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

22 2. MODULES
(5) If R is a field, a module is a vector space and a homomorphism is a linear
map.
(6) A module over Z is an abelian group and an additive map of abelian groups
is a homomorphism.
2.1.5. Proposition. A bijective homomorphism is an isomorphism.
Proof. Let f : M → N be a bijective homomorphism of R-modules and let g :
N → M be the inverse map. For x, y ∈ N write x = f(g(x)), y = f(g(y)) and
get additivity of g, g(x + y) = g(f(g(x)) + f(g(y))) = g(f(g(x) + g(y))) =
g(x) + g(y). Similarly for a ∈ R g(ax) = g(af(g(x))) = g(f(ag(x))) = ag(x),
so g respects scalar multiplication and is a homomorphism.
2.1.6. Lemma. Let a ∈ R and M be a module. The map M → M, x ↦→ ax is a
homomorphism.
Proof. Let f(x) = ax and calculate f(x+y) = a(x+y) = ax+ay = f(x)+f(y)
and f(bx) = a(bx) = (ab)x = (ba)x = b(ax) = bf(x) to get that f is a
homomorphism. Remark that the last calculation uses that R is commutative 1.1.2
(4).
2.1.7. Definition. Let a ∈ R and M be a module.
(1) The scalar multiplication with a is the homomorphism, 2.1.6,
aM : M → M, x ↦→ ax
(2) a ∈ R is a nonzero divisor on M if scalar multiplication aM is injective, i.e.
ax = 0 for all 0 = x ∈ M. Otherwise a is a zero divisor.
2.1.8. Remark. The two notions of nonzero divisor on R : 1.1.6 as element in the
ring and 2.1.4 as scalar multiplication on the ring coincide.
2.1.9. Example. If R is a field, then scalar multiplication on a vector space is either
zero or an isomorphism.
2.1.10. Lemma. Let φ : R → S be a ring homomorphism and N an S-module.
The map
R × N → N, (a, x) ↦→ φ(a)x
is an R-scalar multiplication on N, 2.1.1.
Proof. Let a, b ∈ R, x, y ∈ N and µ(a, x) = φ(a)x. Calculate µ(a + b, x) =
φ(a + b)x = φ(a)x + φ(b)x = µ(a, x) + µ(b, x), µ(a, x + y) = φ(a)(x + y) =
φ(a)x + φ(a)y = µ(a, x) + µ(a, y), µ(1, x) = φ(1)x = 1x = x and µ(ab, x) =
φ(ab)x = φ(a)φ(b)x = µ(a, µ(bx)) showing the conditions 2.1.1.
2.1.11. Definition. Let φ : R → S be a ring homomorphism. The restriction of
scalars of an S-module N is the same additive group N viewed as an R-module
through φ. The scalar multiplication is 2.1.10
R × N → N, (a, x) ↦→ ax = φ(a)x
An S-module homomorphism g : N → N ′ is also an R-module homomorphism.
2.1.12. Example. (1) The scalar multiplication with a Restriction of scalars for
the unique ring homomorphism Z → R give just the underlying abelian group
of a module, 2.1.4 (6).