Topics in algebra Chapter IV: Commutative rings and modules I - 1

Lemma 5.7 Let E be an R-module. E is an injective module if and only if
for every ideal I of R, any R-homomorphism I −→ E may be extended to
an R-homomorphism R −→ E.
Proof. (⇒) Since 0 −→ I −→ R is exact, the assertion follows.
(⇐) Let 0 −→ N g
−→ M be an exact sequence and f : N −→ E be an R
homomorphism; to show that E is injective, we must find a homomorphism
h : M −→ E with h ◦ g = f. Let T be the set of all R-homomorphism
h : M ′ −→ E with h ◦ g = f, where Img ⊆ M ′ ⊆ M. Partially order T by
the natural way and verify the assumptions of Zorn’s Lemma, we conclude
that T contains a maximal element h : M ′ −→ E with h ◦ g = f. We will
show that M ′ = M.
If M ′ = M and x ∈ M − M ′ , then I = {r ∈ R | rx ∈ M ′ } is an ideal of
R. The map i : I −→ E given by i(r) = h(rx) is an R-homomorphism. By
assumption, there is an R-homomorphism j : R −→ E such that j(r) = h(rx)
for all r ∈ I. Let y = j(1R) and define a map ˜ h : M ′ + xR −→ E by
˜h(a + rx) = h(a) + ry. ˜ h is well-defined: If a1 + r1x = a2 + r2x, then
a1 − a2 = (r2 − r1)x ∈ M ′ ∩ xR, so that r2 − r1 ∈ I and
h(a1)−h(a2) = h(a1−a2) = h((r2−r1)x) = j(r2−r1) = (r2−r1)j(1R) = (r2−r1)y.
It is easy to see that ˜ h is an R-homomorphism. This contradicts the maximality
of h and hence M ′ = M. Therefore E is injective.
There is an important fact about injective modules: Every R-module
can be imbedded in an injective R-module. To show this, we need some
preparation.
Definition 5.8 An abelian group G is said to be divisible if for every a ∈ G
and 0 = n ∈ Z there exists b ∈ G such that bn = a.
Example 5.9 Q is a divisible abelian group.
Lemma 5.10 An abelian group D is divisible if and only if D is an injective
Z-module.
Proof. (⇒) Let D be a divisible abelian group and f : I −→ D be a Zmodule
homomorphism. Let I = nZ and f(n) = a; then there is an element
b ∈ D such that nb = a. Let g : Z −→ D be the map given by g(m) = mb,
then it is easy to check that g is a Z-homomorphism and is an extension of
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