Connect polynomial factors with multiplicative orders

Try

Factor X⁶−1 by the exact orders of its complex roots. Roots of order one contribute X−1, order two contributes X+1, order three contributes X²+X+1, and order six contributes X²−X+1. The last factor is the sixth cyclotomic polynomial Φ6. Its value at three is seven, and three has order six modulo seven. This example suggests a bridge between polynomial factorization and primes in particular congruence classes.

Predict

Predict why the condition that a prime divisor not divide the index matters. Although Φ6(2)=3, the order of two modulo three is two, not six. The exceptional prime three divides the index six. In characteristic three, factors of X⁶−1 can share roots. Any proposed order theorem must exclude precisely this failure rather than claiming that every prime divisor of every cyclotomic value has the desired order.

Experiment

Construct the small factors by exact polynomial division.

n	Φn(X)	Degree	Constant term
1	X−1	1	−1
2	X+1	1	1
3	X²+X+1	2	1
4	X²+1	2	1
6	X²−X+1	2	1

Multiply the factors for divisors of six and check X⁶−1. The roots are grouped by exact order, not merely by satisfying the sixth-power equation.

Understand

Define Φn as the monic product of X−ζ over primitive complex nth roots ζ, those with exact order n. Every nth root has an order dividing n, so X^n−1=product over d dividing n of Φd(X). The degree is φ(n), since the primitive roots are exp(2πik/n) with gcd(k, n)=1. To prove integer coefficients, induct on n. Divide X^n−1 by the monic integer product for all proper divisors. Monic long division produces an integer quotient and remainder; the complex root factorization forces the remainder zero and the quotient equal to Φn. Evaluating the product at zero and inducting also gives Φn(0)=1 for n>1.

From integer coefficients to irreducibility

The previous argument proves Φn belongs to Z[X]. It does not yet prove that Φn is irreducible over Q. For this deeper step, use Gauss’s lemma and the repeated-root criterion. A monic irreducible factor over Q of a monic integer polynomial has integer coefficients. Reduction modulo a prime preserves the degree of a monic polynomial, though it need not preserve irreducibility.

Choose a primitive nth root ζ and let f be its monic minimal polynomial over Q. Then f divides Φn and belongs to Z[X]. We first prove that, for every prime p not dividing n, ζᵖ is also a root of f.

Suppose otherwise. Since ζᵖ still has order n, its monic minimal polynomial g is a different irreducible factor of Φn, also in Z[X]. The equality g(ζᵖ)=0 implies f divides g(Xᵖ) over Q; monic polynomial division shows the quotient belongs to Z[X]. Reduce this divisibility modulo p, writing a bar for reduction. In F_p[X] the characteristic-p power rule gives ḡ(Xᵖ)=ḡ(X)ᵖ. Thus f̄ divides ḡᵖ.

The polynomial f̄ is monic of positive degree, so choose one of its irreducible factors h over F_p. Since h divides ḡᵖ, unique factorization in F_p[X] makes h divide ḡ. Therefore h² divides f̄ḡ. Because f and g are distinct irreducible factors of Φn, their product divides Φn over Z; consequently h² divides the reduction of Xⁿ−1. This is impossible: p does not divide n, so the derivative nXⁿ⁻¹ is relatively prime to Xⁿ−1 over F_p. The latter polynomial is squarefree, meaning it has no repeated irreducible factor. The contradiction proves ζᵖ is a root of f.

Apply this result repeatedly. Any positive integer a coprime to n is a product of primes not dividing n. At each step the current root remains primitive and has the same minimal polynomial f, so ζᵃ is a root of f. All φ(n) primitive roots are therefore roots of f. Since f divides the degree-φ(n) monic polynomial Φn, they are equal. This proves cyclotomic irreducibility. The cases n=1 and n=2 also follow directly from X−1 and X+1.

Identify the field automorphisms

The cyclotomic field Q(ζ) contains every nth root as a power of ζ, so it is the splitting field of the separable polynomial Xⁿ−1. Its degree is φ(n), by the irreducibility proof. An automorphism must send ζ to a primitive root ζᵃ with gcd(a,n)=1. Conversely every such image defines a Q-isomorphism through the common minimal polynomial Φn. Its image is Q(ζᵃ)=Q(ζ), since an inverse of a modulo n recovers ζ as a power of ζᵃ.

Thus Gal(Q(ζ)/Q) is the group of invertible residues modulo n: write σ_a(ζ)=ζᵃ. Composition satisfies σ_aσ_b=σ_ab, so the operation is multiplication modulo n, not addition. The group is abelian but need not be cyclic. For n=1 the field is Q and the group is trivial; the residue-group description is used here for n≥2.

For n=8, the residues 1,3,5,7 all square to 1 modulo 8. The group has four elements and every nonidentity element has order two, so it is C₂×C₂, not C₄. Compare n=5: residues 1,2,4,3 are consecutive powers of 2, giving C₄. Equal extension degrees do not determine the Galois group.

Field	Minimal polynomial of a primitive root	Degree	Galois group
Q(ζ₅)	X⁴+X³+X²+X+1	4	C₄
Q(ζ₈)	X⁴+1	4	C₂×C₂
Q(ζ₁₂)	X⁴−X²+1	4	C₂×C₂

For the last row, calculate the units 1,5,7,11 and square them modulo 12. Before using a degree as a construction criterion, remember that a power-of-two degree alone is not a general straightedge-and-compass theorem. Cyclotomic construction criteria require a further field-tower argument.

Apply

Now suppose a prime q divides Φn(a) but not n. The residue a is a root of X^n−1 and is nonzero. This polynomial is squarefree modulo q because its derivative nX^(n−1) has no common root with it. If a had order d< n, with d dividing n, it would also be a root of X^d−1, hence of one of the other cyclotomic factors. That would make a repeated root in X^n−1, a contradiction. Thus its order is n, and Lagrange gives n dividing q−1. For fixed n>1, suppose only finitely many primes were one modulo n. Choose A a sufficiently large multiple of n and all those primes. The monic polynomial Φn has positive degree, so Φn(A)>1. Any prime q dividing this value divides none of A’s chosen factors, because Φn(A)=1 modulo each of them. The order argument produces a new prime one modulo n, contradiction.

For proof practice, explain why the reduction argument chooses an irreducible factor h of f̄ instead of assuming f̄ is irreducible. Then diagnose the forbidden case p dividing n: modulo 2, X⁸−1=(X−1)⁸, so the squarefree contradiction disappears. Finally compute the composition of σ₃ and σ₅ for ζ₈; it is σ₇, since 3·5≡7 modulo 8.

Continue

This proves infinitude in the residue class one without proving the general theorem for every coprime residue class; that stronger statement requires additional methods. The irreducibility theorem proved here is not needed for the residue-class infinitude argument, but it supplies the field degree and automorphism group. The finite polynomial factorizations here connect to generating-function algebra, while orders explain their modular meaning. When using the construction computationally, factoring the resulting large value may be difficult even though the existence proof is decisive.

Return to the number theory unit to choose a targeted retry. Continue with turn unique factorization into a convergent euler product.