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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02
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  • 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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Extensions of line bundles on the projective line split after ordering

Statement

Assume the Axiom of Choice as inherited from the sheaf-cohomology suppliers. Let k be a field and put X=Pk1. Let 0→OX→M→W→0 be a short exact sequence of finite locally free OX-modules (Locally free sheaves of finite rank) in which W is isomorphic to a finite direct sum of twisting sheaves OX(ni) (Twisting sheaf on Proj) with ni≤0 for every i. Then M is isomorphic to OX directly summed with W, M≅OX⊕W. More generally, if 0→OX(b)→E→F→0 is a short exact sequence of finite locally free sheaves with F isomorphic to a direct sum of line bundles OX(bi), bi≤b, then E≅OX(b)⊕F.

Facts & Assumptions

Given: a field k, the scheme X=Pk1, and the two short exact sequences of the statement.

[F1]

X=Pk1≅Proj⁡k[x0,x1]. The twisting sheaves satisfy OX(0)=OX, each OX(d) is invertible, and the multiplication maps OX(m)⊗OX(n)→OX(m+n) are isomorphisms (Projective space is Proj of a polynomial ring, Twisting sheaf on Proj, Invertible twists for degree-one generated rings). The twist of an OX-module is F(d)=F⊗OX(d) with F(0)≅F and F(m)⊗OX(n)≅F(m+n) (Twists of a quasi-coherent sheaf); in particular (F(−c))(c)≅F for every integer c, and twisting is functorial, so it carries isomorphisms to isomorphisms (Tensor product of sheaves of modules).

[F2]

For every d∈Z one has H0(X,OX(d))=0 for d<0 and H0(X,OX(d))≅k d+1 for d≥0 (Global sections of projective twists), and H1(X,OX(d))=0 for d≥−1 (Top cohomology of projective twists). In particular H0(X,OX)≅k with unit section 1.

[F3]

A short exact sequence of OX-modules induces a long exact sequence of cohomology groups (Long exact sequence of sheaf cohomology), and H0(X,F)≅Γ(X,F) is the module of global sections of F (Degree-zero sheaf cohomology is global sections).

[F4]

(Sections as morphisms.) For an OX-module F every global section s∈Γ(X,F) determines a morphism of OX-modules s♯:OX→F by sU♯(a)=a⋅s∣U for opens U⊆X and a∈OX(U), and conversely φ↦φX(1) inverts this assignment; the maps are mutual inverses, so Γ(X,F)≅Hom⁡OX(OX,F), and for a morphism ψ:F→G one has (ψ∘s♯)X(1)=ψX(s). The construction uses only that F(U) is an OX(U)-module with restriction maps linear over the ring maps, and that the unit section 1 generates OX as an OX-module (Modules on a ringed space, Sections, restrictions, and global sections of a presheaf).

[F5]

A sequence of sheaves of modules is exact if and only if all its stalk sequences are exact; stalk formation preserves kernels, commutes with the tensor product of OX-modules and turns an invertible factor into a free module of rank one over the local ring, so tensoring with an invertible sheaf preserves exactness (A sequence of abelian sheaves is exact exactly when it is exact on every stalk, Kernel sheaves are objectwise, while cokernels and images are sheafified, The stalk of a tensor product sheaf is the tensor product of the stalks, Exact sequences of sheaves). For a finite family the direct sum of modules is a coproduct with the coordinate injections and a product with the coordinate projections, and 0→F1→F1⊕F2→F2→0 is exact (The direct sum of an indexed family of modules).

[F6]

The Axiom of Choice is inherited from the Proj and twisting-sheaf suppliers [F1] and the cohomology suppliers [F2] and [F3]; the selection made below is the selection of one lift, and the induction makes finitely many such selections (The Axiom of Choice).

Proof

technique · direct; prove by induction on the number of line-bundle summands of the quotient that an extension of a direct sum of line bundles of degrees at most $c$ by $\mathcal O_X(c)$ splits, splitting off a summand of minimal degree
1.1F1F5

The induction statement. We prove, for every r≥0, the assertion P(r): for every c∈Z and every short exact sequence 0→OX(c)→iE→pF→0 of finite locally free OX-modules with F≅⨁i=1rOX(ci) and ci≤c for all i, one has E≅OX(c)⊕F. Since X≅Proj⁡k[x0,x1] and OX(1) generates the twisting sheaves with OX(m)⊗OX(n)≅OX(m+n) by [F1], the reindexing of the ci and the canonical identifications of direct sums do not change the conclusion, and P(r) for all r and all c gives both assertions of the Statement: the first with c=0.

1.2F5

Base case. For r=0 one has F=0, so p=0 and i is an isomorphism; hence E≅OX(c)≅OX(c)⊕F.

