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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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r one s two intersection of height one localisations

Statement

Assume the Axiom of Choice.

If R is a commutative Noetherian domain satisfying (S2), then inside its fraction field K one has R=⋂ht⁡p=1Rp. For a field the empty intersection is interpreted as K=R.

Facts & Assumptions

Given: The objects and hypotheses in the statement, including the Axiom of Choice (The Axiom of Choice); cited dependent-choice and resolution-existence hypotheses are retained.

[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]

Every submodule of a finite module over a Noetherian ring has a minimal primary decomposition: Assume Dependent Choice. Let R be a Noetherian commutative ring and let M be a finitely generated left R-module. Every submodule N⊆M has a finite primary decomposition. After deleting redundant components and combining equal radicals, one obtains a minimal primary decomposition. When N=M, the decomposition is the empty intersection, interpreted as M. In particular, every ideal of a Noetherian ring has a minimal primary decomposition.

[F3]

The radicals in a minimal primary decomposition are exactly the associated primes of the quotient: Let R be a Noetherian commutative ring, let M be a finitely generated left R-module, and let N=Q1∩⋯∩Qr be a minimal primary decomposition in which each Qi is pi-primary. Assume each pi is a prime ideal. Then Ass⁡R(M/N)={p1,…,pr}.

[F4]

Depth drops by one after quotienting by a regular element: Let R be Noetherian, let M be finite, let I lie in the Jacobson radical, and let x∈I be M-regular. Then depth⁡I(M/xM)=depth⁡I(M)−1.

[F5]

The local depth-zero associated-prime criterion: Let (R,m) be a Noetherian local ring and let M≠0 be a finite R-module. Then depth⁡(M)=0⟺m∈Ass⁡R(M).

[F6]

Associated primes commute with localization for finite modules: Let R be a Noetherian commutative ring, let M be a finitely generated left R-module, and let S⊆R be multiplicative. Then Ass⁡S−1R(S−1M)={S−1p:p∈Ass⁡R(M), p∩S=∅}.

[F7]

A nonzero module over a Noetherian ring has an associated prime: Under Dependent Choice, if R is a Noetherian commutative ring and M a nonzero left R-module, then Ass⁡R(M) is nonempty.

Proof

1.1F6F5F4F1

Let 0≠a∈R be a nonunit. If p∈Ass⁡(R/(a)), localization and the depth-zero criterion make Rp/aRp depth zero. Since a is regular, the depth formula gives depth⁡Rp=1. Condition (S2) forces ht⁡p≤1, and a∈p, a≠0 force equality.

2.1F2F3step 1.1

Choose a minimal primary decomposition (a)=⋂iQi, with radicals pi. Those radicals are associated to R/(a), hence have height one. If b/a belongs to every height-one localization, then for each i there is si∉pi with sib∈(a)⊆Qi. Primaryness gives b∈Qi, hence b∈(a) and b/a∈R.

3.1F7step 1.1step 2.1algebra∎

If a is a unit, membership is immediate without a primary decomposition. The inclusion from R into every localization is automatic. If the height-one family is empty, a nonzero nonunit would yield an associated prime of its nonzero quotient and hence a height-one prime by the preceding argument; thus R is a field and the stipulated empty intersection is correct.

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Sources