and
nonabelian homological algebra
For $R$ a ring, a projective $R$-module is a projective object in the category $R$Mod.
A module $N$ is projective precisely if the hom functor
out of it is an exact functor.
Assuming the axiom of choice, a free module $N \simeq R^{(S)}$ is projective.
Explicitly: if $S \in Set$ and $F(S) = R^{(S)}$ is the free module on $S$, then a module homomorphism $F(S) \to N$ is specified equivalently by a function $f : S \to U(N)$ from $S$ to the underlying set of $N$, which can be thought of as specifying the images of the unit elements in $R^{(S)} \simeq \oplus_{s \in S} R$ of the ${\vert S\vert}$ copies of $R$.
Accordingly then for $\tilde N \to N$ an epimorphism, the underlying function $U(\tilde N) \to U(N)$ is an epimorphism, and the axiom of choice in Set says that we have all lifts $\tilde f$ in
By adjunction these are equivalently lifts of module homomorphisms
Assuming the axiom of choice, the category $R$Mod has enough projectives: for every $R$-module $N$ there exists an epimorphism $\tilde N \to N$ where $\tilde N$ is a projective module.
Let $F(U(N))$ be the free module on the set $U(N)$ underlying $N$. By lemma 1 this is a projective module.
The counit
of the free/forgetful-adjunction $(F \dashv U)$ 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 $U(N)$ above by a projective set $P \twoheadrightarrow U(N)$, giving an epimorphism $F(P) \twoheadrightarrow F(U(N)) \twoheadrightarrow N$ (and $F(P)$ is projective).
We discuss the more explicit characterization of projective modules as direct summands of free modules.
If $N \in R Mod$ is a direct summand of a free module, hence if there is $N' \in R Mod$ and $S \in Set$ such that
then $N$ is a projective module.
Let $\tilde K \to K$ be a surjective homomorphism of modules and $f : N \to K$ a homomorphism. We need to show that there is a lift $\tilde f$ in
By definition of direct sum we can factor the identity? on $N$ as
Since $N \oplus N'$ is free by assumption, and hence projective by lemma 1, there is a lift $\hat f$ in
Hence $\tilde f : N \to N \oplus N' \stackrel{\hat f}{\to} \tilde K$ is a lift of $f$.
An $R$-module $N$ is projective precisely if it is the direct summand of a free module.
By lemma 2 if $N$ is a direct summand then it is projective. So we need to show the converse.
Let $F(U(N))$ be the free module on the set $U(N)$ underlying $N$ as in the proof of prop. 2. The counit
of the free/forgetful-adjunction $(F \dashv U)$ is an epimorphism. Thefore if $N$ is projective, there is a section $s$ of $\epsilon$. This exhibits $N$ as a direct summand of $F(U(N))$.
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 $R$ is a principal ideal domain (in particular $R = \mathbb{Z}$ the integers), then every projective $R$-module is free.
The details are discussed at pid - Structure theory of modules.
For $N \in R Mod$ a projective resolution of $N$ is a chain complex $(Q N)_\bullet \in Ch_\bullet(R Mod)$ equipped with a chain map
(with $N$ 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 $(Q N)_n$ are projective modules.
This means precisely that $Q N \to N$ is an 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 that cofibrant resolutions are the only resolutions that need to be considered.
Every $R$-module has a projective resolution.
See at projective resolution.
If $R$ is the integers $\mathbb{Z}$, or a field $k$, or a division ring, then every projective $R$-module is already a free $R$-module.
projective object, projective presentation, projective cover, projective resolution
injective object, injective presentation, injective envelope, injective resolution
flat object, flat resolution
free module $\Rightarrow$ projective module $\Rightarrow$ flat module $\Rightarrow$ torsion-free module
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