ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-30

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.

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Twist transitions on the projective line

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $k$ be a field and $S = k [x_{0}, x_{1}]$ with $\deg x_{0} = \deg x_{1} = 1$ , so that $P_{k}^{1} = Proj S$ with charts $U_{0} = D_{+} (x_{0})$ and $U_{1} = D_{+} (x_{1})$ , and let $t = x_{1} / x_{0}$ , the coordinate on $U_{0}$ . Then for every integer $n$ the twisting sheaf $O (n) = \tilde{S (n)}$ has frames $e_{0} = x_{0}^{n} on U_{0}, e_{1} = x_{1}^{n} on U_{1},$ and on the overlap $U_{0} \cap U_{1}$ these frames are related by $e_{1} = t^{n} e_{0} .$ Here for $n < 0$ the symbols $x_{i}^{n}$ denote the corresponding units $x_{i}^{n} \in S [x_{0}^{- 1}]$ or $S [x_{1}^{- 1}]$ , and the frames are nowhere-vanishing local generators of the invertible sheaf $O (n)$ (Invertible twists for degree-one generated rings).

Facts & Assumptions

Given: The Axiom of Choice, A field $k$ , the graded ring $S = k [x_{0}, x_{1}]$ with $\deg x_{i} = 1$ , an integer $n$ , and the charts $U_{i} = D_{+} (x_{i})$ of $P_{k}^{1}$ .

[F1]

$P_{k}^{1} = Proj S$ , $U_{i} = D_{+} (x_{i}) = Spec S_{(x_{i})}$ with $S_{(x_{0})} = k [t]$ , $t = x_{1} / x_{0}$ , and $S_{(x_{1})} = k [t^{- 1}]$ ; the overlap is $D_{+} (x_{0} x_{1}) = Spec k [t, t^{- 1}]$ . (Projective space is Proj of a polynomial ring)

[F2]

$O (n) = \tilde{S (n)}$ has sections $Γ (U_{i}, O (n)) = S (n)_{(x_{i})} = (S (n) [x_{i}^{- 1}])_{0}$ , the degree-zero part of the homogeneous localisation, and restrictions are the canonical localisations. (Twisting sheaf on Proj)

[F3]

Since $S$ is generated over $k$ by $S_{1}$ , every $O (n)$ is invertible, with frame $x_{i}^{n}$ on $D_{+} (x_{i})$ : $S (n)_{(x_{i})} = S_{(x_{i})} \cdot x_{i}^{n}$ is free of rank one. (Invertible twists for degree-one generated rings)

[F4]

The assumed Axiom of Choice is the choice-function principle (The Axiom of Choice); it licenses the AC-qualified Proj and associated-sheaf suppliers at step 1.1.

Verification

technique · direct: compute the degree-zero localisations of the shifted modules and compare the resulting local generators on the overlap

1.1F1F2F3F4algebra

The chart modules. The AC premise [F4] licenses the associated-sheaf and Proj charts [F1]–[F3]. For $n \in Z$ the module $S (n)_{(x_{0})} = (S (n) [x_{0}^{- 1}])_{0}$ consists of the classes $a / x_{0}^{k}$ with $a \in S (n)_{k} = S_{n + k}$ homogeneous of degree $n + k$ . Since $x_{0}$ is a unit in the localisation, every such class equals $(a / x_{0}^{n + k}) \cdot x_{0}^{n}$ with $a / x_{0}^{n + k} \in S_{(x_{0})} = k [t]$ ; hence $e_{0} : = x_{0}^{n}$ generates $S (n)_{(x_{0})}$ over $k [t]$ , and symmetrically $e_{1} : = x_{1}^{n}$ generates $S (n)_{(x_{1})}$ over $k [t^{- 1}]$ .

2.1F1F2step 1.1algebra

The overlap. On the overlap $D_{+} (x_{0} x_{1})$ the ring is $k [t, t^{- 1}]$ with $t = x_{1} / x_{0}$ , so $x_{1} = t x_{0}$ and therefore $x_{1}^{n} = t^{n} x_{0}^{n}$ holds in the localisation of $S$ at $x_{0} x_{1}$ for every integer $n$ , positive or negative; under the identifications of step 1.1 this is precisely the frame relation $e_{1} = t^{n} e_{0}$ on $U_{0} \cap U_{1}$ .

3.1

Conclusion. The frame section $e_{i}$ is nowhere vanishing on $U_{i}$ , and the transition relation $e_{1} = t^{n} e_{0}$ is exactly the change of frame of the invertible sheaf $O (n)$ from the $0$ -chart to the $1$ -chart: for $n = 0$ both frames are the constant function $1$ and the relation is $e_{1} = e_{0}$ ; for $n = 1$ it is $e_{1} = t e_{0}$ ; for $n = - 1$ it is $e_{1} = t^{- 1} e_{0}$ with $t^{- 1} = x_{0} / x_{1}$ the coordinate on $U_{1}$ . [F2, F3, step 1.1, step 2.1, cases: n=0 and negative n] \qed

Depends on

Used by

Two graded rings with the same Proj Counterexample

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

The Stacks Project, Constructions of Schemes, Sections 27.8-27.21 (standard reference, not scraped)
Ravi Vakil, The Rising Sea, 29 August 2022, Sections 4.5, 7.4, 9.3, 10.6, 17.4, 17.6, 18.2 (standard reference, not scraped)
Gao-Zhang, Lectures on Algebraic Geometry, Chapter 5 (standard reference, not scraped)