1.3F2F3F5

Vanishing of the relevant H1. Assume r≥1 and that P(r−1) holds. Choose an enumeration with cr=min⁡ici, put F′=⨁i<rOX(ci), so that F≅F′⊕OX(cr), and let q:F→OX(cr) be the projection. Twist the given sequence by OX(−cr): by [F1] and [F5] the result is the short exact sequence 0→OX(c−cr)→E(−cr)→F(−cr)→0, with F(−cr)≅F′(−cr)⊕OX. Since ci−cr≥0 for all i<r, each summand OX(ci−cr) has vanishing H1 by [F2], and H1 vanishes on the finite direct sum F′(−cr) by induction on the number of summands: for a summand split injection 0→G→G⊕OX(cj−cr)→OX(cj−cr)→0 of [F5] the long exact sequence [F3] yields the exact portion H1(X,G)→H1(X,G⊕OX(cj−cr))→H1(X,OX(cj−cr)) with both outer groups zero. Also c−cr≥0≥−1, so H1(X,OX(c−cr))=0 by [F2].

2.1F2F3F4

A lift of the unit section. The long exact sequence [F3] of the twisted sequence begins H0(X,E(−cr))→H0(X,F(−cr))→H1(X,OX(c−cr))=0, so the first map is surjective. The projection q twisted by OX(−cr) is a surjection F(−cr)→OX whose kernel is F′(−cr); by step 1.3 and the long exact sequence of 0→F′(−cr)→F(−cr)→OX→0, the induced map on H0 is surjective. Composing the two surjections there is s∈H0(X,E(−cr)) with image the unit section 1∈H0(X,OX)≅k. By [F4] the section s corresponds to a morphism σ=s♯:OX→E(−cr) with p′∘σ=id, where p′:E(−cr)→OX is the composite q∘p twisted.

3.1F5step 2.1

Splitting off the minimal summand. Let K=ker⁡p′⊆E(−cr). The morphism Ψ:K⊕OX→E(−cr) defined on the summands by the inclusion of K and by σ is an isomorphism. Indeed, on stalks at a point x the argument is the elementary module argument: if k+σ(l)=0 with k∈Kx, l∈OX,x, then applying px′ gives l=0 and then k=0, so Ψx is injective; and for m∈E(−cr)x one has m−σ(px′(m))∈Kx, so Ψx is surjective. By [F5] a morphism of sheaves that is stalkwise bijective is an isomorphism. Hence E(−cr)≅K⊕OX.

4.1F5step 1.3step 3.1

The complement is again an extension of the same shape. Let r′:F(−cr)→F′(−cr) be the projection of step 1.3. The sequence 0→OX(c−cr)→K→F′(−cr)→0 is exact, where the first map is the restriction of the inclusion of OX(c−cr) and the second is the restriction of r′∘p to K. Stalks at x: the sequence 0→OX(c−cr)x→E(−cr)x→F(−cr)x→0 is exact and F(−cr)x=F′(−cr)x⊕OX,x; an element of F′(−cr)x lifted to E(−cr)x can be corrected by an element with the same OX,x-component to lie in Kx, because E(−cr)x→OX,x is surjective, so Kx→F′(−cr)x is surjective; and its kernel is the kernel of E(−cr)x→F(−cr)x, namely OX(c−cr)x, the inclusion being injective. Exactness of the displayed sequence follows from [F5]. Also K is finite locally free: near any point, trivialize the kernel line bundle and the finite locally free quotient F′(−cr), lift the finitely many quotient basis germs to sections of K, and shrink so that their images equal the basis sections. These lifts define a local splitting. Together with a frame of the kernel they identify K locally with a finite free sheaf, as required for P(r−1).

5.1F1step 1.3step 3.1step 4.1

Induction step. In the exact sequence of step 4.1 the quotient F′(−cr)≅⨁i<rOX(ci−cr) has r−1 summands. Since cr=min⁡ici and ci≤c, their exponents satisfy 0≤ci−cr≤c−cr. Thus P(r−1), applied with kernel exponent c−cr, gives K≅OX(c−cr)⊕F′(−cr). Combining with step 3.1 and twisting back by OX(cr), which preserves direct sums and isomorphisms by [F1] and inverts (−cr), E≅E(−cr)(cr)≅K(cr)⊕OX(cr)≅OX(c)⊕F′⊕OX(cr)≅OX(c)⊕F. This proves P(r).

6.1F6step 1.1step 1.2step 2.1step 5.1∎

Conclusion. By steps 1.2 and 5.1 the assertion P(r) holds for every r≥0 and every c∈Z; taking c=0 gives M≅OX⊕W for the first sequence of the Statement, and the general assignment c=b gives E≅OX(b)⊕F whenever all bi≤b. Every selection made was the choice of one lift of a specified element in step 2.1 and finitely many such selections occur, so the Axiom of Choice enters through the Proj, twisting-sheaf and cohomology suppliers recorded in [F6].

Depends on

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