Finitely Generated Modules Over
Noncommutative Chain Rings
by
Greggory M. Scible
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
Decemeber 12, 2011
Keywords: right chain ring, right duo ring
Copyright 2011 by Greggory M. Scible
Approved by
Ulrich Albrecht, Professor of Mathematics and Statistics
Pat Goeters, Professor of Mathematics and Statistics
Georg Hetzer, Professor of Mathematics and Statistics
Abstract
The notion of a right chain ring, a ring whose lattice of right ideals is linearly ordered
by inclusion, is a generalization of a valuation ring. In this work we investigate properties
of right chain rings and the structure of modules over such rings. Several results that have
been established for modules over valuation rings and domains are extended to modules over
non-commutative right chain rings. In this discussion the notion of a right duo ring, a ring
whose right ideals are two-sided, arises naturally.
ii
Acknowledgments
I would like to thank my advisor Dr. Ulrich Albrecht for his patient encouragement
throughout the course of my Graduate studies. He has \lead from the front" as all good
leaders do, and has exempli ed how a professional mathematician should conduct oneself.
Thanks also go to my committee members Dr. Georg Hetzer and Dr. Pat Goeters for their
assistance and encouragement.
I would also like to thank my family members, especially my parents Jim and Joan for
always believing in me and pushing me to explore the limits of my ability. I am deeply
appreciative of their guidance and support throughout my life. Thanks also go to my sister
Jennifer who has always been a source of support and whose belief in me has given me the
freedom to try new things.
Finally, I could not have accomplished anything without my wife Lynette. She gave me
the freedom to pursue my goals and has been a rock of support throughout our marriage.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Background Material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.1 Ring Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Skew Polynomial Rings and Power Series Rings . . . . . . . . . . . . . . . . 9
2.3 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3 Right Chain Rings and Duo Rings . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.1 Examples of Right Chain Rings . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.2 Basic Properties of Right Chain Rings . . . . . . . . . . . . . . . . . . . . . 20
3.3 Right Duo Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4 Modules Over Right Duo Right Chain Rings . . . . . . . . . . . . . . . . . . . . 31
4.1 Finitely Annihilated Modules . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4.2 Ideal Structure of Right Chain Rings . . . . . . . . . . . . . . . . . . . . . . 36
4.3 Modules over Right Chain rings . . . . . . . . . . . . . . . . . . . . . . . . . 43
5 Finitely Generated Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5.1 RD-submodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5.2 RD-Composition Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
5.3 Equivalence of Relative Divisibility and Purity . . . . . . . . . . . . . . . . . 51
5.4 A Jordan-H older Type Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 54
5.5 Essential Pure Submodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
6 Duo Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
6.1 Quasiprojective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
iv
6.2 Duo Modules and Strongly Right Bounded Modules . . . . . . . . . . . . . . 69
6.3 Duo Modules over Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
6.4 Maximal RD-submodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
v
Chapter 1
Introduction
When one undertakes the study of noncommutative rings, it rapidly becomes apparent
that the mathematical landscape has changed signi cantly from that which one encounters
in the commutative setting. Things that are obvious in the commutative setting can often
turn into di cult problems without commutativity. One of the best examples of this is the
construction of the  eld of fractions of a ring R. When R is a commutative integral domain
the process is relatively straightforward. The extension of this concept to noncommutative
rings, however, is highly nontrivial and is indeed still a topic of current research.
In this work, we will investigate the class of noncommutative rings known as right chain
rings, which can be described as rings having the property that, for every a, b2R, either
aR bR or bR aR. These rings are the obvious extension of the concept of a valuation ring
to the noncommutative setting. The study of valuation rings has a very extensive literature.
The structure of modules over these rings has been investigated in detail in [15] and [16].
Much of this interest arises from the fact that many results and techniques arising from the
study of abelian groups can be extended to the study of modules over valuation domains.
Chain rings appear in a surprising number of areas of noncommutative ring theory. In
[5], references are provided showing that such rings appear as the coordinate rings of Hjelm-
slev planes and as building blocks for the localizations of Dedekind prime rings. Further, a
domain having a distributive lattice of right ideals is characterized by the property that their
localizations at maximal right ideals are right chain rings. As a  nal example, in 1968 Osof-
sky [30], extending work of Caldwell [11], established that local rings whose cyclic modules
have cyclic injective hulls are right and left chain rings.
1
Chapter 2 contains a summary of the ring and module theory that is utilized throughout
the work to follow. In Chapter 3, we begin the study of right chain rings and right duo rings.
Smith and Woodward introduced the concept of a  nitely annihilated module in [35]. In this
chapter we introduce cyclically and strongly cyclically annihilated modules and establish the
following result:
Theorem 1.1. Suppose R is a ring. The following statements are equivalent:
1) R is a right chain ring such that all cyclic right R-modules are cyclically annihilated.
2) R is a right duo, right chain ring.
3) Every  nitely generated right R-module is strongly cyclically annihilated.
We utilize this result to obtain conditions on a ring R guaranteeing that R is right
duo and every homomorphic image of R is strongly right bounded. In the remainder of the
Chapter we investigate the ideal structure of right duo right chain rings and show that many
of the results from [15] and [16] concerning ideals of valuation rings can be extended to this
class of rings.
In Chapter 4 we begin the investigation of modules over right duo chain domains. A
natural place to begin studying the structure of modules over any ring R is the class of  nitely
generated modules. In this study, RD-submodules (relatively divisible) play a distinguished
role. A submodule N of a right R-module M is RD if N\Mr = Nr for every r 2 R.
For Abelian groups, this is the usual de nition of a pure subgroup. For modules over more
general rings, we have:
De nition 1.2. Let R be a ring and A a right R-module. A submodule B of A is said to be
pure if every  nite system of equations over B
mX
j=1
xjrij = bi (i = 1;:::;n);
with rij2R and unknowns x1;:::;xm, has a solution in B whenever it is solvable in A.
2
Note that B is an RD-submodule of a right R-module A if and only if, whenever
b2B and the equation xr = b is solvable in A, then it is solvable in B. From this we see
immediately that pure submodules are RD for an arbitrary ring R. The investigation into
the rings for which the converse holds is more di cult. Indeed, we have the following open
question.
Open Question: If R is a right semihereditary ring, then every RD-submodule of a right
R-module is pure.
War eld has shown in [39] that an integral domain R is Pr ufer if and only if the RD-
property is equivalent to purity. We establish in this chapter that over right duo chain
domains, purity and relative divisibility are equivalent.
In [14], Fuchs and Salce show that every  nitely generated module over a valuation
domain has a pure composition series, that is a  nite chain
0 = M0 <   <Mn 1 <Mn = M (1.1)
of submodules of M such that
(i) each Mi is an RD-submodule of M.
(ii) Mi+1=Mi is cyclic for every i.
In Chapter 5, we show that this result extends to modules over right duo chain domains,
and indeed establish a result characterizing right duo rings. The existence of essential pure
submodules in  nitely generated modules over this class of rings is also investigated.
Finally, in Chapter 6 we turn our attention to duo modules, that is, modules having the
property that each submodule is fully invariant. As endomorphisms of RR are left multipli-
cation by elements of R, we see immediately that RR is a duo module if and only if R is a
right duo ring. These modules were introduced in [31] and the structure of  nitely generated
torsionfree duo modules over an integral domain is explored. We show that these results can
be extended to duo Ore domains. In this setting, domains need not be commutative.
3
Chapter 2
Background Material
2.1 Ring Theory
In this section we will gather together the elements of Ring Theory which are used
throughout the work. Standard Assumptions: All rings under consideration are assumed
to be unital and associative. Subrings are assumed to contain the identity. By the term
ideal with no additional quali er, we mean a two-sided ideal. For an ideal I of R, we write
I (R to indicate proper containment, and I R otherwise. The following material can be
found in [18], [26], [41]
Special Elements of Rings
De nition 2.1. Let R be a ring and a2R. Then a is said to be
1) a right zero divisor if ba = 0 for some non-zero b2R and a6= 0.
2) a left zero divisor if ab = 0 for some non-zero b2R and a6= 0.
3) a zero divisor if it is both a left and a right zero divisor.
4) an idempotent if a2 = a.
5) nilpotent if ak = 0 for some positive integer k.
6) central if ab = ba for every b2R.
Annihilators
De nition 2.2. Let A be a non-empty subset of a ring R. Then
4
1) the right annihilator of A is the set annr(a) =fb2Rjab = 0 for all a2Ag.
2) the left annihilator of A is the set annl(A) =fb2Rjba = 0 for all a2Ag.
3) the annihilator of A is the set ann(A) = annr(a)\annl(A).
The right annihilator of a single element will be denoted annr(a) instead of annr(fag).
The right annihilator of any nonempty subset of R is a right ideal of R. If A is a right ideal,
then annl(A) is an ideal of R. A symmetric result holds for left annihilators.
Properties of Ideals
De nition 2.3. A left ideal I of R is
1) minimal if I6= 0 and I does not properly contain any non-zero ideal of R.
2) maximal if I6= R and I is not properly contained in any proper left ideal.
We de ne in a similar way minimal and maximal right ideals.
De nition 2.4. Let P be a proper ideal of R. Then P is a
1) prime ideal if xRy P implies that either x2P or y2P for every x;y2R.
2) completely prime if xy2P implies x2P or y2P for every x;y2R.
3) semi-prime ideal if xRx P implies x2P.
We have the following useful characterizations of prime and semi-prime ideals.
Proposition 2.5. [26, Prop. 10.2] Let R be a ring and P a proper ideal of R. The following
statements are equivalent:
1) P is a prime ideal.
2) For ideals I and J of R, if IJ P, then either I P or J P.
5
Proposition 2.6. [26, Prop. 10.9] Let R be a ring and P a proper ideal of R. The following
statements are equivalent:
1) P is a semi-prime ideal.
2) For an ideal I of R, if I2 P, then I P.
Properties of Rings
De nition 2.7. A ring R is simple if R has no nontrivial ideals. R is regular(in the
sense of von Neumann), if for every a2R, we have a2aRa and strongly regular if for
every a2R, we have a2a2R. If 0 is a prime ideal then R is a prime ring, and if 0 is a
semi-prime ideal then R is a semi-prime ring. R is a domain if R has no left or right
zero divisors and an integral domain if R is a commutative domain. A reduced ring is a
ring R which has no nontrivial nilpotent elements. R is right duo if every right ideal is an
ideal. Finally, a right chain ring is a ring R such that for every a;b2R, either aR bR
or bR aR.
We de ne left chain rings and left duo rings in a similar fashion. If a ring is both a left
and a right chain ring, then we simply refer to it as a chain ring, with a similar convention
for duo rings.
The following theorems characterize regular and strongly regular rings. The proofs may
be found in [41, Prop. 3.10, 3.11].
Proposition 2.8. For a ring R, the following statements are equivalent:
1) R is von Neumann regular.
2) Every principal left ideal is generated by an idempotent.
3) Every principal left ideal is a direct summand of R.
4) Every  nitely generated left ideal is a direct summand.
6
5) Every right R-module is  at.
Proposition 2.9. For a ring R, the following statements are equivalent:
1) R is strongly regular.
2) R is regular and reduced.
3) Every principal left (right) ideal of R is generated by a central idempotent.
4) R is von Neumann regular and every left (right) ideal is an ideal.
Hence from 4), we see that strongly regular rings are duo rings.
Rings Characterized by Homological Properties
In this section we present several classes of rings characterized by homological properties.
De nition 2.10. Let R be a ring. Then
1) R is a right p.p ring if every principal right ideal of R is projective.
2) R is a right semi-hereditary ring if every  nitely generated right ideal is projective
as a right R-module.
3) R is a right hereditary ring if every ideal is projective as a right R-module.
From [18, Background], we have the following results concerning right hereditary and
right semi-hereditary rings.
Theorem 2.11. Let R be a right hereditary ring.
1) Every submodule of a projective R-module is projective.
2) Every factor module of an injective right R-module is injective.
3) Every projective right R-module is isomorphic to a direct sum of copies of right ideals
of R.
7
Theorem 2.12. Let R be a right semihereditary ring.
1) Every  nitely generated submodule of a projective right R-module is projective.
2) Every submodule of a  at right R-module is  at.
3) Every  nitely generated projective right R-module is isomorphic to a direct sum of
copies of right ideals of R.
Jacobson Radical
Let R be a ring. The Jacobson radical of R, is de ned to be the intersection of all
maximal left ideals of R. It is denoted by J(R), or simply J if the ring is clear from the
context. In [26, Corollary 4.2], it is shown that J(R) is also equal to the intersection of all
maximal right ideals of R, and that J(R) is a two sided ideal of R. We have the following
important characterization of J(R).
Proposition 2.13. [26, pages 50{51] Let R be a ring. The following statements are equiv-
alent:
1) y2J(R).
2) 1 xyz is a unit for every x;z2R.
3) My = 0 for every simple right R-module M.
A ring R is local ring if it has a unique maximal right ideal. In [26] it is shown that
this is equivalent to R having a unique maximal left ideal. Hence, in a local ring, J(R) is
the unique maximal ideal. In connection with the Jacobson radical we have Nakayama?s
Lemma. The proof may be found in [26, p.60]:
Theorem 2.14 (Nakayama?s Lemma). Let R be a ring and I a right ideal of R. The
following statements are equivalent:
1) I J(R).
8
2) For any  nitely generated right R-module M, MI = M implies that M = 0.
3) For any right R-modules N  M such that M=N is  nitely generated, N + MI = M
implies N = M.
Corollary 2.15. If MR is  nitely generated and x1;:::;xn are elements of M such that the
cosets xi +MJ generate M=MJ, then x1;:::;xn generate M.
Proof. Apply Nakayama?s Lemma to the module M=(x1R +   +xnR).
2.2 Skew Polynomial Rings and Power Series Rings
Let k be a ring and  a ring endomorphism of k. Consider the set R of all polynomials
in x of the form Pni=0aixi (ai 2k). In contrast to the usual assumption that xa = ax for
every a2k, we stipulate instead that xa =  (a)x for every a2k. By iterating this rule
we see that xia =  i(a)xi for all i 0. De ne addition in R as the normal addition in a
polynomial ring and multiplication by
(
X
i
aixi)(
X
j
bjxj) =
X
i;j
ai i(bj)xi+j =
X
k
(
X
i+j=k
ai i(bj))xk:
It can be shown that with these de nitions R forms a ring, denoted k[x; ], the skew poly-
nomial ring in x.
We can perform a similar construction to de ne k[[x; ]], the skew power series ring in
x, whose elements have the form Pi 0aixi. Using the rules above, it is not hard to see that
U(k[x; ]) = U(k) and U(k[[x; ]]) =fa0 +a1x+   ja02U(k)g (see [26, Ch. 1]).
2.3 Modules
All modules are assumed to be unital right R-modules. Similar to the convention for
ideals, we write N (M for proper containment of a submodule N, and N M otherwise.
9
Chain Conditions
De nition 2.16. Let R be a ring, M a right R-module, and N a submodule of M.
1) M is a simple module if M has no proper nontrivial submodules.
2) N is a maximal submodule if N 6= M and N is a maximal element in the set of
proper submodules of M.
3) N is a fully invariant submodule if f(N) N for every R-endomorphism of M.
If M is a right R-module, then the set Ann(M) =fr2Rj mr = 0 for every m2Mg
is the annihilator of M. As is the usual convention, we write Ann(m) for the annihilator of
a single element. We say that M is a faithful R-module if Ann(M) = 0. Equivalently, M is
faithful if for every 06= r2R there exists m2M such that mr6= 0.
The following material is taken from [19]. Let A be a set and A a collection of subsets
of A. We say that A satis es the ascending chain condition (ACC) if there does not exist
a properly ascending in nite chain A1  A2  ::: of subsets from A. Recall also that an
element B2Ais a maximal element ofAif there does not exist a subset inAthat properly
contains B. A right R-module M is Noetherian if it satis es any of the following equivalent
conditions:
Proposition 2.17. For a module M, the following statements are equivalent:
1) M has the ACC on submodules.
2) Every nonempty family of submodules of M has a maximal element.
3) Every submodule of M is  nitely generated.
De nition 2.18. A ring R is right (left) Noetherian if the module RR (RR) is Noethe-
rian.
10
Similarly, we say that A satis es the descending chain condition (DCC) if there does
not exist a properly descending in nite chain A1  A2  ::: of subsets from A. Recall also
that an element B 2A is a minimal element of A if there does not exist a subset in A
properly contained in B. A right R-module M is Artinian if it satis es any of the following
equivalent conditions:
Proposition 2.19. For a module M, the following statements are equivalent:
1) M has the DCC on submodules.
2) Every nonempty family of submodules of M has a minimal element.
De nition 2.20. A ring R is right (left) Artinian if the module RR (RR) is Artinian.
The following standard results concerning modules satisfying either chain condition may
be found in any standard reference on ring theory, such as [19].
Theorem 2.21. Let N be a submodule of a module M. Then M is Noetherian (Artinian)
if and only if N and M=N are both Noetherian (Artinian)
Corollary 2.22. Any  nite direct sum of Noetherian (Artinian) modules is Noetherian (Ar-
tinian).
Corollary 2.23. If R is a right Noetherian (Artinian) ring, then all  nitely generated right
R-modules are Noetherian (Artinian). A similar statement holds for left Noetherian (Ar-
tinian) rings.
