Alphabeta Math
TheoremStatement: 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.
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The radical of an ideal is the intersection of the prime ideals containing it

Statement

Assume the Axiom of Choice.

Let R be a commutative ring and let IR be an ideal. Then

I=pSpecRIpp,

where the intersection is taken to be R if no prime ideal contains I.

Facts & Assumptions

Given: A commutative ring R, an ideal IR, and the Axiom of Choice.

[L1]

Every prime ideal containing I also contains I (Primes containing an ideal contain its radical).

[L2]

Every element outside I is omitted by some prime ideal containing I (A separating prime for an element outside a radical).

Proof

technique · direct
1.1

Let xI. By [L1], every prime ideal containing I also contains x. Therefore x belongs to the displayed intersection.

L1given
1.2

Let xI. By [L2], there is a prime ideal p containing I with xp. Hence x does not belong to the displayed intersection.

L2given
2.1

Steps 1.1 and 1.2 prove that an element belongs to I exactly when it belongs to every prime ideal containing I, which is the claimed equality.

step 1.1step 1.2

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