A module is projective precisely if the hom functor
out of it is an exact functor.
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.
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)