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

40 3. EXACT SEQUENCES OF MODULES
3.1.4. Proposition. The 0-sequence
0
f
M
g
N
is exact if and only if the following equivalent statements are satisfied.
(1) f is an isomorphism onto Ker g.
(2) Given a homomorphism h : K → N such that g ◦ h = 0 then there is a
unique h ′ : K → M such that h = f ◦ h ′ .
0
f g
M
N
h
K
′
h
Proof. (1) This is clearly equivalent with exactness. (2) Assume the sequence exact.
Im h ⊂ Ker g = Im f, so by (1) put h ′ = f −1 ◦ h. Assume (2) satisfied and
apply it to Ker g → M to see that (1) is satisfied.
3.1.5. Proposition. The 0-sequence
M f
g
N
is exact if and only if the following equivalent statements are satisfied.
(1) The factor homomorphism 2.3.9 g ′ : Cok f → L induced by g is an isomorphism.
(2) Given a homomorphism k : N → K such that k ◦ f = 0 then there is a
unique k ′ : L → K such that k = k ′ ◦ g.
M f
L
g
N
L
k
k
′
K
Proof. (1) The equivalence follows from 2.3.9. (2) Assume the sequence exact. By
2.3.5 there is k ′′ : Cok f → K such that k ′′ ◦ p = k. By (1) put k ′ = k ′′ ◦ g −1 .
Assume (2) satisfied and apply it to N → Cok f to see that (1) is satisfied.
3.1.6. Proposition. Let
Mα
Nα
Lα
L
L
be a family of exact sequences. Then there are exact sequences:
(1) The sum
Mα
Nα
Lα
(2) The product
Mα
Nα
0
0
Lα
Proof. Clear since kernel and image is calculated componentwise.
3.1.7. Definition. An exact sequence
0
f
M
g
N
is a short exact sequence. That is f is injective, Im f = Ker g and g is surjective.
L
0