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Serre splitting above the dimension

PDF p. 1: Serre splitting theorem

Statement

Assume the Axiom of Choice. Let $X$ be a nonempty noetherian affine scheme of finite dimension $d \geq 0$ , and let $E$ be a vector bundle of constant rank $r > d$ . Then $E ≅ O_{X} \oplus E_{1}$ for a vector bundle $E_{1}$ of rank $r - 1$ . More precisely, iteration gives $E ≅ O_{X}^{\oplus (r - d)} \oplus F$ for a vector bundle $F$ of rank $d$ , hence at most $d$ . On the empty affine scheme the corresponding splitting assertion is immediate because all sheaves of modules are zero.

A vector bundle means a finite locally free sheaf, and constant rank $r$ means that its local free rank is $r$ at every point. The dimension of $X$ is the supremum of the lengths of strict chains of irreducible closed subsets.

Facts & Assumptions

Given: $X$ , $d$ , $E$ , $r$ , and the Axiom of Choice as in the statement.

Goal: Split off a trivial line and iterate down to rank $d$ .

[A1]

The assumed Axiom of Choice is used through the affine bundle–module equivalence in [F1] and its finite-projective and rank suppliers [F2]–[F4]. The iteration in [F6] is finite. The Axiom of Choice

[F1]

The affine bundle–module equivalence preserves direct sums and rank. Affine bundles and finite projective modules

[F2]

A finite projective module is finitely presented. Finite projective modules are finitely presented

[F3]

For a nonzero noetherian ring of finite dimension $d$ , the maximal spectrum is noetherian and has dimension at most $d$ . Dimension of the maximal spectrum

[F4]

The global sections of a rank- $r$ affine bundle form a finite projective module whose localizations at primes are free of rank $r$ . Local ranks in the affine bundle correspondence

[F5]

A free rank-one complement of a finite projective constant-rank module is finite projective of rank one less. Projectivity and rank of a free-summand complement

[F6]

A criterion splitting a free line from every finite projective module of rank greater than $d$ iterates down to rank $d$ . Iterating a conditional free-summand criterion

[F7]

The imported Serre module theorem applies to a finitely presented module when the maximal spectrum is noetherian of finite dimension $e$ and every maximal localization has a free direct summand of rank greater than $e$ . Serre free-summand theorem for modules

Proof

1.1F3given

Write $X = Spec A$ . Since $X$ is nonempty, $A$ is nonzero. The ring $A$ is noetherian and its Krull dimension is $d$ : irreducible closed subsets of an affine spectrum are the sets $V (p)$ for prime ideals, with inclusion reversed. By [F3], its maximal spectrum is noetherian of some finite dimension $e \leq d$ .

2.1F2F7givenstep 1.1

Let $P$ be any finite projective $A$ -module of constant rank $s > d$ . It is finitely presented by [F2]. At each maximal ideal its localization is free of rank $s$ , so it has a free direct summand of rank $s > e$ . The imported theorem [F7] therefore gives $P ≅ A \oplus P^{'}$ . Thus the hypothesis of [F6] holds for this ring and $d$ .

3.1A1F1F4F5step 2.1

The module $P = Γ (X, E)$ is finite projective of constant rank $r$ by [F4]. Applying step 2.1 gives a decomposition $P ≅ A \oplus P^{'}$ ; [F5] makes $P^{'}$ finite projective of rank $r - 1$ . The equivalence [F1] turns this into $E ≅ O_{X} \oplus E_{1}$ with $E_{1}$ a rank- $(r - 1)$ bundle.

4.1F1F6step 2.1step 3.1∎

By [F6] and step 2.1, $P ≅ A^{r - d} \oplus Q$ with $Q$ finite projective of rank $d$ . Applying [F1] yields the asserted bundle decomposition. On the empty scheme every sheaf of modules is the zero sheaf, so the stated empty-case splitting is immediate.

Depends on

Used by · 4

Dependency tree · 3 levels

Arrows run from prerequisites to the items that use them. External source nodes link directly to the cited text. The view shows up to three dependency steps; open a prerequisite to continue.

Sources

Jean Fasel and Maria Yakerson, Vector bundles on singular surfaces and threefolds · PDF p. 1, introduction; PDF p. 12, application citing Ser58 §4 Theorem 1
The Stacks Project, More on Algebra, Section 15.129: Splitting off a free module · Theorem 15.129.5; imported through thm-author-3af31ead36fb-serre-module-input