Ring of integers of ℚ(μ∞): non-Noetherian of dimension 1
ProvedNoetherianScheme.ringOfIntegers_cyclotomic_infinitealgebraic-geometrynoetherian
Let be the field obtained from by adjoining all roots of unity (precisely: is generated over by roots of unity and contains a primitive -th root of unity for every ). Let be the integral closure of in . Then
- is not a Noetherian ring, and
- has Krull dimension exactly .
So finite-dimensional affine schemes need not be Noetherian.
Formalization Note The source speaks of a general infinite Galois extension "such as" ; 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 NoetherianSchemeSource
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
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.