4.1 A Few Remarks About Module Theory

Back

Although we won’t embark on a systematic study of modules until Chapter 5,

it will be quite useful for us to gather together a few elementary results

concerning modules for our immediate use.

We start with the appropriate definitions. Let R be a ring (with identity

1) and let M be an abelian group, written additively. Suppose we have a

map R × M ! M, written as scalar multiplication (r,m) 7! r · m (or just

rm), satisfying

(i) (r1 + r2)m = r1m + r2m;

(ii) (r1r2)m = r1(r2m);

(iii) r(m1 + m2) = rm1 + rm2;

(iv) 1 · m = m,

for all r, r1, r2 2 R, and all m,m1,m2 2 M. Then we say that M has

the structure of a left R-module. Naturally, one can analogously define the

concept of a right R-module (scalar multiplications are on the right). In

case R is a commutative ring (as it shall be throughout this chapter) any

left R-module M can be made into a right R-module simply by defining

m · r = r · m, r 2 R, m 2 M. (If R is not commutative, one can’t be

so simple-minded; why?) For the remainder of this chapter since we’ll be

dealing exclusively with commutative rings, we shall simply use the term

87

88 CHAPTER 4. DEDEKIND DOMAINS

“module” without saying “left” or “right,” since the above shows that it

doesn’t matter whether we apply scalar multiplication on the left or on the

right.

The reader will immediately see that an R-module is just like a vector

space, except that the field of scalars is replaced by an arbitrary ring. However,

this comparison is a bit misleading, as vector spaces are really quite

special, with many linear algebraic questions being reducible to questions

about bases and/or dimension. In general, modules don’t have bases, so a

more delicate approach to the theory is necessary.

For now, the most important example of a module over a commutative

ring R is obtained as any ideal I _ R. While this may seem a bit trite, this

viewpoint will eventually pay great dividends.

If M is an R-module and N _ M, then N is said to be an R-submodule

ofM if it is closed addition and under the R-scalar multiplications. If S _ M

is a subset of M, we may set

RhSi = {

X

risi| r1 2 M, si 2 S};

note that RhSi is a submodule of M, called the submodule of M generated

by S. If N _ M is a submodule of the form N = RhSi for some finite

subseteq S _ M, then we say that N is a finitely generated submodule of M.

A map _ : M1 ! M2 of R-modules is called a module homomorphism if _

is a homomorphism of the underlying abelian groups and if _(rm) = r_(m),

for all r 2 R and all m 2 M. If _ : M1 ! M2 is a homomorphism of

R-modules, and if we set ker _ = {m 2 M1| _(m) = 0}, then ker _ is a

submodule of M1. Similarly, one defines the image im _ in the obvious way

as a submodule of M2. In analogy with group theory, if K _! M

_!

N is

a sequence of homomorphisms of R-modules, we say that the sequence is

exact (at M) if im _ = ker _ . An exact sequence of the form 0 ! K !

M ! N ! 0, is called a short exact sequence . Note that if M1,M2 are

R-modules, and if we define the external direct sum M1 _M2 is the obvious

way, then there is always a short exact sequence of the form

0 −! M1 −! M1 _M2 −! M2 −! 0.

If M is an R-module, and if N _ M is a submodule of M, we may give

the quotient group M/N the structure of an R-module exactly as in linear

algebra: r · (m + N) = r · m + N, r 2 R, m 2 M. The reader should

have no difficulty in verifying that the scalar multiplication so defined, is

well-defined and that it gives M/N the structure of an R-module.

4.1. A FEW REMARKS ABOUT MODULE THEORY 89

The following simple result turns out to be quite useful.

Lemma 4.1.1 (Modular Law) Let R be a ring, and let M be an Rmodule.

Assume that M1,M2 and N are submodules of M with M1 _ M2.

Then

M2 + (N \M1) = (M2 + N) \M1.

In analogy with ring theory, an R-module M is said to be Noetherian

if whenever we have a chain

M1 _ M2 _ · · ·

of submodules, then there exists an integer N such that if n _ N, then

Mn = MN. Note that if R is a Noetherian ring, regarded as a module over

itself in the obvious way, then R is a Noetherian R-module. The following

lemma is a direct generalization of Theorem 3.3.1 of Chapter 3.

Proposition 4.1.2 Let R be a ring, and let M be an R-module. The

following are equivalent for M.

(i) M is Noetherian.

(ii) Every submodule of M is finitely generated.

(iii) If S is any collection of submodules of M, then S contains a maximal

element with respect to inclusion.

Proposition 4.1.3 Let 0 ! K ! M ! N ! 0 be a short exact sequence

of R-modules. Then M is Noetherian if and only if K and N both are.

Corollary 4.1.3.1 Let R be a Noetherian ring, and let M be a finitely

generated R-module. Then M is Noetherian.

Exercises

1. Let R be a commutative ring and let M be an R-module. Set

AnnR(M) = {r 2 R| rM = 0}.

(Note that AnnR(M) is an ideal of R.) Prove that the following two

conditions are equivalent for the R-module M.

90 CHAPTER 4. DEDEKIND DOMAINS

(i) AnnR(N) = AnnR(M) for all submodules N _ M, N 6= 0.

(ii) IN = 0 ) IM = 0 for all submodules N _ M,N 6= 0, and

all ideals I _ R. (Here, if I _ R is an ideal, and if M is an

R-module, IM = {finite sums

P

simi| si 2 I, mi 2 M}.)

A module satisfying either of the above conditions is called a prime

module.

2. (i) Show that if P _ R is an ideal, then P is a prime ideal () R/P

is a prime module.

(ii) Show that if M is a prime module then AnnR(M) is a prime ideal.

3. Let M be a Noetherian R-module, and suppose that _ : M ! M

is a surjective R-module homomorphism. Show that _ is injective.

(Hint: for each n > 0, let Kn = ker _n. Then we have an ascending

chain K0 _ K1 _ · · · of R-submodules of M. Thus, for some positive

integer k, Kk = Kk+1. Now let a 2 K1 = ker _. Since _ : M !

M is surjective, so is _k : M ! M. So a = _k(b), for some b 2

M. Now what? Incidently, the above result remains valid without

assuming that R is Noetherian; one only needs that R is commutative,

see Lemma 5.2.8 of Section 5, below.)

4. Let R be a ring and let M be an R-module. If N _ M is an Rsubmodule,

and if N, M/N are finitely generated, show that M is

finitely generated.

5. Let M be an R-module, and let M1,M2 _ M be submodules. If

M = M1+M2 with M1 \M2 = 0, we say that M is the internal direct

sum of M1 and M2. In this case, prove that the map M1 _M2 ! M,

(m1,m2) 7! m1 + m2 is an isomorphism.

6. Let 0 ! K

μ!

M ! N ! 0 be a short exact sequence of R-modules.

Say that the short exact sequence splits if M can be expressed as an

internal direct sum of the form M = μK + M0 for some submodule

M0 _ M. Show that in this case M0 _= N, and so M _= K _ N.

7. 0 ! K

μ!M

_!

N ! 0 be a short exact sequence of R-modules.

Prove that the following conditions are equivalent:

(a) 0 ! K

μ!

M _! N ! 0 splits;

4.2. ALGEBRAIC INTEGER DOMAINS 91

(b) There exists a module homomorphism r : M ! K such that

r _ μ = 1K;

(c) There exists a module homomorphism _ : N ! M such that

_ _ _ = 1N.

8. LetM be an R-module and assume that there is a short exact sequence

of the form 0 ! K ! M ! R ! 0. Show that this short exact

sequence splits.

Навигация