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

38 2. MODULES
(1) The change of ring functor maps an R-module M to the R/I-module M/IM.
The natural isomorphism 2.7.6 is
HomR(M, N) Hom R/I(M/IM, N)
for any R/I-module N.
(2) The induced module functor maps an R-module M to the R/I-module {x ∈
M|I ⊂ Ann(x)}. The natural isomorphism 2.7.10 is
HomR(N, M) Hom R/I(N, HomR(R/I, M))
for any R/I-module N.
2.7.12. Definition. Let R → S, S ′ be ring homomorphisms. The tensor product
ring over R is S ⊗R S ′ with multiplication given by (b ⊗ b ′ )(c ⊗ c ′ ) = bc ⊗ b ′ c ′
extended by linearity. R → S ⊗R S ′ , r ↦→ r ⊗ 1 = 1 ⊗ r is the natural ring
homomorphism.
2.7.13. Proposition. Let φ, φ ′ : R → S, S ′ and ψ, ψ ′ : S, S ′ → T give a commutative
diagram of ring homomorphisms, ψ ◦ φ = ψ ′ ◦ φ ′ . Then b ⊗ b ′ ↦→ ψ(b)ψ ′ (b ′ )
is the unique homomorphism making the following diagram commutative.
Proof. This is clear by 2.6.3.
R
S ′
S
S ⊗R S
′
T
2.7.14. Example. Let R → S be a ring homomorphism. Then
is an isomorphism.
R[X] ⊗R S S[X]
2.7.15. Exercise. (1) Show that the change of rings of a free R-module is a free Smodule.
(2) Let φ : R → S be a ring homomorphism. Show that the change of rings of a
scalar multiplication a : M → M on an R-module is a scalar multiplication φ(a) :
M ⊗R S → M ⊗R S.
(3) Show that the change of rings of the composition of two homomorphisms is the
composition of the change of rings of each homomorphism.
(4) Show the isomorphism
R[X] ⊗R R[Y ] R[X, Y ]