Projective module

In mathematics, particularly in abstract algebra and homological algebra, the concept of projective module over a ring R is a generalisation of the idea of a free module (that is, a module with basis vectors). Various equivalent characterizations of these modules appear below.

Projective modules were first introduced in 1956 in the influential book Homological Algebra by Henri Cartan and Samuel Eilenberg.

Definitions

Lifting property

The usual definition in line with category theory is the property of lifting that carries over from free to projective modules. We can summarize this lifting property as follows: a module P is projective if and only if for every surjective module homomorphism f : NM and every module homomorphism g : PM, there exists a homomorphism h : PN such that fh = g. (We don't require the lifting homomorphism h to be unique; this is not a universal property.)

The advantage of this definition of "projective" is that it can be carried out in categories more general than module categories: we don't need a notion of "free object". It can also be dualized, leading to injective modules.

Split-exact sequences

A module P is projective if and only if for every surjective module homomorphism f : MP there exists a module homomorphism h : PM such that fh = idP. The existence of such a section map h implies that P is a direct summand of M and that f is essentially a projection on the summand P. More explicitly, M = im(h) ⊕ ker(f), and im(h) is isomorphic to P.

The foregoing is a detailed description of the following statement: A module P is projective if every short exact sequence of modules of the form

$0\rightarrow A\rightarrow B\rightarrow P\rightarrow 0\,$

is a split exact sequence.

Direct summands of free modules

A module P is projective if and only if there is a free module F and another module Q such that the direct sum of P and Q is F.

Exactness

An R-module P is projective if and only if the functor Hom(P,-): R-Mod → AB is an exact functor, where R-Mod is the category of left R-modules and AB the category of Abelian groups. When the ring R is commutative, AB is advantageously replaced by R-Mod in the preceding characterization. This functor is always left exact, but, when P is projective, it is also right exact. This means that P is projective if and only if this functor preserves epimorphisms (surjective homomorphisms), or if it preserves finite colimits.

Dual basis

A module P is projective if and only if there exists a set $\{a_i\in P \mid i\in I\}$ and a set $\{f_i\in \mathrm{Hom}(P,R) \mid i\in I\}$ such that for every x in P, fi(x) is only nonzero for finitely many i, and $x=\sum f_i(x)a_i$.

Properties

• Direct sums and direct summands of projective modules are projective.
• If e = e2 is an idempotent in the ring R, then Re is a projective left module over R.
• Submodules of projective modules need not be projective; a ring R for which every submodule of a projective left module is projective is called left hereditary.
• The category of finitely generated projective modules over a ring is an exact category. (See also algebraic K-theory).
• Every module over a field or skew field is projective (even free). A ring over which every module is projective is called semisimple.
• An abelian group (i.e. a module over Z) is projective if and only if it is a free abelian group. The same is true for all principal ideal domains; the reason is that for these rings, any submodule of a free module is free.
• Over a Dedekind domain a non-principal ideal is always a projective module that is not a free module.
• Over a direct product of rings R × S where R and S are nonzero rings, both R × 0 and 0 × S are non-free projective modules.
• Over a matrix ring Mn(R), the natural module Rn is projective but not free. More generally, over any semisimple ring, every module is projective, but the zero ideal and the ring itself are the only free ideals.
• Every projective module is flat.[1] The converse is in general not true: the abelian group Q is a Z-module which is flat, but not projective.[2]
• In line with the above intuition of "locally free = projective" is the following theorem due to Kaplansky: over a local ring, R, every projective module is free. This is easy to prove for finitely generated projective modules, but the general case is difficult.
• A finitely related module is flat if and only if it is projective.[3]

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

The left-to-right implications are true over any ring, although some authors define torsion-free modules only over a domain. 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 polynomial rings over a field: this is Quillen–Suslin theorem.

Projective resolutions

Main article: Projective resolution

Given a module, M, a projective resolution of M is an infinite exact sequence of modules

· · · → Pn → · · · → P2P1P0M → 0,

with all the Pi's projective. Every module possesses a projective resolution. In fact a free resolution (resolution by free modules) exists. The exact sequence of projective modules may sometimes be abbreviated to P(M) → M → 0 or PM → 0. A classic example of a projective resolution is given by the Koszul complex of a regular sequence, which is a free resolution of the ideal generated by the sequence.

