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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
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