# 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

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

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 $S \in Set$ 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

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

By adjunction these are equivalently lifts of module homomorphisms

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

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

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).

### Explicit characterizations

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

###### Lemma

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

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

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

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

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

$id_N : N \to N \oplus N' \to N \,.$

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

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

Hence $\tilde f : N \to N \oplus N' \stackrel{\hat f}{\to} \tilde K$ 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

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

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:

###### 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 $N \in R Mod$ a projective resolution of $N$ is a chain complex $(Q N)_\bullet \in Ch_\bullet(R Mod)$ equipped with a chain map

$Q N \to N$

(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

$\cdots \to (Q N)_1 \to (Q N)_0 \to N$

being an exact sequence;

2. all whose entries $(Q N)_n$ are projective modules.

###### Remark

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.

###### 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.

## References

Lecture notes include

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 January 3, 2014 07:08:10 by Zoran Škoda (77.237.117.98)