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

2.7.4. Proposition. The construction
2.7. CHANGE OF RINGS 37
M ↦→ M ⊗R S
and
f : M → M ′ ↦→ f ⊗ 1S : M ⊗R S → M ′ ⊗R S
is a functor from R-modules to S-modules.
Proof. This is clear from 2.7.2 and 2.6.7.
2.7.5. Proposition. Let R → S be a ring homomorphism, M an R-module and N
an S-modules. Then there is a natural isomorphism of S-modules.
M ⊗R S ⊗S N M ⊗R N, x ⊗ b ⊗ y ↦→ x ⊗ by
Proof. The homomorphism v : S ⊗S N → N, b ⊗ y ↦→ by is an isomorphism,
2.6.10. The homomorphism 1M ⊗ v : M ⊗R S ⊗S N M ⊗R N is an R-module
isomorphism, 2.6.7. The identity x ⊗ bc ⊗ y = x ⊗ b ⊗ cy proves this to be an
S-module homomorphism.
2.7.6. Proposition. Let R → S be a ring homomorphism, M an R-module and N
an S-modules. Then there is a natural isomorphism
HomR(M, N) HomS(M ⊗R S, N), f ↦→ (x ⊗ b ↦→ bf(x))
Proof. A given f is mapped to the composite M⊗R → N ⊗R S → N which is an
S-homomorphism. Given a homomorphism g : M ⊗R S → N then the composite
M → M ⊗R S → N is an R-homomorphism and an inverse to the first given
map.
2.7.7. Lemma. Let R → S be a ring homomorphism, M an R-module and N an
S-module. Then
S × HomR(N, M) → HomR(N, M), (b, f : N → M) ↦→ (y ↦→ f(by))
is an S-scalar multiplication.
Proof. The map (b, f) ↦→ f ◦ bN satisfies the laws 2.1.1.
2.7.8. Definition. Let R → S be a ring homomorphism and M an R-module. The
induced module is the S-module HomR(S, M) with S-scalar multiplication 2.7.7
S × HomR(S, M) → HomR(S, M), (b, f : S → M) ↦→ (c ↦→ f(bc))
2.7.9. Proposition. The induced module
M ↦→ HomR(S, M)
and
f : M → M ′ ↦→ Hom(1S, f) : HomR(S, M) → HomR(S, M ′ )
is a functor from R-modules to S-modules.
Proof. This is clear from 2.7.7 using 2.5.4.
2.7.10. Proposition. Let R → S be a ring homomorphism and M an R-module
and N an S-modules. Then there is a natural isomorphism
HomR(N, M) HomS(N, HomR(S, M)), f ↦→ (y ↦→ [b ↦→ f(by)])
Proof. g ↦→ (y ↦→ g(y)(1)) is an inverse.
2.7.11. Example. Let I ⊂ R be an ideal and R → R/I the projection.