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

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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

1.1

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

L1givenassume-contra

2.1

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 $(a, b)$ .

L2step 1.1choose

3.1

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

L3step 2.1discharge-contradiction∎

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Sources

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