If you start here, you get into a mathematical Vietnam. You can’t get out without defeat […] Good question…but I’m deflecting it. […] We will prove this by the method of prolonged staring, A reddit comment by @rhombomere.

The discriminant of a polynomial is a generalization to arbitrary degree polynomials of the discriminant of a quadratic. If K = F(α) is a Galois extension of a field F and f is the irreducible polynomial of α, then the Galois group Gal(K/F) can be viewed as a subgroup of the group of permutations of the roots of f (S_{n}). The discriminant determines when this subgroup consists solely of even permutations (A_{n}), and **we could use this information to describe the splitting field of a polynomial of degree less or equal than four**.

Definition. Let F be a field, F ⊆ ℂ, f ∈ F[x], deg(f) = n ≥ 1. If there is an extension field K/F such that f factors as a product of linear polynomials (x - α_{1})···(x - α_{n}) over the field K, i.e., K is the splitting field of f over F then Gal(K/F) is called the Galois group of the polynomial f over F.

🔑 Since two splitting fields of f over F are isomorphic by an isomorphism which is the identity on F, the group Gal(K/F) is independent of the choice of K.

Proposition. Let f be an irreducible polynomial of degree n over the field, and let G be the Galois group of f. The Galois group of f is a subgroup of S_{n.}

Proof.

Let G be the Galois group of f. f has n roots, say α_{1},…, α_{n}. If σ ∈ G, σ permutes α_{1},…, α_{n}, so there is a map Φ: G → S_{n}, σ → σ (looking at it as a permutation in S_{n} or S_{{α1,…, αn}}). It is easily seen to be a homomorphism of groups, thus an embedding of the Galois group G.

What is Ker(Φ)? The identity in S_{n} is (1)(2)···(n), that is, σ does not permutes the roots of f ⇒ σ(α_{i}) = α_{i} ∀i, σ(a) = a ∀a ∈ F (because σ ∈ Gal(K/F)) ⇒ σ is the identity on K = [By definition, K is the splitting field of α_{1},…, α_{n}, K = F(α_{1},…, α_{n})] so this is an **injective group homomorphism** ⇒ G is isomorphic to a subgroup of S_{n}∎

Definition. Let S_{n} be the symmetric group on n letters (|S_{n}| = n!). A subgroup G of S_{n} is called transitive if for each i, j ∈ {1, 2, ···, n}, there is a permutation in G (σ ∈ G) sending i to j, σ(i) = j.

Proposition. Let F ⊆ ℂ, let 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}. In other words, for any α, β roots of f, there is a permutation in G sending α to β, σ(α) = β or **the orbit of a root α of f is all the roots of the irreducible polynomial of α**.

Proof.

Gal(K/F) maps roots of f_{i} to other roots, let f(x) = a_{n}x^{n} +··· + a_{1}x + a_{0}, a_{i} ∈ F. Let α be a root of f, f(α) = 0 ⇒ [σ is a F-automorphism] 0 = σ(f(α)) = a_{n}(σ(x))^{n} + ··· + a_{1}σ(x) + a_{0}, which tell that σ(α) is also a root of f ⇒ Gal(K/F) maps roots of f to other roots of f.

⇒) Let α, β ∈ K be two arbitrary roots of f, f irreducible. K is the splitting field of f ⇒ Given any root, F(α)≋F[x]/⟨f(x)⟩≋ F(β), so there exist an isomorphism from F(α) → F(β), α → β.

Recall. Let F be a field and let p(x) ∈ F[x] be an irreducible polynomial over F, deg(p(x)) = n. If a is a root of p(x) in some extension E of F, then F(a) is isomorphic to F[x]/⟨p(x)⟩, F(a)≈F[x]/⟨p(x)⟩.

This isomorphism can be extended to K, σ: K → K, so σ ∈ Gal(K/F) = Gal(f), thus creating an automorphism of K that fixes F. Besides, we see that Gal(K/F) acts transitively on the roots of f because **for every two distinct roots, there exists an automorphism σ ∈ Gal(K/F), that sends one to the other, i.e., α → β**.

Being very formal, the previous isomorphism can be extended to $\bar F$ (algebraic closure of F), $\bar σ: \bar F → \bar F.$ However, we can restrict $\bar σ$ to K since (K/ℚ is Galois -splitting field and F = ℚ- and then normal) ⇒ $\bar σ(K) ⊆ K$. Thus, creating an automorphism of K that fixes F and maps α → β, and this is by definition and element of Gal(K/F).

⇐) Suppose Gal(f) is a transitive subgroup of S_{n}, and for the sake of contradiction, f is reducible. Then, f can be written as product of distinct irreducibles f = f_{1}···f_{r}, where each f_{i} is irreducible with degree at least 1. Let us take any of the f_{i}’s.

We notice that as before Gal(K/F) maps roots of f_{i} to other roots of f_{i}. However, the action cannot be transitive because each root α can only be mapped to another root of f_{i} where f_{i} is the polynomial that has α as a root. ⊥ Observe that if f_{i} and f_{j} has α shared root, then the irreducible polynomial of α must divide both, f_{i} and f_{j}, but f_{i} and f_{j} are irreducible themselves ⊥.

f is irreducible ↭ Gal(f) is a transitive subgroup of S_{n} is **not true if f is not irreducible**. Counterexample: f = (x-3)(x^{2}+1) ∈ ℚ[x]. f has three roots (3, i, -i), but Gal(f) = S_{2} = {id, (23)} < S_{3} and is not transitive. In particular, G has not an automorphism σ that sends 3 → i. On the contrary, f = x^{3} -2 is irreducible ∈ ℚ[x], f has three roots $\sqrt[3]{2},\sqrt[3]{2}w, \sqrt[3]{2}w^2$ then G is a transitive subgroup of S_{3}. Futhermore, G = S_{3}.

