(also nonabelian homological algebra)
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For a ring, a projective -module is a projective object in the category Mod of -modules.
Hence an -module is projective precisely if for all diagrams of -module homomorphisms of the form
where the vertical morphism is an epimorphism, there exists a lift, hence a morphism making a commuting diagram of the form
An -module is projective (def. ) 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 if
is any short exact sequence, hence if is any epimorphism and its kernel inclusion, then is an epimorphism, hence that for any element ,
there exists a such that , hence that
This is manifestly the condition that is projective.
Assuming the axiom of choice, a free module is projective.
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, 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
Assuming the axiom of choice, the category Mod has enough projectives: for every -module there exists an epimorphism where is a projective module.
Let be the free module on the set underlying . By lemma this is a projective module.
The counit
of the free/forgetful-adjunction is an epimorphism.
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 an and such that
then is a projective module.
Let be a surjective homomorphism of modules (which is equivalently an epimorphism in the category of modules) and a homomorphism. We need to show that there is a lift in
By the definition of the direct sum, we can factor the identity on as
Since is free by assumption, and hence projective by lemma , there is a lift in
Hence is a lift of .
An -module is projective precisely if it is a direct summand of a free module.
By lemma , 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. . The counit
of the free/forgetful-adjunction is an epimorphism. Therefore, 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:
If the ring is a principal ideal domain (in particular the integers), then every projective -module is free.
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 finitely generated free modules over .
is finitely generated and projective.
is a dualizable object in the category of -modules (equipped with the tensor product as its 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.
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.
This means precisely that is a cofibrant resolution with respect to the standard model structure on chain complexes (see here) for which the fibrations are the positive-degreewise epimorphisms. Notice that, in this model structure, every object is fibrant, so cofibrant resolutions are the only resolutions that need to be considered.
Every -module has a projective resolution.
Assuming the axiom of choice, by the basis theorem every module over a field is a free module, and hence, in particular, every module over a field is a projective module (by prop. ).
If is the ring of integers , or a field , or a division ring, then every projective -module is a free -module.
projective object, projective presentation, projective cover, projective resolution
injective object, injective presentation, injective envelope, injective resolution
flat object, flat resolution
free module projective module flat module torsion-free module
Early discussion:
Lecture notes include
Charles Weibel, An Introduction to Homological Algebra, section 2.2
Projective modules, Presentations and resolutions (pdf)
Thomas Lam, chapter 6 (pdf)
Original articles include
Last revised on August 19, 2025 at 08:11:03. See the history of this page for a list of all contributions to it.