Injective modules Summary: For ZZ-modules, the inverse p-torsion modules (ZZ[1/p])/ZZ are injective, and every ZZ-module M embeds into a product of these (usually infinite). View any ring R as a ZZ-algebra; then for an injective ZZ-module I, the ZZ-module Hom_ZZ(R, I) becomes an R-module under premultiplication, and is an injective R-module. For an R-module M, view M as a ZZ-module and embed it into an injective ZZ-module I. An inclusion of M into an injective R-module is then provided by the tautological identity Hom_ZZ(M, I) = Hom_R(M, Hom_ZZ(R, I)). This is mostly cribbed from Charles A. Weibel, An introduction to homological algebra, CUP 1994. ==== Definition An R-module I is injective if Hom_R(-, I) is an exact functor. This means that if 0 -> M1 -> M2 is exact (that is, M1 embeds in M2) then any homomorphism e: M1 -> I has some extension f: M2 -> I. Extension means that f gives the same value as e on M1 in M2. Example. A k-vector space V is an injective k-module. This means that for U in W vector spaces, a k-linear map U -> V extends to W. (This requires Zorn's lemma.) Proposition To guarantee that a ZZ-module I is injective, it is enough to prove that any homomorphism e from an ideal J in R to I extends to a homomorphism f: R -> I. (This needs Zorn's lemma.) Proof Let M in N be an inclusion of R-modules and e: M -> I. Suppose that it has been extended to e': M' -> I for some M' with M in M' in N (start from M = M'). If b in N \ M', the extension from ideals allows me to extend e' to e": (M' + bR) -> I. For this, define J = { r in R s.t. br in M'}. Then J is an ideal of R. The homomorphism J -> I given by r |-> br |-> e'(br) is defined on J, so extends to R as a homomorphism f: R -> I. Then e": (M' + bR) -> I is defined as e' on M' and b -> f(1). In more detail, e": (m+br) |-> e'(m) + f(r). [To check well-defined: If some different m1, r1 has m+br = m1+br1 then b(r-r1) = m-m1 in M', so r-r1 in J where the extension f started, so f(r-r1) = e'(m-m1).] QED Corollary An Abelian group (a ZZ-module) is injective iff it is divisible. The modules QQ and (ZZ[1/p])/ZZ are injective, and any injective is a direct product of these. I refer to (ZZ[1/p])/ZZ as the inverse p-torsion module, by analogy with Macaulay's inverse monomials. Check that QQ/ZZ is the direct product of (ZZ[1/p])/ZZ taken over all primes p. An analogous construction works for a PID. Lemma For a ZZ-module M and nonzero m in M, there exists a prime p and a homomorphism f: M -> (ZZ[1/p])/ZZ with f(m) <> 0. Proof The annihilator of m in M is an ideal (n) in ZZ so mZZ iso ZZ/n. Choose any prime p | n and a surjective map ZZ/n ->> ZZ/p (if n = 0 then any prime p works). Compose with an embedding ZZ/p in (ZZ[1/p])/ZZ and extend from mZZ to the whole of M using injectivity of the module. QED Corollary Consider the set of all homomorphisms from M to the injective modules (ZZ[1/p])/ZZ. The direct sum of all these homomorphisms in an embedding of M into an injective ZZ-module. Now for modules over a general ring R (you need a little care about left and right modules if R is noncommutative). R is a ZZ-algebra. If A is a ZZ-module, the ZZ-module Hom_ZZ(R, A) becomes an R-module under premultiplication. Namely, r acts on Hom_ZZ(R, A) by f |-> fr, where fr is the map s -> f(sr) for s in R. That, multiply before applying the map (IMPORTANT). One checks the following points: 1. The identity Hom_ZZ(M, A) = Hom_R(M, Hom_ZZ(R, A)). holds for an R-module M and ZZ-module A. 2. If I is an injective ZZ-module then Hom_ZZ(R, I) is an injective R-module. 3. It follows that for any R-module M, the product of all homomorphisms M -> Hom_ZZ(R, (ZZ[1/p])/ZZ) is an embedding of M into an injective R-module. Finally, let F be a sheaf of OX-modules over a ringed space X. For P in X, the stalk F_P is an OX_P-module. Embed each stalk F_P into an injective OX_P-module I_P. This defines an OX-homomorphism of F into the sheaf of discontinuous sections of DisjointUnion I_P, which is an injective sheaf.