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

3.1. EXACT SEQUENCES 41
3.1.8. Proposition.
quence
(1) Let I ⊂ R be an ideal, then there is a short exact se-
0 I R R/I 0
(2) Let M ⊂ N be a submodule, then there is a short exact sequence
0
M
N
N/M
(3) For scalar multiplication with nonzero divisor a ∈ R on M the sequence
0
aM
M
M
M/aM
is a short exact sequence.
(4) Given a homomorphism f : M → N there are associated two short exact
sequences.
and
0
0
Ker f
Im f
f
M
Im f
N
Cok f
(5) For scalar multiplication with any a ∈ R on M there are associated two short
exact sequences.
and
0
0
Ker aM
aM
M
aM
M
aM
M/aM
3.1.9. Definition. Let f : M → N be a homomorphism.
(1) f has a retraction if there is a homomorphism u : N → M such that u ◦ f =
1M.
(2) f has a section if there is a homomorphism v : N → M such that f ◦v = 1N.
3.1.10. Proposition. Let f : M → N be a homomorphism.
(1) If f has a retraction u : N → M then f is injective, u is surjective and
N = Im f ⊕ Ker u
(2) If f has a section v : N → M then f is surjective, v is injective and
M = Ker f ⊕ Im v
Proof. (1) u(f(x)) = x so f is injective and u is surjective. If y ∈ N then
y = f(u(y)) + (y − f(u(y)) and u(y − f(u(y))) = 0, so N = Im f + Ker u. Let
y ∈ Im f ∩ Ker u. Then y = f(x) gives x = u(f(x)) = u(y) = 0, so y = 0.
Conclude by 2.4.5 that the sum is direct. (2) y = f(v(y)) so f is a retraction of v.
Finish by (1).
3.1.11. Lemma. For a short exact sequence
the following are equivalent
0
(1) f has a retraction.
(2) g has a section.
f
M
g
N
L
0
0
0
0
0
0
0