    # Characterizations of solvability II. Discriminants.

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 (Sn). The discriminant determines when this subgroup consists solely of even permutations (An), 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 - αn) 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 Sn.

Proof.

Let G be the Galois group of f. f has n roots, say α1,…, αn. If σ ∈ G, σ permutes α1,…, αn, so there is a map Φ: G → Sn, σ → σ (looking at it as a permutation in Sn or S1,…, 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 Sn is (1)(2)···(n), that is, σ does not permutes the roots of f ⇒ σ(αi) = αi ∀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,…, αn, K = F(α1,…, αn)] so this is an injective group homomorphism ⇒ G is isomorphic to a subgroup of Sn

Definition. Let Sn be the symmetric group on n letters (|Sn| = n!). A subgroup G of Sn 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 Sn. 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 fi to other roots, let f(x) = anxn +··· + a1x + a0, ai ∈ F. Let α be a root of f, f(α) = 0 ⇒ [σ is a F-automorphism] 0 = σ(f(α)) = an(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 Sn, and for the sake of contradiction, f is reducible. Then, f can be written as product of distinct irreducibles f = f1···fr, where each fi is irreducible with degree at least 1. Let us take any of the fi’s.

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

f is irreducible ↭ Gal(f) is a transitive subgroup of Sn 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 $\sqrt{2},\sqrt{2}w, \sqrt{2}w^2$ 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, ···, αn 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, ···, αn, σ(D) = D.

Example, say n = 3, σ: α1 → α1, α2 → α3, α3 →α2, D = (α12)213)223)2.

σ(D) = (α13)212)232)2. σ(D) = D because we are using the product of the squares of the differences of the polynomial roots, e.g., (α23)2 = (α32)2.

Since σ is a permutation of the roots of f, σ(αij)2 = (ασ(i) - ασ(j))2, so we have two products D and σ(D) with the same terms, except it may differ (ar-as)2 instead of (as-ar)2 but the final product is obviously unaffected.

∀σ ∈ G, σ(D) = D ⇒ D ∈ KG = [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. (α12)2 = (α12)2 -4α1α2 = b2 -4c.

• n = 3, f = x3 +a2x2 + a1x + a0. Disc(f) = (α12)213)223)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 ⊆ An. Otherwise, G $\not\subset A_n$.

Proof.

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

Recall that the set An is called the alternating group, An = {even permutation in Sn} -is a subgroup of Sn- ≤ Sn, |Sn| = n!, |An| = n!2

Suppose σ ∈ Sn 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:

• αi - αj → [σ maps αi → σj] σ(αi - αj) = -(αi - αj)
• αk - αi, αk - αj for k < i (< j) ⇒ σ(αk - αi)·σ(αk - αj) = (αk - αj)·(αk - αi), the final result is unchanged.
• αi - αl, αjl for (i < ) j < l ⇒ σ(αi - αl)·σ(αj - αl) = (αj - αl)·(αi - αl), the final result is unchanged.
• αi - αm, αm - αj for i < m < j ⇒ σ(αi - αm)·σ(αm - αj) = (αj - αm)·(αm - αi) = (αm - αj)·(αi - αm), the final result is again unchanged.

Multiplying all the terms together gives σ(δ) = -δ. Therefore, for an arbitrary σ ∈ Sn, 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 ∉ KG = F

G ⊆ An ↭ ∀σ ∈ G, σ is even ⇒ σ(δ)=δ ⇒ δ ∈ KG = F∎

# Bibliography 