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

3.2. THE SNAKE LEMMA 45
3.2.4. Theorem (snake lemma). Given a commutative diagram of homomorphisms
0
M
u
M ′
f
f ′
N
v
N ′
g
g ′
L
w
L ′
where the rows exact sequences. There is induced a six term long exact sequence
Ker u
f
Ker v
δ
f ′
Cok v
Cok u
g
g ′
Ker w
Cok w
Proof. By construction of δ it is clear that the sequence is a 0-sequence: If y ∈
Ker v then to calculate δ(g(y)) the choice v(y) = 0 gives δ ◦ g = 0. Also
f ′ (δ(z)) = v(y) + Im v shows that f ′ ◦ δ = 0. Exactness at Ker v and Cok v
are clear. Given z ∈ Ker w such that δ(z) = 0. By 3.2.2 choose y, g(y) = z
and x ′ , f ′ (x ′ ) = v(y). Then δ(z) = x ′ + Im u = 0, so choose x, u(x) = x ′ .
Now v(f(x)) = f ′ (u(x)) = v(y) so y − f(x) ∈ Ker v and g(y − f(x)) =
g(y) = z. Therefore exactness at Ker w. Given x ′ + Im u ∈ Cok u such that
f ′ (x ′ ) + Im v = 0 ∈ Cok v. Choose y, v(y) = f ′ (x ′ ) and put z = g(y). Then
w(g(y)) = g ′ (v(y)) = g ′ (f ′ (x ′ )) = 0. Now z ∈ Ker w and δ(z) = x ′ + Im u.
Therefore exactness at Cok u.
3.2.5. Corollary. If f is injective then the f : Ker u → Ker v is injective and the
long exact sequence is
0
Ker u
f
Ker v
δ
f ′
Cok v
Cok u
g
g ′
0
Ker w
Cok w
If g ′ is surjective then g ′ : Cok v → Cok w is surjective and the long exact sequence
is
Ker u
f
Ker v
δ
f ′
Cok v
Cok u
g
g ′
Ker w
Cok w
3.2.6. Corollary. (1) If v is injective and u is surjective, then w is injective.
(2) If v is surjective and w is injective, than u is surjective.
(3) If v is an isomorphism, then w is injective if and only if u is surjective.
3.2.7. Proposition. Given submodules N, L ⊂ M, then there is a short exact
sequence
0
0
M/N ∩ L x↦→(x,x)
M/N ⊕ M/L
(x,y)↦→x−y
M/N + L
0