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

5.3. LOCALIZATION OF MODULES 69
5.3.3. Proposition. Let R be a ring and P a prime ideal. If Mα is a family of
modules, then the homomorphism
(
Mα)P →
(Mα)P
is an isomorphism of RP -modules.
Proof. See 4.2.7.
5.3.4. Corollary. For a homomorphism f : M → N
(1) (Ker f)P Ker fP .
(2) (Im f)P Im fP .
(3) (Cok f)P Cok fP .
Proof. See 4.3.3.
α
5.3.5. Corollary. Let R be a ring and P a prime ideal. For submodules N, L ⊂ M
(1) (M/N)P MP /NP .
(2) (N + L)P NP + LP .
(3) (N ∩ L)P NP ∩ LP .
Proof. See 4.3.4.
5.3.6. Proposition. Let R be a ring, P a prime ideal and M a module.
(1) MP M ⊗R RP .
(2) MP /P RP MP M ⊗R k(P ).
Proof. See 4.4.1.
5.3.7. Proposition. Let R be a ring, P a prime ideal.
(1) For an R module M and an RP -module N there is a natural isomorphism
M ⊗R RP ⊗RP N M ⊗R N
(2) For an R module M, L there is a natural isomorphism
Proof. See 4.2.7 and 2.7.4.
(M ⊗R L)P MP ⊗RP LP
5.3.8. Definition. Let R be a ring. F is a locally free module is FP is a free
RP -module for all prime ideals P .
5.3.9. Lemma. Let R be a ring. F is a locally free module if FQ is a free RQmodule
for all maximal ideals Q.
Proof. A prime ideal P ⊂ Q is contained in a maximal ideal. By 5.3.6 FP
(FQ)PQ is free.
5.3.10. Example. A free module is a locally free module.
5.3.11. Exercise. (1) Let P ⊂ R be a prime ideal and R → S a ring homomorphism.
Show that RP → SP is a ring homomorphism.
(2) Let Q ⊂ S be a prime ideal and R → S a ring homomorphism. Show that RQ∩R →
SQ is a local ring homomorphism.
(3) Let R = K × L be a product of fields. Show that ideal K × {0} is locally free but
not free.
α