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

2
Modules
2.1. Modules and homomorphisms
2.1.1. Definition. Let R be a ring. A module (R-module) is an abelian group
M, addition (x, y) ↦→ x + y and zero 0, together with a scalar multiplication
R × M → M, (a, x) ↦→ ax satisfying
(1) associative : (ab)x = a(bx)
(2) bilinear : a(x + y) = ax + ay, (a + b)x = ax + bx
(3) identity: 1x = x
for all a, b ∈ R, x, y ∈ M. A submodule M ′ ⊂ M is an additive subgroup such
that ax ∈ M ′ for all a ∈ R, x ∈ M ′ . A homomorphism is an additive group
homomorphism f : M → N respecting scalar multiplication
f(x + y) = f(x) + f(y), f(ax) = af(y)
for all a ∈ R, x, y ∈ M. An isomorphism is a homomorphism f : M → N
having an inverse map f −1 : N → M which is also a homomorphism. The
identity isomorphism is denoted 1M : M → M.
2.1.2. Lemma. Let R be a ring and M a module.
(1) a0 = 0 = 0x
(2) (−1)x = −x
(3) (−a)x = −(ax) = a(−x)
for all a ∈ R, x ∈ M.
Proof. (1) Calculate a0 = a(0 + 0) = a0 + a0 and cancel to get a0 = 0. Similarly
0x = 0. (2) By (1) 0 = 0x = (1 + (−1))x = x + (−1)x, so conclude −x =
(−1)x. (3) Calculate (−a)x = ((−1)a)x = (−1)(ax) = −(ax). Similarly
a(−x) = −(ax).
2.1.3. Lemma. Let R be a ring and f : M → N a homomorphism of modules.
(1) f(0) = 0.
(2) f(ax + by) = af(x) + bf(y).
(3) f(−x) = −f(x) for all x ∈ M.
for all a, b ∈ R, x, y ∈ M.
Proof. (1) Calculate f(0) = f(0 + 0) = f(0) + f(0) and conclude f(0) = 0.
(2) Calculate f(ax + by) = f(ax) + f(by) = af(x) + bf(y). (3) By 2.1.2
f(−x) = f((−1)x) = (−1)f(x) = −f(x).
2.1.4. Example. (1) The zero group is the zero module.
(2) Over the zero ring the zero module is the only module.
(3) The zero subgroup of a module is the zero submodule.
(4) The ring R is a module under multiplication. An ideal is a submodule.
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