The degrees of irreducible factors of binomials and multiplicativity of the n -Hartley condition

We prove that the degrees of irreducible factors of a binomial 𝑡 𝑛 -𝛼 are divisible by the least such degree.As a consequence, we deduce that for relatively prime natural numbers 𝑚, 𝑛, a polynomial has the 𝑚𝑛-Hartley condition if and only if it has the 𝑚-Hartley and 𝑛-Hartley conditions. The 𝑛-Hartley conditionLet Q ⊂ C denote the field of algebraic numbers.Let 𝑛 be a positive integer.Write 𝜁 𝑛 = exp(2𝜋𝑖/𝑛), a chosen primitive 𝑛-th root of 1 in Q.If Δ(𝑡) is a polynomial in Z[𝑡], then we say Δ(𝑡) has the 𝑛-Hartley condition or is 𝑛-Hartley for short, if there exists a polynomial 𝑔(𝑡) ∈ Z[𝑡] for whichThe virtue of the 𝑛-Hartley condition is this: if a knot 𝐾 in the 3-dimensional sphere 𝑆 3 has a freely periodic symmetry of order 𝑛, i.e., if the cyclic group of order 𝑛 acts freely on 𝑆 3 by orientation-preserving diffeomorphisms that leave 𝐾 invariant, then the Alexander polynomial Δ 𝐾 (𝑡) of 𝐾 has the 𝑛-Hartley condition.This is [5, Theorem 1.2], slightly modified by [2, Def.1.3, Proposition 2.2]. Example 1.1Although the 𝑛-Hartley condition originates in the study of Alexander polynomials of freely periodic knots, the condition is not restricted to such polynomials.We give some further examples here.• Trivially, if 𝑓 (𝑡) = ℎ(𝑡) 𝑛 for some integer polynomial ℎ(𝑡), then 𝑓 (𝑡) has the 𝑛-Hartley condition: set 𝑔(𝑡) = ℎ(𝑡 𝑛 ).• Products of 𝑛-Hartley polynomials are again 𝑛-Hartley.• The 𝑚-th cyclotomic polynomial Φ 𝑚 (𝑡) has the 𝑛-Hartley condition for precisely those values of 𝑛 that are relatively prime to 𝑚.This is a special case of [2, Proposition 1.5].• Other ad hoc examples may be found.For instance, 𝑓 (𝑡) = 𝑡 3 -4𝑡 2 + 3𝑡 -1 is not self-reciprocal and therefore not the Alexander polynomial of a knot.It is, 2020