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

50 3. EXACT SEQUENCES OF MODULES
3.4.2. Proposition. Given a split exact sequence
0
f
M
g
N
L
and a module K. Then the following sequence is split exact
0
K ⊗R M
K ⊗R N
Proof. This follows from the functor properties 3.1.14.
0
K ⊗R L
3.4.3. Proposition. Let I ⊂ R be an ideal. For any module M, the homomorphism
is an isomorphism.
M ⊗R R/I → M/IM, x ⊗ a + I ↦→ ax + IM
Proof. x + IM ↦→ x ⊗ 1 is an inverse.
3.4.4. Corollary. Let I, J ⊂ R be ideals. Then
is an isomorphism.
R/I ⊗R R/J → R/(I + J), a + I ⊗ b + J ↦→ ab + I + J
3.4.5. Proposition. Let I1, . . . , Ik ⊂ R be pairwise comaximal ideals and M a
module. Then the product of projections
is an isomorphism.
Proof. By Chinese remainders 1.4.2
M/I1 · · · IkM → M/I1M × · · · × M/IkM
R/I1 · · · Ik → R/I1 × · · · × R/Ik
is an isomorphism. Tensor with M and use 3.4.3 to get the isomorphism.
3.4.6. Exercise. (1) Calculate Z/(m) ⊗Z Z/(n) for all integers m, n.
(2) Let I ⊂ R be an ideal. Show that R/I ⊗R R/I R/I.
(3) Let I ⊂ R be an ideal. Show that I ⊗R R/I I/I 2 .
(4) Let 2Z be scalar multiplication. Show that 2Z ⊗ 1 Z/(2) : Z ⊗Z Z/(2) → Z ⊗Z Z/(2)
is not injective.
3.5. Projective modules
3.5.1. Definition. An R-module F is a projective module if for any exact sequence
N → L → 0 the sequence
is exact.
HomR(F, N) → HomR(F, L) → 0
3.5.2. Proposition. A module F is projective if and only if any surjective homomorphism
M → F → 0 has a section.
Proof. Assume F projective and g : M → F surjective. Then HomR(F, M) →
HomR(F, F ) → 0 is exact. So there exists a v : F → M such that g ◦ v = 1F .
v is then a section. Conversely given g : N → L surjective and h : F → L. Let
M = Ker N ⊕F → L, (y, z) ↦→ g(y)−h(z) and pN : M → N, pF : M → F the
0