TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12

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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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A classical affine variety has a domain as its coordinate ring, and conversely

Statement

Let $k$ be an algebraically closed field and let $X \subseteq A_{k}^{n}$ be a nonempty affine algebraic set. Then $X$ is a classical affine variety if and only if its coordinate ring $k [X]$ is an integral domain.

Facts & Assumptions

Given: An algebraically closed field $k$ and a nonempty affine algebraic set $X \subseteq A_{k}^{n}$ .

[L1]

The coordinate ring of $X$ is $k [X] = k [x_{1}, \dots, x_{n}] / I (X)$ (The coordinate ring of an affine algebraic set).

[L2]

A nonempty affine algebraic set is irreducible exactly when its vanishing ideal is prime (Affine algebraic sets correspond to radical ideals, and irreducible ones to prime ideals).

[L3]

For a commutative ring $R$ and ideal $P$ , the quotient $R / P$ is an integral domain if and only if $P$ is prime ( $R / P$ is an integral domain if and only if $P$ is a prime ideal).

Proof

technique · direct

1.1L2given

By [L2], the set $X$ is a classical affine variety exactly when $I (X)$ is a prime ideal of $k [x_{1}, \dots, x_{n}]$ .

1.2L1L3

By [L1] and [L3], the coordinate ring $k [X]$ is an integral domain exactly when the ideal $I (X)$ is prime.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove that $X$ is a classical affine variety if and only if $k [X]$ is an integral domain.

A classical affine variety has a domain as its coordinate ring, and conversely

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources