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

60 4. FRACTION CONSTRUCTIONS
4.2.7. Proposition. If Mα is a family of modules, then the homomorphism
U −1 (
Mα) →
is a natural isomorphism of U −1 R-modules.
α
α
U −1 Mα
Proof. This is the method of common denominators in a finite sum.
xi
=
ui
1
(Πj=iuj)xi
Πiui
i
4.2.8. Exercise. (1) Show that if U contains a nilpotent element, then U −1 M = 0.
(2) Show that
Ker M → U −1 M = {x ∈ M|ux = 0, for some u ∈ U}
(3) Let U = {u1, . . . , um} and u = u1 · · · um. Then show that
U −1 M = {u n } −1 M
(4) Show that U −1 M = 0 if and only if U ∩ Ann(x) = ∅ for all x ∈ M.
(5) Show that the fraction homomorphism of a composition is the composition of the
respective fraction homomorphisms.
(6) Let M be a free R-module. Show that U −1 M is a free U −1 R-module
4.3. Exactness of fractions
4.3.1. Proposition. Let R be a ring and U a multiplicative subset. Given an exact
sequence of R-modules
M f
Then the following sequence is exact
U −1 M
i
g
N
U −1N
L
U −1L Proof. If y
u ∈ U −1N maps to g(y)
u = 0 then there is v ∈ U such that 0 = vg(y) =
g(vy). Choose x ∈ M such that f(x) = vy. Then x
exactness.
4.3.2. Corollary. Given a short exact sequence
0
f
M
Then the following sequence is exact
0
U −1M g
N
U −1N
L
vu
maps to f(x)
vu
0
U −1L If the first sequence is split exact, also the second sequence is split exact.
0
= y
u proving
4.3.3. Corollary. For a homomorphism f : M → N there are natural isomorphisms
of U −1 R-modules.
(1) U −1 Ker f Ker U −1 f.
(2) U −1 Im f Im U −1 f.
(3) U −1 Cok f Cok U −1 f.