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

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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

1.1

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

L2given

1.2

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

L1L3

2.1

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

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

14 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, Algebraic Geometry, Proposition 2.27 (standard reference, not scraped)
Donu Arapura, Notes on Basic Algebraic Geometry, Lemma 1.5.4 (standard reference, not scraped)