Nonsingular Modules
The notion of the singular submodule of a module arises in the attempt to extend the
concept of torsion-freeness from modules over integral domains to general non-commutative
rings. The most obvious extension is to de ne a right R-module M to be torsion-free if and
only if, for every non-zero x2M and all regular r2R, one has xr6= 0. Several problems
11
arise in the use of this de nition, the most notable being that the set of torsion elements
t(M) need not form a submodule for an arbitrary ring R.
The most useful extension of this concept has been found to be the notion of non-
singularity, introduced by Goodearl in [18]. The singular submodule of any right R-module
M over a general ring R always exists, and is equal to t(M) for modules over integral
domains. In this section we de ne the singular submodule, and list several of the more
important results.
De nition 2.24. Let A be a right R-module and B a submodule of A. We say that B is an
essential submodule if every non-zero submodule of A has non-zero intersection with B.
The basic properties of essential submodules may be found in [18, Ch.1] and will be
used without further mention. One of the most important uses of essential submodules is
the formation of the singular submodule.
Remark 2.25. Note that a submodule B of a right R-module A is essential if and only if,
for every non-zero a2A, there exists r2R such that ar6= 0 and ar2B. In practice, this
is often used to show that a submodule is essential.
Notation 2.26. If B is essential in A we write B e A.
De nition 2.27. For a right R-module A, we set
Z(A) =fx2AjxI = 0 for some essential right ideal Ig:
It is shown in [18, p.30] that Z(A) is a submodule of A, the singular submodule. Equiv-
alently, Z(A) =fx2AjAnn(x) e Rg
Considering R as a right R-module, we see that Z(RR) is an ideal of R, the right singular
ideal, denoted Zr(R). Similarly, we de ne the left singular ideal Zl(R). A module A is a
singular module if Z(A) = A. On the other hand, A is a nonsingular module if Z(A) = 0. In
12
a similar fashion, a ring R is right nonsingular if RR is nonsingular. A symmetric de nition
applies for a left nonsingular ring.
The following results will be needed in the sequel.
Proposition 2.28. [18] If A B C, then A e C if and only if A e B e C.
Proof. Suppose  rst that A e B e C, and 06= M C. Since B e C, we have M\B6= 0.
Then A e B implies that A\(M\B)6= 0, that is A\M6= 0. Hence A e C. Conversely,
suppose A e C. Let 0 6= M  B. Then 0 6= M  B  C implies that A\M 6= 0 and
A e B. Further, if 0 6= N  C, then since A is essential in C, we have A\N 6= 0 which
implies that B\N6= 0. It follows that B e C.
De nition 2.29. A uniform module is a non-zero module A such that every non-zero
submodule of A is essential.
De nition 2.30. A module A has Goldie dimension n, written dim(A) = n, if it contains
an essential submodule that is the direct sun of n uniform submodules.
It is shown in [18] that dim(A) = n if and only if A has an independent family of n
non-zero submodules, but no independent family of more than n non-zero submodules.
Proposition 2.31. [18] Let B  A and assume B has  nite Goldie dimension. Then
dim(A) = dim(B) if and only if B e A.
Proof. Let B1     Bn e B, where the Bi are uniform and dim(B) = n. If B e A, then
B1     Bn e A; hence dim(A) = n = dim(B). Conversely, if B is not essential in A, then
there exists a non-zero submodule M of A, such that M\B = 0. Then fB1;:::;Bn;Mg is
an independent family of non-zero submodules of A, whence dim(A) >n = dim(B).
Remark 2.32. Note that dim(A) = 1 if and only if A is uniform.
13
Right Quotient Rings
The material in this section presents an extension of the formation of the  eld of quo-
tients of an integral domain to general rings. The following material is taken from [37].
De nition 2.33. Let R be a ring and S a multiplicatively closed subset of R. Then a right
ring of fractions, or right quotient ring of R with respect to S, is a ring R[S 1] together
with a ring homomorphism ? : R !R[S 1] satisfying:
i) ?(s) is invertible for every s2S,
ii) Every element of R[S 1] has the form ?(a)?(s) 1 with s2S,
iii) ?(a) = 0 if and only if as = 0 for some s2S.
In [37, Ch. II] it is shown that if R[S 1] exists, then it is unique up to isomorphism.
The following is [37, Prop. II.1.4], and is the key result needed.
Theorem 2.34. Let S be a multiplicatively closed subset of a ring R. R[S 1] exists if and
only if S satis es:
1) If s2S and a2R, there exist t2S and b2R such that sb = at.
2) If sa = 0 with s2S, then at = 0 for some t2S.
When R[S 1] exists, it has the form R[S 1] = (R S)= , where  is the equivalence
relation de ned by (a;s)  (b;t) if and only if there exist c;d2R such that ac = bd and
sc = td2S.
In the proof of Theorem 2.34, it is shown that the additive and multiplicative operations
in R[S 1] are given by:
Additive structure: (a;s) + (b;t) = (ac+bd;u) where c;d2R and u = sc = td2S.
Multiplicative structure: (a;s)(b;t) = (ac;tu), where c2R and sc = bu and u2S.
14
In practice, the equivalence class (a;s) is written a=s or as 1.
Similarly, one can de ne the left ring of fractions [S 1]R with respect to a multiplica-
tively closed set S. It is shown in [37, p.51] that when R[S 1] and [S 1]R both exist, then
they are naturally isomorphic. The additive and multiplicative structures of [S 1]R are
similar to those of R[S 1], mutatis mutandis.
One of the most important examples of a multiplicatively closed set is the set S of all
regular elements of a ring R. In this case, the right ring of fractions of R with respect to S
is called the classical right ring of quotients and is denoted Qr(R), or merely Qr if the ring
R is clear from the context. In this case, we have the following important result:
Theorem 2.35. (Ore) Let S be the set of regular elements of a ring R. Then Qr(R) exists if
and only if S satis es the right Ore condition, i.e. for a2R and s2S, we have aS\sR6=;.
If R is a domain, then the right Ore condition reduces to aR\sR6= 0 for all non-zero
elements a;s2R. Clearly this is equivalent to I\J6= 0 for all non-zero right ideals I and
J. Observe that a domain has a classical right ring of quotients if and only if dimRR = 1.
This condition is obvious for a right chain domain R, and therefore the right quotient ring
exists for such rings and is a skew- eld.
Another important example of a multiplicatively closed set is the complement, S, of a
completely prime ideal P. In this case, R[S 1] is referred to as the localization of the ring
R at the completely prime ideal P, and, by a standard abuse of notation, is written RP.
By Lemma 5.2 of [5], if P is a completely prime ideal of a right chain ring R, then the
localization of R at P exists and RP =frs 1jr2R;s2S = RnPg.
Remark 2.36. If R is a chain domain, then R is an Ore domain, hence Qr(R) and Ql(R)
both exist by Theorem 2.35. If P is a completely prime ideal of R, then, by the above
the localization at P is also equal to fs 1a j a 2 R and s 2 RnPg. If it is necessary
to distinguish between the two, we will denote the left localization by PR and the right
localization by RP.
15
Chapter 3
Right Chain Rings and Duo Rings
3.1 Examples of Right Chain Rings
In this section we establish some of the basic properties of right chain rings. A standard
reference for many of these results is [5]. All unde ned terms may be found in [18, 19, 26, 32].
We include most proofs for completeness.
We  rst present several examples of right chain rings.
Example 3.1. Let p be a  xed prime and k2N. Then the ring Zpk is a  nite chain ring.
Example 3.2. By Proposition 3.21 below, any right discrete valuation domain is a right
chain domain.
The next example requires some preliminary work on ordered groups and Malcev-
Neumann rings.
De nition 3.3. A multiplicative group G with identity e is ordered if there exists a total
order < on G such that for any x, y, z2G, we have
x<y implies xz <yz and zx<zy:
Given an ordered group (G;<), its positive cone is the set
P =fx2Gjx>eg:
16
P has the following easily veri ed properties:
P1) P P  P.
P2) Gnfeg= PtP 1.
P3) zPz 1 P for any z2G.
Conversely, given a subset P  G of a multiplicative group G satisfying P1;P2; and P3, we
can de ne an order on G by
x<y () x 1y2P () yx 12P:
The veri cation that the order given above makes G into an ordered group may be found
in [26, pp. 94-95]. As an example, suppose G is an in nite cyclic group fxnjn2Zg. Let
P = fxn j n > 0g. It is easily veri ed that P satis es the properties above, and hence
induces an order on G where xn <xm if and only if m n> 0.
Following [26, 14.5], we detail the construction of the Malcev-Neumann ring. Modi ca-
tions of this construction are used to produce chain rings having various properties. As we
will be utilizing well-ordered subsets of ordered groups, the following technical lemmas are
necessary. The proofs may be found in [26, Section 14.5].
Lemma 3.4. [26, Lemma 14.16] Let (G;<) be a totally ordered set. For any subset S G,
the following are equivalent:
1) S is well-ordered.
2) S satis es the DCC (any sequence s1 s2 s3    in S is eventually constant).
3) Any sequencefs1;s2;s3;:::gin S contains a subsequencefsn(1);sn(2);sn(3);:::g, where
n(1) <n(2) <n(3) <   , such that sn(1) sn(2) sn(3)    .
17
Lemma 3.5. [26, Lemma 14.17] Let S and T be well-ordered subsets of a totally ordered set
(G;<). Then S[T is well-ordered. If (G;<) is an ordered group, then
U = S T =fstjs2S;t2Tg
is also well-ordered. Moreover, for any u2U, there exist only  nitely many ordered pairs
(s;t)2S T such that u = st.
We now present the construction of the Malcev-Neumann ring. Fix a base ring R
and suppose (G;<) is a multiplicative ordered group. Further,  x a group homomorphism
 : G !Aut(R) where the image of g2G is denoted by  g. As a set, the Malcev-Neumann
ring A = R((G; )) consists of all formal sums
 =
X
g2G
 gg ( g2R)
such that supp( ) = fg 2 G j  g 6= 0g is well-ordered. Addition in A is de ned in the
obvious fashion and multiplication by
  = (
X
g2G
 gg)(
X
h2G
 hh) =
X
u
(
X
 g g( h))u;
where the last sum is over all pairs (g;h)2G G such that gh = u. As we may restrict g
and h to supp( ) and supp( ), and since these supports are well-ordered, this last sum is
 nite by Lemma 3.5. It can be shown that with these operations, A forms a ring [26, Ch.
14].
As an example, suppose G =<x> is an in nite cyclic with positive cone P as above.
Then the homomorphism  : G !Aut(R) is speci ed by a single automorphism ! :=  x.
The twist law in this case is x r = !(r)x. As well-ordered subsets of Z consist precisely
of nonempty subsets that are bounded below, we see that the Malcev-Neumann ring in this
18
case is
A = R((<x>;!)) =f
1X
i=n
 ixij i2R; n2Zg;
a twisted Laurent series ring. The interest in Malcev-Neumann rings lies in part in the
following nontrivial result.
Theorem 3.6 (Malcev-Neumann). [26, Thm 14.21] Let R be a division ring and (G;<) and
 as above. Then A = R((G; )) is a division ring.
We consider next valuations of a division ring whose value group is an ordered group
G.
De nition 3.7. For a division ring D and ordered group (G;<), a function
 : D  !G
is called a valuation if the following two properties are satis ed:
V1)  (ab) =  (a) (b) for every a, b2D .
V2)  (a+b) minf (a); (b)g for every a, b2D such that a+b6= 0.
Given a valuation  : D  !G, the set
R =fr2D j (r) eg[f0g
is easily seen to be a subring of D . It is also not di cult to show that R is a duo right chain
ring with unique maximal ideal M =fr2D j (r) >eg[f0g (see [24, pp.216-217]).
Now let R be a division ring, (G;<) an ordered group, and  : G !Aut(R) the trivial
group homomorphism. As the multiplication in the Malcev-Nuemann ring A is induced by
the \twist" g r =  g(r)g, the assumption on  implies that gr = rg for every r2R and
g2G. De ne a map ? : A  !G by ?( ) = min:supp( ).
19
It will be shown that ? is a valuation on the division ring A . Suppose  = ag +   
and  = bh +    , where a, b 2 R , g = min:supp( ), and h = min:supp( ). Then
  = (ab)gh +   . By the properties of an ordered group, gh = min:supp(  ). Therefore
?(  ) = gh = ?( )?( ).
For the sum,  +  = ag + bh +   . The smallest group element in the ordering on G
appearing in the sum obviously cannot be smaller than minfg;hg. Therefore,
?( + ) = min:supp( + ) minfg;hg= minf?( );?( )g:
Hence ? is a valuation. By the above, the subring R =f 2A j?( ) eg[f0gis a duo
right chain domain.
3.2 Basic Properties of Right Chain Rings
In this section, some of the basic results concerning right chain rings are listed. The
most complete reference for results on right chain rings is [5].
Proposition 3.8. [5, Lemma 1.2] If R is a right chain ring, the lattice of right ideals of R
is linearly ordered with respect to inclusion.
Proof. Suppose I1 and I2 are right ideals of R such that I26 I1. Choose a2I2nI1 and note
that if b2I1, then either a = br or b = ar for some r2R. Clearly the  rst case cannot
hold as then a2I1. Hence b = ar2I2. Since b was chosen arbitrarily, I1 I2.
Remark 3.9. Since the right ideals of a right chain ring R are linearly ordered with respect
to inclusion, it is immediate that R is local; that is, R has a unique maximal right ideal
J, which must equal the Jacobson radical of R. Throughout this work J will consistently
represent the Jacobson radical of R, and U = U(R) the multiplicative group of units of R.
Observe that if R is a local ring, then U(R) = RnJ; see Proposition 2.13 on page 8.
Proposition 3.10. [5, Lemma 1.2] A  nitely generated right ideal I of a right chain ring
R is principal.
20
Proof. It su ces to establish the result for I = aR+bR. Since R is a right chain ring, either
aR bR or bR aR. Suppose the latter holds. Then clearly I = aR.
The following Lemma shows that a right ideal I of a right chain ring R is two-sided
exactly if UI = I. This result is fundamental and will be used throughout the following
work.
Lemma 3.11 (Test-units Lemma). [5, Lemma 1.4] Let R be a right chain ring.
1) For every a2R, Ra UaR.
2) A right ideal I of R is two-sided exactly if uI I for every u2U.
Proof. 1): Let x2R. If x is either a unit or xa2aR, then we are done. Suppose x2J
and xa62aR. Since R is a right chain ring we have that aR xaR so a = xas for some
s2R. Note that s must be in J, for otherwise s is a unit and xa2aR. Then 1 + s2U
and xa(1 +s) = xa+xas = xa+a = (1 +x)a. Since x2J by assumption, (1 +x)2U and
it follows that xa = (1 +x)a(1 +s) 12UaR.
2): Clearly the result holds if I is a two-sided ideal. Conversely, suppose I is a right
ideal such that uI  I for every u 2 U. If a 2 I, then Ua  I by assumption. Then
ra2Ra UaR I for every r2R by (1). Therefore I is an ideal.
The following is an easy consequence of the de nition of a right chain ring.
Lemma 3.12. [5, Lemma 1.5] Let R be a right chain ring, A a right ideal and B a two-sided
ideal. Then, AB =fabja2A; b2Bg.
Proof. Clearly the right hand side of the equation above is contained in AB. Conversely, if
Pn
i=1aibi2AB, then since R is a right chain ring we can assume without loss of generality
that aiR a1R for i = 2;:::;n. Then ai = a1ri and Pni=1 aibi = a1Pni=1 ribi = a1b where
b = Pni=1 ribi. Hence Pni=1 aibi2fabja2A; b2Bg and the proof is complete.
21
We will now establish several results concerning prime ideals in right chain rings. Notice
that completely prime ideals are always prime, and prime ideals are semiprime. In general
the converses of both statements are false. See De nition 2.7 on page 6.
Proposition 3.13. Let R be a right chain ring and P a two-sided completely prime ideal of
R. If s62P, then P = sP.
Proof. The inclusion sP  P is immediate as P is an ideal. For the reverse inclusion,
suppose p2P. Since R is a right chain ring, either sR pR or pR sR. The  rst case
cannot hold, as then s 2 P. Therefore pR  sR, so p = sr for some r 2 R. If r 62 P,
then P a completely prime ideal implies p = sr62P, a contradiction. Hence, r2P, and
p = sr2sP.
The next result establishes the equivalence of the notions of semiprime and prime ideals
in right chain rings. Further, we obtain a necessary and su cient condition for an ideal to
be completely prime.
Proposition 3.14. [5, Lemma 1.8] Let R be a right chain ring and P a right ideal of R.
1) P is prime if and only if it is semi-prime.
2) If P is a two-sided ideal, then P is completely prime if and only if x2 2 P implies
x2P for every x2R.
Proof. 1): Necessity is obvious from the de nitions. For su ciency, suppose P is semi-prime
and xRy P for x;y2R. Since R is a right chain ring, we may assume without loss of
generality that xR yR. Then xRxR xRyR P. As P is semi-prime by assumption,
x2P. Thus P is a prime ideal.
