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

2.3. KERNEL AND COKERNEL 25
2.3. Kernel and cokernel
2.3.1. Lemma. Let R be a ring and f : M → N a homomorphism of modules.
Given submodules M ′ ⊂ M, N ′ ⊂ N, then f −1 (N ′ ) ⊂ M and f(M ′ ) ⊂ N are
submodules.
Proof. If x, y ∈ f −1 (N ′ ) then f(x + y) = f(x) + f(y) ∈ N ′ and for a ∈ R
f(ax) = af(x) ∈ N ′ so x+y, ax ∈ f −1 (N ′ ) proving f −1 (N ′ ) to be a submodule.
The same equations prove that f(M ′ ) is a submodule.
2.3.2. Definition. Let f : M → N be a homomorphism of modules. Then there
are submodules, 2.3.1.
(1) The kernel Ker f = f −1 (0).
(2) The image Im f = f(M).
(3) The cokernel Cok f = N/ Im f.
2.3.3. Proposition. Let f : M → N be a homomorphism of modules.
(1) f is injective if and only if Ker f = 0.
(2) f is surjective if and only if Cok f = 0.
(3) f is an isomorphism if and only if Ker f = 0 and Cok f = 0.
Proof. (1) If f is injective and x ∈ Ker f then f(x) = 0 = f(0) so x = 0.
Conversely if Ker f = 0 and f(x) = f(y) then f(x − y) = 0 so x = y. (2) The
factor module N/ Im f = 0 if and only if Im f = N. (3) This follows from (1)
and (2).
2.3.4. Example. Let a ∈ R give scalar multiplication aM : M → M, x ↦→ ax.
(1) Im aM = aM = {ax ∈ M|x ∈ M}.
(2) Ker aM = {x ∈ M|ax = 0}.
(3) Cok aM = M/aM.
2.3.5. Proposition. Let f : M → N be a homomorphism of modules.
(1) Let L ⊂ Ker f be a submodule. Then there is a unique homomorphism f ′ :
M/L → N such that f = f ′ ◦ p.
M
p
M/L
f
(2) The homomorphism f ′ : M/ Ker f → N is a module isomorphism onto the
submodule Im f of N.
M
p
M/ Ker f
f
f ′
f ′
N
Im f
Proof. (1) If x+L = x ′ +L then x−x ′ ∈ L giving f(x) = f(x ′ ). The factor map
f ′ : M/L → N, x + L ↦→ f(x) is therefore well defined and f = f ′ ◦ p. Since
f, p are homomorphisms and p is surjective it follows that f ′ is a homomorphism.
(2) The kernel of f ′ is Ker f/ Ker f = 0 so by 2.3.3 it is an isomorphism.