Try
Work over the two-element field F₂, where 1+1=0. The polynomial h(X)=X³+X+1 has no root: its values at 0 and 1 are both 1. A reducible cubic over a field has a linear factor, so h is irreducible. Its quotient k=F₂[α], with α³=α+1, is a field with eight elements. Each element has a unique form a+bα+cα² with a,b,c∈F₂.
Square α repeatedly. The sequence begins α, α², α⁴=α²+α and then returns to α because α⁸=(α²+α)²=α⁴+α²=α. Why do these three values give exactly the roots of h, rather than just three unrelated field elements?
The answer connects irreducible factors to orbits: an orbit is the list obtained by repeatedly applying one permutation until its starting value returns. In a finite field of characteristic p, the map x↦xᵖ is Frobenius. Here it is squaring. We will prove that an irreducible factor’s degree is the length of its root orbit. This is a statement about finite fields; lifting such a permutation to a characteristic-zero field requires the next lesson’s additional proof.
Predict
Before calculating, decide whether α+1 belongs to the same squaring orbit as α. If it does not, predict the degree of its irreducible polynomial. Then predict how the answer changes for h(X)²: does squaring suddenly produce a cycle of length six, or is multiplicity separate from the permutation on distinct roots?
Review finite-field Frobenius and subfields for the automorphism and fixed-field facts, and polynomial quotients for the field construction. We will supply the orbit-polynomial argument, not assume that the order of an automorphism on an entire field equals every individual orbit length.
Experiment
Use α³=α+1 to reduce every square to the basis 1,α,α². Fill the missing next entries on paper, then compare all eight elements in this orbit table. Every arrow means “square in k.”
| Starting element | Complete orbit, returning to the start | Length | Orbit polynomial |
|---|
| 0 | 0 → 0 | 1 | X |
| 1 | 1 → 1 | 1 | X+1 |
| α | α → α² → α²+α → α | 3 | X³+X+1 |
| α+1 | α+1 → α²+1 → α²+α+1 → α+1 | 3 | X³+X²+1 |
The basis coordinates distinguish every entry, so the four rows partition the field. The nonzero rows account for seven roots of X⁷−1=X⁷+1. Multiply the three corresponding polynomials and check
X⁷+1=(X+1)(X³+X+1)(X³+X²+1).
Both cubics have no roots in F₂ and hence are irreducible. The derivative of X⁷+1 is X⁶, which has no common root with X⁷+1. Thus the factorization is squarefree. The permutation on its seven distinct roots has cycle lengths 1,3,3; it is not a single seven-cycle.
Check the second orbit and its polynomial
Squaring α+1 gives α²+1. Squaring again gives α⁴+1=α²+α+1, and one more square gives α+1. These three values are distinct. If β=α+1, then β³+β²+1=0: expand in characteristic two and use α³=α+1. Hence β is a root of X³+X²+1. The full proof below explains why a root and its whole orbit determine this irreducible polynomial. The table is a worked finite example, not a substitute for that argument.
Understand
Specify the base field before specifying Frobenius
Let F=F_q have q=pʳ elements, where p is prime and r≥1. Let k be a finite extension containing all roots under discussion. A splitting field of a polynomial over a finite field is finite: adjoining finitely many algebraic roots gives a finite-dimensional extension, with q to that dimension many elements. Define φ:k→k by φ(x)=xᑫ. Repeated p-power Frobenius shows it is a field homomorphism, and injectivity plus finiteness makes it an automorphism.
Every element of F is fixed by φ. Conversely the fixed elements of k are roots of Xᑫ−X, which has degree q and already has the q elements of F as distinct roots. Therefore the fixed field of φ is exactly F. This root bound is the coefficient-descent fact needed below. For F_q one must use the q-power map, not automatically the p-power map: the latter can move elements of the coefficient field.
An irreducible polynomial is its root’s orbit polynomial
Let h∈F[X] be monic irreducible of degree n≥1, and let a∈k be a root. Since φ permutes the finite field, a has a finite positive orbit length d. Its orbit consists of the distinct elements a,φ(a),…,φᵈ⁻¹(a). Form the monic polynomial
B(X)=∏ from j=0 to d−1 of (X−φʲ(a)).
Apply φ to coefficients while leaving the formal variable X unchanged. It permutes the factors cyclically, so every coefficient of B is fixed by φ and therefore belongs to F. Since B(a)=0 and h is the minimal polynomial of a over F, polynomial division gives h∣B. In particular n≤d. To recall that minimal-polynomial step: the remainder on division by h would have smaller degree and vanish at a; the kernel of evaluation is generated by the monic irreducible h, so that remainder must be zero.
