# nLab projective module

Contents

### Context

#### Homological algebra

homological algebra

Introduction

diagram chasing

# Contents

## Definition

###### Definition

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

Hence an $R$-module $N$ is projective precisely if for all diagrams of $R$-module homomorphisms of the form

$\array{ && A \\ & & \downarrow^{\mathrlap{epi}} \\ N &\underset{f}{\longrightarrow}& B }$

there exists a lift, hence a morphism $N \overset{\phi}{\to} A$ making a commuting diagram of the form

$\array{ && A \\ & {}^{\mathllap{\phi}}\nearrow & \downarrow^{\mathrlap{epi}} \\ N &\underset{f}{\longrightarrow}& B } \,.$
###### Proposition

An $R$-module $N$ is projective (def. ) precisely if the hom functor

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

(out of $N$) is an exact functor.

###### Proof

The hom-functor in question is a left exact functor for all $N$, hence we need to show that it is a right exact functor precisely if $N$ is projective.

That $Hom_R(N,-)$ is right exact means equivalently that for

$0 \to A \overset{i}{\longrightarrow} B \overset{p}{\longrightarrow} C \to 0 \,,$

any short exact sequence, hence for $p$ any epimorphism and $i$ its kernel inclusion, then $Hom_R(N,p)$ is an epimorphism, hence that for any element $f \in Hom_R(N,C)$,

$\array{ && B \\ && \downarrow^{\mathrlap{p}} \\ N &\underset{f}{\longrightarrow}& C }$

there exists $\phi \colon N \to B$ such that $f = Hom_R(N,p)(\phi) \coloneqq p \circ \phi$, hence that

$\array{ && B \\ &{}^{\mathllap{\phi}}\nearrow& \downarrow^{\mathrlap{p}} \\ N &\underset{f}{\longrightarrow}& C } \,.$

This is manifestly the condition that $N$ is projective.

## 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 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 , 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 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. . 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.

###### Proposition

For an $R$-module $P$, the following statements are equivalent:

1. $P$ 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 $R[f_i^{-1}]$.

2. $P$ is finitely generated and projective.

3. $P$ is a dualizable object in the category of $R$-modules (equipped with the tensor product as monoidal structure).

4. 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 $x \in P$.

###### Proof

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.

### 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

Assuming the axiom of choice, then by the basis theorem every module over a field is a free module and hence in particular every module over a field is a projective module (by prop. ).

###### 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.

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)