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

4.2. MODULES OF FRACTIONS 59
4.2.2. Definition. Let R be a ring and U a multiplicative subset. The module of
fractions U −1M is given by equivalence classes x
u on U × M under the relation
4.2.1
(u, x) ∼ (u ′ , x ′ ) ⇔ there is v ∈ U such that vu ′ x = vux ′
The addition is
x y
+
u v
and the U −1R-scalar multiplication is
a y
·
u v
The canonical homomorphism is
= vx + uy
uv
= ay
uv
M → U −1 M, x ↦→ x
1
4.2.3. Lemma. Let R be a ring and f : M → N a homomorphism of R-modules.
Then there is a homomorphism U −1f : U −1M → U −1N, x f(x)
u ↦→ u of U −1R modules.
Proof. The claims are proved by simple calculations. If (u, x) ∼ (u ′ , x ′ ) then
vu ′ x = vux ′ and therefore vu ′ f(x) = vuf(x ′ ) showing (u, f(x)) ∼ (u ′ , f(x ′ )).
So the map is well defined. The rest is similar.
4.2.4. Proposition. The construction 4.2.2
and 4.2.3
M ↦→ U −1 M
f : M → N ↦→ U −1 f : U −1 M → U −1 N
is a functor from R-modules to U −1 R-modules.
Proof. Follows from the definitions by simple calculations as in the proof of 4.2.3.
For example, U −1 (f + g) = U −1f + U −1g, follows from f(x)+g(x)
u
g(x)
u .
= f(x)
u +
4.2.5. Remark. The induced homomorphism relates to the canonical homomorphism
such that the diagram is commutative.
M
f
N
U −1M U −1f
U −1N That is, the canonical homomorphism is a natural homomorphism.
4.2.6. Proposition. Let U ⊂ R be a multiplicative subset and M a module.
(1) x
1 = 0 in U −1M if and only if Ann(x) ∩ U = ∅.
(2) The canonical homomorphism M → U −1M is injective if and only if U
consists only of nonzero divisors on M.
Proof. x
1 = 0 if and only if vx = 0 for some v ∈ U.