2): Necessity is obvious from the de nition. Suppose the condition holds and ab2P for
some a;b2R. Since R is a right chain ring, either a = br1 or b = ar2 for some r1;r22R. In
the  rst case, a2 = aa = a(br1)2P as ab2P and P is a two-sided ideal. By the assumed
22
condition a2P. In the second case, note that (ba)2 = b(ab)a2P as P is a two-sided ideal.
Hence ba2P by our assumption. Then b2 = bb = bar2 2P which implies b2P. Hence P
is completely prime.
Proposition 3.15. [5, Lemma 3.2] Let R be a right Noetherian, right chain ring. Then R
is right duo.
Proof. Since R is a right Noetherian, every right ideal of R is  nitely generated and thus
right principal. Suppose I = aR is a right ideal of R. Then by Lemma 3.11, it is enough to
show that ua2aR for every unit u2R. Suppose by way of contradiction that there exists
a unit u2R such that ua62aR. Since R is a right chain ring we have that aR  uaR,
so a = uaj for some j 2 J. Then uaR  u2aR     , which contradicts R being right
Noetherian.
3.3 Right Duo Rings
In studying the structure of modules over right chain rings we have found that additional
conditions on the ring are necessary. The condition on the ring that has proven to be of the
greatest importance is that the ring be right duo, i.e. every right ideal is two sided.
In the literature right duo rings are also referred to as right invariant. Observe that R
is right duo if and only if for every a2R we have Ra aR; similarly, R is duo if and only
if aR = Ra for every a2R.
We now present some examples of right duo rings.
Example 3.16. As a trivial example, commutative rings are obviously duo.
Example 3.17. By Proposition 2.9 on page 7, strongly regular rings are right duo.
Example 3.18. By Proposition 3.15, right Noetherian right chain rings are right duo.
Example 3.19. In [24, pp. 214{215] it is shown that there exists a right duo ring which is
not left duo. For the proof, see the next set of results.
23
De nition 3.20. A domain R is a right discrete valuation domain if there exists a
non-zero nonunit  of R such that every non-zero a 2 R can be written in the form  nu
where n 0 and u2U(R).
Proposition 3.21. [24, pp. 214{215] If R is a right discrete valuation domain, then R is a
right duo right chain domain.
Proof. We  rst show that  is neither left nor right invertible. Suppose  is right invertible.
Then  a = 1 for some a2R. We can write a =  nu for some n 0 and unit u. Then
 n+1u =  a = 1 which implies that  is a unit, a contradiction. Next, suppose  is left
invertible. Then b = 1 for some b2R. Write b =  mv for some m 0 and v 2U(R).
Then  mv = b = 1. If m6= 0, this implies that  is right invertible, contradicting the
above. So m = 0, which implies that  2U(R), a contradiction. Hence  is neither left nor
right invertible.
Set M =  R. By the above, M < R. Suppose 0 6= I is a proper right ideal of R and
0 6= a2I. Write a =  nu. Clearly n6= 0, so a2 R = M. Therefore M is the unique
maximal right ideal of R. It follows that R is a local domain with Jacobson radical J = M.
Let I be a non-zero proper right ideal of R. We claim that I =  iR for some positive
integer i. If 0 6= a 2 I, then as above a =  nu for some n > 0. As u is a unit, this
implies that  n 2I. Let i be the least positive integer such that  i 2I. Then n i, so
a =  nu =  i n iu 2  iR. Hence I   iR. As  i 2 I, clearly  iR  I and the result
follows.
By the above, the ideals of R form a chain
0       i+1R  iR      R R
where the inclusions are strict since  is not right invertible. It follows that R is a right
chain ring. Further, R is right noetherian and hence right duo.
24
Using the proof of Proposition 3.21, it is can be shown that the function  : R  !Z
given by  (a) = n if a =  nu, is a valuation. Hence, a right discrete valuation ring is a ring
whose value group is Z[f1g. The next proposition shows that the concept of a duo ring
is not right-left symmetric.
Proposition 3.22. [24, pp. 214{215] There exists a right duo right chain domain that is
not left duo.
Proof. We will exhibit the existence of a noncommutative discrete valuation domain. Let k
be a  eld and  2Endk(k), where  (k)  k. Let R = k[[x; ]], the ring of skew power series
of the form Pi 0xiai (ai2k) with multiplication given by ax = x (a).
Set  = x and suppose  2 R. If an is the least non-zero coe cient of  , then
 = xnan + xn+1an+1 +   , so  = xn(an + xan+1 +   ). Then an + xan +   2 U(R)
as an 6= 0. It follows that every element of R can be written in the form  nu for some
u2U(R). We conclude that R is a right discrete valuation ring, and is therefore a right duo
right chain ring.
Note that since k is a  eld, R is a domain. If R is left duo, then xR  Rx. Let
a2kn (k). Then xa2xR Rx, so xa = bx = x (b) for some b2R. Then x(a  (b)) = 0
and it follows that a =  (b), contradicting the choice of a. Hence R is not left duo.
The following example, presented in [6, Example 6.16], shows the existence of a right
duo right chain ring that is not a domain. The ring is also neither a left chain ring nor a left
duo ring.
Example 3.23. Let k be a division ring and  : k!k a monomorphism such that  (k) (k.
De ne a multiplication on R = k k by
(f;g)(f0;g0) = (ff0; (f)g0 +gf0);
with componentwise addition. It is routine to verify that with this operation R is a ring
with 1R = (1;0) and 0R = (0;0). A direct calculation shows that the Jacobson radical of
25
R is J = f(0;g) 2 R j g 2 kg. As J = (0;g0)R for any 0 6= g0 2 k, the only possible
non-zero right ideals are J and R. Hence R is a right duo right chain ring. R is not a
domain as (0;1)2 = 0R. To see that R is not a left chain ring, choose h;h02kn (k). If
R(0;h)  R(0;h0) then (0;h) = (f;g)(0;h) = (0; (f)h0) = (0; (fh0)). Thus, h2 (k), a
contradiction. By symmetry, R(0;h0) * R(0;h). A similar argument shows that R is also
not left duo.
In [38], Thierrin discussed several properties of duo rings. Several of his results can be
extended to the more general case of right duo rings. The proof of the  rst result requires
modi cations for the right duo case; the other results carry through verbatim for right duo
rings and the proofs are due to Thierrin.
Proposition 3.24. Every idempotent e of a right duo ring R is central.
Proof. In [26] it is shown that an idempotent e2R is central if and only if eRf = 0 = fRe,
where f = 1 e is the complementary idempotent orthogonal to e. Suppose R is right
duo. Then Re  eR and Rf  fR. These relations imply that fRe  feR = 0 and
eRf efR = 0. So we have that eRf = 0 = fRe and by the above, e is central.
Proposition 3.25. [38] Every prime right ideal in a right duo ring is completely prime.
Proof. Suppose R is right duo, P a prime right ideal, and xy2P. Set T =ft2Rjxt2Pg.
Then T is a right ideal. As R is right duo, T is a two-sided ideal. Then y2T implies Ry T.
Hence xRy P. Since P is prime, x2P or y2P, and we conclude that P is completely
prime.
Proposition 3.26. [38] The set of nilpotent elements of a right duo ring R form an ideal
N, which is equal to the intersection of the completely prime ideals of R.
Proof. Let I be the intersection of the completely prime ideals Pi of R and N the set of
nilpotent elements. If an = 0, then an 2 Pi for every i. Since Pi is completely prime, it
follows that a2Pi and thus N I. By Proposition 3.25, I is the intersection of the prime
26
ideals of R. By a result of [29], I is a nil ideal. Then each element of I is nilpotent, and
therefore I N and the proof is complete.
Let N(R) denote the set of all nilpotent elements of a ring R, and P(R) the prime
radical of R. Then R is 2-primal if P(R) = N(R). This class of rings was introduced
by Birkenmeier, Heatherly and Lee [8] in the context of nearrings. The previous result
establishes immediately that:
Corollary 3.27. Right duo rings are 2-primal.
Obviously every right semi-hereditary ring is right p.p. If R is a right chain ring, then
Proposition 3.10 implies that the converse holds. The next result establishes the equivalence
of several conditions on a right duo right chain ring.
Theorem 3.28. Let R be a right duo right chain ring. Then the following conditions are
equivalent.
1) R is semiprime.
2) R is a domain.
3) R is right p:p:
4) R is right semi-hereditary.
5) R is right nonsingular.
Proof. 1) implies 2): Suppose R is semiprime. Since R is a right chain ring, Proposition 3.14
implies that R is a prime ring, so 0 is a prime ideal. Since R is right duo, 0 is completely
prime. Hence R is a domain.
2) implies 3): Let aR be a non-zero principal right ideal of R. Since R has no zero
divisors, the map R!aR given by r7!ar is an isomorphism. Hence R?aR and therefore
aR is projective.
27
3) implies 4): Since R is a right chain ring,  nitely generated right ideals are principal.
Hence by 3),  nitely generated right ideals are projective.
4) implies 5): As R is a right chain ring, it su ces to establish the result for principal
right ideals. Suppose 06= a2R . We have an exact sequence
0 !annr(a) !R f !aR !0;
where f(x) = xa.
As aR is projective by assumption, the sequence splits. Hence, annr(a) is a direct
summand of R and cannot be essential. Thus, Zr(R) = 0, and R is right nonsingular.
5) implies 1): Suppose I is a right ideal of R and I2 = 0. Since R is a right chain ring,
the lattice of right ideals is linearly ordered with respect to inclusion. Hence every right
ideal of R is essential. Then I2 = 0 and R right nonsingular imply that I = 0. Thus, R is
semiprime by Proposition 2.6.
Proposition 2.9 yields that R is strongly regular if and only if R is a von Neumann
regular duo ring. Using this result, we can show that von Neumann regular right chain rings
are duo rings.
Proposition 3.29. Let R be a von Neumann regular right chain ring. Then R is strongly
regular.
Proof. Suppose 0 6= a2R. Since R is von Neumann regular there exists b2R such that
a = aba. Since R is a right chain ring, either b2aR or a2bR. If the  rst case holds then
b = ar for some r2R. Then a = aba = aara = a2ra2a2R.
If b62aR, then a = bj for some j2J. Then a = aba = bjba. From this we see that
(1 bjb)a = 0. By Proposition 2.13, 1 bjb is a unit of R. Therefore a = 0, a contradiction.
Hence the  rst case holds and R is strongly regular.
Corollary 3.30. A von Neumann regular right chain ring is a duo ring.
28
Proof. By Proposition 3.29, von Neumann regular right chain rings are strongly regular. The
result follows by Proposition 2.9.
The following extension of [26, Exercise 10.19] shows that right duo rings behave in
many ways like commutative rings. Consider the following conditions on a ring R:
1) Every ideal of R is semiprime.
2) Every ideal of R is idempotent.
3) R is von Neumann regular.
Then for any ring R we have 3))2))1). The problem as stated is that 1))3) if R is a
commutative ring. This result also holds for right duo rings.
Lemma 3.31. Let R be a right duo ring. Then R=I is reduced for every semiprime ideal I
of R.
Proof. Suppose x2 2I for some x2R. Then Rx xR implies xRx x2R I. As I is
semiprime, x2I.
Proposition 3.32. The following are equivalent for a right duo ring R:
1) Every ideal of R is semiprime.
2) Every ideal of R is idempotent.
3) R is strongly regular.
Proof. For completeness we present the proof in its entirety. 3) implies 2): Since R is
strongly regular, it is von Neumann regular. Suppose I is an ideal of R and a2I. Then
a2aRa I2, so that I I2.
2) implies 1): trivial
1) implies 3). Suppose a2R. Then a2R is a semiprime ideal. Hence R=a2R is reduced
by Lemma 3.31. As a + a2R is a nilpotent element of R=a2R, we have that a2a2R. Thus
R is strongly regular.
29
The following observation is remarkably useful, particularly in the study of  nitely
generated modules over right duo right chain rings. This property of right duo rings allows
us to extend many results established for valuation rings to right chain rings.
Proposition 3.33. Let R be a right duo ring and M a right R-module. Then for every
a2M we have Ann(a) = Ann(aR).
Proof. Suppose a2M. Then Ann(a) is a two-sided ideal of R so that R Ann(a) Ann(a).
Then (aR)Ann(a)  a Ann(a) = 0 which says precisely that Ann(a)  Ann(aR). The
other inclusion is obvious and the result follows.
The following consequence of Proposition 3.33 will be used extensively in the study of
 nitely generated modules over right duo right chain rings.
Proposition 3.34. Let R be a right duo ring and suppose M = x1R+   +xnR is a  nitely
generated right R-module. Then AnnRM = Tni=1AnnR(xi).
Proof. Clearly AnnRM  Tni=1AnnR(xi). Suppose t2Tni=1AnnR(xi) and m2M. Then
m = Pni=1 xiri, for some ri 2 R. Observe that xit = 0 yields xirit = 0. It follows that
mt = Pni=1 xirit = Pni=1 xitsi = 0. Thus, t2AnnRM.
A partial converse to 3.33 also holds; namely, if R is a right chain ring with the property
that every  nitely generated right R-module satis es the conclusion of 3.33, then R is right
duo. To establish this we introduce the concept of a cyclically annihilated module.
30
Chapter 4
Modules Over Right Duo Right Chain Rings
4.1 Finitely Annihilated Modules
In this section we consider  nitely annihilated modules, a notion attributed to P. Gabriel
[17] by Smith and Woodward in [35]. More speci cally, we consider modules that are cycli-
cally annihilated over right duo right chain rings. This leads to a natural generalization of
results obtained by various authors, and a characterization of this class of rings.
De nition 4.1. Let R be a ring. A module MR is  nitely annihilated if there exist
elements m1;:::;mk in M such that AnnR(M) = AnnR(m1;:::;mk). If we have that
AnnR(M) = AnnR(m) for some m2M, then M is cyclically annihilated. A  nitely
generated right R-module M is strongly cyclically annihilated if the equality above holds
for one of the generators of M, i.e., if M = x1R +   + xnR, then AnnR(M) = AnnR(xi)
for some i.
De nition 4.2. A ring R is strongly right bounded if every non-zero right ideal of R
contains a non-zero two-sided ideal.
Remark 4.3. Note that right duo rings are obviously strongly right bounded.
Theorem 4.4. The following statements are equivalent for a ring R:
1) R is a right chain ring such that all cyclic right R-modules are cyclically annihilated.
2) R is a right duo right chain ring.
3) Every  nitely generated right R-module is strongly cyclically annihilated.
31
Proof. 1) implies 2): Suppose R is a right chain ring such that every cyclic right R-module
is cyclically annihilated. Let I be an ideal of R. We claim that if M is a cyclic right R=I-
module, then MR=I is cyclically annihilated. To see this, note that M is a cyclic right R
module via the operation m r = m(r + I). As MR is cyclically annihilated by hypoth-
esis, there exists a 2 MR such that MR AnnR(a) = 0. An easy calculation shows that
AnnR=I(a) = AnnR(a)=I. Hence MR=I AnnR=I(a) = 0 and the claim follows. Therefore,
R=I is a right chain ring such that every cyclic right R=I-module is cyclically annihilated.
We now show that R is strongly right bounded. For 0 6= a 2 R, the cyclic right R-
module R=aR cyclically annihilated. Hence, AnnR(R=aR) = AnnR(x+aR) for some x2R.
Thus J = AnnR(x + aR) = fr2Rjxr2aRg is a two-sided ideal of R which is clearly
contained in aR. Since R is a right chain ring, aR is essential in R. Thus, R=aR is a singular
R-module and Ann(y +aR) is essential in R for all y2R. In particular, J = Ann(x+aR)
is essential, and thus non-zero. So R is a strongly right bounded ring, and this allows us
to establish the existence of a largest non-zero two-sided ideal contained in aR as follows:
set I0 = Pf0 6= K ERjK  aRg, and note that the sum is nonempty as it contains J.
Clearly I0 is the largest non-zero two-sided ideal contained in aR.
Suppose I0 6= aR. By what has been shown, R=I0 is a strongly right bounded right
chain ring. By the argument above, R=I0 is cyclically annihilated. Hence, the right ideal
(a + I0)R=I0 contains a non-zero two-sided ideal I1=I0, where I1 is an ideal of R such that
I0  I1  (a + I0)R. As I0  aR, we have that I0  I1  aR. Since I0 is the largest
two-sided ideal of R contained in aR, we obtain a contradiction. Thus, I0 = aR and aR is
a two-sided ideal of R.
2) implies 3): Suppose that M = x1R +   + xnR. Since R is a right chain ring we
may order the right ideals fAnnR(xi)ji = 1;:::;ng so that AnnR(x1)     AnnR(xn).
It follows from Proposition 3.34 that AnnRM = Tni=1AnnR(xi) = AnnR(x1). Thus M is
strongly cyclically annihilated.
32
3) implies 1): As strongly cyclically annihilated modules are obviously cyclically annihi-
lated, it remains to show that R is a right chain ring. We will show  rst that R is right duo.
Consider the module M = R=aR for some a2R. By 3), AnnR(M) = AnnR(1 +aR) = aR.
Thus aR is a two-sided ideal of R.
For a, b 2 R, consider the module M = R=aR R=bR, which is generated by the
elements f(1 +aR;0);(0;1 +bR)g. By 3) we may assume without loss of generality that
AnnR(M) = AnnR(1 +aR) = aR:
On the other hand, if x2AnnR(M), then (0;0) = (1 +aR;1 +bR)x = (x+aR;x+bR), so
that x2aR\bR. Thus, aR = AnnR(M) aR\bR bR, and R is a right chain ring.
Given the last result, we can establish the following corollary to Proposition 3.33.
Corollary 4.5. Let R be a right chain ring. Suppose every  nitely generated module MR
has the property that AnnR(a) is an ideal for every a2M. Then R is right duo.
Proof. Suppose M = a1R +    + anR for a1;:::;an in M. By hypothesis, Ann(ai) is
an ideal for every i. Hence Ann(M) = TiAnn(ai). Since R is a right chain ring we
may order the annihilators (reindexing if necessary) as Ann(a1)      Ann(an). Then
Ann(M) = Ann(a1), and M is strongly cyclically annihilated. By Theorem 4.4, R is right
duo.
We now consider a condition on a ring similar to the considerations above.
De nition 4.6. A ring R is a right  -ring if every  nitely generated right R-module M
is cyclically annihilated.
If R is a right duo right chain ring, then by Theorem 4.4, every  nitely generated right
R-module is strongly cyclically annihilated. As strongly cyclically annihilated modules are
obviously cyclically annihilated, we have
33
Corollary 4.7. Let R be a right duo, right chain ring. Then R is a right  -ring.
However, right  -rings need not be right chain rings as shown by:
Proposition 4.8. If R and S are right  -rings, then R S is a right  -ring.
Proof. Let K be a  nitely generated T = R S-module. Then there exist  nitely generated
modules MR and NS such that K = MLN and MS = 0 and NR = 0. By hypothesis
there exist x2M and y2N such that AnnR(x) = AnnR(M) and AnnS(y) = AnnS(N).
Clearly (x;y)2M N = K yields AnnT(K) AnnT((x;y)). On the other hand, suppose
(r;s) 2 T = R S satis es (x;y)(r;s) = (0;0). Then, xr = ys = 0, and it follows that
r 2AnnR(x) = AnnR(M), and s2AnnS(y) = AnnS(N). Thus, we see that (r;s) is an
element of AnnT(M N) = AnnT(K) as desired.
In [36], the following characterization of right Artinian rings is established:
Theorem 4.9. The following are equivalent for a ring R.
1) R is right Artinian.
2) Every right R-module is  nitely annihilated.
3) Every countably generated right R-module is  nitely annihilated.
4) R satis es the descending chain condition on two-sided ideals and every cyclic right
R-module is  nitely annihilated.
If the ring in question is taken to be a right chain ring, the above theorem can be
specialized as follows. The proof generally follows that of [36], using the results above to
account for the nature of the ring in question.
Theorem 4.10. Let R be a right chain ring. The following are equivalent:
1) R is right Artinian.
34
2) Every right R-module is cyclically annihilated.
3) Every countably generated right R-module is cyclically annihilated.
4) R satis es the descending chain condition on two-sided ideals and every cyclic right
R-module is cyclically annihilated.
Proof. 1) implies 2): Let MR be a right R-module and choose a 2 M such that Ann(a)
is minimal. Since R is right Artinian, it is right Noetherian. Then by Proposition 3.15, R
is right duo. Thus, Ann(m) is a two-sided ideal of R for every m2M. Then Ann(a)  
T
m2M Ann(m) = Ann(M). It follows that Ann(M) = Ann(a) and M is cyclically annihi-
lated.
2) implies 3): Obvious.
3) implies 4): Suppose I1  I2  I3     is a descending chain of two-sided ideals
of R. Consider the countably generated right R-module M = Ln<!R=In. By hypothesis
there exists x2M such that Ann(M) = Ann(x). There also exists a positive integer k such
that xR R=I1     R=Ik. The hypotheses also guarantee that R is right duo so that
Ann(x) = Ann(xR). Then
Ik = Ann((R=I1)     (R=Ik)) Ann(xR) = Ann(x) = Ann(M) =
\
n<!
In:
Hence, the chain terminates and the claim follows by Theorem 4.4.
4) implies 1): Let J be the Jacobson radical of R. By a result of Bessenrodt, Brungs,
and T orner [5, Prop. 3.16], it is enough to show that J is nilpotent. Since R has the d.c.c.
for ideals, we have that Jn = Jn+1 for some integer n. Set B = Jn and suppose that B6= 0.
Then B2 = (Jn)(Jn) = J2n = Jn = B. Set A = B\Ann(B). Then A  B as A = B
implies B2 = B = 0. Since R is a right chain ring and every cyclic right R-module is
cyclically annihilated, Theorem 4.4 implies that R is a right duo ring. Hence, by a result
of Birkenmeier and Tucci [9, Prop. 6], every homomorphic image of R is strongly right
35
bounded. Thus, B=A contains a non-zero two-sided ideal C=A. By the d.c.c. on ideals, we
may assume that C is the minimal two-sided ideal such that A C B.
Let x 2 CnA and note that xB 6= 0. As x 2 C, a two-sided ideal of R, we have
xB  C. Since R is right duo, xB is also a two-sided ideal. Further, if xB  A, then
xB B\Ann(B) Ann(B). Then 0 = (xB)B = xB2 = xB, a contradiction. So xB*A
and by the choice of C we have xB = C. Then there exists b2B such that xb = x so that
x(1 b) = 0. Hence, b2B J, and therefore 1 b is a unit. Thus, x = 0, a contradiction.
Hence B = 0 and consequently R is right Artinian.
4.2 Ideal Structure of Right Chain Rings
Valuation domains play a key role in many areas of commutative ring theory, especially
in algebraic geometry. The interplay between valuation domains and valuations on a  eld
are well known. A similar thing occurs when one considers chain domains. Notice that a
chain domain R is an Ore domain and hence the quotient ring Q of R exists and is a division
ring. Hence, constructing chain domains gives rise to division rings. The converse question,
i.e.,  nding a right chain domain in a given division ring is also of interest. We will show
below that, for chain domains, the situation is similar to that of valuation domains. Recall
that by a chain domain, with no quali er, we mean a domain that is a right and left chain
ring. A similar convention holds for a chain ring, an Ore domain, etc.
Proposition 4.11. Suppose R is a domain. The following statements are equivalent:
1) R is a chain ring.
2) R is an Ore domain with the property that for every non-zero q2Q, either q2R or
q 12R, where Q is the quotient ring of R.
Proof. 1) implies 2): Since R is a chain domain, RRR is a uniform right and left R-module
and hence an Ore domain. Suppose 06= q = ab 1 2Q and q62R. Then a62Rb. Since R is
36
a left chain ring by assumption this implies that Rb Ra so we may write b = ta for some
t2R. Then q 1 = ba 1 = t2R as claimed.
2) implies 1): As R is a right Ore domain the right quotient ring Qr exists. Suppose
a and b are non-zero elements of R such that Ra * Rb. Then ab 1 62R. By hypothesis
ba 1 = (ab 1) 1 2R. Hence b2Ra and therefore Rb Ra so that R is a left chain ring.
As R is also a left Ore domain Ql also exists and Ql = Qr = Q. By a symmetric argument
with Qr = Q, R is also a right chain ring.
The following, which appears without proof in [6, Lemma 6.1], now follows easily from
Corollary 4.11.
Corollary 4.12. Suppose R is a subring of a division ring Q. The following statements are
equivalent:
1) R is a chain domain with skew  eld of quotients Q.
2) For every non-zero q2Q, either q2R or q 12R.
As noted in [6], Corollary 4.12 immediately implies that every overring of a chain domain
R in Q is also a chain domain. In fact Brungs and T orner show in [10] that each such overring
of R is a localization of R at a prime ideal. Unfortunately these properties do not extend
to right chain domains, in that there exists a right duo right chain domain R with quotient
ring Q, and an overring S of R in Q, such that S is neither a right nor a left chain ring. See
[6, Example 7.1].
We now turn our attention to a class of modules that play a key role in the investigation
of modules over valuation domains.
De nition 4.13. A right R module U is uniserial if its submodules are linearly ordered
with respect to inclusion. Equivalently given a, b2U, either aR bR or bR aR.
Obviously submodules and quotients of uniserial modules are uniserial, uniserial modules
are uniform and therefore indecomposable, and R is a right chain ring if and only if RR is a
uniserial right R-module.
37
Proposition 4.14. Let R be a chain domain and Q the right quotient ring of R. Then
1) Q is a uniserial R module, and
2) every proper R-submodule of Q is isomorphic to a right ideal of R.
Proof. 1): Let U and V be R-submodules of Q and suppose U6 V. Choose a2UnV and
let b2V. Clearly b = 0 implies b2U, so assume b6= 0. Then we claim that a 1b2R.
Suppose to the contrary that a 1b62R. Then by Proposition 4.11, b 1a = (a 1b) 1 2R.
Thus, a = b(b 1a) 2V, a contradiction. Hence a 1b2R, and therefore b2aR U. It
follows that V  U and Q is a uniserial R-module.
2): Suppose M is a proper R-submodule of Q. If M  R, then M is a right ideal of
R and the claim is established. Suppose R is properly contained in M. Since M is proper
there exists a regular element t2R such that t 1 62M. Since Q is uniserial by 1), either
t 1R M or M  t 1R. Clearly the  rst case cannot hold as this forces t 1 2M. So we
have that M <t 1R. Then tM <R and t regular implies that left multiplication by t is a
monomorphism. It follows that M?tM.
De nition 4.15. Let R be a right Ore domain with right quotient ring Q, and suppose K
is an R-submodule of Q. If I is a subset of Q, then the residual of I and K is de ned as
the set (K : I)r =fq2QjIq Kg.
Remark 4.16. If R a right duo ring, then (K : xR)r = (K : x)r for every x2R.
Lemma 4.17. Let R be a right Ore domain with right quotient ring Q, and suppose IR and
KR are R-submodules of Q. Then (K : I)r is an R-submodule of QR.
Proof. Suppose x;y2(K : I)r and a2I. Then ax = k1 and ay = k2 for some k1;k2 2K.
Therefore a(x y) = ax ay = k1  k2 2 K, so x y 2 (K : I)r. If r 2 R then
a(xr) = (ax)r = k1r2K as K is an R-submodule. Hence xr2 (K : I)r. It follows that
(K : I)r is an R-submodule of QR.
38
Proposition 4.18. Let I and K be proper non-zero right ideals of a right duo, right chain
ring R. For 06= x2R, we have I = (K : x)r if and only if K = xI.
Proof. If 0 6= K = xI and a2I, then xa2K, so a2 (K : x)r. Thus, I  (K : x)r. For
the reverse inclusion, suppose a2(K : x)r. Then xa2K = xI so xa = xb for some b2I.
From this we see a b2annr(x). Since xI 6= 0, I *annr(x). As R is a right chain ring,
it follows that annr(x) (I. Therefore a b2I. Thus, b2I implies that a2I. Hence,
(K : x)r I.
Conversely, suppose I = (K : x)r. If a2I, then a2 (K : x)r and xa2K. Hence
xI K. Suppose by way of contradiction that xI  K, and choose a2KnxI. Since R is
a right chain ring, we consider two cases: a = xb0 and x = ab, for some b, b02R. In the  rst
case, since a2K and a = xb0, we have that b02(K : x)r = I. Then a = xb02xI, contrary
to the choice of a. Suppose the second case, x = ab, holds. Since K is a right ideal of R
and a2K, we have R (K : a)r. We obtain a chain of inclusions I < R (K : a)r, and
claim that (K : a)r  (K : x)r as well. To see this, suppose y2(K : a)r. Then ay = k for
some k2K. Since x = ab, it follows that xy = aby. Since R is right duo, by2Ry yR, so
by = yb0 for some b02R. Then xy = aby = ayb0 = kb02K. It follows that y2(K : x)r, as
claimed. Combining these results we have a chain of inclusions I <R (K : a)r (K : x)r.
This is a contradiction as I = (K : x)r by assumption. It follows that xI = K and the proof
is complete.
Corollary 4.19. Suppose I and K are proper non-zero right ideals of a right duo right chain
ring R. Then for every r2R, rI = rK6= 0 implies I = K.
Proof. By Proposition 4.18, I = (rI : r) for every proper non-zero right ideal I. Thus,
I = (rI : r) = (rK : r) = K.
Let I be a non-zero ideal of a right chain ring R. Following Fuchs-Salce [15], we set
Ul(I) =fr2RjrI = Ig and de ne I# as the set RnUl(I). Then I# =fr2RjrI <Ig.
We will show that, under certain conditions, I# is a completely prime ideal containing I.
39
Lemma 4.20. Let R be a right duo right chain domain and I a non-zero proper ideal of R.
Then,
1) I# is a completely prime ideal containing I.
2) If I is a principal left ideal of R, then I# = J.
3) If I is a completely prime ideal of R, then I# = I.
Proof. 1): We will show that I# is a right ideal of R. Suppose r;s 2 I#. Since R is a
right chain ring, we may assume without loss of generality that s = rt for some t2R. Then
r s = r rt = r(1 t). Suppose by way of contradiction that r s62I#. Then r s2Ul(I)
so (r s)I = I. Since r2I#, we may choose an element a2InrI. Then there exists x2I
such that (r s)x = a. So a = (r s)x = r(1 t)x2rI, contradicting the choice of a.
Suppose now that r 2 I# and x 2 R. If rx 62 I#, then rx 2 Ul(I) and (rx)I = I.
Since r2I#, there exists a2InrI. Then there exists y2I such that a = (rx)y2rI, a
contradiction. Thus rx2I# and I# is a right ideal.
The proof that I# is completely prime follows a similar line of argument. Suppose
rs 2 I# and r 62 I#. Then choosing a 2 InrsI, there exists an element b 2 I such
that a = rb since rI = I. If s 62 I#, then there exists c 2 I such that sc = b. Then
a = rb = rsc2rsI, a contradiction. Hence, s2I# and it follows that I# is a completely
prime right ideal.
Finally, if a 2 InI#, then there exists an element b 2 I such that ab = a. Then
a(1 b) = 0. As R is a domain and a6= 0, we have that b = 1, a contradiction as I is a
proper ideal. Hence, I I#.
2): Suppose 06= I = Ra. It is always the case that I#  J, since I# is a proper right
ideal. For the reverse inclusion, if j 2J and j 62I#, then jI = I. So a = jsa for some
s2R. Hence, (1 js)a = 0 and 1 js2U(R). Thus, a = 0, a contradiction. It follows
that J I#.
40
3): Suppose a2RnI. Since R is a right chain ring, this implies that I (aR. If b2I,
then b = ac for some c2R. Since I is completely prime and a62I, we must have c2I.
Then I aI I, so I = aI and therefore a2Ul(I) = RnI#. Hence, RnI RnI#, which
implies I# I. As I I# by 1), it follows that I = I#.
De nition 4.21. Let I be an ideal of a ring R. We say that I is a regular ideal of R if I
contains a regular element.
By [15, Lemma II.4.4], if I is a non-zero ideal of a valuation domain R, then I# is a
RI#-module in a natural way, where RI# is the localization of R at the completely prime
ideal I# (see Theorem 2.34). In order to extend this result to the non-commutative setting,
we will use the left ring of fractions. In what follows we will use the left localization at
I#. Then I#R = fs 1ajs2Ul(I) and a2Rg (see Remark 2.36). We also extend [15,
Proposition I.4.6] to right duo chain domains.
Theorem 4.22. Let R is a right chain ring and I a non-zero regular ideal of R. If L is the
left localization of R at the completely prime ideal I#, then
1) If R is a right duo ring, the set of zero divisors of R is contained in I#.
2) If R is a right duo chain domain, then IR is a left L-module.
3) If R is a right duo chain domain, then EndR(IR)?L.
Proof. 1): Suppose b2R is a right zero divisor. Then there exists a non-zero x in R such
that xb = 0. If b62I#, then bI = I. Let a2I be a regular element, and choose y 2I
such that by = a. Then xa = x(by) = (xb)y = 0, which is a contradiction as a is regular.
Therefore I# contains all right zero divisors.
Similarly, if b2R is a left zero divisor, then bx = 0 for some non-zero x2R. If b62I#,
then bI = I. Choosing a2I regular, there exists y 2I such that by = a. As R is right
duo there exists y02R such that yx = xy0 for some y02R. Then ax = (by)x = (bx)y0 = 0,
41
again contradicting the fact that a is regular. Hence I# contains all left divisors of zero as
well.
2): Since R is a chain domain, R is right and left Ore. As I# is a two-sided completely
prime ideal of R, the left and right localizations at I# exist. Recall that the left localization
of R at the completely prime ideal I# is L =fs 1ajs2Ul(I) and a2Rg. If s2Ul(I) is
non-zero, then s is a regular element of R by 1). Let  s : IR !IR be left multiplication by
s. Since s is in Ul(I) we have that sI = I, so  s is an epimorphism of IR. As s is regular,
we have in addition that  s is injective. Hence  s2AutR(IR) and it follows that s 1I I.
Therefore we may de ne (s 1a) x = s 1(ax) for every s 1a2L and x2I. It will be shown
that IR is a left L-module.
