LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)audited 2026-10-02

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Cohomology of a two-chart double cover

Statement

Assume the Axiom of Choice. Let $k$ be a field, let $g$ be a nonnegative integer, fix polynomials $f \in k [x]$ and $ψ \in k [t]$ , and let $C$ be a separated $k$ -scheme with an affine open cover $U, V$ , where $Γ (U, O_{C}) = k [x, y] / (y^{2} - f (x))$ and $Γ (V, O_{C}) = k [t, w] / (w^{2} - ψ (t))$ . Suppose that $U \cap V = D_{U} (x) = D_{V} (t)$ and the gluing sends $t = x^{- 1}$ and $w = x^{- (g + 1)} y$ . Then $H^{1} (C, O_{C})$ has dimension $g$ over $k$ , with basis given by the classes of $x^{- 1} y, \dots, x^{- g} y$ (an empty basis when $g = 0$ ).

Facts & Assumptions

Given: AC, the field $k$ , integer $g \geq 0$ , separated scheme $C$ , and the two affine charts and gluing in the Statement.

[F1]

Under AC, for a quasi-compact separated scheme, finite affine-cover Čech cohomology of a quasi-coherent module agrees with sheaf cohomology. The structure sheaf is quasi-coherent. (Cech cohomology computes quasi-coherent cohomology on a separated scheme, Quasi-coherent module on a scheme, The Axiom of Choice)

Proof

1.1F1givenalgebra

The two affines make $C$ quasi-compact. Their intersection ring is $T = k [x, x^{- 1}, y] / (y^{2} - f (x))$ , a free $k [x, x^{- 1}]$ -module with basis $1, y$ , since the relation is monic in $y$ . The ordered two-open Čech differential is $(a, b) \mapsto b - a$ , so [F1] gives $H^{1} (C, O_{C}) = T / (A + B)$ as a $k$ -vector space, where $A$ and $B$ are the images of the two chart rings.

2.1F1step 1.1algebra

In $T$ , $A = k [x] \oplus k [x] y$ and $B = k [x^{- 1}] \oplus x^{- (g + 1)} k [x^{- 1}] y$ . The constant-in- $y$ summand is exhausted by $k [x] + k [x^{- 1}]$ . In the $y$ -summand, $A$ contains precisely the powers $x^{m} y$ with $m \geq 0$ , and $B$ precisely those with $m \leq - g - 1$ . The remaining independent Laurent monomials are $x^{- 1} y, \dots, x^{- g} y$ . Thus the quotient has the stated basis and dimension $g$ , including $g = 0$ . AC enters only through [F1]. ∎

Depends on

Used by

Ramification of the double cover y²=f(x) Example

Dependency tree · two levels

15 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; elementary local prerequisite for the Step 5b citation repair (standard reference, not scraped)