nLab projective module



Homological algebra

homological algebra

(also nonabelian homological algebra)



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diagram chasing

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

Hence an RR-module NN is projective precisely if for all diagrams of RR-module homomorphisms of the form

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

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

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

An RR-module NN is projective (def. ) precisely if the hom functor

Hom RMod(N,):RModAb Hom_{R Mod}(N, - ) : R Mod \to Ab

(out of NN) is an exact functor.


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

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

0AiBpC0, 0 \to A \overset{i}{\longrightarrow} B \overset{p}{\longrightarrow} C \to 0 \,,

any short exact sequence, hence for pp any epimorphism and ii its kernel inclusion, then Hom R(N,p)Hom_R(N,p) is an epimorphism, hence that for any element fHom R(N,C)f \in Hom_R(N,C),

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

there exists ϕ:NB\phi \colon N \to B such that f=Hom R(N,p)(ϕ)pϕf = Hom_R(N,p)(\phi) \coloneqq p \circ \phi, hence that

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

This is manifestly the condition that NN is projective.


Existence of enough projective modules


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


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 } \,.

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.


Let F(U(N))F(U(N)) be the free module on the set U(N)U(N) underlying NN. By lemma 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.


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.


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 , 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.


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


By lemma 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. . 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:


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.


For an RR-module PP, the following statements are equivalent:

  1. PP is finite locally free in that there exists a partition 1= if iR1 = \sum_i f_i \in R such that the localized modules P[f i 1]P[f_i^{-1}] are finite free modules over R[f i 1]R[f_i^{-1}].

  2. PP is finitely generated and projective.

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

  4. There exist elements x 1,,x nPx_1,\ldots,x_n \in P and linear forms ϑ 1,,ϑ nHom(P,R)\vartheta_1,\ldots,\vartheta_n \in Hom(P,R) such that x= iϑ i(x)x ix = \sum_i \vartheta_i(x) x_i for all xPx \in P.


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 RR is Noetherian or that PP satisfies some finiteness condition.

Relation to projective resolutions of chain complexes


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.


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.


Every RR-module has a projective resolution.

See at projective resolution.



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. ).


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.


Early discussion:

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)

Last revised on August 23, 2023 at 14:23:46. See the history of this page for a list of all contributions to it.