We  rst show that the multiplication given above is well-de ned. Suppose we have
elements s 1a, t 1b2L such that s 1a t 1b. By Theorem 2.34, there exist c, d2R such
that ca = db and cs = dt 2 Ul(I). Then s 1c 1 = t 1d 1, so s 1 = t 1d 1c. If x 2 I,
then s 1ax = t 1d 1cax = t 1d 1dbx = t 1bx. It follows that the operation is well de ned.
The veri cation of the module axioms is mechanical using Theorem 2.34 and the remarks
following.
3): Note that since R is a right and left chain domain, it is right and left Ore. Hence
the right and left quotient rings Qr and Ql exist. We  rst identify EndR(IR) with a subring
of Ql. Let q2Ql and  q : Ql !Ql be left multiplication by q. If ?2EndR(IR), then since
QR is an injective right R-module, there exists an R-homomorphism  : QR  !QR such
that  jI = ?. As HomR(QR;QR) = HomQ(QR;QR), there exists q02Q such that  =  q0.
Then for i2I, ?(i) =  q0(i) = q0i. De ne  : EndR(IR)  !Q by  (?) = q0. It is easily
veri ed that  is a ring monomorphism and hence EndR(IR) can be viewed as a subring of
Ql.
With this identi cation, we show that EndR(IR) = L. Suppose 06= s 1a2L. If x2I,
then s 1ax2I and we conclude that ax = sy for some y 2I. Computing inside Ql we
obtain that s 1ax = y2I. Hence, left multiplication by s 1a is an endomorphism of IR.
42
For the reverse inclusion, suppose ?2EndR(IR). By the identi cation above, ? =  q for
some q2Ql. If qI = I, then q2L and we are done, so suppose qI <I. As q2Ql, q = s 1a
for some regular s. We consider two cases: aR sR and sR aR. In the  rst case, a = st
for some t2R. Then s 1a = s 1st = t2R L. In the second case, s = at for some t2R.
Then inside Ql we have that s 1 = t 1a 1. Hence q = s 1a = t 1a 1a = t 1 so that q = t 1.
Then qI I implies that t 1I I which implies that I tI I so tI = I. Thus, t2Ul(I)
and therefore q2L.
4.3 Modules over Right Chain rings
Let n be a positive integer. If A is an abelian group, then the set
A[n] =fa2Ajna = 0g
forms a subgroup of A consisting of elements whose order divides n. If R is a commutative
ring and M an R-module, then for any r2R, the set M[r] =fm2Mjmr = 0g is easily
seen to be a submodule of M. For general rings, M[r] need not form a submodule of M. If
R is a right duo ring, however, M[r] is a submodule of M for every r2R.
Lemma 4.23. Let R be a ring. Then M[r] is a submodule of M for every right R-module
M and every r2R if and only if R is right duo.
Proof. Suppose R is right duo and choose m;n2M[r]. Then mr = 0 and nr = 0. Hence
(m n)r = mr nr = 0, so M[r] is an additive subgroup of M. To show that M[r] is a
submodule of M, suppose m2M[r] and s2R. Since R is right duo, sr2Rr rR, so we
can write sr = rt for some t2R. Then (ms)r = m(sr) = m(rt) = (mr)t = 0. Therefore,
ms2M[r] and it follows that M[r] is a submodule of M.
Conversely, suppose the condition holds and a 2 R. Consider the right R-module
M = R=aR. Then M[a] =fr + aR2R=aRjra2aRg. In particular, 1 + aR2M[a]. As
43
M[a] is a submodule of M, R + aR = (1 + aR)R2M[a]. Thus, Ra aR and therefore R
is right duo.
We can also form the set M[I] =fm2M jI Ann(m)g, where I is a right ideal of
R. Once again, if R is commutative, M[I] is a submodule of M. We have:
Lemma 4.24. Let R be right duo and M a right R-module. Then for every right ideal I of
R, M[I] is a submodule of M.
Proof. Clearly M[I] is an additive subgroup of M. To see that M[I] is a submodule of M,
suppose m2M[I] and r2R. Since R is right duo, I is an ideal of R. Then for every a2I,
ra2I. As I Ann(m), it follows that (mr)a = m(ra) = 0. Hence mr2M[I] and M[I] is
a submodule of M.
Consider the set M[I+] =fm2Mj I <Ann(m)g, where I is a right ideal of R. Once
again, if R is a valuation ring, it is not di cult to see that M[I+] is a submodule of M. This
can be extended to right chain rings.
Lemma 4.25. Let R be a right duo right chain ring and M a right R-module. Then for
every right ideal I of R, M[I+] is a submodule of M.
Proof. Suppose m;n 2 M[I+]. Then I < Ann(m) and I < Ann(n), so we may choose
r 2 Ann(m)nI and s 2 Ann(n)nI. Since R is a right chain ring, this implies that I
is properly contained in the right ideals rR and sR. Once again, by the right chain ring
property, we may assume that rR sR. So we have I rR sR. Hence, r = st for some
t2R. Then (m n)r = mr nr = mr nst = 0. Therefore r2Ann(m n), and as r62I,
we see that m n2M[I+]. So M[I+] is an additive subgroup of M.
To see that M[I+] is an R-submodule of M, suppose m 2 M[I+] and r 2 R. Then
I <Ann(m), so we may choose s2Ann(m)nI. As R is right duo, Ann(m) is an ideal of
R. Hence rs2Ann(m). Then (mr)s = m(rs) = 0. As s62I, it follows that mr2M[I+].
Hence M[I+] is a submodule of M.
44
Given the results above, we can easily establish the following.
Proposition 4.26. Let R be a right duo right chain ring and M a right R-module. Then
for every right ideal I of R:
1) M[I] and M[I+] are fully invariant submodules of M.
2) If f2HomR(M;N), then f carries M[I] into N[I] and M[I+] into N[I+].
Proof. 1): Suppose f 2EndR(M). If a2f(M[I]), then a = f(m) where m2M[I]. So if
x2I, then mx = 0. Hence ax = f(m)x = f(mx) = 0, and therefore I Ann(a). It follows
that a2M[I] whence M[I] is fully invariant. The second statement is proven in a similar
fashion.
2): If f(m) 2 f(M[I]), then m 2 M[I]. Hence, I  Ann(m). So for every x 2 I,
mx = 0 implies that f(m)x = f(mx) = 0. Therefore I  Ann(f(m)) and f(m) 2N[I].
The proof of the second statement is similar.
De nition 4.27. Let R be a ring and M a right R-module. The socle of M, written
soc(M), is the sum of all simple submodules of M.
Proposition 4.28. Let R be a right duo right chain ring and M a right R-module. Then
soc(M) = M[J].
Proof. If K is a simple submodule of M, then K ?R=M, where M is a maximal ideal of
R. As R is local, we have that K  M is simple if and only if K ?R=J. So the socle of
M is isomorphic to a sum of copies of R=J. If a + J 2R=J, then J  AnnR(a + J), i.e.
a + J2M[J]. Hence any  nite sum of such elements, that is, any element of soc(M), is in
M[J].
For the reverse inclusion, suppose 06= a2M[J]. Then J Ann(a). As J is maximal,
this implies that Ann(a) = J. Hence aR?R=J, and therefore a2soc(M).
45
Chapter 5
Finitely Generated Modules
In this chapter we begin the investigation of  nitely generated modules over right chain
rings. This work presents a natural extension to the noncommutative setting of results by
Kaplansky, War eld, Fuchs, Salce, and Zanardo concerning  nitely generated modules over
valuation domains.
5.1 RD-submodules
The notion of a pure subgroup of an Abelian group may be extended to modules over
arbitrary rings in several ways. One of the most useful extensions is the concept of an RD-
submodule of a right R-module. This section introduces some of the basic properties of these
submodules. In addition, we provide a characterization of RD-submodules of modules over
right duo chain rings.
De nition 5.1. Let R be a ring and M a right-R module. A submodule N of M is an
RD-submodule if for every r2R, we have Nr = N\Mr. As the inclusion Nr N\Mr
always holds, the RD property amounts to showing that N\Mr Nr for every r2R.
Notation 5.2. We will denote the fact that N is an RD-submodule of M by N RD M.
Example 5.3. If G is an abelian group, then a subgroup H of G is pure if for h 2 H,
h = ng for an integer n and g2G implies h = nh1 for some h1 2H. It is easily seen that
the RD-property for R = Z is equivalent to purity.
The following establishes some of the basic properties of RD-submodules.
Proposition 5.4. Let R be a ring and M a right R-module. If L and N are submodules of
M, then
46
1) If M = NLK, then N RD M.
2) If L RD N and N RD M, then L RD M.
3) If L N M and N RD M, then N=L RD M=L.
4) If L N M and L RD M, then N=L RD M=L implies N RD M.
Proof. 1): Let r2R and suppose mr2N for m2M. Then m = n + k for some n2N,
k2K. Hence mr = (n + k)r = nr + kr. So mr nr2N\K as mr2N by hypothesis.
Since the sum is direct, we have that mr nr = 0. Therefore mr = nr, and N RD M.
2): Let r 2 R and suppose mr 2 L for some m 2 M. Then L  N implies that
mr2N\Mr = Nr as N RD M. Since mr2L also, we have that mr2L\Nr = Lr by
the assumption that L is an RD-submodule of N. It follows that L RD M.
3): Suppose mr + L 2 N=L for some m 2 M;r 2 R. Then mr 2 NTMr = Nr
as N  RD M by hypothesis. Hence mr + L 2 (N=L)r and we have that N=L is an RD
submodule of M=L.
4): Suppose mr2N for some m2M;r2R. Then we have that mr+L = (m+L)r2
N=LT(M=L)r = (N=L)r as N=L RD M=L. Therefore mr2Nr and N RD M.
Remark 5.5. Note that 4) says that preimages of RD-submodules are RD.
Theorem 5.6. Let R be a right duo chain ring. Then the following are equivalent for a right
R module M and every a2M:
1) aR is an RD-submodule of M.
2) If 06= ar = zs for r, s2R and z2M, then Rr Rs.
3) Ann(x) Ann(a) for every x2a+MJ.
Proof. 1) implies 2): Assume that aR is an RD-submodule of M. Suppose r, s2R and
z2M are such that 06= ar = zs. Since aR is RD, we can write ar = ar0s for some r02R.
Assume by way of contradiction that Rs Rr. Then s = jr for some j2R which cannot
47
be a unit of R. Thus, j 2 J and ar = ar0s = ar0jr so that a(1 r0j)r = 0. Note that
u = 1 r0j2U(R) as j2J. So ur2Ann(a) = Ann(aR) as R is right duo. In particular,
ar = au 1(ur) = 0, a contradiction.
2) implies 3): Suppose that (a+y)r = 0 but ar6= 0 for some y2MJ. Then since R is
a chain ring, we can write y = mj for some m2M and j2J. Then 0 6= ar = ( m)(jr).
By 2), Rr  R(jr) which implies that (1 r0j)r = 0 for some r0 2 R. Then r = 0 as
1 r0j2U(R), a contradiction.
3) implies 1): Suppose 0 6= ar = ms for some r, s2R and m2M. If Rr Rs then
r = ts for some t2R. So ms = ar = ats2aRs and we are done. So suppose Rs Rr.
Then s = jr for some j2J. We have 0 6= ar = ms = mjr and therefore (a mj)r = 0.
Then, ar = 0 by 3), a contradiction.
5.2 RD-Composition Series
In this section we characterize right duo rings using the existence of an RD-composition
series for  nitely generated modules over right chain rings. This result naturally extends a
result for commutative rings by Fuchs and Salce in [14].
De nition 5.7. A ring R has the right unique kernel property provided that, whenever
I and K are right ideals of R such that R=I = R=K, then I = K.
Lemma 5.8. A right chain ring R has the right unique kernel property if and only if R is
a right duo ring.
Proof. Suppose that R is right duo, and consider an isomorphism  : R=I!R=K for right
ideals I and K of R. Observe that I and K are two-sided since R is duo. Select r2R such
that  (1 +I) = r+K. For every a2K, we have  (a+I) =  (1 +I)a = ra+K = 0. Since
 is one-to-one, a2I, and thus K I. By symmetry, I = K.
Conversely, suppose that R has the right unique kernel property, and let I be any right
ideal of R. Consider a unit u2R, and de ne  : R=I!R=uI by  (r+I) = ur+uI. Then,
48
r s2I if and only if ur us2uI, and  is a well-de ned monomorphism. Moreover, if
uv = 1, then  (v + I) = 1 + uI, and  is onto. Hence, I = uI by the right unique kernel
property. Since R is a right chain ring, this implies that I is two-sided.
De nition 5.9. A series of submodules 0 = M0 <   <Mn 1 <Mn = M of a module M
satisfying
1) each Mi is an RD-submodule of M and
2) Mi+1=Mi is cyclic for every i,
is an RD-composition series of M. The sequence Ai = Ann(Mi=Mi 1) of right ideals of
R is the annihilator sequence of the RD-composition series. The annihilator sequence is
nondecreasing if Ai Ai+1 for every i = 1;2;:::;n 1.
Theorem 5.10. The following are equivalent for a chain ring R:
1) R is right duo.
2) Every  nitely generated right R-module M admits a  nite chain
0 = M0 <   <Mn 1 <Mn = M
of RD-submodules such that Mi+1=Mi?R=Ii for right ideals Ii of R with the property
that I1      In. Moreover, if 0 = M00 <    < M0n = M is another chain of
RD-submodules of M such that M0i+1=M0i ?R=I0i for right ideals I01      I0n, then
I0i = Ii for every i.
Proof. Suppose R is right duo. Let m+MJ2M=MJ and r+J2R=J. De ne an operation
(m+MJ)(r+J) = mr+MJ. If r r02J, then mr mr0 = m(r r0)2MJ and therefore
mr + MJ = mr0 + MJ. Thus, M=MJ is a right R=J-module. Since M=MJ is  nitely
generated, it is a  nite dimensional vector space over the division ring R=J, and hence has
49
a  nite basis, say fxi + MJji = 1;2;:::;ng. As MJ + (PixiR) = M and M is  nitely
generated, the xi generate M by Nakayama?s Lemma.
As R is right duo, AnnR(M) = Tni=1AnnR(xi). Since R is a chain ring, we can order
the annihilators of the xi (reindexing if necessary) as AnnR(x1)     AnnR(xn). So there
exists at least one k such that AnnR(xk) = AnnR(M). Let 1 j n be chosen largest such
that AnnR(M) = AnnR(x1) =   = AnnR(xj). We claim that there exists an i2f1;:::;jg
such that AnnR(x)  AnnR(xi) for every x2xi + MJ. Suppose by way of contradiction
that for every k j there exists x0k2xk +MJ such that AnnR(xk)  AnnR(x0k).
Then M is generated by the set fx01;:::;x0j;xj+1;:::;xng by another application of
Nakayama?s Lemma. Ordering the annihilators as an increasing chain as above we see that
AnnR(M) = AnnR(xl) for at least one l. By the choice of j we must have l j. Hence,
AnnR(xl) = AnnR(x0l) which contradicts our assumption. Hence the claim is established.
Without loss of generality, we may assume the element so obtained is x1. Then by Theorem
5.6, x1R is an RD-submodule of M.
Set M1 = x1R and induct on the number n of generators. Clearly the result holds for
n = 1. By the induction hypothesis, M=M1 has an RD-composition series
0 = L0=M1 L1=M1 :::Ln 1=M1 = M=M1
where M1 L1 L2 ::: Ln 1 = M are submodules of M. Then
(Li+1=M1)=(Li=M1)?Li+1=Li
is cyclic. As preimages of RD-submodules are RD by Proposition 5.4, we see that the Li
form an RD-composition series of M.
To see that the annihilator condition is satis ed, note
Ann(Li=Li 1) = Ann((Li=Mi)=(Li 1=Mi)) Ann((Li+1=Mi)=(Li=Mi)) = Ann(Li+1=Li):
50
As Mi=Mi 1 is cyclic, Ai satis es R=Ai?Mi=Mi 1 as R is a right duo ring. Set Ii = Ai.
To establish the uniqueness claim, suppose 0 = M00 <    < M0n = M is another RD-
composition series of M, with I0i = Ann(M0i=M0i 1). Then Mi=Mi 1?R=Ii and M0i=M0i 1?
R=I0i, where the right ideals Ii and I0i are unique. Observe that I1     In and I01     
I0n.
Suppose that I1 6= I01. By Lemma 5.8, R=I1 6?R=I01, and hence M1=M0 6?M01=M00. But
then M1=M0?M0j=M0j 1 for some j > 1. Hence, I1 = I0j. In the same way, we can  nd k> 1
such that I01 = Ik. Then I1 Ik = I01 I0j = I1, which implies that I1 = I01, a contradiction.
Thus, I1 = I01. An induction on the length of the composition series establishes the claim.
For the converse, consider a right ideal I of R, and choose a unit u in R. Then R=I =
(u 1 + I)R so that R=I?R=uI. By uniqueness, I = uI. As R is a chain ring, this implies
that I is two-sided and thus, R is right duo.
Let M be a  nitely generated right R-module, where R is a right duo chain ring. Then,
M has a composition series
f0g= U0 U1 ::: Un = M
such that Ui=Ui 1 is cyclic, say Ui=Ui 1  = R=Ii for some right ideal Ii. The last theorem
shows that the factors Ui=Ui 1 determine the \annihilators" I1;:::;In uniquely only if R
is a right duo ring. Therefore, we will assume that R is right duo whenever we consider
composition series for a  nitely generated module.
