Commutative algebra - Department of Mathematical Sciences - old ...

52 3. EXACT SEQUENCES OF MODULES
3.5.11. Proposition. Any module M admits an exact sequence
F → M → 0
where F is a projective module. That is, any module is a factor module of a projective
module.
Proof. Take F free, 2.4.12.
3.5.12. Exercise. (1) Let R = R1 × R2. Show that R1, R2 are projective ideals in R.
(2) Show that the ideal (2)/(6) in the ring Z/(6) is projective.
(3) Show that the ideal (2)/(4) in the ring Z/(4) is not projective.
3.6. Injective modules
3.6.1. Definition. An R-module E is an injective module if for any exact sequence
0 → M → N the sequence
is exact.
HomR(N, E) → HomR(M, E) → 0
3.6.2. Proposition. A module E is an injective module if and only if any injective
homomorphism 0 → E → L has a retraction.
Proof. Assume E injective and f : E → L injective. Then HomR(L, E) →
HomR(E, E) → 0 is exact. So there exists a u : L → E such that u ◦ f = 1E.
Then u is a retraction. Conversely given f : M → N injective and h : M → E.
Let L = Cok M → E ⊕ N, x ↦→ h(x) − f(x) and iE : E → L, iN : N → L the
injections, then iN ◦ f = iE ◦ h. Now iE is injective since f is. Let u : L → E be
a retraction of iE, then h ′ = u ◦ iN satisfies h = h ′ ◦ f.
0
M
3.6.3. Corollary. A short exact sequence
0
f
E
f
N
h
h
′
E iE
L
u
g
N
iN
L
where E is an injective module is a split exact sequence.
3.6.4. Example. Let I ⊂ R be an ideal. If I is injective, then there is a ring
decomposition R/I × R ′ R.
3.6.5. Proposition. A direct summand in an injective module is injective.
Proof. Let E ⊕ E ′ be an injective module and f : E → L an injective homomorphism.
By 3.6.2 there is a retraction u ′ : L ⊕ E ′ → E ⊕ E ′ to (f, 1E ′). Then
u(y) = pE ◦ u ′ (y, 0) is a retraction to f and E is injective.
3.6.6. Proposition. Let Eα be a family of injective modules, then the direct product
α Eα is an injective module
0