Free, Projective and Injective Modules

An R-module F is called a free R-module if F is isomorphic to a direct sum of copies of R: that is, there is a (possibly infinite) index set I with F = i∈I Ri, where Ri = ⟨bi⟩∼= R for all i. We call B = {bi : i ∈I} a basis of F. A free Z-module is a free abelian group, and every commutative ring R, when considered as a module over itself, is itself a free R-module. From our discussion of direct sums, we know that each m ∈F has a unique expression of the form m = i∈I ribi, where ri ∈R and almost all ri = 0. A basis of a free module has a strong resemblence to a basis of a vector space. There is a straightforward generalization of Theorem 3.92 from finite-dimensional vector spaces to arbitrary free modules (in particular, to infinite-dimensional vector spaces).

The number of elements in a basis is called the rank of F. Of course, rank is the analog of dimension. The next proposition shows that rank is well-defined.

We call a map g : F →A with pg = h (in the diagram in Theorem 7.52) a lifting of h. If C is any, not necessarily free, module, then a lifting g of h, should one exist, need not be unique. Since pi = 0, where i : ker p →A is the inclusion, other liftings are g + i f for any f ∈HomR(C, ker p). Indeed, this is obvious from the exact sequence 0 →Hom(C, ker p) i∗ −→Hom(C, A) p∗ −→Hom(C, A′′). Any two liftings of h differ by a map in ker p∗= im i∗⊆Hom(C, A). We now promote this (basis-free) property of free modules to a definition.

A module P is projective if, whenever p is surjective and h is any map, there exists a lifting g; that is, there exists a map g making the following diagram commute: P h g A p A′′ 0 We know that every free module is projective; is every projective R-module free? Note that if projective R-modules happen to be free, then free modules are characterized without having to refer to a basis. We know that the Hom functors are left exact; that is, for any module P, applying HomR(P, ) to an exact sequence 0 →A′ i −→A p −→A′′ gives an exact sequence 0 →HomR(P, A′) i∗ −→HomR(P, A) p∗ −→HomR(P, A′′). Free Modules, Projectives, and Injectives

If A is an R-module, then a subset {ai : i ∈I} ⊆A and a family of R-maps {ϕi : A →R : i ∈I} satisfying the condition in Proposition 7.58 is called a projective basis. Then X is a paracompact space if and only if J is a projective C(X)-module. Remark. The definition of projective module can be used to define a projective object in any category (we do not assert that such objects always exist), if we can translate surjection into the language of categories.

A morphism ϕ : B →C in a category C is an epimorphism if ϕ can be canceled from the right; that is, for all objects D and all morphisms h : C →D and k : C →D, we have hϕ = kϕ implies h = k. But ϕ is not a surjective function if R is not a field. A similar phenomenon occurs in Top. If f : X →Y is a continuous map with im f a dense subspace of Y, then f is an epimorphism, because any two continuous functions agreeing on a dense subspace must be equal. There is a similar problem with monomorphisms, a generalization of injections to arbitrary categories: A category whose objects have underlying sets may have monomorphisms whose underlying function is not an injection. ◀ Let us return to presentations of modules.

If E is a module for which the contravariant Hom functor HomR( , E) is an exact functor—that is, if HomR( , E) preserves all short exact sequences—then E is called an injective module. The next proposition is the dual of Proposition 7.53. Free Modules, Projectives, and Injectives

If R is a domain, then an R-module D is divisible if, for each d ∈D and every nonzero r ∈R, there exists d′ ∈D with d = rd′. Free Modules, Projectives, and Injectives

Let F be a free R-module, and let B = {bi : i ∈I} be a basis of F. If M is any R-module and if γ : B →M is any function, then there exists a unique R-map g : F →M with g(bi) = γ (bi) for all i ∈I. F g B γ M

Every R-module M is a quotient of a free R-module F. Moreover, M is finitely generated if and only if F can be chosen to be finitely generated.

If R is a commutative ring and F is a free R-module, then for every surjection p: A →A′′ and each h : F →A′′, there exists a homomorphism g making the 13A module is called free because it has no entangling relations. following diagram commute: F h g A p A′′ 0

A module P is projective if and only if every short exact sequence 0 →A i→B p→P →0 is split.

Let A be a submodule of a module B. If B/A is projective, then there is a submodule C of B with C ∼= B/A and B = A ⊕C.

An R-module P is projective if and only if P is a direct summand of a free R-module.

An R-module A is projective if and only if there exist elements {ai : i ∈I} ⊆A and R-maps {ϕi : A →R : i ∈I} such that (i) for each x ∈A, almost all ϕi(x) = 0; (ii) for each x ∈A, we have x = i∈I(ϕi x)ai. Moreover, A is generated by {ai : i ∈I} ⊆A in this case.

Given exact sequences 0 →K i→P π→M →0 and 0 →K ′ i′ →P′ π′ →M →0, where P and P′ are projective, then there is an isomorphism K ⊕P′ ∼= K ′ ⊕P.

A module E is injective if and only if a dotted arrow always exists making the following diagram commute whenever i is an injection: E A i f B g In words, every homomorphism from a submodule into E can always be extended to a homomorphism from the big module into E. Remark. Since HomR( , E) is a left exact contravariant functor, the thrust of the proposition is that i∗is surjective whenever i is injective. Injective modules are duals of projective modules in that both of these terms are characterized by diagrams, and the diagram for injectivity is the diagram for projectivity having all arrows reversed. ◀

A module E is injective if and only if every short exact sequence 0 →E i→B p→C →0 is split.

If {Ei : i ∈I} is a family of injective modules, then i∈I Ei is also an injective module.

If R is ascending-chain-finite and {Ei : i ∈I} is a family of injective R-modules, then i∈I Ei is an injective module.

says that, over any ring, every module is a quotient of a projective module (actually, it is a stronger result: Every module is a quotient of a free module). The next result is the dual result for Z-modules: Every abelian group can be imbedded as a subgroup of an injective abelian group.

Let R be a PID, let a ∈R be neither zero nor a unit, and let J = (a). Then R/J is an injective R/J-module.

The ring R = I6 is the direct sum of two ideals: I6 = J ⊕I, where J = {[0], [2], [4]} ∼= I3 and I = {[0], [3]} ∼= I2. Now I6 is a free module over itself, and so J and I, being direct summands of a free module, are projective I6-modules. Neither J nor I can be free, however. After all, a (finitely generated) free I6-module F is a direct sum of, say, n copies of I6, and so F has 6n elements. Therefore, J is too small to be free, for it has only three elements. ◀ Free Modules, Projectives, and Injectives Describing projective R-modules is a problem very much dependent on the ring R. It will then follow from Theorem 7.56 that every projective R-module is free in this case. A much harder result is that if R = k[x1, . . . , xn] is the polynomial ring in n variables over a field k, then every projective R-module is also free; this theorem, implicitly conjectured14 by J.-P. There are domains having projective modules that are not free. For example, if R is the ring of all the algebraic integers in an algebraic number field (that is, an extension of Q of finite degree), then every ideal in R is a projective R-module. Here is another characterization of projective modules. Note that if A is a free R-module with basis {ai : i ∈I} ⊆A, then each x ∈A has a unique expression x = i∈I riai, and so there are R-maps ϕi : A →R given by ϕi : x ↦ri.

In light of Example 7.71, the following abelian groups are injective Z-modules: Q, R, C, Q/Z, R/Z, S1, where S1 is the circle group; that is, the multiplicative group of all complex numbers z with |z| = 1. ◀