Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-sol)audited 2026-09-27
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Finite normalization commutes with principal localization

Statement

Let k be a field, let A be a finite-type integral domain over k, let B be the integral closure of A in Frac⁡(A), and let 0≠f∈A. Then the integral closure of the principal localisation Af inside the field Frac⁡(A) is exactly the localisation Bf, and Bf is a finite Af-module.

Facts & Assumptions

Given: a field k, a finite-type integral domain A over k, its integral closure B in Frac⁡(A), and an element 0≠f∈A.

[L1]

Let A→B be a homomorphism of commutative rings, S⊆A multiplicative and b∈B: if b is integral over A then b/1 is integral over S−1A. Moreover, if A is a domain, S⊆A∖{0}, K is a field extension of Frac⁡(A) and A‾ is the integral closure of A in K, then the integral closure of S−1A in K is exactly S−1A‾ (Integrality and integral closure commute with localisation, Multiplicative subsets and the localisation S−1R as equivalence classes of fractions).

[L2]

For a commutative ring R and f∈R the powers Sf={fn:n∈N} form a multiplicative subset, and the principal localisation is Rf=Sf−1R with elements written r/fn; the element f becomes a unit, and an element r/1 is zero exactly when fnr=0 for some n (Principal localisation Rf={1,f,f2,…}−1R, Multiplicative subsets and the localisation S−1R as equivalence classes of fractions).

[L4]

If A is a domain and 0≠f∈A, then Af is a domain containing A and the identity on A exhibits Frac⁡(A) as a field containing Af, and Frac⁡(Af)=Frac⁡(A): each element of Frac⁡(Af) is a fraction of elements of Af, hence lies in Frac⁡(A), while each a/b with a,b∈A, b≠0, equals (a/1)/(b/1) with a/1,b/1∈Af (The field of fractions Frac⁡(D)=(D∖{0})−1D of an integral domain, Zero divisor, and integral domain: a commutative ring with 1≠0 and no zero divisors, Principal localisation Rf={1,f,f2,…}−1R).

[L5]

If B is a finite A-module with generators b1,…,bm and A⊆Af, then Bf=Af(b1/1)+⋯+Af(bm/1) is a finite Af-module: every element of Bf is b/fn with b=∑iaibi, and b/fn=∑i(ai/fn)(bi/1) (Module finiteness is transitive along a tower of algebras, Generated submodule, cyclic and finitely generated modules, module basis and free module, Principal localisation Rf={1,f,f2,…}−1R).

Proof

technique · direct
1.1

The element f is nonzero in the domain A, so the multiplicative subset Sf={fn} is contained in A∖{0} by [L2] and [L4]. By [L3] the integral closure B of A in Frac⁡(A) is a finite A-module. The localisation Af is a domain containing A, and by [L4] the field Frac⁡(A) contains Af and equals its fraction field, so the phrase "the integral closure of Af inside Frac⁡(A)" is computed inside a field extension of Frac⁡(Af).

L2L3L4given
2.1

Applying the third clause of [L1] with the domain A, the multiplicative subset S=Sf⊆A∖{0}, the field K=Frac⁡(A) and A‾=B: the integral closure of Sf−1A=Af in K=Frac⁡(A) is exactly Sf−1B=Bf. This is a genuine equality inside the fixed field Frac⁡(A), not an isomorphism chosen afterwards, so it is canonical.

L1L2step 1.1
3.1

By [L5], applied to the finite generating list b1,…,bm of the A-module B from step 1.1, the localisation Bf is generated as an Af-module by the finitely many elements b1/1,…,bm/1, so Bf is a finite Af-module. Combined with step 2.1, the integral closure of Af inside Frac⁡(A) is Bf, a finite Af-module. If f is a unit of A, then Sf contains 1 and Af=A, so the statement reduces to Bf=B; the hypothesis f≠0 is exactly what makes Sf a subset of A∖{0}, and it is used nowhere else.

L2L5step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources