LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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 auxiliary ideal is the unit ideal

Statement

Assume the Axiom of Choice.

Let $k$ be an algebraically closed field, let $I \subseteq k [x_{1}, \dots, x_{n}]$ be an ideal, and let $f$ vanish on $V (I)$ . Then the auxiliary ideal $J : = I + (1 - y f) \subseteq k [x_{1}, \dots, x_{n}, y]$ is the unit ideal.

Facts & Assumptions

Given: The Axiom of Choice, an algebraically closed field $k$ , an ideal $I \subseteq k [x_{1}, \dots, x_{n}]$ , and a polynomial $f$ vanishing on $V (I)$ .

[L1]

Every proper ideal lies in a maximal ideal (In a nonzero commutative ring, every proper ideal is contained in a maximal ideal).

[L2]

Over an algebraically closed field, every maximal ideal of a polynomial ring is an evaluation ideal (Over an algebraically closed field, every maximal ideal is an evaluation ideal).

[L3]

The auxiliary ideal $J$ has empty zero locus (The Rabinowitsch auxiliary ideal has no common zero).

Proof

technique · contradiction

1.1L1givenassume-contra

Assume, for contradiction, that $J$ is proper. By [L1], $J$ lies in some maximal ideal $m \subseteq k [x_{1}, \dots, x_{n}, y]$ .

2.1L2step 1.1choose

By [L2], there is a point $(a, b) \in k^{n + 1}$ with $m = (x_{1} - a_{1}, \dots, x_{n} - a_{n}, y - b)$ . Since $J \subseteq m$ , every element of $J$ vanishes at $(a, b)$ .

3.1L3step 2.1discharge-contradiction∎

Step 2.1 contradicts [L3], which says that $J$ has no common zero. Therefore $J$ is not proper, so $J = (1)$ .

Depends on

Used by

Dependency tree · two levels

15 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

J. S. Milne, A Primer of Commutative Algebra, v4.03, Theorem 13.10 (standard reference, not scraped)