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.

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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.