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

28 2. MODULES
(2) Give an example of a homomorphism f : M → N submodules N1, N2 ⊂ N such
that f −1 (N1 + N2) = f −1 (M1) + f −1 (M2).
(3) Let R be a ring and M a module. Show that M may be regarded as an R/ Ann(M)module
in a natural way.
(4) Let L, N ⊂ M be submodules. Show that Ann(L + N) = Ann(L) ∩ Ann(N).
(5) Let f : M → N be a surjective homomorphism. Show that Ann(M) ⊂ Ann(N).
(6) Let f : M → N be an injective homomorphism. Show that Ann(N) ⊂ Ann(M).
2.4. Sum and product
2.4.1.
Lemma. Let R be a ring and (Mα) a family of modules. The product
α Mα is the abelian group of all families (xα), xα ∈ Mα with term wise addition.
The setting
r(xα) = (rxα)
is a scalar multiplication on
α Mα. The direct sum
α Mα is the subgroup
of
α Mα consisting of families with only finitely many nonzero terms. This is a
submodule.
Proof. The laws in 2.1.1 are true for each factor and therefore trivially verified for
the product and sum.
2.4.2. Definition. Let R be a ring and Mα a family of modules. By 2.4.1 there are
modules and homomorphisms
(1) The direct product is
α Mα.
(2) The projections pβ :
α Mα → Mβ are the homomorphisms pβ((xα)) =
xβ.
(3) The direct sum is
α Mα. Elements in
α Mα
are written as finite sums
xα.
(4) The injections iβ : Mβ →
α Mα are the homomorphisms given by iβ(x) =
(xα), where xβ = x and xα = 0, α = β.
2.4.3. Proposition. Let R be a ring and Mα a family of modules.
(1) Given a family of homomorphisms fα : L → Mα, then there exists a unique
homomorphism f : L → Mα such that fα = pα ◦ f.
L
fα
f
Mα
pα
α Mα
(2) Given a family of homomorphisms gα : Mα → L, then there exists a unique
homomorphism g : Mα → L such that gα = g ◦ iα.
α Mα
g
iα
gα
Mα
Proof. (1) f(y) = (fα(y)) is the unique homomorphism. (2) g( xα) = gα(xα)
is well defined since only finitely many xα = 0 and a homomorphism.
L