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Iterating a conditional free-summand criterion

PDF p. 1: Serre splitting theorem

Statement

Assume the Axiom of Choice. Let $A$ be a nonzero commutative ring and let $d$ be a nonnegative integer. Suppose every finite projective $A$ -module of constant rank greater than $d$ has a free rank-one direct summand. Then every finite projective $A$ -module $P$ of constant rank $r > d$ is isomorphic to $A^{r - d} \oplus Q$ for a finite projective module $Q$ of constant rank $d$ . The criterion is an explicit hypothesis of this lemma.

Facts & Assumptions

Given: $A$ , $d$ , $P$ , $r$ , Choice, and the conditional free-summand criterion in the statement.

Goal: Obtain the displayed decomposition and rank.

[A1]

Choice is assumed to invoke the supplied split-complement result, including its associated-bundle clause; the finite iteration introduces no further choice use. The Axiom of Choice

[F1]

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

Proof

1.1F1given

Apply the assumed criterion to $P$ to split off $A$ . By [F1], its complement is finite projective of rank $r - 1$ . If this rank is still greater than $d$ , the same criterion applies again.

2.1F1givenstep 1.1construct∎

Repeat the preceding finite operation exactly $r - d$ times. At the $j$ th application, the current rank is $r - j + 1 > d$ ; [F1] supplies a finite projective complement of rank $r - j$ . Associativity of finite direct sums gives $P ≅ A^{r - d} \oplus Q$ with $Q$ of rank $d$ .

Depends on

Used by · 1

Serre splitting above the dimension

Dependency tree · 3 levels

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