Math 654
Homework #3
February 6, 2013
Due Thursday, February 14, in class.
Justify all of your work.
C1. Modules of fractions. Let R be a commutative ring with 1. Let S ⊆ R be a subset closed
under multiplication with 1 ∈ S. Recall from section III.4 of Hungerford, we can use S to
form the ring of fractions S −1 R. Now let M be a left R-module. We put an equivalence
relation ≈ on the set M × S by defining
m
m0
≈ 0
s
s
⇔
∃ t ∈ S such that t(sm0 − s0 m) = 0.
We then define S −1 M to be the equivalence classes of ≈:
nm
o.
S −1 M =
m ∈ M, s ∈ S
≈,
s
(a) Show that S −1 M is an S −1 R-module in a natural way.
(b) If φ : M → N is an R-module homomorphism, define an S −1 R-module homomorphism
S −1 φ : S −1 M → S −1 N such that
S −1 φ
m
s
=
φ(m)
.
s
Thus show that S −1 (·) defines a covariant functor from left R-modules to left S −1 Rmodules.
(c) Show that S −1 (·) is exact.
C2. Let R be a commutative ring with 1, and let M be an R-module. For a prime ideal p ⊆ R,
let Sp = R − p and let
Rp = Sp−1 R,
Mp = Sp−1 M.
Show that the following are equivalent: (i) M = 0; (ii) Mp = 0 for all prime ideals p of R;
(iii) Mm = 0 for all maximal ideals m of R.
√
√
C3. Let R = Z[ −6] = {a + b −6 | a, b ∈ Z}.
(a) Show that a non-zero ideal I ⊆ R is free as an R-module if and only if it is a principal
ideal.
√
√
(b) Let J = (2, −6) be the ideal generated by 2 and −6. Show that J is not a principal
ideal.
(c) Show that J is projective as an R-module. (Hint: Consider the natural R-module map
R2 → J and then devise a splitting.) Thus J is projective but not free.
1
C4. Let R be a ring with 1 6= 0, and let M be a finitely generated left R-module.
(a) Suppose that M is projective as an R-module. Prove there exist elements m1 , . . . , mk ∈
M and R-module homomorphisms fi : M → R, 1 ≤ i ≤ k, such that for all m ∈ M ,
m=
k
X
i=1
(b) Prove that the converse of (a) is true.
2
fi (m)mi .