Try
Two rational cubics can have very different root symmetries even when neither has a rational root. Compare f=t³−3t+1 and g=t³−t−1. The discriminant formula for a depressed monic cubic t³+pt+q is −4p³−27q². Calculate both discriminants, and test whether either polynomial vanishes at 1 or −1.
The calculations give 81 and −23 respectively, and neither polynomial has either rational-root candidate as a root. We will prove that their splitting-field Galois groups are C₃ and S₃, respectively. A positive discriminant is not the criterion: the relevant condition is being a square in the base field. For example, a positive rational number need not be a rational square.
Review discriminants and repeated roots and symmetric polynomials for the coefficient formula and why the discriminant lies in the base field. This lesson explains the additional information carried by its square root.
Predict
Let K be a field of characteristic different from 2. Let f be monic, separable, of degree n≥1, with distinct roots α₁,…,αₙ in its splitting field L. Define
δ = ∏ over i<j of (αᵢ−αⱼ), D=δ².
Here D is the discriminant and is nonzero. A different ordering of the roots may change δ by a sign, but never changes D. Every K-automorphism of L permutes the roots. Predict what swapping two roots does to δ. Can an automorphism that makes this sign change fix δ when δ≠0?
For n=1 the product is empty and equals 1, so there is no sign obstruction. For larger n the distinction between δ and δ² is exactly the information we will use.
Experiment
For three ordered roots, δ=(α₁−α₂)(α₁−α₃)(α₂−α₃). Swap α₁ and α₂ and reorder the factors to see one minus sign. Now cycle the three roots: a three-cycle is the product of two transpositions, so it gives two minus signs. Fill the action and field columns independently before checking this table.
| Permutation type | Sign | Image of δ | Fixes δ in characteristic different from 2? |
|---|
| Identity | +1 | δ | Yes |
| A transposition | −1 | −δ | No |
| A three-cycle | +1 | δ | Yes |
Apply the table to the two cubics in Try. Both are irreducible over ℚ: a reducible cubic over a field has a linear factor, and the rational-root theorem restricts a rational root of either monic cubic to ±1. Their nonzero discriminants make them separable. For f, D=81=9²; for g, D=−23 is not a rational square. Do not conclude the groups from the table alone: we still need transitivity and a proof that no other transitive cubic subgroup occurs.
Understand
The Vandermonde sign action. Regard ∏(Xᵢ−Xⱼ) as a polynomial in labeled variables. Interchanging adjacent labels reverses their mutual difference and pairs the other affected factors without changing their product. Thus an adjacent interchange multiplies the polynomial by −1. Every permutation is a product of adjacent interchanges; the parity of their number is the permutation sign. Consequently substituting the roots gives
σ(δ)=sign(σ)δ
for each σ in G=Gal(L/K), where G is embedded in Sₙ by its root permutation. The embedding is faithful because the roots generate L. This identity is valid even when different polynomial factors have different G-orbits; irreducibility has not yet been used.
Square discriminant and fixed fields. A splitting field of a separable polynomial is a finite Galois extension, so Lᴳ=K by the proved Galois correspondence. If G consists of even permutations, the sign identity fixes δ under all of G, giving δ∈K and D a square in K. Conversely, if D=c² for c∈K, then (δ−c)(δ+c)=0 in the field L. Hence δ=c or −c, so δ∈K. Since δ≠0 and 2≠0, an odd permutation cannot fix δ. Thus
D is a square in K if and only if G⊆Aₙ.
Let H=G∩Aₙ, the kernel of the sign map restricted to G. The automorphisms fixing δ are exactly H, so the correspondence gives Lᴴ=K(δ). If an odd automorphism exists, H has index two, D is nonsquare, and K(δ) is the quadratic extension obtained by adjoining a root of z²−D. If every automorphism is even, H=G and K(δ)=K instead. The phrase “quadratic sign field” therefore applies only when the sign character is nontrivial.
Why irreducible cubics leave only two groups. Suppose now that f is an irreducible separable cubic. For any two roots α,β, evaluation identifies K(α) and K(β) by a K-isomorphism sending α to β. The extension of embeddings into a splitting field extends this to an automorphism of L. Thus the root action is transitive. Orbit–stabilizer makes 3 divide |G|, while G≤S₃ and Lagrange make |G| divide 6. The only possibilities are 3 and 6. A subgroup of order 6 is S₃. A subgroup of order 3 is cyclic and its nonidentity elements are three-cycles, so it is exactly A₃. This proves the exhaustion, not just examples of two possible groups.
Combining this argument with the square criterion proves: an irreducible separable cubic over K of characteristic different from 2 has group A₃≅C₃ when D is square, and S₃ otherwise. For the Try cubics, the splitting fields consequently have degrees 3 and 6 over ℚ. For g the sign field is ℚ(√−23). For f there is no nontrivial sign field. None of these conclusions asserts that every root of an arbitrary polynomial with square discriminant is already in K.
Keep the hypotheses visible. The reducible cubic t(t²−2) has discriminant 32 and splitting field ℚ(√2), whose group is C₂, not S₃. Its action fixes the root 0 and exchanges the other two, so it is not transitive. The general square/sign criterion remains valid; the two-group cubic classification required irreducibility. Even t(t−1)(t+1), with square discriminant 4, has trivial Galois group rather than A₃.
Characteristic 2 breaks the sign distinction. Over F₂, t²+t+1 has no root, is irreducible, and has derivative 1, so is separable. Its discriminant is 1. Its splitting field F₄ has the nonidentity Frobenius automorphism exchanging the two roots, so its Galois group is C₂=S₂ and is not contained in A₂. Here −δ=δ; the step excluding odd automorphisms is unavailable. Separability alone does not repair that step.
Apply
Change the cubic to h=t³−2 over ℚ. Prove irreducibility, compute its discriminant, identify its full Galois group, and identify the sign field inside its splitting field. Then compare the conclusions with the reducible polynomial t(t²−3). State which step of the cubic classification can and cannot be reused.
Check the changed cubics and their fields
h is irreducible by Eisenstein at 2 (or by checking the rational-root candidates ±1,±2). Its discriminant is −108=36·(−3), which is nonsquare in ℚ. It is separable in characteristic zero, so the theorem gives S₃ and sign field ℚ(√−3). If ρ is the real cube root of 2 and ζ a primitive cube root of unity, its roots are ρ,ρζ,ρζ². Their splitting field is ℚ(ρ,ζ), and ℚ(ζ)=ℚ(√−3) agrees with the sign field. The full subgroup diagram is treated in the existing S₃ example.
For t(t²−3), D=108 and the splitting field is ℚ(√3). Its group is the transposition subgroup C₂; the sign field equals the whole quadratic splitting field. The square/sign proof applies, but transitivity fails because 0 is fixed. Therefore a nonsquare discriminant cannot by itself promote a reducible cubic’s group to S₃.
Continue
Use the complete S₃ fixed-field example to locate the sign kernel among all its subgroups. A group conclusion should record four checks: the base field and its characteristic, separability, irreducibility when transitivity is used, and the square class of D in that field. The sign character detects parity, not the full structure of arbitrary higher-degree Galois groups; quartic and larger classifications need additional information.
