LemmaStatement: AI-generatedProof: AI-generated

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.

The square-cube condition implies reducedness

PDF p. 5: Seminormalization in Lemma 2.4

Statement

If a commutative ring $A$ has a unique $a \in A$ with $a^{2} = x$ and $a^{3} = y$ for every pair $x, y \in A$ with $x^{3} = y^{2}$ , then $A$ is reduced: it has no nonzero nilpotent element.

Facts & Assumptions

Given: A commutative ring satisfying the stated square-cube condition.

Goal: Show that every nilpotent element vanishes.

Proof

1.1given

If $z^{2} = 0$ , then $z^{3} = 0$ as well. Both $0$ and $z$ have square and cube equal to zero, so uniqueness applied to $(x, y) = (0, 0)$ gives $z = 0$ .

2.1step 1.1construct∎

If $t^{n} = 0$ for some positive integer $n$ , choose a power of two at least $n$ . Repeatedly applying step 1.1 to $t$ raised to successively halved exponents shows $t = 0$ . Thus $A$ is reduced.

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Seminormality and seminormalization