nLab
projective module

Context

Homological algebra

homological algebra

and

nonabelian homological algebra

Context

Basic definitions

Stable homotopy theory notions

Constructions

Lemmas

diagram chasing

Homology theories

Theorems

Contents

Definition

Definition

For RR a ring, a projective RR-module is a projective object in the category RRMod.

Proposition

A module NN is projective precisely if the hom functor

Hom RMod(N,):RModAb 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 NR (S)N \simeq R^{(S)} is projective.

Proof

Explicitly: if SSetS \in Set and F(S)=R (S)F(S) = R^{(S)} is the free module on SS, then a module homomorphism F(S)NF(S) \to N is specified equivalently by a function f:SU(N)f : S \to U(N) from SS to the underlying set of NN, which can be thought of as specifying the images of the unit elements in R (S) sSRR^{(S)} \simeq \oplus_{s \in S} R of the S{\vert S\vert} copies of RR.

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

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

By adjunction these are equivalently lifts of module homomorphisms

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

Assuming the axiom of choice, the category RRMod has enough projectives: for every RR-module NN there exists an epimorphism N˜N\tilde N \to N where N˜\tilde N is a projective module.

Proof

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

The counit

ϵ:F(U(N))N \epsilon : F(U(N)) \to N

of the free/forgetful-adjunction (FU)(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)U(N) above by a projective set PU(N)P \twoheadrightarrow U(N), giving an epimorphism F(P)F(U(N))NF(P) \twoheadrightarrow F(U(N)) \twoheadrightarrow N (and F(P)F(P) is projective).

Explicit characterizations

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

Lemma

If NRModN \in R Mod is a direct summand of a free module, hence if there is NRModN' \in R Mod and SSetS \in Set such that

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

then NN is a projective module.

Proof

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

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

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

id N:NNNN. id_N : N \to N \oplus N' \to N \,.

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

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

Hence f˜:NNNf^K˜\tilde f : N \to N \oplus N' \stackrel{\hat f}{\to} \tilde K is a lift of ff.

Proposition

An RR-module NN is projective precisely if it is the direct summand of a free module.

Proof

By lemma 2 if NN is a direct summand then it is projective. So we need to show the converse.

Let F(U(N))F(U(N)) be the free module on the set U(N)U(N) underlying NN as in the proof of prop. 2. The counit

ϵ:F(U(N))N \epsilon : F(U(N)) \to N

of the free/forgetful-adjunction (FU)(F \dashv U) is an epimorphism. Thefore if NN is projective, there is a section ss of ϵ\epsilon. This exhibits NN as a direct summand of F(U(N))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 RR is a principal ideal domain (in particular R=R = \mathbb{Z} the integers), then every projective RR-module is free.

The details are discussed at pid - Structure theory of modules.

Relation to projective resolutions of chain complexes

Definition

For NRModN \in R Mod a projective resolution of NN is a chain complex (QN) Ch (RMod)(Q N)_\bullet \in Ch_\bullet(R Mod) equipped with a chain map

QNN Q N \to N

(with NN 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

    (QN) 1(QN) 0N \cdots \to (Q N)_1 \to (Q N)_0 \to N

    being an exact sequence;

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

Remark

This means precisely that QNNQ 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 RR-module has a projective resolution.

See at projective resolution.

Examples

Proposition

If RR is the integers \mathbb{Z}, or a field kk, or a division ring, then every projective RR-module is already a free RR-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)