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Every finitely generated torsion-free module over a PID is free

Statement

Every finitely generated torsion-free module over a principal ideal domain is finite free.

Facts & Assumptions

Given: A finitely generated torsion-free $R$ -module $M$ , where torsion-free means $Tor (M) = {0}$ (Annihilators, torsion elements and the torsion subset of a module).

[L1]

Every finitely generated PID module is a finite free module direct-summed with cyclic torsion quotients (Invariant-factor decomposition of a finitely generated module over a PID).

Proof

technique · direct

1.1

In the decomposition from [L1], every nonzero quotient $R / (a_{i})$ with $a_{i} \neq 0$ consists of torsion elements, whereas the free summand is torsion-free because $R$ is a domain. Torsion-freeness therefore forces every cyclic quotient summand to be zero.

L1given

2.1

Only the finite free summand remains, so $M$ is finite free. The zero module is the free module on the empty basis, and unit invariant factors already give zero summands.

step 1.1∎

Depends on

Used by

FALSE: every torsion-free module over a PID is free False statement

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Modules over a PID, Corollary 2.6 (standard reference, not scraped)