Every submodule of a free module is free
Over a principal ideal domain the theory becomes as explicit as linear algebra. Submodules of free modules are free, so projective and free coincide, and every module has a projective resolution of length at most one. Finitely generated modules decompose into a free part and a torsion part with a canonical list of invariant factors. This is why abelian groups — modules over ℤ — are the standard proving ground for the whole subject.
Learning objectives
- State the structure theorem for finitely generated modules over a PID.
- Explain why projective and free coincide over a PID.
- Construct the canonical length-one projective resolution.
- Relate invariant factors to elementary divisors.
- Explain the consequence that Extn and Torn vanish for n > 1.
Section 01Submodules of free modules
Over a PID, every submodule of a free module is free, with rank at most that of the ambient module. For finitely generated modules the proof is a straightforward induction; the general case needs transfinite methods but the statement is the same.
First, projective equals free, since a direct summand of a free module is a submodule of it. Second, the kernel of any map from a free module is free — so a projective resolution can be stopped after one step. Both facts fail over general rings, and recovering them is what much of the later theory is about.
Section 02The structure theorem
Every finitely generated module over a PID R decomposes as
with r the rank and the di the invariant factors, unique up to units. Splitting each cyclic factor into prime powers gives the elementary divisors — the same module, presented differently.
| Feature | Meaning |
|---|---|
| r > 0 | The module has a free part; it is not torsion |
| k = 0 | The module is free of rank r |
| r = 0 | The module is torsion — finite when R = ℤ |
| d1 a unit | That factor is trivial and is discarded |
| All di prime powers of one prime | The module is p-primary |
Presenting M by generators and relations gives a matrix over R; its Smith normal form has the invariant factors on the diagonal. The structure theorem and the normal form are two statements of one algorithm.
Section 03Short resolutions and vanishing
- Choose a surjection ε: F0 ↠ M with F0 free.
- Set F1 = ker ε. Free, because submodules of free modules over a PID are free.
- The sequence 0 → F1 → F0 → M → 0 is a projective resolution of length 1.
- Hence Extn(M, −) = 0 and Torn(M, −) = 0 for all n ≥ 2.
Over ℤ everything above holds, so Ext and Tor reduce to a single group each and are computable by hand. Every general theorem in this subject should be tested against ℤ first — if it fails there, it fails everywhere.
ReferenceFrequently asked questions
Does the structure theorem need finite generation?
Yes. Infinitely generated modules over ℤ can be complicated — the additive group of the rationals is torsion-free but not free, and divisible groups behave quite differently. The clean decomposition is a finitely generated phenomenon.
Are invariant factors or elementary divisors preferable?
Invariant factors form a divisibility chain and are what Smith normal form produces directly; elementary divisors separate the primes and are better for questions localised at one prime. They carry identical information.
What is a hereditary ring?
One in which every submodule of a projective module is projective, equivalently of global dimension at most 1. PIDs and Dedekind domains are the standard examples, and hereditary is exactly the hypothesis that makes higher Ext and Tor vanish.
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