Inmathematics, particularly inalgebra, theclass ofprojective modules enlarges the class offree modules (that is,modules withbasis vectors) over aring, keeping some of the main properties of free modules. Various equivalent characterizations of these modules appear below.

Every free module is a projective module, but theconverse fails to hold over some rings, such asDedekind rings that are notprincipal ideal domains. However, every projective module is a free module if the ring is a principal ideal domain such as theintegers, or a (multivariate)polynomial ring over afield (this is theQuillen–Suslin theorem).

Projective modules were first introduced in 1956 in the influential bookHomological Algebra byHenri Cartan andSamuel Eilenberg.

The usualcategory theoretical definition is in terms of the property oflifting that carries over from free to projective modules: a moduleP is projectiveif and only if for everysurjectivemodule homomorphismf :N ↠M and every module homomorphismg :P →M, there exists a module homomorphismh :P →N such thatfh =g. (We don't require the lifting homomorphismh to be unique; this is not auniversal property.)

The advantage of this definition of "projective" is that it can be carried out incategories more general thanmodule categories: we don't need a notion of "free object". It can also bedualized, leading toinjective modules. The lifting property may also be rephrased asevery morphism from${\displaystyle P}$ to${\displaystyle M}$ factors through every epimorphism to${\displaystyle M}$. Thus, by definition, projective modules are precisely theprojective objects in thecategory ofR-modules.

A moduleP is projective if and only if everyshort exact sequence of modules of the form

${\displaystyle 0\to A\to B\to P\to 0}$

is asplit exact sequence. That is, for every surjective module homomorphismf :B ↠P there exists asection map, that is, a module homomorphismh :P →B such thatfh = id_P. In that case,h(P) is adirect summand ofB,h is anisomorphism fromP toh(P), andhf is aprojection on the summandh(P). Equivalently,

${\displaystyle B=\mathrm{Im}(h)\oplus \mathrm{Ker}(f)\text{ where }\mathrm{Ker}(f)\cong A\text{ and }\mathrm{Im}(h)\cong P.}$

A moduleP is projective if and only if there is another moduleQ such that thedirect sum ofP andQ is a free module.

AnR-moduleP is projective if and only if the covariantfunctorHom(P, -):R-Mod →Ab is anexact functor, whereR-Mod is the category of leftR-modules andAb is thecategory of abelian groups. When the ringR iscommutative,Ab is advantageously replaced byR-Mod in the preceding characterization. This functor is alwaysleft exact, but, whenP is projective, it is also right exact. This means thatP is projective if and only if this functor preservesepimorphisms (surjective homomorphisms), or if it preserves finitecolimits.

A moduleP is projective if and only if there exists a set${\displaystyle \{a_i\in P\mid i\in I\}}$ and a set${\displaystyle \{f_i\in \mathrm{H}\mathrm{o}\mathrm{m}(P,R)\mid i\in I\}}$ such that for everyx inP,f_i(x) is only nonzero for finitely manyi, and${\displaystyle x=\sum f_i(x)a_i}$.

The following properties of projective modules are quickly deduced from any of the above (equivalent) definitions of projective modules:

• Direct sums and direct summands of projective modules are projective.
• Ife =e² is anidempotent in the ringR, thenRe is a projective left module overR.

Let${\displaystyle R=R_1\times R_2}$ be thedirect product of two rings${\displaystyle R_1}$ and${\displaystyle R_2,}$ which is a ring with operations defined componentwise. Let${\displaystyle e_1=(1,0)}$ and${\displaystyle e_2=(0,1).}$ Then${\displaystyle e_1}$ and${\displaystyle e_2}$ are idempotents, and belong to thecentre of${\displaystyle R.}$ Thetwo-sided ideals${\displaystyle R{e}_1}$ and${\displaystyle R{e}_2}$ are projective modules, since their direct sum (asR-modules) equals the freeR-moduleR. However, if${\displaystyle R_1}$ and${\displaystyle R_2}$ are nontrivial, then they are not free as modules over${\displaystyle R}$. For instance${\displaystyle \mathbb{Z}/2\mathbb{Z}}$ is projective but not free over${\displaystyle \mathbb{Z}/6\mathbb{Z}}$.

The relation of projective modules to free andflat modules is subsumed in the following diagram of module properties:

The left-to-right implications are true over any ring, although some authors definetorsion-free modules only over adomain. The right-to-left implications are true over the rings labeling them. There may be other rings over which they are true. For example, the implication labeled "local ring or PID" is also true for (multivariate) polynomial rings over afield: this is theQuillen–Suslin theorem.

