Further questions of Louboutin: cyclic prime degree, powers of α, and a quintic unit #
Abstract models; no number field is constructed.
- Cyclic prime degree:
Ka field,σ ∈ RingAut Kwithσ^ℓ = 1,ℓan odd prime,O = ℤ[α]stable underσ,σα ≠ α;N(x) = ∏_{h<ℓ} σʰx,IsU O xfor "x ∈ O^×",d_k = σᵏα − α. The ratios(σⁱα − σʲα)/d₁are units fori ≢ j;η = d₂/d₁satisfiesη − 1 = σ(d₁)/d₁,1 − η ∈ O^×,N(η − 1) = 1,N(1 − η) = −1; for every ring hom fromOto a finite fieldF,ℓ ≤ |F|, and if|F| = pᶠwithp < ℓandf ∣ ℓthenf = ℓ. - Powers of
α:Na commutative ring with a groupGacting by ring automorphisms and an additive valuationv(v x = ⊤ → x = 0); if0 < v(gα)for allg,gα ≠ αfor somegandk ≥ 2, thenℤ[conj αᵏ] ≠ ℤ[conj α]. Forα³ − Cα + C = 0,C = C'² + C' + 7:ℤ[conj αᵏ] = ℤ[conj α]iffk = 1. - A quintic unit:
ε = ζᵃwwithζ⁵ = 1, complex conjugationcwithc·ζ = ζ⁴,c·w = w,5 ∤ a, norm1:ℤ[conj εᵏ] = ℤ[ζ]for everykwith5 ∤ k.
Main results #
Louboutin.SeccionSeis.lem_ratios,Louboutin.SeccionSeis.cor_inert.Louboutin.SeccionSeis.prop_val,Louboutin.SeccionSeis.prop_val_poly,Louboutin.SeccionSeis.cor_q25.Louboutin.SeccionSeis.prop_c27.
References #
- S. R. Louboutin, On the order generated by the conjugates of an algebraic integer, in: Algebra, Logic and Number Theory, Banach Center Publ. 121 (2020), 107–124, doi:10.4064/bc121-10.
- S. R. Louboutin, When is the order generated by a cubic, quartic or quintic algebraic unit Galois invariant: three conjectures, Czechoslovak Math. J. 70 (2020), 905–919, doi:10.21136/CMJ.2020.0019-19.
- C. Marín, Orders generated by conjugates of Galois quartic integers, preprint (2026).
Cyclic fields of odd prime degree: lem:ratios and cor:inert #
The norm N_{K/ℚ}(x) = ∏_{h<ℓ} σʰ(x) of a cyclic extension of degree ℓ generated by σ.
Equations
- Louboutin.SeccionSeis.nrm σ ℓ x = ∏ h ∈ Finset.range ℓ, (σ ^ h) x
Instances For
Generic telescoping for an automorphism τ (later τ = σᵏ), e = τα − α.
The setting of lem:ratios.
lem:ratios, first assertion: (σⁱα − σʲα)/d₁ ∈ O^× for i ≢ j (mod ℓ).
lem:ratios (6.2). O = ℤ[α] σ-stable, σ of odd prime order ℓ on K, σα ≠ α. Then (σⁱα − σʲα)/d₁ ∈ O^× for i ≢ j (mod ℓ); η = d₂/d₁ and η − 1 = σ(d₁)/d₁ lie in O^×, N(η − 1) = 1, and η is exceptional: 1 − η ∈ O^× with N(1 − η) = −1.
cor:inert (6.3), core: the r_i = d_i/d₁ (i < ℓ) lie in O and have unit differences, so their images under any ring hom O → F, F a finite field (the residue field of a prime above p), are distinct: ℓ ≤ |F|.
cor:inert (6.3). If the residue field has pᶠ elements with p < ℓ and f ∣ ℓ (the residue degree divides [K : ℚ] = ℓ), then f = ℓ: p is inert.
Powers of α: prop:val and cor:q25 #
ℤ[conj x]: the subring generated by the G-orbit of x (G the Galois group of the normal closure, acting on its ring of integers N).
Equations
- Louboutin.SeccionSeis.Zconj G x = Subring.closure (Set.range fun (g : G) => g • x)
Instances For
If m ≤ v(s − h·s) for every generator s and every h, the same holds on the whole subring generated (x − h·x lies in the ideal generated by those differences).
prop:val (6.5), valuation core. v a valuation of N with trivial support (𝔓-adic on 𝒪_N), every conjugate of α has positive valuation, α is not G-fixed (α ∉ ℚ), k ≥ 2. Then ℤ[conj αᵏ] ≠ ℤ[conj α]: μ(ℤ[conj αᵏ]) ≥ μ(ℤ[conj α]) + (k − 1) > μ(ℤ[conj α]).
prop:val (6.5) with its hypothesis as in the note: αⁿ ∈ p·N (from Π_α ≡ Xⁿ mod p) and v(p) > 0 (𝔓 lies above p).
Π_α ≡ Xⁿ (mod p) ⇒ αⁿ ∈ p·ℤ[α]: if Π ∈ ℤ[X] is monic of degree n with Π(α) = 0 and p divides every coefficient of degree < n, then αⁿ = p·y.
prop:val (6.5), as stated: Π_α ≡ Xⁿ (mod p) (Π monic of degree n, p ∣ coeff_i for i < n), v(p) > 0, α ∉ ℚ, k ≥ 2 ⇒ ℤ[conj αᵏ] ≠ ℤ[conj α].
cor:q25 (6.6). α³ − Cα + C = 0 with C = C'² + C' + 7, α ∉ ℚ; for each prime p ∣ C a valuation v_p of N with trivial support and v_p(p) ≥ 1 (a prime of 𝒪_N above p). Then for k ≥ 1: ℤ[conj αᵏ] = ℤ[conj α] ↔ k = 1, i.e. S_α = {1}.
prop:c27 (Conjecture 27 of Lou20a): the algebraic core #
prop:c27, core. G finite (Gal(ℚ(ζ₅)/ℚ)) acting on N, c ∈ G (complex conjugation), ζ⁵ = 1, c·ζ = ζ⁴ (= ζ̄), ε = ζᵃw with c·w = w (w = ±((1 − √5)/2)ᵇ is real), 5 ∤ a, N(ε) = ∏g g·ε = 1, G·ζ ⊆ ℤ[ζ] and ε ∈ ℤ[ζ]. Then ℤ[conj ε] = ℤ[ζ]: ε̄⁻¹ = ∏{g≠1} g·ε̄ ∈ A_ε, ε/ε̄ = ζ²ᵃ ∈ A_ε, and 5 ∤ 2a gives ζ ∈ A_ε.
prop:c27 (6.8) ([Lou20a, Conjecture 27]): ℤ[conj εᵏ] = ℤ[ζ] for every k with 5 ∤ k, and hence ℤ[conj εᵏ] = ℤ[conj ε].