# nLab projective module

### Context

#### Homological algebra

homological algebra

and

nonabelian homological algebra

diagram chasing

# Contents

## 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

${\mathrm{Hom}}_{R\mathrm{Mod}}\left(N,-\right):R\mathrm{Mod}\to \mathrm{Ab}$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 $N\simeq {R}^{\left(S\right)}$ is projective.

###### Proof

Explicitly: if $S\in \mathrm{Set}$ and $F\left(S\right)={R}^{\left(S\right)}$ is the free module on $S$, then a module homomorphism $F\left(S\right)\to N$ is specified equivalently by a function $f:S\to U\left(N\right)$ from $S$ to the underlying set of $N$, which can be thought of as specifying the images of the unit elements in ${R}^{\left(S\right)}\simeq {\oplus }_{s\in S}R$ of the $\mid S\mid$ copies of $R$.

Accordingly then for $\stackrel{˜}{N}\to N$ an epimorphism, the underlying function $U\left(\stackrel{˜}{N}\right)\to U\left(N\right)$ is an epimorphism, and the axiom of choice in Set says that we have all lifts $\stackrel{˜}{f}$ in

$\begin{array}{ccc}& & U\left(\stackrel{˜}{N}\right)\\ & {}^{\stackrel{˜}{f}}↗& ↓\\ S& \stackrel{f}{\to }& U\left(N\right)\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ && U(\tilde N) \\ & {}^{\tilde f} \nearrow & \downarrow \\ S &\stackrel{f}{\to}& U(N) } \,.

By adjunction these are equivalently lifts of module homomorphisms

$\begin{array}{ccc}& & \stackrel{˜}{N}\\ & ↗& ↓\\ {R}^{\left(S\right)}& \stackrel{}{\to }& N\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\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 $\stackrel{˜}{N}\to N$ where $\stackrel{˜}{N}$ is a projective module.

###### Proof

Let $F\left(U\left(N\right)\right)$ be the free module on the set $U\left(N\right)$ underlying $N$. By lemma 1 this is a projective module.

The counit

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

of the free/forgetful-adjunction $\left(F⊣U\right)$ 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 $U\left(N\right)$ above by a projective set $P↠U\left(N\right)$, giving an epimorphism $F\left(P\right)↠F\left(U\left(N\right)\right)↠N$ (and $F\left(P\right)$ 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

${R}^{\left(S\right)}\simeq N\oplus N\prime \phantom{\rule{thinmathspace}{0ex}},$R^{(S)} \simeq N \oplus N' \,,

then $N$ is a projective module.

###### Proof

Let $\stackrel{˜}{K}\to K$ be a surjective homomorphism of modules and $f:N\to K$ a homomorphism. We need to show that there is a lift $\stackrel{˜}{f}$ in

$\begin{array}{ccc}& & \stackrel{˜}{K}\\ & {}^{\stackrel{˜}{f}}↗& ↓\\ N& \stackrel{f}{\to }& K\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\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

${\mathrm{id}}_{N}:N\to N\oplus N\prime \to N\phantom{\rule{thinmathspace}{0ex}}.$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 $\stackrel{^}{f}$ in

$\begin{array}{ccccc}& & & & \stackrel{˜}{K}\\ & & & {}^{\stackrel{^}{f}}↗& ↓\\ N& \to & N\oplus N\prime & \to & K\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ && && \tilde K \\ && & {}^{\mathllap{\hat f}}\nearrow & \downarrow \\ N &\to& N \oplus N' &\to& K } \,.

Hence $\stackrel{˜}{f}:N\to N\oplus N\prime \stackrel{\stackrel{^}{f}}{\to }\stackrel{˜}{K}$ 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 $F\left(U\left(N\right)\right)$ be the free module on the set $U\left(N\right)$ underlying $N$ as in the proof of prop. 2. The counit

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

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

In some cases this can be further strenghened:

###### Proposition

If the ring $R$ is a principal ideal domain (in particular $R=ℤ$ 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 $\left(QN{\right)}_{•}\in {\mathrm{Ch}}_{•}\left(R\mathrm{Mod}\right)$ 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

$\cdots \to \left(QN{\right)}_{1}\to \left(QN{\right)}_{0}\to N$\cdots \to (Q N)_1 \to (Q N)_0 \to N

being an exact sequence;

2. all whose entries $\left(QN{\right)}_{n}$ 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 $ℤ$, or a field $k$, or a division ring, then every projective $R$-module is already a free $R$-module.

## References

Lecture notes include

section 2.2 of the textbook

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)

Revised on October 22, 2012 20:13:21 by Urs Schreiber (131.174.189.169)