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

30 2. MODULES
2.4.11. Proposition. Let F be a free module with basis yα. For a module M and
a family of elements xα ∈ M there is a unique homomorphism g : F → M such
that g(yα) = xα given by g( aαyα) = aαxα.
Proof. The basis yα ∈ F gives the isomorphism 2.4.7 f :
α R → F . The
family 1xα : R → M gives a homomorphism g ′ :
R → M by 2.4.3. Then
g = g ′ ◦ f −1 .
2.4.12. Corollary. Let M be an R-module and
M R the free module with basis
ex indexed by x ∈ M. The homomorphism
R → M, axex ↦→ ax x
M
is surjective identifying M as a factor module of a free module in a natural way.
2.4.13. Definition. A module is indecomposable if it is not isomorphic to a direct
sum of two nonzero submodules, otherwise decomposable.
2.4.14. Example. Q is an indecomposable Z-module. Namely if m1
n1
nonzero numbers in two submodules, then n1m2 m1
n1
ber in the intersection.
α
= n2m1 m2
n2
, m2
n2 are
is a nonzero num-
2.4.15. Exercise. (1) Show that if a ring is decomposable as a module, then it is the
product of two nonzero rings.
(2) Let Mα be a finite family of modules. Show that Mα = Mα,
(3) Let Nα ⊂ Mα be a family of submodules modules. Show that
Mα/ Nα Mα/Nα
and that Mα/ Nα Mα/Nα
2.5. Homomorphism modules
2.5.1. Lemma. Let R be a ring and f, g : M → N homomorphisms.
(1) (f + g)(x) = f(x) + g(x) is a homomorphism.
(2) If a ∈ R, then (af)(x) = af(x) is a homomorphism.
Proof. Calculate according to 2.1.1. (1) (f + g)(x + y) = f(x + y) + g(x + y) =
f(x) + f(y) + g(x) + g(y) = (f + g)(x) + (f + g)(y) and (f + g)(ax) =
f(ax) + g(ax) = a(f(x) + g(x)) = a(f + g)(x). (2) (af)(x + y) = af(x + y) =
af(x) + af(y) = (af)(x) + (af)(y) and (af)(bx) = af(bx) = abf(x) =
baf(x) = b(af)(x). The last calculation uses that R is commutative 1.1.2 (4).
2.5.2. Definition. Let R be a ring and M, N modules. By 2.5.1, the homomorphism
module HomR(M, N) is the additive group of all homomorphism with
scalar multiplication
R × HomR(M, N) → HomR(M, N), (a, f) ↦→ af = [x ↦→ af(x)]
2.5.3. Definition. Let a ∈ R be a ring and f : M → M ′ , g : N → N ′ , h, k :
M ′ → N homomorphisms of modules. By 2.5.1
(1) (h + k) ◦ f = h ◦ f + k ◦ f.
(2) (ah) ◦ f = a(h ◦ f).
(3) g ◦ (h + k) = g ◦ h + g ◦ k.
(4) g ◦ (ah) = a(g ◦ h).