5.3 Equivalence of Relative Divisibility and Purity
Purity in the sense of Cohn is an additional extension of the notion of a pure subgroup
of an Abelian group to modules over arbitrary rings. War eld has shown in [39] that if R is
an integral domain, then R is Pr ufer if and only if relative divisibility is equivalent to purity.
This result naturally extends to modules over right duo chain domains. In establishing this
51
equivalence, we utilize the existence of an RD-composition series established in the previous
section.
De nition 5.11. Let R be a ring and A a right R-module. A submodule B of A is said to
be pure if every  nite system of equations over B
mX
j=1
xjrij = bi (i = 1;:::;n);
with rij2R and unknowns x1;:::;xm, has a solution in B whenever it is solvable in A.
Note that B RD A is equivalent to the condition that whenever b2B and the equation
xr = b is solvable in A, then it is solvable in B. From this we see immediately that pure
submodules are RD for an arbitrary ring R. The investigation into the rings for which the
converse holds is more di cult. Indeed, we have the following open question.
Open Question: If R is a right semihereditary ring, then every RD-submodule of a
right R-module is pure.
This is a natural question to ask, given the result of War eld noted above. We need
several preliminary result in order to establish the equivalence for modules over right duo
chain domains.
De nition 5.12. An exact sequence E: 0 !A !B !C !0 of right R-modules is
RD-exact if the map A !B embeds A in B as an RD-submodule. A right R-module M
is cyclically presented if it is of the form R=xR for some x2R.
We have the following Lemma from [39]:
Lemma 5.13. [39] For an exact sequence E : 0  ! A   ! B   ! C  ! 0 of right R-
modules, the following are equivalent:
1) E is RD-exact.
2) The induced map   : HomR(R=rR;B)  ! HomR(R=rR;C) is surjective for every
r2R.
52
3) Then induced map 1  : R=rR RA !R=rR RB is injective for every r2R.
Note that 2) says that cyclically presented modules have the projective property relative
to RD-exact sequences. The following proof follows that of War eld [39] verbatim.
Theorem 5.14. A  nitely presented module MR over a right duo chain domain is a direct
sum of cyclically presented modules.
Proof. Since MR is  nitely presented, by Theorem 5.10 there exists an RD-composition series
of M
0 = M0 <   <Mn 1 <Mn = M:
As Mn 1 is  nitely generated and M  nitely presented, the quotient M=Mn 1 is again  nitely
presented. Hence the quotient M=Mn 1 is  nitely presented and cyclic, so that we can  nd
a  nitely generated right ideal In such that M=Mn 1 ? R=In. Since R is a right chain
domain, In is cyclic. Thus, M=Mn 1 is cyclically presented. By Lemma 5.13 M=Mn 1 has
the projective property relative to RD-exact sequences. Hence, 0  ! Mn 1  ! M  !
M=Mn 1  ! 0 splits and M ?Mn 1 M=Mn 1. As Mn 1 is again  nitely presented, the
result follows by induction.
We have the following characterization for pure exact sequences. The proof may be
found in [25, Theorem 4.89].
Theorem 5.15. For any short exact sequence E : 0  ! A  ! B  ! C  ! 0 of right
R-modules, the following are equivalent:
1) E is pure-exact.
2) E RC0 is exact for any  nitely presented left R-module C0.
3) A is a pure submodule of B.
4) Given a commutative diagram (of right R-modules)
53
Rn     ! Rm?
?yf ??yg
0    ! A    ! B
there exists  2HomR(Rm;A) such that   = f.
5) Any  nitely presented right R-module M has the projective property relative to E.
The following proof is due to War eld [39].
Theorem 5.16. Over right duo chain domains, purity and relative divisibility are equivalent.
Proof. We will show that an RD-exact sequence E: 0  !A !B !C  ! 0 of right
R-modules is pure exact. Let M be a  nitely presented right R-module. By Theorem 5.14,
M is a direct sum of cyclically presented modules. By Lemma 5.13 each of these have the
projective property relative to E, hence M has this same property. By Theorem 5.15, E is
pure-exact.
5.4 A Jordan-H older Type Theorem
Given the existence of an RD-composition series, a natural question to ask is if any
two such series are isomorphic. We show that this is indeed the case for  nitely generated
modules over right duo right chain rings. The results below follow the work of Fuchs and
Salce [16], and Salce and Zanardo [33] closely.
By modifying the proof of Salce and Zanardo in [33, Lemma 1.2], we are able to extend
the following result to modules over right duo chain rings.
Proposition 5.17. Let R be a right duo, right chain ring and M a right R-module. If N is
an RD-submodule of M such that M=N = (a + N)R is cyclic and Ann(M=N) Ann(N),
then M = aR N.
Proof. Clearly M = aR+N. Suppose ar2N. Since N is an RD-submodule, we have that
ar 2 N\Mr = Nr and we can write ar = nr for some n 2 N. Then (a n)r = 0 so
54
r2Ann(a n) = Ann((a n)R) by Lemma 3.33. If we can show that r2Ann(M=N),
then by the assumed condition on annihilators, Nr = 0 and therefore ar = 0 and the sum is
direct.
Let m+N2M=N. Then m+N = as+N for some s2R. Since r2Ann((a n)R),
we have, in particular, that [(a n)s]r = (as ns)r = asr nsr = 0, i.e. asr 2N. It
follows that (m+N)r = (as+N)r = asr +N = N, so r2Ann(M=N).
The next result follows from Proposition 5.17.
Corollary 5.18. Let R be a right duo chain ring and M a right R-module. Suppose
R: 0 = M0 <   <Mn 1 <Mn = M
is a RD-composition series of M of length n, with nondecreasing annihilator sequence
A1      An. If Mk=Mk 1 = (xk + Mk 1)R and A1 =    = Ak for some k < n,
then Mk = Lki=1xiR. In particular, if the annihilator sequence of R is constantly equal to
Ann(M), then M is the direct sum of cyclic submodules.
Proof. The proof is by induction on k. The result is trivial for k = 1 as M1 = x1R is cyclic.
Suppose k = 2. We will show that M2 is the direct sum of cyclic modules using Proposition
5.17. By assumption, we have that A1 = A2. Recall that A1 = Ann(M1) = Ann(x1R) =
Ann(M). So we have that Ann(M) = A1 = A2. Since M1 is an RD-submodule of M,
we obviously have that M1 is RD in M2. Further, M2=M1 = (x2 + M1)R is cyclic. By
hypothesis, A2 = Ann(M2=M1) = A1 = Ann(M1). Hence the hypotheses of Proposition
5.17 are satis ed, and therefore M2 = M1Lx2R. As M1 = x1R is cyclic, the result follows
by induction.
The following result extends [33, Lemma 1.3].
55
Lemma 5.19. Let R be a right duo chain ring. Then every RD-composition series of a
 nitely generated right R-module M is isomorphic to one whose annihilator sequence is non-
decreasing.
Proof. Suppose we have an RD-composition series
R: 0 = M0 <   <Mn 1 <Mn = M
of M with annihilator sequence fAi j i = 1;:::;n 1g. Since R is an RD-composition
series of M, Mi=Mi 1 is cyclic, say Mi=Mi 1 = (xi + Mi 1)R. Suppose there exists an
i2f1;2;:::n 1g such that Ai+1 <Ai. The idea of the proof is to replace the submodule
Mi by a submodule M0i so that Mi=Mi 1 ? M0i=Mi 1 and A0i = Ann(M0i=Mi 1)  Ai+1.
Consider the module K = Mi+1=Mi 1 and its submodule N = Mi=Mi 1. It will be shown
that these modules satisfy the hypotheses of Proposition 5.17.
By Proposition 5.4 N is an RD-submodule of K. Further, K=N = Mi+1=Mi so K=N is
cyclic. With respect to this isomorphism, we see that K = (xi+1 +Mi 1)R +N. As
Ann(K=N) = Ai+1 <Ai = Ann(N);
the hypotheses of Proposition 5.17 are met. Hence K = NL(xi+1 +Mi 1)R.
Set M0i = xi+1R + Mi 1. Note that Mi 1  M0i  Mi+1. We claim that M0i=Mi 1 is
isomorphic to Mi+1=Mi. De ne  : M0i=Mi 1  !Mi+1=Mi by  (xir + Mi 1) = xi+1r + Mi.
To see that  is well de ned, suppose xi+1r + Mi 1 = xi+1s + Mi 1. Then xi+1(r s) 2
Mi 1 Mi.  is clearly an epimorphism. Suppose xi+1r2Mi. Then r2Ai+1 Ai. Hence,
xi+1r + Mi 1 2 (Mi=Mi 1)\(M0i=Mi 1) = 0 as K = NL(xi+1 + Mi 1)R. Hence,  is a
monomorphism, and therefore an isomorphism.
It follows that M0i=Mi 1 is cyclic. We also have that Mi+1=M0i is cyclic, as
Mi+1=M0i ?(Mi+1=Mi 1)=(M0i=Mi 1) = K=(M0i=Mi 1)?N = Mi=Mi 1:
56
Since Mi 1 is an RD-submodule of M and Mi=Mi 1  RD M=Mi 1, Proposition 5.4 implies
that M0i is an RD-submodule of M.
It remains to be shown that Ann(M0i=Mi 1)  Ann(Mi+1=M0i). Suppose we have
that r 2 Ann(M0i=Mi 1). Then M0ir  Mi 1. Therefore xi+1Rr  Mi 1. Recalling
that Mi+1=M0i = xi+1R + M0i, we see that (Mi+1=M0i)r = xi+1Rr + M0i  M0i. Hence
r2Ann(Mi+1=M0i).
In a  nite number of steps we thus obtain a pure composition series with non-decreasing
annihilator sequence isomorphic to P.
De nition 5.20. Let R be a ring and M a  nitely generated R-module. If R is an RD-
composition series of M, say 0 = M0 <M1 <:::<Mn = M, then we call n the length of
R, written ?(M). The minimal number of generators of M is denoted gen(M).
The next three results are due to Salce and Zanardo presented in [33]. The proofs carry
over without change for right duo, right chain rings.
Lemma 5.21. Let R be a right duo chain ring. Then the length of an RD-composition
series R of a  nitely generated right R-module M is equal to gen(M).
Proof. By Nakayama?s Lemma, a set of elements of M generate M if and only if their cosets
modulo MJ generate M=MJ. Hence gen(M) = dimR=J M=MJ. If Mi=Mi 1 = (xi+Mi 1)R,
i = 1;:::;n, then the elements xi generate M; hence gen(M) n. Since M=MJ is a  nite
dimensional vector space over the division ring R=J, the independence of the elements xi
modulo MJ establishes the converse.
Assume by way of contradiction that Pni=1xiri2MJ where ri2R for every i and at
least one of the ri is a unit. Let k be the least index such that rk 62J. Then we have that
Pk
i=1xiri2Mk\MJ = MkJ. It follows that
Pk
i=1xkrk = (
Pk
i=1xisi)j for a suitable j2J.
Hence, xk(rk skj)2Mk 1. Then rk62J implies that rk skj62J, and therefore is a unit.
It follows that xk2Mk 1, a contradiction.
We now examine submodules of M of the form Mr.
57
Lemma 5.22. [33, Lemma 1.5] Let M be a  nitely generated right R-module over a duo
chain ring R. Suppose M has an RD-composition series of length n with non-decreasing
annihilator sequence fAiji = 1;:::;ng. Let r2R. Then
1) If r62[ni=1Ai, then ?(Mr) = ?(M).
2) If r2Ak+1nAk for 0 k n 1, then Mr = Mkr and ?(Mr) = k.
Proof. 1): Note  rst that since R is duo, rR is an ideal of R and Mkr is a submodule of M
for every k = 1;:::;n. Consider the series
0 <M1r<   <Mnr = Mr.
It will be shown that this is a pure composition series of Mr.
Suppose m(rt)2MjrR for some t2R and m2M. Then m(rt)2Mj since Mjr Mj.
Further, m(rt)2M(rt) since m2M. Hence m(rt)2Mj\M(rt). By the purity of Mj in
M, m(rt)2Mj\M(rt) = Mj(rt) = (Mjr)t Mj(rt). Therefore Mjr is pure in Mr.
By assumption, Mj=Mj 1 is cyclic, say Mj=Mj 1 = (xj + Mj 1)R. We claim that
Mjr=Mj 1r = (xjr + Mj 1r)R. Clearly (xjr + Mj 1r)R  Mjr=Mj 1r. For the reverse
inclusion, suppose mj(rs) + Mj 1r 2 Mjr=Mj 1r, where mj 2 Mj. Then we have that
mj + Mj 1r 2 Mj=Mj 1. Since Mj=Mj 1 = (xj + Mj 1)R, there exists t 2 R such that
mj + Mj 1r = xjt + Mj 1. Hence, mj(rs) + Mj 1r = xjtrs + Mj 1r. Since R is right duo,
Rr rR; so there exists t02R such that tr = rt0. Putting the equations above together,
we have that mj(rs) + Mj 1r = xjtrs + Mj 1r = xjrt0s + Mj 1r, which is an element of
(xjr +Mj 1r)R.
We conclude that Mjr=Mj 1r is cyclic. Therefore, the sequence above is a pure compo-
sition series for Mr and ?(Mr) = ?(M).
2): Fix k, 0  k n 1. Suppose j satis es k j n 1. We claim that for such
a j, Mj+1r = Mjr. Since Mj is pure in M, it follows easily that Mj is pure in Mj+1. Since
Mj  Mj+1 for every j = 1;:::;n, we also have Mjr Mj+1r. For the reverse inclusion,
58
note that r2Aj+1nAj implies that Mj+1r Mj. So if mj+1r2Mj+1r, then mj+1r2Mj.
By the purity of Mj in Mj+1, we have that mj+1r2Mjr. We conclude that Mj+1r = Mjr
for k j  n 1. So for k j  n 1, we have Mkr =    = Mnr = Mr. By (1) and
Lemma 5.22, it follows that ?(Mr) = ?(M) = k.
The next result establishes the isomorphism of any two RD-composition series of a
 nitely generated right R module over a duo chain ring.
Theorem 5.23. Let M be a  nitely generated right R-module over a duo chain ring R.
Then any two RD-composition series of M are isomorphic.
Proof. By Lemma 5.21, any RD-composition series of M has length equal to gen(M). Hence
we may assume the series have the same length. We may also assume, by Lemma 5.19, that
the annihilator sequences Ai and Bi are nondecreasing and have the same length.
In the proof of Theorem 5.10, it was shown that the initial non-zero submodule M1 in any
RD-composition series of M was chosen as M1 = x1R, where Ann(x1) = Ann(M). Since R
is right duo, Ann(M) = Ann(x1) = Ann(x1R) = Ann(M1). Therefore, A1 = Ann(M) = B1.
Assume Ak = Bk for some 1 <k<n. Suppose r2Ak+1nAk. By Lemma 5.22, ?(MrR) = k.
Since r62Ak, and Ak = Bk by the induction hypothesis, we have that r62Bk. But then
r must be in Bk+1 and Ak+1  Bk+1. By a symmetric argument, Bk+1  Ak+1, whence
Ak+1 = Bk+1 and the result follows by induction.
5.5 Essential Pure Submodules
The goal of the next set of results is to establish, in any  nitely generated right R-module
M, the existence of an essential pure submodule of M that is the direct sum of cyclic modules
and has the same Goldie dimension as M. We require several technical lemmas, following
[33], with modi cations for the noncommutative case.
Proposition 5.24. Let R be a right duo chain ring and M a  nitely generated right R-
module. Then for every 06= x2M, there exists r2R such that x2MrnMJr.
59
Proof. Let 06= x2M, and suppose 0 = M0 <M1 <:::<Mn = M is an RD-composition
series of M. Then, there exists an index i > 0 such that x 2 MinMi 1. The proof is
by induction on i. If i = 1, then x 2 M1 = x1R, a cyclic module. Hence there exists
an r 2 R such that x = x1r, whence x 2 Mr. Suppose by way of contradiction that
x 2 MJr. Then x = mj0r for some j0 2 J and then x 2 M1 \Mj0r = M1j0r by the
purity of M1 in M. As M1 = x1R we can write x = x1sj0r. As J is an ideal we may set
j = sj02J so that x1r = x = x1jr. Since x1R is an RD-submodule of M, 5.6.3 implies that
AnnR(y) AnnR(x1) for every y2x1+MJ. As x1 x1j2x1+MJ and r2AnnR(x1 x1j),
we have that r2AnnR(x1). But then x = x1r = 0, a contradiction.
Suppose now that i > 1. Set  x = x + Mi 1 and M = M=Mi 1. Then  x is a non-zero
element of Mi=Mi 1. Since Mi=Mi 1 is cyclic with generator xi+Mi 1, we have that  x =  xir
for some r2R. Hence  x2Mr. We claim that x62MJr. If x2MJr, then  x =  mjr for
some j2J, and we can write  x =  xir =  xijr. Arguing as above, noting that the purity of
Mi=Mi 1 in M=Mi 1 implies that  xiR is pure in M, we obtain that r2Ann(  xi). Therefore
x = xir2Mi 1, a contradiction. So we have that x2MrnMJr.
