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

36 2. MODULES
2.6.12. Example. Let R be a ring, F a free module with basis yα and G a free
module with basis zβ. Then F ⊗R G is a free module with basis yα ⊗ zβ.
2.6.13. Proposition. Let R be a ring and M, N, L modules. Then there is a natural
isomorphism
HomR(M ⊗R N, L) HomR(M, HomR(N, L))
(x ⊗ y ↦→ g(x)(y)) ← g
f ↦→ (x ↦→ [y ↦→ f(x ⊗ y)])
Proof. A given f : M ⊗R N → L is mapped to the composite homomorphism
M → HomR(N, M ⊗R N) → HomR(N, L), 2.5.4 and 2.6.2. This is a homomorphism
as map of f by 2.5.4. Given g : M → HomR(N, L) the map M × N →
L, (x, y) ↦→ g(x)(y) induces a homomorphism M ⊗R N → L, x ⊗ y ↦→ g(x)(y)
by 2.6.3. Clearly the maps are inverse to each other and therefore giving an isomorphism
by 2.1.5.
2.6.14. Exercise. (1) Show that Q ⊗Z Q/Z 0.
(2) Show that Z/(m) ⊗Z Z/(n) = 0 if (m, n) = Z.
(3) Let P, Q ⊂ R be different maximal ideals and M a module. Show that M/P M ⊗R
M/QM = 0.
2.7. Change of rings
2.7.1. Proposition. (1) Let φ : R → S be a ring homomorphism and N an S
module. The restriction scalars 2.1.11 viewing N as an R-module through
φ, R × N → N, (a, x) ↦→ ax = φ(a)x, is a functor from S-modules to
R-modules.
(2) Let I ⊂ R be an ideal. Restriction of scalars along R → R/I identifies
R/I-modules M with R-modules such that I ⊂ Ann(M). For R/I-modules
M, N there is a natural isomorphism
HomR(M, N) Hom R/I(M, N)
Proof. This is a restatement of 2.1.11 using 2.5.4.
2.7.2. Lemma. Let R → S be a ring homomorphism, M an R-module and N an
S-module. Then
is an S-scalar multiplication.
S × M ⊗R N → M ⊗R N, (b, x ⊗ y) ↦→ x ⊗ by
Proof. For fixed b ∈ S the map M × N → M ⊗R N, (x, y) ↦→ x ⊗ by induces
the homomorphism µb : M ⊗R N → M ⊗R N, x ⊗ y ↦→ x ⊗ by by 2.6.3. This
gives a well defined scalar multiplication S × M ⊗R N → M ⊗R N, (b, x ⊗ y) ↦→
µb(x ⊗ y).
2.7.3. Definition. Let R → S be a ring homomorphism and M an R-module. The
change of ring S-module is M ⊗R S with S-scalar multiplication 2.7.2
S × M ⊗R S → M ⊗R S, (b, x ⊗ c) ↦→ x ⊗ bc