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O(-1) has no global generator

Counterexample

Assume the Axiom of Choice (The Axiom of Choice). Let $k$ be a field and let $O (- 1)$ be the twisting sheaf on $P_{k}^{1} = Proj k [x_{0}, x_{1}]$ (Invertible twists for degree-one generated rings, Projective space is Proj of a polynomial ring). Then $Γ (P_{k}^{1}, O (- 1)) = 0,$ so the evaluation morphism $Γ (P_{k}^{1}, O (- 1)) \otimes_{Z} O_{P_{k}^{1}} ⟶ O (- 1)$ is the zero morphism, which is not surjective because $O (- 1)$ is a nonzero sheaf. Hence $O (- 1)$ is not globally generated (Global generation by the evaluation map), even though it is invertible.

Facts & Assumptions

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

[F1]

$U_{0} = Spec k [t]$ with $t = x_{1} / x_{0}$ and $U_{1} = Spec k [t^{- 1}]$ ; the overlap is $D_{+} (x_{0} x_{1}) = Spec k [t, t^{- 1}]$ , and restriction of sections is the canonical localisation. (Projective space is Proj of a polynomial ring, Twisting sheaf on Proj)

[F2]

On $U_{0}$ one has $Γ (U_{0}, O (- 1)) = S (- 1)_{(x_{0})} = k [t] \cdot x_{0}^{- 1}$ and on $U_{1}$ one has $Γ (U_{1}, O (- 1)) = S (- 1)_{(x_{1})} = k [t^{- 1}] \cdot x_{1}^{- 1}$ : in the localisation, $x_{0}^{- 1}$ and $x_{1}^{- 1}$ are units of degree $(- 1)$ , with $x_{1}^{- 1} = t^{- 1} x_{0}^{- 1}$ . (Twisting sheaf on Proj)

[F3]

$O (- 1)$ is invertible, in particular nonzero on the nonempty scheme $P_{k}^{1}$ : its restriction to $U_{i}$ is free of rank one with frame $x_{i}^{- 1}$ . (Invertible twists for degree-one generated rings)

[F4]

A global section of a sheaf on $P_{k}^{1}$ is exactly a pair of chartwise sections on $U_{0}$ and $U_{1}$ whose restrictions to $U_{0} \cap U_{1}$ agree; the global section is zero exactly when both chartwise sections are zero. (Twisting sheaf on Proj)

[F5]

A sheaf $M$ is globally generated if the evaluation morphism $Γ (X, M) \otimes_{Z} O_{X} \to M$ is surjective; the zero morphism out of a zero module is not surjective onto a sheaf with a nonzero stalk. (Global generation by the evaluation map)

[F6]

The Axiom of Choice is the choice-function principle (The Axiom of Choice). It licenses the AC-qualified supplier used at step 1.1.

Refutation

technique · direct: compute both chartwise modules of global sections of $\mathcal O(-1)$ and show that agreement on the overlap forces both to vanish

1.1F1F2F6

A global section has two chart expressions. Let $s \in Γ (P_{k}^{1}, O (- 1))$ . Under the AC premise [F6], by [F2] its restriction to $U_{0}$ has the form $s_{0} = a (t) x_{0}^{- 1}$ with $a \in k [t]$ , and its restriction to $U_{1}$ has the form $s_{1} = b (t^{- 1}) x_{1}^{- 1}$ with $b \in k [t^{- 1}]$ ; these are finite polynomials $a (t) = \sum_{m \geq 0} α_{m} t^{m}$ and $b (u) = \sum_{m \geq 0} β_{m} u^{m}$ with $u = t^{- 1}$ .

1.2F1F2algebra

Agreement on the overlap. By [F4] the two expressions agree on $U_{0} \cap U_{1}$ , where $x_{1} = t x_{0}$ is invertible. Substituting $x_{1}^{- 1} = t^{- 1} x_{0}^{- 1}$ turns the agreement into the identity $b (t^{- 1}) = t a (t)$ in $k [t, t^{- 1}]$ . The right-hand side is a finite sum of monomials $t^{m + 1}$ with $m \geq 0$ , so it involves only strictly positive powers of $t$ ; the left-hand side $\sum_{m \geq 0} β_{m} t^{- m}$ involves only nonpositive powers of $t$ . Comparing coefficients in the basis ${t^{j} : j \in Z}$ of $k [t, t^{- 1}]$ gives $α_{m} = 0$ for all $m$ and $β_{m} = 0$ for all $m$ .

2.1F2F4step 1.1step 1.2

Vanishing of all global sections. By step 1.1 every global section is given by its two chart expressions, and by step 1.2 those expressions have $a = 0$ and $b = 0$ ; hence $s_{0} = 0$ and $s_{1} = 0$ , so $s = 0$ by [F4]. Therefore $Γ (P_{k}^{1}, O (- 1)) = 0$ .

3.1

Failure of global generation. With $Γ (P_{k}^{1}, O (- 1)) = 0$ the evaluation morphism of [F5] is the zero morphism; since $O (- 1)$ is invertible and $P_{k}^{1} \neq \emptyset$ , it has a nonzero stalk at every point and the zero morphism is not surjective. Hence $O (- 1)$ is not globally generated, although it is invertible by [F3]. This is the standard contrast with the positive twists: $O (1)$ is generated by its two coordinate sections $x_{0}, x_{1}$ . [F3, F5, step 2.1] \qed

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Sources

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)