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 an $R$-module $P$, the following statements are equivalent:
$P$ is finite locally free in that there exists a partition $1 = \sum_i f_i \in R$ such that the localized modules $P[f_i^{-1}]$ are finite free modules over $R[f_i^{-1}]$.
$P$ is finitely generated and projective.
$P$ is a dualizable object in the category of $R$-modules (equipped with the tensor product as monoidal structure).
There exist elements $x_1,\ldots,x_n \in P$ and linear forms $\vartheta_1,\ldots,\vartheta_n \in Hom(P,R)$ such that $x = \sum_i \vartheta_i(x) x_i$ for all $x \in P$.
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 $R$ is Noetherian or that $P$ satisfies some finiteness condition.
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