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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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  • 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.

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The Noetherian minimal-prime induction split

Statement

Let R be a commutative ring and let IR be a proper radical ideal that is not prime. Then there exist elements x,yRI with xyI. For any such choice of x and y, every prime ideal minimal over I is minimal over I+(x) or minimal over I+(y).

Facts & Assumptions

Given: A commutative ring R and a proper radical ideal IR that is not prime.

[L1]

A prime ideal is proper and contains one factor whenever it contains a product (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct
1.1

Because I is not prime, [L1] gives elements x,yR with xyI but xI and yI.

L1givenchoose
2.1

Let p be a prime ideal minimal over I. Since xyIp and p is prime, [L1] gives xp or yp. If xp and q is a prime ideal with I+(x)qp, then Iqp, so minimality of p over I forces q=p. Thus p is minimal over I+(x). The same argument with y in place of x shows that if yp, then p is minimal over I+(y).

L1step 1.1given
3.1

Therefore every prime ideal minimal over I appears on one side of the split I+(x) or I+(y).

step 2.1

Depends on

Used by

Dependency tree · two levels

5 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