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

4.5. HOMOMORPHISM MODULES OF FRACTIONS 63
Proof. This follows from 2.7.5 and 4.4.1.
4.4.5. Corollary. Let I ⊂ R be an ideal and U a multiplicative subset For a
module M module
U −1 (IM) U −1 IU −1 M
Proof. Use that IM = Im(I ⊗R M → M) and 4.4.4.
4.4.6. Exercise. (1) Let M be a flat R-module. Show that U −1 M is a flat U −1 Rmodule.
(2) Let M be a projective R-module. Show that U −1 M is a projective U −1 R-module.
4.5. Homomorphism modules of fractions
4.5.1. Proposition. Let R be a ring and U a multiplicative subset. For any modules
M, N there is a natural homomorphism
of U −1 R-modules.
U −1 HomR(M, N) → Hom U −1 R(U −1 M, U −1 N)
Proof. Given f : M → N and u ∈ U the setting x
homomorphism U −1 M → U −1 N.
v ↦→ f(x)
uv is a U −1R 4.5.2. Proposition. Let R be a ring and U a multiplicative subset. For any Rmodule
M and any U −1 R-modules N there is a natural isomorphism
of U −1 R-modules.
HomR(M, N) → Hom U −1 R(U −1 M, N)
Proof. This is the change of rings isomorphism 2.7.6 interpreted according to
4.4.2.
4.5.3. Example. Let R be a ring and U a multiplicative subset. R → U −1 R the
canonical homomorphism. The induced module functor maps an R-module M to
the U −1 R-module HomR(U −1 R, M). The natural isomorphism 2.7.10 is
for any U −1 R-module N.
HomR(N, M) Hom U −1 R(N, HomR(U −1 R, M))
4.5.4. Example. The homomorphism 4.5.1 is in general neither injective nor surjective.
(1) Not surjective:
(2) Not injective:
0 = HomZ(Q, Z) → HomQ(Q, Q) = Q
0 = HomZ(Q, Q/Z) ⊗Z Q → HomQ(Q, Q/Z ⊗Z Q) = 0
4.5.5. Exercise. (1) Let M be a Z-module and N a Q-module. Show that there is an
isomorphism HomZ(M, N) HomQ(M ⊗Z Q, N).