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

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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.
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The Rabinowitsch auxiliary ideal has no common zero

Statement

Let $k$ be a field, let $I \subseteq k [x_{1}, \dots, x_{n}]$ be an ideal, and let $f \in k [x_{1}, \dots, x_{n}]$ vanish on every point of $V (I)$ . Then the ideal $J : = I + (1 - y f) \subseteq k [x_{1}, \dots, x_{n}, y]$ has empty zero locus.

Facts & Assumptions

Given: A field $k$ , an ideal $I \subseteq k [x_{1}, \dots, x_{n}]$ , and a polynomial $f$ that vanishes on $V (I)$ .

[L1]

Evaluation at a point is well defined in a polynomial ring (Evaluation and roots of a polynomial in a commutative target ring).

Proof

technique · direct

1.1L1given

Suppose $(a, b) \in k^{n + 1}$ were a common zero of $J$ . Then every element of $I$ vanishes at $a$ , so $a \in V (I)$ . By hypothesis, $f (a) = 0$ .

2.1L1step 1.1algebra∎

But $1 - y f \in J$ , so evaluating at $(a, b)$ gives $1 - b f (a) = 0$ . Using step 1.1 this becomes $1 = 0$ , which is impossible in a field. Therefore $J$ has no common zero.

The Rabinowitsch auxiliary ideal has no common zero

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources