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

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

is a subset of M, we may set

R S = { risi| r1 " M, si " S};

note that R S is a submodule of M, called the submodule of M generated

by S. If N " M is a submodule of the form N = R S 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 " R and all m " M. If : M1 ! M2 is a homomorphism of

R-modules, and if we set ker = {m " 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 " R, m " 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 R-

module. 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 e" 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 " 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 = 0.

(ii) IN = 0 ! IM = 0 for all submodules N " M, N = 0, and

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

R-module, IM = {finite sums simi| si " I, mi " 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 " K1 = ker . Since : M !

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

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 R-

submodule, 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) ! m1 + m2 is an isomorphism.

6. Let 0 ! K ! 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 = K + M for some submodule

M " M. Show that in this case M

= =

7. 0 ! K ! M ! N ! 0 be a short exact sequence of R-modules.

Prove that the following conditions are equivalent:

(a) 0 ! K ! M ! N ! 0 splits;

4.2. ALGEBRAIC INTEGER DOMAINS 91

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

r % = 1K;

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

% = 1N .

8. Let M 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.

4.2 Algebraic Integer Domains

Let " C be an algebraic number. If satisfies a monic polynomial with

integral coefficients, then is called an algebraic integer . More generally,

suppose that R " S are integral domains and that " S. Say that is

integral over R if satisfies a monic polynomial in R[x]. Thus, the algebraic

integers are precisely the complex numbers which are integral over Z.

Lemma 4.2.1 Let R " S be integral domains, and let " S. Then is

integral over R if and only if R[] is a finitely generated R-module.

Note that the proof of the above actually reveals the following.:

Lemma 4.2.2 Let R " S be integral domains. If S is a finitely generated

R-module, then every element of S is integral over R.

Lemma 4.2.3 Let R " S " T be integral domains. If T is a finitely gener-

ated S-module, and if S is a finitely generated R-module, then T is a finitely

generated R-module.

As an immediate consequence we have the following proposition.

Proposition 4.2.4 Let , be algebraic integers. Then and + are

also algebraic integers.

One could consider the ring Zalg " C of all algebraic integers. However,

this ring doesn t have very interesting factorization

"properties. For example,
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