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

82 6. FINITE MODULES
6.5.9. Corollary. Let P ⊂ R be a prime ideal and M a finite presented module.
For any module N there is a natural isomorphism
HomR(M, N)P HomRP (MP , NP )
6.5.10. Proposition. Let R be a ring and F a module. The following conditions
are equivalent.
(1) F is flat.
(2) For any module N and a relation 0 =
i yi ⊗ xi ∈ N ⊗R F , there exist
zj ∈ F and aij ∈ R such that 0 =
i aijyi ∈ N and xi =
j aijzj ∈ F .
(3) For any relation 0 =
i bixi ∈ F , there exist zj ∈ F and aij ∈ R such that
0 =
i aijbi ∈ R and xi =
j aijzj ∈ F .
Proof. (1) ⇒ (2): Let f : Rn → N, ei ↦→ yj, then 0 → Ker f ⊗R F → F n →
N ⊗R F is exact. By assumption (xi) ∈ Ker f ⊗R F , so (xi) =
j aij ⊗ zj
with (aij) ∈ Ker f. (2) ⇒ (3) is clear. (3) ⇒ (1): Let I ⊂ R be an ideal and
bi ⊗ xi ∈ Ker(I ⊗R F → F ). Then 0 =
i aijbi and xi =
j aijzj. Now
calculate bi ⊗ xi =
j i aijbi ⊗ xi = 0. By 3.7.12 F is flat.
6.5.11. Proposition. Let (R, P ) be a local ring and F a finite presented module.
The following conditions are equivalent.
(1) F is free.
(2) F is projective.
(3) F is flat.
(4) P ⊗R F → F is injective.
Proof. (1) ⇒ (2) ⇒ (3) ⇒ (4) are clear. (4) ⇒ (1): Choose xi ∈ F such that
xi ⊗ 1 give a basis for F ⊗R k(P ). The homomorphism f : R n → F, ei ↦→ xi is
surjective by 6.4.5. P ⊗R Ker f → P n → P ⊗R F → 0 is exact, so P Ker f =
Ker(P n → P ⊗R F ) = Ker(P n → P ⊗R F → F ) = Ker f. by the hypothesis.
Ker f is finite 6.5.5 and therefore Ker f = 0 by 6.4.1, so F is free.
6.5.12. Corollary. Let (R, P ) be a local ring and f : F → F ′ a homomorphism
of finite free modules. The following conditions are equivalent.
(1) f has a retraction u : F ′ → F .
(2) f is injective and Cok f is free.
(3) f(P ) is injective.
Proof. (1) ⇒ (2): Cok f is projective, so free by 6.5.11. (2) ⇒ (3) is clear. (3) ⇒
(1): Let f ∨ : F ′∨ → F ∨ be the dual homomorphism. f ∨ (P ) is surjective, so f ∨ is
surjective, 6.4.5. A section v of f ∨ gives a retraction u = v ∨ .
6.5.13. Corollary. Let R be a ring and F a finite presented module. The following
conditions are equivalent.
(1) F is projective.
(2) F is flat.
(3) F is locally free.
(4) FP is free for all maximal ideals P .
Proof. (1) ⇒ (2) ⇒ (3) ⇒ (4) are clear by 6.5.11. (4) ⇒ (1): Let N → L → 0
be exact. By hypothesis HomRP (FP , NP ) → HomRP (FP , LP ) → 0 is exact for
all maximal ideals. By 6.5.9 HomR(F, N)P → HomR(F, L)P → 0 is exact for