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

42 3. EXACT SEQUENCES OF MODULES
For any retraction u there is a unique section v and wise-verse such that
1N = f ◦ u + v ◦ g
Proof. If u is a retraction of f, then Ker g = Im f ⊂ Ker(1N − f ◦ u). By 3.1.5
there is a homomorphism v : L → N such that v ◦ g = 1N − f ◦ u. This is a
section of g. Conversely if v is a section of g then Im(1N − v ◦ g) ⊂ Ker g, so
there is a homomorphism u : N → M such that f ◦ u = 1N − v ◦ g, 3.1.4. u is a
retraction of f. The equation is clearly satisfied.
3.1.12. Definition. Let R be a ring and f : M → N, g : N → L homomorphisms.
A short exact sequence
0
f
M
g
N
is a split exact sequence if equivalently 3.1.11 f has a retraction or g has a section.
3.1.13. Proposition. A sequence
0
f
M
g
N
is a split exact sequence if and only if there are homomorphism u : N → M, v :
L → N satisfying
L
L
0
0
g ◦ f = 0, u ◦ f = 1M, g ◦ v = 1L, f ◦ u + v ◦ g = 1N
If the sequence is split exact then
0
v
L
u
N
N
is split exact and (x, y) ↦→ f(x)+v(y) and z ↦→ u(z)+g(z) gives the isomorphism
M ⊕ L N
Proof. The sequence is a 0-sequence f is injective and g is surjective. From f ◦
u + v ◦ g = 1N follows that z ∈ Ker g ⊂ Im f, so the sequence is short exact. The
rest is contained in 3.1.10.
3.1.14. Corollary. A (contravariant) functor preserves split exact sequences. If
0
f
M
is split exact and T a functor, then
is split exact.
0
g
N
L
T (M) T (f)
T (N) T (g)
T (L)
Proof. By 3.1.13 a split exact sequence is characterized by a set of equations.
These are preserved by the functor, 2.5.4.
3.1.15. Example. A short exact sequence
0
f
M
g
N
where L is a free module is a split exact sequence. Namely let xα ∈ L be a
basis and choose yα ∈ N with g(yα) = xα. The define v : L → N by setting
v(xα) = yα, 2.4.11.
L
0
0
0
0