Determine a quartic group from prime certificates

Try

Can two finite calculations determine a group acting on complex roots? Consider f(X)=X⁴−X−1 in ℤ[X]. We will prove that its splitting-field Galois group over ℚ is S₄, the group of all permutations of its four roots. The finite calculations are certificates; the bridge from them to characteristic zero is a theorem, not a visual resemblance between roots.

First reduce the coefficients modulo 2. The result is X⁴+X+1. Evaluate it at 0 and 1, then decide whether “no roots” proves irreducibility. It does for degrees two and three, but not for degree four: two irreducible quadratic factors remain possible. Write down that missing case before reading further.

Use polynomial division for the factor tests and good-prime cycle lifting for the following established result: a monic integer polynomial with squarefree reduction modulo p has a Galois-group element whose cycle lengths are the degrees of its irreducible factors modulo p. Different primes may produce different elements and different labelings; the theorem does not identify them as a particular generating pair.

Predict

Predict what a single irreducible factor of degree four gives, and what factors of degrees one and three give. Then predict whether either certificate alone determines S₄.

The diagram shows possible labelings of those cycle types. Arrow direction records the permutation, not a motion of numerical complex roots. The labels in the two panels can be chosen independently.

Check the prediction

A four-cycle alone occurs in the proper transitive subgroup generated by (1234), of order 4. A three-cycle occurs in the proper transitive subgroup A₄. We must combine both certificates with a subgroup proof. Also, S₄ is solvable: identifying it will not prove a failure of solvability by radicals.

Experiment

Perform a complete arithmetic audit. Over F₂ the only monic irreducible quadratic is X²+X+1: a monic quadratic with no root must have constant 1, and X²+1 has root 1. Its square is X⁴+X²+1, not our polynomial. Since f modulo 2 has no linear factor, any proper factorization would have to be this product of irreducible quadratics. Thus the reduction is irreducible. Its derivative is 1, so it is squarefree as well.

Over F₇ multiply, rather than merely trusting, the candidate factorization:

(X−3)(X³+3X²+2X+5)=X⁴−7X²−X−15 ≡ X⁴−X−1.

Write h(X)=X³+3X²+2X+5. Its complete value table is below. Recompute at least two entries using integer arithmetic before reduction.

A reducible cubic over a field has a linear factor, so h is irreducible. Its derivative 3X²+6X+2 is nonzero and has lower degree; irreducibility forces gcd(h,h′)=1. Also h(3)=2, so the linear factor is not repeated or shared with h. These facts certify squarefreeness of the entire reduction, not only its cubic factor.

Before proceeding, state separately the arithmetic facts and the theorem they authorize. Modulo 2 gives one four-cycle; modulo 7 gives one three-cycle with one fixed point.

Understand

From reduction to an irreducible rational polynomial

If the monic f factored over ℚ into two positive-degree factors, Gauss’s lemma would give monic positive-degree factors in ℤ[X]. Reducing them modulo 2 preserves their degrees because their leading coefficients are 1. This would contradict the proved irreducibility over F₂. Thus f is irreducible over ℚ. Characteristic zero makes it separable, so its splitting field L has four distinct roots, and its group G=Gal(L/ℚ) acts faithfully on them: an automorphism fixing every root fixes their generated field L. Hence G is a subgroup of S₄. Irreducibility also gives transitivity by the extension-of-embeddings theorem, although the order argument below already suffices once the two cycles are known.

Excluding the only competing group order

By Lagrange’s theorem, 4 and 3 divide |G| because G contains elements of those orders. Therefore 12 divides |G|. Since |G| divides |S₄|=24, its order is 12 or 24.

Here is the needed classification of the order-12 possibility, proved rather than assumed. Any subgroup H of index two in S₄ is normal, since its left and right cosets are H and its complement. The quotient map S₄→S₄/H≅C₂ is nontrivial. All transpositions are conjugate, so their images in the abelian quotient agree. They cannot all map to the identity, because transpositions generate S₄. Thus every transposition maps to the nonidentity element. A product of transpositions maps to the identity exactly when the number of factors is even. The well-defined sign homomorphism therefore shows H=A₄, the even permutations.

But a four-cycle is a product of three transpositions, so it is odd and cannot belong to A₄. This excludes order 12. Consequently |G|=24 and G=S₄. Notice what was not asserted: we never assumed that the two supplied elements have compatible chosen labels, or that they generate G.

The correct radical conclusion

Let V₄ consist of the identity and the three double transpositions (12)(34), (13)(24), (14)(23). Each nonidentity element squares to the identity, and the product of two distinct ones is the third; hence V₄ is abelian. Conjugation preserves the double-transposition cycle type, so V₄ is normal in S₄ and in A₄. The chain

1 ⊲ V₄ ⊲ A₄ ⊲ S₄

has abelian successive factors V₄≅C₂×C₂, A₄/V₄≅C₃ and S₄/A₄≅C₂. Thus S₄ is solvable. The proved characteristic-zero solvability-by-radicals theorem implies that all roots of f lie in a radical tower over ℚ, allowing complex intermediate values. We have proved existence of such a tower, not displayed a particular quartic formula.

Radicals are broader than successive square roots. The earlier constructible-numbers lesson uses the same polynomial’s three root pairings and a cubic resolvent to prove a constructibility obstruction. Reuse that argument for its distinct conclusion; an S₄ group is not a nonsolvability certificate.

Apply

Change the polynomial to g(X)=X⁴+X−1. Give its two prime certificates and its Galois group without copying an unchecked factor table. Then explain a second route through its relation to f. Finally diagnose the claim: “A transitive quartic group containing a three-cycle must be S₄.”

Compare a complete changed-case solution

Modulo 2, g has the same irreducible reduction. Modulo 7, g=(X+3)(X³−3X²+2X−5). The cubic is −h(−X); substitution and multiplication by a nonzero scalar preserve irreducibility. Its values at 0 through 6 are 2,2,2,1,5,6,3. At the linear root 4 the value is 5, so no factor is shared, and the cubic is separable by its nonzero lower-degree derivative. The two squarefree patterns are again 4 and 1+3. The complete order argument gives S₄ and the solvability conclusion as before.

Alternatively g(X)=f(−X). Negation gives a bijection between their roots, so their splitting fields inside a common algebraic closure are identical and the root permutation actions are conjugate. This explains the repeated answer rather than guessing from similar coefficients. Finally A₄ is transitive on four letters and contains three-cycles, but is proper in S₄. The omitted odd four-cycle is a substantive hypothesis.

Continue

Write a certificate with five clearly separated claims: complete modular factorizations; squarefreeness; the lifting theorem; the subgroup exclusion; the solvable series. A wrong arithmetic entry and a missing theorem are different failures and require different repairs.

Return to good-prime cycle lifting for the simultaneous characteristic-zero automorphism, or discriminants and Galois groups for the independent sign criterion. This example does not classify every quartic group, prove that every conjugacy class occurs at some prime, or establish general ramification theory.