroup of S**._{4} - 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 A_{4}or D_{2}. Otherwise, the discriminant is not a square in F ↭ G = S_{4}, D_{4}, or C_{4}.

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

Definition. Let {α_{1}, α_{2}, α_{3}, α_{4}} 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 -β_{1})(x -β_{2})(x -β_{3}) ∈ K[x] where β_{1} = α_{1}α_{2} + α_{3}α_{4}, β_{2} = α_{1}α_{3} + α_{2}α_{4}, β_{3} = α_{1}α_{4} + α_{2}α_{3}, β_{i}∈ K.

Claim: **g(x) ∈ F[x]**

Let σ be an arbitrary automorphism of G, α ∈ G, then σg = (x -σβ_{1})(x -σβ_{2})(x -σβ_{3}) = g ↭ σ permutes {β_{1}, β_{2}, β_{3}}

And this is the case, e.g., σβ_{1} = [β_{1} = α_{1}α_{2} + α_{3}α_{4}] σ(α_{1})σ(α_{2}) + σ(α_{3})σ(α_{4}) = [It follows from G acting transitively on {α_{1}, α_{2}, α_{3}, α_{4}}] 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

Claim. Our irreducible polynomial f and the resolvent have the same discriminant, **Disc(f) = Disc(g). Besides, β _{1}, β_{2}, and β_{3} are all distinct.**

Proof.

β_{1} - β_{2} = (α_{1}α_{2} + α_{3}α_{4}) - (α_{1}α_{3} + α_{2}α_{4}) = α_{1}(α_{2} -α_{3}) -α_{4}(α_{2} -α_{3}) = (α_{1} -α_{4})·(α_{2} -α_{3}) ⇒ [By assumption, α_{i} are all distinct, and therefore α_{1} -α_{4} ≠ 0, α_{2} -α_{3} ≠ 0] β_{1} - β_{2} ≠ 0 ⇒ β_{1} ≠ β_{2}.

β_{1} - β_{3} = (α_{1}α_{2} + α_{3}α_{4}) - (α_{1}α_{4} + α_{2}α_{3}) = α_{1}(α_{2} -α_{4}) -α_{3}(α_{2} -α_{4}) = (α_{1} -α_{3})·(α_{2} -α_{4}) ≠ 0 ⇒ β_{1} ≠ β_{3}.

Analogously, β_{2} - β_{3} = (α_{1} -α_{2})·(α_{3} -α_{4}) ≠ 0 ⇒ β_{2} ≠ β_{3}.

Futhermore, Disc(g) = (β_{1} - β_{2})(β_{1} - β_{3})(β_{2} - β_{3}) = $\prod_{i < j} (α_i-α_j)^2$ = 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 = D_{2} |
G = D_{4} or G = C_{4} |

g is irreducible over F | G = A_{4} |
G = S_{4} |

Proof.

Let B = {β_{1}, β_{2}, β_{3}} ⊆ K, |B| = 3

S_{4} permutes {α_{1}, α_{2}, α_{3}, α_{4}}, S_{3} permutes {β_{1}, β_{2}, β_{3}}, so we can consider the mapping Φ: S_{4} → S_{3}. In other words, the operation on the symmetric group S_{4} on the roots α_{i} defines a transitive operation on B (β_{j}), and the associated permutation representation is a homomorphism from S_{4} to S_{3}.

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

Ker(Φ) = {σ ∈ S_{4}| σ(β_{i}) = β_{i} ∀i -it fixes β_{i} or σ acts trivially on β_{i}-} = [β_{1} = α_{1}α_{2} + α_{3}α_{4}, β_{2} = α_{1}α_{3} + α_{2}α_{4}, β_{3} = α_{1}α_{4} + α_{2}α_{3}] {(), (12)(34), (13)(24), (14)(23)} = D_{2}

(12)(34): β_{1} = α_{1}α_{2} + α_{3}α_{4} → α_{2}α_{1} + α_{4}α_{3} = β_{1}; (12)(34): α_{1}α_{3} + α_{2}α_{4} → α_{2}α_{4} + α_{1}α_{3}; (12)(34): α_{1}α_{4} + α_{2}α_{3} → α_{2}α_{3} + α_{1}α_{4}; (13)(24): β_{1} = α_{1}α_{2} + α_{3}α_{4} → α_{3}α_{4} + α_{1}α_{2} ···

Φ: S_{4} → S_{3}. Let’s consider the restriction of Φ to G. By abuse of our notation, Φ: G → S_{3}.

Let’s assume **g splits in F** ↭ [g is a product of linear polynomials, g(x) = (x -β_{1})(x -β_{2})(x -β_{3})] so β_{1}, β_{2}, β_{3} ∈ F ↭ [σ is an automorphism of K that fixes F] ∀σ ∈ G: σ(β_{i}) = β_{i} ↭ G ⊆ Ker(Φ) = D_{2} ↭ **G = D _{2}**.

Let’s assume g = (x-β_{1})(x-β_{2})(x-β_{3}) is irreducible in F[x] ⇒ [Using the same argument as before G acts transitively on B, so |G|=|Orb(β_{1})||Stab(β_{1})| = 3|Stab(β_{1})|] and therefore 3 divides |G| ⇒ [Possibilities are: |S^{4}| = 24, |A^{4}| = 12, |D_{4}| = 8⊥, |C_{4}| = 4⊥, |D_{2}| = 4⊥] ⇒ **G = S _{4} or A_{4}** ⇒ [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 = A

Suppose 3 divides |G| and therefore, G = S_{4} or A_{4}. 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: g^{n}=e_{G} ⇒ e_{H}=Φ(g^{n})=Φ(g)^{n}] Φ: G → S_{3} = {(), (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) = D_{2} = {(), (12)(34), (13)(24), (14)(23)}. However, |D_{2}| = 4, but |σ| = 3 ɫ 4 = |D_{2}| ⊥

Φ(σ) has order 3 [S_{3} = {() -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)=D_{2}), it permutes the three elements cyclically, and therefore G acts or operates transitively on B ⇒ g = (x-β_{1})(x-β_{2})(x-β_{3}) is irreducible. So we have demonstrated that g = (x-β_{1})(x-β_{2})(x-β_{3}) is irreducible ↭ G = S_{4} or G = A_{4} (3 divides |G|)∎

D square ∈ F | D square ∉ F | |
---|---|---|

g is reducible over F | D_{2} |
G = D_{4} or G = C_{4} |

g is irreducible over F | G = A_{4} |
G = S_{4} |

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).

- If g splits completely in F ↭ G = [G ⊆ Ker(Φ) = D
_{2}⇒] G = D_{2} - If g has exactly one root in F [If g has two roots in G, then it would split completely in F] ↭ G = D
_{4}or C_{4}. - g is irreducible if and only if G = S
_{4}or A_{4}.

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) = D_{2}. So we have a tower of fields, F ⊆ F(δ) ⊆ L ⊆ K ⇒ [Gal(K/L) = D_{2}]. Hence f is solvable over F.

Note: D_{2} 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. D_{2} is solvable follows from the sequence of subgroups {id} ≤_{2} H ≤_{2} D_{4} where H is the subgroup of symmetries of a square that are orientation preserving, i.e., the rotation and the identity.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory.

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