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

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.

All nonempty affine opens of an irreducible affine variety have the same function field

Statement

Assume the Axiom of Choice.

Let $X$ be a classical affine variety. If $U \subseteq X$ is a nonempty affine open subset, then $Frac (k [U]) ≅ k (X)$ canonically. Hence any two nonempty affine opens of $X$ have canonically isomorphic function fields.

Facts & Assumptions

Given: The Axiom of Choice, a classical affine variety $X$ , and a nonempty affine open subset $U \subseteq X$ .

[L1]

On this page, an affine open subset of $X$ is a principal open $D_{X} (\overline{f})$ , and its coordinate ring is $O_{X} (U)$ (Affine open subsets of a classical affine variety).

[L2]

Assuming the Axiom of Choice, for a principal open $D_{X} (\overline{f})$ , the coordinate ring is $k [X]_{\overline{f}}$ (Regular functions on a principal open are the principal localization of the coordinate ring).

[L3]

The coordinate ring $k [X]$ of a classical affine variety is an integral domain (A classical affine variety has a domain as its coordinate ring, and conversely).

[L4]

The function field of $X$ is $Frac (k [X])$ (The function field of an irreducible classical affine variety).

Proof

technique · direct

1.1L1L2L3given

By [L1], write $U = D_{X} (\overline{f})$ . Since $U$ is nonempty, the element $\overline{f}$ is not zero in the domain $k [X]$ of [L3]. By [L2], the coordinate ring of $U$ is $k [X]_{\overline{f}}$ .

2.1L3step 1.1algebra

Every fraction in $k [X]_{\overline{f}}$ is represented by $a / {\overline{f}}^{n}$ with $a \in k [X]$ . Sending that fraction to the same formal quotient in $Frac (k [X])$ defines an injective ring map $k [X]_{\overline{f}} ↪ Frac (k [X])$ . Conversely, if $a / b \in Frac (k [X])$ with $b \neq 0$ , then $b / 1$ is nonzero in the domain $k [X]_{\overline{f}}$ , so the same quotient may be read as an element of $Frac (k [X]_{\overline{f}})$ . Hence each field of fractions contains the other in the obvious way.

3.1L2L4step 2.1∎

Step 2.1 gives $Frac (k [U]) = Frac (k [X]_{\overline{f}}) ≅ Frac (k [X]) = k (X)$ by [L2] and [L4]. Therefore every nonempty affine open of $X$ has the same function field as $X$ , and any two such opens have canonically isomorphic function fields.

Depends on

Used by

Dominant maps pull back function fields functorially Lemma

Dependency tree · two levels

16 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

Donu Arapura, Notes on Basic Algebraic Geometry, Corollary 1.6.4 and §3.1 (standard reference, not scraped)
Michael Artin, Notes for a Course in Algebraic Geometry, Proposition 2.6.1 and Definition 3.4.3 (standard reference, not scraped)