Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)
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normal domain implies r one

Statement

Every commutative Noetherian integrally closed domain satisfies (R1).

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

serre r k and s k conditions: For a commutative Noetherian ring R and an integer j≥0, condition (Rj) means that Rp is regular whenever ht⁡p≤j. Condition (Sj) means that depth⁡Rp≥min⁡{j,dim⁡Rp} for every prime p. A finite module M satisfies (Sj) if depth⁡RpMp≥min⁡{j,dim⁡Supp⁡RpMp} for every prime in its support. Outside the support the condition is vacuous, consistent with depth of the zero module being +∞ and the empty support having no nonnegative dimension. Thus the zero module satisfies all (Sj) conditions, and the zero ring satisfies both families vacuously.

[F2]

Height-one localizations of normal Noetherian domains are DVRs: Let R be a Noetherian integrally closed domain, and let p be a prime ideal of height 1. Then the localisation Rp is a discrete valuation ring.

[F3]

embedding dimension and regular local ring: A nonzero Noetherian local ring (A,m,k) is regular when dim⁡k(m/m2)=dim⁡A.

Proof

1.1F2F3

A height-one localization A=Rp is a DVR by [F2], so it has dimension one and a nonzero principal maximal ideal m=(π). Since A is a domain and π is a nonunit, π∉m2=(π2): otherwise cancellation would make π a unit. Thus m/m2 is a one-dimensional vector space over A/m, generated by the nonzero class of π. Its embedding dimension equals its dimension, so A is regular.

2.1F1step 1.1algebra∎

The only height-zero prime of a domain is (0); its localization is the fraction field, which is regular. These two cases give the definition of (R1), including a field, which has no height-one primes.

Depends on

Used by

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Sources