LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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.

Power extension over a normal affine domain

Statement

Let $X$ be an integral affine scheme over a field $k$ whose coordinate ring $A = O (X)$ is a normal domain, i.e. integrally closed in its fraction field (Affine schemes and their coordinate rings, Global functions on Spec A recover A, Integral closure in an extension ring and integrally closed domains). Let $U \subseteq X$ be a dense open subscheme and let $f \in O (U)$ be such that $f^{d} \in O (X)$ for some integer $d \geq 1$ , where the rings are compared inside $O (X) \subseteq O (U) \subseteq Frac (A)$ . Then $f \in O (X)$ .

Facts & Assumptions

Given: An integral affine scheme $X$ with normal coordinate ring $A = O (X) = Γ (X, O_{X})$ , a dense open subscheme $U \subseteq X$ , an integer $d \geq 1$ and an element $f \in O (U)$ with $f^{d} \in O (X)$ , all inside $Frac (A)$ .

[F1]

Global sections of an affine scheme. The canonical map $A \to Γ (X, O_{X})$ is an isomorphism, so the global sections of $X$ are exactly $A = O (X)$ ; in particular $O (X)$ is a domain. (Global functions on Spec A recover A)

[F2]

Integrality criterion. Let $A \subseteq B$ be a ring extension and let $b \in B$ satisfy a monic polynomial with coefficients in $A$ ; then $b$ is integral over $A$ . (Integral closure in an extension ring and integrally closed domains)

[F3]

Integrally closed domain. Since $A$ is integrally closed in $Frac (A)$ , every element of $Frac (A)$ that is integral over $A$ already belongs to $A$ . (Integral closure in an extension ring and integrally closed domains)

Proof

technique · direct

1.1F1F2given

By hypothesis $f^{d} \in O (X) = A$ , so $f$ is a root of the monic polynomial $T^{d} - f^{d} \in A [T]$ ; hence $f$ is integral over $A$ .

2.1F1F3step 1.1∎

The element $f$ lies in $O (U) \subseteq Frac (A)$ , so step 1.1 and the integral closedness of $A$ give $f \in A = O (X)$ . This includes the cases $U = X$ and $d = 1$ trivially, and $f = 0$ is harmless because $Frac (A)$ is a field.

Remarks

The hypothesis $O (X) \subseteq O (U) \subseteq Frac (A)$ holds for every dense open subscheme of an integral affine scheme: restriction to the generic point injects $O (U)$ into $Frac (A)$ , and restriction from $X$ to $U$ identifies $A$ with a subring of $O (U)$ . This hypothesis is part of the statement because only the extension of rings, not the geometry of $U$ , is used.
The element $f$ need not be assumed nonzero: if $f^{d} = 0$ then $f = 0$ because $O (U) \subseteq Frac (A)$ is contained in a field, so the case $f = 0$ also satisfies the conclusion $f \in O (X)$ .

Depends on

Used by

Every dominant weight of a split semisimple group is a primitive weight Lemma

Dependency tree · two levels

9 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 Groups (corrected 2022 printing, Cambridge University Press) (standard reference, not scraped)
Robert Steinberg, Lectures on Chevalley Groups (Yale University, 1967; notes prepared by J. Faulkner and R. Wilson) (standard reference, not scraped)