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

Proof. Let M be an R-module. Since M is an abelian group, M is embedded
in an injective Z-module D. By the previous lemma, HomZ(R, D) is an
injective R-module. Therefore, to finish the proof, it remains to show that M
is an R-submodule of HomZ(R, D). Notice that the group monomorphism
M −→ D induces the R-monomorphism HomZ(R, M) −→ HomZ(R, D).
Moreover, let i : M −→ HomZ(R, M) be the map given by i(x)(r) = xr for
x ∈ M and r ∈ R; then it is easy to check that i is an R-monomorphism.
We now conclude that M is an R-submodule of HomZ(R, D).
Proposition 5.15 Let R be a ring. Then the following are equivalent for
an R-module E:
(a) E is injective.
(b) Any s.e.s. of the form 0 −→ E f
−→ M g
−→ N −→ 0 splits.
(c) E is a direct summand of any R-module M of which it is a submodule.
Proof. (a) ⇒ (b) Let 0 −→ E f
−→ M g
−→ N −→ 0 be a s.e.s.. Let i : E −→
E be the identity map; then there is a R-homomorphism h : M −→ E such
that h ◦ f = i, so that h ◦ f = idE,i.e., the s.e.s. splits.
(b) ⇒ (c) Suppose E is a submodule of M. Then there is a s.e.s. 0 −→
E i
−→ M π
−→ M/E −→ 0. Since it splits, E is a direct summand of M.
(c) ⇒ (a) Since E can be embedded in an injective R-module E ′ by the
previous corollary and E is a direct summand of E ′ by assumption, therefore
E is an injective module.
Problem set
1. Let R be a field; then every module is free.
2. The following are equivalent for a ring:
(a) Every R-module is projective.
(b) Every s.e.s. splits.
(c) Every R-module is injective.
3. Let E be an R-module; then E is injective if and only if for every ideal
I and R-homomorphism g : I −→ E there is an element x ∈ E such that
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