nLab
projective module

Definition

Definition

For $R$ a ring, a projective $R$-module is a projective object in the category $R$Mod.

Proposition

A module $N$ is projective precisely if the hom functor

$${\mathrm{Hom}}_{R\mathrm{Mod}}(N,-):R\mathrm{Mod}\to \mathrm{Ab}$$Hom_{R Mod}(N, - ) : R Mod \to Ab

out of it is an exact functor.

Properties

Existence of enough projective modules

Lemma

Assuming the axiom of choice, a free module $N\simeq R^{(S)}$ is projective.

Proof

Explicitly: if $S\in \mathrm{Set}$ 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 $\mid S\mid$ 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

$$\begin{array}{ccc}& & U(\tilde{N})\\ & {}^{\tilde{f}}\nearrow & \downarrow \\ S& \stackrel{f}{\to }& U(N)\end{array}.$$\array{ && U(\tilde N) \\ & {}^{\tilde f} \nearrow & \downarrow \\ S &\stackrel{f}{\to}& U(N) } \,.

By adjunction these are equivalently lifts of module homomorphisms

$$\begin{array}{ccc}& & \tilde{N}\\ & \nearrow & \downarrow \\ R^{(S)}& \stackrel{}{\to }& N\end{array}.$$\array{ && \tilde N \\ & \nearrow & \downarrow \\ R^{(S)} &\stackrel{}{\to}& N } \,.

Proposition

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.

Proof

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

$$ϵ:F(U(N))\to N$$\epsilon : F(U(N)) \to N

of the free/forgetful-adjunction $(F⊣U)$ 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).

Explicit characterizations

We discuss the more explicit characterization of projective modules as direct summands of free modules.

Lemma

If $N\in R\mathrm{Mod}$ is a direct summand of a free module, hence if there is $N\prime \in R\mathrm{Mod}$ and $S\in \mathrm{Set}$ such that

$$R^{(S)}\simeq N\oplus N\prime ,$$R^{(S)} \simeq N \oplus N' \,,

then $N$ is a projective module.

Proof

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

$$\begin{array}{ccc}& & \tilde{K}\\ & {}^{\tilde{f}}\nearrow & \downarrow \\ N& \stackrel{f}{\to }& K\end{array}.$$\array{ && \tilde K \\ & {}^{\mathllap{\tilde f}}\nearrow & \downarrow \\ N &\stackrel{f}{\to}& K } \,.

By definition of direct sum we can factor the identity on $N$ as

$${\mathrm{id}}_N:N\to N\oplus N\prime \to N.$$id_N : N \to N \oplus N' \to N \,.

Since $N\oplus N\prime$ is free by assumption, and hence projective by lemma 1, there is a lift $\hat{f}$ in

$$\begin{array}{ccccc}& & & & \tilde{K}\\ & & & {}^{\hat{f}}\nearrow & \downarrow \\ N& \to & N\oplus N\prime & \to & K\end{array}.$$\array{ && && \tilde K \\ && & {}^{\mathllap{\hat f}}\nearrow & \downarrow \\ N &\to& N \oplus N' &\to& K } \,.

Hence $\tilde{f}:N\to N\oplus N\prime \stackrel{\hat{f}}{\to }\tilde{K}$ is a lift of $f$.

Proposition

An $R$-module $N$ is projective precisely if it is the direct summand of a free module.

Proof

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

$$ϵ:F(U(N))\to N$$\epsilon : F(U(N)) \to N

of the free/forgetful-adjunction $(F⊣U)$ is an epimorphism. Thefore if $N$ is projective, there is a section $s$ of $ϵ$. This exhibits $N$ as a direct summand of $F(U(N))$.

In some cases this can be further strenghened:

Proposition

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.

Relation to projective resolutions of chain complexes

Definition

For $N\in R\mathrm{Mod}$ a projective resolution of $N$ is a chain complex $(QN{)}_{•}\in {\mathrm{Ch}}_{•}(R\mathrm{Mod})$ equipped with a chain map

$$QN\to N$$Q N \to N

(with $N$ regarded as a complex concentrated in degree 0) such that

1. this morphism is a quasi-isomorphism (this is what makes it a resolution), which is equivalent to

   $$\cdots \to (QN{)}_1\to (QN{)}_0\to N$$\cdots \to (Q N)_1 \to (Q N)_0 \to N

   being an exact sequence;

2. all whose entries $(QN{)}_n$ are projective modules.

Remark

This means precisely that $QN\to N$ 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.

Proposition

Every $R$-module has a projective resolution.

See at projective resolution.

Examples

Proposition

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.

Related concepts

• projective object, projective presentation, projective cover, projective resolution

• injective object, injective presentation, injective envelope, injective resolution

  • injective module

• free object, free resolution

  • free module, locally free module

• flat object, flat resolution

  • flat module

• free module $\Rightarrow$ projective module $\Rightarrow$ flat module $\Rightarrow$ torsion-free module

References

Lecture notes include

section 2.2 of the textbook

• Charles Weibel, An Introduction to Homological Algebra

or

• Projective modules, Presentations and resolutions (pdf)

or

• Thomas Lam, chpater 6 (pdf)

Original articles include

• Irving Kaplansky, Projective modules, The Annals of Mathematics Second Series, Vol. 68, No. 2 (Sep., 1958), pp. 372-377 (JSTOR)