Definition. Let F be a field with char(F) ≠ 2. Let f(x) ∈ F[x], deg(f) = n. Let α_{1}, α_{2}, ···, α_{n} be the roots of f in some splitting field K of f over f. The discriminant of a polynomial is the product of the squares of the differences of the polynomial roots, Δ = D = Disc(f) = $\prod_{i < j} (α_i-α_j)^2$.

Notice that the discriminant is dependent on the base field F. Besides, the discriminant is zero if and only if the polynomial f has a repeated root. Thus, the discriminant will give us information only when f has no repeated roots. The discriminant D clearly is an element of K.

Proposition. Let F be a field with char(F)≠2, let f(x)∈ F[x] be an irreducible, separable polynomial, and let K be the splitting field of f(x) over F. Then, the discriminant D is in the base field, D ∈ F.

Proof.

Let σ ∈ G = Gal(K/F). Since σ permutes the root of the polynomial α_{1}, α_{2}, ···, α_{n}, σ(D) = D.

Example, say n = 3, σ: α_{1} → α_{1}, α_{2} → α_{3}, α_{3} →α_{2}, D = (α_{1}-α_{2})^{2}(α_{1}-α_{3})^{2}(α_{2}-α_{3})^{2}.

σ(D) = (α_{1}-α_{3})^{2}(α_{1}-α_{2})^{2}(α_{3}-α_{2})^{2}. σ(D) = D because we are using the product of the **squares of the differences of the polynomial roots**, e.g., (α_{2}-α_{3})^{2} = (α_{3}-α_{2})^{2}.

Since σ is a permutation of the roots of f, σ(α_{i}-α_{j})^{2} = (α_{σ(i)} - α_{σ(j)})^{2}, so we have two products D and σ(D) with the same terms, except it may differ (a_{r}-a_{s})^{2} instead of (a_{s}-a_{r})^{2} but the final product is obviously unaffected.

∀σ ∈ G, σ(D) = D ⇒ D ∈ K^{G} = [K is the splitting field of a irreducible, separable polynomial. It will also work if **K is the splitting field of a polynomial over a field F with char(F)** = 0 ⇒ K/F is Galois] F.

- n = 2, f = x
^{2}+ bx + c, D = (α_{1}- α_{2})^{2}= b^{2}-4c.

α_{1} + α_{2} = -b, α_{1}α_{2} = c. (α_{1}-α_{2})^{2} = (α_{1}+α_{2})^{2} -4α_{1}α_{2} = b^{2} -4c.

- n = 3, 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}

Proposition. Let F be a field with char(F)≠2, let f(x)∈ F[x] be an irreducible, separable polynomial, and let K be the splitting field of f(x) over F (or K be a splitting field of an irreducible polynomial over ℚ). If $δ = \prod_{i < j} (α_i-α_j)$ ∈ F (D = Δ is a square in F), then G ⊆ A_{n}. Otherwise, G $\not\subset A_n$.

Proof.

G = Gal(f) = Gal(K/F).

Recall that the set A_{n} is called the alternating group, A_{n} = {even permutation in S_{n}} -is a subgroup of S_{n-} ≤ S_{n}, |S_{n}| = n!, |A_{n}| = ^{n!}⁄_{2}

Suppose σ ∈ S_{n} is a transposition, say σ = (ij) with i < j. Then σ affects only those factors of the product $δ = \prod_{i < j} (α_i-α_j)$ that involve i or j:

- α
_{i}- α_{j}→ [σ maps α_{i}→ σ_{j}] σ(α_{i}- α_{j}) = -(α_{i}- α_{j}) - α
_{k}- α_{i}, α_{k}- α_{j}for k < i (< j) ⇒ σ(α_{k}- α_{i})·σ(α_{k}- α_{j}) = (α_{k}- α_{j})·(α_{k}- α_{i}), the final result is unchanged. - α
_{i}- α_{l}, α_{j}-α_{l}for (i < ) j < l ⇒ σ(α_{i}- α_{l})·σ(α_{j}- α_{l}) = (α_{j}- α_{l})·(α_{i}- α_{l}), the final result is unchanged. - α
_{i}- α_{m}, α_{m}- α_{j}for i < m < j ⇒ σ(α_{i}- α_{m})·σ(α_{m}- α_{j}) = (α_{j}- α_{m})·(α_{m}- α_{i}) = (α_{m}- α_{j})·(α_{i}- α_{m}), the final result is again unchanged.

Multiplying all the terms together gives σ(δ) = -δ. Therefore, for an arbitrary σ ∈ S_{n}, **σ(δ) = δ if and only if σ is a product of an even number of permutations**, and σ(δ) = -δ if and only if σ is a product of an odd number of permutations.

G $\not\subset A_n$ ↭ [By definition] G contains an odd permutation σ ↭ σ can be written as a product of an odd number of transpositions (2-cycles) ↭ [$δ = \prod_{i < j} (α_i-α_j)$] σ(δ) = -δ ≠ δ (char(F)≠2), δ ∉ K^{G} = F

G ⊆ A_{n} ↭ ∀σ ∈ G, σ is even ⇒ σ(δ)=δ ⇒ δ ∈ K^{G} = F∎

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.