CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A submodule of a free module of finite rank over a PID is free of no larger rank

Statement

If $M$ is free of finite rank $n$ over a PID and $N \leq M$ , then $N$ is free of a uniquely determined finite rank $r \leq n$ .

Facts & Assumptions

Given: Invariant basis number for nonzero commutative rings (Every nonzero commutative ring has invariant basis number for finite bases).

[L1]

There is a basis $a_{1} e_{1}, \dots, a_{r} e_{r}$ of the submodule with $r \leq n$ (Simultaneous bases for a submodule of a finite free module over a PID).

Proof

technique · direct

1.1

The family in [L1] is an $R$ -basis of $N$ , so $N$ is free and has a basis of length $r \leq n$ . The zero submodule has the empty basis, and ambient rank zero forces $r = 0$ .

2.1

Invariant basis number makes the finite basis length unique, so $r$ is the rank of $N$ . The endpoints $r = 0$ and $r = n$ , including $N = M$ , are permitted.

step 1.1given∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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, Theorem 2.2 (standard reference, not scraped)
M. Brussel, Finitely Generated Modules over a PID, Theorem 1.0.1 (standard reference, not scraped)