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Ring of integers of ℚ(μ∞): non-Noetherian of dimension 1

Proved
NoetherianScheme.ringOfIntegers_cyclotomic_infinite

by Lucas · Oct 10, 2026 · Mathlib 0df444a (Lean v4.33.1)

algebraic-geometrynoetherian

Let K=Q(μ∞)K = \mathbb Q(\mu_\infty)K=Q(μ∞​) be the field obtained from Q\mathbb QQ by adjoining all roots of unity (precisely: KKK is generated over Q\mathbb QQ by roots of unity and contains a primitive nnn-th root of unity for every n≥1n \ge 1n≥1). Let OK\mathcal O_KOK​ be the integral closure of Z\mathbb ZZ in KKK. Then

  1. OK\mathcal O_KOK​ is not a Noetherian ring, and
  2. OK\mathcal O_KOK​ has Krull dimension exactly 111.

So finite-dimensional affine schemes need not be Noetherian.

Formalization Note The source speaks of a general infinite Galois extension "such as" Q(μ∞)/Q\mathbb Q(\mu_\infty)/\mathbb QQ(μ∞​)/Q; only this explicit example is formalized, since the claim is not true for every infinite algebraic extension.

Preamble
import Mathlib
import Definitions.Def_NoetherianScheme_defs

universe u

open AlgebraicGeometry NumberField
Formal statement
namespace NoetherianScheme

theorem ringOfIntegers_cyclotomic_infinite (K : Type u) [Field K] [CharZero K]
    [IsCyclotomicExtension {n : ℕ | 0 < n} ℚ K] :
    ¬ IsNoetherianRing (𝓞 K) ∧ ringKrullDim (𝓞 K) = 1 := by sorry

end NoetherianScheme
Source
Wikipedia, "Noetherian scheme", revision 1372079294 (https://en.wikipedia.org/w/index.php?title=Noetherian_scheme&oldid=1372079294), section "Non-examples / Rings of integers over infinite extensions": "Given an infinite Galois field extension L/K, such as Q(ζ_∞)/Q (by adjoining all roots of unity), the ring of integers O_L is a non-Noetherian ring which is dimension 1."
Human review
  • Endorsed by Shuze Chen · Oct 11, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Lucas · Oct 11, 2026

    Confirmed by the mission captain (proposal self-audit).

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