If x2Mr, then the proof is complete as x2MJr implies  x2MJr, contradicting the
above. Suppose that x62Mr. We can write x = mr +y for suitable m2M and y2Mi 1.
Then by the induction hypothesis, there exists an s2R such that y2MsnMJs. Note that
y62Mr as x62Mr by hypothesis, and x = mr + y. Since R is a chain ring, we have that
Rr Rs or Rs Rr. As Rs Rr implies that y2Mr, it must be the case that Rr Rs.
Therefore r = ts for some t2R. Hence x = mr +y = m(ts) +y = (mt)s+y2MsnMJs
and the proof is complete.
The following Corollary is an immediate consequence of Proposition 5.24:
Corollary 5.25. If 06= x2M, then there exists r2R and y2MnMJ such that x = yr.
60
In the work to follow, we  nd it necessary to consider conditions on a ring R which
guarantee that aJ = Ja for every a2R, where J is the Jacobson radical of R. This is not
always the case, as witnessed by the following:
Example 5.26. Let R be the 2 2 lower triangular matrix ring with entries from a  eld k.
By [18, Corollary 4.9], R is a right and left Artinian hereditary ring. Since R is Artinian,
the Jacobson radical is equal to the nilradical N of R. Let eij be the standard matrix units.
Then N = re21, where r2k. If a = e11, then a calculation shows that aJ = 0. On the other
hand, again by an easy calculation, Ja = J. Hence, aJ6= aJ.
In the class of rings under consideration, however, we are able to obtain a positive result.
Proposition 5.27. Suppose R is a chain domain. Then aJ = Ja for every a2R if and
only if R is a duo ring.
Proof. Suppose R is a duo ring, that is, aR = Ra for every a2R. Without loss of generality,
we may assume a6= 0. By symmetry, it is enough to show that Ja aJ. If this is not the
case, observe that xar = xr0a2Ja for every x2J and r2R, since R is duo. Hence, Ja
and aJ are right ideals of R. Since Ja*aJ and R is a chain domain, we obtain aJ (Ja.
Thus, Ja=aJ is a non-zero submodule of aR=aJ, for Ja Ra = aR.
Consider  : R=J ! aR=aJ given by  (r + J) = ar + aJ. If r1  r2 2 J, then
ar1 ar22aJ. Thus,  is a well-de ned epimorphism. If  (r+J) = 0, then ar2aJ which
implies ar = ax for some x2J. Thus, a(r x) = 0. If r62J, then rs = 1 for some s2R.
Then 0 = a(r x) = a(rs xs) = a(1 xs), which is a contradiction as 1 xs is a unit of
R. Hence,  is an isomorphism.
Since R is a right chain ring, J is a maximal ideal. Then R=J is a simple R-module.
Hence, aR=aJ is simple which implies Ja=aJ = aR=aJ. We conclude Ja = aR = Ra. Thus,
there exists y2J such that ya = a, which implies (y 1)a = 0. But then a = 0, as y 1 is
a unit. Thus, Ja aJ.
61
Conversely, suppose aJ = Ja for every a2R. Again, without loss of generality, we
may assume a6= 0. It is enough to show that Ra aR. Suppose x2R and consider xa. If
x2J, then xa2Ja = aJ aR. Hence, Ra aR as claimed. Thus, we may assume that x
is a unit of R. Since R is a chain domain, either xaR aR, or aR xaR. In the  rst case,
xa2aR and the result holds. Suppose the second case holds. Then a = xaj for some j2J.
Choose j02J such that aj = j0a. Then a = xaj = xj0a which implies (1 xj0)a = 0. As
1 xj0 is a unit of R, this implies a = 0, a contradiction.
The following technical result is crucial in establishing the main result. The proof is
based on ideas found in [33].
Theorem 5.28. Let R be a duo chain ring, and consider a  nitely generated right R-module
M. Suppose 0 = M0 < M1 < ::: < Mn = M is an RD-composition series of M with
nondecreasing annihilator sequence Ai; i = 1;:::;n. If Mn 1 is not essential in M, then M
has a non-zero cyclic summand.
Proof. Let M = x1R+   +xnR and suppose 06= y2M is such that yR\Mn 1 = 0. Then
y =
nX
i=1
xiai (ai2R): (5.1)
It will be shown that there is no loss of generality in assuming y2MnMJ. Suppose y2MJ;
then by Corollary 5.25, there exists an r2R and x2MnMJ such that y = xr. Moreover,
xR = R=I for some right ideal I of R. As R is a chain ring, R=I is uniserial and therefore
dim(xR) = 1. Hence yR is an essential submodule of xR and xR\Mn 1 = 0.
So assume that y 2 MnMJ and let j be the largest index such that aj 62 J. Then
aj is a unit and we may assume without loss of generality that aj = 1. If j = n then
M = yR Mn 1 and the proof is complete. Assume j <n and set N = Pi6=jxiR. Clearly
M = N +yR, so we need only show that N\yR = 0. Assume that N\yR6= 0. Then we
62
have a relation
06= yr =
X
i6=j
xibi;(bi2R): (5.2)
We claim that if j <h n then there exist a relation
xhrh =
h 1X
i=1
xiah;i with (5.3)
rh2Jr; ah;j2RrnJr (5.4)
The proof is by induction on h and we show  rst that the claim holds for h = n.
From (5.1) and (5.2) we obtain xn(bn anr) = Pn 1i=1 xi(air bi) where bj = 0. Suppose
bn 2 An = Ann(M=Mn 1). Then 0 6= yr = Pi6=jxibi 2 yR\Mn 1, a contradiction.
Similarly, anr 62An. Since R is a chain ring, either rR  bnR or bnR  rR. If the  rst
case holds, then r = bnt for some t 2 R. Recall that an 2 J by our assumption on j.
Hence, anbn 2Jbn = bnJ by Proposition 5.27. Thus, anbn = bna0n for some a0n 2J. Then
bn anr = bn anbnt = bn bna0nt = bn(1 a0nt). As a0n 2 J, (1 a0nt) is a unit of R.
Hence, xnbn(1 a0nt) 2Mn 1, which implies xnbn 2Mn 1. We conclude that bn 2An, a
contradiction. Thus, bnR ( rR, and bn = rs for some non-unit s2R. So bn 2rJ = Jr,
and thus an2J implies that bn anr2Jr.
Set rn = bn anr and note that (5.3) and the  rst part of (5.4) hold for h = n.
To establish the  nal claim in (4), recall that bj = 0 and that aj 62 J. So we have that
ajr2RrnJr. Setting an;j = ajr, we see that the  nal claim in (5.4) holds and the result is
established for h = n.
Assume by induction that the result holds for h>j + 1. So we have a relation
xhrh =
h 1X
i=1
xiah;i (5.5)
such that rh2Jr and ah;j2RrnJr.
63
Since rh 2Jr, we have that xhrh 2MJr\Mh 1 = Mh 1Jr by the purity of Mh 1.
Therefore we can write xhrh = Ph 1i=1 xiciqr where q2J;ci 2R. Subtracting, we see that
xh 1(ah;h 1 ch 1qr) = Ph 2i=1 xi(ciqr ahi). By the same argument as above, using the fact
that r62Ah 1, we will show that ah;h 1 ch 1qr2Jr.
Observe  rst that ch 1qr2Jr as q2J. Therefore our claim reduces to showing that
ah;h 1 2 Jr. If Rah;h 1 ( Rr, then ah;h 1 = jr for some j 2 J and the claim follows.
Suppose Rr Rah;h 1. Then r = sah;h 1 for some s2R. Utilizing Proposition 5.27 in an
identical fashion as above, we obtain that xhah;h 1 2Mh 1. Hence, ah;h 1 2Ah 1. Thus,
r = sah;h 1 2 Ah 1, a contradiction. So we have that ah;h 1  ch 1qr 2 Jr. Moreover,
ah;j 2 RrnJr implies that cjqr ah;j 2 RrnJr. Setting rh 1 = ah;h 1  ch 1qr and
ah 1;j = cjqr ah;j for i h 2, the claim is established for h 1 and the result follows by
induction.
Now set h = j + 1. By the result above, we have a relation xj+1rj+1 = Pji=1xiaj+1;j
where rj+1 2Jr and aj+1;j 2RrnJr. Repeating the argument above, using the fact that
rj+1 2Jr and the relative divisibility of Mj, we can write xj(aj+1;j djqr) 2Mj 1 where
dj2R and q2J. Then r62Aj 1 implies that aj+1;j djqr2Jr. Thus, q2J implies that
djqr2Jr. Hence, aj+1;j2Jr, a contradiction. It follows that N\yR = 0.
Now that these technical results are established, the proof of the following is an easy
consequence of Lemma 5.28. The proofs of the following results are due to Fuchs and Salce
[16].
Theorem 5.29. Over a right duo, chain ring R, every  nitely generated right R-module
M contains an essential pure submodule which is the direct sum of cyclic modules such that
dim(B) = dim(M).
Proof. We induct on n = gen(M). If n = 1, then the result is trivial, so assume n> 1. By
induction, Mn 1 has an essential pure submodule B0 that is the direct sum of dim(Mn 1)
non-zero cyclic submodules. If Mn 1 is essential in M, then B0 is essential in M. Set B = B0
64
and note that B has  nite Goldie dimension. Then by Proposition 2.31, B  e M implies
that dim(B) = dim(M) and the result follows. If Mn 1 is not essential in M, then by Lemma
5.28, there exists 06= y2M and a submodule N of M such that M = yR N. Note that
gen(N) = n 1, so by the induction hypothesis, N contains an essential pure submodule
B00, which is the direct sum of dim(N) non-zero cyclic submodules. Set B = yR B00. Then
B is an essential pure submodule of M that is the direct sun of non-zero cyclic submodules
and the proof is complete.
Recall that the minimal number of elements needed to generate M is denoted gen(M)
and we have shown that gen(M) = dimR=JM=MJ. Also, the number of factors in a pure
composition series of M is written ?(M) and ?(M) = gen(M) for a  nitely generated right
R-module M over a right duo, chain ring. As a result of Theorem 5.29 and the observations
above, we can obtain an upper estimate on the Goldie dimension of a  nitely generated right
R-module.
Corollary 5.30. For a  nitely generated right R-module over a right duo chain ring R, the
following holds:
dim(M) gen(M):
Proof. Let B be an essential pure submodule which is the direct sum of non-zero cyclic
submodules. Since B is pure, we have that BJ = B\MJ. Then (B + MJ)=MJ is a
submodule of M=MJ, and B=BJ = B=B\MJ = (B +MJ)=MJ. Since M=MJ is a  nite
dimensional vector space over the division ring R=J, dimR=JB=BJ  dimR=JM=MJ . By
the observations above, we have
dim(M) = dim(B) = dimR=JB=BJ dimR=JM=MJ = gen(M):
The  nal result of this section is a criteria for a  nitely generated right R-module over
a right duo, chain ring to be a direct sum of cyclic modules.
65
Corollary 5.31. A  nitely generated right R-module over a right duo, chain ring R is the
direct sum of cyclic modules if and only if gen(M) = dim(M).
Proof. If M is the direct sum of cyclic modules, the clearly gen(M) = dim(M). Conversely,
assume gen(M) = dim(M). By Theorem 5.29, M contains an essential pure submodule B
that is the sum of non-zero cyclic submodules and dim(B) = dim(M) = gen(M). Suppose
B is a proper submodule of M and dim(B) = gen(M). Then
B=BJ?B=(B\MJ) = (B +MJ)=MJ (M=MJ:
Then B + MJ (M, a contradiction. Hence, gen(M) > gen(B) = dim(B) = gen(M), an
obvious contradiction. Thus M = B and the result follows.
66
Chapter 6
Duo Modules
6.1 Quasiprojective Modules
Quasiprojective modules are a generalization of the familiar concept of projective mod-
ules. One of the  rst studies of quasiprojective modules was undertaken by Wu and Jans in
[42] and was motivated by work on the dual concept of quasiinjective modules.
De nition 6.1. Let M be a right R-module. A right R-module U is M-projective if every
diagram
U
f
  
M  //M=T // 0
can be embedded in a commutative diagram
U
f
  
f0
||M
 //M=T // 0
where  is the canonical map. If M is M-projective, then M is said to be a quasiprojective
module.
In [42] the following Proposition is established.
Proposition 6.2. [42, Proposition 2.1] Let M be a right R-module. If
0 !T  !P  !M !0
is exact with P projective and T fully invariant in P, then M is quasiprojective.
67
As a corollary, we obtain the following.
Corollary 6.3. Let R be a right duo ring. Then every cyclic right R-module is quasiprojec-
tive.
Proof. Suppose aR is a cyclic right R-module and ? : R !aR is left multiplication by
a. Then we have an exact sequence 0  ! annr(a)  ! R ? ! aR  ! 0. As R is right
duo, annr(a) is a fully invariant submodule. Since R is projective, the result follows by
Proposition 6.2.
The following basic properties of M-projective modules are established in Anderson-
Fuller [2, Proposition 16.2].
Theorem 6.4. [2] Let U be a right R-module.
1) If 0 !K !M  !N  !0 is a short exact sequence of right R-modules, then U
is M-projective if and only if U is projective relative to both K and N.
2) If U is projective relative to each of M1;:::;Mn, then U is  ni=1Mi-projective. More-
over, if U is  nitely generated and M -projective ( 2A), then U is projective relative
to  AM .
Herrmann has investigated quasiprojective modules in [21], however a complete classi -
cation of such modules remains open (see [16, Problem 21]). There seems to have been little
work done concerning quasiprojective modules over more general rings. As a start, we o er
the following which extends a result presented in [21]. The proof follows that of Herrmann
closely.
Theorem 6.5. Let R be a chain domain and I a proper right ideal of R. If Q is the quotient
ring of R, then I is quasiprojective if and only if it is Q-projective.
Proof. Suppose I is quaisprojective and f : I  !Q=J, where J (Q. We claim that f is
not an epimorphism. Suppose to the contrary that f(I) = Q=J. Then there exists an ideal
68
L such that I=L ? Q=J, which is a divisible R-module. Hence, (I=L)r = I=L for every
non-zero r 2R. Hence, I = Ir + L for every non-zero r 2R. Choose 0 6= c2I. Then
c = c1c+l for some c1 2I and l2L. This implies c c1c2L and therefore (1 c1)c2L.
As (1 c1) is a unit of R, it follows that c2L. Hence, I = L and therefore Q=J = 0, a
contradiction.
Thus im(f) = V=J, where V is a proper R-submodule of Q. Then V is isomorphic to
a right ideal K of R, so without loss of generality we set V = K and im(f) = K=J. There
is an embedding R  ! I whose restriction embeds K in I. As I is quasiprojective it is
K-projective by Theorem 6.4.
Then we have the following commutative diagram.
I
f
  
g
}}K //
incl:
  
K=J
incl:
  
Q  //Q=J
where  is the canonical map. Then f =  g and I is Q-projective. As I <Q, the converse
follows from Theorem 6.4.
6.2 Duo Modules and Strongly Right Bounded Modules
De nition 6.6. Let R be a ring and M a right R-module. We say that M is a duo module
if every submodule N of M is fully invariant.
Example 6.7. The following are some examples of duo modules.
1. Since elements of the endomorphism ring of RR are precisely left multiplication by
elements of R, the module RR is a duo module if and only if R is a right duo ring.
69
2. For a ring R, a right R-module M is called a multiplication module if for every
submodule N of M, there exists an ideal I of R such that N = MI. If f is an
endomorphism of such a module, then
f(N) = f(MI) = f(M)I MI = N:
Thus, multiplication modules are duo modules.
3. By [12], Noetherian uniserial modules are duo.
As a consequence of (3) above, we obtain the well known result that right Noetherian
right chain rings are right duo rings. A simple observation is given by the following Lemma.
Lemma 6.8. [31] A right R-module M is a duo module if and only if for every m2M and
f2EndR(M), we have that f(m)2mR.
A ring is strongly right bounded if every non-zero right ideal contains a non-zero
two-sided ideal. Right duo rings are obvious examples of strongly right bounded rings. The
following extends this concept to right R-modules.
De nition 6.9. A right R-module is strongly right bounded if every non-zero submodule
of M contains a non-zero submodule that is fully invariant in M.
Thus, RR is strongly right bounded if and only if R is a strongly right bounded ring.
In [13] it is shown that a strongly right bounded ring has the property that every non-zero
right ideal is an essential extension of a two-sided ideal. The following shows that this result
extends to strongly right bounded modules as well.
Lemma 6.10. Let M be a strongly right bounded, right R-module. Then every non-zero
submodule is an essential extension of a fully invariant submodule.
Proof. Let 06= N be a submodule of M and
T =
X
fL MjL is fully invariant and L Ng:
70
Then T is fully invariant in M and T  N. Suppose K is a non-zero submodule of N
and T\K = 0. Since M is strongly right bounded, there exists a non-zero fully invariant
submodule K0 contained in K. But then 0 6= K0 T\K, a contradiction. Thus N is an
essential extension of T.
The following Lemma contains some well known properties of fully invariant submodules.
The proof is included for completeness. We will write N EM to denote that N is fully
invariant in M.
Lemma 6.11. Let M be a right R-module. Then
1) The sum and intersection of any family of fully invariant submodules of M is again
fully invariant.
