CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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Top cohomology of projective twists

Statement

Assume the Axiom of Choice as inherited from the cited theorem (The Axiom of Choice). Let $A$ be a commutative ring with $1$ (Commutative ring), let $n \geq 0$ , let $d \in Z$ and let $O_{X} (d)$ be the twisting sheaf on $X = P_{A}^{n}$ (Relative projective space from standard charts, Twisting sheaf on Proj). If $n \geq 1$ , then $H^{n} (X, O_{X} (d))$ is the free $A$ -module on the Laurent monomials $x_{0}^{e_{0}} \dots x_{n}^{e_{n}}$ with $e_{i} < 0$ for every $i$ and $\sum_{i} e_{i} = d$ . It is zero for $d > - n - 1$ , and for $d \leq - n - 1$ it has rank $(\binom{- d - 1}{n})$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ) whenever $A \neq 0$ , while it is the zero module for $A = 0$ . For $n = 0$ the top group is $H^{0} (P_{A}^{0}, O (d)) ≅ A$ for every $d \in Z$ .

Facts & Assumptions

Given: The Axiom of Choice as inherited, a commutative ring $A$ with $1$ , integers $n \geq 0$ and $d$ , and the twisting sheaf $O (d)$ on $P_{A}^{n}$ .

[F1]

Cohomology of twists on projective space: for every commutative ring $A$ , every $n \geq 1$ and every $d$ , $H^{n} (P_{A}^{n}, O (d))$ is the free $A$ -module on the Laurent monomials $x_{0}^{e_{0}} \dots x_{n}^{e_{n}}$ with $e_{i} < 0$ for all $i$ and $\sum_{i} e_{i} = d$ , and it is nonzero precisely when $d \leq - n - 1$ and $A \neq 0$ (for $n > 0$ ); for $n = 0$ , $H^{0} (P_{A}^{0}, O (d)) ≅ A$ for every $d$ . (Cohomology of O(d) on projective space)

[F2]

Compositions with positive parts: for integers $N \geq 1$ and $k \geq 1$ the number of compositions of $N$ into exactly $k$ positive parts is $(\binom{N - 1}{k - 1})$ , and there are none when $k > N$ . (Compositions of $n$ into $k$ positive parts are counted by $(\binom{n - 1}{k - 1})$ )

[F3]

The free module over a ring $R$ on an indexed set is the direct sum of copies of $R$ indexed by that set; when $R = 0$ every such direct sum is the zero module, whatever the index set. (The direct sum of an indexed family of modules)

Proof

technique · direct: read the top group off the full computation and count its stated monomial basis by converting the all-negative exponent vectors into compositions of $-d$ into $n+1$ positive parts

1.1F1

For $n \geq 1$ apply [F1]: $H^{n} (P_{A}^{n}, O (d))$ is the free $A$ -module on the Laurent monomials $x^{e}$ with $e_{i} < 0$ for all $i$ and $\sum_{i} e_{i} = d$ . This is the first assertion.

1.2F2F3algebra

The all-negative exponent vectors correspond bijectively to tuples $g_{i} = - e_{i} \geq 1$ with $\sum_{i} g_{i} = - d$ , and these are the compositions of $- d$ into exactly $n + 1$ positive parts when $- d \geq 1$ . If $d > - n - 1$ , then $- d < n + 1$ . For $- d \leq 0$ , a sum of positive $g_{i}$ cannot equal $- d$ ; for $1 \leq - d < n + 1$ , the no-compositions clause of [F2] applies with $N = - d$ and $k = n + 1$ . Thus in either case the module is zero. If $d \leq - n - 1$ , then $- d \geq n + 1 \geq 2 \geq 1$ and [F2] with $N = - d$ and $k = n + 1$ counts the tuples as $(\binom{- d - 1}{n})$ , so the free module has that rank when $A \neq 0$ and is the zero module when $A = 0$ by [F3].

1.3F1

For $n = 0$ , [F1] gives $H^{0} (P_{A}^{0}, O (d)) ≅ A$ for every $d$ ; the top group in dimension zero is $H^{0}$ , so the last assertion follows.

2.1F1F2F3∎

Boundaries and choice accounting. The endpoint $d = - n - 1$ is the first value with a basis monomial, namely $g_{i} = 1$ for all $i$ and hence $e_{i} = - 1$ ; for $d = - n$ one has $k = n + 1 > n = - d$ , so by the second clause of [F2] there is no composition and the group is zero. The case $d = 0$ and all positive $d$ give the zero group. The zero ring $A = 0$ is handled by [F3]. The Axiom of Choice is inherited from [F1] and nothing further is selected.

Depends on

Used by

Generator cocycle for H1 of O(-2) Example

Dependency tree · two levels

48 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, Cohomology of Schemes, Lemma 30.8.2 (Tag 01XV) (standard reference, not scraped)
Ravi Vakil, The Rising Sea (29 August 2022), Section 19.2 (standard reference, not scraped)