Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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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.

Associated primes localize forward

Statement

Let R be a commutative ring, let M be a left R-module, let S⊆R be multiplicative, and let p∈Ass⁡R(M) with p∩S=∅. Then S−1p∈Ass⁡S−1R(S−1M).

Facts & Assumptions

Given: A commutative ring R, a left R-module M, a multiplicative subset S⊆R, and a prime ideal p∈Ass⁡R(M) with p∩S=∅.

[L1]

A prime ideal is associated to a module exactly when its residue module embeds in that module (Associated primes are exactly primes of embedded cyclic residue modules).

[L2]

Injective module maps remain injective after localisation (Injective module maps remain injective after localisation).

[L3]

Localisation commutes with quotient modules, so S−1(R/p)≅(S−1R)/(S−1p) (Localisation commutes with quotient modules and arbitrary direct sums).

[L4]

If p∩S=∅, then S−1p is a prime ideal of S−1R (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

Proof

technique · direct
1.1L1L2

By [L1], there is an injective R-module map R/p↪M. Applying [L2] yields an injective S−1R-module map S−1(R/p)↪S−1M.

2.1L1L3L4step 1.1

By [L3], the source identifies with (S−1R)/(S−1p), and by [L4] the ideal S−1p is prime. Therefore [L1], applied over the ring S−1R, shows that S−1p is associated to S−1M.

3.1step 2.1∎

Hence associated primes localize forward.

Depends on

Used by

Dependency tree · two levels

14 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