2) If X Y are submodules of M such that Y EM and XEY, then XEM.
3) If M = i2IMi and N EM, then N = i2I(Mi\N).
4) Suppose N  K are submodules of M. If N is a fully invariant submodule of M and
K=N is fully invariant in M=N, then K is fully invariant in M.
Proof. 1): This is clear.
2): Let f 2 EndR(M). Then f(Y)  Y as Y is fully invariant. Thus, fjY is an
endomorphism of Y so that fjY (X) X. As X Y we have f(X) = fjY (X) X.
3): For each j2I let pj : M  !Mj and ij : Mj  !M be the canonical projections
and injections respectively. Then ijpj is an endomorphism of M. Thus, ijpj(N)  N for
every j2I. Then N  jijpj(N)  j (Mj\N) N, whence N = j (Mj\N).
4): Let f 2 EndR(M). De ne f : M=N ! M=N by f (m + N) = f(m) + N. As
N is a fully invariant submodule of M, f is a well-de ned endomorphism of M=N. Then
f(K) + N = f (K=N)  K=N as K=N is fully invariant. Thus, f(K)  K and K is fully
invariant in M.
71
Clearly any homomorphic image of a right duo ring is right duo. However, there exist
strongly right bounded rings that have homomorphic images that are not strongly right
bounded [9]. In [9] it is also shown that a ring R is right duo if and only if every homomorphic
image of R is strongly right bounded. The following result extends this to quasiprojective
duo modules.
Theorem 6.12. Let M be a quasiprojective right R-module. The following are equivalent:
1) [31] M is a duo module.
2) Every homomorphic image of M is strongly right bounded.
Proof. 1) implies 2): Suppose M is quasi-projective and duo. As duo modules are obviously
strongly right bounded, it is enough to show that every homomorphic image of M is duo.
Suppose N and K are submodules of M such that N  K, and f 2EndR(M=N). Since
M is quasi-projective, there exists g 2 EndR(M) such that g(m) + N = f(m + N) for
every m 2 M. As M is a duo module, we have that g(K)  K. Then it follows that
f(K=N) = g(K) +N K=N. Hence, M=N is a duo module.
2) implies 1): Suppose there exists 06= N  M such that N is not fully invariant. By
hypothesis, M is strongly right bounded, and therefore N properly contains a non-zero fully
invariant submodule K. Consider the following sum:
T =
X
fL MjL is fully invariant and L Ng:
Then T is a fully invariant submodule of M, and by assumption T  N. Thus N=T is
a non-zero submodule of M=T. Since M=T is strongly right bounded, there exists a non-
zero fully invariant submodule K=T  N=T. Then T  K  N. By Lemma 6.11, K is
a fully invariant submodule of M, and by the choice of T we must have that K = N, a
contradiction.
We obtain as a corollary the result of Birkenmeier and Tucci.
72
Corollary 6.13. Let R be a ring. Then R is right duo if and only if every homomorphic
image of R is strongly right bounded.
6.3 Duo Modules over Domains
In [31] the following Theorem is established.
Theorem 6.14. [31, Th. 3.3] Suppose R is an integral domain. The following are equivalent
for a torsion-free uniform module U.
i) U is duo.
ii) U contains a non-zero cyclic fully invariant submodule.
iii) Ol(U) = R, where Ol(U) =fq2QjqU Ug.
The following example shows that this Theorem cannot be extended to general rings.
Example 6.15. Let R = M2(Z) be the ring of 2 2 matrices with integer entries and
U = e1R, where e1 =
0
B@ 1 0
0 0
1
CA. Then U is a torsion-free, uniform right R-module. A
calculation shows that Ol(U) =
0
B@ Z Q
0 Q
1
CA6= R.
We now classify the right and left Ore domains for which this equivalence remains valid.
Theorem 6.16. Let R be a right and left Ore domain. The following are equivalent:
1) R is a right and left duo ring.
2) The equivalence of i), ii), and iii) in Theorem 6.14 is valid for all torsion-free uniform
right R-modules U.
Proof. 1) implies 2): Suppose U is a torsion-free uniform right R-module. We will show that
i), ii), and iii) above are equivalent.
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i) implies ii): Trivial.
ii) implies iii): We will  rst show that U is a left R-submodule of Q. Let r2R and
u2U. Then u = ab 1 for some a, b2R. So ru = rab 1 = ar1b 1 since Ra aR. As R is
left duo, we also have that bR Rb so that br1 = r2b, or equivalently, r1b 1 = b 1r2. Thus,
ru = rab 1 = ar1b 1 = ab 1r2 = ur22U. It follows that R Ol(U).
Conversely, suppose q2Ol(U), that is, qU U. By hypothesis there exists x2U such
that xR is non-zero and fully invariant. We claim that xR Rx. To see this, write x = ab 1
for a, b2R and suppose r2R. Then
xr = ab 1r = ar1b 1 = r2ab 1 = r2x2Rx:
As left multiplication by q is an R-endomorphism of U we have qx = xr = sx for r, s2R.
As U is torsion-free, it follows that q = s2R.
iii) implies i): Suppose N is a non-zero submodule of U and ?2EndR(UR). Since U
is uniform, ? is left multiplication by some q 2 Q. Hence, q 2Ol(U) = R, so ? is left
multiplication by some r2R. Then for every x2N we have ?(x) = rx2Rx = xR N
which implies that N is fully invariant.
2) implies 1): Since R is a right Ore domain, RR is uniform. By hypothesis RR is a duo
module, i.e., R is a right duo ring. If 0 6= c2R, then cR?R and thus cR is a right duo
ring as well. Then (cR)R is a duo module so that Ol(cR) = R. An easy calculation shows
that Ol(cR) = cRc 1. Thus, R =Ol(cR) = cRc 1 for every non-zero c2R. It follows that
cR = Rc for every c2R, whence R is a right and left duo ring.
We now consider  nitely generated torsionfree duo modules. Our result extends [31,
Thm 3.7]. We need a result of Levy in the proof.
Theorem 6.17. [27, Theorem 6.1] Let R be a right semihereditary ring having a semisimple
two-sided quotient ring. Then every  nitely generated right R-module is the direct sum of its
torsion submodule and a  nite set of right ideals of R.
74
Given this result we can establish the following.
Theorem 6.18. Let R be a duo chain domain. The following are equivalent for a  nitely
generated torsionfree right R-module M.
1) M is duo.
2) M contains a non-zero cyclic fully invariant submodule.
3) M is a uniform module and Ol(M) = R.
Proof. 1) implies 2): trivial
2) implies 3): Note  rst that R is right semihereditary by Theorem 3.28. Further, R
is right and left Goldie and thus has right and left quotient rings that coincide. Choose
0 6= m2M such that mR is fully invariant in M. Then there exists a submodule L of M
such that mR L e M so that M=(mR L) is a  nitely generated torsion right R-module.
Thus there exists a regular element a2R such that Ma mR L.
As M is torsionfree, it follows from Theorem 6.17 that M is a  nitely generated pro-
jective right R-module. By [28, Lemma 5.2], we have that MM = End(M). De ne
f : mR L  ! R by f(mr + x) = r, and suppose u 2 L. Then f 2 M so that
uf 2MM = End(M). It follows that ua = uf(ma) 2mR\L = 0 as mR is fully in-
variant. Thus, ua = 0 and M torsionfree imply that u = 0. Then L = 0 so that mR is an
essential submodule of M. As mR?R, a uniform module, it follows that M is uniform and
Ol(M) = R by Theorem 6.16.
3) implies 1): Follows from Theorem 6.16.
6.4 Maximal RD-submodules
This section is motivated by results obtained by Goldsmith and Zanardo [20] for tor-
sionfree modules over valuation domains. We show that several results may be extended
to a certain class of noncommutative rings. The proofs are motivated by their results.
75
Throughout the following section R is a semiprime right and left Goldie ring and M a right
R-module.
We  rst note that the set of regular elements CR(0) of R is right Ore; hence the set
t(M) =fa2Mjax = 0 for some x2CR(0)g
is a submodule of M. Recall that a right R-module is torsion-free if t(M) = 0 and torsion
if t(M) = M. The following result is well known, but we could not locate a proof so one is
provided.
Proposition 6.19. Let R be a semiprime right Goldie ring and M a torsion-free right
R-module. Then a submodule N of M is RD if and only if M=N is torsion-free.
Proof. Suppose  rst that N is an RD-submodule of M and let m + N 2t(M=N). Then
there is a regular element r2R such that mr2N. Since N is RD, mr2N\Mr = Nr,
so mr = nr for some n2N. Then (m n)r = 0 so that m n2t(M) = 0. Thus, m2N
and t(M=N) = 0.
Conversely, suppose t(M=N) = 0 and mr 2 N for some m 2 M and r 2 R. Then
(m + N)r = 0 in M=N. If r2CR(0), then m + N 2t(M=N) = 0 so that m2N. In this
case, it follows that N is an RD-submodule of M. If r62CR(0), then rR is not an essential
right ideal of R. So there exists a non-zero right ideal K of R such that rR\K = 0. Then
the sum K +rK +r2K +   is direct, contradicting dimR(RR) <1.
Suppose MR is torsion-free and N is a submodule of M. The set
fx2Mjxr2N for some r2CR(0)g
is the RD-hull of N, denoted N .
Lemma 6.20. If MR is torsion-free and N a submodule of M, then
76
1) N is an RD-submodule of M.
2) N N , and if K is an RD-submodule of M containing N, then N  K.
Proof. 1): We will show  rst that N is a submodule of M. Suppose x1, x2 2N . Then
there exist r1, r22CR(0) such that xiri = 0 for i = 1;2. SinceCR(0) is right Ore, there exists
y2x1R\x2R\CR(0). Then (x1 x2)y2N and x1 x22N . If r2R, then rz = r1s for
some z2CR(0) and s2R. So x1rz = x1r1s2N, and z regular implies that x1r2N .
Suppose that m + N 2 t(M=N ). Then there exists a regular element r 2 R such
that mr2N . By the de nition of N , this implies that there is a regular s2R such that
mrs2N. As R is semiprime right Goldie and r and s are regular, we have that rs is regular.
Thus, m2N and it follows that t(M=N ) = 0. By the preceding Proposition, N is an
RD-submodule of M.
2): Clearly N  N . Suppose K is an RD-submodule of M containing N and let
x2N . Then xr2N K for some regular r, so that xr2K\Mr = Kr. But then there
exists k2K such that (x k)r = 0, and therefore x k2t(M). As M is torsion-free, it
follows that x2K. Thus, N  K.
Proposition 6.21. Let R be semiprime right Goldie and M a torsionfree right R-module.
Then a submodule N of M is a maximal RD-submodule if and only if dimR(M=N) = 1.
Proof. Suppose  rst that dim(M=N) = 1 so that M=N ?JR  QrR. Then J is a uniform
right R-module so every proper non-zero submodule K of J is essential. Let a2J. The
right ideal I = fr 2Rjar 2Kg is an essential right ideal of R. Since R is semiprime
right Goldie, there exists a regular element y2I. Then ay2K and it follows that J=K is
a non-zero torsion module. Thus no proper non-zero submodule of J is RD. It follows that
N is a maximal RD-submodule of M.
Conversely, suppose N is a maximal RD-submodule. Then M=N has no proper non-
zero RD-submodules. If dimR(M=N)6= 1, then there exist non-zero submodules U1=N and
U2=N such that U1=N\U2=N = 0. Choose 0 6=  a 2 U1=N. Then the RD-hull of the
77
cyclic submodule generated by  a is a proper non-zero RD-submodule of M=H, which is a
contradiction.
Proposition 6.22. Every right R-module M contains a maximal RD-submodule.
Proof. Let E be an injective hull of M. Since R is semiprime right Goldie, E = Li2ISi,
where each Si is a simple right Qr-module. Note that each Si has Goldie dimension 1. Fix
j2I and let E0 = Li6=jSi. If N = M\E0, then
M=N = M=(M\E0)?M +E0=E0 E=E0?Sj:
Thus, dim(M=N) = 1 and N is a maximal RD-submodule of M by Proposition 6.21.
We recall the following de nition from [20]:
De nition 6.23. Let R be a ring. A cardinal  is the level of coherency of R if  is the
smallest cardinal such that, for every short exact sequence
0 !N !X !J !0;
where J is a right ideal of R and X is  nitely generated, we have genN   . Suppose R
has a two-sided quotient ring Q. Then we de ne  R = supfgenJg where J ranges over the
R-submodules of Q.
The next three results establish several cardinality arguments that are needed in the
proof of the main result.
Proposition 6.24. Let R be a semiprime Goldie ring and  the level of coherency of R. If
N is an RD-submodule of a  nitely generated torsionfree R-module X, then genN  .
Proof. Assume, without loss of generality, that N 6= X. The proof is by induction on
dimR(X). If X has Goldie dimension 1, then dimR(X=N)  1, so X=N is isomorphic to
78
a  nitely generated right ideal of R. As  is the level of coherency of R and X is  nitely
generated, it follows that genN  .
Now assume genN > 1. Choose a maximal RD-submodule H of X containing N. Then
dimR(X=H) = 1 and therefore genH =    . If  is  nite then H is  nitely generated, N
is RD in H, and dimH <dimX. Therefore by the induction hypothesis, dimN  .
If  is an in nite cardinal, choose a set of generators fx g < of H indexed by  . For
every  nite subset F of  , let NF = N\(P 2F x R). Then NF is relatively divisible in
P
 2F x R and
dim(
X
 2F
x R) dimH <dimX:
By the induction hypothesis genNF   . Then N = PF NF implies
genN 
X
F
genNF    =  
and the proof is complete.
Theorem 6.25. Let R be a semiprime Goldie ring and  the level of coherency of R. If N
is an RD-submodule of the torsionfree right R-module M, then
genN  (genM):
Proof. Let genM =  . Choose a set of generators fx g < of M indexed by  . For
every  nite subset F of  , let NF = N \(P 2F x R). Since NF is an RD-submodule
of P 2F x R, by Proposition 6.24 genNF   . Then N = PF NF and it follows that
genN PF genNF    , as desired.
Corollary 6.26. Suppose R is a semiprime Goldie ring, M a torsionfree right R-module
with gen(M)  R, and N is an RD-submodule of M. Then gen(N)  R as well.
79
Proof. Let  be the level of coherency of R and consider an exact sequence
0 !N !X !J !O
where J is a right ideal of R and X is  nitely generated. As gen(M)  R by assumption,
6.25 implies that gen(N)    gen(M)     R. We complete the proof by showing that
   R. As  is the smallest cardinal such that gen(N)   we need only show that
gen(N)  R.
Since X is a  nitely generated nonsingular module over R, [18, Proposition 2.12] implies
that X can be embedded into a  nite direct sum of copies of Q. As N is isomorphic to a
submodule of X, we may assume without loss of generality that there is an embedding
0  ! N  ! Qn for some natural number n. The proof is by induction on n. If n = 1
then N is isomorphic to an R-submodule of Q and the result is clear. Suppose we have an
embedding 0  ! N  ! Qn+1. Set N0 = N\Qn  Qn. The the induction hypothesis
implies that gen(N0)  R. We also have N=N0 = N=(Qn\N)?(N +Qn)=Qn Q. Hence
gen(N=N0)  R and it follows that gen(N)  R
Given these preliminary results, we can now establish the main theorem.
Theorem 6.27. Let R be a semiprime Goldie ring with classical quotient ring Q such that
R6= Q. If M is a torsionfree right R-module such that
1) Every maximal RD-submodule of M is a direct sum of rank 1 projective modules.
2) genM > R.
Then M is a direct sum of rank 1 projective modules.
Proof. Let H be a maximal RD-submodule of M. Then dim(M=H) = 1 which implies that
M=H is isomorphic to an R-submodule of Q. Thus genM=H   R. Let  : M !M=H
be the canonical epimorphism. For each generator xi of M=H, choose yi 2 M such that
80
 (yi) = xi and let A be the submodule of M generated by the yi. Clearly genA  R as
well.
We claim that M = A + H. To see this, suppose x 2 M. Then we have x + H =
(Piyiri) +H so that x (Piyiri) H. But Piyiri2A and thus x2A+H. An obvious
cardinality argument shows that genH > R.
Now, A=(A\H) ? (A + H)=H = M=H so that dim(A=A\H) = 1 and A\H is
an RD-submodule of A. Since genA   R, 6.26 implies that gen(A\H)   R. Write
H = Li2IPi where each Pi is a rank 1 projective module. Decompose H as H = H0LH1,
where H0 and H1 are rank 1 projective, A\H H0, and genH1 >  R. Let N = A + H0.
Then M = A + H = A + H0 + H1 = N + H1. We claim that the sum N + H1 is direct.
Suppose a+h0 = h1, where a2A, h02H0 and h12H1. Then a = h1 h02A\H H0,
say h1 h0 = h00. Thus, h1 = h00 +h02H1\H0 whence h1 = 0.
So M = NLH1. Now decompose H1 as H1 = H2LPi0, where Pi0 is rank 1 projective.
Then M = NLH1 = NLH2LPi0, and M=(NLH2)?Pi0. Thus, NLH2 is a maximal
RD-submodule of M and hence a direct sum of rank 1 projective modules. It follows that
M is the direct sum of rank 1 projectives.
81
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