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.

Square-cube roots in a normal domain

PDF p. 5: Seminormalization in Lemma 2.4

Statement

Let $A$ be an integrally closed domain. For every $x, y \in A$ with $x^{3} = y^{2}$ there is a unique $a \in A$ with $a^{2} = x$ and $a^{3} = y$ . An integrally closed domain is a domain containing every element of its fraction field that satisfies a monic polynomial over it.

Facts & Assumptions

Given: An integrally closed domain $A$ and $x, y \in A$ satisfying $x^{3} = y^{2}$ .

Goal: Produce the unique simultaneous square and cube root.

[F1]

Equal squares and cubes in a reduced ring determine an element uniquely (Uniqueness from squares and cubes).

Proof

1.1givenalgebra

If $x = 0$ , then $y = 0$ because $A$ is a domain, and $a = 0$ works. If $x \neq 0$ , put $a = y / x$ in the fraction field. The relation gives $a^{2} = x$ and $a^{3} = y$ .

2.1givenF1step 1.1∎

In the second case $a$ satisfies the monic equation $T^{2} - x = 0$ , so integral closedness puts $a$ in $A$ . A domain is reduced, so [F1] gives uniqueness in both cases.

Depends on

Uniqueness from squares and cubes

Used by · 1

Seminormality and seminormalization

Dependency tree · 1 level

Arrows run from prerequisites to the items that use them. External source nodes link directly to the cited text.