Characterizations of solvability II. Discriminants.

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  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 - α₁)···(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 α₁,…, α_n. If σ ∈ G, σ permutes α₁,…, 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 α₁,…, α_n, K = F(α₁,…, 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_nxⁿ +··· + a₁x + a₀, 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 + ··· + a₁σ(x) + a₀, 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 = f₁···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)(x²+1) ∈ ℚ[x]. f has three roots (3, i, -i), but Gal(f) = S₂ = {id, (23)} < S₃ and is not transitive. In particular, G has not an automorphism σ that sends 3 → i. On the contrary, f = x³ -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 S₃. Futhermore, G = S₃.

Definition. Let F be a field with char(F) ≠ 2. Let f(x) ∈ F[x], deg(f) = n. Let α₁, α₂, ···, 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 α₁, α₂, ···, α_n, σ(D) = D.

Example, say n = 3, σ: α₁ → α₁, α₂ → α₃, α₃ →α₂, D = (α₁-α₂)²(α₁-α₃)²(α₂-α₃)².

σ(D) = (α₁-α₃)²(α₁-α₂)²(α₃-α₂)². σ(D) = D because we are using the product of the squares of the differences of the polynomial roots, e.g., (α₂-α₃)² = (α₃-α₂)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.

Examples

• n = 2, f = x² + bx + c, D = (α₁ - α₂)² = b² -4c.

α₁ + α₂ = -b, α₁α₂ = c. (α₁-α₂)² = (α₁+α₂)² -4α₁α₂ = b² -4c.

• n = 3, f = x³ +a₂x² + a₁x + a₀. Disc(f) = (α₁ -α₂)²(α₁ -α₃)²(α₂ -α₃)² = -4a₂³a₀ + a₂²a₁² + 18a₂a₁a₀ -4a₁³ -27a₀²

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∎

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.