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

54 3. EXACT SEQUENCES OF MODULES
Proof. By 2.4.12 choose a surjection F → HomZ(M, Q/Z) where F is a free
R-module. Then
0 → M → HomZ(F, Q/Z)
is exact by 3.6.11. Since F
α R the module
HomZ(F, Q/Z)
HomZ(R, Q/Z)
is injective, 3.6.6 and 3.6.10.
3.6.13. Proposition. A homomorphism f : M → N is injective if and only if
is surjective for any injective module E.
HomR(N, E) → HomR(M, E) → 0
Proof. Assume that Ker f = 0 and choose 0 → Ker f → E, 3.6.12. The sequence
HomR(N, E) → HomR(M, E) → HomR(Ker f, E) → 0
is exact. So HomR(N, E) → HomR(M, E) is not surjective.
3.6.14. Definition. A submodule N ⊂ M is an essential extension if any nonzero
submodule L ⊂ M has nonzero intersection N ∩ L = 0. An essential extension
M ⊂ E with E injective is an injective envelope of M.
3.6.15. Proposition. Any module M has an injective envelope. If M ⊂ E, E ′ are
two injective envelopes, then there is an isomorphism f : E → E ′ fixing M.
Proof. By 3.6.12 choose M ⊂ E ′ with E ′ injective. By Zorn’s lemma choose
M ⊂ E ⊂ E ′ maximal among the essential extensions of M. If E = E ′ then the
set of modules N ′ ⊂ E ′ such that E ∩ N ′ is nonempty by maximality of E. Let
N be maximal among these by Zorn’s lemma. It follows that E ⊕ N = E ′ and E
is injective by 3.6.6. Given two envelopes, let f : E → E ′ be any homomorphism
fixing M. Then f is injective, since M ⊂ E is essential. If f is not surjective, then
E ′ f(E) ⊕ E ′′ contradicting that M ⊂ E ′ is essential.
3.6.16. Exercise. (1) Let R be a domain. Show that the fraction field is an injective
module.
(2) Let R be a domain. Show that the torsion free divisible module injective.
(3) Show that for a ring that all modules are projective if and only all modules are injective.
α
3.7. Flat modules
3.7.1. Definition. An R-module F is a flat module if for any exact sequence 0 →
M → N the sequence
0 → M ⊗R F → N ⊗ F
is exact.
3.7.2. Example. If a ∈ R is a nonzero divisor and M is a flat modules, then aM
is injective and a is a nonzero divisor on M.
3.7.3. Proposition. A direct summand in a flat module is flat.
Proof. Let F ⊕ F ′ be flat and M → N injective. Then M ⊗R (F ⊕ F ′ ) →
N ⊗R (F ⊕ F ′ ) is injective. Conclusion by 2.6.11.