Every surjection onto a projective splits
A free module has a basis, so a homomorphism out of it is determined by arbitrary choices on the basis elements — which means any map into a quotient lifts. A projective module keeps the lifting property while giving up the basis, and it is exactly a direct summand of a free module. That is all resolutions need: every module is a quotient of a free one, so projective resolutions always exist, and Hom(P, −) is exact precisely when P is projective.
Learning objectives
- Define free modules by their universal property.
- State the lifting property defining projectivity.
- Prove that projective is equivalent to being a direct summand of a free module.
- Construct a projective resolution of an arbitrary module.
- Give an example of a projective module that is not free.
Section 01Free modules
The free module on a set S is the direct sum of copies of Λ indexed by S. Its universal property is that a homomorphism out of it is exactly a function from S to the target:
Every module is a quotient of a free module — take S to be any generating set, or the underlying set of the module itself. This single fact is what guarantees that projective resolutions exist for every module over every ring.
Defining a map out of a free module requires no compatibility checking whatsoever: pick images for the basis and extend. Every lifting argument in the subject ultimately reduces to this freedom.
Section 02Projectivity
P is projective when every diagram with a surjection B ↠ C and a map P → C admits a lift P → B.
- P is projective: every map out of P lifts along every surjection.
- Every short exact sequence 0 → A → B → P → 0 splits. Apply the lifting property to the identity of P.
- P is a direct summand of a free module. Take a free F with F ↠ P and split it.
- Hom(P, −) is an exact functor.
- P satisfies the dual basis lemma: there are elements and functionals reproducing each element as a finite sum.
- All five conditions are equivalent.
Over ℤ/6ℤ, the ideal generated by 2 is a direct summand and hence projective, but it is not free — it has the wrong number of elements. Over a Dedekind domain, non-principal ideals are projective and non-free, and the class group measures precisely how many such modules exist.
Section 03Rings where the distinction collapses
| Ring | Projective ⇒ free? | Note |
|---|---|---|
| Field | Yes | Every module is free — a vector space |
| Principal ideal domain | Yes, for finitely generated | Submodules of free are free |
| Local ring | Yes | Kaplansky's theorem, even without finite generation |
| Polynomial ring over a field | Yes | Quillen–Suslin, resolving Serre's problem |
| Dedekind domain | No | Ideals are projective; class group counts the failure |
| ℤ/nℤ for composite n | No | Idempotents split off summands |
| Group ring ℤ[G] | Generally no | Relevant to modular representation theory |
Section 04Projective resolutions
- Choose a surjection ε: P0 ↠ M with P0 projective — a free module always works.
- Set K0 = ker ε, the first syzygy.
- Choose a surjection P1 ↠ K0 and compose with the inclusion to get d1: P1 → P0.
- Repeat on ker d1, and so on. The process never obstructs, because every module is a quotient of a free one.
- The result … → P2 → P1 → P0 → M → 0 is exact.
M itself may behave badly under a functor, but each Pn behaves perfectly. Applying the functor to the resolution rather than to M, and taking homology, produces invariants that agree with the functor in degree 0 and record its failure above.
ReferenceFrequently asked questions
Is every projective module a direct sum of finitely generated ones?
Over many rings yes, and Kaplansky proved that any projective module is a direct sum of countably generated ones. Over an arbitrary ring the structure can be complicated, and this is one reason finitely generated hypotheses appear so often.
Why not always use free resolutions?
They exist and are perfectly valid, and many texts use them exclusively. Projective resolutions are preferred because they can be much smaller, and because the projective property is what the proofs actually use — freeness is a convenience, not a necessity.
What is a syzygy?
The kernel at each stage of a resolution — the module of relations among the chosen generators. Iterating gives higher syzygies, and Hilbert's syzygy theorem bounds how long the process continues over a polynomial ring.
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