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Equality in affine K zero and stable isomorphism

PDF p. 12: Cancellation reduction in Theorem A

Statement

Assume the Axiom of Choice. Let $X = Spec A$ be a noetherian affine scheme, where $A$ is a commutative ring with unit, and let $E$ and $F$ be vector bundles on $X$ . Their classes in the vector-bundle Grothendieck group $K^{0} (X)$ are equal if and only if there is an integer $N \geq 0$ and an isomorphism $E \oplus O_{X}^{N} ≅ F \oplus O_{X}^{N}$ .

A vector bundle is a finite locally free $O_{X}$ -module: each point has a neighbourhood on which the module is isomorphic to a finite direct sum of copies of $O_{X}$ . The displayed isomorphism is called a stable isomorphism. The group $K^{0} (X)$ is the abelian group generated by isomorphism classes $[V]$ of vector bundles, with the relation $[V] = [V^{'}] + [V^{''}]$ for every short exact sequence $0 \to V^{'} \to V \to V^{''} \to 0$ of vector bundles. In particular, it is the vector-bundle group, not the Grothendieck group of coherent sheaves.

Facts & Assumptions

Given: The Axiom of Choice, a noetherian affine scheme $X = Spec A$ , and vector bundles $E, F$ on $X$ .

Goal: Prove that $[E] = [F]$ in $K^{0} (X)$ exactly when $E$ and $F$ are stably isomorphic as stated.

[A1]

The assumed Axiom of Choice is used through the affine bundle–module equivalence and splitting in [F1] and the Grothendieck-group convention in [F3]. The complement construction in [F2] itself uses only finite choices. The Axiom of Choice

[F1]

Under the Axiom of Choice, every short exact sequence of vector bundles on an affine scheme splits. Associated sheaf and global sections give an equivalence between vector bundles on $X$ and finite projective $A$ -modules, preserving finite direct sums; the sheaf associated to $A^{N}$ is $O_{X}^{N}$ . A finite projective module is a finitely generated module with the lifting property for maps through surjective module homomorphisms. Affine bundles and finite projective modules

[F2]

Every finite projective $A$ -module $P$ has a finite projective complement $Q$ with $P \oplus Q ≅ A^{N}$ for some $N \geq 0$ . Finite projective complements

[F3]

Since $X$ is locally noetherian, the vector-bundle group has the generators and exact-sequence relations in the statement. Its bundle isomorphism classes form a set. Grothendieck groups of coherent sheaves and of vector bundles on a scheme

Proof

technique · Construct the group completion of the direct-sum monoid and then replace a common bundle summand by a free one

1.1F3construct

Let $M$ be the set of isomorphism classes of vector bundles, with addition induced by direct sum and zero the class of the zero bundle. Direct sum is associative and commutative up to isomorphism, so $M$ is a commutative monoid. On $M \times M$ define $(a, b) \sim (c, d)$ when there is $t \in M$ such that $a + d + t = b + c + t$ . This is an equivalence relation: reflexivity and symmetry are immediate; for transitivity, if $a + d + t = b + c + t$ and $c + f + u = d + e + u$ , adding $f + u$ to the first equality and $b + t$ to the second yields $a + f + (d + t + u) = b + e + (d + t + u)$ . Componentwise addition respects the relation. Thus the quotient is an abelian group, with zero represented by $(0, 0)$ and inverse of $(a, b)$ represented by $(b, a)$ ; indeed $(a + b, a + b) \sim (0, 0)$ .

2.1A1F1F3step 1.1

The assignment sending a bundle class $a$ to the class of $(a, 0)$ is additive. By [F1], each short exact sequence in the presentation of $K^{0} (X)$ splits, so its middle bundle is isomorphic to the direct sum of its ends. Consequently this assignment respects every relation in [F3] and induces a homomorphism from $K^{0} (X)$ to the group constructed in step 1.1. If $[E] = [F]$ in $K^{0} (X)$ , their images are equal, so by the definition of $\sim$ there is a vector bundle $H$ with $E \oplus H ≅ F \oplus H$ .

3.1F1F2step 2.1construct

By [F1], $H$ corresponds to a finite projective $A$ -module $P$ . Choose $Q$ and $N$ as in [F2], and let $H^{'}$ be the vector bundle associated to $Q$ . The direct-sum compatibility in [F1] gives $H \oplus H^{'} ≅ O_{X}^{N}$ . Adding $H^{'}$ to both sides of the isomorphism in step 2.1 and using associativity therefore gives $E \oplus O_{X}^{N} ≅ F \oplus O_{X}^{N}$ .

4.1F3step 3.1algebra∎

Conversely, if such an isomorphism holds, the split exact-sequence relations in [F3] give $[E] + [O_{X}^{N}] = [F] + [O_{X}^{N}]$ in the abelian group $K^{0} (X)$ . Adding the inverse of $[O_{X}^{N}]$ proves $[E] = [F]$ . Together with step 3.1 this establishes the equivalence.

Depends on

Used by · 4

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