For a ring, a projective -module is a projective object in the category Mod.
Hence an -module is projective precisely if for all diagrams of -module homomorphisms of the form
there exists a morphism making a commuting diagram of the form
An -module is projective (def. 1) precisely if the hom functor
(out of ) is an exact functor.
The hom-functor in question is a left exact functor for all , hence we need to show that it is a right exact functor precisely if is projective.
That is right exact means equivalently that for
any short exact sequence, hence for any epimorphism and its kernel inclusion, then is an epimorphism, hence that for any element ,
there exists such that , hence that
This is manifestly the condition that is projective.
Existence of enough projective modules
Explicitly: if and is the free module on , then a module homomorphism is specified equivalently by a function from to the underlying set of , which can be thought of as specifying the images of the unit elements in of the copies of .
Accordingly then for an epimorphism, the underlying function is an epimorphism, and the axiom of choice in Set says that we have all lifts in
By adjunction these are equivalently lifts of module homomorphisms
Actually, the full axiom of choice is not necessary here; it is enough to have the presentation axiom, which states the category of sets has enough projectives (whereas the axiom of choice states that every set is projective). Then we can replace above by a projective set , giving an epimorphism (and is projective).
We discuss the more explicit characterization of projective modules as direct summands of free modules.
If is a direct summand of a free module, hence if there is and such that
then is a projective module.
Let be a surjective homomorphism of modules and a homomorphism. We need to show that there is a lift in
By definition of direct sum we can factor the identity on as
Since is free by assumption, and hence projective by lemma 1, there is a lift in
Hence is a lift of .
By lemma 2 if is a direct summand then it is projective. So we need to show the converse.
Let be the free module on the set underlying as in the proof of prop. 2. The counit
of the free/forgetful-adjunction is an epimorphism. Thefore if is projective, there is a section of . This exhibits as a direct summand of .
This proposition is often stated more explicitly as the existence of a dual basis, see there.
In some cases this can be further strengthened:
The details are discussed at pid - Structure theory of modules.
For an -module , the following statements are equivalent:
is finite locally free in that there exists a partition such that the localized modules are finite free modules over .
is finitely generated and projective.
is a dualizable object in the category of -modules (equipped with the tensor product as monoidal structure).
There exist elements and linear forms such that for all .
The equivalence of 2., 3., and 4. is mostly formal. For the equivalence with 1., see this math.SE discussion for good references. Note that the equivalences are true without assuming that is Noetherian or that satisfies some finiteness condition.
Relation to projective resolutions of chain complexes
For a projective resolution of is a chain complex equipped with a chain map
(with regarded as a complex concentrated in degree 0) such that
this morphism is a quasi-isomorphism (this is what makes it a resolution), which is equivalent to
being an exact sequence;
all whose entries are projective modules.
Every -module has a projective resolution.
See at projective resolution.
If is the integers , or a field , or a division ring, then every projective -module is already a free -module.
Lecture notes include
Original articles include
- Irving Kaplansky, Projective modules, The Annals of Mathematics Second Series, Vol. 68, No. 2 (Sep., 1958), pp. 372-377 (JSTOR)