The length of a finite resolution is the subscript n such that Pn is nonzero and Pi=0 for i greater than n. If M admits a finite projective resolution, the minimal length among all finite projective resolutions of M is called its projective dimension and denoted pd(M). If M does not admit a finite projective resolution, then by convention the projective dimension is said to be infinite. As an example, consider a module M such that pd(M) = 0. In this situation, the exactness of the sequence 0 → P0M → 0 indicates that the arrow in the center is an isomorphism, and hence M itself is projective.[4]

Projective modules over commutative rings

Projective modules over commutative rings have nice properties.

The localization of a projective module is a projective module over the localized ring. A projective module over a local ring is free. Thus a projective module is locally free (in the sense that its localization at every prime ideal is free over the corresponding localization of the ring).

The converse is true for finitely generated modules over Noetherian 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 which are locally free and not projective. For instance, a Boolean ring has all of its localizations isomorphic to F2, 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 is R/I where R is a direct product of countably many copies of F2 and I is the direct sum of countably many copies of F2 inside of R. The R-module R/I is locally free since R is Boolean (and it's finitely generated as an R-module too, with a spanning set of size 1), but R/I is not projective because I is not a principal ideal. (If a quotient module R/I, for any commutative ring R and ideal I, is a projective R-module then I is principal.)

However, it is true that for finitely presented modules M over a commutative ring R (in particular if M is a finitely generated R-module and R is noetherian), the following are equivalent.[5]

1. $M$ is flat.
2. $M$ is projective.
3. $M_\mathfrak{m}$ is free as $R_\mathfrak{m}$-module for every maximal ideal $\mathfrak{m}$ of R.
4. $M_\mathfrak{p}$ is free as $R_\mathfrak{p}$-module for every prime ideal $\mathfrak{p}$ of R.
5. There exist $f_1,\ldots,f_n \in R$ generating the unit ideal such that $M[f_i^{-1}]$ is free as $R[f_i^{-1}]$-module for each i.
6. $\widetilde{M}$ is a locally free sheaf on $\operatorname{Spec}R$.

Moreover, if R is a noetherian integral domain, then, by Nakayama's lemma, these conditions are equivalent to

• The dimension of the $k(\mathfrak{p})$–vector space $M \otimes_R k(\mathfrak{p})$ is the same for all prime ideals $\mathfrak{p}$ of R.[6] That is to say, M has constant rank ("rank" is defined in the section below).

Let A be a commutative ring. If B is a (possibly non-commutative) A-algebra that is a finitely generated projective A-module containing A as a subring, then A is a direct factor of B.[7]

Rank

Let P be a finitely generated projective module over a commutative ring R and X be the spectrum of R. The rank of P at a prime ideal $\mathfrak{p}$ in X is the rank of the free $R_{\mathfrak{p}}$-module $P_{\mathfrak{p}}$. It is a locally constant function on X. In particular, if X is connected (that is if R or its quotient by its nilradical is an integral domain), then P has constant rank.

Vector bundles and locally free modules

A basic motivation of the theory is that projective modules (at least over certain commutative rings) are analogues of vector bundles. This can be made precise for the ring of continuous real-valued functions on a compact Hausdorff space, as well as for the ring of smooth functions on a smooth manifold (see Serre–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 a smooth vector bundle).

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

Projective modules over a polynomial ring

The Quillen–Suslin theorem, which solves Serre's problem is another deep result; it states that if K is a field, or more generally a principal ideal domain, and R = K[X1,...,Xn] is a polynomial ring over K, then every projective module over R is free. This problem was first raised by Serre with K a field (and the modules being finitely generated). Bass settled it for non-finitely generated modules and Quillen and Suslin 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: if R is a commutative ring such that every (finitely generated) projective R-module is free, then is every (finitely generated) projective R[X]-module is free? The answer is no. A counterexample occurs with R equal to the local ring of the curve y2 = x3 at the origin. So Serre's problem can not be proved by a simple induction on the number of variables.

Notes

1. ^ Hazewinkel, et. al. (2004), Corollary 5.4.5, p. 131.
2. ^ Hazewinkel, et. al. (2004), Remark after Corollary 5.4.5, p. 131–132.
3. ^ Cohn 2003, Corollary 4.6.4
4. ^ A module isomorphic to a projective module is of course projective.
5. ^ Exercises 4.11 and 4.12 and Corollary 6.6 of David Eisenbud, Commutative Algebra with a view towards Algebraic Geometry, GTM 150, Springer-Verlag, 1995. Also, Milne 1980
6. ^ Here, $k(\mathfrak{p})=R_\mathfrak{p}/\mathfrak{p}R_\mathfrak{p}$ is the residue field of the local ring $R_\mathfrak{p}$.
7. ^ Bourbaki, Algèbre commutative 1989, Ch II, §5, Exercise 4