Any free module is projective. The converse is true in the following cases:

• ifR is a field orskew field:any module is free in this case.
• if the ringR is aprincipal ideal domain. For example, this applies toR =Z (theintegers), so anabelian group is projective if and only if it is afree abelian group. The reason is that anysubmodule of a free module over a principal ideal domain is free.
• if the ringR is alocal ring. This fact is the basis of the intuition of "locally free = projective". This fact is easy toprove forfinitely generated projective modules. In general, it is due toKaplansky (1958); seeKaplansky's theorem on projective modules.

In general though, projective modules need not be free:

• Over adirect product of ringsR ×S whereR andS arenonzero rings, bothR × 0 and0 ×S are non-free projective modules.
• Over aDedekind domain a non-principalideal is always a projective module that is not a free module.
• Over amatrix ring M_n(R), the natural moduleRⁿ is projective but is not free whenn > 1.
• Over asemisimple ring,every module is projective, but a nonzero proper left (or right) ideal is not a free module. Thus the only semisimple rings for which all projectives are free aredivision rings.

The difference between free and projective modules is, in a sense, measured by thealgebraicK-theorygroupK₀(R); see below.

Every projective module isflat.^[1] The converse is in general not true: the abelian groupQ is aZ-module that is flat, but not projective.^[2]

Conversely, afinitely related flat module is projective.^[3]

Govorov (1965) andLazard (1969) proved that a moduleM is flat if and only if it is adirect limit offinitely-generatedfree modules.

In general, the precise relation between flatness and projectivity was established byRaynaud & Gruson (1971) (see alsoDrinfeld (2006) andBraunling, Groechenig & Wolfson (2016)) who showed that a moduleM is projective if and only if it satisfies the following conditions:

• M is flat,
• M is a direct sum ofcountably generated modules,
• M satisfies a certainMittag-Leffler-type condition.

This characterization can be used to show that if${\displaystyle R\to S}$ is afaithfully flat map of commutative rings and${\displaystyle M}$ is an${\displaystyle R}$-module, then${\displaystyle M}$ is projective if and only if${\displaystyle M{\otimes }_{R}S}$ is projective.^[4] In other words, the property of being projective satisfiesfaithfully flat descent.

Submodules of projective modules need not be projective; a ringR for which every submodule of a projective left module is projective is calledleft hereditary.

Quotients of projective modules also need not be projective, for exampleZ/n is a quotient ofZ, but nottorsion-free, hence not flat, and therefore not projective.

The category of finitely generated projective modules over a ring is anexact category. (See alsoalgebraic K-theory).

Given a module,M, aprojectiveresolution ofM is an infiniteexact sequence of modules

⋅⋅⋅ →P_n → ⋅⋅⋅ →P₂ →P₁ →P₀ →M → 0,

with all theP_i s projective. Every module possesses a projective resolution. In fact afree resolution (resolution by free modules) exists. The exact sequence of projective modules may sometimes be abbreviated toP(M) →M → 0 orP_• →M → 0. A classic example of a projective resolution is given by theKoszul complex of aregular sequence, which is a free resolution of theideal generated by the sequence.

Thelength of a finite resolution is the indexn such thatP_n isnonzero andP_i = 0 fori greater thann. IfM admits a finite projective resolution, the minimal length among all finite projective resolutions ofM is called itsprojective dimension and denoted pd(M). IfM does not admit a finite projective resolution, then by convention the projective dimension is said to be infinite. As an example, consider a moduleM such thatpd(M) = 0. In this situation, the exactness of the sequence 0 →P₀ →M → 0 indicates that the arrow in the center is an isomorphism, and henceM itself is projective.

Projective modules overcommutative rings have nice properties.

Thelocalization of a projective module is a projective module over the localized ring.A projective module over alocal ring is free. Thus a projective module islocally free (in the sense that its localization at everyprime ideal is free over the corresponding localization of the ring). The converse is true forfinitely generated modules overNoetherian rings: a finitely generated module over a commutative Noetherian ring is locally free if and only if it is projective.