For the reverse inequality, h(φ(a))=φ(h(a))=0 because all coefficients of h lie in the fixed field F. Repeating gives d distinct roots of h. The root bound gives d≤n. Hence d=n, and h and B are equal because they are monic, have equal degree, and h divides B.
This proves at once that h has distinct roots and that they form a single q-power orbit. The proof uses finite orbit length, coefficient descent, division and the polynomial root bound. Neither degree inequality alone proves the result; both are required.
General factorizations and their boundaries
Write a nonzero monic polynomial as f=∏hᵢᵉⁱ with distinct monic irreducibles hᵢ. Each hᵢ contributes one orbit of length deg hᵢ. Distinct irreducibles have disjoint root sets, since a common root would give the same minimal polynomial. If f is squarefree, every exponent is 1, so the cycles partition all deg f roots and their lengths are exactly the factor degrees, counted once per distinct factor.
For repeated factors, the permutation still acts on distinct roots. For example (X³+X+1)² has degree six but only the same three roots as h, with one length-three orbit; each root has multiplicity two. Multiplicity is polynomial data, not another orbit or a longer cycle. The nonzero constant polynomial 1 has no roots or factors and an empty list of cycles. The zero polynomial is excluded: a factorization into monic irreducibles is not defined for it.
Reduction from integer coefficients has another boundary. The degree-two polynomial 2X²+X+1 becomes X+1 modulo 2. Its single fixed root describes the reduced polynomial of degree one; it cannot be treated as a cycle list on two original roots. A leading coefficient divisible by p can erase degree. Monic integer input prevents this loss, but monicity alone does not prevent repeated roots modulo p: (X−1)(X+1) reduces to (X+1)² modulo 2. Good-prime lifting must check squarefreeness separately.
Changing the base changes the orbits
Take k=F₂[θ] with θ⁴+θ+1=0. This polynomial is irreducible: it has no roots in F₂, and the only monic irreducible quadratic there is X²+X+1, whose square is X⁴+X²+1, not our polynomial. Thus k has 16 elements. The squaring orbit of θ has length four, by the proved theorem.
Put ω=θ²+θ. Then ω²=θ⁴+θ²=θ²+θ+1=ω+1. The element ω is neither 0 nor 1, since otherwise θ would satisfy a polynomial of degree at most two over F₂. Hence F₂[ω] is a four-element subfield. Relative to this field use φ(x)=x⁴, not squaring. Now θ⁴=θ+1 and (θ+1)⁴=θ, so the orbit is θ,θ+1. Its polynomial is
(X−θ)(X−θ−1)=X²+X+ω.
The remaining two roots θ²,θ²+1 form the other fourth-power orbit, with polynomial X²+X+ω². Both factors are irreducible over F₄ by the orbit theorem (or because their roots are not fixed by fourth powers). Multiplying gives X⁴+X+1, since ω+ω²=1 and ωω²=1. A degree-four factor over F₂ has become two degree-two factors over F₄. Neither polynomial nor field element changed; the fixed coefficient field did.
Apply
Change to F₃ and f(X)=X⁴−1. Construct k=F₃[i] with i²=−1. Prove this is a field, list the cube-map orbits of the roots, and recover the complete factorization over F₃. Then repeat the cycle interpretation for f², keeping multiplicities separate.
Compare the changed field and repeated case
The squares in F₃ are 0 and 1, so X²+1 has no root and the quotient is a field with nine elements. Its cube map sends a+bi to a−bi. The roots of f are 1,−1,i,−i, with orbits (1), (−1), (i,−i). Thus f=(X−1)(X+1)(X²+1), giving lengths 1,1,2. The factors are distinct, and the derivative is X³ in characteristic three, relatively prime to f. For f² the polynomial degree is eight but the distinct-root permutation has exactly those same three cycles; all four roots now have multiplicity two. Over k itself the relevant q is 9, and all four roots are fixed, so the polynomial splits into four linear factors.
Continue
Return to separability to connect repeated roots with a derivative gcd, or splitting fields for existence of the finite field holding all roots. The next arithmetic step must prove that one characteristic-zero automorphism acts on all reduced roots as Frobenius simultaneously. The finite-field orbit theorem alone does not justify that lifting, identify a whole characteristic-zero Galois group, or assert that every cycle type occurs at some prime.
