Answer 6
(1) Let R be a commutative ring and S a multiplicatively closed set
in R and without loss of generality, we will assume that 1 ∈ S.
As usual, we will denote for any R-module M , the natural map
from M → S −1 M by j. So, j(m) = [(m, 1)].
(a) Show that the kernel of j : M → S −1 M is the set of all
elements m ∈ M such that sm = 0 for some s ∈ S.
If sm = 0, then j(m) = [(m, 1)] = [(sm, s)] = [(0, s)] = 0.
Conversely, if [(m, 1)] = 0 = [(0, 1)] and thus we have some
s ∈ S such that s(m · 1 − 0 · 1) = sm = 0.
i
π
(b) Show that if 0 → M → N → P → 0 is an exact sequence
i0
π0
of R-modules, 0 → S −1 M → S −1 N → S −1 P → 0 is an
exact sequence of S −1 R-modules.
This is routine checking after observing that i0 ([(m, s)]) =
[(i(m), s)] and π 0 ([(n, s)]) = [(π(n), s)].
We should show that i0 is injective, π 0 is surjective and the
kernel of π 0 is the image of i0 . If i0 ([(m, s)]) = 0, we have
[(i(m), s)] = 0 and this implies there is some t ∈ S such
that 0 = ti(m) = i(tm). Now, injectivity of i says tm = 0
and then [(m, s)] = [(tm, ts)] = [(0, ts)] = 0. All other
parts are equally straight forward.
(c) Show that if an R-module M is Noetherian, then S −1 M is
a Noetherian S −1 R-module.
If K ⊂ S −1 M is an S −1 R submodule, L = j −1 (K) ⊂ M is
an R-submodule of M . We first show that S −1 L = K. If
[(l, s)] ∈ S −1 L, where l ∈ L, we see that j(l) = [(l, 1)] ∈
K and since [(1, s)] ∈ S −1 R, we see that [(1, s)][(l, 1)] =
[(l, s)] ∈ K. Thus S −1 L ⊂ K. Now, let [(m, s)] ∈ K.
Then so does [(s, 1)][(m, s)] = [(sm, s)] = [(m, 1)] = j(m)
and thus m ∈ L. But, then [(m, s)] ∈ S −1 L and so we are
done.
Now, let K be any submodule as before. Then L as before is a submodule of M is finitely generated, since M is
Noetherian. Then so is K = S −1 L as an S −1 R module.
(d) Show that if M is an R-module, then M = 0 if and only if
Mm = 0 for all maximal ideals m ⊂ R.
If M = 0, Mm = 0 is obvious.
For the converse, let m ∈ M . For any maximal ideal m, we
have the natural map j(m) : M → Mm and since the latter
is assumed to be zero, m is in the kernel of this map. So,
1
2
there exists s(m) ∈ R − m such that s(m)m = 0. Let I be
the annihilator of m, that is, the set of elements s ∈ R such
that sm = 0. Easy to check that this is an ideal and m = 0
if and only if I = R. So, if I 6= R, we can find a maximal
ideal m such that I is contained in it. But, we also have
an s(m) 6∈ m which is in I. this is a contradiction.
(e) If M, N are R-modules, show that there is a natural homomorphism of S −1 R-modules, S −1 HomR (M, N ) → HomS −1 R (S −1 M, S −1 N ).
Can you think of some conditions which will make this map
an isomorphism?
The first part is routine. The correct condition for the
latter is that M is finitely presented. This means, M is
finitely generated and if we take a surjection from a free
finitely generated module to M , the kernel is also finitely
generated. In this situation, isomorphism can be proved
by basically using the fact that the map is an isomorphism
for M = R.
(2) Recall that for a topological space X and a subset A, A is
said to be discrete if for any point a ∈ A, there exists an open
neighbourhood U of a such that U ∩ A = {a}. Show that if
A ⊂ Rn is a discrete subgroup, then A is a free abelian group.
(Hint: Discrete subsets of compact sets are finite.)
Since A is a subgroup of Rn , clearly it is torsion free. So,
suffices to prove that it is finitely generated. We may replace
Rn with the vector subspace generated by A without loosing any
information and thus may assume that A generates Rn . Pick a
basis v1 , . . . , vn ∈ A for Rn . Then Zv1 ⊕ · · · ⊕ Zvn ⊂ A ⊂ Rn
and taking quotients, we have the image of A, which we call B
in Rn /Zn = (R/bZ)n = (S 1 )n which is compact. One can easily
check that B is discrete in this compact space and then one
checks that B is finite. Thus we get A/Zn finite which implies
A is finitely generated.
(3) If R is a Noetherian ring and M a module over R, show that
M is Noetherian if and only if M is finitely generated.
If M is Noetherian, M is finitely generated whether R is Noetherian or not. Conversely assume that M is finitely generated.
Then we have a surjection of R-modules, Rn → M for some
n. Since R is Noetherian, so is Rn (finite direct sum) and a
quotient of Noetherian module is Noetherian.
(4) Let R be a commutative ring and I an ideal. For a module
M , as usual we denote by IM , the submodule of M generated
by elements of the form am, where a ∈ I, m ∈ M . If M is
3
finitely generated (very important) and IM = M , show that
there exists an element f ∈ R such that 1 − f ∈ I and f M = 0.
(This is one version of Nakayama’s lemma).
Let m1 , . . . , mn ∈ M be a finite set of generators. IM = M ,
we see that there exists aij ∈ I, 1 ≤ i, j ≤ n such that mi =
P
j aij mj . Consider the n × n matrix A = Id − (aij ). The above
equations say that A[m1 , . . . , mn ]t = 0. Multiplying on the left
by the adjoint of A, we get that det AId[m1 , . . . , mn ]t = 0 and
thus det Ami = 0 for all i. So, let f = det A, then f M = 0.
Further det A ≡ 1 mod I, since aij ∈ I for all i, j.
(5) Let R be a Noetherian ring and let f : R → S be a ring homomorphism. Then, we can view S as an R-module naturally and
assume that S is a finitely generated R-module. Then show that
for any s ∈ S, there exists a monic polynomial P (X) ∈ R[X]
such that P (s) = 0.
Since S is finitely generated as a module over a Noetherian
ring R, it is Noetherian as a module. Given s ∈ S, consider
the subring B of S given by B = ⊕n Rsn . Then, this isPan Rsubmodule of S and hence finitely generated. Let αi = aij sj
be a finite set of generators, where aij ∈ R and these are finite
sums. So, these are all combinations of 1, s, s2 , . . . sm for some
finite m and thus we may take these powers of s as our generators. But then we can write sm+1 = a0 + a1 s + · · · + am sm . So,
take P (X) = X m+1 − am X m − · · · − a0 .
(6) Let the assumptions be as in the previous problem. Show that
S is Noetherian.
We already observed that S is Noetherian as an R-module.
If I ⊂ S is an ideal, in particular it is an R-submodule of S and
hence finitely generated as an R-module and clearly the same
generators generates it as an ideal of S.