However, there are examples of finitely generated modules over a non-Noetherian ring that are locally free and not projective. For instance, aBoolean ring has all of its localizations isomorphic toF₂, the field of two elements, so any module over a Boolean ring is locally free, but there are some non-projective modules over Boolean rings. One example isR/I whereR is a direct product of countably many copies ofF₂ andI is the direct sum of countably many copies ofF₂ inside ofR.TheR-moduleR/I is locally free sinceR is Boolean (and it is finitely generated as anR-module too, with a spanning set of size 1), butR/I is not projective becauseI is not a principal ideal. (If a quotient moduleR/I, for any commutative ringR and idealI, is a projectiveR-module thenI is principal.)

However, it is true that forfinitely presented modulesM over a commutative ringR (in particular ifM is a finitely generatedR-module andR is Noetherian), the following are equivalent.^[5]

1. ${\displaystyle M}$ is flat.
2. ${\displaystyle M}$ is projective.
3. ${\displaystyle M_{\mathfrak{m}}}$ is free as${\displaystyle R_{\mathfrak{m}}}$-module for everymaximal ideal${\displaystyle \mathfrak{m}}$ ofR.
4. ${\displaystyle M_{\mathfrak{p}}}$ is free as${\displaystyle R_{\mathfrak{p}}}$-module for every prime ideal${\displaystyle \mathfrak{p}}$ ofR.
5. There exist${\displaystyle f_1,\dots ,f_n\in R}$ generating theunit ideal such that${\displaystyle M[f_i^{-1}]}$ is free as${\displaystyle R[f_i^{-1}]}$-module for eachi.
6. ${\displaystyle \tilde{M}}$ is alocally free sheaf on theaffine scheme${\displaystyle \mathrm{Spec}R}$ (where${\displaystyle \tilde{M}}$ is thesheaf associated toM.)

Moreover, ifR is a Noetherianintegral domain, then, byNakayama's lemma, these conditions are equivalent to

• Thedimension of the${\displaystyle k(\mathfrak{p})}$-vector space${\displaystyle M{\otimes }_{R}k(\mathfrak{p})}$ is the same for all prime ideals${\displaystyle \mathfrak{p}}$ ofR, where${\displaystyle k(\mathfrak{p})}$ is the residue field at${\displaystyle \mathfrak{p}}$.^[6] That is to say,M has constant rank (as defined below).

LetA be a commutative ring. IfB is a (possibly non-commutative)A-algebra that is a finitely generated projectiveA-module containingA as asubring, thenA is a direct factor ofB.^[7]

LetP be a finitely generated projective module over a commutative ringR andX be thespectrum ofR. Therank ofP at a prime ideal${\displaystyle \mathfrak{p}}$ inX is the rank of the free${\displaystyle R_{\mathfrak{p}}}$-module${\displaystyle P_{\mathfrak{p}}}$. It is a locally constant function onX. In particular, ifX is connected (that is ifR has no other idempotents than 0 and 1), thenP has constant rank.

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A basic motivation of the theory is that projective modules (at least over certain commutative rings) are analogues ofvector bundles. This can be made precise for the ring ofcontinuousreal-valued functions on acompactHausdorff space, as well as for the ring ofsmooth functions on asmooth manifold (seeSerre–Swan theorem that says a finitely generated projective module over the space of smooth functions on a compact manifold is the space of smooth sections of asmooth vector bundle).

Vector bundles arelocally free. If there is some notion of "localization" that can be carried over to modules, such as the usuallocalization of a ring, one can define locally free modules, and the projective modules then typically coincide with the locally free modules.

TheQuillen–Suslin theorem, which solves Serre's problem, is anotherdeep result: ifK is a field, or more generally aprincipal ideal domain, andR =K[X₁,...,X_n] is apolynomial ring overK, then every projective module overR is free.This problem was first raised by Serre withK a field (and the modules being finitely generated).Bass settled it for non-finitely generated modules,^[8] andQuillen andSuslin independently and simultaneously treated the case of finitely generated modules.

Since every projective module over a principal ideal domain is free, one might ask this question: ifR is a commutative ring such that every (finitely generated) projectiveR-module is free, then is every (finitely generated) projectiveR[X]-module free? The answer isno. Acounterexample occurs withR equal to the local ring of the curvey² =x³ at the origin. Thus the Quillen–Suslin theorem could never be proved by a simpleinduction on the number of variables.

The WikibookCommutative Algebra has a page on the topic of:Torsion-free, flat, projective and free modules

• Projective cover
• Schanuel's lemma
• Bass cancellation theorem
• Modular representation theory

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